1

Welcome!Welcome to CRUX MATHEMATICORUM with MATHEMATICAL

MAYHEM. We hope that those of you new to any part of our journal will

enjoy all the material. We see this merger as bene�cial to all our readers.

We now o�er a broader range of problems. High School students and teach-

ers will �nd more material at that level. Please note that we will accept

solutions to problems in the MAYHEM section only from students. These

are the people that we wish to stimulate. When they �nd they are successful

with the MAYHEM and SKOLIAD problems, we hope they will then �nd a

further challenge with the problems in the OLYMPIAD corner, and, eventu-

ally, with the PROBLEMS section. Good luck, and we look forward to hearing

from you.

You will note that the majority of the members of the Crux Editorial

Board are continuing in their previous jobs. Sadly, Colin Bartholomew has

taken early retirement fromMemorial University, and I shallmiss him. How-

ever, we are pleased to welcome Clayton Halfyard as Associate Editor. There

will be a short pro�le of him elsewhere in this issue. We are also delighted

to welcome Naoki Sato as Mayhem Editor and Cyrus Hsia as Assistant May-

hem Editor. These two Canadian IMO medallists bring with them a wealth

of experience of problem solving, and many years of experience of writing for

MAYHEM.

There are some minor administrative changes: the table of contents

(which is larger) is now on the back outside cover; and the ordering of the

various sections has been changed. There is no change in the quantity of

material that has been recently published in CRUX. Whereas previously

MAYHEM appeared �ve times per year, it now appears eight times per year.

The annual quantity of material will be approximately the same.

Here are some details that will help you in cross-referencing previous

material:

1. Please note that the volume number is consecutive for CRUX.

2. Material occurring in previous volumes of CRUX and material occuring

in CRUX with MAYHEM will be referenced as [year: page no].

For example, the last page of the last issue of CRUX is [1996: 384], and

the �rst page of CRUX with MAYHEM is [1997: 1].

3. Material occurring in previous volumes ofMAYHEM will be referenced

as [MAYHEM volume: issue, page no: year].

For example, [MAYHEM 8: 5, 28: 1996] is the last page of the last issue

of MAYHEM.

Bruce Shawyer

Editor-in-Chief

2

Bienvenue!Bienvenue au CRUX MATHEMATICORUM with MATHEMATICAL

MAYHEM. Nous esp �erons que ceux qui ne connaissent pas l'une des

nouvelles parties de notre journal en appr �ecieront tout le contenu. �A notre

avis, la fusion de ces deux magazines sera tout �a l'avantage de nos lecteurs.

La gamme des probl �emes publi �esy est plus vaste qu'auparavant, et les �el �eves

et les professeurs du secondaire y trouveront davantage de probl �emes �a leur

niveau. Nous vous signalons au passage que nous accepterons uniquement

les solutions aux probl �emes de la sectionMAYHEM provenant des �etudiants;

apr �es tout, c'est eux que nous voulons stimuler! Lorsqu'ils pourront r �eussir

les probl �emes des sections MAYHEM et SKOLIAD, nous esp �erons qu'ils

s'attaqueront �a ceux de la partie OLYMPIAD, et meme, un peu plus tard,

�a ceux de la section PROBLEMS. Bonne chance et au plaisir de recevoir de

vos nouvelles!

Vous aurez remarqu �e que lamajorit �e desmembres de l' �equipe �editoriale

du Crux sont demeur �es en poste. Malheureusement, Colin Bartholomew a

pris une retraite anticip �ee de l'Universit �e Memorial; il me manquera. Nous

sommes toutefois heureux d'accueillir Clayton Halfyard au poste de r �edacteur

adjoint. Vous trouverez son pro�l ailleurs dans ce num�ero. Nous sommes

�egalement enchant �es d'avoir parmi nous Naoki Sato et Cyrus Hsia, qui

occupent respectivement les postes de r �edacteur et de r �edacteur adjoint du

Mayhem. Ces deux m�edaill �es canadiens de l'OIM apportent avec eux tout un

bagage d'expertise en r �esolution de probl �emes, et des ann �ees d'exp �erience �a

la r �edaction du MAYHEM.

Vous constaterez en outre quelques petits changements d'ordre admin-

istratif : la table des mati �eres ( �elargie) est d �esormais en quatri �eme de couver-

ture et les sections ne sont plus dans lememe ordre. Le volume d'information

du CRUX sera comparable �a la quantit �e d'information qu'il contenait avant

la fusion. LeMAYHEM, qui paraissait cinq fois l'an, sera publi �e huit fois par

ann �ee; ainsi, le volume annuel de contenu demeurera approximativement le

meme.

Voici quelques d �etails sur la fa�con de noter les renvois :

1. Les num�eros de volume du CRUX continuent dans la meme s �equence.

2. On notera ainsi les renvois �a des volumes �pr �e-fusion� du CRUX et aux

volumes du CRUX with MAYHEM : [ann �e: no de page].

Par exemple, la derni �ere page du dernier num�ero du CRUX sera [1996:

384], et la premi �ere page du CRUX with MAYHEM sera [1997: 1].

3. On notera ainsi les renvois �a des volumes �pr �e-fusion� duMAYHEM :

[volume du MAYHEM: no, no de page: ann �ee].

Par exemple, [MAYHEM 8: 5, 28: 1996] est la derni �ere page du dernier

num�ero du MAYHEM.

Le r �edacteur en chef

Bruce Shawyer

3

THE ACADEMY CORNERNo. 8

Bruce Shawyer

All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

In this issue, courtesy of Waldemar Pompe, student, University of War-

saw, Poland, we print an international contest paper for university students.

Please send me your nice solutions.

INTERNATIONAL COMPETITION

FOR UNIVERSITY STUDENTS IN MATHEMATICS

July 31 { August 1996, Plovdiv, Bulgaria

First day | August 2, 1996

1. Let for j = 0; 1; : : : ; n, aj = a0 + jd, where a0, d are �xed real

numbers. Put

A =

0BBBB@a0 a1 a2 : : : ana1 a0 a1 : : : an�1

a2 a1 a0 : : : an�2

: : : : : : : : : : : : : : :

an an�1 an�2 : : : a0

1CCCCA :

Calculate detA | the determinant of A.

2. Evaluate the integral Z �

��

sinnx

(1 + 2x) sinxdx ;

where n is a natural number.

3. A linear operator A on a vector space V is called an involution if

A2 = E, where E is the identity operator on V .

Let dimV = n <1.

(i) Prove that for every involution A on V there exists a basis of V

consisting of eigenvectors of A.

(ii) Find the maximal number of distinct pairwise commuting involu-

tions on V .

4

4. Let a1 = 1, an = 1n

n�1Pk=1

akan�k for n � 2.

Show that

(i) limsupn!1

janj1=n < 2�1=2 ;

(ii) limsupn!1

janj1=n � 2=3.

5. (i) Let a, b be real numbers such that b � 0 and 1 + ax+ bx2 � 0 for

every x 2 [0; 1].

Prove that

limn!1

n

Z 1

0

(1 + ax + bx2)n dx =

��1=a if a < 0;

+1 if a � 0:

(ii) Let f : [0; 1]! [0;1) be a function with continuous second deriva-

tive and f 00(x) � 0 for every x 2 [0; 1]. Suppose that

L = limn!1

n

Z 1

0

(f(x))n dx

exists and 0 < L < +1.

Prove that f 0 has a constant sign and L =

�minx2[0;1]

jf 0(x)j��1

:

6. Upper content of a subset E of the plane R�R is de�ned as

C(E) = inf

(nXi=1

diam (Ei)

)

where the in�mum is taken over all �nite families of setsE1; E2; : : : ; En

in R� R such that E �nSi=1

Ei.

Lower content of E is de�ned as

K(E) = supflength (L)g

such that L is a closed line segment onto which E can be contracted.

Show that

(i) C(L) = length (L), if L is a closed line segment.

(ii) C(E) � K(E).(iii) equality in (ii) need not hold even if E is compact.

Hint: IfE = T [T 0 where T is the triangle with vertices (�2; 2), (2;2)and (0; 4), and T 0 is its re ection about the x-axis, then C(E) = 8 >

K(E).

5

Remarks:

All distances used in this problem are Euclidean.

Diameter of a set E is diam (E) = supfdist (x; y) j x; y 2 Eg.Contraction of a set E to a set F is a mapping f : E ! F such that

dist (f(x); f(y))� dist (x; y) for all x; y 2 E.A set E can be contracted onto a set F if there is a contraction f of E

to F which is onto, that is such that f(E) = F .

Triangle is de�ned as the union of the three line segments joining its

vertices, so that it does not contain the interior.

Problems 1 and 2 are worth 10 points, problems 3 and 4 are worth 15 points,problems 5 and 6 are worth 25 points.

You have 5 hours.

Please write the solutions on separate sheets of paper. Good luck!

6

THE OLYMPIAD CORNERNo. 179

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

Another year has passed, and the �rst with Bruce Shawyer as Editor-

in-Chief. He has made the transition pleasant and easy. Special thanks go

to Joanne Longworth whose TEX skills have made changed fonts and formats

easy to incorporate. Thanks also go to the many contributors to the two

Corners including:

Miguel Amengual Covas Cyrus C. Hsia Dieter Ruo�

S�efket Arslanagi�c Murray Klamkin Toshio Seimiya

Mansur Boase Derek Kisman Michael Selby

Seung-Jin Bang Ted Lewis D.J. Smeenk

Christopher Bradley Joseph Ling Daryl Tingley

Francisco Bellot Rosado Beatriz Margolis Panos E. Tsaoussoglou

Paul Colucci Stewart Metchette Ravi Vakil

Hans Engelhaupt Richard Nowakowski Edward T.H. Wang

Tony Gardiner Michael Nutt Hoe Teck Wee

Solomon Golomb Siu Taur Pang Chris Wildhagen

Gareth Gri�th Bob Prielipp Siming Zhan

Georg Gunther Chandan Reddy

Let me open with a major apology. In the November 1996 number of

the corner we gave twelve of the problems proposed to the jury but not used

at the 36th International Mathematical Olympiad held in Canada. Those

familiar with the process of selection will know that the problems do not

initiate with the host country. They come from proposers in other countries,

and the responsibility of the host country selection committee is to re�ne

and select from these submissions the o�cial list of problems proposed to

the jury. Over the years I have loosely referred to the problems proposed

to the jury at the Xth International Olympiad in Y as the \Y-problems" for

short. However, when that became part of a longer more o�cial sounding

sub-title \Canadian Problems for consideration by the International Jury," I

should have seen that it read as if the original proposers are from Canada,

thus insulting the creators. In retrospect I do not understand why that in-

terpretation did not jump o� the page. To the many creative non-Canadians

who submitted problems for possible use at the 36th IMO my sincere apolo-

gies.

7

As an Olympiad Contest this issue we give the problems of the 44th

Mathematical Olympiad from Latvia. My thanks go to Richard Nowakowski,

Canadian Team Leader to the IMO in Hong Kong for collecting the problems

for me.

LATVIAN 44MATHEMATICAL OLYMPIAD

Final Grade, 3rd RoundRiga, 1994

1. It is given that cosx = cos y and sinx = � siny. Prove that

sin 1994x+ sin1994y = 0.

2. The plane is divided into unit squares in the standard way. Consider

a pentagon with all its vertices at grid points.

(a) Prove that its area is not less than 3=2.

(b) Prove that its area is not less than 5=2, if it is given that the pentagon

is convex.

3. It is given that a > 0, b > 0, c > 0, a+ b+ c = abc. Prove that at

least one of the numbers a, b, c exceeds 17=10.

4. Solve the equation 1!+2!+3!+ � � �+n! = m3 in natural numbers.

5. There are 1994 employees in the o�ce. Each of them knows 1600

of the others. Prove that we can �nd 6 employees, each of them knowing all

5 others.

1st SELECTION ROUND

1. It is given that x and y are positive integers and 3x2+x = 4y2+y.

Prove that x� y, 3x+ 3y + 1 and 4x+ 4y+ 1 are squares of integers.

2. Is it possible to �nd 21994 di�erent pairs of natural numbers (ai; bi)

such that the following 2 properties hold simultaneously:

(1)1

a1b1+

1

a2b2+ � � �+

1

a21994b21994= 1,

(2) (a1 + a2 + � � �+ a21994) + (b1 + b2 + � � �+ b21994) = 31995?

3. A circle with unit radius is given. A system of line segments is

called a cover i� each line with a common point with the circle also has some

common point with some of the segments of the system.

(a) Prove that the sum of the lengths of the segments of a cover is more

than 3,

(b) Does there exist a cover with this sum less than 5?

4. A natural number is written on the blackboard. Two players move

alternatively. The �rst player's move consists of replacing the number n on

the blackboard by n=2, by n=4 or by 3n (�rst two choices are allowed only

if they are natural numbers). The second player's move consists of replacing

the number n on the blackboard by n+1 or by n�1. The �rst player wants

the number 3 to appear on the blackboard (no matter who writes it down).

Can he always achieve his aim?

8

5. Three equal circles intersect at the point O and also two by two at

the points A, B, C. Let T be the triangle whose sides are common tangents

of the circles; T contains all the circles inside itself. Prove that the area of T

is not less than 9 times the area of ABC.

2nd SELECTION ROUND

1. It is given that 0 � xi � 1, i = 1; 2; : : : ; n. Find the maximum of

the expression

x1

x2x3 : : : xn + 1+

x2

x1x3x4 : : : xn + 1+ � � �+

xn

x1x2 : : : xn�1 + 1:

2. There are 2n points on the circle dividing it into 2n equal arcs. We

must draw n chords having these points as endpoints so that the lengths of

all chords are di�erent. Is it possible if:

(a) n = 24,

(b) n = 1994?

3. A triangle ABC is given. From the vertex B, n rays are constructed

intersecting the sideAC. For each of the n+1 triangles obtained, an incircle

with radius ri and excircle (which touches the side AC) with radius Ri is

constructed. Prove that the expression

r1r2 : : : rn+1

R1R2 : : : Rn+1

depends on neither n nor on which rays are constructed.

3rd SELECTION ROUND

1. A square is divided into n2 cells. Into some cells \1" or \2" is

written so that there is exactly one \1" and exactly one \2" in each row and

in each column. We are allowed to interchange two rows or two columns;

this is called a move. Prove that there is a sequence of moves such that after

performing it \1"-s and \2"-s have interchanged their positions.

2. Let aij be integers, jaij j < 100. We know that the equation

a11x2 + a22y

2 + a33z2 + a12xy + a13xz + a23yz = 0

has a solution (1234;3456;5678). Prove that this equation also has a solu-

tion with x, y, z pairwise relatively prime which is not proportional to the

given one.

3. Let ABCD be an inscribed quadrilateral. Its diagonals intersect

at O. Let the midpoints of AB and CD be U and V . Prove that the lines

through O, U and V , perpendicular to AD, BD and AC respectively, are

concurrent.

9

Next we give �ve Klamkin Quickies. My thanks go to Murray Klamkin,

the University of Alberta, for supplying them to us. Next issue we will give

the \quick" solutions along with another �ve of his special teasers.

FIVE KLAMKIN QUICKIES

October 21, 1996

1. For x; y; z > 0, prove that

(i) 1 +1

(x+ 1)��1 +

1

x(x+ 2)

�x,

(ii) [(x+y)(x+z)]x[(y+z)(y+x)]y[(z+x)(z+y)]z � [4xy]x[4yz]y[4zx]z.

2. If ABCD is a quadrilateral inscribed in a circle, prove that the four

lines joining each vertex to the nine point centre of the triangle formed by

the other three vertices are concurrent.

3. How many six digit perfect squares are there each having the prop-

erty that if each digit is increased by one, the resulting number is also a

perfect square?

4. Let ViWi, i = 1; 2; 3; 4, denote four cevians of a tetrahedron

V1V2V3V4 which are concurrent at an interior point P of the tetrahedron.

Prove that

PW1 + PW2 + PW3 + PW4 � maxViWi � longest edge:

5. Determine the radius r of a circle inscribed in a given quadrilateral

if the lengths of successive tangents from the vertices of the quadrilateral to

the circle are a; a; b; b; c; c; d; d, respectively.

We now turn to solutions from the readers to problems posed in the

May 1995 number of the corner on the Sixth Irish Mathematical Olympiad,

May 8, 1993 [1995: 151{152].

SIXTH IRISH MATHEMATICAL OLYMPIAD

May 8, 1993 | First Paper(Time: 3 hours)

1. The real numbers �, � satisfy the equations

�3 � 3�2 + 5�� 17 = 0; �3 � 3�2 + 5� + 11 = 0:

Find �+ �.

Solutions by �Sefket Arslanagi�c, Berlin, Germany; by Beatriz Margolis,Paris, France; by Vedula N. Murty, Andhra University, Visakhapatnam, In-dia; by D.J. Smeenk, Zaltbommel, the Netherlands; by Panos E. Tsaous-soglou, Athens, Greece; and comment by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give Margolis's solution.

10

De�ne f(x) = x3� 3x2 +5x. We show that if f(�)+ f(�) = 6, then

�+ � = 2. Since f(x) = (x� 1)3 + 2(x� 1) + 3, we have

f(�)� 3 = (�� 1)3 + 2(�� 1)

f(�)� 3 = (� � 1)3 + 2(� � 1)

Adding gives

0 = (�� 1)3 + (� � 1)3 + 2(�+ � � 2)

= (�+ � � 2)[(�� 1)2 + (�� 1)(�� 1) + (� � 1)2 + 2]

and, since the second factor is positive, we obtain the result. (See Olympiad

Corner 142 and its solution.)

[Wang's Comment:] The problem is strikingly similar to problem 11.2

of the XXV Soviet Mathematical Olympiad, 11th Form [1993: 37]. The

method used by Bradley given in the published solution [1994: 99] works for

the present problem as well and, in fact, yields the same answer: �+� = 2.

3. The line l is tangent to the circle S at the point A; B and C are

points on l on opposite sides of A and the other tangents from B, C to S

intersect at a point P . If B, C vary along l in such a way that the product

jABj � jACj is constant, �nd the locus of P .

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;and by D.J. Smeenk, Zaltbommel, the Netherlands. We use Smeenk's solu-tion.

B CAH

I

P

D

E

p q

r

rr

pq

t t

S

�=2

�=2

=2

=2

Let S be the incircle s(I; 2) of 4BCP . We denote \PBA = �,

\PCA =

AB = pAC = q

11

with pq = k2, a constant.

Let S touch BP and CP at D and E respectively. For4PEI we have\EIP = 1

2(� + ). Thus

t = r tan1

2(� + ) =

(p+ q)r2

pq� r2:

The semiperimeter of 4BCP is

p+ q + t = p+ q+(p+ q)r2

pq� r2=pq(p+ q)

pq � r2:

The area, F , of 4BCP is

rpq(p+ q)

pq� r2=

1

2(p+ q)PH;

where PH is the altitude to BC. It follows immediately that

PH =2pqr

pq � r2=

2k2r

k2 � r:

So the locus of P is a line parallel to BC.

4. Let a0; a1; : : : ; an�1 be real numbers, where n � 1, and let

f(x) = xn + an�1xn�1 + � � � + a0 be such that jf(0)j = f(1) and each

root � of f is real and satis�es 0 < � < 1. Prove that the product of the

roots does not exceed 1=2n.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

Let f(x) = (x� �1)(x� �2) : : : (x� �n) where the �i denote the n

real roots of f , i = 1; 2; : : : ; n. Then from jf(0)j = f(1)we getQ(1��i) =Q

�i. (All products are over i = 1; 2; : : : ; n.) Using the Arithmetic Mean{

Geometric Mean Inequality we then get

�Y�i

�2=Y�i(1� �i) �

Y��i + (1� �i)

2

�2

=1

22n

from whichQ�i � 1

2nfollows. Equality holds if and only if �i =

12for all

i = 1; 2; : : : ; n.

May 8, 1993 | Second Paper(Time: 3 hours)

3. For non-negative integers n, r the binomial coe�cient�n

r

�denotes

the number of combinations of n objects chosen r at a time, with the con-

vention that�n

0

�= 1 and

�n

r

�= 0 if n < r. Prove the identity

1Xd=1

�n� r + 1

d

��r � 1

d� 1

�=

�n

r

�

12

for all integers n, r with 1 � r � n.

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

We use a combinatorial argument to establish the obviously equivalent

identitykX

d=1

�n� r + 1

d

��r � 1

r � d

�=

�n

r

�(�)

where k = minfr;n � r + 1g. It clearly su�ces to demonstrate that the

left hand side of (�) counts the number of ways of selecting r objects from

n distinct objects (without replacements). Let jS2j = r � 1. For each �xed

d = 1; 2; : : : ; k, any selection of d objects from S1 (SnS2) together with any

selection of r�d objections from S2 would yield a selection of r objects from

S. The total number of such selections is�n�r+1

d

��r�1

r�d�. Conversely, each

selection of r objects from S clearly must arise in this manner. Summing

over d = 1; 2; : : : , (�) follows.4. Let x be a real number with 0 < x < �. Prove that, for all natural

numbers n, the sum

sinx+sin3x

3+

sin5x

5+ � � �+

sin(2n� 1)x

2n� 1

is positive.

Solutions by �Sefket Arslanagi�c, Berlin, Germany; and by Vedula N.Murty, Andhra University, Visakhapatnam, India. We give Murty's solu-tion.

We use mathematical induction. Let

Sn(x) =

nXk=1

sin(2k� 1)x

(2k� 1):

S1(x) = sinx > 0 for x 2 (0; �). Thus the proposed inequality is true

for n = 1. Let Sr(x) > 0 for r = 1; 2; : : : ; n � 1. We will deduce that

Sn(x) > 0 for x 2 (0; �). Suppose that Sn(x0) � 0 for some x0 2 (0; �),

and that Sn(x) attains its minimum at x = x0. Hence ddx[Sn(x)]x=x0 = 0.

That is

S0n(x0) =nX

k=1

cos((2k� 1)x0) = 0;

so that

2 sinx0S0n(x0) =

nXk=1

2 cos((2k� 1)x0) sinx0

=

nXk=1

[sin(2kx0)� sin((2k� 2)x0)]

= sin2nx0:

13

Thus S0n(x0) =sin2nx02 sinx0

= 0 implying sin2nx0 = 0. Hence

x0 2��

2n;2�

2n;3�

2n; : : : ;

(2n� 1)

2n

�:

It is easily veri�ed that at each of these values Sn(x0) > 0, a contradiction.

Hence Sn(x) > 0 for x 2 (0; �).

Editor's Note: Both solutions used the calculus. Does anyone have a

more elementary solution?

We complete this number of the Corner with solutions by our readers

to problems of the 1992 Dutch Mathematical Olympiad, Second Round given

in the June 1995 number of the Corner [1995; 192{193].

1992 DUTCH MATHEMATICAL OLYMPIAD

Second RoundSeptember 18, 1992

1. Four dice are thrown. What is the chance that the product of the

numbers equals 36?

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,Ontario.

There are four di�erent kinds of outcomes in which the product is 36:

each of f1; 1; 6; 6g and f2;2; 3; 3g can occur in 4!

2!2!= 6 ways; f1; 4; 3; 3g can

occur in 4!2!

= 12 ways; and f1; 2; 3; 6g can occur in 4! = 24 ways. Hence the

probability that the product equals 36 is 4864

= 127.

2. In the fraction and its decimal notation (with period of length 4)

every letter represents a digit. Di�erent letters denote di�erent digits. The

numerator and denominator are mutually prime. Determine the value of the

fraction:ADA

KOK= :SNELSNELSNELSNEL : : :

[Note. ADA KOK is a famous Dutch swimmer. She won gold in the

1968 Olympic Games in Mexico. SNEL is Dutch for FAST.]

Solution by the Editor.Let x = ADA

KOK= :SNEL. Then 104x = SNEL:SNELSNEL : : :

and 104x� x = SNEL. So

x =SNEL

9999=

SNEL

11� 909=

SNEL

33� 303=

SNEL

99� 101:

Taking KOK = 909 we obtain SNEL = 11� ADA, and A = L, which is

impossible.

TakingKOK = 303, we obtain SNEL = 33�ADA, so 3A < 10, as

the product has four digits and 3A = L. Because S 6= L 6= 0, 3 � 3A < 9,

14

giving A = 1 or A = 2. Now A = 1 gives L = 3 = K, which is impossible,

so A = 2. This gives L = 6, and D � 2, so there is a carry. This gives

D � 4, as A = 2, K = 3.

For D = 4, ADAKOK

= :SNEL is 242303

= :7986, a solution.

For D = 5, ADA = 252 is not coprime to KOK = 303.

For D = 6, SNEL = 8646 and N = L.

For D = 7, SNEL = 8976 and D = E.

For D = 8, SNEL = 9036 and O = N .

For D = 9, SNEL = 9636 and N = L.

Taking KOK = 101 gives SNEL = 99 � ADA forcing A = 1 for

SNEL to have four digits, but then A = K.

Thus the only solution is 242303

= :7986.

3. The vertices of six squares coincide in such a way that they enclose

triangles; see the picture. Prove that the sum of the areas of the three outer

squares (I, II and III) equals three times the sum of the areas of the three

inner squares (IV, V and VI).

�����BBBBB���""

""TTTT""""aa

aaaa�

�����aaaaaa

@@@BBBBB

���PPP�����

PPP

TTTT����

�����

���@@@

III

III

VIV

VI

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;and by Vedula N. Murty, Andhra University, Visakhapatnam, India. Wegive Murty's solution.

Let the �gure on the next page be labelled as shown:

Let MN = x1, LU = x3, UY = x2, AC = x4, AB = x5, BC = x6,

\MBN = �, \LAU = �, \V CY = , \BAC = A, \ACB = C and

\ABC = B.

Then we have �+ � = �, � + A = �, + C = �

x21 = x26 + x25 � 2x5x6 cos�

x22 = x24 + x26 � 2x4x6 cos

x23 = x24 + x25 � 2x4x5 cos�

9>>>>=>>>>;: : : (1)

15

S

T

U

V

W

X

YN

P

Q

M

L

A

B C

����

���BBBBBBB����""

""""TTTTTT""""""aa

aaaa

aa��������aaaaaaaa

@@@@BBBBBBB

����PPPP�������

PPPP

TTTTTT�

�����������

����@

@@@

III

III

VIV

VI

x24 = x25 + x26 � 2x5x6 cosB

x24 = x24 + x26 � 2x4x6 cosC

x26 = x24 + x25 � 2x4x6 cosA

9>>>>=>>>>;: : : (2)

From (2), we have

x24 + x25 + x26 = 2x4x5 cosA+ 2x5x6 cosB + 2x4x6 cosC

= �2x4x5 cos� � 2x5x6 cos�� 2x4x6 cos : : : (3)

From (1), we have

x21 + x22 + x23 = 2(x24 + x25 + x26)� 2x5x6 cos�

�2x4x5 cos� � 2x4x6 cos

using (3)

= 2(x24 + x25 + x26) + x24 + x25 + x26

= 3(x24 + x25 + x26):

That is, Area of (I + II + III) = 3 Area of (IV + V + V I).

4. For every positive integer n, n? is de�ned as follows:

n? =

(1 for n = 1

n

(n�1)?for n � 2

Provep1992 < 1992? < 4

3

p1992.

16

Solutions by Vedula N. Murty, Andhra University, Visakhapatnam, In-dia; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-tario. We give Wang's solution and his remark.

Using the more convenient notation f(n) for n?, we show that in gen-

eral pn+ 1 < f(n)<

4

3

pn (�)

for all even n � 6. In particular, for n = 1992, we would getp1993 <

f(1992)< 43

p1992.

First note that f(n) = nf(n�1)

= nn�1

f(n�2) for all n � 3. IfN = 2k

where k � 2, then multiplying f(2q) = 2q

2q�1f(2q� 2) for q = 2; 3; : : : ; k,

we get

f(2k) =4

3�6

5� � �

2k

2k� 1� f(2)

=

�2

1

���4

3

���6

5

�� � ��

2k

2k � 1

�>

�3

2

��5

4

��7

6

�� � ��2k + 1

2k

�:

Hence

(f(2k))2 >2 � 4 � 6 � � � 2k

1 � 3 � 5 � � � (2k� 1)�3 � 5 � 7 � � � (2k+ 1)

2 � 4 � 6 � � � 2k= 2k+ 1;

from which it follows that

f(n) = f(2k)>p2k+ 1 =

pn+ 1: (1)

On the other hand, for k � 3 we have

2(2k) =

�2

3

��4

5

��6

7

�� � ��2k� 2

2k� 1

�� 2k

<

�2

3

��5

6

��7

8

�� � ��2k� 1

2k

�2k:

Hence

(f(2k))2 <

�2

3

�2

�4 � 6 � � � (2k� 2)

5 � 7 � � � (2k� 1)�5 � 7 � � � (2k� 1)

6 � 8 � � � 2k� (2k)2

=

�2

3

�2

� 4 � 2k;

from which it follows that

f(n) = f(2k)<4

3

p2k =

4

3

pn: (2)

The result follows from (1) and (2).

17

Remark: Using similar arguments, upper and lower bounds for f(n)

when n is odd can also be easily derived. In fact, if we set P =1�3�5���(2k�1)

2�4�6���2k(usually denoted by

(2k�1)!!

(2k)!!) then various upper and lower bounds for P

abound in the literature; for example, it is known that

1

2

s5

4k+ 1� P �

1

2

s3

2k+ 1

and1q

(n+ 1

2)�

< P �1

pn�

:

(Compare, for example, x3.1.16 on p. 192 of Analytic Inequalities by D.S.

Mitronovi�c.)

Clearly, each pair of these double inequalities would yield correspond-

ing upper and lower bounds for the function f(n) considered in the given

problem.

5. We consider regular n-gons with a �xed circumference 4. We call

the distance from the centre of such a n-gon to a vertex rn and the distance

from the centre to an edge an.

a) Determine a4; r4; a8; r8.

b) Give an appropriate interpretation for a2 and r2.

c) Prove: a2n = 1

2(an + rn) and r2n =

pa2nrn.

an rn

Let u0; u1; u2; u3; : : : be de�ned as follows:

u0 = 0; u1 = 1; un =1

2(un�2 + un�1) for n even and

un =pun�2 � un�1 for n odd:

d) Determine: limn!1 un.

Solution by Vedula N. Murty, Andhra University, Visakhapatnam, In-dia.

Let O be the centre of the regular n-gon. Let A1A2 denote one side of

the regular n-gon

18

A1 A2

O

rnan

Then we have \A1OA2 = 2�n, \OA1A2 = \OA2A1 = �

2� �

n. Thus

j���!A1A2j =

rr2n + r2n � 2r2n cos

2�

n

=

r2r2n(1� cos

2�

n)

=

r4r2n sin

2 �

n= 2rn sin

�

n:

The circumference of the regular n-gon is 2nrn sin�n= 4 whence

rn =2

n sin �n

;

an = rn sin

��

2��

n

�= rn cos

�

n=

2

ncot

�

n:

In particular

r4 =1

2

1

sin �4

=

p2

2; a4 =

2

4cot

�

4=

1

2;

r8 =2

8 sin �8

=1

4 sin �8

:

Now, cos �4= 1p

2= 1� 2 sin2 �

8gives

sin�

8=

1

2

q2�

p2;

so

r8 =1

4

2p2�

p2=

1

2�

1p2�

p2;

and

a8 = r8 cos�

8=

1

4

s2 +

p2

2�p2=

1

4

1

2�p2

p2;

since cos �4= 2 cos2 �

8� 1.

For (b), r2 = 1, a2 = 0 as the 2-gon is a straight line with O lying at

the middle of A1 and A2.

19

For (c), we have

an + rn = rn

�1 + cos

�

n

�= 2rn cos

2�

2n

=4

n sin �n

cos2�

2n

=4

2n sin �2n

cos �2n

cos2�

2n=

2

ncot

�

2n:

Thus 1

2(an + rn) =

1

ncot( �

2n) = a2n, and

a2nrn =1

n

cos �2n

sin �2n

�2

n sin �n

=1

n2

cos �2n

sin2 �2n

cos �2n

=1

n2 sin2 �2n

;

sopa2nrn = 1

n sin �2n

= r2n.

For (d), note u0 = 0, u1 = 1, and u2 = 12. For n � 2 we have

that un is either the arithmetic or geometric mean of un�1 and un�2 and

in either case lies between them. It is also easy to show by induction that

u0; u2; u4; : : : form an increasing sequence, and u1; u3; u5; : : : form a de-

creasing sequence with u2l � u2s+1 for all l; s � 0. Let limk!1 u2k = P

and limk!1 u2k+1 = I. Then P � I. We also have from u2n = 12(u2n�1 +

u2n�2) that P = 12(I + P ) so that I = P and limn!1 un exists. Let

limn!1 un = L.

With a2 = 0 and r2 = 1, let u2k = a2k+1 and u2k+1 = r2k+1 , for

k = 0; 1; 2; : : : . From (c), u0 = a21 = a2 = 0 and u1 = r21 = r2 = 1.

Also for n = 2k + 2, u2k+2 = a2k+1+1 = a2�2k+1 = 12(a2k+1 + b2k+1) =

12(u2k + u2k+1); that is un = 1

2(un�2 + un�1) and for n = 2k+ 3

u2k+3 = u2(k+1)+1 = r2k+1+1 = r2(2k+1)

=pa2(2k+1) � r2k+1 =

pa2k+1+1 � r2k+1

=pu2(k+1) � u2k+1

so un =pun�1 � un�2. Thus un and un satisfy the same recurrence and it

follows that L = limk!1 a2k+1 = limk!1 r2k+1. Now, from the solution

to (c),

rn =2

n sin �n

=2

�

�n

sin �n

;

so limn!1 rn = 2�since �

n! 0. Therefore limn!1 un = 2

�.

That completes the column for this issue. Olympiad season is approach-

ing. Send me your contests and nice solutions.

20

BOOK REVIEWS

Edited by ANDY LIU

Shaking Hands in Corner Brook and other Math Problems,edited by Peter Booth, Bruce Shawyer and John Grant McLoughlin,

published by the Waterloo Mathematics Foundation, Waterloo, 1995,

153 pages, paperback, ISBN 0-921418-31-0.

Reviewed by Robert Geretschl�ager and Gottfried Perz.

� One representative from each of six regions met in Corner Brook, New-

foundland, to discuss math problems. Each of these delegates shook

hands with each other delegate. How many handshakes were there?

This is the �rst of many problems that can be found in this collec-

tion of problems from the Newfoundland and Labrador Teachers' Association

(NLTA) SeniorMathematics League, which conveniently o�ers an explanation

as to the whereabouts of Corner Brook.

The NLTA Math League was started in 1987, and has since developed

into a very interesting competition at the regional level. A number of aspects

make this competition di�erent from most math competitions. First of all,

it is purely a team competition, with four students from each participating

school comprising a team. There is no individual ranking, and so students

are motivated to work together at �nding solutions. For each of ten ques-

tions posed, a team can receive �ve points for a correct team answer. If

the members of a team cannot agree on the correct answer, they can submit

individual answers, for which their team can get one point each, if correct.

Finally, there is a relay question, made up of four parts. In the relay, each

part yields an answer, which is necessary to be able to solve the next part

(much as in the American Regions Mathematics League (ARML), which may

be better known to many readers). The relay section can yield a maximum of

15 points (made up of �ve points for the solution and extra points for solving

the problems in a short time), for a possible total of 65 points. If teams end

up with the same point sum, a tie breaker question is posed.

The concept behind this competition is geared to fostering cooperative

problem solving, something that is generally ignored in olympiad-style com-

petitions. The level of di�culty of the problems posed is adequate to the

time allowed (usually from 3 to 10 minutes per question) and the intentions

of the competition, and ranges from fairly easy to pre-olympiad level. The

book is divided into sections covering regular questions, relay questions, tie

breakers and solutions. The problems are in a random order, and no indica-

tion is given of which questions were posed at which competition. Perhaps

at least one example of ten speci�c questions posed at one competition, and

the order they were posed in, might have been of interest.

21

Here are a few problems to whet your appetite:

� How many three digit numbers include at least one seven but have no

zeros?

� Al, Betty, Charles, Darlene and Elaine play a game in which each is

either a frog or a moose. A frog's statement is always false while a

moose's statement is always true.

Al says that Betty is a moose.

Charles says that Darlene is a frog.

Elaine says that Al is not a frog.

Betty says that Charles is not a moose.

Darlene says that Elaine and Al are di�erent kinds of animals.

How many frogs are there?

� Triangle ABC is isosceles, with \ABC = \ACB. There are points

D; E and F onBC; CA andAB, respectively, that form an equilateral

triangle. Given that \AFE = x� and \CED = y�, calculate \BDFin terms of x and y.

The book has a very pleasing layout, with the cover showing the densest

packing of seven circles in an equilateral triangle. The solutions are nicely

presented, and in several cases, alternate solutions are given, occasionally

labeled the \routine way" and the \subtle" or \smart way".

ShakingHands inCorner Brook shouldbe of interest to anyone involvedwith high school mathematics, either in competitions, or simply seeking en-

richment material for the interested student.

Copies of the above reviewed book may be obtained from:

Canadian Mathematics Competition

Faculty of Mathematics, University of Waterloo

Waterloo, Ontario, Canada. N2L 3G1

The cost is $12 in Canadian funds (plus 7% GST for shipping to Canadian

addresses). Cheques or money orders in Canadian funds should be made

payable to: Canadian Mathematics Competition. All pro�ts from the sale of

this book are for the Newfoundland Mathematics Prizes Fund.

22

Teachers interested in providing a lively and stimulating high school

mathematics competition for their students may be interested in participat-

ing in a NLTA Senior Mathematics League in their own area. Shawn Godin,

St. Joseph Scollard Hall, North Bay, Ontario, a regular contributor to CRUX,

is already a participant. Meetings can be held within individual schools, or

between teams from more than one school.

Sample games and other information on how the NLTA Senior Mathe-

matics League is organised may be obtained free from:

Dr. Peter Booth

Department of Mathematics and Statistics

Memorial University of Newfoundland

St. John's, Newfoundland, Canada. A1C 5S7

Tel: int+ 709{737{8786

Fax: int+ 709{737{3010

email: pbooth@fermat.math.mun.ca

Schools that participate on a regular basis will be sent questions and

detailed solutions �ve times per year (October, November, February, March

and May). There is an annual fee of $50 (Canadian funds) for each group of

schools participating. Cheques or money orders should be made payable to

Newfoundland Mathematics Prizes Fund.

23

A Probabilistic Approach

to Determinants

with Integer Entries

Theodore Chronis

It is well known that the probability for an integer number to be odd

is equal to the probability for the number to be even. What about determi-

nants? What is the probability for a rectangular matrix with integer entries

to have odd determinant? More generally, ifm is a natural number, what is

the probability for which detA � mi modm, where mi is chosen from the

set f0; 1; 2; : : : ;m� 1g?I have the following problem to propose; I hope you will �nd it inter-

esting.

Let A be an n � n matrix whose elements are integers. What is the

probability the determinant of A is an odd number?

Solution:

Let A = [aij ] ; i; j = 1; : : : ; n. It is obvious that

detA � det ([aij mod 2]) mod 2:

So the problem is to �nd the probability that the determinant of an n � n

matrix with elements from the set f0; 1g is an odd number. Let An be an

n� n matrix with elements from the set f0; 1g. Let also N(detAn) be the

number of odd n � n determinants, and P (detAn) be the corresponding

probability. Let

f(K = fk1; k2; : : : ; kng) = N

0BBBB@

����������

k1 k2 :: :: kna21 a22 :: :: a2n: : :

: : :

an1 an2 :: :: ann

����������

1CCCCA ;

where k1; k2; : : : ; kn 2 f0; 1g are �xed.

Then N(detAn) =P(f(K)), where the sum is calculated for all

the 2n � 1 possible permutations K = fk1; k2 : : : ; kng with ki 2 f0; 1g,i = 1; : : : ; n and k1; k2; : : : ; kn not all zero. (Note that f(0;0; : : : ; 0) = 0:)

24

Lemma: f(K) = 2n�1N(detAn�1), where k1 + k2 + : : :+ kn 6= 0.

Proof. It is well known that a determinant remains unchanged if from the

elements of one of its columns we substract the corresponding elements of

another column. It is also obvious that the same is true for N(detAn).

Additionally, a determinant just changes sign if we interchange two columns,

while N(detAn) remains unchanged. So

N

0BBBB@

����������

k1 k2 :: :: kna21 a22 :: :: a2n: : :

: : :

an1 an2 :: :: ann

����������

1CCCCA = N

0BBBB@

����������

1 0 :: :: 0

a21 a22 :: :: a2n: : :

: : :

an1 an2 :: :: ann

����������

1CCCCA

() f(K) = 2n�1N(detAn�1):

Hence N(detAn) = 2n�1N(detAn�1)(2n� 1).

Of course N(detA1) = 1 and so

N(detAn) = 2n(n�1)=2

nYi=1

(2i� 1):

Finally, P (detAn) =N(detAn)

2n2 () P (detAn) =

Qn

i=1(1� 2�i).

Note: The in�nite sequenceQn

i=1(1� 2�i); n = 1; 2; : : : is decreasing and

bounded below by 0, so limn!1P (detAn) exists. UsingMathematica, we

found that

limn!1

P (detAn) �= 0:288788095086602421278899721929 : : : :

Ayras 15, Ki�sia

Athens 14562, GREECE

e-mail: tchronis@egnatia.ee.auth.gr

25

THE SKOLIAD CORNERNo. 19

R.E. Woodrow

The problem set we give in this issue comes to us with our thanks from

Tony Gardiner of the UK Mathematics Foundation, School of Mathematics,

University of Birmingham. The Nat West JuniorMathematical Challengewas

written Tuesday, April 26, 1994 by about 105,000 students. Students from

England and Wales must be in school year 8 or below. The use of calculators,

calendars, rulers, and measuring instruments was forbidden.

1994 NAT WEST UK JUNIOR MATHEMATICAL

CHALLENGE

Tuesday, April 26, 1994 | Time: 1 hour

1. 38 + 47 + 56 + 65 + 74 + 83 + 92 equals

A. 425 B. 435 C. 445 D. 456 E. 465.

2. What is the largest possible number of people at a party if no two of

them have birthdays in the same month?

A. 11 B. 12 C. 13 D. 23 E. 334.

3. I have $500 in 5p coins. How many 5p coins is that?

A. 100 B. 500 C. 1000 D. 2500 E. 10000.

4. What was the precise date exactly sixty days ago today? (No calendars!)

A. Friday 25th February B. Saturday 26th February

C. Friday 26th February D. Saturday 27th February

E. Tuesday 26th February.

5. You have to �nd a route fromA toB moving hor-

izontally and vertically only, from one square to

an adjacent square. Each time you enter a square

you add the number in that square to your total.

What is the lowest possible total score for a route

from A to B?

3 9 B

8 5 6

9 11 7

A 8 10

A. 28 B. 29 C. 30 D. 31 E. 34.

26

6. On a clock face, how big is the angle between the lines joining the centre

to the 2 O'clock and the 7 O'clock marks?

A. 160� B. 150� C. 140� D. 130� E. 120�.

7. Gill is just six and boasts that she can count up to 100. However, she

often mixes up nineteen and ninety, and so jumps straight from nine-

teen to ninety one. How many numbers does she miss out when she

does this?

A. 70 B. 71 C. 72 D. 78 E. 89.

8. In how many ways can you join the two shapes

shown here to make a �gure with a line of

symmetry?

A. 0 B. 1 C. 2 D. 3 E. 4.

9. If you divide 98765432 by 8, which non-zero digit does not appear in

your answer?

A. 2 B. 4 C. 6 D. 8 E. 9.

10. How many numbers between 20 and 30 (inclusive) cannot be written

as a multiple of 5, or as a multiple of 7, or as the sum of a multiple of

5 and a multiple of 7?

A. 1 B. 2 C. 3 D. 5 E. 6.

11. Four children are arguing over a broken toy. Alex says Barbara broke

it. Barbara says Claire broke it. Claire and David say they do not know

who broke it. Only the guilty child was lying. Who broke the toy?

A. Alex B. Barbara C. Claire D. David E. can't be sure.

12. The diagram is made up of one circle

and two semicircles. Which of the three

regions | the black region, the white

region and the large circle | has the

longest perimeter?

27

A. the black region B. the circle C. the white region

D. black and white are equal and longest

E. all three perimeters are equal.

13. From my house the church spire is in the direction NNE. If I face in this

direction and then turn anticlockwise through 135� I can see the Town

Hall clock. In which direction am I then facing?

A. WSW B. due West C. SW D. due South E. SSE.

14. Samantha bought seven super strawberry swizzles and ten tongue twist-

ing to�ees for $1.43. Sharanpal bought �ve super strawberry swizzles

and ten tongue twisting to�ees for $1.25. How much is one tongue

twisting to�ee?

A. 7p B. 8p C. 9p D. 10p E. 18p.

15. I am forty eight years, forty eight months, forty eight weeks, forty eight

days and forty eight hours old. How old am I?

A. 48 B. 50 C. 51 D. 52 E. 53.

16. The numbers 1 to 12 are to be placed

so that the sum of the four numbers

in each of the six rows is the same.

Where must the 7 go?

4

9

C E

3 1

6 D

A B

8 5

A. at A B. at B C. at C D. at D E. at E

17. Three hedgehogs | Roland, Spike and Percival | have a leaf collecting

race. Roland collects twice as many as Percival, who collects one and

a half times as many as Spike. (Spike is moulting, and so has fewer

prickles for her to stick the leaves onto.) Between them they collect

198 leaves. How many did Spike manage to collect?

A. 18 B. 22 C. 36 D. 44 E. 66.

28

18. The four digits 1, 2, 3, 4 are writtin in increasing order. You must insert

one plus sign and one minus sign between the 1 and the 2, or between

the 2 and the 3, or between the 3 and the 4, to produce expressions

with di�erent answers. For example,

1� 23 + 4 gives the answer� 18:

How many di�erent positive answers can be obtained in this way?

A. 2 B. 3 C. 4 D. 5 E. 6.

19. Roger Rabbit has twice as many sisters as brothers. His sister Raquel

notices that 2=5 of her brothers and sisters are boys. How many Rabbit

children are there in the family?

A. 2 B. 4 C. 8 D. 16 E. 32.

20. The population of a new town in 1990 was 10; 000. It has since doubled

every year. If it kept on doubling every year for ten years, what would

its population be in the year 2000?

A. 100; 000 B. 200; 000 C. 1; 000; 000 D. 2; 000; 000 E. 10; 000; 000.

21. LMNO is a square. P is a point inside the square such that NOP is

an equilateral triangle. How big is the angle PMN?

A. 75� B. 70� C. 60� D. 45� E. 30�.

22. If the perimeter of a rectangle is 16x+18 and its width is 2x+6, what

is its length?

A. 18x+ 24 B. 7x+ 6 C. 12x+ 6 D. 6x+ 3 E. 14x+ 12.

23. In a group of �fty girls each one is either blonde or brunette and is either

blue-eyed or brown-eyed. Fourteen are blue-eyed blondes, thirty one

are brunettes and eighteen are brown-eyed. How many are brown-eyed

brunettes?

A. 5 B. 7 C. 9 D. 13 E. 18.

24. A bottle of JungleMonster Crush (JMC)makes enough drink to �ll sixty

glasses when it is diluted in the ratio 1 part Crush to 4 parts water. How

many glasses of drink would a bottle of JMC make if it is diluted in the

ratio 1 part Crush to 5 parts water?

A. 48 B. 60 C. 72 D. 75 E. 80.

29

25. A 3 by 3 by 3 cube has three holes, each

with a 1 by 1 cross section running from

the centre of each face to the centre of

the opposite face. What is the total sur-

face area of the resulting solid?

PPPP

PP

����

��PPPPPP

������

PPPP

PP

����

��

��

��PP

PP

PP��

��PP��PP

A. 24 B. 48 C. 72 D. 78 E. 84.

That completes the Skoliad Corner for this issue. Send me your con-

tests, suggestions, and recommendations to improve this feature.

Introducing the new Associate Editor-in-Chief

For those of you who do not know Clayton, here is a short pro�le:

Born: Haystack, Newfoundland 1

Educated Haystack School School, Newfoundland

Thornlea School, Newfoundland

Memorial University of Newfoundland

Queen's University, Kingston, Ontario

Employment Random South School Board,

Trinity Bay, Newfoundland

Memorial University of Newfoundland

Mathematical Mathematical Education

Interests Commutative Algebra

1 You may have di�culty �nding Haystack on a map of Newfoundland | it is not lost

| it no longer exists!

30

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

the Mayhem Editor, Naoki Sato, Department of Mathematics, University of

Toronto, Toronto, ON Canada M5S 1A1. The electronic address is

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Editorial

It gives me great pleasure to unveil the premiere issue of \Crux Mathemati-

corum with Mathematical Mayhem". This merger has been in the works for

quite some time, and it has �nally been successfully realized.

For the bene�t of those who have not heard ofMayhem, I will provide a

brief description. Mayhem was founded in 1988 by two high school students

Ravi Vakil and Patrick Surry, who wished to establish a journal speci�cally

oriented towards students, and totally operated by students. Although the

journal has been passed down through many hands, and though it has not

always been easy, this mandate has always been resolutely upheld; it has

made Mayhem a unique and exceptional journal. And rest assured, we will

still be running our share of \Crux with Mayhem".

Our features include articles, olympiads, and a problems section. The

material is generally focused towards contests and olympiads, and how to

prepare for them, and the topics range from high school mathematics to un-

dergraduate material. We cannot emphasize enough that we are a journal

dedicated to mathematics students. I myself am a fourth-year student at the

University of Toronto, and Cyrus Hsia (the Mayhem Assistant Editor) is a

third-year student, also at the University of Toronto.

Our fearless sta� also consists of undergraduate and high school stu-

dents.

This brings me to my next point. After considerable discussion, \May-

hem" has decided to restrict itself to publishingsolutions only from students.

The rationale behind this move is that the Crux problems already draw many

solutions, and if the same people were to respond to our problems, which are

considerably easier, it would simply overwhelm the section. We know that

31

there are many non-students who have contributed to the problems sections

over the years, who have our full gratitude, and hope they understand our

position. We are, however, prepared to make exceptions in, well, exceptional

cases.

However, we warmly welcome submissions for articles and problems

from all people. Back issues are available; the information is inside the back

cover. Any correspondence about Mayhem should be sent to Mathemat-

ical Mayhem, c/o Naoki Sato, Department of Mathematics, University of

Toronto, M5S 1A1, or at the e-mail address <mayhem@math.toronto.edu>.

Subscriptions, however, should be sent to the Canadian Mathematical So-

ciety o�ces, as mentioned on the inside back cover.

Well, I think that's about it. For people who have subscribed to May-

hem, welcome back, I know it's been a long wait. I look forward to working

with Bruce Shawyer on our new project (I think he will bring a certain \dis-

cipline" to Mayhem, but we will resist it as much as possible. Don't tell him

I said that.) Here's to a new year and a new era for Mayhem.

Naoki Sato

Mayhem Editor

Shreds and Slices

Positive Matrices and Positive Eigenvalues

Theorem. An n � n matrix M with positive real entries has at least one

positive eigenvalue.

Proof. Let S = f(x1; x2; : : : ; xn) j x1; x2; : : : ; xn � 0; x21+x22+ � � �+x2n =

1g; that is, S is the portion of the unit sphere inRn with all coordinates non-

negative.

De�ne the map f : S ! S by f(~v) = M~vjM~vj . Since M has all positive real

entries, for any ~v 2 S, M~v also has all non-negative coordinates, so f is

well-de�ned, and does indeed map S into S.

Note that S is a closed, simply connected set. Then by Brouwer's Fixed

Point Theorem, there is a �xed point of f ; that is, for some ~v 2 S, f(~v) =

~v = M~vjM~vj ) M~v = �~v, for some positive value � (since � cannot be zero),

namely � = jM~vj. This � is a positive eigenvalue ofM .

32

Newton's Relations

Given n reals a1, a2, : : : , an, let Sk be the sum of the products of the

ai taken k at a time, and let Pk = ak1+ak2+� � �+akn. Consider the generating

functions

S0 + S1x+ S2x2 + � � �+ Snx

n

= 1+ (a1 + a2 + � � �+ an)x+ (a1a2 + a1a3 + � � �+ an�1an)x2

+ � � �+ a1a2 � � � anxn

= (1 + a1x)(1 + a2x) � � � (1 + anx)

and

P0 � P1x+ P2x2 � � � �

= (1� a1x+ a21x2 � � � � ) + (1� a2x+ a22x

2 � � � � )+ � � �+ (1� anx+ a2nx

2 � � � � )

=1

1 + a1x+

1

1 + a2x+ � � �+

1

1 + anx:

Their product is

(n� P1x+ P2x2 � � � � )(1 + S1x+ S2x

2 + � � � )= (1 + a1x)(1 + a2x) � � � (1 + anx)

��

1

1 + a1x+

1

1 + a2x+ � � �+

1

1 + anx

�;

since P0 = n and S0 = 1.

We claim the expression is equal to n+ (n� 1)S1x+ (n� 2)S2x2 + � � �+

Sn�1xn�1. To see this, consider the coe�cient of xk. Since the expression

is symmetric, the coe�cient is some multiple of Sk. How many times does

the term a1a2 � � � akxk appear? It must have appeared in the product (1 +

a1x)(1 + a2x) � � � (1 + anx) as a1a2 � � � akal, where k < l � n, before

having the term al divided out. There are n� k choices for l, and hence the

coe�cient is (n� k)Sk.

Hence,

(n� P1x+ P2x2 � � � � )(1 + S1x+ S2x

2 + � � � )= n+ (n� 1)S1x+ (n� 2)S2x

2 + � � �+ Sn�1xn�1;

and equating coe�cients:

nS1 � P1 = (n� 1)S1

nS2 � S1P1 + P2 = (n� 2)S2

nS3 � S2P1 + S1P2 � P3 = (n� 3)S3

� � �nSn�1 � Sn�2P1 + Sn�3P2 � � � �+ (�1)n�1Pn�1 = Sn�1;

33

or

P1 � S1 = 0

P2 � S1P1 + 2S2 = 0

P3 � S1P2 + S2P1 � 3S3 = 0

� � �Pn�1 � S1Pn�2 + S2Pn�3 � � � �+ (�1)n�1(n� 1)Sn�1 = 0;

and

Pm � S1Pm�1 + S2Pm�2 � � � �+ (�1)nPm�nSn = 0

form � n:

The last equation is the well-known recursion sequence for the Pi, and the

previous equations (known as Newton's relations) can help pin down the

values of P1, P2, : : : , Pn�1, or vice-versa.

Problem. If

x+ y + z = 1;

x2 + y2 + z2 = 2;

x3 + y3 + z3 = 3;

determine the value of x4 + y4 + z4.

Solution. Newton's relations become

P1 � S1 = 1� S1 = 0;

P2 � S1P1 + 2S2 = 2� S1 + 2S2 = 0;

P3 � S1P2 + S2P1 � 3S3 = 3� 2S1 + S2 � 3S3 = 0;

which imply that S1 = 1, S2 = �1=2, and S3 = 1=6. Also, P4 � 3S1 +

2S2 � S3 = 0 ) P4 = 25=6.

Here is the last problem of the 1995 Japan Mathematical Olympiad, Final

Round.

Problem. Let 1 � k � n be positive integers. Let a1, a2, : : : , ak be complex

numbers satisfying

a1 + a2 + � � �+ ak = n

a21 + a22 + � � �+ a2k = n

� � �ak1 + ak2 + � � �+ akk = n

Show that (x+a1)(x+a2) � � � (x+an) = xk+�n1

�xk�1+

�n2

�xk�2+� � �+

�nk

�.

Solution. Given P1 = P2 = � � � = Pk = n, we must �nd S1, S2, : : : , Sk. We

will prove that Sm =�n

m

�by induction. Clearly S1 = P1 = n =

�n

1

�.

34

Now, for some m, assume S1 =�n

1

�, S2 =

�n

2

�, : : : , Sm�1 =

�n

m�1

�. Then

by the equations above, Pm�S1Pm�1+S2Pm�2�� � �+(�1)m�1Sm�1P1+

m(�1)mSm = 0, or

n� n

�n

1

�+ n

�n

2

�� � � �+ (�1)m�1n

�n

m� 1

�+m(�1)mSm = 0

so that

m

nSm =

�n

m� 1

���

n

m� 2

�+

�n

m� 3

�� � � � =

�n� 1

m� 1

�;

and further,

Sm =n

m

�n� 1

m� 1

�=

�n

m

�:

So by induction, we are done.

Mathematically Correct Sayings

[The following shred/slice appeared in the newsgroup rec.humor.funny.]

After applying some simple algebra to some trite phrases and cliches, a new

understanding can be reached of the secret to wealth and success. Here it

goes.

Knowledge is Power,

Time is Money,

and as everyone knows, Power is Work divided by Time.

So, substituting algebraic equations for these time worn bits of wisdom, we

get:

K = P (1)

T = M (2)

P = W=T (3)

Now, do a few simple substitutions. PutW=T in for P in equation (1), which

yields:

K = W=T (4)

PutM in for T into equation (4), which yields:

K = W=M (5)

Now we've got something. Expanding back into English, we get: Knowledge

equals Work divided by Money.

What this MEANS is that:

35

1. The More You Know, the More Work You Do, and

2. The More You Know, the Less Money You Make.

Solving for Money, we get:

M = W=K (6)

Money equals Work divided by Knowledge.

From equation (6) we see that Money approaches in�nity as Knowledge ap-

proaches 0, regardless of the Work done.

What THIS MEANS is: The More you Make, the Less you Know.

Solving for Work, we get

W =M �K (7)

Work equals Money times Knowledge

From equation (7) we see that Work approaches 0 as Knowledge approaches

0.

What THIS MEANS is: The stupid rich do little or no work.

Working out the socioeconomic implications of this breakthrough is left as

an exercise for the reader.

Contest Dates

Here are some upcoming (or in some cases, already past) contest dates to mark

on your calendar.

Contest Grade Date

Gauss Grades 7 & 8 Wednesday, May 14, 1997

Pascal Grade 9 Wednesday, February 19, 1997

Cayley Grade 10 Wednesday, February 19, 1997

Fermat Grade 11 Wednesday, February 19, 1997

Euclid Grade 12 Tuesday, April 15, 1997

Descartes Grades 12 & 13 Wednesday, April 16, 1997

CIMC Grades 10 & 11 Wednesday, April 16, 1997

AJHSME Grades 7 & 8 Thursday, November 21, 1996

AHSME High School Thursday, February 13, 1997

AIME High School Thursday, March 20, 1997

USAMO High School Thursday, May 1, 1997

AJHSME Grades 7 & 8 Thursday, November 20, 1997

COMC High School Wednesday, November 27, 1996

CMO High School Wednesday, March 26, 1997

APMO High School March, 1997

IMO High School July 18 { 31, 1997

36

A Journey to the Pole | Part I

Miguel Carri �on �Alvarezstudent, Universidad Complutense de Madrid

Madrid, Spain

For those of us who can not seem to get a strong grip on synthetic ge-

ometry, analytic geometry comes in handy. Even though polar coordinates

can be superior to rectangular coordinates in some situations, they are sys-

tematically ignored by instructors and students alike. The purpose of this

series is to introduce their uses with the idea that, as is always happening

in mathematics, with a little ingenuity, the concepts central to polar coordi-

nates can be applied elsewhere. This �rst article uses polar coordinates in

elementary geometry.

De�nition

In polar coordinates, the position of a point P is determined by the

distance r from a point O called the pole and the angle � between OP and a

semi-in�nite line called the polar axis. By convention, the polar axis is taken

O

P

r

Polar axis�

to be the positive x-axis, and the

transformation from polar to carte-

sian coordinates is given by x =

r cos �, y = r sin �. The in-

verse change of coordinates is not

so straightforward; The obvious ex-

pressions are r =px2 + y2, � =

arctan(yx), but the equation for �

is not single-valued, even if � is re-

stricted to [0; 2�), and � is unde-

�ned at the origin. Fortunately, we

need not worry about this: when

handling curves in polar coordinates, the change from rectangular to polar co-

ordinates is of little use, and it is convenient to allow r and � to take on all

real values. With this provision, a point can be referred to by an in�nite set of

coordinate pairs: (r; �) = ((�1)nr; �+n�). Unless you want to do multiple

integrals, this is not a problem, but rather something to exploit!

Polar Curves

Polar curves are usually written in the form r = r(�), and unlike curves

of the form y = y(x), they can be closed and need not be simple (they

can intersect themselves). Implicit curves of the form f(r; �) = 0 can be

even more general. From a cartesian equation, the substitution x = r cos �,

y = r sin � yields a polar expression. Throughout this article, I have tried

to avoid this whenever possible, and it turns out that it is always possible.

37

Some symmetries of the curves can be detected by checking the functions r(�)

or f(r;�) above for the following simple properties (this is not a complete

list):

� The curve is symmetric about the pole if r(�) = r(� + �) or

f(�r; �) = f(r;�)

� The curve has n-fold symmetry about the pole if r(�) = r(� + 2�n)

� The curve is symmetric about the polar axis if r(�) = r(��)

� The curve is symmetric about a line at an angle � to the polar axis if

r(�) = r(2�� �)

The following transformations are also useful:

� Any curve can be rotated through � by substituting � � � for �

� The x- and y-axes can be permuted by substituting �2� � for �

Example 1. The equation of a circle of radius a centered at the origin is

r = a.

Example 2. The equation of a line passing through the origin at an angle

� to the polar axis is � = �.

Exercise 1. Find the equation of a line at an angle � to the polar axis

passing at a distance d to the pole.

Exercise 2. Identify the curve r = 2a cos �.

The Cosine Law

More often than not, when working in polar coordinates, one uses

nothing but trigonometry, and the cosine law is the starting point of many

derivations. If you think about it, it comes closest to being a `vector addi-

tion rule' to use if you need to translate a curve, although this is best done

in rectangular coordinates. I will not give a translation rule, because it is

cumbersome and is of little use. Instead, I will use the cosine law to derive

the equation of a circle of radius � centered at (R;�) (see �gure). Applying

the cosine law to side � of4OCP , we have

�2 = R2 + r2 � 2Rr cos(� � �)

= [r � R cos(� � �)]2 +R2 � R2 cos2(� � �)

=) [r � R cos(� � �)]2 = �2 � R2 sin2(� � �):

38

� �

R

r

P

C

�

O

From this equation, it is evident that

if R � �, then the curve is de-

�ned for a limited range of � given by

� �

R� sin(���) � �

R, as we would

expect when the origin lies outside

the circle. The squared length of the

tangent from O, when sin(�� �) =�

R, is P = R2 cos2(���) = R2(1�

�2

R2 ) = R2 � �2; this is called the

potence of the origin w.r.t. the circle. Note that this formula is correct, even

when R < � and sin � = �

Rhas no solution. Incidentally, the solution to

Exercise 2 can be obtained easily by noting that if the origin is on the circle,

then R = �, and

r � R cos(� � �) = R cos(� � �) =) r = 2R cos(� � �):

Example 3. Polar equation of the el-

lipsewith one focus at the origin and

the main axis at an angle � to the

polar axis. The cosine law applied

to side F 0P of triangle FF 0P gives

�

F

�

F 0

P

2c

r

2a � r

(2a� r)2 = r2 + 4c2 � 4rc cos(� � �)

=) a2 � ar = c2 � rc cos(� � �)

=) r[a� c cos(� � �)] = a2 � c2

=) r =b2=a

1� e cos(� � �):

A Catalogue of Important Curves

The following curves are all important in their own right, but since their

polar expressions are particularly simple, they make good examples of the use

of polar coordinates.

39

Conic sections

We already have the equation for

the ellipse. The polar equation of

the parabola is even easier to derive.

In the �gure, we have the focus at

the origin, the axis at an angle � to

the polar axis and a distance d from

the focus to the directrix. From the

�gure on the previous page, we

have

q

q� �

FV

r

d + r cos � P&

r = d+ r cos(� � �) =) r =d

1� cos(� � �):

Exercise 3. In a similar way, derive the equation for the hyperbola, not-

ing how both branches are handled. Hence, deduce that the general equation

of the conic is r = de1�e cos(���) , where e is the eccentricity. This equation

can be obtained immediately from the de�nition of a conic as the locus of the

points whose distances to a line (called the directrix) and a point (called the

focus) are at a constant ratio e.

The Cardioid

The cardioid is the trajectory of a

point on a circle that rolls on another

circle of the same radius. So de-

�ned, it is a special case of the epicy-

cloid, which is the curve described

by a point on a circle rolling on an-

other circle with no restriction on

the radii; the general equation of the

epicycloid is best expressed in para-

metric form.

Rr

�

�

C0

P 0

O

P

�2

"

#

�2

C"�

In the �gure, the two circles have radius R. The condition that the one cen-

tered at C0 rolls on the one centered at C implies that triangles OCP 0 andPC0P 0must be congruent. In trianglesOCP 0 andPC0P 0, � = 2R cos(�=2).

Similarly, in triangle OP 0P , r = 2� cos(�=2). Putting all together, we have

r = 4R cos2(�=2) =) r = 2R(1 + cos �):

The cardioid is also a special case of Pascal's Lima�con, of the equation

r = b + a cos �. The cusp at O becomes a loop if b < a, and a smooth

indentation if b > a. The lima�con can be de�ned as the locus of the feet

40

of perpendiculars dropped from the origin to tangent lines to a circle. The

radius of the circle is b and the distance from O to its centre is a.

The Lemniscate

The lemniscate is the locus of the

points such that the product of their

distances to two points 2a apart

is a2. In the �gure, the cosine law

gives

�

O

P

a

a=tatr

a2t2 = r2 + a2 + 2ar cos �; a2=t2 = r2 + a2 � 2ar cos �

=) a4 = (r2 + a2)2 � 4a2r2 cos2 �

=) r4 = 2a2r2(2 cos2 � � 1)

=) r2 = 2a2 cos(2�):

The lemniscate is a special case of Cassini's ovals, which are de�ned in the

same way, but with no restriction on the product of the distances.

O�

aP

The Rose

Loosely related to the lemniscate are

the roses, of equation

r = a cos[n(� � �)]

with integer n. For odd n, the curve

has n `leaves' and it is traced com-

pletely when � varies from 0 to �.

For even n, the curve has 2n

`leaves', and it is traced completely only when � varies from 0 to 2� (see the

�gure).

More general curves can be obtained if n is rational or irrational. In

the �rst case there is an integer number of overlapping lobes and the curve

is closed, but in the latter case the curve never closes, and in fact it is dense

in the disc r � a.

The Spirals

Polar coordinates are particularly suited to spirals. The two most fa-

mous are the Archimedean Spiral r = a�, which is the trajectory of a point

whose angular and radial velocities are proportional, and the Logarithmic

Spiral r = ea�. In fact, any continuous monotonic function that goes to in-

�nity as � goes to in�nity de�ned in a semi-in�nite interval will give rise to

a spiral, like r = a ln�. A related feature of polar curves is the limit cycle,occurring when r(�) has a �nite limit r0 at in�nity. In that case, the curve

winds around the origin in�nitelymany times, approaching the circle r = r0.

Recognizing a limit cycle makes it easier to sketch a polar curve.

41

Straight Lines

We �nish by deriving the equation

of a straight line not passing through

the origin. From the �gure, we have

d = r cos(� � ), where d is the

distance from the line to the origin

and is the direction of the clos-

est point. An alternative form is

d = r sin(�� �), where � is the di-

rection of the line and 0 > ��� > �

(see the �gure).

� �O

r

d

Additional Problems

Problem 1. Considering r(�0 +�) and r(�0��), derive an expression

for a secant line to a conic. Passing to the limit �! 0, write an equation for

the tangent line at �0.

Problem 2. PQ is a chord through the focus F of a conic, and the

tangents at P , Q meet at T . Prove that T lies on the directrix corresponding

to F and that FT ? PQ.

Problem 3. Let s be the tangent line at the vertex of a parabola and t

be the tangent at P . If r and s meet at Q, prove that FQ bisects the angle

between FP and the axis of the parabola, and that FQ ? s.

IMO Report

Richard Hoshinostudent, University of Waterloo

Waterloo, Ontario.

After ten days of intensive training at the Fields Institute in Toronto,

the 1996 Canadian IMO team travelled to Mumbai, India to participate in the

37th International Mathematical Olympiad. For the �rst time in our team's

history, every team member brought home a medal, with three silver and

three bronze.

This year's team members were: Sabin \Get me a donut" Cautis, Adrian

\Da Chef" Chan, Byung Kyu \Spring Roll" Chun, Richard \YES! WE'VE GOT

BAGELS!"Hoshino, Derek \Leggomy Eggo"Kisman, and Soroosh \Mr. Bean"

Yazdani. Our team leaders were J.P. \Radishes" Grossman and Ravi \Oli"

Vakil (no, he's not Italian). Special thanks go out to our coaches, Naoki

\Dr. Cow" Sato and Georg \Where's my Ethanol" Gunther.

42

This year's paper was one of the most di�cult ever, and thus, the cuto�s

for medals were among the lowest in history, 28 for gold, 20 for silver and

12 for bronze. Only one student, a Romanian, received a perfect score of 42.

Our team's scores were as follows:

CAN 1 Sabin Cautis 13 Bronze Medal

CAN 2 Adrian Chan 14 Bronze Medal

CAN 3 Byung Kyu Chun 18 Bronze Medal

CAN 4 Richard Hoshino 22 Silver Medal

CAN 5 Derek Kisman 22 Silver Medal

CAN 6 Soroosh Yazdani 22 Silver Medal

Some weird coincidences: all the silver medallists got the exact same

score, are graduating and are headed to the University of Waterloo in Septem-

ber, and all the bronze medallists are eligible to return to Argentina for next

year's IMO. Overall, Canada �nished sixteenth out of seventy-�ve countries,

one of our highest rankings ever. A main reason for our success was our

combined team score of 36 out of 42 on question #6, a problem that many

countries answered very poorly (in fact, only two countries had more points

on that problem than we did, and they both got 37 out of 42). Unfortu-

nately, question #2 was answered very poorly by Canada, with only one 7,

even though the problem was created by our own team leader, J.P. Gross-

man.

We all owe special thanks to Dr. Graham Wright of the CanadianMath-

ematical Society, Dr. Bruce Shawyer of Memorial University and Dr. Richard

Nowakowski of Dalhousie University for their hard work and organization in

making our trip possible and Dr. Ed Barbeau of the University of Toronto

for all his commitment and dedication to training all the IMO team hopefuls

with his year-long correspondence program.

Overall, the experience was memorable for all of us, although we could

have done without the cockroaches in our rooms. Best of luck to all the

students who will be working hard to make the 1997 IMO team, which will

be held in Chapadmalal, Argentina near Mar del Plata.

Mayhem Problems

A new year brings new changes and new problem editors. Cyrus Hsia now

takes over the helm as Mayhem Advanced Problems Editor, with RichardHoshino �lling his spot as the Mayhem High School Problems Editor, andveteran Ravi Vakil maintains his post asMayhem Challenge Board ProblemsEditor. Note that all correspondence shouldbe sent to the appropriate editor| see the relevant section.

In this issue, you will �nd only problems | the next issue will feature

only solutions. We intend to have problems and solutions in alternate issues.

43

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions be submitted

by 1 June 1997, for publication in the issue 5 months ahead; that is, issue 6.

We also request that only students submit solutions (see editorial), but we

will consider particularly elegant or insightful solutions for others. Since this

rule is only being implemented now, you will see solutions from many people

in the next few months, as we clear out the old problems from Mayhem.

High School Problems

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,

Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>

H217. Let a1, a2, a3, a4, a5 be a �ve-term geometric sequence satis-

fying the inequality 0 < a1 < a2 < a3 < a4 < a5 < 100, where each term

is an integer. How many of these �ve-term geometric sequences are there?

(For example, the sequence 3, 6, 12, 24, 48 is a sequence of this type).

H218. A Star Trek logo is inscribed inside a

circle with centre O and radius 1, as shown. Points

A, B, and C are selected on the circle so that AB =

AC and arc BC is minor (that is ABOC is not a

convex quadrilateral). The area of �gure ABOC is

equal to sinm�, where 0 < m < 90 andm is an in-

teger. Furthermore, the length of arc AB (shaded as

shown) is equal to a�=b, where a and b are relatively

prime integers. Let p = a+ b+m.

A

O

������������SSSSBBBBBBBB

CB

��������������������������������������������������������������

(i) If p = 360, andm is composite, determine all possible values for m.

(ii) Ifm and p are both prime, determine the value of p.

H219. Consider the in�nite sum

S =a0

100+

a1

102+

a2

104+

a3

106+ � � � ;

where the sequence fang is de�ned by a0 = a1 = 1, and the recurrence

relation an = 20an�1 + 12an�2 for all positive integers n � 2. IfpS can

be expressed in the form apb, where a and b are relatively prime positive

integers, determine the ordered pair (a; b).

44

H220. Let S be the sum of the elements of the set

f1; 2; 3; : : : ; (2p)n� 1g:

Let T be the sum of the elements of this set whose representation in base 2p

consists only of digits from 0 to p� 1.

Prove that 2n� TS= (p� 1)=(2p� 1).

Advanced Problems

Editor: Cyrus Hsia, 21 Van Allen Road, Scarborough, Ontario, Canada.

M1G 1C3 <hsia@math.toronto.edu>

A193. If f(x; y) is a convex function in x for each �xed y, and a convex

function in y for each �xed x, is f(x; y) necessarily a convex function in x

and y?

A194. Let H be the orthocentre (point where the altitudes meet) of a

triangle ABC. Show that if AH : BH : CH = BC : CA : AB, then the

triangle is equilateral.

A195. Compute tan 20� tan 40� tan 60� tan 80�.

A196. Show that r2 + r2a + r2b + r2c � 4K, where r, ra, rb, rc, andK

are the inradius, exradii, and area respectively of a triangle ABC.

Challenge Board Problems

Editor: Ravi Vakil, Department of Mathematics, One Oxford Street,

Cambridge, MA, USA. 02138-2901 <ravi@math.harvard.edu>

C70. Prove that the group of automorphisms of the dodecahedron

is S5, the symmetric group on �ve letters, and that the rotation group of

the dodecahedron (the subgroup of automorphisms preserving orientation)

is A5.

C71. Let L1, L2, L3, L4 be four general lines in the plane. Let pij be

the intersection of lines Li and Lj. Prove that the circumcircles of the four

triangles p12p23p31, p23p34p42, p34p41p13, p41p12p24 are concurrent.

C72. A �nite group G acts on a �nite set X transitively. (In other

words, for any x; y 2 X, there is a g 2 G with g � x = y.) Prove that there

is an element of G whose action on X has no �xed points.

45

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.

In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.

To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81

2"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 September 1997. They mayalso be sent by email to cruxeditor@cms.math.ca. (It would be appreciatedif email proposals and solutionswere written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication.

2201. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a convex quadrilateral, and O is the intersection of its diag-

onals. Let L;M;N be the midpoints ofDB;BC;CA respectively. Suppose

that AL;OM;DN are concurrent. Show that

either AD k BC or [ABCD] = 2[OBC];

where [F ] denotes the area of �gure F.

2202. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Suppose than n � 3. Let A1; : : : ; An be a convex n-gon (as usual with

interior angles A1; : : : ; An).

Determine the greatest constant Cn such that

nXk=1

1

Ak

� Cn

nXk=1

1

� � Ak

:

Determine when equality occurs.

46

2203. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Let ABCD be a quadrilateral with incircle I. Denote by P , Q, R

and S, the points of tangency of sides AB, BC, CD and DA, respectively

with I.Determine all possible values of \(PR;QS) such that ABCD is cyclic.

2204. Proposed by �Sefket Arslanagi�c, Berlin, Germany.For triangle ABC such that R(a+ b) = c

pab, prove that

r <3

10a:

Here, a, b, c, R, and r are the three sides, the circumradius and the

inradius of4ABC.

2205. Proposed by V�aclav Kone �cn �y, Ferris State University, Big Rapids,Michigan, USA.

Find the least positive integer n such that the expression

sinn+2A sinn+1B sinnC

has a maximum which is a rational number (where A, B, C are the angles of

a variable triangle).

2206. Proposed by Heinz-J �urgen Sei�ert, Berlin, Germany.

Let a and b denote distinct positive real numbers.

(a) Show that if 0 < p < 1, p 6= 1

2, then

1

2

�apb1�p + a1�pbp

�< 4p(1� p)

pab+ (1� 4p(1� p))

a+ b

2:

(b) Use (a) to deduce P �olya's Inequality:

a� b

loga� log b<

1

3

�2pab+

a+ b

2

�:

Note: \log" is, of course, the natural logarithm.

2207. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.

Let p be a prime. Find all solutions in positive integers of the equation:

2

a+

3

b=

5

p:

47

2208. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

1. Find a set of positive integers fx; y; z; a; b; c; kg such that

y2z2 = a2 + k2

z2x2 = b2 + k2

x2y2 = c2 + k2

2. Show how to obtain an in�nite number of distinct sets of positive inte-

gers satisfying these equations.

2209. Proposed by Miguel Amengual Covas, Cala Figuera, Mallorca,Spain.

Let ABCD be a cyclic quadrilateral having perpendicular diagonals crossing

at P . Project P onto the sides of the quadrilateral.

1. Prove that the quadrilateral obtained by joining these four projections

is inscribable and circumscribable.

2. Prove that the circle which passes through these four projections also

passes through the mid-points of the sides of the given quadrilateral.

2210?. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,

Madrid, Spain.

Given a0 = 1, the sequence fang (n = 1; 2; : : : ) is given recursively by�n

n

�an �

�n

n� 1

�an�1 +

�n

n� 2

�an�2 � : : :�

�n

bn2c

�abn

2 c = 0:

Which terms have value 0?

2211. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.

Several people go to a pizza restaurant. Each person who is \hungry" wants

to eat either 6 or 7 slices of pizza. Everyone else wants to eat only 2 or 3

slices of pizza each. Each pizza in the restaurant has 12 slices.

It turns out that four pizzas are not su�cient to satisfy everyone, but that

with �ve pizzas, there would be some pizza left over.

How many people went to the restaurant, and howmany of these were \hun-

gry"?

48

2212. Proposed by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario.

Let S = f1; 2; : : : ; ng where n � 3.

(a) In how many ways can three integers x, y, z (not necessarily distinct)

be chosen from S such that x + y = z? (Note that x + y = z and

y + x = z are considered to be the same solution.)

(b) What is the answer to (a) if x, y, z must be distinct?

2213. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

A generalization of problem 2095 [1995: 344, 1996: 373].

Suppose that the function f(u) has a second derivative in the interval (a; b),

and that f(u) � 0 for all u 2 (a; b). Prove that

1. (y � z)f(x) + (z � x)f(y) + (x� y)f(z) > 0 for all x; y; z 2 (a; b),

z < y < x

if and only if f 00(u) > 0 for all u 2 (a; b);

2. (y � z)f(x) + (z � x)f(y) + (x� y)f(z) = 0 for all x; y; z 2 (a; b),

z < y < x

if and only if f(u) is a linear function on (a; b).

Correction

2137. [1996: 124, 317]Proposed by Aram A. Yagubyants, Rostov naDonu, Russia.

Three circles of (equal) radius t pass through a point T , and are each

inside triangle ABC and tangent to two of its sides. Prove that:

(i) t =rR

R+ r,

[NB: r instead of 2]

(ii) T lies on the line segment joining the

centres of the circumcircle and the incircle

of4ABC.

49

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2101. [1996: 33] Proposed by Ji Chen, Ningbo University, China.Let a; b; c be the sides and A;B;C the angles of a triangle. Prove that

for any k � 1, X ak

A�

3

�

Xak;

where the sums are cyclic. [The case k = 1 is known; see item 4.11, page 170

of Mitrinovi�c, Pe�cari�c, Volenec, Recent Advances in Geometric Inequalities.]

I. Solution by Hoe Teck Wee, Singapore.

Let f(x) =sink x

xfor 0 < x < �. Then, for 0 < k � 1, we have

f 0(x) =(kx cosx� sinx) sink�1 x

x2:

For x � �=2, we have cosx � 0, so that kx cosx� sinx � 0.

For x < �=2, we have cosx > 0 and tanx > x � kx, so that kx cosx �sinx � 0.

Therefore f 0(x) =(kx cosx� sinx) sink�1 x

x2� 0.

Without loss of generality, we may assume that A � B � C. For

0 < k � 1, we have that f(x) is a non-increasing function, so that f(A) �f(B) � f(C). Thus, by Tchebyshev's inequality, we have sin

kA

A+

sinkB

B+

sinkC

C

!(A+B+C) � 3

�sinkA+ sink B + sinkC

�:

By the Sine Rule, we have a = 2R sinA, b = 2R sinB and c = 2R sinC,

where R is the circumradius of 4ABC. Multiply the inequality by (2R)k

and substitute A+B + C = � to get

X ak

A�

3

�

Xak: (1)

For k � 0, we have

a � b � c =) ak � bk � ck =)ak

A�bk

B�ck

C:

Thus, by using Tchebyshev's inequality again, (??) holds for k � 0.

50

In conclusion, (??) holds for any k � 1.

II. Solution by Kee-Wai Lau, Hong Kong. (Edited)We let, without loss of generality, a � b � c, so that A � B � C:

Next, put f(k) =�ak=A+ bk=B + ck=C

� �ak + bk + ck

��1, then

f 0(k) = �X akbk(A� B)(loga� log b)

AB

�Xak��2

� 0;

where the sums are cyclic. Since we are given that f(1) � 3=n, it follows

that f(k)� 3=n for k � 1.

Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. One incorrect solu-tion was received.

Janous establishes the following generalization.

Let k and ` be real numbers. Then:

1. X ak

A`��3

�

�`Xak;

for the cases

� 0 � k � ` � 1,

� k � 0 � `,

� k � 0 and ` � �1:

2. X ak

A`��3

�

�`Xak;

where

k � 0 and � 1 � ` � 0:

2102. [1996: 33] Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is a triangle with incentre I. Let P and Q be the feet of the

perpendiculars from A to BI and CI respectively. Prove that

AP

BI+AQ

CI= cot

A

2:

Solution by Panos E. Tsaoussoglou, Athens, Greece, and by eight oth-ers!

51

In4APB, we have sin(B=2) =AP

AB,

in4AQC, we have sin(C=2) =AQ

AC,

in4ABI, we haveBI

sin(A=2)=

AB

cosC=2,

in4ACI. we haveCI

sin(A=2)=

AC

cosB=2,

so that

AP

BI+AQ

CI=

sin(B=2) cos(C=2)

sin(A=2)+

sinC=2 cosB=2

sinA=2

=sin(B=2 + C=2)

sin(A=2)=

cos(A=2)

sin(A=2)= cot(A=2):

Editor's comment. Our featured solution is based on the property:

\AIB = � � A=2� B=2 = �=2 + C=2, so that sin(\AIB) = cos(C=2).

Other solvers used the equally e�cient relations: BI = r= sin(B=2), or

BI = 4R sin(A=2) sin(C=2).

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain (two solutions); FEDERICO ARDILA, student, Massachusetts Insti-tute of Technology, Cambridge, Massachusetts, USA; FRANCISCO BELLOTROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CARL BOSLEY, student,Washburn Rural High School, Topeka, Kansas, USA; CARL BOSLEY, stu-dent, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHERJ. BRADLEY, CliftonCollege, Bristol,UK;MIGUELANGEL CABEZ �ONOCHOA,Logro ~no, Spain; THEODORE CHRONIS, student, AristotleUniversity of Thes-saloniki, Greece; HAN PING DAVIN CHOR, Student, Cambridge, MA, USA;DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLO-RIAN HERZIG, student, Perchtoldsdorf, Austria; JOHN G. HEUVER, GrandePrairie Composite High School, Grande Prairie, Alberta; CYRUS HSIA, stu-dent, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursul-inengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Uni-versity, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba Tonka Schoolof Mathematics, Rousse, Bulgaria; P. PENNING, Delft, the Netherlands;GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; WALDEMARPOMPE, student, University of Warsaw, Poland; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; GEORGETSAPAKIDIS, Agrinio, Greece; MELITIS D. VASILIOU. Elefsis, Greece;CHRIS WILDHAGEN, Rotterdam, the Netherlands; and the proposer.

52

2103. [1996: 33] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle. Let D be the point on side BC produced beyond

B such that BD = BA, and letM be the mid-point of AC. The bisector of

\ABC meets DM at P . Prove that \BAP = \ACB.

Solution by Hans Engelhaupt, Franz{Ludwig{Gymnasium, Bamberg,Germany.

Let PX be parallel to AC with X lying on the line BC. Let Y be the

intersection of PX with AD. P is the midpoint of XY because M is the

mid-point of AC.

ThenB is the mid-point ofDX [PB is parallel to AD since 2\DAB =

\DAB + \BDA = \ABC = 2\PBA]:

Hence BX = BD = AB: TriangleBPA is congruent to triangleBPX

[PB = PB;AB = XB;\ABP = \XBP .]

Therefore, \BAP = \BXP = \BCA [PXkAC].

AB

C

MY

X

P

D

�=2

�=2

�=2

Also solved by FEDERICO ARDILA, student,Massachusetts Institute ofTechnology, Cambridge, Massachusetts,USA; �SEFKET ARSLANAGI �C, Berlin,Germany; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MARIAASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol,UK; HAN PINGDAVINCHOR, Student, Cambridge, MA, USA; TIM CROSS, King Edward's School,Birmingham, England; DAVID DOSTER, Choate Rosemary Hall, Wallingford,Connecticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MITKO

53

CHRISTOV KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bul-garia; KEE-WAI LAU, Hong Kong; VASILIOU MELETIS, Elelsis, Greece;P. PENNING, Delft, the Netherlands; WALDEMAR POMPE, student, Uni-versity of Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands;HOE TECK WEE, Singapore; and the proposer.

2104. [1996: 34] Proposed by K.R.S. Sastry, Dodballapur, India.In how many ways can 111 be written as a sum of three integers in geometric

progression?

Solution by Zun Shan, Nanjing Normal University, Nanjing, China.The answer is seventeen or sixteen depending on whether we allow the

common ratio of the G.P. (geometric progression) to be zero or not.

Suppose 111 = a+ar+ar2 where a is an integer and r is a rational number.

If r = 0, then we get the trivial solution

111 = 111 + 0 + 0: (1)

Suppose r = nm6= 0 where m and n are nonzero integers. Without loss

of generality, we may also assume that m > 0 and (m;n) = 1. Since the

reverse of the G.P. a; ar; ar2 is another G.P. ar2; ar; a, we may also assume

that jrj � 1 and so 0 < m � jnj.From a(1+ r+ r2) = 111 we get a(m2+mn+n2) = 111m2. Since clearly

(m2 +mn + n2;m2) = 1 we have m2ja. Letting a = km2 where k is an

integer we then get k(m2 + mn + n2) = 111 which implies kj111. Since

m2 +mn+ n2 > 0 and 111 = 3 � 37, we have k = 1; 3; 37, or 111. Note

that m2 +mn+ n2 = m2 + jnj(�m+ jnj) � m2.

Case [1] If k = 1, thenm2 +mn+ n2 = 111 =) m2 � 111 =) m � 10.

Whenm = 1, a = 1 and from n2+n = 110 we get n = 10;�11. Thusr = 10, �11 and we obtain the solutions:

111 = 1 + 10 + 100 = 1� 11 + 121: (2)

For 2 � m � 9 it is easily checked that the resulting quadratic equation

in n has no integer solutions.

When m = 10, a = 100 and from n2 + 10n = 11 we get n = 1, �11.Since m � jnj, n = �11 and r = �11

10yielding the solution:

111 = 100� 110 + 121: (3)

Case [2] If k = 3, then m2 +mn + n2 = 37 =) m2 � 37 =) m � 6.

Quick checkings reveal that there are no solutions for m = 1, 2, 5, 6.

Whenm = 3, a = 27 and from n2 + 3n = 28 we get n = 4, �7. Thusr = 4

3;�7

3yielding the solutions:

111 = 27 + 36 + 48 = 27� 63 + 147: (4)

54

Whenm = 4, a = 48 and from n2 + 4n = 21 we get n = 3, �7.

Since m � jnj, n = �7 and r = �74yielding the solution:

111 = 48� 84 + 147: (5)

Case [3] If k = 37, then m2 +mn+ n2 = 3 =) m2 � 3 =) m = 1 =)a = 37 and from n2 + n = 2 we get n = 1, �2. Thus r = 1 or �2yielding the solutions:

111 = 37 + 37 + 37 = 37� 74 + 148: (6)

Case [4] If k = 111, thenm2 +mn+n2 = 1 =) m2 � 1 =) m = 1 =)a = 111, and from n2 +n = 0 we get n = �1 as n 6= 0. Thus r = �1and we get the solution

111 = 111� 111 + 111: (7)

Reversing the summand in (2) � (7) and noting that two of them are

\symmetric", we obtain seventeen solutions in all:

111 = 111 + 0 + 0 = 1 + 10 + 100 = 100 + 10 + 1

= 1� 11 + 121 = 121� 11 + 1 = 100� 110 + 121

= 121� 110 + 100 = 27 + 36 + 48 = 48 + 36 + 27

= 27� 63 + 147 = 147� 63 + 27 = 48� 84 + 147

= 147� 84 + 48 = 37 + 37 + 37 = 37� 74 + 148

= 148� 74 + 37 = 111� 111 + 111:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain, (who assumed r 6= 0 and found sixteen solutions); F.J. FLANIGAN,San Jose State University, San Jose, California, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; (both of them found all seventeensolutions). There were also two incomplete and twenty-three incorrect so-lutions submitted!

Among these twenty-three, thirteen submissions claimed six solutions; sixsubmissions claimed �ve solutions; two submissions claimed six or nine so-lutions; one submission claimed two solutions, and one submission claimedone solutiononly. Most of the errors were the result of assuming by mistakethat a(1 + r + r2) = 111 =) 1 + r + r2 must be an integer.

2105. [1996: 34] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK

Find all values of � for which the inequality

2(x3 + y3 + z3) + 3(1 + 3�)xyz � (1 + �)(x+ y + z)(yz+ zx+ xy)

holds for all positive real numbers x; y; z.

55

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-berta.

On setting x = y = 1 and z = 0, we obtain 4 � (1+�)2 and thus �nd

that � must be � 1. We now show that the inequality holds for all � � 1.

First, if � = 1, the inequality reduces to

x3 + y3 + z3 + 6xyz � (x+ y + z)(yz+ zx+ xy);

which is equivalent to the special case n = 1 of the known Schur inequality

xn(x� y)(x� z) + yn(y� z)(y� x) + zn(z � x)(z � y) � 0;

true for all real n, and which has come up many times in this journal. The

rest will follow by showing that for all � < 1,

(1 + �)(x+ y+ z)(yz+ zx+ xy)� 3(1 + 3�)xyz

� (1 + 1)(x+ y + z)(yz+ zx+ xy)� 3(1 + 3)xyz: (1)

[Editor's note: rewrite the original inequality as

2(x3 + y3 + z3) � (1 + �)(x+ y + z)(yz+ zx+ xy)� 3(1 + 3�)xyz;

then (1) says that the right hand side is largest when � = 1, so doing the

case � = 1 would be enough.] But (1) can be written

(1� �)[(x+ y + z)(yz+ zx+ xy)� 9xyz] � 0

which [after cancelling the positive factor 1� �] is a known elementary in-

equality, equivalent to Cauchy's inequality

(x+ y + z)

�1

x+

1

y+

1

z

�� 9:

Also solved by CARL BOSLEY, student, Washburn Rural High School,Topeka, Kansas, USA; THEODORE CHRONIS, student, Aristotle Universityof Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf, Aus-tria; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens,Greece; and HOE TECK WEE, Singapore. There were three incorrect solu-tions sent in.

2106. [1996: 34] Proposed by Yang Kechang, Yueyang University,Hunan, China.

A quadrilateral has sides a; b; c; d (in that order) and area F . Prove that

2a2 + 5b2 + 8c2 � d2 � 4F:

When does equality hold?

56

Solution by Federico Ardila, student, Massachusetts Institute of Tech-nology, Cambridge, Massachusetts, USA.

Let ABCD be the quadrilateral with AB = a, BC = b, CD = c, and

DA = d. We can assume, without loss of generality, that AC = 1. There-

fore, we can locate the quadrilateral in a system of Cartesian coordinates

where

A = (0; 0); B = (p; q); C = (1; 0); D = (r; s):

We assume that ABCD is simple so that its area is well-de�ned. If ABCD

is not convex we can make it convex and keep the side lengths the same

while increasing the area. This means that we will be done if we can show

that the result is true for convex quadrilaterals. It's also clear from this that

if the result is true for convex quadrilaterals, then equality cannot hold for

non-convex quadrilaterals. Therefore, assume q < 0 and s > 0. Now note

that

2a2 + 5b2 = 2(p2 + q2) + 5((p� 1)2 + q2)

= 7p2 � 10p+ 5+ 7q2

= 7�p� 5

7

�2 � 257+ 5+ 7q2

2a2 + 5b2 � 7q2 + 107; (2)

and

8c2 � d2 = 8((r� 1)2 + s2)� (r2 + s2)

= 7r2 � 16r + 8 + 7s2

= 7�r � 8

7

�2 � 647+ 8+ 7s2

8c2 � d2 � 7s2 � 87: (3)

Combining (??) and (??), we get

2a2 + 5b2 + 8c2 � d2

� 7q2 + 7s2 + 27

=�7q2 + 1

7

�+�7s2 + 1

7

�=

�7�jqj � 1

7

�2+ 2jqj

�+�7�jsj � 1

7

�2+ 2jsj

�� 2

�jqj+ jsj

�(4)

= 4(ABC) + 4(CDA):

2a2 + 5b2 + 8c2 � d2 � 4F;

as we wished to prove (where ABC and CDA refer to the areas of the

two triangles ABC and CDA respectively). For equality to hold (when

A = (0; 0) and C = (1; 0)), it must hold in steps (??), (??), and (??). There-

fore p = 57, r = 8

7, q = �1

7, and s = 1

7. Thus, in general, equality holds if

and only if ABCD is directly similar to quadrilateral A0B0C0D0, where

A0 = (0; 0); B0 = (57;�1

7); C0 = (1; 0); D0 = (8

7; 17):

57

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA; KEE-WAI LAU, Hong Kong; and the proposer.

KONE �CN �Y makes the observation that the quadrilateral is cyclic andthat \ABD = 135�. The proposer makes the early observation that themaximum area for a quadrilateral with �xed sides occurs when it is cyclic anduses properties of cyclic quadrilaterals in the proof. He also generalizes theresult to an inequality which JANOUS uses in his proof and which appears inthe Addenda to theMonograph \Recent Advances in Geometric Inequalities"by D. S. Mitrinovi�c et al. in I. Journal of Ningbo University 4, No. 2 (Dec.1991), 79{145.

2107. [1996: 34] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

Triangle ABC is not isosceles nor equilateral, and has sides a; b; c. D1

and E1 are points of BA and CA or their productions so that

BD1 = CE1 = a. D2 andE2 are points ofCB andAB or their productions

so that CD2 = AE2 = b. Show that D1E1 k D2E2.

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.Let S be the intersection of AB and D2E1.

[Editor's comment by Chris Fisher. Even though there seem to be two choices

for each Di and Ei, no solver had any trouble choosing the positions that

make the result correct; furthermore, it must have been \obvious" to evey-

one but me that AB is not parallel to D2E1, so that S exists. Alas, perhaps

I need stronger glasses.]

Then CS is the bisector of \ACB, since CE1 = CB and CA = CD2.

Therefore

D1S

SE2

=BD1 � BS

AE2� AS= a = BSb� AS =

a

b;

sinceBS

AS=a

b. It then follows that D1E1kD2E2, since

E1S

SD2

=CE1

CD2

=a

b=

D1S

SE2

:

Also solved by FEDERICO ARDILA, student,Massachusetts Institute ofTechnology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnas-ium, Bamberg, Germany; JOHN G. HEUVER, Grande Prairie CompositeHighSchool, Grande Prairie, Alberta; CYRUSHSIA, student,University of Toronto,Toronto, Ontario; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA;P. PENNING, Delft, the Netherlands; JOEL SCHLOSBERG, student, Hunter

58

College High School, New York NY, USA; GEORGE TSAPAKIDIS, Agrinio,Greece; MELITIS D. VASILIOU. Elefsis, Greece; and the proposer.

Janous adds the observation that D1E1 and D2E2 are not only par-allel, but their lengths are in the ratios a : b (as is clear from the featuredsolution).

2108. [1996: 34] Proposed by Vedula N. Murty, Andhra University,Visakhapatnam, India.

Prove that

a+ b+ c

3�

1

4

3

s(b+ c)2(c+ a)2(a+ b)2

abc;

where a; b; c > 0. Equality holds if a = b = c.

Solution by Florian Herzig, student, Perchtoldsdorf, Austria, (modi�edslightly by the editor).

By the arithmetic-geometric mean inequality we have

a2b+ ab2 + b2c+ bc2 + c2a+ ca2 � 66pa6b6c6 = 6abc;

which implies

9(a2b+ ab2 + b2c+ bc2 + c2a+ ca2 + 2abc)

� 8(a2b+ ab2 + b2c+ bc2 + c2a+ ca2 + 3abc);

or

9(a+ b)(b+ c)(c+ a) � 8(a+ b+ c)(ab+ bc+ ca)

= 4(a+ b+ c)(a(b+ c) + b(c+ a) + c(a+ b)):

Using the arithmetic-geometric mean inequality again, we then have

3

4(a+ b)(b+ c)(c+ a)

� (a+ b+ c) �a(b+ c) + b(c+ a) + c(a+ b)

3

� (a+ b+ c)pabc(a+ b)(b+ c)(c+ a) (1)

From (1) it follows immediately that

1

4

3

s(a+ b)2(b+ c)2(c+ a)2

abc�a+ b+ c

3:

Clearly, equality holds if a = b = c. [Ed. In fact, if equality holds, then

from (1) we have a(b+ c) = b(c+ a) = c(a+ b). The �rst equality implies

59

a = b and the second one implies b = c. Thus, equality holds in the given

inequality if and only if a = b = c. This was observed by about half of the

solvers.]

Also solved by FEDERICO ARDILA, student,Massachusetts Institute ofTechnology, Cambridge, Massachusetts, USA; CHRISTOPHER J. BRADLEY,Clifton College, Bristol, UK; HAN PING DAVIN CHOR, Student, Cambridge,MA, USA; THEODORE CHRONIS, student, Aristotle University of Thessa-loniki, Greece; TIM CROSS, King Edward's School, Birmingham, England;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.KLAMKIN, University of Alberta, Edmonton, Alberta; KEE-WAI LAU, HongKong; JUAN-BOSCO ROMERO M�ARQUEZ, Universidad de Valladolid, Val-ladolid, Spain; JOEL SCHLOSBERG, student, Hunter College High School,New York NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; PANOSE. TSAOUSSOGLOU, Athens, Greece; and the proposer.

Janous commented that upon the transformation

a!1

a; b!

1

b; c!

1

c;

the given inequality can be shown to be equivalent to

rab+ bc+ ca

3� 3

s(a+ b)(b+ c)(c+ a)

8

which is known in the literature as Carlson's inequality (cf. eg. P.S. Bullen,D.S. Mitrinovic and P.M. Vasic, \Means and Their Inequalities", Dordrechf,1988. An anonymous reader commented that in this form, the inequalitywas problem 3 of the 1992 Austrian-Polish Mathematics Competition andhas appeared in Crux before (see [1994:97; 1995:336-337]). Several solversshowed that the given inequality is equivalent to various other trigonometricinequalities involving a triangle XY Z, for example

cos

�X

2

�cos

�Y

2

�cos

�Z

2

��

3p3

8

or

sinX + sinY + sinZ �3p3

2;

etc. These inequalities can be found in \Geometric Inequalities" byO. Bottema et al.

2109. [1996: 34] Proposed by Victor Oxman, Haifa, Israel.In the plane are given a triangle and a circle passing through two of the

vertices of the triangle and also through the incentre of the triangle. (The

incentre and the centre of the circle are not given.) Construct, using only an

unmarked ruler, the incentre.

60

Solution by P. Penning, Delft, the Netherlands.

Let the triangle be ABC, and � the circle passing through B;C and

the incentre. The angles of the triangle are denoted by the symbol for the

corresponding vertex A;B or C.A

B

C

B0

S

C0

X

..

�

ANALYSIS:

Let point S be the intersection of the circumcircle and the angular bisec-

tor through A. The arcs SB and SC of the circumcircle are now equal and so

are the chords SB and SC. Introduce X on AS such that SX = SB = SC:

\BSA = \BCA = C;

4XBS is isosceles with SX = SB, so \XBS = 90� � C=2:

\CBS = \CAS = A=2; \XBC = 90� � C=2�A=2 = B=2:

So BX is the angular bisector at vertex B, andX must be the incentre. The

circle � apparently has the point S as centre. [It does, see Roger A. Johnson,Modern Geometry, 292].

[Editor's note: If either AB or AC is tangent to �, then they bothare and AB = AC. Suppose AB is tangent to �. Then \SBA = �

2, so

\BSA + \SAB = �2. Since SC = SB; \BCS = \SBC = \SAC =

\SAB. Therefore, \ACS = \ACB+\BCS = \BSA+\SAB = �2and

AC is tangent to �. In addition, tangents to a circle from an exterior pointare equal, so AB = AC:]

So ifAB 6= AC, neither line is tangent to�. LetB0 andC0 be the otherintersections of AB, respectively AC, with �. There is mirror-symmetry

with respect to the line AS : � remains �; AB re ects into AC and AC

re ects into AB; B and C0 are mirror-images and so are C and B0. The sideBC becomes B0C0; as a consequence they must intersect on the mirror-line

AS.

CONSTRUCTION:

Find the other two intersections, B0 and C0, of AB and AC with the

circle �. The intersection of BC and B0C0 isM . The incentre is the inter-

61

section of AM and �.

COMMENT:

The construction fails if ABC is isosceles, with AB = AC. In that

case � touches both AB and AC in B and C respectively. M is now the

midpoint of BC, but that cannot be found with unmarked ruler.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; RICHARD I. HESS, Rancho Palos Verdes, California, USA; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAILAU, Hong Kong; and the proposer.

2110. [1996: 35] Proposed by Jordi Dou, Barcelona, Spain.Let S be the curved Reuleaux triangle whose sides AB, BC and CA

are arcs of unit circles centred at C, A andB respectively. Choose at random

(and uniformly) a pointM in the interior and let C(M) be a chord of S for

which M is the midpoint. Find the length ` such that the probability that

C(M) > ` is 1=2.

Solution by the proposer.Let� be the locus of the mid-pointM of segments of constant length�,

whose ends S1 and S2 move on the boundary of S. For the pointsM inside

� the chords bisected byM are greater than �.

(?) The area contained between S and � is �

4�2. It is su�cient to show that

�

4`2 =

[S]

2:

Since [S] = �

2� 2

p3

4, we will have

` =

� �

p3

�

! 12

' 0:67:

Brief proof of the assertion (?) (Special Case of Holditch's Theorem)

The ends S1, S2 of the chord of constant length move along the contour

of the closed curve S. The mid-pointM describes the curve �.

Let S1 = (x1; y1); S2 = (x2; y2), and M = (x1+x22

; y1+y22

). Suppose

that x1, y1, x2, y2 are functions of t such that for t0 � t � t1, we have S1,

S2 describing S.

[S] =

Z t1

t0

y1 dx1 =

Z t1

t0

y2 dx2 =

Z t1

t0

1

2(y1 dx1 + y2 dx2)

[�] =

Z t1

t0

y dx =

Z t1

t0

1

4(y1 + y2)(dx1 + dx2):

Then

[S]� [�] =1

4

Z t1

t0

(y2 � y1)(dx2 � dx1):

62

Substitute X = x2 � x1, Y = y2 � y1, giving

[S]� [�] =1

4

Z t1

t0

Y dX =�

4�2;

since (X;Y ) describes a circle of radius �.

2111. [1996: 35] Proposed by Hoe Teck Wee, student, Hwa ChongJunior College, Singapore.

Does there exist a function f : N �! N (where N is the set of positive

integers) satisfying the three conditions:

(i) f(1996) = 1;

(ii) for all primes p, every prime occurs in the sequence

f(p), f(2p), f(3p); : : : ; f(kp); : : : in�nitely often; and

(iii) f(f(n)) = 1 for all n 2 N ?

I. Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, On-tario.

Yes, a function does exist that satis�es the three conditions. It is:

f(x) =

�pi if condition � holds;

1 otherwise;

where condition � is: if the prime factorization of x is x =Qpeii , there exists

a power ei such that ei > 2 and ei > ej for all j 6= i.

For example, 109850 has condition �, since 109850 = 2 � 52 � 133 and

the power of 13 is bigger than 2 and bigger than all other powers in the

factorization; thus f(109850) = 13.

Now

� f satis�es condition (i) since f(1996) = f(22� 499) = 1;

� f satis�es condition (ii) because for any two primes p and q, f(xi) = q

for every xi = qip, i = 3; 4; : : : ;

� f satis�es condition (iii) since for all n either f(n) = 1 or f(n) = pifor some prime pi, and in either case f(f(n)) = 1.

II. Solution by Chris Wildhagen, Rotterdam, the Netherlands.For each n 2 N let qn be the nth prime and b(n) be the number of 1's

in the binary representation of n. De�ne f : N! N as follows:

f(m) =

�qb(n) ifm = pn with p prime and n � 2;

1 else:

63

Clearly f satis�es the three given conditions.

Also solved by FEDERICO ARDILA, student, Massachusetts Instituteof Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student,St. Paul's School, London, England; CARL BOSLEY, student,Washburn RuralHigh School, Topeka, Kansas, USA; CHARLES R. DIMINNIE, Angelo StateUniversity, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla, Washing-ton, USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUSHSIA, student, University of Toronto, Toronto, Ontario; DAVID E. MANES,State University of New York, Oneonta, NY, USA; ROBERT P. SEALY,MountAllison University, Sackville, New Brunswick; and the proposer. There werethree incorrect solutions sent in.

Most solvers gave a variation of Solution I.

2112. [1996: 35] Proposed by ShawnGodin, St. Joseph Scollard Hall,North Bay, Ontario.

Find a four-digit base-ten number abcd (with a 6= 0) which is equal to

aa + bb + cc + dd.

Solution by Cyrus Hsia, student, University of Toronto, Toronto, On-tario (modi�ed slightly by the editor).

We �rst stipulate that 00 = 1. Let m = abcd, s = aa + bb + cc + dd

and assume that m = s. Clearly, 103 � m < 104. If x � 6 for any

x 2 fa; b; c; dg then s � 66 > 104 which is a contradiction. So a; b; c; d � 5.

If x < 5 for all x 2 fa; b; c; dg, then s � 4 � 44 = 1024 and furthermore

a = b = c = d = 4 is the only combination for which s � 103. However, in

this case, s = 1024 6= 4444 = m. Hence x = 5 for some x 2 fa; b; c; dg.We cannot have more than one 5 or else s � 2 � 55 = 6250 would imply

that some digit of m is at least 6. Hence, we have exactly one 5.

Since s > 55 = 3125 and s � 55 + 3 � 44 = 3893 < 4000, we must have

a = 3. Thus, s = 55+33+xx+yy = 3152+xx+yy where x; y 2 fa; b; c; dg.Without loss of generality, we may assume that 0 � y � x � 4.

If x = 0, then y = 0 and s = 3154 which has no 0 among its digits.

If x = 1, then y = 0; 1 and s = 3154 whilem has no 4 among its digits.

If x = 2, then s = 3156 + yy and it is easily veri�ed that s has no 2 among

its digits for y = 0; 1; 2.

If x = 3, then s = 3179 + yy and it is easily veri�ed that s has no 5 among

its digits for y = 0; 1; 2; 3.

If x = 4, then s = 3408+ yy and, again, it is readily checked that s has no 5

among its digits when y = 0; 1; 2; 4. However, when y = 3, s = 3435 which

is clearly a solution.

To summarize, m = 3435 is the only solution.

64

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; FEDERICO ARDILA, student, Massachusetts Institute of Technol-ogy, Cambridge, Massachusetts, USA; MANSUR BOASE, student, St. Paul'sSchool, London, England; CHRISTOPHER J. BRADLEY, Clifton College,Bristol, UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;TIM CROSS, King Edward's School, Birmingham, England; JEFFREY K.FLOYD, Newnan, Georgia, USA; FLORIAN HERZIG, student, Perchtolds-dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; LUKE LAMOTHE, stu-dent, St. Joseph Scollard Hall S.S., North Bay, Ontario, KATHLEEN E.LEWIS, SUNY Oswego. Oswego, NY, USA; DAVID E. MANES, State Univer-sity of New York, Oneonta, NY, USA; JOHN GRANT McLOUGHLIN, Okana-gan University College, Kelowna British Columbia; P. PENNING, Delft, theNetherlands; CORY PYE, student, Memorial University of Newfoundland,St. John's,Newfoundland; JOEL SCHLOSBERG, student,Hunter CollegeHighSchool, New York NY, USA; ROBERT P. SEALY, Mount Allison University,Sackville, New Brunswick; HEINZ-J�URGEN SEIFFERT, Berlin, Germany;DIGBY SMITH, Mount Royal College, Calgary, Alberta; CHRISWILDHAGEN, Rotterdam, the Netherlands; and the proposer.

Of the twenty-six solvers (including the proposer), nine of them only gavethe answer 3435. About half of all the solvers claimed, with or withoutproof, that 3435 is the only solution. Chronis, Hess, and Janous found theanswer by computer search. Hess remarked that no other solutions werefound for the present problem and the corresponding problem on 5{digitintegers. Janous investigated the corresponding n{digit problem of �nding

all n{digit integers an�1an�2: : :a1a0 which equal

n�1Xk=0

aakk . He showed that

a necessary condition is n � 10. For n > 1, he conducted an extensive, butnot exhaustive, computer search, which revealed no solutions other than theone found by all the solvers! He made a guess that 1 and 3435 are the onlyintegers with the desired property. Can any reader prove or disprove this?

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

65

THE ACADEMY CORNERNo. 9

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

In this issue, we feature a university undergraduate mathematics com-petition. We invite your solutions, especially from university students. Pleasesend me your nice solutions.

Undergraduate Mathematics Competition

September 18, 1996

Answer as many questions as you can. Complete solutions carry more credit

than scattered comments about many problems.

1. If n is any integer, show that n5 � n is divisible by 5.

2. A line l with slopem = 2 cuts the parabola y2 = 8x to form a chord.

Find the equation of l if the midpoint of the chord lies on x = 4.

3. Show that three tangents can be drawn from the origin to the curvegiven by

y = x3 � 13x2 + 10x� 36.

4. Prove that

�n

0

�2+

�n

1

�2+ : : : +

�n

n

�2=

�2n

n

�, for all positive

integers n.

5. Show that Z �=2

0

sin13 x

sin13 x+ cos13 xdx =

�

4:

6. In triangle ABC, \B = 3\A.

If the sides opposite to the angles A;B;C have lengths a; b; c, respec-tively, prove that

ac2 = (b� a)2 (b+ a).

66

THE OLYMPIAD CORNERNo. 180

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

We begin this number of the Corner with the problems of the 3rd Math-ematical Olympiad of the Republic of China (Taiwan). The contest was writ-ten April 14 and 15, 1994. Many thanks go to Richard Nowakowski, Cana-dian Team Leader to the IMO in Hong Kong for collecting these problemsand many others.

3rd MATHEMATICAL OLYMPIAD OF THEREPUBLIC OF CHINA (Taiwan)

First Day | April 14, 1994

1. Let ABCD be a quadrilateral with AD = BC and \A + \B =120�. Three equilateral triangles 4ACP , 4DCQ and 4DBR are drawnon AC, DC and DB away from AB. Prove that the three new vertices P ,Q and R are collinear.

2. Let a, b, c be positive real numbers, � be a real number. Supposethat

f(�) = abc(a� + b� + c�);

g(�) = a�+2(b+ c� a) + b�+2(a� b+ c) + c�+2(a+ b� c):

Determine jf(�)� g(�)j.3. Let a be a positive integer such that (51994 � 1) j a. Show that

the expression of the number a in the base 5 contains at least 1994 digitsdi�erent from zero.

Second Day | April 15, 1994

4. Prove that there are in�nitely many positive integers n havingthe following property: for every arithmetic progression a1; a2; : : : ; an ofintegers with n terms, both the mean and standard deviation of the setfa1; a2; : : : ; ang are integers. (Note: For any set fx1; x2; : : : ; xng of realnumbers, the mean of the set is de�ned to be the number

x =x1 + x2 + � � �+ xn

n

67

and the standard deviation of the set is de�ned to be the numbersP(xi � x)2

n:

5. Let X = f0; a; b; cg and M(X) = ff j f : X ! Xg be the setof all functions from X into itself. Here, an addition table of X is given asfollows:

� 0 a b c

0 0 a b c

a a 0 c b

b b c 0 a

c c b a 0

(1) If S = ff 2M(X) j f(x� y � x) = f(x)� f(y)� f(x); 8x; y 2 Xg,determine the number of elements of S.(2) If I = ff 2M(X) j f(x� x) = f(x)� f(x); 8x 2 Xg, determine thenumber of elements of I.

6. For �1 � x � 1 de�ne

Tn(x) =1

2n

h�x+

p1� x2

�n+�x�

p1� x2

�ni(1) Prove that, for �1 � x � 1, Tn(x) is a monic polynomial of degree n inthe x-variable and the maximum value of Tn(x) is

1

2n�1.

(2) Suppose that p(x) = xn+an�1xn�1+� � �+a1x+a0 is amonic polynomialwith real coe�cients such that for all x in�1 � x � 1, p(x) > � 1

2n�1. Prove

that there exists x� in �1 � x � 1 such that p(x�) � 12n�1

.

Last issue we gave Five Klamkin Quickies. Here we give his \Quick"solutions plus another 5 problems. Many thanks to Murray S. Klamkin, theUniversity of Alberta.

ANOTHER FIVE KLAMKIN QUICKIESOctober 21, 1996

6. Determine the four roots of the equation x4 + 16x� 12 = 0.

7. Prove that the smallest regular n-gon which can be inscribed in agiven regular n-gon is one whose vertices are the midpoints of the sides ofthe given regular n-gon.

8. If 311995 divides a2 + b2, prove that 311996 divides ab.

9. Determine the minimum value of

S =p(a + 1)2 + 2(b� 2)2 + (c+ 3)2 +

p(b+ 1)2 + 2(c� 2)2 + (d + 3)2) +p

(c+ 1)2 + 2(d� 2)2 + (a+ 3)2 +p(d+ 1)2 + 2(a� 2)2 + (b+ 3)2

68

where a, b, c, d are any real numbers.

10. A set of 500 real numbers is such that any number in the set isgreater than one-�fth the sum of all the other numbers in the set. Determinethe least number of negative numbers in the set.

We will give the solutions to these in the next issue so you can havesome fun looking for the answers. Next we give his solutions to the �veQuickies we gave last issue.

FIVE KLAMKIN QUICKIESOctober 21, 1996

1. For x; y; z > 0, prove that

(i) 1 +1

(x+ 1)��1 +

1

x(x+ 2)

�x,

(ii) [(x+y)(x+z)]x[(y+z)(y+x)]y[(z+x)(z+y)]z � [4xy]x[4yz]y[4zx]z.

Solution. Both inequalities will follow by a judicious application of theweighted arithmetic-geometric mean inequality (W{A.M.{G.M.) which forthree weights is

uavbwc ��au+ bv+ cw

a+ b+ c

�a+b+c;

where a; b; c; u; v;w � 0.

(i) The inequality can be rewritten in the more attractive form�1 +

1

x

�x��1 +

1

x+ 1

�x+1;

and which now follows by the W{A.M.{G.M.

�1 +

1

x

�x�(1 + x

�1 + 1

x

�1 + x

)x+1=

�1 +

1

x+ 1

�x+1:

(ii) Also, the inequality here can be rewritten in the more attractiveform �

2x

z + x

�z+x � 2y

x+ y

�x+y � 2z

y + z

�y+z� 1:

But this follows by applying the W{A.M.{G.M. to

1 =X

[z + x]

�2x

z + x

��X

[z + x]:

2. If ABCD is a quadrilateral inscribed in a circle, prove that the fourlines joining each vertex to the nine point centre of the triangle formed bythe other three vertices are concurrent.

69

Solution. The given result still holds if we replace the nine point centresby either the orthocentres or the centroids.

A vector representation is particularly �a propos here, since (with thecircumcentre O as an origin and F denoting the vector from O to any pointF ) the orthocentre Ha, the nine point centre Na, the centroid Ga of4BCDare given simply by Ha = B + C +D, Na = (B + C +D)=2, Ga =(B + C +D)=3, respectively, and similarly for the other three triangles.Since the proofs for each of the three cases are practically identical, we justgive the one for the orthocentres. The vector equation of the line La joiningA toHa is given byLa = A+�a[B + C +D �A] where �a is a real param-eter. By letting �a = 1=2, one point on the line is [A+ B + C +D]=2 andsimilarly this point is on the other three lines. For the nine point centres, thepoint of concurrency will be 2[A+ B + C +D]=3, while for the centroids,the point of concurrency will be 3[A+B + C +D]=4.

3. How many six digit perfect squares are there each having the prop-erty that if each digit is increased by one, the resulting number is also aperfect square?

Solution. If the six digit square is given by

m2 = a � 105 + b � 104 + c � 103 + d � 102 + e � 10 + f;

then

n2 = (a+1)�105+(b+1)�104+(c+1)�103+(d+1)�102+(e+1)�10+(d+1);

so that

n2 �m2 = 111;111 = (111)(1;001) = (3 � 37)(7 � 11 � 13):

Hence,n+m = di and n�m = 111; 111=di

where di is one of the divisors of 111; 111. Since 111; 111 is a product of �veprimes it has 32 di�erent divisors. But since we must have di > 111; 111=di,there are at most 16 solutions given by the form lm = 1

2(di� 111; 111=di).

Then sincem2 is a six digit number, we must have

632:46 � 200p10 < 2m < 2; 000:

On checking the various divisors, there are four solutions. One of them cor-responds to di = 3 � 13 � 37 = 1; 443 so that m = 1

2(1; 443� 7 � 11) = 683

and m2 = 466; 489. Then, 466; 489 + 111; 111 = 577; 600 = 7602. Theothers are given by the table

di m m2 n2 n

3 � 7 � 37 = 777 317 100; 489 211; 600 4603 � 11 � 37 = 1; 221 565 319; 225 430; 336 6567 � 11 � 13 = 1; 001 445 198; 025 309; 136 556

70

4. Let viwi, i = 1; 2; 3; 4, denote four cevians of a tetrahedron v1v2v3v4which are concurrent at an interior point P of the tetrahedron. Prove that

pw1 + pw2 + pw3 + pw4 � max viwi � longest edge:

Solution. We choose an origin, o, outside of the space of the tetrahe-dron and use the set of 4 linearly independent vectors Vi = ovi as a basis.Also the vector from o to any point q will be denoted by Q. The interiorpoint p is then given by P = x1V1 + x2V2 + x3V3 + x4V4 where xi > 0and

Pi xi = 1. It now follows that Wi =

P�xiVi1�xi (for other properties of

concurrent cevians via vectors, see [1987: 274{275]) and then that

pwi =

����P � xiVi

1� xi� P

���� =����xi(P � Vi)

1� xi

���� =������xiXj

xjVj � Vi

1� xi

������ ;

viwi =

����P � xiVi

1� xi� Vi

���� =����P � Vi

1� xi

���� =������Xj

xjVj � Vi

1� xi

������ :Summing

Xi

pwi =Xi

������xiXj

xjVj � Vi

1� xi

������ =Xi

xi(viwi) � maxi

viwi;

and with equality only if viwi is constant. Also,

viwi �Xj 6=i

�xj

1� xi

�maxrjVr � Vij = max

rjVr � Vij:

Finally, Xpwi � max

iviwi � max

r;sjVr � Vsj:

Comment: In a similar fashion, it can be shown that the result generalizesto n-dimensional simplexes. The results for triangles are due to Paul Erd }os,Amer. Math. Monthly, Problem 3746, 1937, p. 400; Problem 3848, 1940,p. 575.

5. Determine the radius r of a circle inscribed in a given quadrilateralif the lengths of successive tangents from the vertices of the quadrilateral tothe circle are a; a; b; b; c; c; d; d, respectively.

Solution. Let 2A, 2B, 2C, 2D denote the angles between successivepairs of radii vectors to the points of tangency and let r be the inradius. Then

r =a

tanA=

b

tanB=

c

tanC=

d

tanD:

71

Also, since A+ B + C +D = �, we have tan(A+B) = tan(C +D) = 0;or

tanA+ tanB

1� tanA tanB+

tanC + tanD

1� tanC tanD= 0;

so thatr(a+ b)

r2 � ab+r(c+ d)

r2 � cd= 0:

Finally,

r2 =abc+ bcd+ cda+ dab

a+ b+ c+ d:

Now we turn our attention to solutions by our readers to problemsgiven in the September 1995 number of the Corner where we gave the 16thAustrian Polish Mathematics Competition [1995: 221{222].

16TH AUSTRIAN POLISH MATHEMATICSCOMPETITION

First Day | June 30, 1993Time: 4.5 hours (individual competition)

1. Determine all natural numbers x; y � 1 such that 2x � 3y = 7.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;�Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina;

and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

We give Amengual's solution.First of all we note that both numbers x and y must be even. Suppose

to the contrary that one of the numbers is odd.

If x is odd, the number 2x + 1 (which we have from the factorizationAx + 1 = (A+ 1)(Ax�1 � Ax�2 + � � � � A+ 1) that 2x + 1) is a multipleof 3. Consequently 2x� 3y � �1 mod 3, while 7 � 1 mod 3, and the givenequation is invalid mod 3.

If y is odd, we will use the modular technique as in the previous case,but this time modulo 8. We have 32 � 1 mod 8. It follows that 32k+1 �3 mod 8 for k = 0; 1; 2; : : : .

Consequently 3y + 7 � 0 mod 8 and since 2x � 3y + 7 we must havex � 2. If x = 1, we have 2� 3y = 7, which is impossible. If x = 2, we have22 � 3y = 7, which is also impossible.

Suppose that the numbers x and y are even. So x = 2l, y = 2m, with landm natural numbers. The given equation can then be written in the form(2l + 3m)(2l � 3m) = 7, where 2l + 3m and 2l � 3m are natural numbers,which implies that 2l + 3m = 7 and 2l � 3m = 1. These two equationsdetermine the values of l,m, namely l = 2,m = 1 for which we have x = 4,y = 2.

72

The only two natural numbers x; y � 1 such that 2x � 3y = 7 aretherefore x = 4 and y = 2.

5. Determine all real solutions x, y, z of the system of equations:

x3 + y = 3x+ 4;

2y3 + z = 6y+ 6;

3z3 + x = 9z + 8:

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;

and by Panos E. Tsaoussoglou, Athens, Greece. We give the solution of

Tsaoussoglou.From x3 + y = 3x+ 4, we have

x3 � 1� 1� 3x = 2� y;

or

(x� 2)(x+ 1)2 = 2� y: (1)

From 2y3� 2� 2� 6y = 2� z, we have

2(y� 2)(y+ 1)2 = (2� z); (2)

and 3z3 � 3� 3� 9z = 2� x gives

3(z � 2)(z+ 1)2 = (2� x); (3)

so that

(x� 2)(x+ 1)2 = �(y� 2);

2(y� 2)(y+ 1)2 = �(z � 2);

3(z � 2)(z+ 1)2 = �(x� 2);

and

(x� 2)(y� 2)(z� 2)

�(x+ 1)2(y + 1)2(z + 1)2 +

1

6

�= 0:

As the last factor is always positive for real x; y; z, we have

(x� 2)(y� 2)(z� 2) = 0:

This gives at least one of x = 2, y = 2, z = 2. In conjunction with (1), (2)and (3) this gives the unique solution x = y = z = 2.

6. Show: For all real numbers a; b � 0 the following chain of inequal-ities is valid p

a+pb

2

!2� a+

3pa2b+

3pab2 + b

4

� a+pab+ b

3�

vuut 3pa2 +

3pb2

2

!3

:

73

Also, for all three inequalities determine the cases of equality.

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosnia

and Herzegovina; and by Panos E. Tsaoussoglou, Athens, Greece. We give

Tsaossoglou's solution.

1.

pa+

pb

2

!2

�3pa2( 3pa+

3pb) +

3pb2( 3pa+

3pb)

4

is equivalent to

(pa+

pb)2 � (

3pa2 +

3pb2)( 3

pa+

3pb);

which holds by the Cauchy inequality.

Let 6pa = A,

6pb = B, (A3 + B3)2 � (A4 + B4)(A2 + B2).

2. 3(a+ b) + 33pab( 3pa+

3pb) � 4(a+

pab+ b);

or equivalently

a+ 33pa2b+ 3

3pab2 + b � 2(a+ 2

pab+ b);

or( 3pa+

3pb)3 � 2(

pa+

pb)2;

or �A2 + B2

2

�3��A3 + B3

2

�2;

(with A, B as above), a known inequality.

3.a+

pab+ b

3�

vuut 3pa2 +

3pb2

2

!3

.

With A and B as above this is equivalent to�A6 + A3B3 + B6

3

�2��A4 + B4

2

�3:

For this it is enough to prove that�A4 + B4

2

�3��A6 + A3B3 +B6

3

�2� 0;

or

9(A4 + B4)3 � 8(A6 + A3B3 + B6)2

= A12 � 16A9B3 + 27A8B4 � 24A6B6 + 27A4B8 � 16A3B9 + B12

= (A� B)4�A8 + 4A7B + 10A6B2 + 4A5B3 � 2A4B4

+4A3B5 + 10A2B6 + 4AB7 + B8�

� (A� B)4(A3B3(A� B)2) � 0:

74

9. Let4ABC be equilateral. On sideAB produced, we choose a pointP such that A lies between P andB. We now denote a as the length of sidesof 4ABC; r1 as the radius of incircle of 4PAC; and r2 as the exradius of4PBC with respect to side BC. Determine the sum r1 + r2 as a functionof a alone.

Solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain;

and by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosnia and Herze-

govina. We give Amengual's solution and comment.Looking at the �gure, we see that \T1O1R = 60� since it is the supple-

ment of\T1AR = 120� (as an exterior angle for4ABC). Hence, \AO1R =30�. Similarly, we obtain \BO2S = 30�.

O1

O2

T1 A B T2

C

T 01

T 0

2

R

Sr1

r2

P

Since tangents drawn to a circle from an external point are equal, wehave

T1T2 = T1A+AB + BT2 = RA+ AB + SB

= r1 tan 30� + a+ r2 tan30

� =r1 + r2p

3+ a;

and

T 01T02 = T 01C +CT 02 = CR+CS = (a�RA) + (a� SB) = 2a� r1 + r2p

3:

Since common external tangents to two circles are equal, T1T2 = T 01T02.

Hence,r1 + r2p

3+ a = 2a� r1 + r2p

3;

whence we �nd that

r1 + r2 =ap3

2:

Comment. This problem is identical to problem 2.1.11, page 25, of H.Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, (The CharlesBabbage Research Centre, 1989).

75

We next give the solution to one problem of the VII Nordic Mathemat-ical Contest.

2. [1995: 223] VII Nordic Mathematical Contest

A hexagon is inscribed in a circle with radius r. Two of its sides havelength 1, two have length 2 and the last two have length 3. Prove that r is aroot of the equation

2r3 � 7r� 3 = 0:

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain.

Equal chords subtend equal angles at the centre of a circle; if each ofsides of length i subtends an angle �i (i = 1; 2; 3) at the centre of the givencircle, then

2�1 + 2�2 + 2�3 = 360�;

whence�1

2+�2

2= 90� � �3

2;

and

cos

��1

2+�2

2

�= cos

�90� � �3

2

�= sin

�3

2:

Next we apply the addition formula for the cosine:

cos�1

2cos

�2

2� sin

�1

2sin

�2

2= sin

�3

2; (1)

where (see �gures)

sin�1

2=

1=2

r; cos

�1

2=

p4r2 � 1

2r;

sin�2

2=

1

r; cos

�2

2=

pr2 � 1

r;

sin�3

2=

3=2

r:

3=2

r

�32

1

r

�22

1

2

r

�12

A B C

We substitute these expressions into (1) and obtain, after multiplyingboth sides by 2r2, p

4r2 � 1 �pr2 � 1� 1 = 3r:

76

Now write it in the formp(4r2 � 1)(r2� 1) = 3r + 1;

and square, obtaining

(4r2 � 1)(r2� 1) = 9r2 + 6r+ 1;

which is equivalent tor(2r3 � 7r � 3) = 0:

Since r 6= 0, we have2r3 � 7r� 3 = 0;

which was to be shown.

To �nish this number of the Corner we turn to two problems from the32nd Ukrainian Mathematical Olympiad given in the October 1995 numberof the Corner [1995 : 266].

32nd UKRAINIANMATHEMATICAL OLYMPIADMarch 1992 | Selected Problems

2. (8) There are real numbers a, b, c, such that a � b � c > 0. Provethat

a2 � b2

c+c2 � b2

a+a2 � c2

b� 3a� 4b+ c:

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosnia

and Herzegovina; Panos E. Tsaoussoglou, Athens, Greece; and by Edward

T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give the solu-

tion of Arslanagi�c.From a � b � c > 0 we have

a+ b

c� 2; 0 <

b+ c

a� 2 and

a+ c

b� 1:

Now we geta2 � b2

c� 2(a� b); because a � b;

c2 � b2

a� 2(c� b); because c � b;

anda2 � c2

b� a� c; because a � c:

After addition of these inequalities, we have

a2 � b2

c+c2 � b2

a+a2 � c2

b� 2(a� b) + 2(c� b) + (a� c);

77

that is,a2 � b2

c+c2 � b2

a+a2 � c2

b� 3a� 4b+ c:

The equality holds if and only if a = b = c > 0.

5. (10) Prove that there are no real numbers x, y, z, such that

x2 + 4yz+ 2z = 0;

x+ 2xy + 2z2 = 0;

2xz + y2 + y+ 1 = 0:

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosnia

and Herzegovina; by D.J. Smeenk, Zaltbommel, the Netherlands; by Panos

E. Tsaoussoglou, Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier

University, Waterloo, Ontario. We give Wang's solution.Label the three given equations (1), (2) and (3), (in that order), respec-

tively. If x = 0 or z = 0, then from (3), y2 + y + 1 = 0, which has noreal solutions. Hence we may assume that xz 6= 0. From (1) and (2), we getx2 = �2z(2y + 1) and 2z2 = �x(2y + 1), which, when multiplied gives2x2z2 = 2xz(2y+ 1)2 or xz = (2y+ 1)2. Substituting into (3) we get

2(2y+ 1)2 + y2 + y + 1 = 0

or3y2 + 3y+ 1 = 0;

which has no solution.

That completes the Corner for this issue. Olympiad season is upon us.Send me your Olympiads, as well as your nice solutions.

78

BOOK REVIEWS

Edited by ANDY LIU

Leningrad Mathematical Olympiads 1987{1991, by Dmitry Fomin andAlexey Kirichenko,published by MathPro Press, Westford, MA, USA, 1994.(Contests in Mathematics Series, vol. 1.), paperbound, 202 + xxii pages,ISBN 0{9626401{4{X, US$24.Reviewed by J �ozsef Pelik �an, E �otv �os Lor �and University, Budapest, Hun-gary.

The regulations of the InternationalMathematical Olympiad (IMO) stip-ulate that a country can send a team to the IMO consisting of at most sixstudents. Still, there was a city in the 1965 IMO which participated withnine students: �ve in the Russian team, two in the Israeli team, and one ineach of the USA and German teams. This city was Leningrad, now known byits old name St. Petersburg again.

If a city has such an output of talented young mathematicians, you canimagine that the city olympiad organized there should be of an exceptionallyhigh quality. And so it is. Let me brie y describe the system of the LeningradOlympiads. Until 1989 Soviet schools comprised grades 1{10, and the compe-tition was held for 5{10 graders. Since that time the Russian schools comprisegrades 1{11, and the competition is held for 6{11 graders. The competitionis organized in four rounds. The �rst round is at the school level in Decemberand January and the second round held at the regional level (that is, in the 22regions of Leningrad) in February, some 10,000 to 12,000 students taking partin the second round. Both of these rounds are written examinations. It is im-portant to emphasize this, because the last two rounds | quite unusually fora mathematics competition | are oral examinations. This requires of coursea huge number of able and devoted jurors, and few countries | let alonecities | can produce the number of mathematicians necessary for this task.In the third round (called \main round") about 90{130 students participatein each grade. In the �nal round (called somewhat misleadingly \eliminationround" | this term normally means the initial phase of a competition, notthe �nal one) which is held only for 9{11 (earlier 8{10) graders some 100 stu-dents take part altogether, although in 1991 there were only 34. The mainround is held in February or March, and the elimination round in March.In the period 1962{1983 and also in 1991 the elimination round was usedonly to pick the city team at the All-Union Olympiad (this might explain thestrange name), while in the period 1984{1990 the result of the competitionwas decided by the elimination round (not by the main round).

A few words about the earlier history of the Leningrad (and other)Olympiads in the Soviet Union. The Leningrad Olympiad started in 1934(the Moscow Olympiad one year later). The All-Russia Olympiad started

79

in 1961 and the All-Union (i.e. Soviet Union) Olympiad in 1967. These arerespectable ages, although there are a few national mathematics contestswhich are much older (notably the Hungarian E �otv �os | later K �ursch �ak |competitions which started in 1894). But I fully agree with the authors thatthe Leningrad competition is quite unique in being an oral one. On theother hand I must contest another statement of the authors, namely thatthe Leningrad competition would be quite unique in having only new andoriginal problems. In fact the majority of the competitions I know of havenew and original problems.

Besides the main part containing the problems and solutions the bookcontains useful appendices: statistical data about the number of participants,the names of each year's winners, a glossary and the names of the authorsof each problem. Also, there are valuable comments on the book by thepublisher, Stanley Rabinowitz, and a very good evaluation of the reasons ofthe excellence of Soviet mathematics by Mark Saul.

To give a sample of the problems (and solutions) I give you two exam-ples from the book. Both come from the 1991 Olympiad and were given tograde 11 students.

Problem A black pawn is placed in the top right square of an 8 � 8chess-board. One may place a white pawn on any empty square of the board,having repainted all pawns in adjacent squares so that black pawns becomewhite, and vice versa. (Two squares are called adjacent if they have a commonvertex.) Can one place pawns in this manner so that all 64 squares of theboard would be �lled with white pawns?

Solution Connect the centres of each pair of adjacent free squares bya segment. Then, when we place the next white pawn on some square ofthe board, we erase all segments from the centre of this square. Thus, thenumber of erased segments coincides each time with the number of free ad-jacent squares for the square on which each white pawn is placed. To makea certain pawn be white at the end, it must be placed on a square having aneven number of free adjacent squares because the number of repaintings foreach pawn is equal to the number of pawns that will be placed on adjacentsquares after this one. So, we conclude that each time we have to erase aneven number of segments. But this is impossible because the initial numberof segments is odd (we have lost at the beginning three segments caused bythe black pawn in the corner) and the �nal number is even | it is equal tozero. Thus, we see that in the �nal position there is at least one black pawn.

The next problem has a rather unusual solution.

Problem One may perform the following two operations on a naturalnumber:

(a) multiply it by any natural number;

(b) delete zeros in its decimal representation.

80

Prove that for any natural number n, one can perform a sequence of theseoperations that will transform n to a one-digit number.

Solution We use the following fact:

Lemma For any integer n that is not divisible by 2 or 5 one can �nd anumber consisting of digits 1 only that is a multiple of n :

Proof Consider n numbers 1; 11; : : : ; 11 : : : 1 : If one of these num-bers is a multiple of n, we are done. Otherwise, their remainders modulo ncan haven�1 possible values 1; : : : ; n�1 ; and therefore (by the pigeonholeprinciple) at least two of those numbers have the same remainders modulon :Then their di�erence is divisible by n and looks like 11 : : : 1 � 10k : But n iscoprime with 10, and we conclude that the �rst factor is divisible by n :

Now let n be an arbitrary natural number. Multiplyingn by 2 and 5 anddeleting zeros, we can transform n to a number that is coprime with 10. Thenmultiplying the result by the appropriate number, we can obtain a numbercontaining only digits 1 in its decimal representation (we use the assertion ofthe lemma). Now the chain of the following operations leads to the desiredresult:

(a) Multiplying by 82, we obtain the number 911 : : : 102 :

(b) Delete zero in the last obtained number and multiply it by 9. This givesthe number 8200 : : : 08 ; which is transformed into 828.

(c) 828 � 25 = 20700 :

(d) 27 � 4 = 108 :

(e) 18 � 5 = 90 ; and we can obtain the single-digit number 9.

All in all this book is very well written, full of interesting problems andI warmly recommend it to anyone interested in mathematical competitions,or just in nice problems.

81

Folding the Regular Heptagon

Robert Geretschl�ager, Bundesrealgymnasium, Graz, Austria

Introduction

Ever since Greek antiquity, mathematicians have been considering construc-tions that can be done with straight-edge and compass only, the so-calledEuclidean constructions. A number of famous problems, such as squaringthe circle, trisecting angles and doubling the cube, were unsolvable for theGreeks, and later shown to be theoretically unsolvable by Euclidean meth-ods. The reason for this is that only such problems that can be reduced al-gebraically to combinations of linear and quadratic equations are solvable inthis sense. We now know that these three problems, as well as many others,cannot be represented by combinations of such equations.

One speci�c problem the Greeks attempted to solve in this way was theconstruction of regular n{gons for small n. They were successful in �ndingconstructions for n = 3, 4, 5, 6, 8, 10 and 12, but not for n = 7, 9 or 11.Since 7 is the smallest n for which no construction could be found, it was ofspecial interest why this particular problem should prove so stubborn. Asit turned out, the construction of the regular heptagon by Euclidean meth-ods is impossible for the same reason that angle trisection and doubling thecube are, in that each of these problems requires the graphic solution of anirreducible cubic equation in its algebraic representation.

As shown in \Euclidean Constructions and the Geometry of Origami"

([1]), all cubic equations can be solved graphically using elementary meth-ods of origami1. This is especially interesting in light of the fact that regularn{gons are commonly used in the development of origami folding bases. Aheptagon could conceivably �nd use in developing models of insects for in-stance, since six legs + one head = seven corners. In this article, I present atheoretically precise method of folding the regular heptagon from a square,derived from the results established in the above-mentioned article. Thefolding method is presented in standard origami notation, and the mathe-matical section is cross-referenced to the appropriate diagrams.

The Cubic Equation

The seven corners of a regular heptagon can be thought of as the seven solu-tions of the equation

z7 � 1 = 0 (1)

1 Origami is, of course, the art of paper folding. For readers not yet familiar with this ancient

art, but interested in becoming so, there is a large amount of introductory literature easily

available. I was personally introduced to origami by the books of Robert Harbin ([2]). A �ne

introduction to the geometry of origami is the classic \Geometric Exercises in Paper Folding" by

T. Sundara Row ([3]).

82

in the complex plane. This implies that the unit circle is the circumcircle ofthe heptagon, and that one corner of the heptagon is the point z1 = 1 on thereal axis (Fig. 1.1.). Since one solution of (1) is known, the other six are theroots of

z7 � 1

z � 1= z6 + z5 + z4 + z3 + z2 + z1 + 1 = 0: (2)

For any speci�c z satisfying this equation, the conjugate z is also a solution,since the real axis is an axis of symmetry of the regular heptagon. Also, since

jzj = jzj = 1;

we have z = 1z. Therefore we can de�ne

� = z +1

z= z + z = 2 � Re z: (3)

r

r

r

r

r

r

r

r

6

-z1=1

x

z2

iy

z7

aOb

z3

z6

z4

z5

c

Fig. 1.1.

z1 = 1 + i � 0 z2 = cos 2�7+ i � sin 2�

7z3 = cos 4�

7+ i � sin 4�

7

z4 = cos 6�7+ i � sin 6�

7z5 = z4 = cos 8�

7+ i � sin 8�

7

z6 = z3 = cos 10�7

+ i � sin 10�7

z7 = z2 = cos 12�7

+ i � sin 12�7

a = cos 2�7

= Re z2 = Re z7 b = cos 4�7

= Re z3 = Re z6c = cos 6�

7= Re z4 = Re z5

Dividing by z3, we see that equation (2) is equivalent to

z3 + z2 + z + 1+1

z+

1

z2+

1

z3= 0

since 0 is not a root, and since

�3 =

�z +

1

z

�3= z3 + 3z +

3

z+

1

z3

= z3 +1

z3+ 3

�z +

1

z

�= z3 +

1

z3+ 3�

() �3 � 3� = z3 +1

z3

83

and

�2 =

�z +

1

z

�2= z2 + 2+

1

z2

() �2 � 2 = z2 +1

z2;

substituting yields�z3 +

1

z3

�+

�z2 +

1

z2

�+

�z +

1

z

�+ 1 = 0

() �3 � 3� + �2 � 2 + � + 1 = 0

() �3 + �2 � 2� � 1 = 0:

From (3) ,we see that each root of the equation

�3 + �2 � 2� � 1 = 0 (4)

is real, and is equal to twice the common real component of two conjugatecomplex solutions of (1). It is therefore possible to �nd the six complex rootsof (1) in the complex plane by �nding the roots of (4), taking half their val-ues, �nding the straight lines parallel to the imaginary axis and at preciselythese distances from it, and �nally �nding the points of intersection of theseparallel lines with the unit circle. We shall now proceed to utilize these stepsin folding the regular heptagon.

A Step-by-step Description of the Folding Process

As is usually the case in origami, we assume a square of paper to be given. Weconsider the edge-to-edge folds in step 1 as the x{ and y{axes of a system ofcartesian coordinates, and the edge-length of the given square as four units.The mid-point of the square is then the origin M(0;0), and the end-pointsof the folds have the coordinates (�2; 0) and (2;0), and (0;�2) and (0;2)respectively. For readers not familiar with origami notation, it should bementioned that dashed lines represent so-called \valley" folds (folding up),and dot-dashed lines represent so-called \mountain" folds (folding down).Thin lines represent visible creases in the paper generated by previous folds.

As shown in [1], the solutions of the cubic equation

x3 + px2 + qx+ r = 0

are the slopes of the common tangents of the parabolas p1 and p2 de�ned bythe foci

F1

��p2+r

2;q

2

�and F2

�0;

1

2

�and directrices

l1 : x = �p2� r

2and l2 : y = �1

2

84

respectively.The solutions of (4) can therefore be obtained by �nding the common

tangents of the parabolas with foci

F1(�1;�1) and F2

�0;

1

2

�

and directrices

l1 : x = 0 and l2 : y = �1

2

respectively. Since the slope of the common tangents is not altered by trans-lating the parabolas parallel to the y{axis we can, for convenience, use

F1

��1;�1

2

�and F2(0; 1)

andl1 : x = 0 and l2 : y = 0:

This is precisely what is done in steps 2 to 5. F1 is the point A, and F2 is thepoint B. The fold in step 4 is then the only common tangent of the parabolaswith positive slope, and thus twice the real component of the solutions of (1)which lie to the right of the imaginary axis and are not equal to 1. In otherwords, the slope of this fold is 2 � cos 2�

7. Step 4, by the way, is the only step

that cannot be replaced by a straight-edge and compass construction.In steps 6 to 8, the unit-length is then transferred in such a way that

point E in step 8 has y{coordinate �2 � cos 2�7. Since the distance from M

to point 1 in step 9 is 2 units, the distances from M to points 2 and 7 arealso 2 units, and so points 7, 1 and 2 are three consecutive corners of theregular heptagon. (We assume that point 1 with coordinates (0;�2) is the�rst corner, and continue from there.)

Step 10 thus yields two sides of the heptagon, and steps 11 to 13 yieldthe remaining sides of the heptagon by making use of its radial symmetry,until �nally step 14 shows us the completed regular heptagon. The foldingprocess is shown at the end of the article.

Conclusion

Unlike other regular n{gons with small n, the regular heptagon is not verycommon in popular culture or graphics. Apart from the seven-sided star onecomes across in astrology, the heptagon does not seem to show up much inpublic, unlike its close relatives. We come across the octagon at many a streetcorner, and the pentagon and hexagon can be seen on most soccer balls, justto name a few. I do not know if this (relatively) easy generation of the regularheptagon will lead to its mass popularization, but an ardent Heptagonist cancertainly dream.

It should be mentioned that a similar folding method for the regularheptagon is described in the article \Draw of a Regular Heptagon by the

85

Folding" by Benedetto Scimemi ([4]) in the relatively hard to �nd Proceed-ings of the First International Meeting of Origami Science and Technology.(I have only recently gained access to a copy myself.) This volume o�ers agreat many ideas for further research for anyone interested in the geometryof origami, and is certainly worth searching for.

References

[1] R. Geretschl�ager, Euclidean Constructionsand the Geometry of Origami,Mathematics Magazine, 68 No. 5 December 1995, pp. 357-371

[2] R. Harbin, Origami, The Art of Paper-Folding, Vols. 1-4, Hodder Paper-backs, Norwich (1968)

[3] T. Sundara Row, Geometric Exercises in Paper Folding, Dover Publica-tions, Inc., Mineola, NY (1966) reprint of 1905 edition

[4] B. Scimemi, Draw of a Regular Heptagon by the Folding, Proceedingsof the First International Meeting of Origami Science and Technology,Ferarra, Italy (1989)

The Folding Process

1.

Fold and unfold twice.

2.

Fold back twice.

86

3.

qA

Fold and unfold, making acrease mark at point A(bisecting the side).

4.

q

q

A

B

Fold such that A and Bcome to lie on the creases.

5.

q

q

A

B

Unfold everything.

6.

q qC

D

Fold C to D.

87

7.

Fold and unfold both layers atcrease, then unfold everything.

8.

q

E

Fold horizontally through E,then unfold.

9.

q

q q

q

7 2

1

M

Fold throughM , such that 1lies on crease, resulting in 2

and 7 (M is the mid-point of theheptagon, 1, 2 and 7 are corners).

10.

q

q qq q

q q

Fold back twice, so that themarked points come to lie onone another; resulting folds are�rst two sides of the heptagon.

88

11.

q

q

2

M

Fold throughM and 2.

12.

Fold back lower layers using edgesof upper layer as guidelines;

resulting folds are twomore sides of the heptagon;open up fold from step 12 andrepeat 11 and 12 on left side.

13.

q

q5 q 4

Fold back �nal edge of theheptagon through 4 and 5.

14.

The �nished heptagon.

89

THE SKOLIAD CORNERNo. 20

R.E. Woodrow

This issue we give the problems of the Mathematical Association Na-tional Mathematics contest, written November 18, 1994. The contest waswritten by about 30,000 students in the United Kingdom. My thanks go toTony Gardiner, School ofMathematics, University of Birmingham for sendingme the contest.

THE MATHEMATICAL ASSOCIATION NATIONALMATHEMATICS CONTEST 1994

Friday, November 18, 1994 | Time: 90 minutes

1. The average of x and 8x is 18. What is the value of x?

A. 1 12

B. 2 C. 4 D. 4 12

E. 9.

2. Which of the following does not have six lines of symmetry?

A. �g 2a B. �g 2b C. �g 2c D. �g 2d E. �g 2e.

3. I write out the numbers from 1 up to 30 in words. If N denotesthe number of times I write the letter \n", M denotes the number of timesI write the letter \m", and C denotes the number of times I write the letter\c", then N +M + C equals

A. 27 B. 28 C. 29 D. 30 E. 31.

4. Which of the following �ve numbers has a prime factor in commonwith exactly one of the other four numbers?

A. 91 B. 52 C. 39 D. 35 E. 24.

5. ABCD is a quadrilateral with AB = AD = 25 cm, CB = CD =52 cm and DB = 40 cm. How long in AC in cm?

A. 32:5 B. 48 C. 52 D. 60 E. 63.

90

6. The number of pounds of pickled peppers that Peter Piper purchasedfor $59 is equal to the number of pounds Peter would pay for two hundredand thirty six pounds of peppers. How much would he pay for twenty poundsof pickled peppers?

A. $5 B. $10 C. $20 D. $40 E. $80.

7. Which expression has the smallest value when x = �0:5?A. 21=x B. �1

xC. 1

x2D. 2x E. 1p�x .

8. Over an average lifetime in the UK, roughly how many times doesa person's heart beat?

A. 4� 107 B. 5� 107 C. 2� 108 D. 3� 109 E. 2� 1010.

9. What is the sum of the reciprocals of the �rst six triangular numbers1, 3, 6, 10, etc.?

A. 10 B. 127

C. 56 D. 32

E. 4921.

10. A rope 15 m long and 5 cm indiameter is coiled in a at spiral as shown.What is the best estimate for the diameterof the \circle" (in cm)?

q

A. 10 B. 100 C. 150 D. 200 E. 300.

11. If a b = (ab+a+b+1)

a, then 19 94 equals

A. 95 B. 100 C. 208 D. 1882 E. 1994.

12. The diagram shows a semicirclewith radius 1 cm and with centre O. IfC isan arbitrary point on the semicircle, whichof the following statements may be false?

A BO

C

A. \ACB is a right angle. B.4OAC is isosceles.C. the area of4ABC is � 1 cm2 D. 4AOC is equal in area to 4OBC

E. AO2 + OB2 = AC2 + BC2.

91

13. A giant marrow in my garden weighed 50 pounds and was 98%water. Suppose that during a rainy day it absorbed water so that it became99% water. What would its new weight be (in pounds)?

A. 50:01 B. 50:5 C. 98 D. 99 E. 100.

14. A solid cuboid has edges of lengths a, b, c. What is its surfacearea?

A. (a+ b+ c)2 � (a2 + b2 + c2) B. abc C. 2(a2 + b2 + c2)D. (a+ b+ c)2 E. ab+ bc+ ca.

15. Given two copies of an isosce-les right-angled triangle ABC, squaresBDEF and PQRS are inscribed in dif-ferent ways as shown. What is the ratio

area PQRS

area BDEF?

@@@@@@

B C

A

B C

A

F E

D

@@@@@@@

@��@

@��

S

R

P

Q

A. 89

B. 23

C. 1 D.

q23

E. 98.

16. What is the last digit of 1994(1995+1996+1997+1998+1999+2000)?

A. 0 B. 2 C. 4 D. 6 E. 8.

17. When two dice are thrown the probability that the total score isa multiple of 2 is 1

2. For how many other values of n is it true that, when

two dice are thrown, the probability that the total score is a multiple of n isequal to 1

n?

A. 1 B. 2 C. 3 D. 4 E. 5.

18. How many digits are there in the smallest number which is com-posed entirely of �ves (eg. 5555) and which is divisibly by 99?

A. 9 B. 10 C. 18 D. 36 E. 45.

19. The price of a secondhand car is displayed (in pounds) on four cardson the windscreen. Each card shows one digit. If the card with the thousandsdigit blew o� in the wind, the apparent price of the car would drop to oneforty-ninth of the intended value. What number is on that card?

A. 5 B. 6 C. 7 D. 8 E. 9.

92

20. The graph of y � x against y + x is as shown.

-

6

O y + x

y � x

����

The same scale has been used on each axis. Which of the following shows thegraph of y against x?

-

6

O x

y

-

6

O x

y

����

-

6

O x

y

����

-

6

O x

y

AAAA

-

6

O x

y

A. �g 20a B. �g 20b C. �g 20c D. �g 20d E. �g 20e.

21. Which is smallest?

A. 5 + 6p7 B. 7 + 6

p5 C. 6 + 5

p7 D. 7 + 5

p6 E. 6 + 7

p5.

22. A train leaves London at 0600 and arrives in Newcastle at 0930.Another train leaves Newcastle at 0700 and arrives in London at 0930. Ifboth used the same route and each travelled at a constant speed, at whattime would they meet?

A. 075712

B. 080212

C. 080712

D. 082712

E. more information required.

23. The triangle ABC has a right-angle at A. The hypotenuse BCis trisected at M and at N so that BM = MN = NC. If AM = x andAN = y, then MN is equal to

A. x+y2

B.

p(y2�x2)2

C.p(y2� x2) D.

p(x2+y2)

3E.

qx2+y2

5.

24. If Susan drives to work at x mph she will be one minute late; ifshe drives at y mph she will be one minute early. How far does she drive towork (in miles)?

A. yx

30(y�x) B. 2yx

y�x C. x+yy�x D. x+y

2E. x+y

60(y�x).

93

25. The octagonal �gure is obtainedby �tting eight congruent isosceles trapeziatogether. If the three shorter sides of eachtrapezium have length 1, how long is eachouter edge?

@@��

@@ ��

@@@�

��

@@@ �

��

!!��

LL

aa

!!

��

LL

aa

!!

A. 1 +p2 B. 1+

p2

2C.p2 D. 2

E. 1 +p2�p2.

Last number we gave the 1994 Nat West U.K. Junior MathematicalChallenge. It was written Tuesday, April 26, 1994. Here are the answers.

1. 455 2. B 3. E 4. A 5. A6. B 7. B 8. D 9. D 10. A11. B 12. E 13. A 14. B 15. E16. E 17. C 18. C 19. D 20. E21. A 22. D 23. D 24. C 25. C

That completes the Skoliad Corner for this issue. Sendme your suitablecontests and solutions. Also sendme any suggestions for improvement of thisfeature.

94

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

the Mayhem Editor, Naoki Sato, Department of Mathematics, University of

Toronto, Toronto, ON Canada M5S 1A1. The electronic address is

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Richard Hoshino (University of Waterloo), WaiLing Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Matrix Exponentials: An Introduction

Donny Cheungstudent, University of Waterloo

Waterloo, Ontario.

Let us start with the well-known power series for the exponential func-tion,

ex = 1+ x+x2

2!+x3

3!+x4

4!+ � � � =

1Xk=0

xk

k!;

which works for all x 2 C . Likewise, we will de�ne the exponential functionfor matrices (with complex entries) as

eM = I +M +M2

2!+M3

3!+M4

4!+ � � � =

1Xk=0

Mk

k!

for all n� n matrices M .

Matrix Exponentials Really Do Exist

First of all, we must show that this sum actually converges. Otherwise,the answer we get would be meaningless, and that would be bad. To startus o�, we will de�ne the norm of an n�nmatrix, which we will denote jAj,as follows:

jAj = max1�i�n

0@ nXj=1

jaijj1A :

That is, for each row, take the sum of the absolute values of the componentsin that row, and the norm will be the largest of these row sums.

95

Lemma 1 For n� n matrices A and B,

jABj � jAjjBj:

Proof We label the entries of A and B in the standard way, aij and bij,respectively.

For any row in A, de�ne the row sum as

nXj=1

jaij j. Since (AB)ij =

nXk=1

aikbkj, the row sum of row i in AB is

nXj=1

�����nXk=1

aikbkj

����� �nXj=1

nXk=1

jaikbkj j =

nXk=1

nXj=1

jaikj jbkj j

=

nXk=1

0@jaikj nX

j=1

jbkjj1A �

nXk=1

(jaikj jBj)

= jBjnXk=1

jaikj � jAjjBj:

Now, jABj, the maximum row sum, is still a row sum, thus jABj � jAjjBj.�

Lemma 2 For n� n matrices A and B,

jA+ Bj � jAj + jBj:

The proof is left as an exercise to the reader.

Theorem 1 The sum eM converges for all n� n matrices M .

Proof Using the two lemmas, we get

eM = I +M +M2

2!+M3

3!+M4

4!+ � � � ;

so that

��eM �� = jI +M +M2

2!+M3

3!+M4

4!+ � � � j

� 1 + jM j+ jM j22!

+jM j33!

+jM j44!

+ � � �

= ejMj:

Since jM j is a real number, ejMj is �nite, and jeM j is bounded. Thus, eM

converges. �

96

Exercises

1. Prove Lemma 2.

Some Linear Algebra Lingo

Recall that two n� n matrices A and B are similar if

B = C�1AC

for some invertible n � n matrix C. (It is usually easier, when verifyingthat two matrices are similar, to show that CB = AC instead.) Recall alsothat a diagonalmatrix is a matrix with all its non-zero entries along the maindiagonal. (But entries along the main diagonal are not necessarily non-zero).For example: 0

BB@5 0 0 0

0 �4 0 0

0 0 0 0

0 0 0 �1

1CCA :

If a matrix A is similar to a diagonal matrix, we say that A is diagonalizable.

Calculating Matrix Exponentials

Suppose D is a diagonal matrix, with diagonal elements �1, �2, : : : ,�n. We have

D =

0BBBB@

�1 0 � � � 0

0 �2 � � � 0

.

.

....

. . ....

0 0 � � � �n

1CCCCA ; D

k=

0BBBB@

�k1 0 � � � 0

0 �k2 � � � 0

.

.

....

. . ....

0 0 � � � �k

1CCCCA ;

and

eD

= I + D +D

2

2!+ � � �

=

0BBBBB@

1 + �1 +�21

2!+ � � � 0 � � � 0

0 1 + �2 +�22

2!+ � � � � � � 0

.

.

....

. . ....

0 0 � � � 1 + �n +�2n2!

+ � � �

1CCCCCA

=

0BBBB@

e�1 0 � � � 0

0 e�2

� � � 0

.

.

....

. . ....

0 0 � � � e�n

1CCCCA :

97

Jordan Matrices

A Jordan block matrix is an n� n matrix of the form0BBBBBB@

� 1 0 � � � 0 0

0 � 1 � � � 0 0

.

.

....

.

.

.. . .

.

.

....

0 0 0 � � � � 1

0 0 0 � � � 0 �

1CCCCCCA

with �'s all down the main diagonal and 1's directly above them (except forthe �rst column), and 0's everywhere else. Examples include

(5),

�5 1

0 5

�, and

0BB@

�3 1 0 0

0 �3 1 0

0 0 �3 1

0 0 0 �3

1CCA :

A Jordan matrix is an n� n matrix with Jordan blocks down the diag-onal. An example containing our three examples above:0

BBBBBBBB@

5 0 0 0 0 0 0

0 �3 1 0 0 0 0

0 0 �3 1 0 0 0

0 0 0 �3 1 0 0

0 0 0 0 �3 0 0

0 0 0 0 0 5 1

0 0 0 0 0 0 5

1CCCCCCCCA:

Notice that all diagonal matrices are also Jordan matrices.

Exercises

1. Verify that0BBBBBB@

� 1 0 � � � 0 0

0 � 1 � � � 0 0

.

.

....

.

.

.. . .

.

.

....

0 0 0 � � � � 1

0 0 0 � � � 0 �

1CCCCCCA

k

=

0BBBBBBB@

�k

�k1

��k�1

�k2

��k�2

: : :�k

n�1

��k�n+1

�kn

��k�n

0 �k

�k1

��k�1

: : :�k

n�2

��k�n+2

�k

n�1

��k�n+1

.

.

....

.

.

.. . .

.

.

....

0 0 0 : : : �k

�k1

��k�1

0 0 0 : : : 0 �k

1CCCCCCCA:

2. Compute eB when B is a Jordan block matrix.

3. Compute eJ when J is a Jordan matrix.

98

So why did we just go through all that?

Here's the reason:

Theorem 2 For any n� n matrix A, there exists a Jordan matrix J which issimilar, that is, C�1AC = J for some C.

J is known as a Jordan canonical form of A. Some matrices have morethan one Jordan canonical form.

The proof is beyond the scope of this article, but can be found in anygood advanced linear algebra textbook. But we will prove the following:

Theorem 3 If J is a Jordan canonical form of n � n matrix A, withA = CJC�1, then eA = C(eJ)C�1.

Proof

Ak = (CJC�1)k

= (CJC�1)(CJC�1) � � � (CJC�1)| {z }k

= CJ(C�1C)J(C�1C)J(C�1C) � � � (C�1C)JC�1

= CJkC�1:

Now, eA = I + A+A2

2!+ � � �

= CIC�1 + CJC�1 + CJ2

2!C�1 + � � �

= C

�I + J +

J2

2!+ � � �

�C�1

= C(eJ)C�1:

Thus, eA = C(eJ)C�1. �

The process of �nding a Jordan canonical form J of a matrix A is alsobeyond the scope of this article. Once again, I refer you to a good textbook.However, we have shown that it is possible to calculate eA for any n � n

matrix A.

And now for something less theoretical...

Here, we discuss an interesting application ofmatrix exponentials: �rst-order systems of linear di�erential equations. By that, we mean systems ofthe form:

x01(t) = a11x1(t) + a12x2(t) + � � �+ a1nxn(t);

x02(t) = a21x1(t) + a22x2(t) + � � �+ a2nxn(t);

.

.

.

x0n(t) = an1x1(t) + an2x2(t) + � � �+ annxn(t):

99

Letting

x(t) =

0BBBB@

x1(t)x2(t)

.

..xn(t)

1CCCCA ;

we de�ne the derivative x(t) as

x0(t) =

0BBBB@

x01(t)x02(t)...

x0n(t)

1CCCCA :

We will also let A be the matrix of coe�cients for the system. Now wecan rewrite the system of equations as x0(t) = Ax(t).

Exercises

1. Verify that x(t) = eAtc is a solution to x0(t) = Ax(t) for every constantvector c.

Further Exploration

This is where I become too lazy to show you all the neat stu� you can dowith matrix exponentials, and where I encourage you, the reader, to exploreon your own.

Exercises

1. We don't have to stop at matrix exponentials. We can de�ne matrixequivalents for functions like sin(x) and cos(x) in a very similar fash-ion. What other types of functions can we extend to the matrices.

2. Prove that sin(A) sin(A) + cos(A) cos(A) = I for any arbitrary n�n

matrix A.

3. When is eA+B = eAeB? The fact that matrix multiplication isn't com-mutative causes some problems.

4. As a corollary to the above problem, show that e2A = eAeA.

5. Prove or disprove: there exist a 2 � 2 matrix A with real entries suchthat

sinA =

�1 19960 1

�:

[1996 Putnam, B4]

100

Open problems

1. As we have de�ned eA, it is very easy to de�ne xA for x 2 R. Can wede�ne BA for n � n matrices A and B? What would their properties

be? Would ABC =�AB

�C?

2. Can we de�ne a log function? Would it have the similar properties tothe real log function?

3. Can we solve other types of vector di�erential equations with matrixexponentials or (more generally) matrix functions like sin and cos?

Mayhem Problems

A new year brings new changes and new problem editors. Cyrus Hsia nowtakes over the helm as Mayhem Advanced Problems Editor, with Richard

Hoshino �lling his spot as the Mayhem High School Problems Editor, andveteran Ravi Vakil maintains his post asMayhem Challenge Board Problems

Editor. Note that all correspondence shouldbe sent to the appropriate editor| see the relevant section.

In this issue, you will �nd only solutions | the next issue will featureonly problems. We intend to have problems and solutions in alternate issues.

We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from theprevious issue be submitted by 1 June 1997, for publication in the issue 5months ahead; that is, issue 6. We also request that only students submitsolutions (see editorial [1997: 30]), but we will consider particularly elegantor insightful solutions for others. Since this rule is only being implementednow, you will see solutions from many people in the next few months, as weclear out the old problems from Mayhem.

101

High School Problems | Solutions

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,Canada. L3S 3E8 rhoshino@undergrad.math.uwaterloo.ca

H205. A circular billiard table is given with a cue ball at the circum-ference. It is shot at an angle of � to the line from the ball to the centre ofthe table. For what angles � will the ball come back to this point, assumingthe ball keeps going inde�nitely?

Solution by Samuel Wong, Mary Ward Catholic Secondary School,

Toronto.

A1

A2

A3

O

�

�

��

We let the ball start at A1 as shown. Then \OA1A2 = \A1A2O = �

(equal radii) and \A1OA2 = 180��2� (angles sum to 180�). Since the angleof incidence equals the angle of re ection, we have \A3A2O = \A1A2O =� = \A2A3O by equal radii so \A3OA2 = \A2OA1 = 180� � 2�. Thus,the angles at the centre are always equal. The cue ball returns to A1 i�

k(180� � 2�) = 360�c =) k(90� � �) = 180�c;

for some positive integers c and k. Since � must be real, 90� � � must beeither a rational or irrational number.

Case I: If 90�� � is rational, it can be expressed in the form mn, where

m and n are positive, relatively prime integers. Thus k � mn= 180c. We may

take k = 180n and thenm = c, so for all � rational, the cue ball will returnto A1.

Case II: If 90��� is irrational, let x = 90���. Then kx = 180�c =)x = 180�c

k. But x is irrational, so we have a contradiction.

Also solved byMIGUEL CARRI �ON �ALVAREZ, Universidad Complutense

de Madrid, Spain.

102

H206. For what values of n is an n-digit natural number uniquelydetermined from the sum and product of its digits?

SolutionbyMiguel Carri �on �Alvarez, Universidad Complutense deMad-

rid, Spain.

The sum and product of the digits are two conditions, and so only twodigits can be determined in general. This means n = 1 or 2, but whenn = 2, the order cannot be determined (for example, consider 12 and 21), sothe answer is n = 1.

H207. Is there a natural number n, such that �(n) = p, where p is anodd prime number?

Solution by Bob Prielipp, University of Wisconsin-Oshkosh, WI.

We shall show that there is NO positive integer n such that �(n) isan odd prime number. This is an immediate consequence of the fact that�(1) = 1 = �(2) and the result established below.

Theorem: If n is a positive integer, n � 3, then 2 divides �(n).Proof: Let n be a positive integer, n � 3.Case I: n = 2a, where a is an integer, a � 2. Then �(n) = 2a�1,

where a� 1 is a positive integer, so 2 divides �(n).Case II: n has an odd prime factor p. If the prime factorization of n is

pa11 pa22 � � � pakk , where pk is odd, then

�(n) = pa1�11 p

a2�12 � � � pak�1k (p1 � 1)(p2� 1) � � � (pk � 1):

Because pk is odd, pk � 1 is even so, 2 divides �(n)

Also solved byMIGUEL CARRI �ON �ALVAREZ, Universidad Complutense

de Madrid, Spain, SAMUEL WONG, Mary Ward Catholic Secondary School,

Toronto.

Advanced Problems | Solutions

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.M1G 1C3 hsia@math.toronto.edu

A185. Let bn be the highest power of 3 dividing

�3k

n

�, 0 � n � 3k.

Calculate

3kXn=0

1

bn.

Solution by Wai Ling Yee, student, University of Waterloo, Waterloo,

Ontario.

The exponent of the highest power of 3 dividing t! is�t

3

�+

�t

32

�+

�t

33

�+ � � � :

103

Thus, the exponent of the highest power of 3 dividing

�3k

n

�=

3k!

n!(3k� n)!is

kXj=1

$3k

3j

%�$n

3j

%�$3k � n

3j

%!:

Sincen

3j+

3k � n

3j= 3k�j , an integer,

$3k

3j

%�$n

3j

%�$3k � n

3j

%

can only take two possible values: 0 and 1.When 3j j n, the value is 0. Otherwise, it is 1. Therefore, the value of bn is3k divided by the highest power of 3 dividing n.

Consider the numbers divisible by 3j ; 0 � j � k � 1. Ignoring 0,we see that 3k�j of the numbers are divisible by 3j. These numbers can bewritten as 3j � (3m);3j � (3m + 1), or 3j � (3m + 2). Therefore, there are2 � 3k�j�1 numbers in f1; 2; : : : ; 3kg for which 3j is the highest power of 3dividing them.

Each of the 2 � 3k�j�1 numbers contributes1

3k�jto the sum

3kXn=0

1

bnfor a

total of 23. Therefore,

3kXn=0

1

bn=

3k�1Xn=1

1

bn+ 2 =

2

3k + 2:

Also solved by Edward Wang,Wilfred Laurier University, Waterloo, On-

tario.

A186. Let an be the sequence de�ned by a1 = 1, and

an+1 =1 + a21 + � � �+ a2n

n; n � 1:

Prove that every an is an integer.

Solution

An April Fool's joke. It turns out that a1, a2, : : : , a43 are all integers,but a44 is not an integer.

[Ed: can you prove this?]

104

A187. Let Sk(n) be the polynomial in n, such that Sk(n) =

nXi=1

ik for

all positive integers n, e.g. S0(n) = n, S1(n) = (n2 + n)=2.Prove that for k � 1, n(n+ 1) j Sk(n).Furthermore, prove that for k odd, k � 3, n2(n+ 1)2 j Sk(n).

Solution by Wai Ling Yee, student, University of Waterloo, Waterloo,

Ontario.

Note that n(n+1) j S1(n) =n(n+ 1)

2. Assume that n(n+1) divides

all Sk(n) up to k = t. Consider:

(n+ 1)t+2

= nt+2

+

t+ 2

1

!nt+1

+

t+ 2

2

!nt+ � � �+

t+ 2

t+ 1

!n+ 1;

(n� 1 + 1)t+2

= (n� 1)t+2

+

t+ 2

1

!(n� 1)

t+1+

t+ 2

2

!(n� 1)

t

+ � � �+

t+ 2

t+ 1

!(n� 1) + 1;

(n� 2 + 1)t+2

= (n� 2)t+2

+

t+ 2

1

!(n� 2)

t+1+

t+ 2

2

!(n� 2)

t+

� � �+

t+ 2

t+ 1

!(n� 2) + 1;

.

.

.

(2 + 1)t+2

= 2t+2

+

t+ 2

1

!2t+1

+

t+ 2

2

!2t+ � � �+

t+ 2

t+ 1

!2 + 1;

(1 + 1)t+2

= 1t+2

+

t+ 2

1

!1t+1

+

t+ 2

2

!1t+ � � �+

t+ 2

t+ 1

!1 + 1:

Adding all of the equations together, we have:

n+1Xi=2

it+2 =

nXi=1

it+2 +

t+1Xi=0

nXj=1

�t+ 2

t+ 2� i

�ji

if and only if

(n+ 1)t+2 � n� 1 =

�t+ 2

1

�St+1(n) +

tXi=1

�t+ 2

t+ 2� i

�Si(n):

By the induction hypothesis, we need only to prove that n(n + 1) divides(n+ 1)t+2 � n� 1 = n(n+ 1)t+1, which is true.

105

By induction, n(n+ 1) j Sk(n) for k � 1.

Now, S3(n) =n2(n+1)2

4; therefore n2(n+ 1)2 divides S3(n). Assume

that n2(n+1)2 divides Sk(n) for all odd k up to k = t�1, t even. Consider:

(n� 1)t+2

= nt+2

�

t+ 2

1

!nt+1

+

t+ 2

2

!nt� � � � �

t+ 2

t+ 1

!n+ 1;

(n� 1� 1)t+2

= (n� 1)t+2

�

t+ 2

1

!(n� 1)

t+1+

t+ 2

2

!(n� 1)

t

� � � � �

t+ 2

t+ 1

!(n� 1) + 1;

(n� 2� 1)t+2

= (n� 2)t+2

�

t+ 2

1

!(n� 2)

t+1+

t+ 2

2

!(n� 2)

t

� � � � �

t+ 2

t+ 1

!(n� 2) + 1;

.

.

.

(2� 1)t+2

= 2t+2

�

t+ 2

1

!2t+1

+

t+ 2

2

!2t� � � � �

t+ 2

t+ 1

!2 + 1;

(1� 1)t+2

= 1t+2

�

t+ 2

1

!1t+1

+

t+ 2

2

!1t� � � � �

t+ 2

t+ 1

!1 + 1:

Subtracting the sum of all the equations in the above group from the sum ofall the equations in the �rst group, we have:

(n+ 1)t+2 + nt+2 � 1 = 2

��t+ 2

1

�St+1(n) +

�t+ 2

3

�St�1(n)

+ � � �+�t+ 2

t� 1

�S3(n) +

�t+ 2

t+ 1

�S1(n)

�:

By the induction hypothesis, we have to prove only that n2(n+ 1)2 divides

(n+ 1)t+2 + nt+2 ��t+ 2

t+ 1

�(n2 + n)� 1:

106

nt+2 ��t+ 2

t+ 1

�(n2 + n)� 1 = (n+ 1� 1)t+2 �

�t+ 2

t+ 1

�(n2 + n)� 1

= (n+ 1)t+2 � � � �+�t+ 2

t

�(n+ 1)2 �

�t+ 2

t+ 1

�(n+ 1)1 + 1

��t+ 2

t+ 1

�(n2 + n)� 1

= (n+ 1)t+2 � � � �+�t+ 2

t

�(n+ 1)2 �

�t+ 2

t+ 1

�(n2 + 2n+ 1)

= (n+ 1)t+2 � � � �+�t+ 2

t

�(n+ 1)2 �

�t+ 2

t+ 1

�(n+ 1)2:

Therefore (n+ 1)2 divides (n+ 1)t+2 + nt+2 ��t+ 2

t+ 1

�(n2 + n)� 1.

(n+ 1)t+2 ��t+ 2

t+ 1

�n� 1

= nt+2 + � � �+�t+ 2

t

�n2 +

�t+ 2

t+ 1

�n+ 1�

�t+ 2

t+ 1

�n� 1

= nt+2 + � � �+�t+ 2

t

�n2:

Therefore n2 divides (n+ 1)t+2 + nt+2 � �t+2t+1

�(n2 + n)� 1.

Therefore n2(n+ 1)2 divides St+1(n).

By induction, n2(n+ 1)2 divides Sk(n) for odd k � 3.

Challenge Board Problems | Solutions

Editor: Ravi Vakil, Department of Mathematics, One Oxford Street,Cambridge, MA, USA. 02138-2901 ravi@math.harvard.edu

We begin by dredging up an old favourite of ours. Well, maybe notso old { a solution appeared in [Mayhem 8: 4, 25: 1996]. We have sincereceived a new solution that is worth printing.

C64. The numbers x1, x2, : : : , xn are such that x1+x2+ � � �+xn = 0and x21+x22+ � � �+x2n = 1. Prove that there are two numbers among themwhose product is no greater than �1=n.

(1991 Tournament of Towns)

107

Solution by Naoki Sato, student, University of Toronto, Toronto, On-

tario.Choose i such that x1 � x2 � � � � � xi � 0 � xi+1 � � � � � xn. Then

x21 + x22 + : : :+ x2i � x1 � x1 + x1 � x2 + � � �+ x1 � xi= x1(x1 + x2 + � � �+ xi)

= �x1(xi+1 + � � �+ xn)

� �(n� i)x1 � xn:

Similarly,

x2i+1 + � � �+ x2n � xn(xi+1 + � � �+ xn)

= �xn(x1 + x2 + � � �+ xi) � �ix1 � xn:Thus 1 = x21 + � � �+ x2n � �nx1 � xn.(As always, please send us new solutions to old problems.)

C68. Proposed by Vin de Silva, student, Oxford University, Oxford,

England.LetM be an n�n orthogonal matrix. (In other words, the rows are n

vectors in n-space that are of length 1 and mutually perpendicular.) Let Abe the k�k matrix in the upper-left corner ofM , and let B be the (n�k)�(n� k) matrix in the lower-right corner ofM . Prove that detA = detB.

Solution by Eric Wepsic, D.E.Shaw and Co., New York, NY, USA.Let ci be the ith column of the orthogonal matrix Q, and let ri be the

ith row. Let xi be the ith basis vector. Let M1 be the top k � k submatrixof Q, and let M2 be the bottom (n� k)� (n� k) submatrix. Then

detM1 = c1 ^ c2 ^ � � � ^ ck ^ xk+1 ^ � � � ^ xn= Qx1 ^ Qx2 ^ � � � ^ Qxk ^ xk+1 ^ � � � ^ xn:

As Qt has determinant 1, this is equal to:

QtQx1 ^QtQx2 ^ � � � ^ QtQxk ^ Qtxk+1 ^ � � � ^Qtxn

= x1 ^ x2 ^ � � � ^ xk ^ rk+1 ^ � � � ^ rn= detM2:

Solution by Sam Vandervelde, University of Chicago, Illinois, USA.Let Q1 be the matrix whose �rst k columns are the �rst k columns ofQ,

and whose remainingn�k columns are the last n�k columns of the identitymatrix. Let Q2 be the same sort of thing with the �rst k columns identicalto I, and the last n�k columns those of Qt. Then an easy calculation (usingQtQ = I, QtI = Qt) shows that QtQ1 = Q2. Taking determinants of bothsides yields the desired result.

The two solutions presented were in some sense identical, although

the perspectives were di�erent.

108

C69. Let Q be a polyhedron in R3. Let ~n1, : : : , ~nk and A1, : : : , Ak

be the normal vectors and areas of the faces of Q respectively. Prove that

kXi=1

Ai~ni = 0:

SolutionbyMiguel Carri �on �Alvarez, student,Universidad Complutense

de Madrid, Spain.We have

Pk

i=1Ai~ni =Hd~s, where d~s is the element of area (pointing

outwards) and the integral is over the faces of the polyhedron. ThenId~s =

�Idsx;

Idsy;

Idsz

�=

�I~ux � d~s;

I~uy � d~s;

I~uz � d~s

�;

for constant �elds ~ux, ~uy, ~uz. We recall the divergence theorem:H~A � d~s =R

V(~r � ~A)dV . But ~ux, ~uy, ~uz are constant �elds, so ~r � ~ui = 0. ThusHd~s = (0; 0; 0).

Comments.

1. If � is the pressure �eld in a uid, the total force on a (three-dimensional)object is

F = �I� d~s = �

Z(~r�)dV:

(Prove this!) In this case, we have constant pressure. What this result

says is that a polyhedron sitting in a uid with constant pressure from

all sides (and no other forces) will not move.

2. This is true in all dimensions.

3. Many other interesting results follow from this one, including the fol-

lowing variant of the Pythagorean theorem. Consider a tetrahedron

ABCD with AB, AC, AD mutually perpendicular. Then

(ABC)2 + (ACD)2 + (ADB)2 = (BCD)2

where the brackets denote the area of the triangle. This is also true

in other dimensions. If you can think of any other interesting conse-

quences (including other variants of Pythagoras), please send them in!

109

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 812"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed to

the Editor-in-Chief, to arrive no later than 1 October 1997. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication.

2214. Proposed by Walther Janous, Ursulinengymnasium, Inns-

bruck, Austria.Let n � 2 be a natural number. Show that there exists a constant C = C(n)such that for all real x1; : : : ; xn � 0 we have

nXk=1

pxk �

vuut nYk=1

(xk + C):

Determine the minimum C(n) for some values of n.[For example, C(2) = 1.]

2215?. Proposed by Theodore Chronis, student, Aristotle University

of Thessaloniki, Greece.Let P be a point inside a triangle ABC. It is known how to determine Psuch that PA + PB + PC is a minimum (known as Fermat's Problem forTorricelli).

Determine P such that PA+ PB + PC is a maximum.

110

2216. Proposed by Walther Janous, Ursulinengymnasium, Inns-

bruck, Austria.Suppose that � � 1 is a natural number.

1. Determine the set of all �'s such that the diophantine equationx� + y2 = z2 has in�nitely many solutions.

2.?For any such �, determine all solutions of this equation.

2217. Proposed by Bill Sands, University of Calgary, Calgary, Al-

berta.

(a) Prove that for every su�ciently large positive integer n, there are arith-metic progressions a1; a2; a3 and b1; b2; b3 of positive integers such thatn = a1b1 + a2b2 + a3b3.

(b) What happens if we require a1 = b1 = 1?

(This is a variation of problem 3 of the 1995/96 Alberta High School Mathe-matics Competition, Part II, which will appear in a future Skoliad Corner.)

2218. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Suppose that a, b, c are positive real numbers and that

abc = (a+ b� c)(b+ c� a)(c+ a� b):

Clearly a = b = c is a solution. Determine all others.

2219. Proposed by Christopher J. Bradley, Clifton College, Bristol,

UK.

Show that there are an in�nite number of solutions of the simultaneousequations:

x2 � 1 = (u+ 1)(v� 1)

y2� 1 = (u� 1)(v+ 1)

with x; y; u; v positive integers and x 6= y.

2220. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,

Madrid, Spain.

Let V be the set of an icosahedron's twelve vertices, which can be par-titioned into four classes of three vertices, each one in such a way that thethree selected vertices of each class belong to the same face.

How many ways can this be done?

111

2221. Proposed by �Sefket Arslanagi�c, University of Sarajevo, Sara-

jevo, Bosnia and Herzegovina.

Find all members of the sequence an = 32n�1 + 2n�1; (n 2 N) whichare the squares of any positive integer.

2222. Proposed by Shawn Godin, St. Joseph Scollard Hall, North

Bay, Ontario.

Find the value of the continued root:vuut4 + 27

s4 + 29

r4 + 31

q4 + 33

p: : ::

NOTE: This was inspired by the problems in chapter 26 \Ramanujan, In�nity

and the Majesty of the Quattuordecillion", pp 193{195, in \Keys to In�nity"by Cli�ord A. Pickover, John Wiley and Sons, 1995.

2223. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,

Madrid, Spain.

We are given a bag with n identical bolts and n identical nuts, whichare to be used to secure the n holes of a gadget.

The 2n pieces are drawn from the bag at random one by one. Throughoutthe draw, bolts and nuts are screwed together in the holes, but if the numberof bolts exceeds the number of available nuts, the bolt is put into a hole untilone obtains a nut, whereas if the number of nuts exceed the number of bolts,the nuts are piled up, one on top of the other, until one obtains a bolt.

Let L denote the discrete random variable which measures the height of thepile of nuts.

Find E[L] +E[L]2.

2224. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

Point P lies inside triangle ABC. Triangle BCD is erected outwardlyon sideBC such that \BCD = \ACP and\CBD = \ABC. Prove that ifthe area of quadrilateral PBDC is equal to the area of triangle ABC, thentriangles ACP and BCD are similar.

2225. Proposed by Kenneth Kam Chiu Ko, Mississauga, Ontario.

(a) For any positive integer n, prove that there exists a unique n-digitnumber

N such that:(i) N is formed with only digits 1 and 2; and(ii) N is divisible by 2n.

(b) Can digits \1" and \2" in (a) be replaced by any other digits?

112

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

We have identi�ed \anonymous" in the solutions of problems 2064,2065, 2067, 2068, 2069, 2071 and 2077 as MARIA MERCEDES S �ANCHEZBENITO, I.B. Luis Bu ~nuel, Madrid, Spain.

The name of VICTOR OXMAN, University of Haifa, Haifa, Israel, shouldbe added to the list of solvers of problem 2091. The name of ROBERTP. SEALY, Mount Allison University, Sackville, New Brunswick, should beadded to the list of solvers to problem 2104.

2113. [1996: 35] Proposed byMarcin E. Kuczma, Warszawa, Poland.

Prove the inequality

nXi=1

ai

! nXi=1

bi

!�

nXi=1

(ai + bi)

! nXi=1

aibi

ai + bi

!

for any positive numbers a1; : : : ; an; b1; : : : ; bn.

I. Solution by Vedula N. Murty, Andhra University, Visakhapatnam, India.

The given inequality follows from the easily veri�ed identity:

nXi=1

ai

! nXi=1

bi

!�

nXi=1

(ai + bi)

! nXi=1

aibi

ai + bi

!

=X

1�i<j�n

(aibj � ajbi)2

(ai + bi)(aj + bj):

II. Solution by Federico Ardila, student,Massachusetts Institute of Technol-

ogy, Cambridge, Massachusetts, USA.

We know from Cauchy's inequality that

nXi=1

(ai � bi)

!2�

nXi=1

(ai + bi)

! nXi=1

(ai � bi)2

ai + bi

!:

113

Therefore nXi=1

ai

! nXi=1

bi

!=

1

4

0@ nX

i=1

ai +

nXi=1

bi

!2

�

nXi=1

ai �nXi=1

bi

!21A

=1

4

0@ nX

i=1

(ai + bi)

!2�

nXi=1

(ai � bi)

!21A

� 1

4

0@ nX

i=1

(ai + bi)

!2

�

nXi=1

(ai + bi)

! nXi=1

(ai � bi)2

ai + bi

!!

=

nXi=1

(ai + bi)

! nXi=1

(ai + bi)2 � (ai � bi)

2

4(ai + bi)

!

=

nXi=1

(ai + bi)

! nXi=1

aibi

ai + bi

!;

which completes the proof.

III. Solution by Kee-Wai Lau, Hong Kong.

For n = 1, the result is clear. Suppose that the inequality holds forn = k. Then for n = k + 1, we have

k+1Xi=1

ai

! k+1Xi=1

bi

!� k+1Xi=1

(ai + bi)

! k+1Xi=1

aibi

ai + bi

!

=

kXi=1

ai

! kXi=1

bi

!�

kXi=1

(ai + bi)

! kXi=1

aibi

ai + bi

!

+ak+1

kXi=1

bi

!+ bk+1

kXi=1

ai

!

�(ak+1 + bk+1)kXi=1

aibi

ai + bi� ak+1bk+1

ak+1 + bk+1

kXi=1

(ai + bi)

�kXi=1

�ak+1bi + bk+1ai � (ak+1 + bk+1)

aibi

ai + bi

� ak+1bk+1

ak+1 + bk+1(ai + bi)

�by the induction hypothesis

=1

ak+1 + bk+1

kXi=1

(ak+1bi � bk+1ai)2

ai + bi� 0;

completing the induction.

114

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-

jevo, Bosnia and Herzegovina; HAN PING DAVIN CHOR, Student, Cam-

bridge, MA, USA; THEODORE CHRONIS, student, Aristotle University of

Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;

RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN,

University of Alberta, Edmonton, Alberta; MITKO KUNCHEV, Baba Tonka

School of Mathematics, Rousse, Bulgaria; DAVID E. MANES, State Univer-

sity of New York, Oneonta, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin,

Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; CHRIS

WILDHAGEN, Rotterdam, the Netherlands; and the proposer. There were

also one anonymous solution and one incorrect submission.

Klamkin remarked that this is a known inequality due to E.A. Milne

[1], that came up in establishing an integral inequality. It also appeared as

problem 67 in [2]. This was also noticed by Janous and Tsaoussoglou. Ar-

slanagi�c, Herzig, Hess, Manes and Tsaoussoglou all pointed out that equality

holds in the proposed inequality if and only if the vectors (a1; a2; : : : ; an)and (b1; b2; : : : ; bn) are linearly dependent. This is obvious from proof I

above.

References

[1.] E.A. Milne, Note on Rosseland's Integral for the Stellar Absorption Co-e�cient, Monthly Notices Royal Astronomical Soc. 85 (1925) 979{984.

[2.] G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge Uni-versity Press, London, 1934, pp. 61{62.

2114. [1996: 75] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a square with incircle �. A tangent ` to � meets the sidesAB and AD and the diagonal AC at P , Q and R respectively. Prove that

AP

PB+AR

RC+AQ

QD= 1:

Solution by Francisco Bellot Rosado, I.B. Emilio Ferrari, and Maria

Ascensi �on L �opez Chamorro, I.B. Leopoldo Cano, Valladolid, Spain.

We suppose that the equation of � is x2+ y2 = 1, and the coordinatesof A and C are (1; 1) and (�1;�1) respectively. Let T (cos t; sin t) be thecoordinates of the point of tangency of line ` with �, with 0 < t < �

2:

Then the equation of PQ is

y � sin t+ x � cos t = 1;

and the coordinates of the involved points are

R

�1

sin t+ cos t;

1

sin t+ cos t

�; P

�1� cos t

sin t

�; Q

�1� sin t

cos t; 1

�;

115

and soAP

PB=

sin t+ cos t� 1

sin t� cos t+ 1;

AR

RC=

sin t+ cos t� 1

sin t+ cos t+ 1;

AQ

QD=

cos t+ sin t� 1

cos t� sin t+ 1;

from which an easy calculation shows that

AP

PB+AQ

QD+AR

RC= 1:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; CLAUDIO

ARCONCHER, Jundia��, Brazil; FEDERICO ARDILA, student, Massachusetts

Institute of Technology, Cambridge, Massachusetts, USA; �SEFKET

ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina;

SAM BAETHGE, Science Academy, Austin, Texas, USA; CARL BOSLEY, stu-

dent, Washburn Rural High School, Topeka, Kansas, USA; CHRISTOPHER

J. BRADLEY, CliftonCollege, Bristol,UK;MIGUELANGEL CABEZ �ONOCHOA,

Logro ~no, Spain; TIM CROSS, King Edward's School, Birmingham, England;

JORDI DOU,Barcelona, Spain; HANS ENGELHAUPT, Franz{Ludwig{Gymnas-

ium, Bamberg, Germany; DAVID HANKIN, Hunter College Campus Schools,

New York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;

RICHARD I.HESS, Rancho Palos Verdes, California, USA; PETER HURTHIG,

Columbia College, Burnaby, BC; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,

Michigan, USA; MITKO KUNCHEV, Baba Tonka School of Mathematics,

Rousse, Bulgaria; P. PENNING, Delft, the Netherlands; WALDEMAR

POMPE, student, University ofWarsaw, Poland; D.J. SMEENK, Zaltbommel,

the Netherlands; MELETIS VASILIOU, Elefsis, Greece; and the proposer.

2115. [1996: 75] Proposed by Toby Gee, student, the John of Gaunt

School, Trowbridge, England.

Find all polynomials f such that f(p) is a prime for every prime p.

Solution by Luis V. Dieulefait, IMPA, Rio de Janeiro, Brazil.

Let f be a polynomial such that f(p) is prime for every prime valueof p. We will prove that f is a constant polynomial (the constant being, ofcourse, a prime number) or f is the identity; that is f(x) = x, for everyinteger x. Suppose that f is not the identity. Then f(x) = x possesses onlya �nite number of solutions.

Therefore, there exist p, q primes, p 6= q such that f(p) = q. Apply-ing Dirichlet's Theorem, we know that the arithmetical progression p+ kq,k = 0; 1; 2; : : : , contains in�nitely many primes. Besides, for every k � 0,

116

f(p+ kq) � f(p) � 0 (mod q), and so for the in�nitely many primes ri ofthe form p + kq we must have f(ri) = q. But f being a polynomial thattakes the value q in�nitely many times means that f(x) = q for all x.

Also solved by FEDERICO ARDILA, student, Massachusetts Institute

of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student,

St. Paul's School, London, England; CARL BOSLEY, student,Washburn Rural

High School, Topeka, Kansas, USA; RICHARD I. HESS, Rancho Palos Verdes,

California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-

tria; MICHAEL JOSEPHY, Universidad de Costa Rica, San Jos �e, Costa Rica;

JOEL SCHLOSBERG, student, Hunter College High School, New York NY,

USA; and the proposer. There were 3 incorrect submissions.

2116. [1996: 75] Proposed by Yang Kechang, Yueyang University,

Hunan, China.

A triangle has sides a; b; c and area F . Prove that

a3b4c5 � 25p5(2F )6

27:

When does equality hold?

Several solvers pointed out that this proposal is just a special case of[1984: 19], proposed by M.S. Klamkin, which asks for the maximum value

of P � sin�A: sin� B: sin C, where A; B; C are the angles of a triangleand �; �; are given positive numbers. The featured solution by WaltherJanous [1985: 908] establishes the maximum to be

Pmax =

��(�+ � + )

(� + �)(� + )

��=2:

��(�+ � + )

(� + )(� + �)

��=2:

� (�+ � + )

( + �)( + �)

� =2:

In the humble opinion of the editor, nothing in the submissions addedanything substantive to [1985: 908].

Solved by PETER HURTHIG, Columbia College, Burnaby, BC;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.

KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y,

Ferris State University, Big Rapids, Michigan, USA; KEE-WAI LAU, Hong

Kong; DIGBY SMITH, Mount Royal College, Calgary, Alberta; and PANOS

E. TSAOUSSOGLOU, Athens, Greece.

2117. [1996: 76] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle withAB > AC, and the bisector of\AmeetsBC atD. Let P be an interior point of the sideAC. Prove that \BPD < \DPC.

Two solutions by Shailesh Shirali, Rishi Valley School, India.

[Editor's comment: Shirali strengthens the result by replacing the con-dition AB > AC by the less stringent condition \ABC < 90�. (That seems

117

to be su�cient for the inequality in his �rst proof: 90� > \ABC > \PBC

implies sin\ABC > sin\PBC.) Neither he nor the other solvers explicitlystated that they had stronger results.]

Solution I. Let E be the point where the internal bisector of \P of4PBC meets side BC. Then

BE

EC=PB

PC=

sin\PCB

sin\PBC>

sin\ACB

sin\ABC=AB

AC=BD

DC;

so that E lies between D and C. It follows that \BPD < \DPC.

[Editor's further comment: The above argument clearly continues towork for some positions of P on the side AC even when \ABC � 90�. Aset of conditions on P and4ABC under which\BPD < \DPC is implicitin Shirali's second argument.]

Solution II. The set of points X in the plane of the triangle for which\BXD = \DXC is, the circle of Apollonius throughA with respect toB

and C. [Equivalently, is the locus of points X for whichXB

XC=BD

DC; it is

the circle through A and D whose centre lies on BC; inversion in this circle�xesA andD, and interchangesB withC; see, for example, H.S.M. Coxeter,Introduction to Geometry, x6.6 pages 88-89.] When AB > AC [so thatBD > DC], the set of points X for which \BXD < \DXC is the set ofpoints interior to . In this case the centre of lies on the extension of BCbeyond C, so the segment AC lies entirely within . It follows that for allpoints P on AC, we have \BPD < \DPC. [When AB < AC, the centreof lies on the part of BC beyond B, so that now we seek points P outside; when \B � 90�, the entire line segment AC lies outside , which agreeswith the �rst proof. When \B is obtuse, then some points of AC lie inside { for those points, the desired inequality no longer holds.]

Also solved by FEDERICO ARDILA, student,Massachusetts Institute of

Technology, Cambridge, Massachusetts, USA; �SEFKET ARSLANAGI �C, Uni-

versity of Sarajevo, Sarajevo, Bosnia and Herzegovina; CHRISTOPHER J.

BRADLEY, Clifton College, Bristol, UK; JORDI DOU, Barcelona, Spain;

V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;

KEE-WAI LAU, Hong Kong; WALDEMAR POMPE, student, University of

Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E.

TSAOUSSOGLOU, Athens, Greece; GEORGE TSAPAKIDIS, Agrino, Greece;

MELETIS VASILIOU, Elefsis, Greece; and the proposer.

2118. [1996: 76] Proposed by Paul Yiu, Florida Atlantic University,

Boca Raton, Florida, USA.

The primitive Pythagorean triangle with sides 2547 and 40004 and hy-potenuse 40085 has area 50945094, which is an 8{digit number of the formabcdabcd. Find another primitive Pythagorean triangle whose area is of thisform.

118

Solution by the proposer.

There are three such primitive Pythagorean triangles:

m n m2 � n2 2mn m2 + n2 Area

(i) 146 137 2547 40004 40085 50945094

(ii) 146 9 21235 2628 21397 27902790

(iii) 105 32 10001 6720 12049 33603360

8{digit numbers of the form abcdabcd can be factored as 10001 �N =73 � 137 � N for an integer N in the range 1000 � N < 10000. The sidesof a primitive Pythagorean triangle are given by m2 � n2, 2mn, m2 + n2,with relatively prime integers m and n of di�erent parity. The area of thetriangle is 4 = mn(m� n)(m+ n). Note that the factors m, n, m + n,and m � n are pairwise relatively prime, and exactly one of them is even.These numbers are in the order

(i) m+ n > m > n > m� n, or

(ii) m+ n > m > m� n > n.

[Clearly, none ofm;n orm�n can be 10001r. Alsom+n 6= 10001;otherwise, mn � (10000)(1) and 4 > 108.] Each of 137 and 73 dividesexactly one of the four numbers m; n;m � n. We list the twelve di�erenttypes in the table below, in which h and k are positive integers.

Type m + n m n m � n conditions

(1) 2(137h) + 73k 137h+ 73k 137h 73k k odd

(2) 137h+ 2(73)k 137h+ 73k 73k 137h h odd

(3) 137h+ 73k 73k 137h 73k � 137h h + k odd

(4) �137h + 2(73)k 73k �137h+ 73k 137h h odd

(5) 73k �137h+ 73k 137h 73k � 2(137)h k odd

(6) 73k 1

2(137h+ 73k) 1

2(�137h+ 73k) 137h h � k odd

(7) 137h+ 73k 137h 73k 173h� 73k h + k odd

(8) 2(137h)� 73k 137h 137h� 73k 73k k odd

(9) 137h 137h� 73k 73k 137h� 2(73k) h odd

(10) 137h 1

2(137h+ 73k) 1

2(137h� 73k) 73k h � k odd

(11) 137h 73k 137h� 73k �137h+ 2(73k) h odd

(12) 73k 137h �137h+ 73k 2(137)h� 73k k odd

In Types (1) and (2), 4 > ((137h)(73k))2 = (hk)2(10001)2 > 108.

In each of Types (3), (4), (5), (6), note that 137h > 100. This cannotbe the smallest of the four numbersm;n;m�n, and must therefore be thethird in order. The area of the triangle is greater than (137h)3 � 1. We need

therefore consider only h satisfying (137h)3 < 108; h < 108

3

137� 3:388; in

other words, h = 1, 2, 3. In each of these types, the smallest number is alinear function�p 137h+ q 73k, where each of p and q is 1, 2 or 1

2.

119

Note that �p 137h + q 73k < 1h73 since the product of the two smallest

measurements cannot exceed 10001.

It follows that 0 < �p 137h+ q 73k < 1h73: From this,

p

q� 13773

� h < k <p

q� 13773

� h+1

qh:

In each of Types (7), (8), (9), (10), either n or m � n is 73k, so4 > (73k)3. In other words, k � 6. Also, from 0 < p137h�q73k < 1

k137,

we haveq

p� 73

137� k < h <

q

p� 73

137� k +

1

pk:

Type p q possible (h; k) relevant (h; k) triangle

(3) 1 1 (1; 2), (2; 4) (1; 2) (i)

(4) 1 1 (1; 2), (2; 4) (1; 2) (ii)

(5) 2 1 (1; 4), (2; 8) none

(6) 1

2

1

2(1; 2), (1; 3), (2; 4), (3; 6) (1; 3) m = 178, n = 41

(7) 1 1 (1; 1) none

(8) 1 1 (1; 1) (1; 1) m = 137, n = 64

(9) 1 2 (2; 1) none

(10) 1

2

1

2(1; 1), (2; 1), (2; 2), (1; 1) (iii)

(2; 3), (3; 5) (3; 5) m = 388, n = 23

Type (11). We consider (h; k) in either of the parallelograms

137h� 73k = 0; 137h� 73k = 100;

� 137h+ 2(73k) = 0; �137h+ 2(73k) = 464;

137h� 73k = 0; 137h� 73k = 464;

� 137h+ 2(73k) = 0; �137h+ 2(73k) = 100:

(Note: 464 = [3p108]). These are the points

(h; k) = (1;1); (2; 3); (t; t) for t = 2; : : : ; 8;

and we need only consider (h; k) = (1; 1) (since h is odd and gcd (h; k) = 1).But this triangle, with m = 73, n = 64, is too small; it has area 5760576.

Type (12). We consider (h; k) in either of the parallelograms

�137h+ 73k = 0; �137h+ 73k = 100;

2(137)h� 73k = 0; 2(137)h� 73k = 464;

120

�137h+ 73k = 0; �137h+ 73k = 464;

2(137)h� 73k = 0; 2(137)h� 73k = 100:

These are the points

(h; k) = (1; 2); (1;3); (2; 7); (3; 11); (2;4); (3;6);

and we need only consider (h; k) = (1;3); (2; 7) and (3,11). But the corre-sponding triangles are all too big.

Remark Enlarging the primitive triangle with m = 73; n = 64 bythe factors 2, 3, 4, we obtain non-primitive Pythagorean triangles of areas23042304, 51845184, and 92169216 respectively. Enlargements of the prim-itive triangles (i), (ii), (iii) all have areas exceeding 108: The non-primitivetriangle from m = 137, n = 73 has area 13441344.

Both the additional triangles were also found by TIM CROSS, King

Edward's School, Birmingham, England; HANS ENGELHAUPT, Franz{Lud-

wig{Gymnasium, Bamberg, Germany; and RICHARD I. HESS, Rancho Palos

Verdes, California, USA.

The following readers each found one of the triangles: CHARLES

ASHBACHER, Cedar Rapids, Iowa, USA; SAM BAETHGE, Science Academy,

Austin, Texas, USA; MANSUR BOASE, student, St. Paul's School, London,

England; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; CHARLES

R. DIMINNIE, Angelo State University, San Angelo, TX, USA; SHAWN

GODIN, St. Joseph Scollard Hall, North Bay, Ontario; DAVID HANKIN,

Hunter College Campus Schools, New York, NY, USA; FLORIAN HERZIG,

student, Perchtoldsdorf, Austria; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,

Michigan, USA; DAVID E. MANES, State University of New York, Oneonta,

NY, USA; P. PENNING, Delft, the Netherlands; CORY PYE, student,

Memorial University of Newfoundland, St. John's, Newfoundland; JOEL

SCHLOSBERG, student, Hunter College High School, New York NY, USA;

HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal

College, Calgary, Alberta; DAVID R. STONE, Georgia Southern University,

Statesboro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece.

There was one incorrect submission.

2119. [1996: 76] Proposed by Hoe Teck Wee, student, Hwa Chong

Junior College, Singapore.

(a) Show that for any positive integer m � 3, there is a permutation of m1's,m 2's andm 3's such that

(i) no block of consecutive terms of the permutation (other than theentire permutation) contains equal numbers of 1's, 2's and 3's; and

121

(ii) there is no block ofm consecutive terms of the permutation whichare all equal.

(b) For m = 3, how many such permutations are there?

Solution by P. Penning, Delft, the Netherlands.

(b) I found two:

aabbcbcca and abbcbccaa;

where a; b; c is any permutation of 1; 2; 3. (In fact these are mirror-imagesof one another.) Thus there are 3! � 2 = 12 such permutations of 1's, 2'sand 3's.

(a) To generate a solution for m > 3, start with three blocks of m equalterms, and to make it satisfy condition (ii) let the last term of each blockchange places with the �rst term of the next block (and do the same with the�rst and last block):

1

m�2z }| {22 : : : 2 3 2

m�2z }| {33 : : : 31 3

m�2z }| {11 : : : 1 2:

Note that for m = 3 this permutation violates condition (i). [It is easy tocheck that it works wheneverm > 3.|Ed.] Also note that if the permutationis considered as a cycle then there are solutions only ifm > 3.

Also solved by FEDERICO ARDILA, student, Massachusetts Institute

of Technology, Cambridge, Massachusetts, USA; MANSUR BOASE, student,

St. Paul's School, London, England; HANS ENGELHAUPT, Franz{Ludwig{

Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtolds-

dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

DAVID E. MANES, State University of New York, Oneonta, NY, USA; JOEL

SCHLOSBERG, student, Hunter College High School, New York, NY, USA;

DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;

and the proposer. Part (b) onlywas solved by CHARLES ASHBACHER, CedarRapids, Iowa, USA.

Note that Penning's solution shows that for m > 3, condition (ii) canbe strengthened just a teensy bit to:

(ii)0 there is no block ofm� 1 consecutive terms of the permutation which

are all equal.

However, there is an even better result. For anym � 3 there is a permuta-

tion ofm 1's,m 2's and m 3's which satis�es condition (i) and also

(ii)? there is no block of three consecutive terms of the permutation which

are all equal.

(Thanks to expert colleague James Currie for suggesting examples of such

permutations, which readers might enjoy �nding for themselves!)

122

2120. [1996: 76] Proposed byMarcin E. Kuczma, Warszawa, Poland.

Let A1A3A5 and A2A4A6 be non-degenerate triangles in the plane.For

i = 1; : : : ; 6 let `i be the perpendicular from Ai to line Ai�1Ai+1

(where of course A0 = A6 and A7 = A1). If `1; `3; `5 concur, prove that`2; `4; `6 also concur.

Solution by Christopher J. Bradley, Clifton College, Bristol, UK.

Let the vector position of the vertex Ai relative to the point of concur-rency of `1; `3; `5 be ~ai. Then ~a1 �( ~a2� ~a6) = ~a3 �( ~a4� ~a2) = ~a5 �( ~a6� ~a4) =0. In particular, the sum is also zero, so

~a2 � ( ~a1 � ~a3) + ~a4 � ( ~a3 � ~a5) + ~a6 � ( ~a5 � ~a1) = 0: (1)

Now suppose that the perpendicular from A2 to A1A3 meets the perpendic-ular from A4 to A3A5 at B, with position vector ~b. (These perpendicularscannot be parallel.) Then

(~b� ~a2) � ( ~a1 � ~a3) = 0 and (~b� ~a4) � ( ~a3 � ~a5) = 0:

Adding gives ~b � ( ~a1 � ~a5) = ~a2 � ( ~a1 � ~a3) + ~a4( ~a3 � ~a5), which, by (1), isequal to � ~a6 � ( ~a5 � ~a1).

Hence (~b� ~a6) � ( ~a1 � ~a5) = 0, which means that BA6 ? A1A5.

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; FEDERICO

ARDILA, student, Massachusetts Institute of Technology, Cambridge, Mas-

sachusetts, USA; JOHN G. HEUVER, Grande Prairie Composite High School,

Grande Prairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Inns-

bruck, Austria; P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA,

Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; MELETIS

VASILIOU, Elefsis, Greece; and the proposer.

Problem 2120 can be found in many reference texts. Dan Pedoe pro-

vides two proofs in his Geometry, A comprehensive Course (Dover, 1998).Clerk Maxwell's remarkable treatment of the problem, reducing it to an

observation about the radical axes of appropriate circle, appears in x28.4,

page 115. Earlier, in x6.1, pages 35-37, Pedoe uses barycentric coordinates

for a proof of a closely related problem involving reciprocal triangles. He

comments that whole books have been written on the subject of reciprocal

�gures.

Seimiya uses Steiner's theorem in his solutions (Jacob Steiner, Gesam-

melte Werke I, p. 189): A necessary and su�cient condition in order for`1; `3; `5 to be concurrent is

A1A22 + A3A

24 + A5A

26 = A2A

23 + A4A

25 +A6A

21:

Since (by permuting the subscripts) this is also a necessary and suf-

�cient condition for `2; `4; `6 to concur, our problem follows immediately.

Arconcher refers to this condition as Carnot's Theorem, but he provides no

123

reference. Ardila discovered the theorem for himself and showed that it fol-

lows fairly quickly from Pythagoras' Theorem, somaybe we should refer to it

as the Pythagoras-Ardila-Steiner-Carnot Theorem. Seimiya added that the

problem may also be found in Aref and Wernick, Problems and Solutions inEuclidean Geometry (Dover), p. 55 problems 2.32, He includes two of his

interesting generalizations that have appeared in Japanese, one in 1928 and

one in 1967.

Finally, Heuver found the problem as Ex. 7 of x54 of George Salmon's

A Treatise on Conic Sections, 6th ed. His solution exploits Salmon's imagi-

native use of pencils of lines in Cartesian coordinates.

2121. [1996: 76] Proposed by Krzysztof Chelmi �nski, Technische

Hochschule Darmstadt, Germany; and Waldemar Pompe, student, Univer-

sity of Warsaw, Poland.

Let k � 2 be an integer. The sequence (xn) is de�ned by x0 = x1 = 1and

xn+1 =xkn + 1

xn�1for n � 1:

(a) Prove that for each positive integer k � 2 the sequence (xn) is a se-quence of integers.

(b) If k = 2, show that xn+1 = 3xn � xn�1 for n � 1.

(c)? Note that for k = 2, part (a) follows immediately from (b). Is there ananalogous recurrence relation to the one in (b), not necessarily linear,which would give an immediate proof of (a) for k � 3?

I Solution ((a) and (b) only) by Walther Janous, Ursulinengymnasium,

Innsbruck, Austria.

(a) We immediately get x2 = 2 and x3 = 2k + 1. Now we use mathematicalinduction for the proof. Assume that x0; x1; : : : ; xn are all natural numbers.We must show that xn+1 2 N. First we note that since xn�2 �xn = xkn�1+1

it follows that xn�2 and xn�1 are relatively prime. Using xn = (xkn�1 +1)=xn�2 we infer that

xn+1 =xkn + 1

xn�1=

(xkn�1 + 1)k + xkn�2xkn�2xn�1

:

Thus obviously xkn�2 dividesN = (xkn�1 +1)k + xkn�2 since xn is a naturalnumber. Furthermore, modulo xn�1 we have:

N � 1 + xkn�2 = xn�3 � xn�1 � 0:

That is, xn�1 also dividesN and we are done.

124

(b) Now,

xn+1 =x2n + 1

xn�1() xn�1 � xn+1 � x2n = 1:

That is, the sequence fyng = fxn�1 � xn+1 � x2ng is constant. Settingyn+1 = yn we have

xn � xn+2 � x2n+1 = xn�1 � xn+1 � x2n

() xn(xn + xn+2) = xn+1(xn�1 + xn+1)

() xn + xn+2

xn+1=

xn�1 + xn+1

xn:

That is, the sequence fzng = f(xn�1+xn+1)=xng is constant. From z1 = 3we get (xn�1 + xn+1)=xn = 3; that is, xn+1 = 3xn � xn�1 for all n � 1,as claimed.

II Solution by Christopher J. Bradley, Clifton College, Bristol, UK.

We �rst establish the following lemma:

Lemma. Suppose a and b have highest common factor 1 and (ak+bk+1)=abis an integer (where k is an integer, k � 2), then(i) c = (bk + 1)=a is an integer;(ii) (bk + ck + 1)=bc is an integer;(iii) b and c have highest common factor 1.Proof. Let ak + bk + 1 = �(a; b)ab where �(a; b) is an integer. Then sincea divides �(a; b)ab� ak it also divides bk + 1; that is, (bk + 1)=a = c is aninteger, which proves (i). Also

bk + ck + 1

bc=ac+ ck

bc=a+ ck�1

b=ak + (bk + 1)k�1

ak�1b:

Now bk + 1 = ac has a as a factor, so ak + (bk+ 1)k�1 is divisible by ak�1

and hence the numerator is divisible by ak�1. Also ak + 1 is divisible by b(from part (i)), so multiplying out the numerator by the Binomial Theoremwe see that the numerator is divisible by b. But a and b have highest commonfactor 1, so the numerator is divisible by ak�1b. Hence

�(b; c) =bk + ck + 1

bc

is an integer, which proves (ii). Since bk+ ck+1 = �(b; c)bc, if b and c havea common factor h, then h divides bk + ck � �(b; c)bc; that is, h divides 1.Hence b and c have highest common factor 1, which proves (iii), and thelemma is proved.

We claim that the recurrence relation sought is

xn+1 = �(xn�1; xn)xn � xk�1n�1 (�)

125

with �(xn�1; xn) = (xkn+xkn�1+1)=(xnxn�1), which is but one step away

from xn+1 = (xkn+1)=xn�1. That �(xn�1; xn) is always an integer followsby induction, using the lemma, with the start of induction satis�ed sincex0 = x1 = 1 have highest common factor 1 and �(x0; x1) = 3 is an integer.Relation (�) then establishes that fxng is a sequence of integers. This givesus parts (c) and (a). The term (x2n+x

2n�1+1)=(xnxn�1) is known from work

on alternate terms of the Fibonacci sequence to be equal to 3 for all n � 2and is also 3 for n = 1, which proves (b).

Parts (a) and (b) together were also solved by FEDERICO ARDILA,

student,Massachusetts Institute of Technology, Cambridge, Massachusetts,

USA; FLORIANHERZIG, student, Perchtoldsdorf,Austria; and the proposers.

Part (b) alone was solved by �SEFKET ARSLANAGI �C, University of Sarajevo,

Sarajevo, Bosnia and Herzegovina; and TIM CROSS, King Edward's School,

Birmingham, England. There was one incorrect solution.

JANOUS comments that this was posed as a problem of the �nal round

of the third Austrian Mathematical Olympiad held in 1972, and refers the in-

terested reader to the book \ �Osterreichische Matematik-Olympiaden" 1970-1989, G. Baron & E. Windischbacher, Innsbruck 1990, problem 42.

2122.[1996: 77] Proposed by Shawn Godin, St. Joseph Scollard Hall,North Bay, Ontario.

Little Sam is a unique child and his math marks show it. On four teststhis year his scores out of 100 were all two-digit numbers made up of eightdi�erent non-zero digits. What's more, the average of these scores is thesame as the average if each score is reversed (so 94 becomes 49, for example),and this average is an integer none of whose digits is equal to any of the digitsin the scores. What is Sam's average?

I. Solution by Mansur Boase, student, St. Paul's School, London, England.

Let the four marks be 10a+ b, 10c+ d, 10e+ f and 10g+ h. Then

10a + b + 10c + d+ 10e + f + 10g + h = 10b + a+ 10d + c+ 10f + e+ 10h+ g;

so

9(a+ c+ e+ g) = 9(b+ d+ f + h)

and

a+ c+ e+ g = b+ d+ f + h: (1)

The average of the four marks is therefore

10a + b + 10b + a+ 10c + d+ 10d + c + 10e + f + 10f + e+ 10g + h+ 10h + g

4 � 2

=11(a+ b+ c+ d+ e+ f + g+ h)

8: (2)

126

The average must consist only of k's and 0's, the digits not equal to a, b, c,d, e, f , g or h. Thus (2) is

11(1 + 2 + � � �+ 9� k)

8=

11(45� k)

8;

and so 8 j (45 � k). Since k � 9, the only solution for k is k = 5, andtherefore Sam's average is 11(40)=8 = 55. An example of four such marks is98; 76; 34; 12.

II. Solution by Tara McCabe, student,Mount Allison University, Sack-

ville, New Brunswick.

[Editor's note: McCabe �rst obtained equation (1) as above, using thesame notation.] Since a; : : : ; h are all di�erent and non-zero, their totalmust lie between 1 + � � � + 8 = 36 and 2 + � � � + 9 = 44 inclusively. From(1),

36

2= 18 � a+ c+ e+ g � 22 =

44

2: (3)

Letting the average be the two digit number xy,

10a+ b+ 10c+ d+ 10e+ f + 10g+ h

4= 10x+ y;

which by (1) means

11(a+ c+ e+ g) = 4(10x+ y):

Consequently, a+c+e+gmust be divisible by 4, and from (3), a+c+e+g =20. Therefore 11(20) = 4(10x + y), and so 55 = 10x + y. Little Sam hasan average of 55.

A natural addition to this problem is to try to �nd Little Sam's four testscores ab; cd; ef and gh. We know that a+ c+ e+ g = 20, and the problembecomes: how many sets of test scores are possible?

First, choose four digits from f1; 2; 3; 4; 6; 7; 8; 9g to be a; c; e and g (0and 5 are not allowed). Notice that for a sum of 20 to be possible, two digitsmust be chosen from f1; 2; 3; 4g and two from f6; 7; 8; 9g. There are �4

2

�= 6

ways to choose two digits from f1; 2; 3; 4g. The sum S1 of these two digits issuch that 3 � S1 � 7, and the only sum that can occur twice is 5. Similarly,there are 6 ways to choose two digits from f6; 7; 8; 9g, the sum S2 of thesetwo digits is such that 13 � S2 � 17, and the only sum that can occur twiceis 15. There are 2 � 2 = 4 combinations when S1 = 5 (and S2 = 15) andone combination for each other sum S1. Therefore, there are eight ways tochoose four digits from f1; 2; 3; 4; 6; 7; 8; 9g such that the sum is 20.

Once a; c; e and g are chosen, b; d; f andh are simply the four remainingdigits. There are 4! = 24 di�erent ways to assign these four digits to be b; d; fand h. Therefore there are 8 � 24 = 192 di�erent sets of four test scorespossible for Little Sam. He's not such a unique child after all!

127

Also solved by SAMBAETHGE, Nordheim, Texas, USA; CHRISTOPHER

J. BRADLEY, CliftonCollege, Bristol,UK; TIMCROSS, King Edward's School,

Birmingham, UK; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-

berg, Germany; NOEL EVANS and CHARLES DIMINNIE, Angelo State Uni-

versity, San Angelo, Texas, USA; FLORIAN HERZIG, student, Perchtolds-

dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KATHLEEN

E. LEWIS, SUNYOswego,Oswego, New York, USA; DAVID E.MANES, SUNY

at Oneonta, Oneonta, NY, USA; JOHN GRANT MCLOUGHLIN, Okanagan

University College, Kelowna, B.C.; P. PENNING, Delft, the Netherlands;

CORY PYE, student,Memorial University ofNewfoundland, St. John's,New-

foundland; JOEL SCHLOSBERG, student, Hunter College High School, New

York, NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DAVID R.

STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E.

TSAOUSSOGLOU, Athens, Greece; and the proposer. Three other solvers did

not prove that 55 is the only possible answer. There was also one incorrect

answer sent in.

At the end of his solution, the proposer asks for the number of sets

of test scores satisfying the problem, but only McCabe was able to read his

mind and answer this too!

2123. [1996: 77] Proposed by Sydney Bulman{Fleming and Edward

T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

It is known (e.g., exercise 23, page 78 of Kenneth H. Rosen's Elemen-

tary Number Theory and its Applications, Third Edition) that every natu-ral number greater than 6 is the sum of two relatively prime integers, eachgreater than 1. Find all natural numbers which can be expressed as the sumof three pairwise relatively prime integers, each greater than 1.

Solutionby Shawn Godin, St. Joseph ScollardHall, North Bay, Ontario.

Since every number greater than 6 is the sum of two relatively primeintegers, then each even number greater than 6 can be expressed as the sumofof two relatively prime odd numbers. If we add 2 to each of these expressionswe see that every even number greater than 8 can be expressed as the sumof three pairwise relatively prime integers (two relatively prime odd integersand 2).

All odd numbers can be expressed in the form 18N + k, where N is a non-negative integer and k is an odd number less than 18. Note that:

18N + 1 = (6N � 3) + (6N � 1) + (6N + 5) for N � 1

18N + 3 = (6N � 1) + (6N + 1) + (6N + 3) for N � 1

18N + 5 = (6N � 1) + (6N + 1) + (6N + 5) for N � 1

128

18N + 7 = (6N � 1) + (6N + 3) + (6N + 5) for N � 1

18N + 9 = (6N + 1) + (6N + 3) + (6N + 5) for N � 1

18N + 11 = (6N + 1) + (6N + 3) + (6N + 7) for N � 1

18N + 13 = (6N + 1) + (6N + 5) + (6N + 7) for N � 1

18N + 15 = (6N + 3) + (6N + 5) + (6N + 7) for N � 0

18N + 17 = (6N + 1) + (6N + 7) + (6N + 9) for N � 1

Clearly each of the pairs of terms in each expression is relatively prime, sinceif there is a number which divides each term in a pair it must divide thedi�erence. The restriction on N ensures that each term is greater than 1.Putting all this together shows that the numbers which can be so written are10, 12, 14, 15, 16, and all the natural numbers greater than 17.

[Ed: it is easily veri�ed that all other integers do not have the desiredproperty.]

Also solved by RICHARD I. HESS, Rancho Palos Verdes, California,

USA; PETERHURTHIG, Columbia College, Burnaby, BC;WALTHER JANOUS,

Ursulinengymnasium, Innsbruck, Austria; MICHAEL JOSEPHY, Universidad

de Costa Rica, San Jos �e, Costa Rica; KEE-WAI LAU, Hong Kong; DAVID

E. MANES, State University of New York, Oneonta, NY, USA; MICHAEL

PARMENTER, Memorial University of Newfoundland, St. John's, Newfound-

land; P. PENNING, Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzi-

gymnasium, Graz, Austria; ROBERT P. SEALY, Mount Allison University,

Sackville, New Brunswick; DAVID R. STONE, Georgia Southern University,

Statesboro, Georgia, USA; and the proposers. There was 1 incorrect submis-

sion.

Most solvers used mod 18 or mod 12 arithmetic to handle the odd

values of n. MANES pointed out that both this problem and the problem in

Rosen's book can be found in Sierpi �nski's book, \250 Problems in ElementaryNumber Theory", American Elsevier, New York, 1970, (problems 47 and 48)where they are both solved.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

129

THE ACADEMY CORNERNo. 10

Bruce Shawyer

All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

This month, we present a university entrance scholarship examination

paper from the 1940's. Thanks to Georg Gunther, Sir Wilfred Grenfell Col-

lege, Corner Brook, Newfoundland, for providing this. We challenge today's

university students to solve these problems | send me your nice solutions!

1. Find all the square roots of

1� x+p22x� 15� 8x2:

2. Find all the solutions of the system of equations:

x+ y+ z = 2;

x2 + y2 + z2 = 14;

xyz = �6:

3. Suppose that n is a positive integer and that Ck is the coe�cient of xk

in the expansion of (1 + x)n. Show that

nXk=0

(k+ 1)C2k =

(n+ 2) (2n� 1)!

n! (n� 1)!:

4. (a) Suppose that a 6= 0 and c 6= 0, and that ax3 + bx+ c has a factor

of the form x2 + px+ 1. Show that a2 � c2 = ab.

(b) In this case, prove that ax3 + bx + c and cx3 + bx2 + a have a

common quadratic factor.

5. Prove that all the circles in the family de�ned by the equation

x2 + y2 � a(t2 + 2)x� 2aty � 3a2 = 0

(a �xed, t variable) touch a �xed straight line.

6. Find the equation of the locus of a point P which moves so that the

tangents from P to the circle x2 + y2 = r2 cut o� a line segment of

length 2r on the line x = r.

130

7. If the tangents at A, B and C to the circumcircle of triangle 4ABCmeet the opposite sides at D, E and F , respectively, prove that D, E

and F are collinear.

8. Find the locus of P which moves so that the polars of P , with respect

to three non-intersecting circles, are concurrent.

9. Suppose that P is a point within the tetrahedron OABC. Prove that

\AOB + \BOC + \COA is less than \APB + \BPC + \CPA.

10. Two unequal circles of radii R and r touch externally, and P and Q are

the points of contact of a common tangent to the circles, respectively.

Find the volume of the frustum of a cone generated by rotating PQ

about the line joining the centres of the circle.

11. Prove that

sin2(�+�)+sin

2(�+�)�2 cos(���) sin(�+�) sin(�+�) = sin

2(���):

12. Three points A, B and C are on level ground. B is east of A, C is

N. 49� E. of A, and C is N. 11�300 W. of B.

Find the direction of C as seen from the mid-point of AB.

13. With each corner of a square of side r as a centre, four circles of radius

r are drawn.

Show that the area of the central curvilinear quadrilateral formed inside

the square by the intersection of the four circles is

r2�1�

p3 +

�

3

�:

14. An observer on a boat is vertically beneath the centre of a bridge, which

crosses a straight canal at right angles. Looking upwards, the observer

sees that the angle subtended by the length of the bridge is 2�. The

observer then rows a distance � along the middle of the canal, and then

�nds that the length of the bridge now subtends an angle of 2�.

Show that the length of the bridge is

2�pcot2 � � cot2 �

:

131

THE OLYMPIAD CORNERNo. 181

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

We begin this number with two contests. Thanks go to Richard

Nowakowski, Canadian Team Leader to the 35th IMO in Hong Kong, for

collecting them and forwarding them to us.

SELECTED PROBLEMS FROM THE ISRAELMATHEMATICAL OLYMPIADS, 1994

1. p and q are positive integers. f is a function de�ned for positive

numbers and attains only positive values, such that f(xf(y)) = xpyq. Prove

that q = p2.

2. The sides of a polygon with 1994 sides are ai =p4 + i2, i =

1; 2; : : : ; 1994. Prove that its vertices are not all on integer mesh points.

3. A \standard triangle" in the plane is a (�lled) isosceles right triangle

whose sides are parallel to the x and y axes. A �nite family of standard

triangles, containing at least three, is given. Every three of this family have

a common point. Prove that there is a point common to all triangles in that

family.

4. A shape c0 is called \a copy of the planar shape c" if the following

conditions hold:

(i) There are two planes � and �0 and a point P that does not belong

to either of them.

(ii) c 2 � and c0 2 �0.(iii) A point X0 satis�esX0 2 c0 i� X0 is the intersection of �0 with the

line passing through X and P .

Given a planar trapezoid, prove that there is a square which is a copy

of this trapezoid.

5. Find all polynomials p(x), with real coe�cients, satisfying

(x� 1)2p(x) = (x� 3)2p(x+ 2)

for all x.

132

PROBLEMS FROM THE BI-NATIONALISRAEL-HUNGARY COMPETITION, 1994

1. a1; : : : ; ak; ak+1; : : : ; an are positive numbers (k < n). Suppose

that the values of ak+1; : : : ; an are �xed. How should one choose the values

of a1; : : : ; ak in order to minimizeP

i;j;i6=jaiaj?

2. Three given circles pass through a common point P and have the

same radius. Their other points of pairwise intersections are A, B, C. The

3 circles are contained in the triangle A0B0C0 in such a way that each side of

4A0B0C0 is tangent to two of the circles. Prove that the area of 4A0B0C0is at least 9 times the area of4ABC.

3. m, n are two di�erent natural numbers. Show that there exists a

real number x, such that 13� fxng � 2

3and 1

3� fxmg � 2

3, where fag is

the fractional part of a.

4. An \n-m society" is a group of n girls andm boys. Show that there

exist numbers n0 andm0 such that every n0-m0 society contains a subgroup

of �ve boys and �ve girls in which all of the boys know all of the girls or none

of the boys knows none of the girls.

Last issue we gave �ve more Klamkin Quickies. Next we give his

\Quicky" solutions to these problems. Many thanks to Murray S. Klamkin,

the University of Alberta, for creating the problems and solutions.

ANOTHER FIVE KLAMKIN QUICKIESOctober 21, 1996

6. Determine the four roots of the equation x4 + 16x� 12 = 0.

Solution. Since

x4 +16x� 12 = (x2+2)2� 4(x� 2)2 = (x2 +2x� 2)(x2� 2x+6) = 0;

the four roots are �1�p3 and 1� i

p5.

7. Prove that the smallest regular n-gon which can be inscribed in a

given regular n-gon is one whose vertices are the midpoints of the sides of

the given regular n-gon.

Solution. The circumcircle of the inscribed regular n-gonmust intersect

each side of the given regular n-gon. The smallest that such a circle can be

is the inscribed circle of the given n-gon, and it touches each of its sides at

its midpoints.

133

8. If 311995 divides a2 + b2, prove that 311996 divides ab.

Solution. If one calculates 12; 22; : : : ; 302 mod 31 one �nds that the

sum of no two of these equals 0 mod 31. Hence, a = 31a1 and b = 31b1 so

that 311993 divides a21 + b21. Then, a1 = 31a2 and b1 = 31b2. Continuing in

this fashion (with p = 31), we must have a = p998m and b = p998n so that

ab is divisible by p1996.

More generally, if a prime p = 4k + 3 divides a2 + b2, then both a

and b must be divisible by p. This follows from the result that \a natural nis the sum of squares of two relatively prime natural numbers if and only ifn is divisible neither by 4 nor by a natural number of the form 4k+ 3" (see

J.W. Sierpi �nski, Elementary Theory of Numbers, Hafner, NY, 1964, p. 170).

9. Determine the minimum value of

S =p(a+ 1)2 + 2(b� 2)2 + (c+ 3)2 +

p(b+ 1)2 + 2(c� 2)2 + (d+ 3)2)

+p(c+ 1)2 + 2(d� 2)2 + (a+ 3)2 +

p(d+ 1)2 + 2(a� 2)2 + (b+ 3)2

where a, b, c, d are any real numbers.

Solution. Applying Minkowski's inequality,

S �p(4 + s)2 + 2(s� 8)2 + (s+ 12)2 =

p4s2 + 288

where s = a+ b+ c+ d. Consequently, minS = 12p2 and is taken on for

a = b = c = d = 0.

10. A set of 500 real numbers is such that any number in the set is

greater than one-�fth the sum of all the other numbers in the set. Determine

the least number of negative numbers in the set.

Solution. Letting a1; a2; a3; : : : denote the numbers of the set and S

the sum of all the numbers in the set, we have

a1 >S � a1

5; a2 >

S � a2

5; : : : ; a6 >

S � a6

5:

Adding, we get 0 > S � a1 � a2 � � � � � a6 so that if there were six or less

negative numbers in the set, the right hand side of the inequality could be

positive. Hence, there must be at least seven negative numbers.

Comment. This problem where the \5" is replaced by \1" is due to

Mark Kantrowitz, Carnegie{Mellon University.

134

First a solution to one of the 36th IMO problems:

2. [1995: 269] 36th IMOLet a, b, and c be positive real numbers such that abc = 1. Prove that

1

a3(b+ c)+

1

b3(c+ a)+

1

c3(a+ b)� 3

2:

Solution by Panos E. Tsaoussoglou, Athens, Greece.By the Cauchy{Schwartz inequality

[a(b+ c) + b(c+ a) + c(a+ b)]

�1

a3(b+ c)+

1

b3(c+ a)+

1

c3(a+ b)

�

��1

a+

1

b+

1

c

�2;

or

2(ab+ ac+ bc)

�1

a3(b+ c)+

1

b3(c+ a)+

1

c3(a+ b)

�

� (ab+ ac+ bc)2

(abc)2;

or1

a3(b+ c)+

1

b3(c+ a)+

1

c3(a+ b)� ab+ ac+ bc

2;

because abc = 1.

Alsoab+ ac+ bc

3� 3pa2b2c2 = 1:

Therefore1

a3(b+ c)+

1

b3(c+ a)+

1

c3(a+ b)� 3

2

holds.

135

Now we turn to some of the readers' solutions to problems proposed

to the jury but not used at the 35th IMO in Hong Kong [1995: 299{300].

PROBLEMS PROPOSED BUT NOT USEDAT THE 35th IMO IN HONG KONG

Selected Problems

3. A semicircle � is drawn on one side of a straight line `. C and D

are points on �. The tangents to � at C and D meet ` at B and A respec-

tively, with the center of the semicircle between them. Let E be the point

of intersection of AC and BD, and F be the point on ` such that EF is

perpendicular to `. Prove that EF bisects \CFD.

Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk,Zaltbommel, the Netherlands. We give Seimiya's write-up.

� �

T FQ O

E

D

C

P

AB `

Let P be the intersection of AD and BC. Then \PCO = \PDO =

90�, \CPO = \DPO and PC = PD. Let Q be the intersection of PE

with AB. Then by Ceva's Theorem, we get

BQ

QA� ADDP

� PCCB

=BQ

QA� ADCB

= 1:

Thus we get

BQ

QA=BC

AD: (1)

Since \BPO = \APO we get

PB

PA=BO

AO: (2)

We put \PAB = �, \PBA = �.

136

Let T be the foot of the perpendicular from P to AB. Then from (1)

and (2) we have

BC

AD=BO cos�

AO cos�=PB cos�

PA cos�=PT

TA: (3)

From (1) and (3) we have

BQ

QA=PT

TA:

Hence Q coincides with T so that P , E, F are collinear. [See page 136.]

Because \PCO = \PDO = \PFO = 90�, P , C, F , O, D are con-

cyclic. Hence \CFE = \CFP = \CDP = \DCP = \DFP = \DFE.

Thus EF bisects \CFD.

AB OF

C

D

P

E

`

4. A circle ! is tangent to two parallel lines `1 and `2. A second circle

!1 is tangent to `1 at A and to ! externally at C. A third circle !2 is tangent

to `2 at B, to ! externally at D and to !1 externally at E. AD intersects

BC at Q. Prove that Q is the circumcentre of triangle CDE.

Solutions by Toshio Seimiya, Kawasaki, Japan; and by D.J. Smeenk,Zaltbommel, the Netherlands. We give Smeenk's solution.

We denote the three circles as !(O;R), !1(O1; R1), !2(O2; R2). Now

let ! touch `1 at F and `2 at F 0. Let the line through O2 parallel to `1intersect FF 0 at G and the production of AO1 at H.

137

F 0 M B`2

`1 F L A

O1

O2

E

C

Q

DK

G

R

R2

R1

R2

R1

N

O

Let the line through D parallel to `1 intersect FF0 at K.

Let the line through D parallel to FF 0 intersect `1 at L, `2 at M and

GO2 at N . Now AF is a common tangent of ! and !1, so

AF = 2pRR1 (1)

and

BF 0 = 2pRR2 = GO2: (2)

It follows that

HO2 = j2pRR2 � 2

pRR1j;

HO1 = 2R� R1 � R2:

In right triangle O1HO2,

(2pRR2 � 2

pRR1)

2 + (2R� R1 � R2)2 = (R1 +R2)

2:

After some reduction R = 2pR1R2.

Next consider triangle GOO2.

GO = R� R2; GO2 = 2pRR2; DO2 = R2; DO = R; KDkGO2:

We �nd that GN = FL =R

R+ R2

�GO2 =2RpRR2

R+ R2

.

With (1) we have

AL = 2pRR1 �

2RpRR2

R+ R2

: (3)

138

Furthermore DN =R2

R+R2

�GO =R2(R� R2)

R+R2

and

DL = 2R�R2 �R2(R� R2)

R+ R2

=2R2

R+ R2

: (4)

Now AD2 = AL2 +DL2. With (3) and (4),

2pRR1 �

2RpRR2

R1 + R2

!2

+

�2R2

R1 + R2

�2= 4RR1 = AE2:

So AD = AF .

That means that AD touches ! atD andAD is a common tangent and

the radical axis of ! and !2.

In the same way BC is the radical axis of ! and !1 andQ is the radical

point of !, !1 and !2.

So QC = QD = QE, as required.

5. A line ` does not meet a circle ! with center O. E is the point on

` such that OE is perpendicular to `. M is any point on ` other than E.

The tangents from M to ! touch it at A and B. C is the point on MA such

that EC is perpendicular to MA. D is the point on MB such that ED is

perpendicular to MB. The line CD cuts OE at F . Prove that the location

of F is independent of that ofM .

Solution by Toshio Seimiya, Kawasaki, Japan.

AsMA, MB are tangent to ! at A, B respectively, we get \OAM =

\OBM = 90� and OM ? AB. Let N , P be the intersections of AB with

OM and OE respectively.

Since M , E, P , N lie on the circle with diameter MP we get ON �OM = OB2 = r2 where r is the radius of !. Hence P is a �xed point. (P

is the pole of `.)

Let G be the foot of the perpendicular from E to AB. As \OBM =

\OAM = \OEM = 90�, O, B,M , E, A are concyclic, so that by Simson's

Theorem C, D, G are collinear.

139

!

`EM

D

B

O

A

G

N P

C F

r

Since A, C, E, G lie on the circle with diameter AE we get

\EGF = \EGC = \EAC = \EAM: (1)

As O, M , E, A are concyclic and OM is parallel to EG we have

\EAM = \EDM = \DEG = \FEG: (2)

From (1) and (2) we get

\EGF = \FEG: (3)

Since \EGP = 90� we get

\FGP = \FPG: (4)

From (3) and (4) we have EF = FG = FP . Thus F is the midpoint of

EP . Hence F is a �xed point.

140

Next, we give a counterexample to the �rst problem of the set of prob-

lems proposed to the jury, but not used at the 35th IMO in Hong Kong given

in the December 1995 number of the corner.

1. [1995: 334] Problems proposed but not used at the 35th IMO inHong Kong.

ABCD is a quadrilateral with BC parallel to AD. M is the midpoint

of CD, P that of MA and Q that of MB. The lines DP and CQ meet at

N . Prove that N is not outside triangle ABM .

Counterexamples by Joanna Jaszu �nska, student, Warsaw, Poland; andby Toshio Seimiya, Kawasaki, Japan. We give Jaszu �nska's example.

A D

M

CB

P

N

X

Q

t

We draw a triangle ADM and denote the midpoint of MA by P . Let

C be a point on the half-lineDM such that M is the midpoint of CD.

Let N be any point of the segment PD, inside triangle ADM .

We construct a parallelogramMCBX such thatMX andBC are par-

allel to AD and X lies on the segment CN .

Let us denote the point where the diagonal MB of this quadrilateral

meets CN by Q. Q is then the midpoint of MB.

Connect points A and B. We have thus constructed a quadrilateral

ABCD with BC parallel to AD, M is the midpoint of CD, P that ofMA

and Q that of MB. Lines DP and CQ meet at N .

N is inside triangle ADM ; hence it is outside triangle ABM .

141

Next we look back to some further solutions to problems of the Sixth

Irish Mathematical Olympiad given in [1995: 151{152] and for which some

solutions were given in [1997: 9{13]. An envelope from Michael Selby ar-

rived which I mis�led. It contains solutions to problems 1, 2, and 4 of Day 1,

and problems 1, 2, 3 and 4 of Day 2.

1. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.

Given �ve points P1, P2, P3, P4, P5 in the plane having integer coor-

dinates, prove that there is at least one pair (Pi; Pj) with i 6= j such that

the line PiPj contains a point Q having integer coordinates and lying strictly

between Pi and Pj.

Solution by Michael Selby, University of Windsor, Windsor, Ontario.

The points can be characterized according to the parity of their x and

y coordinates. There are only four such classes: (even, even), (even, odd),

(odd, even), (odd, odd).

Since we are given �ve such points, at least two must have the same

parity of coordinates by the Pigeonhole Principle. Suppose they are Pi and

Pj , Pi = (xi; yi), Pj = (xj ; yj). Then xi + xj is even and yi + yj is even,

since the xi, xj have the same parity and yi, yj have the same parity. Hence

the midpoint

Q =

�xi + xj

2;yi + yj

2

�

has integral coordinates.

2. [1995: 152] Second Paper, Sixth Irish Mathematical Olympiad.

Let a1; a2; : : : an, b1; b2; : : : bn be 2n real numbers, where

a1; a2; : : : ; an are distinct, and suppose that there exists a real number �

such that the product

(ai + b1)(ai+ b2) : : : (ai + bn)

has the value � for all i (i = 1; 2; : : : ; n). Prove that there exists a real

number � such that the product

(a1 + bj)(a2 + bj) : : : (an + bj)

has the value � for all j (j = 1; 2; : : : ; n).

Solution by Michael Selby, University of Windsor, Windsor, Ontario.

De�ne

Pn(x) = (x+ b1)(x+ b2) � � � (x+ bn)� �: (1)

Then Pn(ai) = 0 for i = 1; 2; : : : ; n.

Therefore Pn(x) = (x� a1)(x� a2) � � � (x� an) by the Factor Theorem.

142

Now (�1)nPn(�x) = (x+ a1)(x+ a2) � � � (x+ an). So

(�1)nPn(�bi) = (bi + a1)(bi+ a2) � � � (bi+ an)

= (�1)n+1� by (1):

Hence (bi+ a1)(bi+ a2) � � � (bi + an) = (�1)n+1� for i = 1; 2; : : : ; n.

Thus, the result is true with � = (�1)n+1�.

That completes the Corner for this number. We are in high Olympiad

season. Send me your nice solutions and contests.

Do you believe what occurs in print?

The last sentence of the quoted passage, taken from The Daughters ofCain by Colin Dexter (Macmillan, 1994), contains two factual erors. What are

they?

`Have you heard of \Pythagorean Triplets"?'

`We did Pythagoras Theorem at school.'

`Exactly. The most famous of all the triplets, that is |

\3, 4, 5" 32 + 42 = 52. Agreed?'

`Agreed.'

`But there are more spectacular examples than that.

The Egyptians, for example, knew all about \5961, 6480, 8161".'

Contributed by J.A. McCallum, Medicine Hat, Alberta.

143

BOOK REVIEWS

Edited by ANDY LIU

The Lighter Side of Mathematics,

edited by Richard K. Guy and Robert E. Woodrow,

Mathematical Association of America, Washington DC,

1994, ISBN 0-88385-516-X, 376+ pages, softcover, US $38.50,

reviewed byMurray S. Klamkin, University of Alberta.

This book is the proceedings of the Eugene StrensMemorial Conference

on RecreationalMathematics and its Historyheld at the University of Calgary

in August 1986 to celebrate the founding of the Strens Collection which is now

the most complete library of recreational mathematics in the world.

I had been invited to attend this conferences but unfortunately had a

previous committment. To make up for missing this conference, the next best

thing was reviewing this proceedings book which on doing so made me real-

ize what I had missed, for example, some very interesting talks plus getting

together with the leading practitioners of recreational mathematics, some of

whom were long time colleagues.

One does not normally include a list of contents in a book review, but by

doing so, it will give the reader a good indication of the wealth of recreational

material here . So if you have any interest in recreational mathematics, this

is a book for you. And even if you do not have such an interest, reading this

book may give you one.

Contents

Preface

The Strens Collection 1

Eugene Louis Charles Marie Strens 5

Part 1: Tiling & Coloring

Frieze Patterns, Triangulated Polygons and Dichromatic Symmetry,

H. S. M. Coxeter & J. F Rigby 15

Is Engel's Enigma a Cubelike Puzzle? J. A. Eidswick 28

Metamorphoses of Polygons, Branko Gr �unbaum 35

SquaRecurves, E-Tours, Eddies, and Frenzies:

Basic Families of Peano Curves on the Square Grid,

Douglas M. McKenna 49

Fun with Tessellations, John F. Rigby 74

Escher: A Mathematician in Spite of Himself, D. Schattschneider 91

Escheresch, Athelstan Spilhaus 101

The Road Coloring Problem, Daniel Ullman 105

Fourteen Proofs of a Result About Tiling a Rectangle, Stan Wagon 113

Tiling R3 with Circles and Disks, J. B. Wilker 129

144

Part 2: Games & Puzzles

Introduction to Blockbusting and Domineering, Elwyn Berlekamp 137

AGenerating Function for the Distribution of the Scores of all Possible Bowling Games,

Curtis N. Cooper & Robert E. Kennedy 149

Is the Mean Bowling Score Awful?

Curtis N. Cooper & Robert E. Kennedy 155

Recreation and Depth in Combinatorial Games, Aviezri Fraenkel 159

Recreational Games Displays

Combinatorial Games, Aviezri S. Fraenkel 176

Combinatorial Toys, Kathy Jones 195

Rubik's Cube | application or illumination of group theory?

Mogens Esrom Larsen 202

Golomb's Twelve Pentomino Problems, Andy Liu 206

A New Take-Away Game, Jim Propp 212

Confessions of a Puzzlesmith, Michael Stueben 222

Puzzles Old & New: Some Historical Notes, Jerry Slocum 229

Part 3: People & Pursuits

The Marvelous Arbelos, Leon Banko� 247

Cluster Pairs of an n-Dimensional Cube of Edge Length Two,

I. Z. Bouwer & W. W. Cherno� 254

The Ancient English Art of Change Ringing, Kenneth J. Falconer 261

The Strong Law of Small Numbers, Richard K. Guy 265

Match Sticks in the Plane, Heiko Harborth 281

Misunderstanding My Mazy Mazes May Make Me Miserable,

Mogens Esrom Larsen 289

Henry Ernest Dudeney: Britain's Greatest Puzzlist, Angela Newing 294

From Recreational to Foundational Mathematics, Victor Pambuccian 302

Alphamagic Squares, Lee C. F. Sallows 305

Alphamagic Squares: Part II, Lee C. F. Sallows 326

The Utility of Recreational Mathematics, David Singmaster 340

The Development of Recreational Mathematics in Bulgaria,

Jordan Stoyanov 346

V � E + F = 2, Herbert Taylor 353

Tracking Titanics, Samuel Yates 355

List of Conference Participants 363

Postscript:

There is a typographical error in Andy Liu's article on page 206. The number

of tetrominoes is �ve and not four. This is corrected in the reprint of this

article as Appendix C in the new edition of Golomb's \Polyominoes".

145

In Memoriam | Leon Banko�

We were saddened to learn of the recent death of Dr. Leon Banko�,

who has been a contributor to CRUX over many years. Sadly, he was not

able to contribute recently, and some of our more recent readers may not

know so much about him. We refer you to an excellent article in the March

1992 issue of the College Mathematics Journal, entitled A Conversation with

Leon Banko�, written by G.L. Alexanderson.

One of Leon's long time friends, Dr. Clayton Dodge, has written the

appreciation printed below.

Leon Banko� practiced dentistry for sixty years in Beverly Hills, Cali-

fornia, until his retirement just a few years ago. His patients included many

Hollywood personalities whose names are household words. Among his sev-

eral other interests, such as piano, guitar, calculators, and computers, he

lectured and wrote papers both on dentistry and mathematics. His specialty

was geometry, and the �gure he loved best was the arbelos, or shoemaker's

knife, which consists of three semicircles having a common diameter line.

The two smaller semicircles are externally tangent to each other and inter-

nally tangent to the largest semicircle.

It is said that the test of a mathematician is not what he himself has

discovered, but what he inspired others to do. Leon discovered a third circle

congruent to the twin circles of Archimedes and published that result in the

September 1974 issue of Mathematics Magazine (\Are the Twin Circles of

Archimedes Really Twins?", pp. 214-218.) This revelation motivated the

discovery by Leon and by others of several other members of that family of

circles. An article on those circles is in progress.

Dr. Banko� edited the Problem Department of the Pi Mu Epsilon

Journal from 1968 to 1981, setting and maintaining a high standard of ex-

cellence in the more than 300 problems he included in its pages. Although

the Journal has a relatively small circulation, its Problem Department grew

to have a large number of regular contributors. He became acquainted with

Crux Mathematicorum early in its history, when it was called Eureka, andmade many contributions to its pages over the years, maintaining a close

friendship with its founder and �rst editor Leo Sauv �e. Like Leo, who started

Crux to add some spice to his mathematical life of teaching basic post high

school courses, Leon worked in mathematics for mental exercise and recre-

ation, making friendswith and earning the respect of many well knownmath-

ematicians.

Leon and I became good friends, �rst through correspondence regarding

the PiMu Epsilon Journal Problem Department, and later throughmany per-

sonal meetings, including the August 1979 meeting of problemists in

Ottawa, sponsored by Leo Sauv �e and Fred Maskell of Crux. Following the

146

formal sessions in Ottawa, seven of us drove to Quebec City for an enjoyable

weekend of sightseeing and fellowship: Leo and Carmen Sauv �e, Leon and

Francine Banko�, Charles and Avetta Trigg, and I. Since his retirement from

his dentistry, Leon has worked on the manuscript for a proposed book on the

properties of the arbelos, carrying on a monumental task started by him and

the late Victor Th �ebault. Much material has been collected for this project

and much remains to be done on it. Indeed, he asked me to �nish the job.

At one time some years ago a schoolgirl wrote to Albert Einstein about

a mathematical question she had. Apparently Einstein misinterpreted her

question and gave an incorrect answer. Banko� pointed out this error and in

his mathematical museum he now has a copy of the Los Angeles Times with

the front page headline \Local Dentist Proves Einstein Wrong."

Leon developed many physical problems in his later years. He was a

�ghter and he won several physical battles. When I last visited him at his

home in Los Angeles in October 1996, he was �ghting liver cancer, but still

working on the Th �ebault material, in spite of failing eyesight. On Sunday

afternoon, February 16, 1997, the cancer overtook him and he died at his

home at the age of 88.

He was a gentleman, a scholar, and a true friend.

Clayton W. Dodge

University of Maine

Orono, ME 04469-5752

Heronian Triangles with

Associated Inradii in Arithmetic Progression

Paul YiuDepartment of Mathematics, Florida Atlantic University

In memory of Dr. Leon Banko�

1. The area of a triangle is given in terms of its sides a, b, c by the Heron

formula

4 =ps(s� a)(s� b)(s� c);

where s := 12(a + b + c) is the semiperimeter. A triangle (a; b; c;4) is

called Heronian if its sides and area are all integers. L. Banko� [1] has made

an interesting observation about the Heronian triangle (13;14; 15; 84). The

147

height on the side 14 being 12, this triangle can be decomposed into two

Pythagorean components, namely, (5; 12; 13) and (9; 12; 15). The inradii of

these Pythagorean triangles, and that of the Heronian triangle, are respect-

ively 2, 3, 4, three consecutive integers! (See Figure 1). Noting that the sides

of the Heronian triangle are themselves three consecutive integers, Banko�

remarked that \no other Heronian triangle can claim that distinction".

Actually, apart from this, there are exactly two other Heronian trian-

gles with three consecutive integers for the associated inradii. Each of these

two Heronian triangles is decomposable into two Pythagorean components,

namely,

(15;20; 25; 150) = (9; 12; 15) [ (16; 12; 20); (2)

(25;39; 56; 420) = (20; 15; 25) [ (36; 15; 39): (3)

The three inradii in these two cases are 3, 4, 5, and 5, 6, 7 respectively. The

Heronian triangle (15; 20; 25; 150) in (2) has an interesting property that no

other Heronian triangle shares. Here, if the smaller Pythagorean component

(9; 12; 15) is excised from the larger one (16;12; 20), another Heronian tri-

angle results, namely,

(15;20; 7; 42) = (16; 12; 20) n (9; 12; 15): (4)

This has inradius 2. (See Figure 2). Note that the four inradii are consecutive

integers! The same construction applied to Banko�'s example (13; 14;15; 84)

gives (13; 15; 4; 24), with inradius 32, not an integer. For the Heronian tri-

angle (25;39; 56; 420) in (3), this fourth inradius is 3, albeit not consecutive

with the other three inradii 5, 6, and 7. In x 4 below, we shall show that,

up to similarity, the con�guration in Figure 2 is the only one with the four

associated inradii in arithmetic progression.

q q q

q

q

q

5 9

2

43

1213 15

q

q

q

q

q

q

q

q

9 9 7

3

5 4

2

122015

Figure 1. Figure 2.

2. Consider two right triangles with a common side y and opposite acute

angles � and � juxtaposed to form a triangle �(+). We shall assume � >

�, so that �1 can be excised from �2 to form another triangle �(�). (See

Figures 3 and 4). The inradius of a triangle with area4 and semiperimeter s

is given by r = 4s. (See, for example, Coxeter [2, p. 12]). For a right triangle

with legs a, b, and hypotenuse c, this is also given by the simpler formula

148

r = s�c = 12(a+b�c). We shall determine the similarity classes of triangles

for which the inradii of the triangle and its two Pythagorean components are

in arithmetic progression. The calculations can be made simple by making

use of the fact that in a right triangle with an acute angle �, the sides are

in the ratio 2t : 1 � t2 : 1 + t2, where t = tan �2. Let t1 := tan �

2and

t2 := tan �

2.

�1 �2

a1 a2y

x1 x2

� �

�1 �(�)

y

x1 a0

3

a1 a2

� �

�(+) = �1 [ �2 �(�) = �2n�1Figure 3. Figure 4.

By choosing y = 2t1t2, we have, in Figures 3 and 4,

a1 = t2(1 + t21); a2 = t1(1 + t22);

x1 = t2(1� t21); x2 = t1(1� t22);

a3 = x1 + x2 = (t1 + t2)(1 � t1t2); a03 = x2 � x1 = (t1 � t2)(1 + t1t2):

From these, we determine the inradii of the four triangles �1, �2, �(+) and

�(�):

r1 = t1t2(1� t1); r2 = t1t2(1� t2);

r+ = t1t2(1� t1t2); r� = t1t2(1� t2t1):

(5)

Since r1 < r2, r� < r2, and r� < r+, there are only three cases in

which three of these inradii can be in arithmetic progression:

(i) r1, r2, r+ are in A.P. if and only if t1, t2, t1t2 are in A.P., that is,

t1 + t1t2 = 2t2. From this, t2 = t12�t1 , and

r1 : r2 : r+ = 2� t1 : 2 : 2 + t1: (6)

(ii) r�, r1, r2 are in A.P. if and only if t2t1, t1, t2 are in A.P., that is,

t2t1

+ t2 = 2t1. From this, t2 =2t211+t1

, and

r� : r1 : r2 = 1 : 1 + t1 : 1 + 2t1: (7)

(iii) r1, r�, r2 are in A.P. if and only if t1,t2t1, t2 are in A.P., that is,

t1 + t2 = 2t2t1. From this, t2 =

t212�t1 , and

r1 : r� : r2 = 2� t1 : 2 : 2 + t1: (8)

149

In each of these cases, with t := t1, the proportions of the sides of �(�)are as follows. These triangles are all genuine for 0 < t < 1.

A.P. a1 : a2 : a3 (or a03)

(i) r1; r2; r+ (2� t)(1 + t2) : 2(2� 2t+ t2) : (3� t)(1� t)(2 + t)

(ii) r�; r1; r2 2t(1 + t)(1 + t2) : 1 + 2t+ t2 + 4t4 : (1� t)(1 + t+ 2t3)

(iii) r1; r�; r2 t(2� t)(1 + t2) : 4� 4t+ t2 + t4 : 2(1� t)(2� t+ t3)

3. Among the triangles constructed above with three associated inradii

in A.P., the only cases in which the three sides also are in A.P. are tabu-

lated below. In each case, we give the smallest Heronian triangle with three

associated inradii in integers.

(t1; t2) Triangle with decomposition inradii

(1) (23; 12) (13;15; 14;84) = (5; 12;13) [ (9;12; 15) (r1; r2; r+) = (2;3; 4)

(2) (12; 13) (15;20; 25;150) = (9;12; 15) [ (16;12; 20) (r1; r2; r+) = (3;4; 5)

(3) (12; 16) (15;37; 26;156) = (35;12; 37) n (9; 12; 15) (r1; r�; r2) = (3;4; 5)

(4) (15; 115

) (39;113; 76;570) = (112;15; 113) n (36;15; 39) (r�; r1; r2) = (5;6; 7)

Banko�'s observation [1] on the Heronian triangle (13;14; 15; 84) is case (1)

in this table.

4. Finally, we consider the possibility for the four inradii r�, r1, r2, r+,to be in A.P. First, assume r1; r�; r2 in A.P. By (8), r1 : r� : r2 = 2� t1 :

2 : 2 + t1; indeed, t2 =t21

2�t1 . From (5), we have, after simpli�cation,

r1 : r� : r2 : r+ = 2� t1 : 2 : 2 + t1 : 2 + t1 + t21:

Now, these four inradii are in A.P. if and only if 2 + t1 + t21 = 2+ 2t1. This

is clearly impossible for 0 < t1 < 1.

It remains, therefore, to consider the possibility that r�, r1, r2, r+ be

in A.P. This requires, by (7), r1 : r2 = 1 + t1 : 1 + 2t1, and also by (6),

r1 : r2 = 2 � t1 : 2. It follows that 1 + t1 : 1 + 2t1 = 2 � t1 : 2, from

which t1 = 12. Consequently, r1 : r2 = 3 : 4. Also, r1 : r+ = 3 : 5 from (6),

and r� : r1 = 2 : 3 from (7). Thus, we have the con�guration in Figure 2, in

which the four inradii are in the ratio

r� : r1 : r2 : r+ = 2 : 3 : 4 : 5:

The author thanks the referee for valuable comments and suggestions.

References

1. L. Banko�, An Heronian oddity, Crux Math. 8 (1982) p.206.

2. H.S.M. Coxeter, Introduction to Geometry, Wiley, New York, 1961.

150

THE SKOLIAD CORNERNo. 21

R.E. Woodrow

This number we give the problems of the Manitoba Mathematical Con-

test for 1995. This is a two hour contest aimed primarily at grade 12 students,

and sponsored by the Actuaries Club of Winnipeg, The Manitoba Association

of Mathematics Teachers, The Canadian Mathematical Society and The Uni-

versity of Manitoba. My thanks go to Diane and Roy Dowling, organizers of

the contest for supplying us with it.

THE MANITOBAMATHEMATICAL CONTEST 1995For Students in Grade 12

Wednesday, February 22, 1995 | Time: 2 hours

1. (a) If a and b are real numbers such that a+ b = 3 and a2+ab = 7

�nd the value of a.

(b) Noriko's average score on three tests was 84. Her score on the �rst

test was 90. Her score on the third test was 4 marks higher than her score

on the second test. What was her score on the second test?

2. (a) Find two numbers which di�er by 3 and whose squares di�er

by 63.

(b) Find the real number which is a root of the equation

27(x� 1)3 + 8 = 0:

3. (a) Two circles lying in the same plane have the same centre. The

radius of the larger circle is twice the radius of the smaller circle. The area

of the region between the two circles is 7. What is the area of the smaller

circle?

(b) The area of a right triangle is 5. Also, the length of the hypotenuse

of this triangle is 5. What are the lengths of the other two sides?

4. (a) The parabola whose equation is 8y = x2 meets the parabola

whose equation is x = y2 at two points. What is the distance between these

two points?

(b) Solve the equation 3x3 + x2 � 12x� 4 = 0.

5. (a) Find the real number a such that a4 � 15a2 � 16 = 0 and

a3 + 4a2 � 25a� 100 = 0.

(b) Find all positive numbers x such that xxpx = (x

px)x.

151

6. If x, y and z are real numbers prove that�xjyj � yjxj

� �yjzj � zjyj

� �xjzj � zjxj

�= 0:

7. x and y are integers between 10 and 100. y is the number obtained

by reversing the digits of x. If x2 � y2 = 495 �nd x and y.

8. Three points P , Q and R lie on a circle. If PQ = 4 and \PRQ =

60� what is the radius of the circle?

9. Three points are located in the �nite region between the x-axis and

the graph of the equation 2x2 + 5y = 10. Prove that at least two of these

points are within a distance 3 of each other.

10. Three circles pass through the origin. The centre of the �rst circle

lies in the �rst quadrant, the centre of the second circle lies in the second

quadrant, and the centre of the third circle lies in the third quadrant. If P

is any point that is inside all three circles, show that P lies in the second

quadrant.

Last number we gave the problems of the Mathematical Association

National Mathematics Contest 1994 from the United Kingdom. Here are the

answers.

1. C 2. E 3. C 4. D 5. E

6. B 7. A 8. D 9. B 10. B

11. B 12. E 13. E 14. A 15. A

16. C 17. D 18. C 19. B 20. B

21. C 22. B 23. E 24. B 25. E

That completes the Skoliad Corner for this issue. I need suitable contest

materials and welcome your suggestions for the evolution of this feature.

152

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

the Mayhem Editor, Naoki Sato, Department of Mathematics, University of

Toronto, Toronto, ON Canada M5S 1A1. The electronic address is

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

A Journey to the Pole | Part II

Miguel Carri �on �Alvarezstudent, Universidad Complutense de Madrid

Madrid, Spain

In this second (and last, for your relief) article, we look at some ad-

vanced topics like inversion or the applications of calculus to the theory of

curves.

Inversion

Inversion is a transformation determined by a point called the centreof inversion O and an inversion ratio �k2. The image of a point P is a

point P 0 such that P 0 is on line OP and jOP 0j = �k2jOP j. It is evident

that a curve r = f(�) can be inverted by letting r = � k2

f(�). Inversion is

a conformal transformation, meaning that the angles between intersecting

curves are preserved. We will prove this in a later section.

Exercise 1. Prove that the inverse curve of a straight line is itself if it passes

through the origin or a circle through the origin if it does not.

Example 1. By inspection of the equations of the conic r =de

1� e cos(� � �)and the lima�con r = b + a cos �, it is evident that the inverse of a conic

about its focus is a lima�con. I imagine that trying to prove this theory with

synthetic geometry would result in a severe headache.

This last result provides a di�erent de�nition of conics as loci if we

invert the de�nition of the lima�con given above. Consider a circle or straight

153

line and a point O not on it. Draw a circle through O tangent to the circle or

line. The diameter through O intersects the circle at P . The locus of all P 's

is a conic with O at one focus.

Tangent Lines

We leave the realm of elementary geometry to enter calculus, where we

will teach the same old dog new tricks. The �rst is how to �nd the tangent

line to a polar curve.

Our starting point will be the equation of the straight line d = r sin(���). The tangent line at �0 is a �rst-order approximation to the curve involv-

ing r(�0) anddr

d�

�����0

. Di�erentiating the equation of the straight line with

respect to � at �0, we get 0 =dr

d�

�����0

sin(�� �0)� r(�0) cos(�� �0), which

implies that tan(�� �0) =r

(dr=d�)

����0. This quantity can sometimes be use-

ful in itself, as �� �0 represents the angle between the radius vector and the

tangent line. We will make use of it in the next example.

The orientation � is determined from its tangent, and

tan(�� �0 + �0) =tan(�� �0) + tan �0

1� tan(�� �0) tan �0

implies that

tan� =r(�0) +

drd�j�0 tan �0

drd�j�0 � r(�0) tan �0

:

The parameter d in the equation of the tangent line is given (after some

trigonometric manipulations) by

d =r(�0) tan(�� �0)p1 + tan2(�� �0)

;

which gives

d =r2(�0)p

r2(�0) + (dr=d�)2j�0:

Example 2. Proof that inversion preserves angles. Let r = f(�) and r = g(�)

be two curves that intersect at �0. Their directions at �0 are �1 and �2, and

the angle between them satis�es

tan(�2��1) = tan(�2� �0+ �0��1) =tan(�2� �0)� tan(�1 � �0)

1 + tan(�2 � �0) tan(�1 � �0):

Now, the inverted curves, r0 = 1=f(�) and r0 = 1=g(�), also intersect at �0and their directions �01 and �

02 satisfy

tan(�01 � �0) =1=f

�1f2

�df

d�

� =�f

(df=d�)= tan(�0 � �1)

154

(and similarly for �02). Hence, tan(�2 � �1) = tan(�01 � �02) and we are

done.

Tangent lines through the origin

An interesting special case is when

r(�0) = 0. In that case, tan� =

tan��0 or � = �0. In words, if the

curve crosses the origin for a given

�0, the equation of the tangent line

at the origin is � = �0.

-

6

�

r

-

6

�

r

When sketching curves, a useful result is that if r(�) has an odd-order

root at �0, then the curve is smooth at the origin, but if the root is even-order,

then there is a cusp (see the �gure).

Asymptotes

d

O

�0

In certain cases r tends to in�nity for

some �nite value of �, signalling the

possibility of an asymptote. In han-

dling asymptotes, it is convenient to

consider s(�) = 1=r(�), in which

case tan(� � �0) =�s

(ds=d�)j�0

and tan� =

ds

d�j�0 tan �0 � s(�0)

ds

d�j�0 + s(�0) tan �0

.

There is a possible asymptote if

s(�0) = 0 and its equation is:

s(�) =sin(� � �0)

d, or

� = �0 if d = 0.

This is because, as in the case of tangent lines through the origin, the slope

of the tangent line is tan� = tan �0. The parameter d is given by

d = lim�!�0

1ps2(�) + (ds=d�)2

:

If this does not diverge, there is an asymptote.

When sketching curves, it is useful to know from which side of the

asymptote the curve approaches in�nity. This is achieved by studying the

sign of s(�) � 1

dsin(� � �0), which tends to 0 at � = �0. If it tends to 0+,

the curve is closer to the origin than the asymptote (see the �gure on the

last page), and if it tends to 0�, the curve is farther from the origin than the

asymptote.

Exercise 2. Sketch the curve r = ln �, its asymptote and the tangent line at

the origin.

155

Example 3. The parabola r =1

1� cos �satis�es lim

�!0r(�) = 1, but it has

no asymptotes since1p

(1� cos �0)2 + (sin�0)2=

1p2(1� cos �0)

, which

diverges at �0.

Arc Length

Another application of calculus is the computation of curve lengths.

Usually one would take the expression for the line element in cartesian coor-

dinates, dl2 = dx2 + dy2 and transform it to polar coordinates. To use only

polar coordinates, one could apply Pythagoras' Theorem to (dr) and (rd�).

Although this gives the right answer, it is not rigorous. A rigorous argument

that does not rely on rectangular coordinates follows.

Applying the cosine rule to side PP 0 of POP 0 (see �gure) we have

dl2 = r2(� + d�) + r2(�)� 2r(�)r(� + d�) cos(d�):

Expanding each term in a Taylor series up to the second order in d�, we get

dl2 = r2 + 2rdr

d�+

"�dr

d�

�2+ r

d2r

d�2

#d�2

+r2 � 2r

�r +

dr

d�+

1

2

d2r

d�2

��1� 1

2d�2

�:

Keeping terms up to second order in d� we have

dl2 =

"r2 +

�dr

d�

�2#d�2:

We can thus write the expression for the

length of a curve in polar coordinates

as follows: l =

Z �

�0

sr2 +

�dr

d�

�2d� =

Z �

�0

1

s2

ss2 +

�ds

d�

�2d�, where s = 1

r.

�

d�r

r� � rd� r d�

dl

dr

O

P

P 0

Exercise 3. Derive the polar expression for arc length from the cartesian

expression dl2 = dx2 + dy2.

Curvature

It seems tautological to say that curvature is an important feature of

curves, but the fact is that a planar curve is uniquely determined (up to trans-

lations and rotations) if its curvature is known as a function of arc length.

This is generally of little practical importance, since the resulting di�eren-

tial equations can only be solved if you know the solution! We will give the

formula for curvature in terms of s(�) = l=r(�) and some applications.

156

Curvature, �, can be de�ned as the rate of change of the direction of

the tangent line per unit arc length. We have

� =d�

ds=d�

ds� d�

d�=

s2qs2 +

�dsd�

�2 ��1 +

d(�� �)

d�

�:

Now, tan(�� �) =�s

(ds=d�), so that

d(�� �)

d�=

d�arctan �s

(ds=d�)

�d�

=1

1 + s2

(ds=d�)

���dsd�

�2+ s

�d2sd�2

�(ds=d�)2

giving

� =s2q

s2 +�dsd�

�2 �241 + � �ds

d�

�2+ s

�d2sd�2

�s2 +

�dsd�

�235

=s2q

s2 +�ds

d�

�2 �264s+

�d2sd�2

�2s2 +

�ds

d�

�2375

=s+

�d2sd�2

�h1 +

�12dsd�

�2i3=2 :

In terms or r, we have

� =r2 + 2(dr=d�)2 � r(d2r=d�2)

[r2 + (dr=d�)2]3=2:

Example 4. If curvature is zero we obtain the di�erential equation s+s00 = 0,

with general solution s = d cos(� � �0), that is, the equation of a straight

line.

Exercise 4. Check that the curvature at each point of a lemniscate is pro-

portional to the distance to the origin. (Hint: to simplify the algebra, divide

numerator and denominator by r3 in the curvature formula above.)

157

Area Enclosed by Polar Curves

d��max

O

r + dr

r

Q0

Q

P0

P

�

The last application of calculus is

the calculation of areas. The natu-

ral surface element is the `triangle'

de�ned by a segment of curve and

the radius vectors at the endpoints

(OPP 0 in the �gure). The area of

this triangle is, to �rst order in d�,

dA =1

2r2d�.

If the origin does not lie inside the curve the equation r = r(�) will have

more than one branch, as shown, and the sign of the `enclosed area' dA

depends on the orientation given to the curve. In the �gure the curve is

traversed counterclockwise, and so the outer branch (PP 0) has positive sign(the positive sense of d� coincides with the direction of the curve) and the

inner branch (QQ0) has negative sign (the positive sense of d� as opposed

to the direction of the curve). The same applies when calculating the area

enclosed by two intersecting curves.

Example 5. As our last example, we will evaluate the area enclosed by the

circle (r� R cos �)2 = �2 � R2 sin2 �. The area is given by

A =

Z sin �=�=R

sin�=��=R

1

2(r2+ � r2�)d�;

where r� = R cos � �p�2 � R2 sin2 �. We have

A =

Z1

2(r+ + r�)(r+ � r�)d� =

Z2(R cos �)

q�2 �R2 sin2 �d�:

Letting R sin � = � sin� and R cos �d� = � cos�d�, we have

A =

Z �=2

�=��=22�2 cos2 �d� = ��2

as expected.

158

A Pattern in Permutations

John Linnellstudent, University of Massachusetts

Boston, Massachusetts, USA

Let pn(k) be the number of permutations on n elements (say, n large ant-

eaters) with exactly k �xed points (i.e. a permutation which takes k elements

to themselves), so for example, p3(0) = 2, p3(1) = 3, p3(2) = 0, and

p3(3) = 1. It should be clear that

nXk=0

pn(k) = n!, but may be not so

obvious that

nXk=0

kpn(k) = n! (this was problem 1 on the 1987 IMO). It may

be even more surprising to learn that for n � 2,

nXk=0

k2pn(k) = 2n!. What

kind of pattern ensues? As my old analysis prof would no doubt say, \this is

good exercise," and it is kind of fun to follow a trail like this and see where

it leads.

Based on the above results, for n � 1 and t � 0, let

Q(n; t) =1

n!

nXk=0

ktpn(k):

Then Q(n;0) = Q(n; 1) = 1 for all n � 1 andQ(n; 2) = 2 for all n � 2 (not

n � 1, and we will see why soon). Our �rst conjecture would probably be

then that indeed each Q(n; t) is an integer. But we will have to see a little

more before we can prove anything.

Take n = 5. Then we can make the following table:

k p5(k) kp5(k) k2p5(k) k3p5(k) k4p5(k) k5p5(k) k6p5(k)

0 44 0 0 0 0 0 0

1 45 45 45 45 45 45 45

2 20 40 80 160 320 640 1280

3 10 30 90 270 810 2430 7290

4 0 0 0 0 0 0 0

5 1 5 25 125 625 3125 15625P120 120 240 600 1800 6240 24240P

=5! 1 1 2 5 15 52 202

Table 1.

159

Thus, Q(5;0) = 1, Q(5; 1) = 1, Q(5; 2) = 2, Q(5;3) = 5, and so on.

So far so good. We can make a second table with the actual Q(n; t) values:

nnt 0 1 2 3 4 5 6

1 1 1 1 1 1 1 1

2 1 1 2 4 8 16 32

3 1 1 2 5 14 41 122

4 1 1 2 5 15 51 187

5 1 1 2 5 15 52 202

6 1 1 2 5 15 52 203

Table 2.

Now we are getting somewhere. Notice how the rows seem to converge

to a single sequence of integers, with one new term kicking in with each row.

This sequence begins1, 1, 2, 5, 15, 52, 203, : : : . I could not �nd this sequence

in any of my references, so I did what any enterprising student would do. I

sent an e-mail to sequences@research.att.com, with \lookup 1 1 2 5 15

52 203" in the body. For those not familiar, it is an on-line sequence server

that tries to solve or match any sequence you might send it; I should also

add that they ask that you send at most one request per hour. Soon enough,

I had a response, which indicated that this was a sequence known as the Bell

numbers, which satisfy

B(0) = 1; B(n+ 1) =

nXk=0

�n

k

�B(k):

This recursion is striking, because if one looks at the second row of

table 2, it may remind one of the identity 2n =

nXk=0

�n

k

�. Could it be? Yes,

in fact the same recursion that generates the Bell numbers is what generates

successive rows of table 2. Now we really have something.

Claim. For all n � 1 and t � 0, Q(n+ 1; t+ 1) =

tXi=0

�t

i

�Q(n; i).

Proof. First, note that pn(k) =

�n

k

�pn�k(0) [why?]. Then

Q(n+ 1; t+ 1) =1

(n+ 1)!

n+1Xk=0

kt+1pn+1(k)

=1

(n+ 1)!

n+1Xk=0

kt+1

�n+ 1

k

�pn+1�k(0)

160

=n+ 1

(n+ 1)!

n+1Xk=1

n!

(k� 1)!(n+ 1� k)!

kt+1

kpn+1�k(0)

=1

n!

n+1Xk=1

kt�

n

k� 1

�pn+1�k(0)

=1

n!

n+1Xk=1

ktpn(k� 1)

=1

n!

nXk=0

(k+ 1)tpn(k)

=1

n!

n+1Xk=1

tXi=0

�t

i

�kipn(k)

=

tXi=0

�t

i

� 1

n!

nXk=0

kipn(k)

!

=

tXi=0

�t

i

�Q(n; i):

So we have proven quite a bit actually, including:

1. Each Q(n; t) is an integer (i.e.,

nXk=0

ktpn(k) is divisible by n!), and

2. For �xed t, Q(n; t) eventually becomes B(t) for su�ciently high n.

The claim looks complicated, but we know what we want to prove, and

it turns out to be just a little algebraic manipulation. So in the end, we have

a nice result from a simple observation.

161

IMO CORRESPONDENCE PROGRAM

Canadian students wishing to participate in this program should �rst

contact Professor Edward J. Barbeau, Department of Mathematics, Univer-

sity of Toronto, Toronto, Ontario. Please note that there is a fee for par-

ticipation in the program: $12. Please make the cheque payable to Edward

J. Barbeau.

PROBLEM SET 1

Algebra

1. Solve the system of equations

x2 + 2yz = x;

y2 + 2xz = z;

z2 + 2xy = y:

2. Let m be a real number. Solve, for x, the equation

jx2 � 1j+ jx2 � 4j = mx:

3. Let fx1; x2; : : : ; xn; : : : g be a sequence of nonzero real numbers. Show

that the sequence is an arithmetic progression if and only if, for each

integer n � 2,

1

x1x2+

1

x2x3+ � � �+ 1

xn�1xn=n� 1

x1xn:

4. Suppose that x and y are two unequal positive real numbers. Let

r =

�x2 + y2

2

�1=2g = (xy)1=2

a =x+ y

2h =

2xy

x+ y:

Which of the numbers r�a, a�g, g�h is largest and which is smallest?

5. Simplify

x3 � 3x+ (x2 � 1)px2 � 4� 2

x3 � 3x+ (x2 � 1)px2 � 4 + 2

to a fraction whose numerator and denominator are of the form upv

with u and v each linear polynomials. For which values of x is the

equation valid?

162

6. Prove or disprove: if x and y are real numbers with y � 0 and

y(y+ 1) � (x+ 1)2, then y(y� 1) � x2.

7. X is a collection of objects upon which the operation of addition, sub-

traction and multiplication are de�ned so as to satisfy the following

axioms:

(1) if x; y belong to X, then x+ y and xy both belong to X;

(2) for all x; y in X, x+ y = y+ x;

(3) for all x; y; z inX, x+ (y+ z) = (x+ y) + z and x(yz) = (xy)z;

(4) for all x; y; z in x, x(y + z) = xy + xz;

(5) there is an element 0 such that 0+x = x+0 = x and for each x inX,

there exists a unique element denoted by �x for which x+ (�x) = 0;

(6) x� y = x+ (�y) for each pair x; y of elements of X;

(7) x3 � x = x+ x+ x = 0 for x in X.

Note that these axioms do not rule out the possibility that the product

of two non-zero elements of X may be zero, and so it may not be valid

to cancel terms.

On X, we de�ne a relation � by the following condition:

x � y if and only if x2y � xy2 � xy + x2 = 0.

Prove that the following properties obtain:

(i) x � x for each element x of X;

(ii) if x � y and y � x, then x = y;

(iii) if x � y and y � z, then x � z.

8. Let n be a positive integer and suppose that u and v are positive real

numbers. Determine necessary and su�cient conditions on u and v

such that there exist real numbers a1; a2; : : : ; an satisfying

a1 � a2 � � � � � an � 0

u = a1 + a2 + � � �+ an

v = a21 + a22 + � � �+ a2n:

When such a representation is possible, determine the maximum and

minimum values of a1.

9. Suppose that x + 1

y= y + 1

z= z + 1

x= t, where x; y; z are not all

equal. Determine xyz.

10. Let a � 0. The polynomial x3 � ax + 1 has three distinct real roots.

For which values of a does the root u of least absolute value satisfy1a< u < 2

a?

163

11. Determine the range of values of cd subject to the constraints ab = 1,

ac+ bd = 2, where a, b, c, d are real.

12. Find polynomials p(x) and q(x) with integer coe�cients such that

p(p2 +

p3 +

p5)

q(p2 +

p3 +

p5)

=p2 +

p3:

Mayhem Problems

The Mayhem Problems editors are:

Cyrus Hsia Mayhem Advanced Problems Editor,Richard Hoshino Mayhem High School Problems Editor,Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only problems | the

next issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions to the new

problems in this issue be submitted by 1 August 1997, for publication in the

issue 5 months ahead; that is, issue 8. We also request that only students

submit solutions (see editorial [1997: 30]), but we will consider particularly

elegant or insightful solutions from others. Since this rule is only being im-

plemented now, you will see solutions from many people in the next few

months, as we clear out the old problems from Mayhem.

164

High School Problems

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,

Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>

There is a correction forH220; the expression 2n� T

Sshould be 2n� T

S.

H221. Let P = 195 +6605 +13165. It is known that 25 is one of the

forty-eight positive divisors of P . Determine the largest divisor of P that is

less than 10; 000.

H222. McGregor becomes very bored one day and decides to write

down a three digit number ABC, and the six permutations of its digits. To

his surprise, he �nds thatABC is divisible by 2,ACB is divisible by 3,BAC

is divisible by 4, BCA is divisible by 5, CAB is divisible by 6, and CBA is

a divisor of 1995. Determine ABC.

H223. There are n black marbles and two red marbles in a jar. One by

one, marbles are drawn at random out of the jar. Jeanette wins as soon as

two black marbles are drawn, and Fraserette wins as soon as two red marbles

are drawn. The game continues until one of the two wins. Let J(n) and F (n)

be the two probabilities that Jeanette and Fraserette win, respectively.

1. Determine the value of F (1) + F (2) + � � �+ F (3992).

2. As n approaches in�nity, what does J(2)� J(3)� J(4)� � � � � J(n)

approach?

H224. Consider square ABCD with side length 1. Select a point M

exterior to the square so that \AMB is 90�. Let a = AM and b = BM .

Now, determine the point N exterior to the square so that CN = a and

DN = b. Find, as a function of a and b, the length of line segmentMN .

N

C B

M

AD

b

a

165

Advanced Problems

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.

M1G 1C3 <hsia@math.toronto.edu>

A197. Calculate Z �

2

� �

2

sin(2N + 1)�

sin�d�;

where N is a non-negative integer.

A198. Given positive real numbers a, b, and c such that a+ b+ c = 1,

show that aabbcc + abbcca + acbacb � 1.

A199. Let P be a point inside triangle ABC. Let A0, B0, and C0 bethe re ections of P through the sides BC, AC, and AB respectively. For

what points P are the six points A, B, C, A0, B0, and C0 concyclic?

A200. Given positive integers n and k, for 0 � i � k � 1, let

Sn;k;i =X

j�i (mod k)

�n

j

�:

Do there exist positive integers n, k > 2, such that Sn;k;0, Sn;k;1, : : : ,

Sn;k;k�1 are all equal?

Challenge Board Problems

Editor: Ravi Vakil, Department of Mathematics, One Oxford Street,

Cambridge, MA, USA. 02138-2901 <ravi@math.harvard.edu>

There are no new Challenge Board Problems this month | we reprint

those from issue 1 this year [1997: 44]'

C70. Prove that the group of automorphisms of the dodecahedron

is S5, the symmetric group on �ve letters, and that the rotation group of

the dodecahedron (the subgroup of automorphisms preserving orientation)

is A5.

C71. Let L1, L2, L3, L4 be four general lines in the plane. Let pij be

the intersection of lines Li and Lj. Prove that the circumcircles of the four

triangles p12p23p31, p23p34p42, p34p41p13, p41p12p24 are concurrent.

C72. A �nite group G acts on a �nite set X transitively. (In other

words, for any x; y 2 X, there is a g 2 G with g � x = y.) Prove that there

is an element of G whose action on X has no �xed points.

166

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.

In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.

To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81

2"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed tothe Editor-in-Chief, to arrive no later than 1 November 1997. They may alsobe sent by email to crux-editors@cms.math.ca. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication.

2226. Proposed by K.R.S. Sastry, Dodballapur, India.

An old man willed that, upon his death, his three sons would receive

the u'th, v'th, w'th parts of his herd of camels respectively. He had uvw�1

camels in the herd when he died. Obviously, their sophisticated calculator

could not divide uvw � 1 exactly into u, v or w parts. They approached a

distinguished CRUX problem solver for help, who rode over on his camel,

which he added to the herd and then ful�lled the old man's wishes, and took

the one camel that remained, which was, of course, his own.

Dear CRUX reader, how many camels were there in the herd?

2227. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.

Evaluate Yp

" 1Xk=0

�2k

k

�(2p)2k

#:

where the product is extended over all prime numbers.

167

2228. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

Let A be the set of all real numbers from the interval (0; 1) whose dec-

imal representation consists only of 1's and 7's; that is, let

A =

( 1Xk=1

ak

10k: ak 2 f1; 7g

):

Let B be the set of all reals that cannot be expressed as �nite sums of mem-

bers of A. Find supB.

2229. Proposed by Kenneth Kam Chiu Ko, Mississauga, Ontario.

(a) Letm be any positive integer greater than 2, such that x2 � 1 (mod m)

whenever (x;m) = 1.

Let n be a positive integer. Ifmjn+1, prove that the sum of all divisors

of n is divisible bym.

(b)?

Find all possible values ofm

2230. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

Triangles BCD and ACE are constructed outwardly on sides BC and

CA of triangle ABC such that AE = BD and \BDC + \AEC = 180�.The point F is chosen to lie on the segment AB so that

AF

FB=DC

CE:

Prove thatDE

CD+ CE=

EF

BC=

FD

AC:

2231. Proposed byHerbert G �ulicher,WestfalischeWilhelms-Univer-sit�at, M�unster, Germany.

In quadrilateral P1P2P3P4, suppose that the diagonals intersect at the

point M 6= Pi (i = 1; 2; 3; 4). Let \MP1P4 = �1, \MP3P4 = �2,

\MP1P2 = �1 and \MP3P2 = �2.

Prove that

�13 :=jP1M jjMP3j

=cot�1 � cot�1

cot�2 � cot�2;

where the+(�) sign holds if the line segment P1P3 is located inside (outside)

the quadrilateral.

168

2232. Proposed by �Sefket Arslanagi�c, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina.

Find all solutions of the inequality:

n2 + n� 5 <

�n

3

�+

�n+ 1

3

�+

�n+ 2

3

�< n2 + 2n� 2; (n 2 N):

(Note: If x is a real number, then bxc is the largest integer not exceeding x.)2233. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,

Austria.

Let x, y, z be non-negative real numbers such that x+ y+ z = 1, and

let p be a positive real number.

(a) If 0 < p � 1, prove that

xp + yp+ zp � Cp

�(xy)p+ (yz)p+ (zx)p

�;

where

Cp =

(3p if p � log 2

log 3�log 2 ;2p+1 if p � log 2

log 3�log 2 :

(b)?

Prove the same inequality for p > 1.

Show that the constant Cp is best possible in all cases.

2234. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Given triangle ABC, its centroid G and its incentre I, construct, using

only an unmarked straightedge, its orthocentre H.

2235. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Triangle ABC has angle \CAB = 90�. Let �1(O;R) be the circum-

circle and �2(T; r) be the incircle. The tangent to �1 at A and the polar line

of A with respect to �2 intersect at S. The distances from S to AC and AB

are denoted by d1 and d2 respectively.

Show that

(a) STkBC,

(b) jd1 � d2j = r.

[For the bene�t of readers who are not familiar with the term \polar line", we

give the following de�nition as in, for example,Modern Geometries, 4th Edi-tion, by James R. Smart, Brooks/Cole, 1994:

The line through an inverse point and perpendicular to the line joiningthe original point to the centre of the circle of inversion is called the polar ofthe original point, whereas the point itself is called the pole of the line.]

169

2236. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Let ABC be an arbitrary triangle and let P be an arbitrary point in the

interior of the circumcircle of 4ABC. Let K, L, M , denote the feet of the

perpendiculars from P to the lines AB, BC, CA, respectively.

Prove that [KLM ] � [ABC]

4.

Note: [XY Z] denotes the area of 4XY Z.2237. Proposed by Meletis D. Vasiliou, Elefsis, Greece.

ABCD is a square with incircle �. Let ` be a tangent to �. Let A0, B0,C0, D0 be points on ` such that AA0, BB0, CC0, DD0 are all perpendicular

to `.

Prove that AA0 � CC0 = BB0 �DD0.

Correction

2173. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

Let n � 2 and x1; : : : ; xn > 0 with x1 + : : :+ xn = 1.

Consider the terms

ln =

nXk=1

(1 + xk)

s1� xk

xk

and

rn = Cn

nYk=1

1 + xkp1� xk

where

Cn = (pn� 1)n+1(

pn)n=(n+ 1)n�1:

[Ed: there is no x in the line above.]

1. Show l2 � r2.

2. Prove or disprove: ln � rn for n � 3.

170

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

1940. [1994: 108; 1995: 107; 1995: 205; 1996: 321] Proposed by JiChen, Ningbo University, China.

Show that if x; y; z > 0,

(xy + yz+ zx)

�1

(x+ y)2+

1

(y+ z)2+

1

(z + x)2

�� 9

4:

Solution by Marcin E. Kuczma, Warszawa, Poland.Let F be the expression on the left side of the proposed inequality.

Assume without loss of generality x � y � z � 0, with y > 0 (not excluding

z = 0), and de�ne:

A = (2x+ 2y � z)(x� z)(y� z) + z(x+ y)2;

B = (1=4)z(x+ y � 2z)(11x+ 11y+ 2z);

C = (x+ y)(x+ z)(y+ z);

D = (x+ y + z)(x+ y � 2z) + x(y � z) + y(x� z) + (x� y)2;

E = (1=4)(x+ y)z(x+ y+ 2z)2(x+ y� 2z)2:

It can be veri�ed that

C2(4F � 9) = (x� y)2�(x+ y)(A+ B + C) + (x+ z)(y+ z)D=2

�+ E:

This proves the inequality and shows that it becomes an equality only

for x = y = z and for x = y > 0, z = 0.

Comment.

The problem is memorable for me! It was my \solution" [1995: 107]

that appeared �rst. According to someone's polite opinion it was elegant,

but according to the impolite truth, it was wrong. I noticed the fatal error

when it was too late to do anything; the issue was in print already.

In [1995: 205] a (correct) solution by Kee-Wai Lau appeared. Mean-

while I found two other proofs, hopefully correct, and sent them to the ed-

itor. Like Kee-Wai Lau's, they required the use of calculus and were lacking

\lightness", so to say, so the editor asked [1995:206] for a \nice" solution.

I became rather sceptical about the possibility of proving the result by those

techniques usually considered as \nice", such as convexity/majorization ar-

guments | just because the inequality turns into equality not only for

x = y = z, but also for certain boundary con�gurations.

In response to the editor's prompt, VedulaMurty [1996: 321] proposed

a short proof avoiding hard calculations. But I must frankly confess that I do

171

not understand its �nal argument: I do not see why the sum of the �rst two

terms in [1996: 321(3)] must be non-negative. While trying to clarify that, I

arrived at the proof which I present here.

This proof can be called anything but \nice"! Decomposition into sums

and products of several expressions, obviously nonnegative, and equally ugly,

has the advantage that it provides a proof immediately understood and ver-

i�ed if one uses some symbolic calculation software (with some e�ort, the

formula can be checked even by hand). But the striking disadvantage of such

formulas is that they carefully hide from the reader all the ideas that must

have led to them; they take the \background mathematics" of the reasoning

away. In the case at hand I only wish to say that the equality I propose here

has been inspired by Murty's brilliant idea to isolate the polynomial that

appears as the third term in [1996: 321(3)] and to deal with the expression

that remains.

I once overheard a mathematician problemist claiming lack of sympathy

to inequality problems. In the ultimate end, he said, they all reduce to the

only one fundamental inequality, which is x2 � 0!

2124. [1996: 77] Proposed by Catherine Shevlin, Wallsend, England.

Suppose that ABCD is a quadrilateral with \CDB = \CBD = 50�

and \CAB = \ABD = \BCD. Prove that AD ? BC.

A B

C

D

I. Solutionby FlorianHerzig, student, Perchtoldsdorf, Austria. (Essentiallyidentical solutionswere submitted by Jordi Dou, Barcelona, Spain and HansEngelhaupt, Franz{Ludwig{Gymnasium, Bamberg, Germany. The solutionby Carl Bosley, student, Washburn Rural High School, Topeka, Kansas, USAwas very similar.)

Let F1 and F2 be the feet of the perpendiculars from D and A to BC

respectively. Let p = BC = CD and q = AC. Then, applying the Sine Rule

to 4ABC, we have

CF1 = p cos 80�; CF2 = q cos 70� =p sin 30�

sin80�=p cos 70�

2 sin 80�:

172

Thus we have

CF1

CF2=

cos80�cos 70�

2 sin 80�

=2 sin80� cos 80�

cos 70�=

sin160�

sin20�= 1:

Thus, F1 = F2, and this point is the intersection of AD and BC, whence

AD ? BC.

II. Solution by Federico Ardila, student, Massachusetts Institute ofTechnology, Cambridge, Massachusetts, USA.

Consider a regular 18{gon P1P2 : : : P18.

q

q

q

q

qq

q

q

q

q

q

q

q

q q

q

q

q

P1

P2

P3

P4

P5 P6P7

P8

P9

P10

P11

P12

P13P14P15

P16

P17

P18

Qq

We will �rst show that P1P10,

P2P12 and P4P15 concur.

By symmetry, P1P10, P4P15 and

P5P16 are concurrent. Thus it

is su�cient to prove that P1P10,

P2P12 and P5P16 are concurrent.

Using the angles version of Ceva's

theorem in triangle 4P1P5P12, itif su�cient to prove that

sin(\P1P12P2)

sin(\P2P12P5)� sin (\P12P5P16)sin(\P16P5P1)

� sin (\P5P1P10)

sin(\P10P1P12)= 1;

orsin(10�)

sin(30�)� sin(40

�)

sin(30�)� sin(50

�)

sin(20�)= 1:

But this is true since

sin 10� sin40� sin 50� = sin10� sin 40� cos 40�

= sin10��sin80�

2

�=

sin10� cos 10�

2

=sin20�

4= (sin30�)2 sin20�:

So, P1P10, P2P12 and P4P15 concur at, say, Q.

Using this, it is easy to check that

\P2P4Q = \P4QP2 = 50�;

and

\P2P1Q = \P4QP1 = \QP2P4 (= 80�):

This information clearly determines the quadrilateral P1P2P4Q up to simi-

larity, so P1P2P4Q � ACDB.

173

Since P1P4 ? P2Q, it follows that AD ? BC.

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina;SAM BAETHGE, Science Academy, Austin, Texas, USA; CHRISTOPHER J.BRADLEY, Clifton College, Bristol, UK; TIM CROSS, King Edward's School,Birmingham, England; CHARLES R. DIMINNIE, Angelo State University,San Angelo, TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford,Connecticut,USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA;PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba TonkaSchool of Mathematics, Rousse, Bulgaria; KEE-WAI LAU, Hong Kong;P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan;D.J. SMEENK, Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU,Athens, Greece; MELETIS VASILIOU, Elefsis, Greece (two solutions); andthe proposer.

The proposer writes: The genesis of this problem lies in a question

asked by Junji Inaba, student, William Hulme's Grammar School, Manch-

ester, England, inMathematical Spectrum, vol. 28 (1995/6), p. 18. He gives

the diagram in my question, with the given information:

\CDA = 20�; \DAB = 60�;\DBC = 50�; \CBA = 30�;

and asks the question: \can any reader �nd \CDB without trigonometry?"

In fact, such a solution was given in the next issue of Mathematics Spec-

trum by Brian Stonebridge, Department of Computer Science, University of

Bristol, Bristol, England.

The genesis of the diagram is much older, if one produces BD and AC to

meet at E. See Mathematical Spectrum, vol. 27 (1994/5), pp. 7 and 65{

66. In one reference, the question of �nding \CDA is called \Mahatma's

Puzzle", but no reference was available. Can any reader enlighten me on the

origin of this puzzle?

2125. [1996: 122] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

At Lake West Collegiate, the lockers are in a long rectangular array,

with three rows of N lockers each. The lockers in the top row are numbered

1 to N , the middle row N +1 to 2N , and the bottom row 2N +1 to 3N , all

from left to right. Ann, Beth, and Carol are three friends whose lockers are

located as follows:

174

: : : : : :

��@@

��@@

��@@

By the way, the three girls are not only friends, but also next-door

neighbours, with Ann's, Beth's, and Carol's houses next to each other (in that

order) on the same street. So the girls are intrigued when they notice that

Beth's house number divides into all three of their locker numbers. What is

Beth's house number?

Solution by Han Ping Davin Chor, student, Cambridge, MA, USA.

From the diagram, it can be observed that the lockers have numbers

x+ 3; N + x+ 5 and 2N + x;

where 1 � x � N , x a positive integer. Here locker x+3 is in the �rst row,

locker N +x+5 is in the second row, and locker 2N +x is in the third row.

Let y be Beth's house number, where y is a positive integer. Since y divides

into x+ 3, N + x+ 5 and 2N + x, y must divide into

2(N + x+ 5)� (2N + x)� (x+ 3) = 7:

Therefore y = 1 or 7. However, Beth's house is in between Ann's and

Carol's. Assuming that 0 is not assigned as a house number, it means that

Beth's house number cannot be 1 (else either Ann or Carol would have a

house number of 0). Therefore Beth's house number is 7.

Also solved by SAM BAETHGE, Science Academy, Austin, Texas, USA;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; TIM CROSS, KingEdward's School, Birmingham, England; CHARLES R. DIMINNIE, AngeloState University, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Lud-wig{Gymnasium, Bamberg, Germany; J. K. FLOYD, Newnan, Georgia, USA;IAN JUNE L. GARCES, Ateneo deManila University,Manila, the Philippines,and GIOVANNI MAZZARELLO, Ferrovie dello Stato, Firenze, Italy; SHAWNGODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIAN HERZIG,student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes,California, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, New York, USA;DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JOHN GRANTMCLOUGHLIN, Okanagan University College, Kelowna, B. C.; P. PENNING,Delft, the Netherlands; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Aus-tria; CORY PYE, student,Memorial University of Newfoundland, St. John's,Newfoundland; JOEL SCHLOSBERG, student, Hunter College High School,New York NY, USA; ROBERT P. SEALY, Mount Allison University, Sackville,New Brunswick; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DAVIDSTONE, Georgia Southern University, Statesboro, Georgia, USA; EDWARD

175

T.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M.WILKE, Topeka, Kansas, USA; and the proposer.

Two solvers eliminated 1 as a possible answer, because the problemsaid that the girls were \intrigued" that Beth's house number divided alltheir locker numbers, which would hardly be likely if Beth's house numberwere just 1! Thus they didn't need the information about the location ofBeth's house at all. Another solver, to whom the editor has therefore giventhe bene�t of the doubt,merely stated that \the locationof Ann's and Carol'shouses doesn't enter into the problem".

2126. [1996: 123] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

At Lake West Collegiate, the lockers are in a long rectangular array, with

three rows ofN lockers each, whereN is some positive integer between 400

and 450. The lockers in the top row were originally numbered 1 to N , the

middle row N + 1 to 2N , and the bottom row 2N + 1 to 3N , all from left

to right. However, one evening the school administration changed around

the locker numbers so that the �rst column on the left is now numbered 1

to 3, the next column 4 to 6, and so forth, all from top to bottom. Three

friends, whose lockers are located one in each row, come in the next morning

to discover that each of them now has the locker number that used to belong

to one of the others! What are (were) their locker numbers, assuming that

all are three-digit numbers?

Solution by Ian June L. Garces, Ateneo de Manila University, Manila,the Philippines, and GiovanniMazzarello, Ferrovie dello Stato, Firenze, Italy.

The friends' locker numbers are 246, 736 and 932.

To show this, �rst consider any particular locker. Then the original

(before the change) number of this locker can be written as iN + j, where

0 � i � 2 (the row) and 1 � j � N (the column). With respect to this

original locker number, this particular locker has a new (after the change)

number 3(j � 1) + (i+ 1) = 3j + i� 2.

Consider now the three friends' lockers. Since the three lockers are

located one in each row, we can let them be j1, N + j2 and 2N + j3 where

1 � j1; j2; j3 � N . For each of these lockers, the corresponding new locker

numbers will be 3j1�2, 3j2�1 and 3j3. Then there will be two possibilities

for how their original locker numbers and their new locker numbers were

\properly" interchanged:

Possibility 1. The �rst possibility is when

j1 = 3j3; (1)

N + j2 = 3j1 � 2; (2)

176

2N + j3 = 3j2 � 1: (3)

Substituting (1) into (2) and solving for j2, we have j2 = 9j3 � 2 � N .

Substituting this last equality into (3) and solving for j3, we have

j3 =5N + 7

26

which implies that N � 9 mod 26. Choosing N between 400 and 450, we

have the uniqueN = 425 and thus j3 = 82, j2 = 311 and j1 = 246. Hence

the original locker numbers are 246, 736 and 932 which, after the change,

will respectively be 736, 932 and 246 which satisfy what we want.

Possibility 2. The other possibility is when

j1 = 3j2� 1; N + j2 = 3j3; 2N + j3 = 3j1 � 2:

Similar computation as in Possibility 1 yields N = 425, j2 = 115, j3 = 180

and j1 = 344. But this means that one of the lockers will have number 1030

which is contrary to the assumption.

Therefore, the only possible locker numbers of the three friends are

246, 736 and 932.

Also solved by SAM BAETHGE, Science Academy, Austin, Texas, USA;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JOSEPHCALLAGHAN, student, University of Waterloo, Waterloo, Ontario; HANPING DAVIN CHOR, student, Cambridge, MA, USA; TIM CROSS, King Ed-ward's School, Birmingham, England; CHARLES R. DIMINNIE, Angelo StateUniversity, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gym-nasium, Bamberg, Germany; SHAWNGODIN, St. Joseph ScollardHall, NorthBay, Ontario; RICHARD I. HESS, Rancho Palos Verdes, California, USA;PETER HURTHIG, Columbia College, Burnaby, BC; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; KATHLEEN E. LEWIS, SUNYOswego, Oswego, New York, USA; DAVID E. MANES, SUNY at Oneonta,Oneonta, NY, USA; P. PENNING, Delft, theNetherlands; GOTTFRIED PERZ,Pestalozzigymnasium, Graz, Austria; ROBERT P. SEALY, Mount Allison Uni-versity, Sackville, New Brunswick; DAVID STONE, Georgia Southern Uni-versity, Statesboro, Georgia, USA; and the proposer.

Many solvers mentioned that the other set of locker numbers arisingfrom the problem is 344, 540 and 1030. Some remarked that the value ofNwas 425 in both cases. However, apparently nobody noticed that these twotriples of numbers enjoy a curious relationship:

246 + 1030 = 736 + 540 = 932 + 344 !

So now readers are challenged to �gure out why this relationship is true.

When N = 425, the problem says that the numbers 246; 736;932 areinterchanged when the lockers are renumbered. So let's call this set of num-bers a \swapset" for N = 425; that is, for a particular N , a swapset is

177

any set of numbers which get swapped among each other by the renumber-ing. We want true swapping; so we don't allow the sets f1g or f3Ng (orthe \middle" locker f(3N+1)=2g whenN is odd), which are obviously un-changed by the renumbering, to be in swapsets. Lots of problems concerningswapsets could be looked at. For example, one of the solvers (Stone) pointsout that there are no swapsets of two numbers when N = 425, but thereare when N = 427: lockers 161 and 481 get swapped. Which values of Nhave swapsets of size two? Here's another problem. It's clear that the setof all numbers from 1 to 3N , minus the two or three numbers that stay thesame, will be a swapset for every N . But are there any numbers N whichhave no other swapsets? If so, can you describe all suchN?

2127. [1996: 123] Proposed by Toshio Seimiya, Kawasaki, Japan.ABC is an acute triangle with circumcentre O, andD is a point on the

minor arc AC of the circumcircle (D 6= A;C). Let P be a point on the side

AB such that \ADP = \OBC, and let Q be a point on the side BC such

that \CDQ = \OBA. Prove that\DPQ = \DOC and\DQP = \DOA.

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.First I prove that B is an excentre of4PDQ.

\ABC = 180� � \ADC

= 180� � (\ADP + \CDQ+ \PDQ)

= 180� � (\CBO+ \ABO + \PDQ)

= 180� � \ABC � \PDQ; (1)

) \ABC = 90� � \PDQ

2;

\PDB = \ADB � \ADP = \ACB � \OCB = \ACO;

and

\QDB = \CDB � \CDQ = \CAB � \OAB = \CAO:

Since 4OAC is isosceles, we have that \PDB = \QDB and thus BD is

the internal angle bisector of \PDQ. (2)

What is more, we know that, in any4XY Z, the excentre, M , (whose

excircle touches Y Z), is exactly the point on the internal angle bisector of

\Y XZ outside the triangle for which

\YMZ = 180� � \MZY � \MY Z

=\Y

2+\Z

2=

180� � \X

2= 90� � \X

2:

Therefore B is an excentre of4PDQ because of (1) and (2). Then BP

and BQ are the external angle bisectors of \DPQ and \DQP , respectively,

whence

\APD = \BPQ and \CQD = \BQP: (3)

178

Starting with

\BOC = 2\BDC

we obtain

180� � 2\OBC = 2\BDC;

90� � \OBC = \BDC;

180� � \BDC = 90� + \OBC;

\BCD + \DBC = 90� + \ADP;

(180� � \DAP ) + \DBC = 90� + (180� � \DAP � \APD);

\DBC = 90� � \APD;

\DOC = 180� � (\APD + \BPQ)

[ because of (3) ]

= \DPQ;

and analogously \DOA = \DQP:

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; HAN PING DAVIN CHOR, student, Cambridge, MA, USA; P.PENNING, Delft, the Netherlands; WALDEMAR POMPE, student, Univer-sity of Warsaw, Poland; D.J. SMEENK, Zaltbommel, the Netherlands; andthe proposer.

2128. [1996: 123] Proposed by Toshio Seimiya, Kawasaki, Japan.ABCD is a square. Let P and Q be interior points on the sides BC

and CD respectively, and let E and F be the intersections of PQ with AB

and AD respectively. Prove that

� � \PAQ+ \ECF <5�

4:

Solution by Heinz-J �urgen Sei�ert, Berlin, Germany.In cartesian coordinates, let A = (0; 0), B = (1; 0), C = (1; 1),

D = (0; 1), P = (1; p) and Q = (q;1), where 0 < p; q < 1:

Then E =�1�pq1�p ; 0

�and F =

�0; 1�pq

1�q�, tan\PAB = p, tan\DAQ = q,

tan\DCF = FD =q(1�p)1�q , tan\BCE = BE =

p(1�q)1�p .

Since

\PAQ =�

2� \PAB � \DAQ and \ECF =

�

2+ \DCF + \BCE;

it follows that

\PAQ+ \ECF

= � + arctanq(1 � p)

1� q� arctan q + arctan

p(1 � q)

1� p� arctan p

= � + arctan

�(1� pq)(p� q)2

(1� p)(1� q)(1� pq)2 + (p(1 � q)2 + q(1� p)2)(p+ q)

�

179

by the addition formula for arctangents. Since 0 < p; q < 1, it su�ces to

show that

0 � (1� pq)(p� q)2 < (p(1� q)2 + q(1� p)2)(p+ q):

The left inequality is obviously true, while the right follows from the identity

(p(1�q)2+q(1�p)2)(p+q) = (1�pq)(p�q)2+2pq((1�p)2+(1�q)2):

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina; NIELS BEJLEGAARD, Stavanger, Norway;FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; JOSEPHCALLAGHAN, student, University of Waterloo; RICHARD I. HESS, RanchoPalos Verdes, California, USA; VICTOR OXMAN, University of Haifa, Haifa,Israel; and the proposer.

2130. [1996: 123] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

A and B are �xed points, and ` is a �xed line passing through A. C is a

variable point on `, staying on one side of A. The incircle of�ABC touches

BC at D and AC at E. Show that lineDE passes through a �xed point.

SolutionbyMitko Kunchev, Baba Tonka School ofMathematics, Rousse,Bulgaria.

We choose the point P on ` with AP = AB. Let C be an arbitrary

point of `, di�erent from P but on the same side ofA. The incircle of4ABCtouches the sides BC, AC, AB in the points D, E, F respectively. Let

ED meet PB in the point Q. According to Menelaus' Theorem applied to

4CBP and the collinear points E, D, Q, we get

PE

EC� CDDB

� BQQP

= 1: (1)

We have EC = CD (because they are tangents from C). Similarly,

AF = AE, so that FB = EP (since AB = AP ). But also, FB = DB, so

that DB = PE. Setting EC = CD and DB = PE in (1), we conclude that

BQ = QP ; therefore Q is the mid-point of BP . Hence the line DE passes

through the �xed point Q.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHERJ. BRADLEY, CliftonCollege, Bristol, UK; FLORIANHERZIG, student, Perch-toldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan;PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA (two sol-utions); and the proposer.

180

Seimiya and Yiu used the same argument as Kunchev. Seimiya men-tions that the result is easily shown to hold also when C coincides with P(even though the featured argument breaks down). Yiu extends the resultto include excircles: The line joining the points where an excircle touches the

segment BC and the line ` also passes through Q.

2131. [1996: 124] Proposed by Hoe Teck Wee, Singapore.

Find all positive integers n > 1 such that there exists a cyclic permuta-

tion of (1; 1; 2; 2; : : : ; n; n) satisfying:

(i) no two adjacent terms of the permutation (including the last and �rst

term) are equal; and

(ii) no block of n consecutive terms consists of n distinct integers.

Solution by the proposer.

It is clear that 2 does not have the desired property.

Suppose 3 has the speci�ed property. So there exists a permutation of

(1; 1; 2; 2; 3; 3) satisfying the two conditions. WLOG assume that the �rst

term is 1. From (ii) we know that the second term is not 1, say it is 2.

From (i) the third term must be 1. From (i) and (ii) the fourth term must

be 2. This leaves the two 3s as the last two terms, contradicting (i).

Suppose 4 has the speci�ed property. So there exists a permutation of

(1; 1; 2; 2; 3; 3; 4; 4) satisfying the two conditions. Arrange these eight (per-

muted) numbers in a circle in that order so that they are equally spaced. Then

the two conditions still hold. Now consider any four consecutive numbers

on the circle. If they consist of only two distinct integers, we may assume

by (i) that WLOG these four numbers are 1; 2; 1; 2 in that order, and that the

other four numbers are 3; 4; 3; 4. Then (ii) does not hold. If they consist of

three distinct integers, by (i) and (ii) we may assume WLOG that these four

numbers are (a) 1; 2; 3; 1 or (b) 1; 2; 1; 3 or (c) 1; 2; 3; 2, in these orders. By

reversing the order, (c) reduces to (b). Next consider (a). If the next number

is 2, then by (ii) we have 1; 2; 3; 1; 2; 3, and the two 4s are adjacent, contra-

dicting (i). If the next number is 3, reversing the order to obtain 3; 1; 3; 2; 1

reduces it to (b). Finally consider (b). By (i) and (ii) the next number must

be 2, followed by 3, so the two 4s are adjacent, contradicting (ii).

Next consider the following permutation for n > 4:

(4; 5; : : : ; n; 1; 2; 3; 2; 3; 4; 5; : : : ; n):

Clearly, (i) is satis�ed. (ii) follows from the fact that there does not exist a

set of four consecutive terms which is a permutation of (1;2; 3; 4).

In conclusion, the answer is: n > 4.

181

Also solved by HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bam-berg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;RICHARD I. HESS, Rancho Palos Verdes, California, USA; and DAVID E.MANES, SUNY at Oneonta, Oneonta, NY, USA. There was one incompletesolution.

2132. [1996: 124] Proposed by �Sefket Arslanagi �c, Berlin, Germany.

Let n be an even number and z a complex number.

Prove that the polynomial P (z) = (z + 1)n � zn � n is not divisible by

z2 + z + n.

I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.

Let Q(z) = z2+ z+n. For n = 0 or 1, we have that P (z) = 0, which

is clearly divisible by Q(z). For any n > 1, suppose that P (z) is divisible

by Q(z). Then Q(n) divides P (n).

ButQ(n) = n(n+2)� 0 (mod n), whileP (n) = (n+1)n�n2�n �1 (mod n). Thus P (z) is not divisible by Q(z).

II. Composite solution by F.J. Flanigan, San Jose State University, SanJose, California, USA and Edward T.H. Wang, Wilfrid Laurier University, Wa-terloo, Ontario.

Let D(z) = z2 + z+ n. If n = 0; 1, then P (z) = 0, which is divisible

by D(z). If n = 2, then P (z) = 2z � 1, which is clearly not divisible by

z2 + z + 2.

For n > 2, suppose thatD(z) dividesP (z). Then, sinceD(z) is monic,

P (z) = Q(z)D(z), where Q(z) is a polynomial of degree n� 3 with integer

coe�cients. Thus P (0) = Q(0)D(0), or 1 � n = nQ(0), which is clearly

impossible.

III. Solution and generalization by Heinz-J �urgen Sei�ert, Berlin, Ger-many.

Let n � 2 be an even integer. We shall prove that if a, b, c, are complex

numbers such that a 6= 0, then the polynomial

P (z) = (z + b)n� zn � a

is not divisible by z2 + bz + c.

The proposer's result, which does not hold for n = 0, is obtained when

a = c = n and b = 1.

Let z1 and z2 denote the (not necessarily distinct) roots of z2+ bz+ c.

The z1+ z2 = �b, so that P (z1) = zn2 � zn1 � a, and P (z2) = zn1 � zn2 � a.

Since P (z1) + P (z2) = �2a 6= 0, our result follows.

The example (z+ 1)6� z6 = (z2+ z+1)(6z3+9z2+ 5z+1) shows

that the condition a 6= 0 cannot be dropped.

182

Also solved by: CHARLES R. DIMINNIE, Angelo State University, SanAngelo, TX, USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Con-necticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAILAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA;NORVALD MIDTTUN (two solutions), Royal Norwegian Naval Academy,Norway; ROBERT P. SEALY,MountAllisonUniversity, Sackville, New Bruns-wick; DAVID R. STONE, Georgia Southern University, Statesboro, Georgia,USA; and the proposer.

Besides the solvers listed in Solutions I and II above, only Janous ob-served and showed that the assertion holds for all n � 2.

2134?. [1996: 124] Proposed by Waldemar Pompe, student, Univer-sity of Warsaw, Poland.

Let fxng be an increasing sequence of positive integers such that the

sequence fxn+1 � xng is bounded. Prove or disprove that, for each inte-

ger m � 3, there exist positive integers k1 < k2 < : : : < km, such that

xk1 ; xk2 ; : : : ; xkm are in arithmetic progression.

Solution by David R. Stone, Georgia Southern University, Statesboro,Georgia, USA, and Carl Pomerance, University of Georgia, Athens, Georgia,USA.

An old and well-known result of van der Waerden [4] is that if the

natural numbers are partitioned into two subsets, then one of the subsets has

arbitrarily long arithmetic progressions. It is not very di�cult to show [1]

that van der Waerden's theorem has the following equivalent formulation:

for every number B and positive integer m, there is a number

W (m;B) such that if n � W (m;B) and 0 < a1 < a2 < : : : <

an are integers with each ai+1� ai � B, thenm of the ai's form

an arithmetic progression.

Thus, for the problem as stated, if we let B be the bound on the di�er-

ences xn+1 � xn, then for any given m � 3, there exists a W (m;B) with

the property stated above. Then, for any n � W (m;B), any �nite sub-

sequence of length n will have an arithmetic progression of length m as a

sub-subsequence. That is, the original sequence contains in�nitely many

arithmetic progressions of lengthm.

In 1975, Sz �emeredi [3] proved a conjecture of Erd }os and Tur �an which im-

proves on van der Waerden's Theorem, relaxing the condition that the se-

quence's di�erences have a uniform upper bound, requiring only that the

sequence have a positive upper density. Hence the problem posed here also

follows from the theorem of Sz �emeredi, who, we believe, received (for this

result) the highest cash prize ever awarded by P �al Erd }os | $1,000.

183

Comment by the solvers.Do we know how Pompe became interested in this problem?

References

[1] T.C. Brown, Variations on van der Waerden's and Ramsey's theorems,

Amer. Math. Monthly 82 (1975), 993{995.

[2] Carl Pomerance, Collinear subsets of lattice point sequences| an analog

of Szemeredi's Theorem, J. Combin. Theory 28 (1980), 140{149.

[3] E. Szemeredi, On sets of integers containing no k elements in arithmetic

progression, Acta Arith. 27 (1975), 199{245.

[4] B.L. van der Waerden, Beweis einer Baudetschen Vermutung, Nieuw.Arch. Wisk. 15 (1928), 212{216.

Also solved by THOMAS LEONG, Staten Island, NY, USA; and JOELSCHLOSBERG, student, Hunter College High School, New York NY, USA;both using van der Waerden's theorem or its variation. Leong gave the ref-erence: Ramsey Theory by R.L. Graham, B.L. Rothschild and J.H. Spencer.Schlosberg remarked that van der Waerden's theorem was discussed in theJuly 1990 issue of Scienti�c American.

The proposer showed that van der Waerden's theorem follows easilyfrom the statement of his problem. His intention (and hope) was to �nd aproof independent of van der Waerden's theorem. This would establish anew \proof" of the latter. In view of his comment and the solution above, itshould be obvious that the two statements are equivalent, and hence such aproof is unlikely.

2135. [1996: 124] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.

Let n be a positive integer. Find the value of the sum

bn=2cXk=1

(�1)k(2n� 2k)!

(k+ 1)!(n� k)!(n� 2k)!:

Solution by Florian Herzig, student, Perchtoldsdorf, Austria. [Modi-�ed slightly by the editor.]

Let Sn denote the given summation. Note that S1 is an \empty" sum,

which we shall de�ne to be zero. We prove that Sn = ��

2n

n+ 2

�.

Since

�2

3

�= 0, this is true for n = 1. Assume that n � 2. Follow-

ing standard convention, for k = 0, 1, 2, : : : , let�xk�(f(x)) denote the

184

coe�cient of xk in the series expansion of the function f(x). Let

P (x) =�1� x2

�n+1(1� x)�(n+1):

Then, by the binomial expansion, its generalization (by Newton), and the

well-known fact that

��nk

�= (�1)k

�n+ k� 1

k

�, we have:

�x2k+2

� ��1� x2

�n+1�

=h�x2�k+1

i ��1� x2

�n+1�

= (�1)k+1

�n+ 1

k+ 1

�;

for k = �1; 0; 1; 2; : : : , and

�xn�2k

� �(1� x)

�(n+1)�

= (�1)n�2k��n� 1

n� 2k

�

=

�2n� 2k

n� 2k

�

for k � n=2. Hence�xn+2

�(P (x)) =

�x2k+2 � xn�2k� (P (x))

=

bn=2cXk=�1

(�1)k+1

�n+ 1

k+ 1

��2n� 2k

n� 2k

�:

On the other hand, since P (x) = (1 + x)n+1, we have�xn+2

�(P (x)) = 0.

Thereforebn=2cXk=�1

(�1)k+1

�n+ 1

k + 1

��2n� 2k

n� 2k

�= 0:

Since

Sn = � 1

n+ 1

bn=2cXk=1

(�1)k+1

�n+ 1

k + 1

��2n� 2k

n� 2k

�;

we get

Sn = � 1

n+ 1

��n+ 1

1

��2n

n

���n+ 1

0

��2n+ 2

n+ 2

��

=(2n+ 2)!

(n+ 1)(n+ 2)!n!� (2n)!

n!n!

=f2(2n+ 1)� (n+ 2)(n+ 1)g

(n+ 2)!n!� (2n)!

=�n(n� 1) (2n)!

(n+ 2)!n!=

�(2n)!

(n+ 2)!(n� 2)!

= ��

2n

n+ 2

�:

185

Also solved by WALTHER JANOUS, Ursulinengymnasium, Innsbruck,Austria; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the proposer. Thecorrect answer, without a proof, was sent in by RICHARD I. HESS, RanchoPalos Verdes, California, USA.

If, in the given summation, one lets k start from zero (this was, in fact,the proposer's original idea), then it is easy to see that the answer becomes

1

n+ 2

�2n+ 2

n+ 1

�;

the (n+ 1){th Catalan number.

2136. [1996: 124] Proposed by G.P. Henderson, Campbellcroft, On-tario.

Let a; b; c be the lengths of the sides of a triangle. Given the values of

p =Pa and q =

Pab, prove that r = abc can be estimated with an error

of at most r=26.

Solution by P. Penning, Delft, the Netherlands.Scale the triangle down by a factor (a+ b+ c). The value of p then be-

comes 1, the value of q becomes Q =q

(a+ b+ c)2, and R =

r

(a+ b+ c)3.

Introduce s =a+ b

2and v =

a� b

2:

a = s+ v; b = s� v; c = 1� 2s;

Q = 2s� 3s2 � v2; R = (1� 2s)(s2� v2):

Since a, b, c, represent the sides of a triangle, we must require

0 < c < a+ b and � c < a� b < c:

[Ed: in other words, the triangle is not degenerate | a case which must be

discarded as inappropriate.]

This translates to

1

4< s <

1

2and jvj < 1

2� s:

Lines of constant Q are ellipses in the s{v plane, with centre s = 13, v = 0.

So we write:

s =1

3+ A cos(x); v =

p3A sin(x);

with A =

p(1� 3Q)

3replacing Q.

Very symmetrical expressions are now obtained for a, b, c:

a =1

3�2A cos(120�+x); b =

1

3�2A cos(120��x); c =

1

3�2A cos(x):

186

R = abc� 1

27� A2 � 2A3 cos(3x):

Now, R is minimal for x = 0:

Rmin =1

27� A2 � 2A3:

R is maximal for x = 60� provided that a � 12, A � 1

12:

Rmax =1

27� A2 + 2A3:

For 112� A � 1

6, we have cos(120�+ xmax) = � 1

12A, since the maximum of

a is 12. So

cos(3xmax) = � 4

(12A)3+

3

(12A);

Rmax =1

27� A2 � 2A3

�� 4

(12A)3+

3

(12A)

�=

1

24� 3A2

2:

We must determine the reciprocal of the relative spread in R:

F =Rmax +Rmin

Rmax �Rmin

:

For A � 112, we have

F =

127� A2

2A3:

The minimum in F is reached at A = 112, so that Fmin = 26.

For 1

12� A � 1

6, both Rmax and Rmin are zero for A = 1

6. So

Rmin =

�1

6� A

��2

9+

4A

3+ 2A2

�;

Rmax =

�1

6� A

��1 + 6A

4

�:

The minimum in F is also at A = 1

6and yields the same value for Fmin = 26.

Also solved by NIELS BEJLEGAARD, Stavanger, Norway; and the pro-poser. One incorrect submission was received in that the sender assumedthat a degenerate triangle disproved the proposition.

187

2137. [1996: 124, 317; 1997: 48] Proposed by Aram A. Yagubyants,Rostov na Donu, Russia.

Three circles of (equal) radius t pass through a point T , and are each

inside triangle ABC and tangent to two of its sides. Prove that:

(i) t =rR

R+ r, (ii) T lies on the line segment joining the centres

of the circumcircle and the incircle of 4ABC.

Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.

A B

C

X Y

Z

I

T

r r

r

r

r

r

rr

We denote the centres of the three

circles by X, Y and Z. Since the

three circles pass through a common

point T and have equal radius t, it

follows that X, Y and Z lie on the

circle with centre T and radius t.

Since each of the circles is tangent to

two sides of 4ABC, it follows that

X, Y andZ lie on the internal bisec-

tors of \A, \B and\C. SinceAB is

a common tangent of two intersect-

ing circles with radius t, it follows

that ABkXY , and analogously,

we have Y ZkBC and ZXkAC.

This implies that the lines AX, BY and CZ are bisectors of the angles of

4XY Z as well, and so 4ABC and4XY Z have the same incentre I.

Thus we conclude that triangles4ABC and4XY Z are homothetic with I

as centre of similitude. This implies that:

(i) the ratio of the radii of the circumcircles of4ABC and4XY Z equals

the ratio of the radii of the incircles of the triangles; that is

R : t = r : (r � t) Rr � Rt = rt

t(R+ r) = Rr t =Rr

R+ r;

(ii) as corresponding points in the homothety, T (the circumcentre of

4XY Z) and the circumcentre of 4ABC lie collinear with I, as de-

sired.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina; NIELS BEJLEGAARD, Stavanger, Norway;FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol,UK; HAN PINGDAVINCHOR, student, Cambridge, MA, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtolds-dorf, Austria; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,USA;

188

P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan;D.J. SMEENK, Zaltbommel, the Netherlands; PAUL YIU, Florida AtlanticUniversity, Boca Raton, Florida, USA; and the proposer.

Janous has seen both parts of the problem before; although unable toprovide a reference to part (i), he reconstructed the argument that he hadseen, which was much like our featured solution. He, among several others,noted that (ii) is essentially problem 5 of the 1981 IMO [1981: 223], solutionon pp. 35{36 ofM.S. Klamkin, International Mathematical Olympiads 1979{

1985, MAA, 1986. See also the \generalization" 694 [1982: 314] and therelated problem 1808 [1993: 299].

2138. [1996: 169] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

ABC is an acute angle triangle with circumcentre O. AO meets the

circle BOC again at A0, BO meets the circle COA again at B0, and CO

meets the circle AOB again at C0.Prove that [A0B0C0] � 4[ABC], where [XY Z] denotes the area of

triangle XY Z.

Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.There is an even sharper inequality:

[A0B0C0] � 3 3

vuut Ycyclic

cos2(B � C)

sinA sin 2A[ABC]:

For this, we �rst represent [A0B0C0] as a function of A, B, C and R

(the circumradius).

A B

A0

C

O

We have: \AOB = 2C, so that

\BOA0 = 180� � 2C and \BA0O =

\BCO = 90� �A.

[Both angles subtend the line BO on

circle BOC!] Thus,

\A0BO

= 180� � (180� � 2C)� (90� � A)

= A+ 2C � 90�

= 180� � B +C � 90�

= 90� � (B � C):

Hence, using the law of sines in4OBA0, we get

jOA0jsin(90� � (B � C))

=R

sin(90� �A);

that is, jOA0j = R cos(B � C)

cosA.

189

Similarly, jOB0j = R cos(C � A)

cosBand jOC0j = R cos(A�B)

cosC.

Now, since \A0OB0 = \AOB = 2C, we get, via the trigonometric area

formula of triangles, that

[A0B0O] = 1

2jOA0j � jOB0j � sin(\A0OB0)

=R2

2

cos(B � C) cos(C � A)

cosA cosBsin 2C;

and similarly for [B0C0O] and [C0A0O]. Thus

[A0B0C0] = [A0B0O] + [B0C0O] + [C0A0O]

=

0@ Xcyclic

cos(C � A) cos(C � B)

cosA cosBsin2C

1A� R2

2: (1)

Next, we recall the formula

[ABC] = 2R2Ycyclic

sinA (2)

From (1), we get, via the arithmetic-geometric-mean inequality:

Xcyclic

cos(C � A) cos(C �B)

cosA cosBsin2C

� 3

24 Ycyclic

�cos2(B �C)

cos2A� sin 2A

�3513

= 6

24 Ycyclic

cos2(B � C)

cosA� sinA

3513

= 6

24 Ycyclic

cos2(B � C)

sin2A cosA

3513

�Ycyclic

sinA

= 12

24 Ycyclic

cos2(B � C)

sinA sin 2A

3513

�Ycyclic

sinA;

so that, using (1) and (2),

[A0B0C0] � 3

24 Ycyclic

cos2(B � C)

sinA sin2A

3513

� [ABC]

190

as claimed.

Finally, we recall the angle inequality:

Ycyclic

cos2(B �C) � 512

27

Ycyclic

�sin2A cosA

� 24= 64

27

Ycyclic

(sinA sin2A)

35

which is valid for all triangles (but interesting only for acute triangles) with

equality if and only if A = B = C = 60�, or the degenerate cases with two

of A, B, C being right angles. This immediately yields

24 Ycyclic

cos2(B � C)

sinA sin2A

3513

� 3

r64

27=

4

3;

and the original inequality follows.

Also solved by D.J. SMEENK, Zaltbommel, the Netherlands.

2139. [1996: 169, 219] Proposed by Waldemar Pompe, student, Uni-versity of Warsaw, Poland.

Point P lies inside triangle ABC. Let D, E, F be the orthogonal pro-

jections from P onto the lines BC, CA, AB, respectively. Let O0 and R0

denote the circumcentre and circumradius of the triangleDEF , respectively.

Prove that

[ABC] � 3p3R0

qR02 � (O0P )2;

where [XY Z] denotes the area of triangle XY Z.

Solution by the proposer.

Let C denote the circumcircle of DEF . Let P 0 be the symmetric point

to P with respect to O0. Let E be the ellipse with foci P and P 0 tangent(internally) to C. The diameter of the ellipse E is 2R0, and its area is equal

to �R0qR02 � (O0P )2. Since the locus of the orthogonal projections from

P onto tangents to the ellipse E is the circle C, the sides of ABC must be

tangent to E. Thus E is inscribed in the triangle ABC. Let L be an a�ne

mapping which takes E to some circle of radiusR, and let it take the triangle

ABC to the triangleA0B0C0. Since L preserves the ratio of areas, we obtain

[ABC]

�R0qR02 � (O0P )2

=[ABC]

area of E =[A0B0C0]

�R2� 3

p3R2

�R2; (1)

since among all triangles circumscribed about the given circle, the one of

smallest area is the equilateral triangle. Thus (1) is equivalent to the desired

inequality, so we are done.

191

Remarks: The same proof works also for an n-gon which has an inte-

rior point whose projections onto the sides of the n-gon are concyclic. The

analogous inequality will be

[A1 : : : An] � n tan

��

n

�R0qR02 � (O0P )2 :

Note that as a special case, when the n-gon has the incircle and P = O0 weobtain the well-known result that among all n-gons circumscribed about agiven circle, the one of smallest area is the regular one, though it is used in

the proof.

2140. [1996: 169] Proposed by K.R.S. Sastry, Dodballapur, India.Determine the quartic f(x) = x4+ ax3+ bx2+ cx� c if it shares two

distinct integral zeros with its derivative f 0(x) and abc 6= 0.

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let the zeros of f(x) be the integers p and q; without loss of generality

p > q. It is a well-known theorem that if a polynomial Q(x) divides the

polynomial P (x) as well as the derivative P 0(x), then (Q(x))2divides P (x).

Applying the theorem for this problem, we obtain

f(x) = (x� p)2(x� q)2 = x4 + axx + bx2 + cx� c:

Comparing coe�cients of x and the constant term yields

0 = c+ (�c) = �2(p+ q)pq+ p2q2:

As pq = 0 implies abc = 0, we may divide by pq

pq � 2p� 2q = 0

(p� 2)(q� 2) = 4 = 4 � 1 = (�1)(�4)

Hence (p; q) = (6;3) _ (1;�2) (since p 6= q) and the two possible polyno-

mials are

f1(x) = (x� 6)2(x� 3)2 = x4 � 18x3 + 117x2 � 324x+ 324;

f2(x) = (x+ 2)2(x� 1)2 = x4 + 2x3 � 3x24x+ 4:

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; THEODORE CHRONIS, student, Aristotle University of Thessaloniki,Greece; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX,USA; DAVIDDOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA;HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany;F.J. FLANIGAN, San Jose State University, San Jose, California, USA; SHAWNGODIN, St. Joseph Scollard Hall, North Bay, Ontario; WALTHER JANOUS,Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State

192

University, Big Rapids, Michigan, USA; BEATRIZ MARGOLIS, Paris, France;L. RICE, Woburn Collegiate, Scarborough, Ontario; HARRY SEDINGER,St. Bonaventure University, St. Bonaventure, NY, USA; HEINZ-J�URGENSEIFFERT, Berlin, Germany; SKIDMORECOLLEGE PROBLEMGROUP, Sara-toga Springs, New York, USA; DIGBY SMITH,MountRoyal College, Calgary,Alberta; and the proposer. There were eight incorrect or incomplete solu-tions.

Do you know the equation of this graph?

Contributed by Juan-Bosco Romero M�arquez, Universidad de Valladolid, Valladolid, Spain.

-

6

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

193

THE ACADEMY CORNERNo. 11

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Bernoulli Trials 1997

Christopher SmallFaculty of MathematicsUniversity of WaterlooWaterloo, Ontario

At the University of Waterloo, we recently held a new mathematicscompetition called the Bernoulli Trials. The Bernoulli Trials came about asthe result of a brainstorming session between Ian Goulden, Ken Davidsonand myself. We felt that a winter competition was most desirable, and that ithad to be fundamentally di�erent from the competitions that the Universityof Waterloo holds in the Fall. So we decided to hold a double-knockoutcompetition with \true" and \false" as the response categories for each roundof the competition.

Students registered ahead of time so that we had an idea of the atten-dance �gures. This gave us a chance to draw up a master list of participants,which we put on a transparency for the overhead projector. As the roundsof the competition progressed, the students were able to see their resultsposted on one of the screens.

Students were given a supplyof \ballots" on which to place their \votes"for each round. At the beginning of each round, we posted a mathematicalconjecture on the overhead projector. Ten minutes were allotted to workingon each conjecture, and at the end of that time all students had to submit acompleted, signed ballot with a \true" or \false" response. When all ballotswere in, we announced the correct answer and a hint for solving the problem.(There was no time for a full solution during the competition.) The next �veminutes were used to record the ballots. During this time, the students wereable to stretch or chat about the problem and help themselves to food pro-vided at the back of the room. Then it was on to the next round and the nextconjecture. The double-knockout format meant that everyone had a chance

194

to make one mistake without being eliminated from the competition. Thosewho were eliminated at any stage were encouraged to stay and cheer on theremaining students.

In order to make the event work, we needed to have many more ques-tions available than we actually used. The questions varied from quite easyto extremely di�cult. Our planwas to adjust the level of di�culty based uponthe performance of the students. The hope was that in the early stages wecould keep the students interested without knocking them out too quickly.Over all, we were quite satis�ed with the results. The competition startedshortly after 9:00 a.m., and were �nished for a late lunch around 1:00 p.m.

There were 24 students participating. The �nal two participants, �rstyear students, Richard Hoshino and Soroosh Yazdani, who both lasted a full15 rounds, were declared co-winners of the competition. Tied for third andfourth place were Kevin Hare and Fr �ed �eric Latour, who lasted 11 rounds. In�fth place was David Kennedy, who lasted 10 rounds.

The prizes were given on the spot at the end of the competition. Inkeeping with the name Bernoulli Trials, the prizes were handed out on thespot at the end of the competition in batches of 100 coins each: quartersfor third, fourth and �fth places, loonies for second place, and toonies for�rst. In a spirit of generosity, Richard Hoshino and Soroosh Yazdani invitedeveryone to lunch with the prize money they had earned.

THE BERNOULLI TRIALS

True or False?

1. The point

(x; y) =

p2 + 1p2� 1

;

p3 + 1p3� 1

!

is outside the closed curve x = 1+ 5 cos �; y = �1 + 5 sin �, where� 2 [0; 2�).

2. Let p and q be real numbers with p�1 � q�1 = 1 and 0 < p � 1=2.Then

p+1

2p2 +

1

3p3 + ::: = q� 1

2q2 +

1

3q3� 1

4q4 + :::

3. There are at least two positive integers n for which 2n � 1 and 2

n+ 1

are primes.

4. When expressed in the decimal system, 1000! ends in exactly 249 zeros.

That is, 1000! has the form n� 10249, where n is not divisible by 10.

195

5. For any set M � R3, there exists a plane � � R3 such that eitherM \ � has in�nitely many points or M \ � is empty.

6. Suppose Johann Bernoulli tosses 1997 fair coins and Jakob Bernoullitosses 1996 fair coins.

The probability that Johann has more heads than Jakob is more than1=2.

7. Let x(p) be de�ned by x(1) = x and x(p+1)= xx

(p)

.

Then limx!0+ x(1997)

= 1.

8. If the elementary symmetric functions of a set of real numbers are allpositive, then the real numbers themselves must be positive.

9. Let a; b; c > 0. Then abc+

acb+

bca> 2(a+ c� b).

10. There does not exist a Fibonacci number with a decimal representationending in four zeros.

11.25Xk=1

k

k4 + k2 + 1=

325

651

12. There exists a unique pair of positive integers n;m with n > m suchthat 1997 = n2

+m2.

13. Z �=2

0

x dx

1 + sin x=

1p2

14. The equation 7px � 5

px =

3px � p

x has exactly three solutions innonnegative real x.

15. For some a > 0, let f : R! R satisfy

f(x+ a) =1

2+

pf(x)� [f(x)]2

for all x.

Then f is periodic.

In the next issue, we will give the hints and the answers!

196

THE OLYMPIAD CORNERNo. 182

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

For your problem solving pleasure over the next few weeks, we givethree contests which were collected and forwarded to us by RichardNowakowski, when he was Canadian Team Leader to the IMO in Hong Kong.

SWEDISH MATHEMATICS CONTEST 1993

FinalNovember 20

1. The integer x is such that the sum of the digits of 3x is the same asthe sum of the digits of x. Prove that 9 is a factor of x.

2. A railway line is divided into 10 sections by the stations A, B, C,D, E, F , G, H, I, J and K. The distance between A and K is 56 km. Atrip along two successive sections never exceeds 12 km. A trip along threesuccessive sections is at least 17 km. What is the distance between B andG?z }| {

A B C D E F G H I J K

3. Assume that a and b are integers. Prove that the equation a2+b2+

x2 = y2 has an integer solution x, y if and only if the product ab is even.

4. To each pair of real numbers a and b, where a 6= 0 and b 6= 0, thereis a real number a � b such that

a � (b � c) = (a � b) � cand a � a = 1:

Solve the equation x � 36 = 216.

5. A triangle with perimeter 2p has the sides a, b and c. If possible,a new triangle with the sides p� a, p� b and p� c is formed. The processis then repeated with the new triangle. For which original triangles can theprocess be repeated inde�nitely?

6. Let a and b be real numbers and let f(x) = (ax+b)�1. For which aand b are there three distinct real numbers x1, x2, x3 such that f(x1) = x2,f(x2) = x3 and f(x3) = x1?

197

DUTCH MATHEMATICAL OLYMPIADSecond RoundSeptember 1993

1. Suppose that V = f1; 2; 3; : : : ; 24; 25g.Prove that any subset of V with 17 or more elements contains at least twodistinct numbers the product of which is the square of an integer.

2. Given is a triangle ABC, \A = 90�. D is the midpoint of BC, F

is the midpoint of AB, E the midpoint of AF and G the midpoint of FB.AD intersects CE, CF and CG respectively in P , Q and R. Determine theratio PQ

QR.

A E F G B

C

P QR

D

q

q

qq

q qq

q q q3. A series of numbers is de�ned as follows: u1 = a, u2 = b, un+1 =

1

2(un + un�1) for n � 2. Prove that limn!1 un exists. Express the value

of the limit in terms of a and b.

4. In a plane V a circle C is given with centre M . P is a point not onthe circle C. q

qqq

P

M

A

B

V

(a) Prove that for a �xed point P , AP 2+ BP 2 is a constant for every

diameter AB of the circle C.(b) Let AB be any diameter of C and P a point on a �xed sphere S

not intersecting V . Determine the point(s) P on S such that AP 2+BP 2 is

minimal.

5. P1; P2; : : : ; P11 are eleven distinct points on a line. PiPj � 1 forevery pair Pi, Pj. Prove that the sum of all (55) distances PiPj , 1 � i < j �11 is smaller than 30.

198

PROBLEMS OF THE 11th BALKANMATHEMATICAL OLYMPIAD

Novi Sad, YugoslaviaMay 8{14, 1994 (Time: 4.5 hours)

1. [Cyprus] Given an acute angle \XAY and a point P in its interior,construct a straight line through P and intersecting the sidesAX andAY atthe points B and C respectively, so that the triangle ABC has area equal to(AP )

2.

(Success rate: 26:66%)

2. [Greece] Show that the polynomial

x4 � 1994x3 + (1993+m)x2 � 11x+m; m 2Zhas at most one integral root. (Success rate: 39:66%)

3. [Romania] Compute (for n � 2)

max

8><>:

n�1Xk=1

j�k+1 � �kj j (�k)nk=1 is a permutation

of the numbers 1; 2; : : : ; n

9>=>; :

(Success rate: 30%)

4. [Bulgaria] Find the smallest number n > 4 such that there is a setof n people with the following properties:

(i) any two people who know each other have no common acquain-tances;

(ii) any two people who do not know each other have exactly two com-mon acquaintances.

Note: Acquaintance is a symmetric relation. (Success rate: 19%)

OVERALL SUCCESS RATE: 28:83%; Number of Students: 30,Mean: 11:33, Median: 7:5, S.D.: 12:3.

Recently I made time to clean o� my desktop and sort a backlog ofpaper that piled up over the last year while I was on secondment to a uni-versity committee and only in my o�ce for a short time each week. Thereappeared a small cache of solutions to problems from the 1995 numbers ofthe Corner which I had put down in haste (and in error), letting them becomecovered by other layers of incoming material. Let me try to set the recordstraight by acknowledging the readers' e�orts now. My sincere apologies!Miguel Amengual Covas, Cala Figuera, Spain, submitted solutions to prob-lem 2 of Part I of the Turkish Mathematical Olympiad Committee Final Se-lection Test, for which a solution was given in the December 1996 number

199

[1996: 349{350], and to problem 2 of part II of that test, for which a solutionwas given [1996: 352{353]. Mansur Boase, student, St. Paul's School, Lon-don, England, submitted solutions to problems 1 and 2 of Part I of the TurkishMathematical Olympiad Final Selection Test, for which solutions were given[1996: 349{350], [1996: 349{350] and for questions 1 and 3 of Part II, forwhich solutions were given [1996: 351{352], [1996: 353-354]. In anothersubmission he gave answers to problems 1, 2 and 4 of the 1992 Dutch Mathe-matical Olympiad, for which solutions were given [1997: 13], [1997: 13{14]and [1997: 15{17], respectively. In a third letter he provided solutions tofour problems of the 16th Austrian-Polish Mathematics Competition. Twosolutions of these appeared earlier this year: problem 1 in [1997: 71{72] andproblem 6 in [1997: 72{73]. In yet a fourth letter he gave solutions to threeof the problems proposed to the jury of the 35th IMO but not used. A solu-tion to number 3 was given in [1997: 135]. Finally a fat envelope of solutionsby George L. Evagelopoulos to problems proposed but not used at the 35thIMO appeared. Below we give some of the solutions submitted by MansurBoase but missed before. First back to problems of the 2nd MathematicalOlympiad of the Republic of China.

1. [1995: 80] 2nd Mathematical Olympiad of the Republic of China.A sequence fang of positive integers is de�ned by

an = [n +pn + (1=2)], n 2 N, where N is the set of all positive integers.

Determine the positive integers which belong to the sequence.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.Now an =

�n+

pn+

1

2

�, and an+1 =

�n+ 1+

pn+ 1 +

1

2

�so

an+1 � an = 2 ifpn+ 1 >

hpn+ 1

i+

1

2= a+

1

2

andpn <

�pn�+

1

2= a+

1

2or a� 1

2if n = a2 � 1

and an+1 � an = 1 otherwise.For an+1 � an = 2 we have n+ 1 > a2 + a+

1

4and n < a2 + a+ 1

4,

since for a � 2,pa2 � 1 6< a� 1

2. Thus n+ 1 = a2 + a+ 1. Now

�a2 + a+

pa2 + a+

1

2

�= a2 + 2a

�a2 + a+ 1 +

pa2 + a+ 1 +

1

2

�= a2 + 2a+ 2:

Thus the only numbers left out are of the form a2 +2a+1 = (a+1)2; that

is, the squares. Thus all positive non-squares belong to the sequence.

200

Now to problems given in the September number.

3. [1995: 221] 16th Austrian Polish Mathematical Competition.Let the function f be de�ned as follows:

If n = pk > 1 is a power of a prime number p, then f(n) := n+ 1.

If n = pk11 � � � pkrr (r > 1) is a product of powers of pairwise di�erent

prime numbers, then f(n) := pk11 + � � �+ pkrr .

For every m > 1 we construct the sequence fa0; a1; : : : g such that a0 = m

and aj+1 = f(aj) for j � 0. We denote by g(m) the smallest element ofthis sequence. Determine the value of g(m) for allm > 1.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.It is evident that g(2) = 2, g(3) = 3, g(4) = 4, g(5) = 5, g(6) = 6,

g(7) = 7.Lemma. For m > 6, g(m) > 6.Proof. Considerm > 6, and its associated sequence.

6 = 1 + 5 = 2 + 4:

Now 2+4 is not permissible since (2; 4) 6= 1, and the only way 6 can be partof the sequence is if 5 occurs earlier.

5 = 1 + 4 = 2 + 3:

So 5 is only part of the sequence if 4 or 6 occur earlier.

4 = 1 + 3 = 2 + 2;

so 4 is part of the sequence only if 3 occurs earlier.

3 = 1 + 2;

so 3 is part of the sequence only if 2 occurs earlier, which is impossible witha0 > 6.

Consider the next several values of g(m). By calculation g(8) = 7 =

g(9) = g(10) = g(11) = g(12) = g(13) = g(14) = g(15) = g(16). Wenext prove by induction that if m � 16 then g(m) = 7. We may assumem > 16.

Observe that at least one of m, m+ 1, m+ 2, m+ 3, m+ 4, m+ 5

is not a power of a prime because it is congruent to 0 mod 6. Letting n bethe �rst value which is not a power of a prime m � n � m+ 5. The resultfollows immediately from the following since g(m)� g(n)� f(n).

201

Lemma. Supposen > 16 is not a power of a prime. Then f(n)< n�5.Proof. Write n = ab with 1 < a < b and (a; b) = 1. Then b � 5. Now

a+ b < ab� 5

is equivalent to 1 +ab+

5

b< a which is immediate unless a = 2 because

ab< 1, 5

b� 1. But if a = 2 then b > 8 as ab > 16 and

1 +a

b+

5

b< 1 +

2

8+

5

8< 2:

The weaker inequality a + b < ab for 1 < a < b is immediate. It followsthat

f(n)< a+ b < ab� 5

if n > 16.This completes the proof.

4. [1995: 221] 16th Austrian Polish Mathematical Competition.The Fibonacci numbers are de�ned by F0 = F1 = 1 and Fn+2 =

Fn+1 + Fn for n � 0. Let A and B be natural numbers such that A19

divides B93 and also B19 dividesA93. Prove: for all natural numbers n � 1

the number (A4+B8

)Fn+1 is divisible by (AB)

Fn.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.Now A19 j B93 and B19 j A93. Thus any prime, p, which divides A

must divide B, and vice versa.Also if p�1 j A then p[(93=19)�1] is the largest power of p which can

divide B and vice versa. (�)First we must prove that

AB j (A4+ B8

)2: (1)

LetA = p�11 p�22 : : : p�nn ;

B = p�11 p

�22 : : : p�nn :

Consider

p�1+�11 j (p4�11 + p

8�11 )

2= p8�1

1 + p16�11 + 2p

4�1+8�11 :

Because of (�), �1 < 5�1, and �1 < 5�1, so �1 + �1 < 8�1, 16�1 and4�1 + 8�1. From this (1) follows.

Next we argue that

(AB)2 j (A4

+ B8)3= A12

+ B24+ a multiple of (AB)

2: (2)

So we must prove that (AB)2 j A12

+ B24; that is

p2�1+2�11 j p12�11 + p

24�11 :

202

Again because of (�), 2�1 + 2�1 < 2�1 + 2(5�1) = 12�1 and 2�1 +

2�1 < 24�1. Thus (2) holds.We have therefore proved the assertion for n = 1 and n = 2.Assume for a proof by induction that

(AB)Fk j (A4

+B8)Fk+1

and(AB)

Fk+1 j (A4+ B8

)Fk+2 :

Multiplying, we get

(AB)Fk+Fk+1 j (A4

+ B8)Fk+1+Fk+2 ;

that is(AB)

Fk+2 j (A4+ B8

)Fk+3 :

Thus the statement is true for n = k + 2. The result follows by induction.

We conclude this \catch-up" part of the Corner with two solutions toproblems posed to the jury but not used at the 35th IMO.

1. [1995: 299] Problems Proposed But Not Used, 35th IMO in Hong

Kong, Selected Problems

M is a subset of f1; 2; 3; : : : ; 15g such that the product of any threedistinct elements of M is not a square. Determine the maximum number ofelements inM .

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.Let n denote the maximum number of elements in such a subset M .

It is easy to check that there are no three elements in the following subsetwhose product is a square

M = f3; 4; 5; 6; 7; 9; 10; 11; 13; 14g:

Hence n � 10.We shall prove that n > 10 is impossible. We split the problem into

three cases: (i) 2 2M ; (ii) 2 62M , 3 62M ; (iii) 2 62M , 3 2M .Case (i). 2 2M .We now list (x; y), x; y 6= 2, 1 � x < y � 15 such that 2xy is a perfect

square and hence we cannot have x 2M and y 2M :

(1; 8); (3; 6); (4; 8); (5;10); (6; 12); (7; 14) (8; 9):

In order to have n > 10 we can have at most one of 5 or 10 and one of 7or 14. Thus 8 62M . Now it follows that 6 62M as well. To have n � 11 werequire 1; 4; 9 2M but 1 � 4 � 9 = 36 = 6

2, a contradiction.

203

Case (ii). 2 62M , 3 62M .One member from each of the following triples could not belong to M :

(1;4; 9); (7;8; 14); (5; 12; 15):

Thus n � 10.Case (iii). 2 62M , 3 2M .Each of the following triples must contain a number which does not

belong to M :

(3; 1; 12); (3; 4; 12); (3; 9; 12); (3;6; 8); (3; 5; 15); (1; 4; 9):

We must have 12 62 M , one of 6 or 8 =2 M , 5 62 M or 15 62 M and 1 or 4or 9 62M . But as 2 62M we have at least 5 elements which do not belong.

Thus n = 10.

6. [1995: 335] Problems Proposed But Not Used, 35th IMO in Hong

Kong, More Selected Problems

N is an arbitrary point on the bisector of \BAC. P and O are pointson the lines AB and AN , respectively, such that \ANP = 90

�= \APO.

Q is an arbitrary point on NP , and an arbitrary line through Q meets thelines AB and AC at E and F respectively. Prove that \OQE = 90

� if andonly if QE = QF .

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; and by George Evagelopoulos, Athens, Greece. We give the \o�cial"

solution which is very similar to that of Evagelopoulos.First assume that \OQE = 90

�. Extend PN to meet AC at R. NowOEPQ and ORFQ are cyclic quadrilaterals. Hence \OEQ = \OPQ =

\ORQ = \OFQ. It now follows that we have 4OEQ � 4OFQ andQE = QF .

q q

q

q qqqq

q

B C

A

E

PQ

NR

F

O

Now suppose that \OQE 6= 90�. Let the perpendicular through O to

EF meet NP at Q0 6= Q. Draw the line through Q0 parallel to EF , meetingthe lines AB and AC at E0 and F 0, respectively. Then Q0E0 = Q0F 0 asbefore. Let AQ0 meet EF atM 6= Q. ThenME = MF so that QE 6= QF .It follows that if QE = QF , then \OQE = 90

�. (See next page for thediagram.)

204

q q

q

q qqq

q

q

qq

qB C

A

E0

P

Q0

E

F

Q

M

N

F 0

O

To conclude this number of the Olympiad Corner we give solutions bythe readers to problems given in the February 1996 number with the NationalRound of the 29th Spanish Mathematical Olympiad, 1992 [1996: 22{23].

1. At a party there are 201 people of �ve di�erent nationalities. Ineach group of six, at least two people have the same age. Show that thereare at least �ve people of the same country, of the same age and of the samesex.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; and by

Derek Kisman, student, Queen Elizabeth High School, Calgary, Alberta. We

give Boase's solution.By the Pigeonhole Principle there are at least 41 of the same nationality

and of these at least 21 must be of the same gender. Let these 21 ages beai, 1 � i � 21.

Consider a1, a2, a3, a4, a5, a6 at least 2 must be equal. Without lossof generality, let a1 = a2.

Similarly with a3; a4; : : : ; a8, without loss of generality, let a3 = a4.Continuing as above we can go as far as a15 = a16. The sequence now reads

a2a2a4a4a6a6a8a8a10a10a12a12a14a14a16a16a17a18a19a20a21

Consider a2a4a6a8a10a12. Two must be equal. Without loss of generality,let a2 = a4. Similarly we have from a6; : : : ; a16 we may assume that a6 =

a8.Now consider a2a6a10a12a14a16. If a2 = ai or a6 = ai for some

other i then we have �ve equal terms and we are done.Therefore we must have a pair equal among a10, a12, a14, a16; without

loss of generality, let a10 = a12. The sequence now reads

a2a2a2a2a6a6a6a6a10a10a10a10a14a14a16a17a18a19a20a21:

205

If any ai, 17 � i � 21 is among a2, a6, a10, we are done.Otherwise, consider a2a6a10a14a16x where x is any one of the last �ve

terms. Each x must be equal to a14 or a16. Without loss of generality, letthree of them be equal to a14 (by the Pigeonhole Principle).

This provides �ve terms equal to a14, and hence �ve with the same age,gender and nationality.

2. Given the number triangle

0 1 2 3 4 : : : : : : 1991 1992 1993

1 3 5 7 : : : : : : 3983 3985

4 8 12 : : : : : : 7968

in which each number equals the sum of the two above it, show that the lastnumber is a multiple of 1993.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by

Derek Kisman, student, Queen Elizabeth High School, Calgary, Alberta; and

by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. First,

we give Kisman's solution.Form a new triangle of numbers by adding each element of this triangle

and its (horizontally ipped) reverse. It obeys the same addition rule as theoriginal, and every element in the top row is 1993. Thus every element inthe new triangle is a multiple of 1993. Let n be the last number in the newtriangle. The last number of the original triangle is just n=2. Since it is aninteger, it must also be a multiple of 1993.

Next we give Wang and Yu's generalization.We show more generally that if the given row is 0; 1; 2; : : : ; n, where

n is a natural number, then the last number in the triangle is n2n�1 and soit is a multiple of n. Note that the completed triangle will have n+ 1 rows.If we denote the ith row by Ri then Ri contains n + 2 � i numbers, i =

1; 2; : : : ; n+1. LetRi;j denote the jth number inRi, j = 1; 2; : : : ; n+2�i.We show by induction on i that

Ri;j = (i+ 2j � 3)2i�2: (�)

This is the case when i = 1 since R1 = 0; 1; 2; : : : ; n implies R1;j = j � 1

and (i+ 2j � 3)2i�2

= (2j � 2)2�1

= j � 1. Suppose (�) holds for somei � 1. Then from the described construction, we have

Ri+1;j = Ri;j + Ri;j+1

= (i+ 2j � 3)2i�2

+ (i+ 2j � 1)2i�2

= (2i+ 4j � 4)2i�2

= (i+ 1+ 2j � 3)2i+1�2

for j = 1; 2; : : : ; n+1� i, completing the induction. Letting i = n+1 andj = 1 in (�) we �nd that the last number in the triangle is Rn+1;1 = n2n�1.

206

3. Show that in any triangle, the diameter of the incircle is not biggerthan the circumradius.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-

tario. We give Wang's solution and remark.Let a, b, c, r, R and � denote the side lengths, the inradius, the cir-

cumradius and the area of the given triangle, respectively. Then it is well

known that r =�

sand R =

abc

4�where s =

1

2(a + b + c) denotes the

semiperimeter of the triangle. Hence 2r � R is equivalent to 8�2 � sabc,

which by virtue of Heron's formula

� =

ps(s� a)(s� b)(s� c)

becomes 8(s� a)(s� b)(s� c) � abc;

or (b+ c� a)(c+ a� b)(a+ b� c) � abc: (1)

To show (1), note that a2 � (b� c)2 � a2, b2 � (c� a)2 � b2, andc2 � (a� b)2 � c2. Multiplying, we obtain

(b+ c� a)2(c+ a� b)2(a+ b� c)2 � a2b2c2: (2)

Since b+ c� a, c+ a� b and a+ b� c are all positive, (1) follows from (2)by taking the square roots of both sides.

Clearly, equality holds if and only if a = b = c, that is, when thetriangle is equilateral.

Remark. This is a classical problem which has been known for at least230 years. See, for example. x5.1 of Geometric Inequalities by O. Bottemaet al. In fact, the inequality 8(s� a)(s� b)(s� c) � abc mentioned in theproof above can also be found in x1.3 of this book.

4. Show that each prime number p (di�erent from 2 and from 5) hasan in�nity of multiples which can be written as 1111 : : : 1.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; by

Derek Kisman, student, Queen Elizabeth High School, Calgary, Alberta; and

by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We

give Godin's solution.

It su�ces to show that, for each prime p (other than 2 and 5), thereexists a number n of the form 11 : : : 1, such that p j n. Since if p j 11 : : : 1(where the number is made up of k 1's), then p clearly divides all numbersof the form

(11 : : : 1)

NXj=0

10kj

for any nonnegative integer which has the desired property.

207

By Fermat's Little Theorem, 10p�1 � 1 mod p for all primes except 2and 5.

Thus p j (10p�1 � 1), but 10p�1 � 1 = 9(11 : : : 1).Thus for p 6= 2; 3; 5, p3 j 111 : : : 1, where there are (p� 1) 1's in the

number. It is easy to check that 3 j 111 so we are done.

6. A game-machine has a screen in which the �gure below is shown.At the beginning of the game, the ball is at the point S.

With each impulse from the player, the ball moves up to one of the neigh-bouring circles, with the same probability for each. The game is over whenone of the following events occurs:

(1) The ball goes back to S, and the player loses.

(2) The ball reaches G, and the player wins.

Determine:

(a) The probability for the player to win the game.

(b) The mean time for each game.

S

A

C

B

G

D����

����

��������

��������

� bbbbb

����

�����HHHHH

Solutions by Shawn Godin, St. Joseph Scollard Hall, North Bay, On-

tario; and by Derek Kisman, student, Queen Elizabeth High School, Calgary,

Alberta. We give Kisman's solution.

(a) From C or D any path to S has a corresponding exact opposite toG, so when at C or D the chance of winning is exactly 1=2. From S we mustgo to A where we have a 1=3 chance of losing and a 2=3 chance of going toC or D, where there is a 1=2 chance of winning.

So the probability is 2

3� 12=

1

3.

(b) To �nd the mean we sum the products of the number of moves andthe probability of getting that number. The number of moves to win or loseis always even. For a given number of moves 2n, we can �nd the probabilityof lasting exactly that long:

1 move is entering from S to A, probability 1

n� 1 moves are switching from A or B to C or D, probability 2

3

n� 1 moves are switching from C or D to A, or B, probability 11 move is moving from A to B or S or G, probability 1

3.

208

Thus for 2n moves the probability is

1

3

�2

3

�n�1=

�2

3

�n� 12:

Now, the in�nite sum to �nd the mean is thus

2 ��2

3

�1� 12+ 4 �

�2

3

�2

� 12+ � � �+ 2n

�2

3

�n� 12+ � � �

= 1 ��2

3

�+ 2 �

�2

3

�2+ 3 �

�2

3

�3

+ � � �+ n

�2

3

�n+ � � �

=

2

3+

�2

3

�2+

�2

3

�3+ � � �

!+

2

3

2

3+

�2

3

�2+

�2

3

�3+ � � �

!

+

�2

3

�2 2

3+

�2

3

�2

+ � � �!+ � � �

=

�2

3

�1

1� 2

3

+

�2

3

�21

1� 2

3

+

�2

3

�31

1� 2

3

+ � � �

= 3

2

3+

�2

3

�2+ � � �

!

= 3 ��2

3

�1

1� 2

3

= 6:

So the mean time is 6.

That completes the issue of the Corner. Please send me your nice so-lutions along with your national and regional Olympiads.

209

BOOK REVIEWS

Edited by ANDY LIU

Cien Problemas de Matem�aticas: Combinatoria, �Algebra, Geometr��a,One Hundred Mathematics Problems: Combinatorics, Algebra, Geometry

by Francisco Bellot Rosado and Mar��a Ascensi �on L �opez Chamorro,publishedby Instituto de Ciencias de la Educaci �on, Universidadde Valladolid,ISBN 84-7762-405-4, 1994, softcover, 146+ pages.reviewed byMar��a Falk de Losada.

There are very few problem-solving books written in Spanish and evenfewer good problem-solving books written in other languages and trans-lated into Spanish. One notable exception has been several Russian bookspublished in Spanish by MIR. However, Francisco Bellot Rosado and Mar��aAscensi �on L �opez Chamorro have recently addressed that needwith their bookCien Problemas de Matem�aticas: Combinatoria, �Algebra y Geometr��a andthey have selected their problems from several important sources. From theIMO, the OME (Spanish Mathematics Olympiad) and the OIM (OlimpiadaIberoamericana deMatem�aticas) there are both problems that have appearedand also problems that were proposed but unused. From Mathesis andGazeta Matematica we �nd many classical problems that every interestedproblem-solver should know. From several other competitions and journalsthere is a wide-ranging selection of problems. These were, in general, un-available to the Spanish-speaking student previously.

Many of the solutions come from the authors themselves, while oth-ers are those given by Spanish participants in the IMO, OIM and still otherscome from a variety of sources. It is a truism that it is important for stu-dents working in the area of problem solving to develop con�dence in them-selves and their own abilities. For students from countries that do not have atime-honoured tradition in mathematics competitions and serious problem-solving, as is true in many Spanish-speaking countries, having access to so-lutions given by their own fellow students brings the possibility of successcloser and helps to build such con�dence.

210

Folding the Regular Nonagon

Robert Geretschl�ager, Bundesrealgymnasium, Graz, Austria

Introduction

In the March 1997 of Crux [1997: 81], I presented a theoretically precisemethod of folding a regular heptagon from a square of paper using origamimethods in an article titled \Folding the Regular Heptagon". The methodwas derived from results established in \Euclidean Constructions and the

Geometry of Origami" [2], where it is shown that all geometric problems thatcan be reduced algebraically to cubic equations can be solved by elementarymethods of origami. Speci�cally, the corners of the regular heptagon werethought of as the solutions of the equation

z7 � 1 = 0

in the complex plane, and this equation was then found to lead to the cubicequation

�3 + �2 � 2� � 1 = 0;

which was then discussed using methods of origami. Finally, a concretemethod of folding the regular heptagon was presented, as derived from thisdiscussion.

In this article, I present a precise method of folding the regular nonagonfrom a square of paper, again as derived from results established in [2].However, as we shall see, the sequence of foldingsused is quite di�erent fromthat used for the regular heptagon. As for the heptagon, the folding methodis once again presented in standard origami notation, and the mathematicalsection cross-referenced to the appropriate diagrams.

Angle Trisection

For any regular n-gon, the angle under which each side appears as seen fromthe mid-point is 2�

n. Speci�cally, for n = 9, the sides of a regular nonagon

are seen from its mid-point under the angle 2�9. As is well known, this angle

cannot be constructed by Euclidean methods. Three times this angle (or 2�3)

can, however, and we note that it would be possible to construct a regularnonagon by Euclidean methods, if it were possible to trisect an arbitrary an-gle, or at least the speci�c angle 2�

3. If this were the case, all that we would

have to do would be to construct an equilateral triangle, trisect the anglesfrom its mid-point to its corners, and intersect these with the triangle's cir-cumcircle. Unfortunately however, as generations of mathematicians havebeen forced to accept (although there are a few hold-outs still out there),

211

angle trisection by Euclidean methods of straight-edge and compass is im-possible, as is the construction of a regular nonagon.

The underlying reason for the impossibility of angle trisection by Euclid-ean methods is the fact that straight-edge and compass constructions onlyallow the solution of problems that reduce algebraically to linear or quadraticequations. Angle trisection, however, involves the irreducible cubic equation

x3 � 3

4x� 1

4cos 3� = 0;

which derives from the well established fact (see for instance [1]) that

cos 3� = 4cos3 �� 3 cos�:

For the speci�c case at hand, where 3� =2�3, the cubic equation in question

is

x3 � 3

4x+

1

8= 0:

As shown in [2], the solutions of this equation are the slopes of the commontangents of the parabolas p1 and p2, whereby p1 is de�ned by its focus

F1�1

16;�3

8

�and its directrix

`1 : x = � 1

16;

and p2 is de�ned by its focusF2�0; 1

2

�and its directrix

`2 : y = �1

2:

(It is not too di�cult to prove that this is indeed the case. Interested readersmay like to try their hand at doing the necessary calculations themselves.)Finding the common tangents of two parabolas de�ned by their foci and di-rectrices is quite straight-forward in origami, as it merely means making onefold, which places two speci�c points (the foci) onto two speci�c lines (thecorresponding directrices). By this method, we will therefore now show howto fold a regular nonagon.

A Step-by-step Description of the Folding Process

As is usually the case in origami, we assume a square of paper to be given. Weconsider the edge-to-edge folds in step 1 as the x{ and y{axes of a system ofcartesian coordinates, and the edge-length of the given square as two units.The mid-point of the square is then the origin M(0;0), and the end-pointsof the folds have the coordinates (�1; 0) and (1;0), and (0;�1) and (0;1)

respectively.

212

Steps 1 through 8 yield the foci and directrices of the parabolas dis-cussed at the end of the second section of this article. The point C is thefocus F1 of parabola p1, B is the focus F2 of parabola p2, and the creasesonto which these two points are folded in step 9 are the directrices `1 and`2. Since the coordinates involved are all arrived at by halving certain linesegments, it is quite easy to see that this is indeed the case.

The fold made in step 9 is then a common tangent of the parabolas,and its slope is therefore cos

2�9. (This step, by the way, is the only one that

cannot be replaced by Euclidean constructions.) The point of steps 10 to 13is then to �nd the horizontal line represented by the equationy = � cos

2�9. This is the horizontal fold through point E, as the distance

between the vertical folds through points D and E is equal to 1.

We then obtain corners 2 and 9 of the nonagon (assuming the point withcoordinates (0;�1) to be corner 1) on this horizontal line by folding the unitlength onto this line from mid-point M in step 14. Step 15 therefore yieldsthe �rst two sides of the nonagon, and steps 16 and 17 complete the fold,making use of both the radial symmetry of the �gure, and its axial symmetrywith respect to the vertical lineM1. Step 18, �nally, shows us the completedregular nonagon.

The Folding Process

The diagrams follow the bibliography.

Conclusion

In the conclusion of Folding the Regular Heptagon, I declared myself as anardent Heptagonist. I have no qualms or reservations about declaring myselfan equally ardent Nonagonist now. Perhaps I will �nd some similar-mindedfolk out there willing to join me in my quest of popularizing these heretoforesadly neglected polygons.

References

[1] B. Bold, Famous Problems of Geometry and How to Solve Them, DoverPublications, Inc., Mineola, NY (1969).

[2] R. Geretschl�ager, Euclidean Constructionsand the Geometry of Origami,Mathematics Magazine, Vol. 68, No. 5 December 1995.

213

1.

Fold and unfold twice.

2.

qA

Fold and unfold three times,making crease marks each time.

�nal crease yields point A.

3.

qA

Fold edge to point A and unfold.

4.

Refold edge to edge.

214

5.

Fold and unfold both layersat crease, then unfold.

6.

q B

Fold and unfold twice, makingcrease mark at point B.

7.

q B

qC

Fold upper edge to crease, unfold,then fold lower edge to new crease,making crease mark at point C.

8.

q B

qC

Mountain fold along creases.

215

9.

q

q

B

C

Fold so that B comes to lie oncrease, and previous fold on C.

10.

qC qB

Unfold everything.

11.

qD

Fold along crease and unfold, thenfold vertically through point D.

12.

qE

Fold edge to edge and unfoldeverything.

216

13.

qE

Fold horizontally throughpoint E and unfold.

14.

q M

q q9 2

1

Fold throughM such that point 1lies on crease, unfold a repeat onother side (points 1, 2, 9 arecorners of the nonagon).

15.

q qq q

qM

q q9 2

1

Fold back twice such that markedpoints come to lie on each other,resulting folds are sides of the

nonagon.

16.

q M

q2

Fold throughM and 2.

217

17.

Mountain fold lower layer usingedges of upper layer as guide lines.Resulting folds are two more sidesof the nonagon. Open up fold fromstep 16 and repeat steps 16 and 17on left side, then fold throughMand 2 once more. New folds are

new guide lines. Repeating processcompletes the nonagon.

18.

The �nished nonagon.

218

THE SKOLIAD CORNERNo. 22

R.E. Woodrow

We begin this number by giving the problems of the Twelfth W.J. Blun-don Contest, which was written February 22, 1995. This contest is preparedat the Memorial University of Newfoundland and is sponsored by the Cana-dian Mathematical Society.

THE TWELFTH W.J. BLUNDON CONTESTFebruary 22, 1995

1. (a) From a group of boys and girls, 15 girls leave. There are then lefttwo boys for each girl. After this, 45 boys leave. There are then 5 girls foreach boy. How many boys and how many girls were in the original group?

(b) A certain number of students can be accommodated in a hostel. Iftwo students share each room then two students will be left without a room.If 3 students share each room then two rooms will be left over. How manyrooms are there?

2. How many pairs of positive integers (x; y) satisfy the equation

x

19+

y

95= 1?

3. A book is to have 250 pages. How many times will the digit 2 beused in numbering the book?

4. Without using a calculator

(a) Show thatp7 +

p48 +

p7�p48 is a rational number.

(b) Determine the largest prime factor of 9919.

5. A circle is inscribed in a circular sectorwhich is one sixth of a circle of radius 1, andis tangent to the three sides of the sector asshown. Calculate the radius of the inscribedcircle.

6. Determine the units digit of the sum

2626

+ 3333

+ 4545:

219

7. Find all solutions (x; y) to the system of equations

x+ y+x

y= 19

x(x+ y)

y= 60:

8. Find the number of di�erent divisors of 10800.

9. Show that n4�n2 is divisible by 12 for any positive integer n > 1.

10. Two clocks now indicate the correct time. One gains a secondevery hour, and the other gains 3 seconds every 2 hours. In how many dayswill both clocks again indicate the correct time?

Last issuewe gave the 1995Manitoba Mathematical Contest. Here arethe solutions.

THE MANITOBAMATHEMATICAL CONTEST 1995For Students in Grade 12

Wednesday, February 22, 1995 | Time: 2 hours

1. (a) If a and b are real numbers such that a+ b = 3 and a2+ ab = 7

�nd the value of a.

Solution. a2 + ab = a(a+ b) = a � 3 = 7 so a = 7=3.

(b) Noriko's average score on three tests was 84. Her score on the �rsttest was 90. Her score on the third test was 4 marks higher than her scoreon the second test. What was her score on the second test?

Solution. 90+s+(s+4)

3= 84 so 2s+ 94 = 3� 84 = 252, 2s = 158 and

the score on the second test was 79.

2. (a) Find two numbers which di�er by 3 and whose squares di�erby 63.

Solution. Let the smaller number be x. The larger number is thenx+3.Now j(x+3)

2� x2j = 63. Thus jx2 +6x+ 9� x2j = 63 or j6x+9j = 63.Now 6x+ 9 = 63 OR 6x+ 9 = �63. Thus x = 7 OR x = �8.

(b) Find the real number which is a root of the equation:27(x� 1)

3+ 8 = 0.

Solution. 27(x � 1)3+ 8 = 0 is equivalent to (3(x � 1))

3= �8 =

(�2)3. As �8 has only one real cube root 3(x� 1) = �2 so x = 1=3.

220

3. (a) Two circles lying in the same plane have the same centre. Theradius of the larger circle is twice the radius of the smaller circle. The areaof the region between the two circles is 7. What is the area of the smallercircle?

Solution. Let the radius of the smaller circle be r. Then the radius of thelarger circle, R = 2r. The areas of the circles are �r2 and �R2 respectively.Thus the area of the region between them is

�R2 � �r2 = 7;

thus �(2r)2 � �r2 = 7;

(4� 1)�r2 = 7;

�r2 =7

3:

The area of the smaller circle is 7=3.

r

2r

(b) The area of a right triangle is 5. Also, the length of the hypotenuseof this triangle is 5. What are the lengths of the other two sides?

Solution.

y5

x

Let the lengths of the two legs be x andy, with x � y. Then 1

2xy = 5 and x2 +

y2 = 52= 25 from the formula for the

area, and Pythagoras. So 2xy = 20 and(x+y)2 = x2+y2+2xy = 45, (x�y)2 =5, giving x + y = 3

p5, x � y =

p5, so

x = 2p5 and y =

p5. The lengths of the

other two sides arep5 and 2

p5.

4. (a) The parabola whose equation is 8y = x2 meets the parabolawhose equation is x = y2 at two points. What is the distance between thesetwo points?

Solution. For the intersection points 8y = x2 and x = y2 giving8y = (y2)2 or y(y3� 8) = 0, so y = 0 and x = 0 OR y = 2 and x = 4. Theintersection points are (0; 0) and (4; 2). The distance is

p42 + 22 = 2

p5.

(b) Solve the equation 3x3 + x2 � 12x� 4 = 0.

Solution.

3x3 + x2 � 12x� 4 = 0;

x2(3x+ 1)� 4(3x+ 1) = 0;

(3x+ 1)(x2 � 4) = 0:

The three solutions are x = �2;�1=3; 2.

221

5. (a) Find the real number a such that a4 � 15a2 � 16 = 0 anda3 + 4a2 � 25a� 100 = 0.

Solution.

a4 � 15a2 � 16 = 0;

(a2 � 16)(a2 + 1) = 0:

Now for real a, a2 + 1 6= 0 so a2 � 16 = 0 and a = �4.Substituting these values in the other equation gives:

for a = �4; (�4)3 + 4(�4)2 � 25)� 4)� 100 = 0;

and a = �4 is a solution;

for a = 4; (43+ 4(4)

2� 25(4)� 100 = 138� 200 6= 0:

The only real solution is a = �4.(b) Find all positive numbers x such that xx

px= (x

px)x.

Solution. Now

xxpx= (x

px)x; so xx

px= (x3=2)x = x

32x:

Equating exponents:

xpx =

3

2x; so

px =

3

2and x =

�3

2

�2

=9

4;

since x > 0.

6. If x, y and z are real numbers prove that

(xjyj � yjxj)(yjzj � zjyj)(xjzj � zjxj) = 0:

Solution. Consider the left hand side. If any of x, y, z equal 0, say xthen (xjyj � yjxj) = (0� 0) = 0, and the equation holds. If none of x, y,z are zero then two must have the same sign. Without loss of generality wemay consider the two cases x > 0 and y > 0 and x < 0, y < 0.

If x; y > 0 then xjyj � yjxj = xy � yx = 0.

If x; y < 0 then xjyj � yjxj = x(�y)� y(�x) = �xy + xy = 0.

In any case the left hand side has one of the terms 0, so the productis 0 and the identity holds.

222

7. x and y are integers between 10 and 100. y is the number obtainedby reversing the digits of x. If x2 � y2 = 495 �nd x and y.

Solution. Let the digits of the numbers be a, b, with x = 10a+ b andy = a+ 10b (so 9 � a > b � 1). Now

x2 � y2 = (10a+ b)2� (a+ 10b)2

= ((10a+ b) + (a+ 10b))((10a+ b)� (a+ 10b))

= 11(a+ b) � 9(a� b)

= 99(a2 � b2) = 495:

Thus a2 � b2 = (a + b)(a� b) = 5 and a + b = 5, a � b = 1 so a = 3,b = 2. The numbers x and y are 32 and 23, respectively.

8. Three points P , Q and R lie on a circle. If PQ = 4 and \PRQ =

60� what is the radius of the circle?

Solution.

P

Q

R

O

M

r

r

rrr

r

r Let the radius of the circle be denotedby r, and the centre O. Now \POQ =

2\PRQ = 120�. Let the bisector of

PQ be at M . Now OM ? MQ

and \MOQ =1

2\POQ = 60

�, so

MQ

OQ=

2

r= sin60

�=

p3

2.

Thus r =4p3. The radius is

4p3

3.

9. Three points are located in the �nite region between the x-axis andthe graph of the equation 2x2 + 5y = 10. Prove that at least two of thesepoints are within a distance 3 of each other.

Solution.

-

6(0; 2)

(p5; 0)(�

p5; 0)

rr r

Consider the rectangles R1, R2 with ver-tices (�p5; 2), (�p5; 0), (0;0), (0;2)

and (0; 2), (0; 0), (p5; 0), (

p5; 2) respec-

tively. Every point of the region lies inone of the two boxes (or both). The di-

ameter of each box is

q(p5)2 + 22 = 3.

If three points in the region are given, twomust lie in one ofR1 or R2 by the Pigeon-hole Principle and cannot be further than3 units apart.

223

10. Three circles pass through the origin. The centre of the �rst circlelies in the �rst quadrant, the centre of the second circle lies in the secondquadrant, and the centre of the third circle lies in the third quadrant. If Pis any point that is inside all three circles, show that P lies in the secondquadrant.

Solution.

q

6-

The key is to consider the third circle. Ifits centre is in the third quadrant and itpasses through the origin then the onlypoint in common with the �rst quadrant isthe origin itself, so no point inside couldbe in the �rst quadrant. Similarly no pointinside the �rst circle can be in the thirdquadrant and no point in the second circlelies in the fourth quadrant. The result isimmediate.

That completes the Skoliad Corner for this issue. Please send me con-test material, submissions, and suggestions for the future direction of thisfeature.

Sports Writer's Math Oddity

In the issue of Sports Illustrated dated March 10, 1997, we �nd a longarticle about the coach of the University of Kansas Men's Basketball Coach.We quote:

Williams is straighter than an Arrow shirt, so square that he's di-

visible by four, and cornier than a corncob pipe.

Pity that the writer could not even get this little bit of math right!

224

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to

the Mayhem Editor, Naoki Sato, Department of Mathematics, University of

Toronto, Toronto, ON Canada M5S 1A1. The electronic address is

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).The rest of the sta� consists of Richard Hoshino (University of Waterloo), WaiLing Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Fibonacci Residues

Almost everyone is familiar with the Fibonacci sequence 0, 1, 1, 2, 3,5, 8, 13, : : : , where the �rst two terms are 0, 1, and each subsequent termis the sum of the previous two terms. This sequence, for those who don'tknow it, is very important and comes up in many problems and solutionsunexpectedly.

Now consider each term of the sequence modulo 2. We obtain thesequence 0, 1, 1, 0, 1, 1, : : : . This is not very interesting, as we know thatthe only values of integers modulo 2 are 0 and 1, so we expect something likethis which eventually repeats. Notice that the sequence that repeats (namely0, 1, 1) has three terms, and that the pairs (0; 1), (1; 0), and (1;1) appear inthe sequence, but the pair (0; 0) does not.

Likewise, if we consider the sequence modulo 3, we get the followingsequence:

0; 1; 1; 2; 0; 2; 2; 1; 0; 1; 1; 2; : : : :

The only values this sequence can have are 0, 1, and 2. This sequence repeatsafter eight terms. It contains all pairs (0; 1), (0; 2), : : : , (2; 2), except (0;0).

At this point, it might be reasonable to make a conjecture. Is it truethat upon taking the Fibonacci sequence modulo n, the sequence repeatsafter n2�1 terms and contains all the pairs (0; 1), (0;2), : : : , (n�1; n�1),except the term (0;0)? Maybe this is just true for prime integers n.

Before plunging headlong into any proofs, we should at least check thevalidity of the conjectures with more values of n, and it turns out that theyaren't true. Let r(n) be the length of the repeating part of the Fibonacci

225

sequence modulo n, known as the period. Here are the values of r(n) forthe �rst few positive integers n.

n r(n) n r(n) n r(n) n r(n) n r(n) n r(n) n r(n) n r(n)

2 3 8 12 14 48 20 60 26 84 32 48 38 18 44 30

3 8 9 24 15 40 21 16 27 72 33 40 39 56 45 120

4 6 10 60 16 24 22 30 28 48 34 36 40 60 46 48

5 20 11 10 17 36 23 48 29 14 35 80 41 40 47 32

6 24 12 24 18 24 24 24 30 120 36 24 42 48 48 24

7 16 13 28 19 18 25 100 31 30 37 76 43 88 49 112

We see that our conjecture was wrong. In fact, it appears that

r(2m) = 3� 2m�1; r(3m) = 8� 3

m�1; and r(5m) = 4� 5m:

Can this be proved? Are there patterns for other numbers other than powerof primes? Is there a closed form for the function r(n) in terms of the fac-torization of n?

Perhaps to answer this question we must look at other repeating se-quences other than that generated by (0; 1)modulo n. Let us call the repeat-ing sequence generated by (a; b) modulo n the residue-cycle of (a; b), anddenote it by the symbol ha; bin. For example, the residue-cycle h0;1i4 isf0, 1, 1, 2, 3, 1g, which we can verify has length six. There are also threeother residue-cycles: h0;2i4 = f0, 2, 2g, h0; 3i4 = f0, 3, 3, 2, 1, 3g and thetrivial sequence h0;0i4 = f0g. For n = 4, all possible residue-cycles areh0; 0i4, h0; 1i4, h0; 2i4, and h0; 3i4. Note h1; 1i4 = h0;1i4, since these se-quences are understood to be cyclic, so any pair of consecutive terms in aresidue-cycle can generate it.

Some observations:

(1) ha; bin need not be hb; ain. For example, h1; 3i4 and h3; 1i4 are di�er-ent residue-cycles.

(2) It might seem that the residue-cycles form a group structure under theoperation ha; bin + hc; din = ha+ c; b+ din, where addition of a+ c

and b+ d is taken modulo n. This is not true. For example, assumingsuch a group operation could exist, consider the residue-cycles h0;1i5= f0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1g = h0; 2i5 andh1; 3i5 = f1, 3, 4, 2g.

We have then h0; 1i5+h0; 1i5 = h0; 2i5 = h0;1i5, and h0; 1i5 = h1;2i5so h0;1i5 + h0; 1i5 = h0;1i5 + h1; 2i5 = h1; 3i5, contradiction.

(3) For any n, if there are k distinct residue-cycles, with the ith one con-

taining li elements, then we have the relationPk

i=1 li = n2.

As a corollary of this fact, we see that the length of the residue-cycleh1; 1in is at most n2 � 1.

(4) If a residue-cycle has length one, it must be h0; 0in for some n.

We see that answering the question of what the length of the residue-cycle generated by (0;1) modulo n is, is a speci�c case of knowing what allthe possible lengths are.

226

Givenn, wemay list the lengths of the distinct residue-cycles modulon.

n lengths of residue-cycles modulo n

2 1, 3

3 1, 8

4 1, 3, 6, 6

5 1, 4, 20

7 1, 16, 16, 16

8 1, 3, 6, 6, 12, 12, 12, 12

Is there a pattern in these numbers? Do the residue-cycles for a givennform some kind of structure? (group, ring, etc.)

Acknowledgements

Anand Govindarajan, First Year student, University of Toronto at Scar-borough, for conjecturing that r(n) is related to the factorization of n.

What is the next term?

Naoki Satostudent, University of Toronto

Toronto, Ontario

There are many puzzles in which the �rst few terms of a sequence are given,and the next term is asked to be determined. Supposedly, there will be aunique rule for the sequence. A similar type of problem appears in mathe-matics, where we are also told that the sequence is given by a polynomial.Since polynomials have nice properties, so do these sequences, some of whichwe will explore here. First, we will lay down a few important and motivatingresults.

Result 1. Given any distinct n+1 reals, x0, x1, : : : , xn, and any n+1

reals, y0, y1, : : : , yn, there exists a unique polynomial p of degree at most n,such that p(xi) = yi, 0 � i � n.

Proof. Assume p(x) = cnxn+ cn�1xn�1 + � � � + c1x + c0, for some

constants ci. The condition p(xi) = yi, 0 � i � n, is equivalent to thefollowing linear system in the variables ci:

cnxn0 + cn�1x

n�10 + � � �+ c1x0 + c0 = y0;

cnxn1 + cn�1x

n�11 + � � �+ c1x1 + c0 = y1;

.

.

.

cnxnn + cn�1x

n�1n + � � �+ c1xn + c0 = yn:

227

The matrix�xji

�0�i;j�n

has non-zero determinant, since it is Vander-

monde's determinant. By Cramer's Rule, there is a unique solution for the ci,and hence for p. �

The next result is simply a corollary, sometimes called the Identity The-orem for polynomials.

Result 2. If two polynomials of degree at most n agree at n+1 points,then they are identical.

This is interesting stuff, but what does it have to do with extrapolatingsequences? In these problems, we will �t a polynomial to the given data.By the Identity Theorem, this polynomial will be the one we seek. What isparticularly nice about this method is that we will not have to appeal to theLagrange Interpolation Formula or any other tedious equations to actually�nd the polynomial we seek (note that the �rst result only assures us of theexistence of a polynomial p; little mention is made of what the coef�cientsmight be).

Problem 1. Let p be the polynomial of least degree satisfyingp(k) = 1=(k+ 1) for k = 0; 1; : : : ; n. Find p(n+ 1).

Solution. Let q(x) = (x+ 1)p(x)� 1. Then q(k) = 0 for 0 � k � n,so q(x) = cx(x� 1) � � � (x�n) for some constant c. Note q(�1) = �1, but

q(�1) = c(�1)(�2) � � � (�n� 1) = c � (�1)n+1(n+ 1)!;

so c = (�1)n=(n+ 1)!, and

q(x) = (�1)nx(x� 1) � � � (x� n)=(n+ 1)!;

givingq(n+ 1) = (n+ 2)p(n+ 1)� 1 = (�1)n;

so thatp(n+ 1) = (1 + (�1)n)=(n+ 2):

Hence,

p(n+ 1) =

�2=(n+ 2) if n is even,0 if n is odd:

Problem 2. Let p be the polynomial of least degree satisfying p(k) = 1

for k = 0; 1; : : : ; n� 1, and p(n) = 0. Find p(n+ 1).

Solution. Let q(x) = p(x)� 1, so that

q(x) = cx(x� 1)(x� 2) � � � (x� (n� 1))

for some constant c. Then

q(n) = c � n! = p(n)� 1 = �1;

228

givingc = �1=n!;

so that

p(n+ 1) = q(n+ 1) + 1 = c � (n+ 1)! + 1 = �n� 1 + 1 = �n:

Problem 3. Let p be the polynomial of least degree satisfyingp(k) = 2

k, for k = 0; 1; : : : ; n. Find p(n+ 1).

Solution. Let

q(x) =

�x

0

�+

�x

1

�+ � � �+

�x

n

�:

(Note that

�x

k

�is simply a polynomial in x, and that upon expanding, it is

easy to see that�x

k

�= 0 for 0 � x < k, where x is an integer.) It is easy to

verify that q(k) = 2k for k = 0; 1; : : : ; n. Hence, p � q. Furthermore,

p(n+ 1) =

�n+ 1

0

�+

�n+ 1

1

�+ � � �+

�n+ 1

n

�

= 2n+1 �

�n+ 1

n+ 1

�= 2

n+1 � 1:

Sometimes though, these methods are not su�cient. We turn to an-other method, one that is probably well-known among people who haveplayed around with sequences.

Given a polynomial p, we list p(0), p(1), p(2), : : : , in sequence. Underthis row, write the sequence of differences between consecutive terms, andthen repeat. We thus form a difference table.

For example, for

p(x) = (2x3 � 15x2 + 19x+ 6)=6;

the table will be:

1 2 0 �3 �5 �4 2 15 � � �

1 �2 �3 �2 1 6 13 � � �

�3 �1 1 3 5 7 � � �

2 2 2 2 2 � � �

.

.

.

We can in fact do the same for any function, but only polynomials havethis property:

229

Result 3. The difference table of p will eventually have a constant rowif and only if p is a polynomial.

To see this, we only need to realize that a row represents a polynomialof degree k if and only if the row below represents a polynomial of degreek�1. One direction is trivial. To see the other direction, we use a telescopingsum, the same that is used to �nd the sum of the kth powers of 1, 2, : : : , n(which is in fact a polynomial of degree k + 1). Since a constant is simply apolynomial of degree 0, the result follows.

Now, several nice properties follow from this result. First, if the valueof a polynomial p, of degree at most n, is an integer for n+1 consecutive in-tegers, then p(x) is an integer for all integers x. Note that it is not necessaryfor p to have integer coe�cients (as seen in the example above). Second, ifwe label the �rst row 0, and the second 1, etc., then we �nd that the degreeof the polynomial is the index of the �rst row to become constant. In theabove example, p is a cubic, and indeed the third row is the �rst to becomeconstant. We will exploit this property in extrapolating sequences. (Note: Inthe example, the �rst terms in each row, reading down, are 1, 1, �3, and 2.A nice result in this subject says that

p(x) = 1

�x

0

�+ 1

�x

1

�+ (�3)

�x

2

�+ 2

�x

3

�:

A proof will not be provided here, but should not be too hard to construct.)

Problem 4. Let p be the polynomial of least degree satisfyingp(k) = (�1)k, for k = 0; 1; : : : ; n. Find p(n+ 1).

Solution. Suppose n is odd. Writing out the difference table,

1 �1 1 �1 � � � �1

�2 2 �2 � � � �2

4 �4 � � � �4

� � �

2n�1

�2n�1

�2n

We know that the last row, the nth row, is constant. Hence, the tablecan be extended. See Figure 1 on page 230.

Hence, p(n + 1) = �2n+1+ 1. Note that this is simply the sum of

the rightmost elements of the �rst difference table. Similarly, for n even,p(n+ 1) = 2

n+1 � 1.

Problem 5. Let Fn denote the nth Fibonacci number. Let p be the poly-

nomial of least degree satisfying p(k) = Fk, for k = 0; 1; : : : ; n. Calculatep(n+ 1).

230

1 �1 1 �1 : : : �1 �2n+1 + 1

�2 2 �2 : : : �2 �2n+1 + 2

4 �4 : : : �4 �2n+1 + 4

: : :

2n�1 �2n�1 �2n � 2

n�1

�2n �2n

Figure 1.

Solution. Forming the difference table,

F0 F1 F2 F3 � � � FnF�1 F0 F1 � � � Fn�2

F�2 F

�1 � � � Fn�4� � �

F�n+1 F

�n+2

F�n

Then p(n+ 1) is simply the sum of the right-most values, or

F�n + F�n+2 + � � �+ Fn�2 + Fn:

Now, F�k = (�1)k+1Fk for all k. Therefore, if n is even, all the terms willcancel, and the sum is 0. If n is odd, the sum is

2(F1 + F3 + � � �+ Fn) = 2(F2 � F0 + F4 � F2 + � � �+ Fn+1 � Fn�1)

= 2(Fn+1 � F0) = 2Fn+1:

Problems

A. Find the monic polynomial f of minimal degree such that

f(x) � 0 (mod 100):

B. Find the value of the polynomial p of least degree satisfying the follow-ing:

(a) p(k) = ak, k = 0, 1, : : : , n,

(b) p(k) =�n

k

�, k = 0, 1, : : : , n.

(Note that part (a) generalizes Problems 2, 3, and 4.)

231

C. Let p be a polynomial of degree at most n. Show that

n+1Xk=0

(�1)k�n+ 1

k

�p(k) = 0:

Hint: Use the difference table, or realize that the values p(0), p(1),: : : , p(n+ 1) must satisfy a recurrence relation. Which one?

D. Assume p is a polynomial of degree at most n such that p(k2) is an inte-ger for k = 0; 1; : : : ; n. Show that p(k2) is an integer for all integers k.

Euler's and DeMoivre's Theorem

Soroosh Yazdanistudent, University of Waterloo

Waterloo, Ontario

A fundamental relation in mathematics is given by Euler's Theorem,which states:

ei� = cos � + i sin �: (1)

Using the Taylor expansion, one can check the identity as follows:

ei� = 1 +i�

1!+

(i�)2

2!+

(i�)3

3!+ � � �

= 1 + i�

1!� �2

2!� i

�3

3!+ � � �

=

�1� �2

2!+�4

4!� � � �

�+ i

��

1!� �3

3!+�5

5!� � � �

�= cos � + i sin�:

A result, known as Demoivre's Theorem immediately follows:

(cos � + i sin �)n = cosn�+ i sinn�:

We will use this result to derive several identities quickly and elegantly.

Now using the fact that cos(��) = cos � and sin(��) = � sin �, we get:

e�i� = ei(��) = cos(��) + i sin(��) = cos � � i sin �:

232

Adding this result with (1) we get:

ei� + e�i� = 2 cos � =) ei� + e�i�

2= cos �:

Similarly,ei� � e�i�

2i= sin�:

Combining these identities, we get

tan � =sin �

cos �=

ei� � e�i�

i(ei� + e�i�):

This means that we can write trigonometric functions as pure exponentialequations. From this we can prove many identities.

Problem: Prove that

2 sin

��+ �

2

�cos

��� �

2

�= sin�+ sin�:

Solution:

L.H.S. = 2 sin

��+ �

2

�cos

��� �

2

�

= 2

ei

�+�2 � e�i

�+�2

2i

! ei

���2 + e�i

���2

2

!

=ei

�+�2 ei

���2 + ei

�+�2 e�i

���2 � e�i

�+�2 ei

���2 � e�i

�+�2 e�i

���2

2i

=ei� + ei� � e�i� � e�i�

2i= sin�+ sin�

= R.H.S.

Problems

1. Prove 2 cos(�+�

2) cos(

���2

) = cos�+ cos�:

2. Prove 2 sin(�+�

2) sin(

���2

) = cos� � cos�:

3. Prove sin(�+ �) = sin� cos� + cos� sin�:

Here is another neat problem: What is the expansion of cosn�?We will need the Binomial Theorem.

233

cosn� =1

2(ein� + e�in�)

=1

2

�(ei�)n + (e�i�)n

�=

1

2

�(cos � + i sin�)n + (cos � � i sin�)n

�=

1

2

0@ nXj=0

�n

j

�(cos

n�j �)(ij)(sinj �)

+

nXj=0

�n

j

�(cos

n�j �)(�i)j(sinj �)1A

=1

2

nXj=0

�n

j

�(cos

n�j � sinj �)�ij + (�i)j�

=

�n

0

�cos

n � ��n

2

�cos

n�2 � sin2 � +

�n

4

�cos

n�4 � sin4 � � � � �

Some readers will notice at this point that we got the real part of(cos � + i sin�)n. Although we could have come to the same conclusionby equating real and imaginary parts of the expansion, I thought that thispath might be a bit more interesting. A similar expression for sinn� exists,which is left as an exercise for the reader to derive.

Here is a problem that takes advantage of the exponential form oftrigonometric equations.

Problem: Evaluate

nXj=0

sin j�.

Solution:

nXj=0

sin j� =1

2i

nXj=0

�eij� � e�ij�

�=

1

2i

nXj=0

�(ei�)j � (e�i�)j

�

=1

2i

(ei�)n+1 � 1

(ei� � 1)� (e�i�)n+1 � 1

(e�i� � 1)

!

=((ei�)n+1 � 1)(e�i� � 1)� ((e�i�)n+1 � 1)(ei� � 1)

2i(ei� � 1)(e�i� � 1)

=(ein� � e�i� � ei(n+1)�

+ 1)� (e�in� � ei� � e�i(n+1)�+ 1)

2i(2� ei� � e�i�)

=sinn� + sin� � sin(n+ 1)�

2� 2 cos �:

234

Now we can simplify this equation using our previous identities.

sinn� + sin� � sin(n+ 1)�

2� 2 cos �

=(sinn�+ sin(�(n+ 1)�)) + (sin� + sin0)

2(cos 0� cos �)

=2cos(n+

1

2)� sin(��

2) + 2 sin

�

2cos

�

2

4 sin�

2sin

�

2

=� cos(n+

1

2)� + cos

�2

2 sin�2

=2sin(

n+1

2)� sin(n

2)�

2 sin�2

=sin(

n+1

2)� sin(n

2)�

sin�2

:

We can also evaluatenXj=0

cos j�, which is left as an exercise for the

reader.

The last identity that we will derive is the expansion of tann�. We

have seen that tan � =ei� � e�i�

i(ei� + e�i�).

Therefore, i tan � =ei� � e�i�

ei� + e�i�. Now we will evaluate i tann�:

i tann� =ein� � e�in�

ein� + e�in�=

�ei��n � �e�i��n

(ei�)n+ (e�i�)n

=

�2ei�

�n � �2e�i��n(2ei�)

n+ (2e�i�)n

=

�2ei�

ei�+e�i�

�n��

2e�i�

ei�+e�i�

�n�

2ei�

ei�+e�i�

�n+

�2e�i�

ei�+e�i�

�n=

(1 + i tan �)n � (1� i tan �)n

(1 + i tan �)n + (1� i tan �)n

235

=

nXj=0

�n

j

�(i tan �)j �

nXj=0

�n

j

�(�i tan �)j

nXj=0

�n

j

�(i tan �)j +

nXj=0

�n

j

�(�i tan �)j

=

nXj=0

�n

j

�[(i tan �)j � (�i tan �)j ]

nXj=0

�n

j

�[(i tan �)j + (�i tan �)j ]

=

i

��n

1

�tan

1 � ��n

3

�tan

3 � + � � ��

�n

0

�tan

0 � ��n

2

�tan

2 � + � � �;

which implies that

tann� =

�n

1

�tan

1 � � �n3

�tan

3 � + � � ��n

0

�tan

0 � � �n2

�tan

2 � + � � � :

Mayhem Problems

The Mayhem Problems editors are:

Cyrus Hsia Mayhem Advanced Problems Editor,Richard Hoshino Mayhem High School Problems Editor,Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |see the relevant section. In this issue, you will �nd only solutions | thenext issue will feature only problems.

We warmly welcome proposals for problems and solutions. With thenew schedule of eight issues per year, we request that solutions from theprevious issue be submitted by 1 October 1997, for publication in the issue5 months ahead; that is, issue 2 of 1998. We also request that only students

submit solutions (see editorial [1997: 30]), but we will consider particularlyelegant or insightful solutions from others. Since this rule is only being im-plemented now, you will see solutions from many people in the next fewmonths, as we clear out the old problems from Mayhem.

236

High School Problems | Solutions

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,Canada. L3S 3E8 rhoshino@undergrad.math.uwaterloo.ca

H208. In a triangle ABC with incentre I and angles 2a, 2b, and 2c

at vertices A, B, and C respectively, and circumradius R, show thatCI = 4R sina sin b.

Solution by Bob Prielipp, University of Wisconsin-Oshkosh, WI, USA.To avoid confusion, let the angles a, b, and c be denoted by �, �, and .

Then the result follows immediately from the following theorem:

Theorem. r = 4R sin� sin� sin , where r is the in-radius.Proof. By the extended Sine Law,

a

2R= sin2� = 2 sin� cos�:

Also, sin� =r

AIand cos� =

s� a

AIin the small triangle with vertices A, I

and the point of tangency on side AB with the incircle.

Putting all this together gives

sin2 � =

ar

4R(s� a);

and likewise we have similar results for the other two angles. Their productgives

sin2 � sin

2 � sin2 =

abcr3

(4R)3(s� a)(s� b)(s� c):

Let K be the area of triangle ABC. Using the relations

abc

4R= K; rs = K; and K =

ps(s� a)(s� b)(s� c)

(Heron's formula), the result follows.

Armed with this, we use sin =r

CI, so that

CI =r

sin =

4R sin� sin� sin

sin = 4R sin� sin�;

as required.

H209. In an acute triangle with angles a, b, and c, show that thefollowing inequality holds:

sinc

cos(a� b)+

sina

cos(b� c)+

sinb

cos(c� a)� sina+ sinb+ sin c

sin2a+ sin2b+ sin2c:

237

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.

Without loss of generality, assume that b � a. Note that

sin(a� �) + sin(b+ �) = cos�(sina+ sin b)� sin�(cosa� cos b):

(1)

We have 0 � cos� � 1, sin� � 0, and cos a � cos b � 0. Thus, for0 � � � �=2, we have that (1) is less than or equal to sina+ sin b.

Setting � = �=2� b > 0, we have

sin(a+ b� �=2) + sin�=2 � sina+ sin b:

Since the triangle is acute, a+ b > �=2 and

1 < sina+ sinb < sina+ sinb+ sin c:

Using this inequality and the Weighted AM-HM Inequality with

x1 = cos(b� c); x2 = cos(c� a); x3 = cos(a� b);

and weights

w1 =sinaPsina

; w2 =sin bPsina

; w3 =sincPsina

;

respectively, we get

x1w1 + x2w2 + x3w3 �1

w1

x1+

w2

x2+

w3

x3

;

so that

w1

x1+w2

x2+w2

x3� 1

w1x1 +w2x2 +w3x3:

Let

D = sin(a+ c� b) + sin(b+ c� a) + sin(a+ b� c)

+ sin(a+ c� b) + sin(b+ c� a) + sin(a+ b� c):

238

Thus

sin c

cos(a� b)+

sina

cos(b� c)+

sin b

cos(c� a)

� (sina+ sinb+ sin c)2

sinc cos(a� b) + sina cos(b� c) + sinb cos(c� a)

>sina+ sin b+ sinc

sinc cos(a� b) + sina cos(b� c) + sinb cos(c� a)

=2(sina+ sin b+ sin c)

2 sinc cos(a� b) + 2 sina cos(b� c) + 2 sin b cos(c� a)

=2(sina+ sin b+ sin c)

D

=2(sina+ sinb+ sin c)

sin2b+ sin2a+ sin2c+ sin2b+ sin2a+ sin2c

=sina+ sinb+ sin c

sin2a+ sin2b+ sin2c:

H210. A \shu�e" of 2n cards labelled a1, a2, : : : , a2n in that orderconsists of taking the �rst n cards and merging it with the last n so that a1,a2, : : : , an will be in that order somewhere in the new pile and likewise foran+1, an+2, : : : , a2n. Prove that if we start with a pile of 2n cards labelled1, 2, : : : , 2n, then it is possible after a �nite number of shu�es to get thecards in the reverse order.

Solution. De�ne a \step" to be two shu�es de�ned in the problem.We will show by induction on k that after k steps, it is possible to place thecards a2n, a2n�1, : : : , a2n�k+1 in the �rst k positions, the cards ak, ak�1,: : : , a1 in the last k positions, and leave the remaining (n � 2k) cards intheir original positions. Call the position that the card labelled ai originallyoccupies position i.

Base Case: k = 1. We show that in one step (or two shu�es) cards a2nand a1 can be in the �rst and last positions respectively and the other cardsleft in their original positions. For the �rst shu�e, put a1 at position n anda2, a3, : : : , an at positions n + 2, n + 3, : : : , 2n respectively, and an+1,an+2, : : : , a2n�1 in positions 1, 2, : : : , n�1 respectively, leaving card a2n togo into positionn+1. For the second shu�e just take the �rstn cards and putit after the last n cards to give the desired con�guration. Diagrammatically,this is as follows:

a1 a2 : : : an an+1 : : : a2n�1 a2n#

an+1 an+2 : : : a2n�1 a1 a2n a2 : : : an#

a2n a2 : : : , an an+1 : : : a2n�1 a1

239

Induction Step: Suppose after k steps, 1 � k < n, that the hypothesisholds, that is, cards a2n, : : : , a2n�k+1 and ak, : : : , a1 are in positions 1, 2,: : : , k, and 2n� k + 1, : : : , 2n respectively and the other (n� 2k) cardsare in their original positions. Now for the �rst shu�e on the (k+1)

th stepleave a2n, : : : , a2n�k+1 and ak, : : : , a1 in their respective positions. Placecard ak+1 in position n+1 and cards ak+2, : : : , an in positions n+1, : : : ,2n � k � 1. Then place cards an+1, : : : , a2n�k+1 in positions k + 1, : : : ,n�2 and �nally a2n�k must go in positionn+1. For the second shu�e againleave a2n, : : : , a2n�k+1 and ak, : : : , a1 alone. Place cards at positions k+1,: : : , n in positions n+ 1, : : : , 2n� k and those cards that were at positionn+1, : : : , 2n�k down into positions k+1, : : : , n. Diagrammatically, thisis as follows:

a2n : : : a2n�k+1 ak+1 : : : an an+1 : : : a2n�k ak : : : a1#

a2n : : : a2n�k+1 an+1 : : : a2n�k+1 ak+1 a2n�k ak+2 : : : a2n�k ak : : : a1#

a2n : : : a2n�k+1 a2n�k ak+2 : : : an an+1 : : : a2n�k�1 ak+1 ak : : : a1

Induction is complete, so when k = n we have all the cards in thereverse order and we are done.

Advanced Problems | Solutions

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.M1G 1C3 hsia@math.toronto.edu

A184. Let A be a 3� 3matrix with rational entries such that A8= I.

Show that A4= I.

Solution

Let mA(x) be the minimal polynomial of A (over the rationals). ThenmA(x) is a polynomial with rational coe�cients, with degree at most 3. Wealso know mA(x) divides x

8 � 1, whose factorization over the rationals is(x� 1)(x+ 1)(x2 + 1)(x4 + 1). The polynomialmA(x) must be a productof some of these factors, and x4 + 1 cannot be one of them. Hence, mA(x)

divides (x� 1)(x+ 1)(x2 + 1) = x4 � 1) A4= I.

Ravi Vakil and Nicolas Guay both correctly pointed out that the orig-

inal statement of the problem, which had \real" instead of \rational", was

awed, as seen in the counterexamples:

A =

0@p2� 1 1 0p2� 2 1 0

0 0 1

1A and A =

0@p2 �1 0

1 0 0

0 0 1

1A ;

which they provided respectively.

240

Also solved byMIGUEL CARRI �ON �ALVAREZ, Universidad Complutense

de Madrid, Spain; DONNY CHEUNG, University of Waterloo, Waterloo, On-

tario; EDWARD WANG, Wilfrid Laurier University, Waterloo, Ontario.

A188. Let ABCD and A0B0C0D0 be squares with opposite orienta-tion, with A = A0. Let M be the midpoint of DD0. Show that AM ? BB0

and 2 � AM = BB0.

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.A 90

� rotation about A brings B0 to D0 and D to B. Therefore,

��!AB0 � �!AD =

��!AD0 � �!AB:

Hence,

2��!AM � ��!BB0 =

��!AD +

��!AD0

�����!AB0 ��!

AB�

=�!AD � ��!AB0 ��!

AD � �!AB +��!AD0 � ��!AB0 ���!AD0 � �!AB

=�!AD � ��!AB0 � 0 + 0���!

AD0 � �!AB = 0;

and so, AM ? BB0.

Now,�����!AB0��� = �����!AD0���, ����!AB��� = ����!AD���, and

\BAB0 + \DAD0 = 180�;

so that �!AB � ��!AB0 = ��!AD � ��!AD0:

Therefore,

4

�����!AM ���2 = 4��!AM � ��!AM =

��!AD +

��!AD0

����!AD +

��!AD0

�=

����!AD���2 + 2�!AD � ��!AD0 +

�����!AD0���2=

����!AB���2 � 2�!AB � ��!AB0 +

�����!AB0���2=

�����!BB0���2 :Hence, 2 � AM = BB0.

Also solved byMIGUEL CARRI �ON �ALVAREZ, Universidad Complutense

de Madrid, Spain.

241

A189. Prove that no seven positive integers, not exceeding 24, canhave sums of all subsets distinct.

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.Assume that it is possible to �nd a set of seven natural numbers, such

that all sums of subsets are distinct. The numbers must be distinct, andhence there is a maximum value, x, and a minimum value, y. The set has27 � 1 = 127 non-empty subsets, and hence the di�erence between the

largest sum and the smallest sum must be at least 126. The smallest summust be y. The largest sum cannot exceed

x+ (x� 1) + (x� 2) + (x� 4) + (x� 5) + (x� 7) + y = 6x� 19 + y;

since there cannot be four consecutive terms of an arithmetic progression(�rst plus last equals sum of middle two terms). Therefore the di�erencebetween the largest and the smallest sum cannot exceed 6x� 19. This takesits maximum value when x = 24. However, 6 � 24 � 19 = 125 < 126,contradiction. Therefore no seven positive integers, not exceeding 24, canhave sums of all subsets distinct.

Also solved by EDWARD WANG, Wilfrid Laurier University, Waterloo,

Ontario.

242

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 81

2"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed to

the Editor-in-Chief, to arrive no later than 1 December 1997. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication.

2238. Proposed byWaldemar Pompe, student,University ofWarsaw,

Poland.

A four-digit number abcd is said to be faulty if it has the followingproperty:

The product of the two last digits c and d equals the two-digitnumber ab, while the product of the digits c� 1 and d� 1 equalsthe two digit number ba.

Determine all faulty numbers!

2239. Proposed by Kenneth Kam Chiu Ko, Mississauga, Ontario.

Suppose that 1 � r � n and consider all subsets of r elements of theset f1, 2, 3, : : : , ng. The elements of these subsets are arranged in ascendingorder of magnitude. For i from 1 to r, let ti denote the ith smallest elementin the subset. Let T (n; r; i) denote the arithmetic mean of the elements ti.

Prove that T (n; r; i) = in+ 1

r + 1.

243

2240. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Let ABC be an arbitrary triangle with the pointsD, E, F on the sides

BC, CA, AB respectively, so thatBD

DC� BF

FA� 1 and

AE

EC� AF

FB.

Prove that [DEF ] � [ABC]

4with equality if and only if two of the

three points D, E, F , (at least) are mid-points of the corresponding sides.

Note: [XY Z] denotes the area of triangle4XY Z.2241. Proposed by Toshio Seimiya, Kawasaki, Japan.

Triangle ABC (AB 6= AC) has incentre I and circumcentre O. Theincircle touches BC at D. Suppose that IO ? AD.

Prove that AD is a symmedian of triangle ABC. (The symmedian isthe re ection of the median in the internal angle bisector.)

2242. Proposed by K.R.S. Sastry, Dodballapur, India.

ABCD is a parallelogram. A point P lies in the plane such that

1. the line through P parallel to DA meets DC at K and AB at L,

2. the line through P parallel to AB meets AD at M and BC at N , and

3. the angle between KM and LN is equal to the non-obtuse angle ofthe parallelogram.

Find the locus of P .

2243. Proposed by F.J. Flanigan, San Jose State University, San Jose,

California, USA.

Given f(x) = (x� r1)(x� r2) : : : (x� rn) and f0(x) = n(x� s1)(x�

s2) : : : (x � sn�1), (n � 2), consider the harmonic mean h of the n(n� 1)

di�erences ri � sj.

If f(x) has a multiple root, then h is unde�ned, because at least oneof the di�erences is zero.

Calculate h when f(x) has no multiple roots.

2244. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle and D is a point on AB produced beyond B suchthat BD = AC, and E is a point on AC produced beyond C such thatCE = AB. The perpendicular bisector of BC meets DE at P .

Prove that \BPC = \BAC.

244

2245. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,

Madrid, Spain.

Prove that3n+ (�1)(n2)

2� 2

n is divisible by 5 for n � 2.

2246. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Suppose thatG, I and O are the centroid, the incentre and the circum-centre of a non-equilateral triangle ABC.

The line throughB, perpendicular toOI intersects the bisector of\BACat P .

The line through P , parallel to AC intersects BC at M .

Show that I, G andM are collinear.

2247. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,

Austria.

(a) Suppose that n � 3 is an odd natural number.

Show that the only polynomial P 2 R[x] satisfying the functional equa-tion:

(P (x+ 1))n= (P (x))

n+

n�1Xk=0

�n

k

�xk; for all x 2 R;

is given by P (x) = x.

(b)? Suppose that n � 1 is a natural number.

Show that the only polynomial P 2 R[x] satisfying the functional equa-tion:

(P (x+ 1))n= (P (x))

n+

n�1Xk=0

�n

k

�xk; for all x 2 R;

is given by P (x) = x.

(c)? Suppose that n � 1 is a natural number.

Show that the only polynomial P 2 R[x] satisfying the functional equa-tion:

P ((x+ 1)n) = (P (x))

n+

n�1Xk=0

�n

k

�xk; for all x 2 R;

is given by P (x) = x.

245

2248. Proposed by Shawn Godin, St. Joseph Scollard Hall, North

Bay, Ontario.

Find the value of the sum

1Xk=1

d(k)

k2; where d(k) is the number of posi-

tive integer divisors of k.

2249. Proposed by K.R.S. Sastry, Dodballapur, India.

How many distinct acute angles � are there such that

cos� cos 2� cos 4� =1

8?

2250. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a scalene triangle with incentre I. Let D, E, F be the pointswhere BC, CA, AB are tangent to the incircle respectively, and let L, M ,N be the mid-points of BC, CA, AB respectively.

Let l,m, n be the lines throughD,E, F parallel to IL, IM , IN respectively.

Prove that l, m, n are concurrent.

Historical Titbit

Taken from a 1950's University Scholarship Paper.

1. A, B, C, D are four points on a circle.

The tangent at B meets CD at L, the tangent at C meets AB at M ,and AD meets BC at N .

Prove that L, M , N are collinear.

2. If ABCDEF are concyclic and A(CDEF ) is a harmonic pencil, provethat B(CDEF ) is also a harmonic pencil.

246

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2129?. [1996: 123]Proposedby Stanley Rabinowitz,Westford,Massa-

chusetts, USA.For n > 1 and i =

p�1, prove or disprove that

1

4i

4nXk=1

gcd (k;n)=1

ik tan

�k�

4n

�

is an integer.

Solutionby G.P. Henderson, Campbellcroft, Ontario; (modi�ed slightly

by the editor).

Let S(n) denote the given sum. We show that S(n) is always an inte-ger. First we need a lemma:

Lemma. Let n be a positive integer, and let g(n) =

nXj=1

tan

�4j � 3

4n

�.

Then g(n) = (�1)n+1n.

Proof. We �rst obtain an expansion of tan(n�). [Ed: see also [1997: 233].]Using the following well-known formulae:

sin(n�) =

n

1

!sin� cos

n�1��

n

3

!sin

3� cos

n�3�+ : : : ; (1)

cos(n�) = cosn� �

n

2

!sin

2� cos

n�2� +

n

4

!sin

4� cos

n�4� � : : : : (2)

Dividing both the numerator and the denominator of the fraction(1)

(2)by

cosn �, and setting x = tan �, we get

tan(n�) =P (x)

Q(x);

where P (x) and Q(x) are polynomials in x. Speci�cally, if n is odd, then

P (x) =

�n

1

�x�

�n

3

�x3 + : : :+ (�1)(n�1)=2

�n

n

�xn;

Q(x) = 1��n

2

�x2 + : : :+ (�1)(n�1)=2

�n

n� 1

�xn�1;

and if n is even, then the last terms of P (x) and Q(x) are

�(�1)n=2�

n

n� 1

�xn�1 and (�1)n=2

�n

n

�xn;

247

respectively.

Now consider the equation

tan(n�) = 1 =P (x)

Q(x): (3)

The roots of tan(n�) = 1 are clearly given by � = (4j � 3)=4n; that isx = tan[(4j � 3)=4n], j = 1, 2, : : : , n. Thus g(n) is simply the sum of theroots of tan(n�) = 1.

On the other hand, the roots ofP (x)

Q(x)= 1 are the roots of

P (x)� Q(x) = 0. Hence, when n is odd, the sum of the roots is

� �coe�cient of xn�1 in (�Q(x))�[coe�cient of xn in P (x)]

= n = (�1)n+1n;

and when n is even, the sum of the roots is�coe�cient of xn�1 in (P (x))

�[coe�cient of xn in �Q(x)] = �n = (�1)n+1n:

Therefore, g(n) = (�1)n+1n. The proof of the lemma is complete.

Continuing, set A = fk j 1 � k � 4n; gcd(k; n) = 1g. Note that ifk 2 A, then 4n� k 2 A and 4n� k 6= k. If k is even, then the sum of theterms corresponding to k and 4n� k is zero. If k is odd, then one of k and4n� k is congruent to 1 and the other to �1, (all congruences are modulo 4,unless otherwise stated), and the corresponding terms are equal. Therefore

S(n) =1

2

Xk2B

tan

�k�

4n

�;

where B = fk j 1 � k � 4n; gcd(k; n) = 1 and k � 1 (mod 4)g.Let n = 2

ma where a is odd. If a = 1, then n = 2m, m � 1, and

k 2 B if and only if k = 4j � 3, j = 1, 2, : : : , n. Hence

S(n) =g(n)

2= �n

2;

which is an integer.

If a � 3, let p1, p2, : : : , pr, denote the distinct (odd) prime divisorsof a, and suppose that pj � bj, where bj = �1 for j = 1, 2, : : : , r.

To get S(n) from g(n), we must subtract the terms with k divisible byat least one of p1, p2, : : : , pr.

Set h = h(j1; j2; : : : ; js) =

Xk

tan

�k�

4n

�, where the summation is

over all k such that k � 1, and k is a multiple of c = pj1pj2 : : : pjs , where1 � s � r and all the p's are distinct.

248

If c � 1, then the values of k in the sum are c, 5c, : : : , (4d�3)c, whered = n=c. Hence

h =

dXj=1

tan

�(4j � 3)�

4d

�= g(d) = (�1)d+1d

= (�1)n+1n

pj1pj2 : : : pjs; (4)

since n = cd and c being odd, together imply that d � n (mod 2).

If c � �1, then the values of k in the sum are 3c, 7c, : : : , (4d � 1)c

and

h =

dXj=1

tan

�(4j � 1)�

4d

�:

Replacing j by d� j + 1, we �nd that

h = �g(d) = �(�1)d+1d

= �(�1)n+1n

pj1pj2 : : : pjs: (5)

Since

bj1bj2 : : : bjs =

�1 if c � 1;

�1 if c � �1;the answers in formulae (4) and (5) can be combined in the formula:

h = (�1)n+1nbj1bj2 : : : bjs

pj1pj2 : : : pjs:

It then follows from the Inclusion{Exclusion Principle that:

S(n) =1

2

0@g(n)� nX

j=1

h(j)+Xj1<j2

h(j1; j2)� : : :

1A

=(�1)n+1n

2

0@1� nX

j=1

bj

pj+bj1bj2

pj1pj2� : : :

1A

=(�1)n+1n

2

�1� b1

p1

��1� b2

p2

�: : :

�1� br

pr

�

=(�1)n+1n(p1 � b1)(p2 � b2) : : : (pr � br)

2p1p2 : : : pr

which is clearly an integer. (In fact, it is divisible by 2m+2r�1 since pj � bj

is divisible by 4.) This completes the proof.

249

Also solved by KURT GIRSTMAIR, University of Innsbruck, Innsbruck,

Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA, and

V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA.

Girstmair, using character coordinates, which play an important role in

the study of cyclotomic �elds, shows that

S(n) =

� �T (n) if n is even;T (n)=2 if n is odd;

where T (n) = nY 1

p

Yp�1

(p � 1)

Yp�3

(p + 1) with p running through all

prime divisors ofn and the congruences are taken modulo 4. It is not di�cult

to show that his answer is equivalent to the one given in the solution above.

2133. [1996: 124] Proposed by K.R.S. Sastry, Dodballapur, India.

Similar non-square rectangles are placed outwardly on the sides of aparallelogram �. Prove that the centres of these rectangles also form a non-square rectangle if and only if � is a non-square rhombus.

As Florian Herzig and V�aclav Kone �cn �y both noted, the similar non-

square rectangles must be placed \nicely", that is, they must either all have

the longest side or all have the shortest side in common with the parallelo-

gram. Otherwise, the conclusion of the problem does not hold!

Solution by Christopher J. Bradley, Clifton College, Bristol, UK.

-

6

������

@@

@@@I

i

j

i�

j�

�

�

Let i; j be unit vectors making an angle � withone another, and let i� and j

� be the imagesof i and j under a 90� anticlockwise rotation.

Then

i � j = cos � i� � i = 0 i

� � j = sin�

i� � j� = cos � j

� � j = 0 j� � i = sin �

(�)

(The proof is trivial, but these relations will be much used in what follows.)

Let the parallelogram be OACB with�!OA = 2ai and

�!OB = 2bj and

let W;X; Y;Z be the centres of the non-square similar rectangles (see the�gure).

250

-�����

2ai

2bj

q

qq qO

A

CBX

Z

WY

�

Those centred at X and Z have sides 2a; 2kaand the other two 2b;2kb, where k 6= 1, sincethey are non-square. Let the position vectorsof W;X; Z relative to O be w; x; z, respec-tively. Then it is easy to see that

w = bj + kbj�; x = 2bj + ai = kai�

and z = ai� kai�:

Hence ��!ZW = w � z = bj + kbj� � ai + kai� and��!WX = x� w = bj + ai + kai� � kbj�:

From which, using relation (�) repeatedly, we see that��!ZW � ��!WX = (b2 � a2)(1� k2) and

ZW 2= WX2 if and only if 2ab cos �(1� k2) = 0:

But k 6= 1, so ZW and WX are at right angles if and only if a = b andZW = WX if and only if � = 90

�.

So ZWXY is a non-square rectangle if and only if OACB is a non-square rhombus.

The proposer notes that this is related to two well-known results.

Napoleon's Theorem: The centres of equilateral triangles, placed on thesides of any triangle, form an equilateral triangle.

Th�ebault's Theorem: The centres of squares placed on any parallelogramform a square.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-

jevo, Bosnia and Herzegovina; HANS ENGELHAUPT, Franz{Ludwig{Gym-

nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,

Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN

G. HEUVER, Grande Prairie CompositeHigh School, Grande Prairie, Alberta;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV

KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;

P. PENNING, Delft, the Netherlands; TOSHIO SEIMIYA, Kawasaki, Japan;

D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

2141. [1996: 170] Proposed by Toshio Seimiya, Kawasaki, Japan.A1A2A3A4 is a quadrilateral. Let B1, B2, B3 and B4 be points on the

sides A1A2, A2A3, A3A4 and A4A1 respectively, such that

A1B1 : B1A2 = A4B3 : B3A3 and A2B2 : B2A3 = A1B4 : B4A4:

251

Let P1, P2, P3 and P4 be points on B4B1, B1B2, B2B3 and B3B4 respec-tively, such that

P1P2kA1A2; P2P3kA2A3 and P3P4kA3A4:

Prove that P4P1kA4A1.

A3

A2

A4

A1

B2

B3

B4

B1

P1

P4P3

P2

Solution by D.J. Smeenk, Zaltbommel, the Netherlands.

We denote:

A1A2 = a1; A2A3 = a2; A3A4 = a3; A4A1 = a4;

A1B1 = �a1; B1A2 = �a1; �+ � = 1;

A2B2 = �a2; B2A3 = �a2; �+ � = 1;

A3B3 = �a3; B3A4 = �a3;

A4B4 = �a4; B4A1 = �a4;

\A4A1A2 = �1; \A1A2A3 = �2;

\A2A3A4 = �3; \A3A4A1 = �4:

The distance from P1 and P2 to A1A2 is d1; from P2 and P3 to A2A3

is d2; from P3 and P4 to A3A4 is d3; from P4 to A4A1 is d4 and from P1 toA4A1 is d

04. It su�ces to show: d04 = d4 which implies P4P1kA4A1.

Consider4B1A2B2. We see that [B1A2P2] + [A2B2P2] = [B1A2B2]

or

�a1d1 + �a2d2 = ��a1a2 sin�2; (1)

Similarly, for 4B2A3B3, 4B3A4B4 and 4B4A1B1:

�a2d2 + �a3d3 = ��a2a3 sin�3; (2)

�a3d3 + �a4d4 = ��a3a4 sin�4; (3)

�a4d04 + �a1d1 = ��a1a4 sin�1: (4)

Eliminating d2 out of (1) and (2), we �nd:

��a1d1 � ��a3d3 = ��� (a1a2 sin�2 � a2a3 sin�3): (5)

252

Eliminating d3 out of (3) and (5):

�a1d1 + �a4d4 = ��(a1a2 sin�2 � a2a3 sin�3 + a3a4 sin�4): (6)

We rewrite (4):

�a1d1 + �a4d04 = ��a1a4 sin�1: (7)

(6) and (7) imply

�a4(d4�d04) = ��(a1a2 sin�2+a3a4 sin�4)���(a2a3 sin�3+a1a4 sin�1):

(8)

Then

2[A1A2A3A4] = a1a2 sin�2 + a3a4 sin�4 = a2a3 sin�3 + a1a4 sin�4:

(9)

(8) and (9) imply d04 = d4 and P4P1jjA4A1.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK (for a parallelogram); WALTHER JANOUS, Ursulinengymnasium, Inns-

bruck, Austria; and the proposer.

2142. [1996: 170] Proposed by Victor Oxman, Haifa, Israel.

In the plane are given an arbitrary quadrangle and bisectors of threeof its angles. Construct, using only an unmarked ruler, the bisector of thefourth angle.

Solution by Toshio Seimiya, Kawasaki, Japan.

ABCD is a given quadrangle. We denote the bisectors of anglesA, B, C,D, by a, b, c, d, respectively. We assume that a, b, c are given. Weshall construct d using only an unmarked ruler.

Case I. ABCD is not a parallelogram. We may assume that AB is notparallel to CD.

Construction: Let O be the intersection of AB;CD, and let P be theintersection of b and c. Join O and P , and let Q be the intersection of a andOP . Join D and Q; then DQ is d.

Proof: P is either the incentre or the excentre of 4OBC [dependingon whether or not the given quadrangle is convex and how its angles arearranged], so that OP is a bisector of \BOC. Thus Q is the excentre orincentre of4OAD so that QD bisects \ADC.

Case II. ABCD is a parallelogram. If ABCD is a rhombus then b

coincides with d and is the bisector of \D. We shall assume that ABCD isnot a rhombus.

Construction. Let P be the intersection of b and c. Draw AC and BD,and let O be their intersection. Join P , O, and let Q be the intersection ofPO with a. Join Q and D; then QD is the bisector of \D.

253

[Editor's comment. Although Seimiya provides a brief argument that hisconstruction is correct, it simply proves that a parallelogram is symmetricabout its centre, which can be left to the reader.]

Also solved by CLAUDIO ARCONCHER, Jundia��, Brazil; NIELS

BEJLEGAARD, Stavanger, Norway; CHRISTOPHER J. BRADLEY, CliftonCol-

lege, Bristol, UK (Case I only); and the proposer.

2143. [1996: 170] Proposed by B. M???y, Devon, Switzerland.My lucky number, 34117, is equal to 166

2+ 81

2 and also equal to159

2+ 94

2, where j166� 159j = 7 and j81� 94j = 13; that is,

it can be written as the sum of two squares of positive integers in

two ways, where the �rst integers occurring in each sum di�er by

7 and the second integers di�er by 13.

What is the smallest positive integer with this property?

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let N = a2 + b2 = (a� 13)2+ (b+ 7)

2, where a � 14 and b � 1, bethe smallest natural number having the desired property. Simplifying yields

a2 + b2 = a2 � 26a+ 169 + b2 + 14b+ 49;

then 13a = 7b+ 109 and �nally

b =13a� 109

7:

As b is a natural number, 13a� 109 � 0 mod 7 which implies a � 3 mod 7

[for example, since 109 � 39 mod 7]. Hence the smallest a is 17, and thenb = 16 and

N = 172+ 16

2= 4

2+ 23

2= 545:

We get all solutions by setting a = 10 + 7k for k � 1:

(10 + 7k)2 + (3 + 13k)2 = (7k� 3)2+ (10+ 13k)2 = 109(2k2 + 2k+ 1):

The solution mentioned in the proposal is obtained for k = 12.From the above equation we see that 109 divides all solutions. And

incidentally, the smallest solution which is also a square is

8852+ 1628

2= 872

2+ 1635

2= 1853

2:

Also solved by SAMBAETHGE, Nordheim, Texas, USA; CHRISTOPHER

J. BRADLEY, Clifton College, Bristol, UK; THEODORE CHRONIS, student,

Aristotle University of Thessaloniki, Greece; TIM CROSS, King Edward's

School, Birmingham, England; CHARLES R. DIMINNIE, Angelo State Uni-

versity, San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnas-

ium, Bamberg, Germany; IAN JUNE L. GARCES, Ateneo de Manila Uni-

versity, Manila, The Philippines, and GIOVANNI MAZZARELLO, Ferrovie

254

dello Stato, Firenze, Italy; SHAWN GODIN, St. Joseph Scollard Hall, North

Bay, Ontario; DAVID HANKIN, Hunter College High School, New York, NY,

USA; RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Val-

ley Catholic High School, Beaverton, Oregon, USA; V �ACLAV KONE �CN �Y, Fer-

ris State University, Big Rapids, Michigan, USA; DAVID E.

MANES, SUNY at Oneonta, Oneonta, NY, USA; GOTTFRIED PERZ, Pesta-

lozzigymnasium, Graz, Austria; JOEL SCHLOSBERG, student, Hunter Col-

lege High School, New York NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin,

Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVID

R. STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS

E. TSAOUSSOGLOU, Athens, Greece; KENNETHM.WILKE, Topeka, Kansas,

USA; and the proposer. Two other readers sent in incorrect solutions prob-

ably due to misunderstanding the problem.

Janous notes that 545 is a palindrome and wonders if there are any

other palindromic solutions! Kone�cn �y replaces the digits 5 and 4 in the

solution by the corresponding letters E and D respectively, and therefore

wonders if the proposer's name is MEDEY. (It is not | in fact, each star

represents two letters.)

2144. [1996: 170] Proposed by B. M???y, Devon, Switzerland.My lucky number, 34117, has the interesting property that 34 = 2 � 17

and 341 = 3 � 117� 10; that is,

it is a 2N + 1-digit number (in base 10) for some N , such that

(i) the number formed by the �rst N digits is twice the number

formed by the last N , and

(ii) the number formed by the �rst N + 1 digits is three times

the number formed by the last N + 1, minus 10.

Find another number with this property.

I Solution by David Hankin, Hunter College High School, New York,

NY, USA.

Let A be the number formed by the �rst N digits of the required num-ber, letB be the number formed by the lastN digits, and let x be the middledigit. Then

A = 2B and 10A+ x = 3(x � 10N + B)� 10:

From these, we obtain

17B = 3x � 10N � x� 10; (1)

so x + 10 � 3x � 10N mod 17. Since A = 2B has N digits, B < 5 � 10N�1and so 17B < 8:5 � 10N. Thus by (1), 1 � x � 2.

255

When x = 1, we have 3 � 10N � 11 mod 17, so 10N � 15 mod 17.

This is satis�ed by N = 2, and since 1016 � 1 mod 17 [by Fermat's little

theorem], it is satis�ed by N = 2 + 16k for non-negative k. Note thatx = 1, N = 2 yields the lucky number referred to in the problem.

When x = 2, we have 6 � 10N � 12 mod 17, so 10N � 2 mod 17. A

little arithmetic yields N = 10 + 16k this time, so the next smallest num-ber with the required property is the one given by x = 2 and N = 10.These values yield 17B = 6 � 1010 � 12, from which B = 3 529 411 764,A = 7 058 823 528, and a lucky number of

705 882 352 823 529 411 764

(which coincidentally happens to be my lucky number too).

II Solution by Shawn Godin, St. Joseph Scollard Hall, North Bay, On-

tario.

[Godin �rst obtained equation (1) and that x must be 1 or 2, and thenhis solution, with Hankin's notation, continues as follows.]

When x = 1, we need to �nd a natural number of the form

B =3 � 10N � 11

17:

Now we can calculate

3

17= 0:1764705882352941;

and if we look in the long division for the �rst occurrence of a remainderof 11, we can construct a solution. It occurs after the �rst 7, which yieldsB = 17, A = 34 and thus the lucky number 34117 which is the proposer'slucky number. It is also the �rst in an in�nite family of solutions. We cancreate new ones by appending a full period of the repeating decimal for 3=17to the left of the B of our last solution. Thus the next solution in this familyhas

B = 1764705882352941 17; A = 3529411764705882 34;

and thus is

352941176470588234 1 176470588235294117:

There is another family of solutions with x = 2, obtained the sameway. In this case

B =6 � 10N � 12

17and

6

17= 0:3529411764705882

with a remainder of 12 occurring �rst at the second 4, so

B = 3529411764; A = 7058823528;

256

and the lucky number is

7058823528 2 3529411764:

Similarly this is the �rst solution in an in�nite family of solutions. The othersare obtained by appending a full period of 6=17 to the left of the B of thelast solution.

Also solved by SAMBAETHGE, Nordheim, Texas, USA; CHRISTOPHER

J. BRADLEY, CliftonCollege, Bristol,UK;MIGUELANGEL CABEZ �ONOCHOA,

Logro ~no, Spain; THEODORE CHRONIS, student, AristotleUniversity of Thes-

saloniki, Greece; TIM CROSS, King Edward's School, Birmingham, England;

CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA;

HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany;

FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,

Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-

nasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley CatholicHigh School,

Beaverton, Oregon, USA; AMIT KHETAN, Troy, Michigan, USA; V �ACLAV

KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; DAVID E.

MANES, SUNY at Oneonta, Oneonta, NY, USA; HEINZ-J�URGEN SEIFFERT,

Berlin, Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta;

DAVID R. STONE, Georgia Southern University, Statesboro, Georgia, USA;

PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer. One incor-

rect solution was sent in.

Notice from Godin's solution that the fractionswith denominator 17 all

have the same digits in their repeating parts, just \cycled around". This is

the same behaviour that readers will likely know from the fractions1=7, 2=7,

etc. In both cases the reason for this behaviour is that the repeating parts

have the largest possible number of digits, 16 in the case of denominator

17. The consequence for this problem is that the solutions have \repeated

blocks" in them, which some solvers remarked upon.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

257

THE ACADEMY CORNERNo. 12

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

In the last issue, we printed the Bernoulli Trials. Here are the hints

and answers.

Bernoulli Trials 1997

HINTS | 1

1. The curve is a circle centred at (1;�1) with radius 5.

2. Write the two series in terms of the log function.

3. For any integer m, one of the numbers m, m+ 1 and m+ 2 must be

divisible by three.

4. Find the highest power of 5 which divides 1000!

5. A real polynomial of odd degree has a positive and �nite number of

roots.

6. Show that, in fact, the probability equals 1=2.

7. Show that limx!0+

x(2p) = 1 and limx!0+

x(2p+1) = 0.

8. Expand (x� a1)(x� a2) : : : (x� an).

9. Show that equality can be obtained.

10. By an elementary application of the Pigeonhole Principle, there will

eventually be two pairs of adjacent Fibonacci numbers, say fm; fm+1

and fn; fn+1 whose last four digits respectively are the same. That is,

fm and fn are congruent modulo 104, and similarly, so are fm+1 and

fn+1.

11. Factor the denominator as (k2 + k + 1)(k2 � k + 1) and use partial

fractions.

258

12. The number 1997 is prime and congruent to 1 modulo 4.

13. There is more than one way to skin a cat: Multiply top and bottom by

1� sinx.

14. Make the substitution x = yp to turn this into a polynomial.

15. Show that f has period 2a.

HINTS | 2

1. Show that the Euclidean distance from (1;�1) to the given point is

greater than 5.

2. Show that � log(1� p) = log(1 + q).

3. Note that 2n is never divisible by three.

4. Use Legendre's formula for the prime factorization of n!

5. Consider the set f(t; t3; t5) : �1 < t < +1g.6. Write the number of heads that Johann gets in 1997 tosses as x and the

number of heads that Jakob gets in 1996 tosses as y. Compare the dis-

tribution of coins for the two Bernoulli sequences with the distribution

with heads and tails reversed.

Note that x > y if and only if 1997� x � 1996� y.

7. Show that xp log x! 0 as x! 0 for all p > 0. Use induction.

8. Expanding out, we see that the elementary symmetric functions appear

as the coe�cients of the polynomial. Suppose the ai's are distinct.

Show that if any ai is negative, then x > 0 can be chosen so that the

polynomial is negative.

9. Symmetry suggests trying the substitution a = c to �nd a counterexam-

ple. [This does not work!] To investigate this question further, prove

the inequality is true when strict inequality is relaxed to �. The in-

equality is equivalent to [h(a; b; c)]2 � 0 for an appropriate function h.

10. Having found two pairs of adjacent Fibonacci numbers with the same

last four digits, now work backwards. Compare fm�1 and fn�1 etc.

Keep going.

11. Note that

2k

k4 + k2 + 1=

1

k2 � k + 1� 1

k2 + k+ 1

Using partial fractions, show that the sum can be written as a telescop-

ing sum.

259

12. Apply Fermat's Great Theorem.

13. There is more than one way to skin a cat: use integration by parts and

the transformation t = tan(x=2).

14. Descartes Rule of Signs tells us that if z is the number of positive zeros

of a polynomial, and c is the number of changes of sign of the sequence

of coe�cients, then c� z is a nonnegative even number.

15. Show that

[f(x+ a)]2 � f(x+ a) = �[f(x)� 1=2]2:

Then write out a formula for f(x+2a) using the original equation given

in the question.

ANSWERS

1. True 2. True 3. False 4. True 5. False

6. False 7. False 8. True 9. False 10. False

11. True 12. True 13. False 14. True 15. True

Comment

For the information of readers, to the knowledge of the editor, other

versions of the Bernoulli Trials have been held at least in his home institu-

tion and at the Canadian IMO Residential Training Program. In the latter

instance, local high school students were added to the list of competitors.

On both occasions, all the competitors reported that it was an enjoyable ex-

perience, and a pleasant social event as well.

260

THE OLYMPIAD CORNERNo. 183

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

We begin with the 1997 CanadianMathematical Olympiad which we re-

produce with the permission of the Canadian Mathematical Olympiad Com-

mittee of the Canadian Mathematical Society. Thanks go to Daryl Tingley,

University of New Brunswick, and Chair of the CanadianMathematical Olym-

piad Committee of the Canadian Mathematical Society, for forwarding the

questions to me.

PROBLEMS FROM THE 1997 CANADIANMATHEMATICAL OLYMPIAD

1. How many pairs of positive integers x; y are there, with x � y, and

such that gcd (x; y) = 5! and lcm(x; y) = 50! ?

Note:

gcd (x; y) denotes the greatest common divisor of x and y,

lcm(x; y) denotes the least common multiple of x and y,

and n! = n� (n� 1)� � � � � 2� 1.

2. The closed interval A = [0; 50] is the union of a �nite number

of closed intervals, each of length 1. Prove that some of the intervals can

be removed so that those remaining are mutually disjoint and have total

length � 25.

Note: For a � b, the closed interval [a; b] := fx 2 R : a � x � bg haslength b� a; disjoint intervals have empty intersection.

3. Prove that

1

1999<

1

2� 34� 56� � � 1997

1998<

1

44:

4. The point O is situated inside the parallelogram ABCD so that

\AOB + \COD = 180�:

Prove that \OBC = \ODC.

261

5. Write the sum

nXk=0

(�1)k�n

k

�k3 + 9k2 + 26k+ 24

in the formp(n)

q(n), where p and q are polynomials with integer coe�cients.

The next set of problems are from the twenty-sixth annual United States

of America Mathematical Olympiad written May 1, 1997. These problems

are copyrighted by the Committee on the American Mathematical Competi-

tions of the Mathematical Association of America andmay not be reproduced

without permission. Solutions, and additional copies of the problems may

be obtained from Professor Walter E. Mientka, AMC Executive Director, 917

Oldfather Hall, University of Nebraska, Lincoln, NE, USA 68588{0322. As al-

ways, we welcome your original, \nice" solutions and generalizations which

di�er from the published solutions.

26th UNITED STATES OF AMERICAMATHEMATICAL OLYMPIAD

Part I 9 a.m. - 12 noonMay 1, 1997

1. Let p1; p2; p3; : : : be the prime numbers listed in increasing order,

and let x0 be a real number between 0 and 1. For positive integer k, de�ne

xk =

8<:

0 if xk�1 = 0;�pk

xk�1

�if xk�1 6= 0;

where fxg denotes the fractional part of x. (The fractional part of x is given

by x� bxc where bxc is the greatest integer less than or equal to x.) Find,

with proof, allx0 satisfying0 < x0 < 1 for which the sequence x0; x1; x2; : : :

eventually becomes 0.

2. Let ABC be a triangle, and draw isosceles triangles BCD, CAE,

ABF externally toABC, withBC,CA,AB as their respective bases. Prove

that the lines through A, B, C perpendicular to the lines !EF;

!FD;

!DE,

respectively, are concurrent.

3. Prove that for any integer n, there exists a unique polynomial Q

with coe�cients in f0; 1; : : : ; 9g such that Q(�2) = Q(�5) = n.

262

Part II 1 p.m. - 4 p.m.

4. To clip a convex n-gon means to choose a pair of consecutive sides

AB, BC and to replace them by the three segments AM , MN , and NC,

where M is the midpoint of AB and N is the midpoint of BC. In other

words, one cuts o� the triangle MBN to obtain a convex (n + 1)-gon.

A regular hexagon P6 of area 1 is clipped to obtain a heptagon P7. Then

P7 is clipped (in one of the seven possible ways) to obtain an octagon P8,

and so on. Prove that no matter how the clippings are done, the area of Pnis greater than 1=3, for all n � 6.

5. Prove that, for all positive real numbers a; b; c,

(a3 + b3 + abc)�1 + (b3 + c3 + abc)�1 + (c3 + a3 + abc)�1 � (abc)�1:

6. Suppose the sequence of nonnegative integers a1; a2; : : : ; a1997 sat-is�es

ai + aj � ai+j � ai + aj + 1

for all i; j � 1 with i + j � 1997. Show that there exists a real number x

such that an = bnxc (the greatest integer � nx) for all 1 � n � 1997.

As a third Olympiad for this issue we give selected problems of the

3rd Ukrainian Mathematical Olympiad, written March 26 and 27, 1994. My

thanks go to Richard Nowakowski, Canadian Team leader to the IMO inHong

Kong for collecting the set.

3rd UKRAINIANMATHEMATICAL OLYMPIADSelected ProblemsMarch 26{27, 1994

1. (class 9) A convex quadrangle ABCD is given. Bisectors of external

angles of ABCD form a new quadrangle PQRS. Prove that sum of diago-

nals of PQRS is more than the perimeter of ABCD.

2. (9{10) A convex polygon and point O inside it are given. Prove that

for any n > 1 there exist points A1; A2; : : : ; An on the sides of the polygon

such that��!OA1 +

��!OA2 + : : :+

��!OAn =

�!O .

3. (10) A sequence of natural numbers ak, k � 1, such that for each k,

ak < ak+1 < ak + 1993 is given. Let all prime divisors of ak be written for

every k. Prove that we receive an in�nite number of di�erent prime numbers.

4. (10) Inside the triangle ABC the point D is given such that angles

\ABC and \DBC are equal. Let K and L be the projections of D on lines

(AC) and (BD) respectively. Prove that the midpoints of AB, CD andKL

are collinear.

263

5. (11) 19942 squares 1 � 1 of k di�erent colours are given. Find

all natural numbers k for which it is always possible to construct a 1994 �1994 square coloured symmetrically with respect to the diagonal from any

collection of 19942 such squares.

6. (11) A mathematical olympiad consists of two days. The Jury intends

to place the participants in rooms so that any two students will be in rooms

with di�erent numbers of participants in one of the days. Is it possible for

(a) 9 participants?

(b) 14 participants?

(The Jury has the necessary number of rooms.)

7. (11) Three points P , Q and S in space are given. Two rays from P

and two rays from Q are taken such that each ray from P intersects two rays

from Q. Let A, B, C, D be the intersection points of these four rays. It

is known that pyramid SABCD has a section KLMN of rectangular form

(K 2 SA, L 2 SB, M 2 SC, N 2 SD). The area of ABCD equals 1.

Prove that the volume of SABCD is not larger than PQ=6.

8. (11) Two circles with radii R and r (R > r) in the plane are tangent

at pointM with one circle inside the other. Chord AB of the bigger circle is

tangent to the smaller circle. Find the largest possible value of the perimeter

of ABM .

Eagle-eyed readers Dan Velleman, Stan Wagon and Andy Liu spotted

trouble with problem 4 of the SwedishMathematics Contest 1993 as we gave

it in the last number of the Corner. Indeed, as they point out, setting a�b = 1

for all a, b satis�es the conditions of the problem, but does not give a unique

solution! I must not have been seeing stars when I proof-read that one. Here

is the correct version to try.

4. Swedish Mathematics Contest 1993 [1997: 196]

To each pair of real numbers a and b, where a 6= 0 and b 6= 0, there is

a real number a � b such that

a � (b � c) = (a � b)c

a � a = 1:

Solve the equation x � 36 = 216.

264

In somewhat of a departure from past practice we now give the solu-

tions given to the Canadian Mathematical Olympiad given earlier this issue!

(I hope you have solved them by now.) The \o�cial solutions" provided for

me by Daryl Tingley, University of New Brunswick, and Chair of the Canadian

Mathematical Olympiad Committee of the Canadian Mathematical Society,

feature solutions by contestants in the CMO itself. Each solution is credited

to one of the contestants.

PROBLEM SOLUTIONS FOR THE 1997 CANADIANMATHEMATICAL OLYMPIAD

1. How many pairs of positive integers x; y are there, with x � y, and

such that gcd (x; y) = 5! and lcm (x; y) = 50! ?

Note:

gcd (x; y) denotes the greatest common divisor of x and y,

lcm (x; y) denotes the least common multiple of x and y,

and n! = n� (n� 1)� � � � � 2� 1.

Solution by Deepee Khosla, Lisgar Collegiate Institute, Ottawa, ON

Let p1; : : : ; p12 denote, in increasing order, the primes from 7 to 47.

Then

5! = 23 � 31 � 51 � p01 � p02 � � � � p012

and

50! = 2a1 � 3a2 � 5a3 � pb11 � pb22 � � � � pb1212 :

Note that 24; 32; 52; p1; : : : ; p12 all divide 50!, so all its prime powers di�er

from those of 5!

Since x; yj50!, they are of the form

x = 2n1 � 3n2 � � � � � pn1512

y = 2m1 � 3m2 � � � � � pm15

12 :

Then max(ni;mi) is the ith prime power in 50! and min(ni; mi) is the ith

prime power in 5!

Since, by the above note, the prime powers for p12 and under di�er in

5! and 50!, there are 215 choices for x, only half of which will be less than y.

(Since for each choice of x, we have that y is forced and either x < y or

y < x.) So the number of pairs is 215=2 = 214.

2. The closed interval A = [0; 50] is the union of a �nite number

of closed intervals, each of length 1. Prove that some of the intervals can

be removed so that those remaining are mutually disjoint and have total

length greater than or equal to 25.

Note: For a � b, the closed interval [a; b] := fx 2 R : a � x � bg haslength b� a; disjoint intervals have empty intersection.

265

Solution by Byung-Kuy Chun, Harry Ainlay Composite High School,

Edmonton, AB.

Look at the �rst point of each given unit interval. This point uniquely

de�nes the given unit interval.

Lemma. In any interval [x; x + 1) there must be at least one of these �rst

points (0 � x � 49).

Proof. Suppose the opposite. The last �rst point before x must be x� " for

some " > 0. The corresponding unit interval ends at x � " + 1 < x + 1.

However, the next given unit interval cannot begin until at least x+ 1.

This implies that points (x� "+ 1; x+ 1) are not in set A, a contra-

diction.

Therefore there must be a �rst point in [x; x+ 1).

Note that for two �rst points in intervals [x; x+ 1) and [x+ 2; x+ 3)

respectively, the corresponding unit intervals are disjoint, since the intervals

are in the range [x; x+ 2) and [x+ 2; x+ 4) respectively.

Thus, we can choose a given unit interval that begins in each of

[0; 1) [2; 3) : : : [2k; 2k+ 1) : : : [48; 49):

Since there are 25 of these intervals, we can �nd 25 points which correspond

to 25 disjoint unit intervals.

Solution by Colin Percival, Burnaby Central Secondary School, Burn-

aby, BC.

I prove the more general result, that if [0; 2n] =SiAi; jAij = 1; Ai

are intervals then 9 a1 : : : an, such that Aai \ Aaj = ;.Let 0 < " � 2

n�1and let bi = (i� 1)(2 + "); i = 1 : : : n. Then

minfbig = 0;maxfbig = (n� 1)(2 + ") � (n� 1)

�2 +

2

n� 1

�

= (n� 1)

�2n

n� 1

�= 2n:

So all the bi are in [0; 2n].

Let ai be such that bi 2 Aai . SinceSAi = [0; 2n], this is possible.

Then since (bi�bj) = (i�j)(2+")� 2+" > 2 (i > j), and theAi are

intervals of length 1,minAai�maxAaj > 2�1�1 = 0, soAai \Aaj = ;.Substituting n = 25, we get the required result.

3. Prove that

1

1999<

1

2� 34� 56� � � 1997

1998<

1

44:

266

Solution by Mihaela Enachescu, Dawson College, Montr �eal, PQ.

Let P =1

2� 34� : : : � 1997

1998. Then

1

2>

1

3[because 2 < 3],

3

4>

3

5

[because 4 < 5], : : : ; : : :1997

1998>

1997

1999[because 1998 < 1999].

So

P >1

3� 35� : : : � 1997

1999=

1

1999: (1)

Also 1

2< 2

3[because 1 � 3 < 2 � 2], 3

4< 4

5[because 3 � 5 < 4 � 4], : : : ,

1997

1998<

1998

1999[because 1997 � 1999 = 19982 � 1 < 19982].

So P <2

3� 45� : : : � 1998

1999=

�2

1� 43� 65� : : : � 1998

1997

�| {z }

1=P

1

1999.

Hence P 2 <1

1999<

1

1936=

1

442and P <

1

44: (2)

Then (1) and (2) give 1

1999< P < 1

44.

4. The point O is situated inside the parallelogram ABCD so that

\AOB + \COD = 180� :

Prove that \OBC = \ODC.

Solution by Joel Kamnitzer, Earl Haig Secondary School, North York,

ON.

D C

BA

O

O0

Consider a translation which maps D to A. It will map O ! O0 with��!OO0 =

�!DA, and C will be mapped to B because

�!CB =

�!DA.

267

This translation keeps angles invariant, so \AO0B = \DOC =

180� � \AOB.

Therefore AOBO0 is a cyclic quadrilateral. And further,

\ODC = \O0AB = \O0OB. But, since O0O is parallel to BC,

\O0OB = \OBC:

Therefore

\ODC = \OBC:

Solution by Adrian Chan, Upper Canada College, Toronto, ON.

D C

BA

�

�

180� � �

180� � �

O

Let \AOB = � and \BOC = �. Then \COD = 180� � � and

\AOD = 180� � �.

Since AB = CD (parallelogram) and sin � = sin(180� � �), the sine

law on4OCD and4OAB gives

sin\CDO

OC=

sin(180�� �)

CD=

sin �

AB=

sin\ABO

OA

so

OA

OC=

sin\ABO

sin\CDO: (1)

Similarly, the Sine Law on4OBC and4OAD gives

sin\CBO

OC=

sin�

BC=

sin(180� � �)

AD=

sin\ADO

OA

so

OA

OC=

sin\ADO

sin\CBO: (2)

Equations (1) and (2) show that

sin\ABO � sin\CBO = sin\ADO � sin\CDO:

268

Hence

1

2[cos(\ABO+ \CBO)� cos(\ABO � \CBO)]

=1

2[cos(\ADO+ \CDO)� cos(\ADO � \CDO)]:

Since \ADC = \ABC(parallelogram) and \ADO+\CDO = \ADC and

\ABO + \CBO = \ABC it follows that

cos(\ABO � \CBO) = cos(\ADO � \CDO):

There are two cases to consider.

Case (i): \ABO � \CBO = \ADO � \CDO.

Since\ABO+\CBO = \ADO+\CDO, subtracting gives 2\CBO =

2\CDO so \CBO = \CDO, and we are done.

Case (ii): \ABO � \CBO = \CDO � \ADO.

Since we know that \ABO+\CBO = \CDO+\ADO, adding gives

2\ABO = 2\CDO so \ABO = \CDO and \CBO = \ADO.

Substituting this into (1), it follows that OA = OC.

Also, \ADO + \ABO = \CBO+ \ABO = \ABC.

Now, \ABC = 180� � \BAD since ABCD is a parallelogram.

Hence \BAD + \ADO + \ABO = 180� so \DOB = 180� and

D;O;B are collinear.

We now have the diagram

D C

BA

�

180� � �

O

Then \COD+ \BOC = 180�, so \BOC = � = \AOB.

4AOB is congruent to4COB (SAS,OB is common, \AOB = \COB

and AO = CO), so \ABO = \CBO. Since also \ABO = \CDO we con-

clude that \CBO = \CDO.

269

Since it is true in both cases, then \CBO = \CDO.

5. Write the sum

nXk=0

(�1)k�n

k

�k3 + 9k2 + 26k+ 24

;

in the formp(n)

q(n), where p and q are polynomials with integer coe�cients.

Solutionby Sabin Cautis, Earl Haig Secondary School, North York, ON.

We �rst note that

k3 + 9k2 + 26k + 24 = (k+ 2)(k+ 3)(k+ 4):

Let S(n) =

nXk=0

(�1)k�n

k

�k2 + 9k2 + 26k+ 24

.

Then

S(n) =

nXk=0

(�1)kn!

k!(n� k)!(k+ 2)(k+ 3)(k+ 4)

=

nXk=0

(�1)k(n+ 4)!

(k+ 4)!(n� k)!

!��

k + 1

(n+ 1)(n+ 2)(n+ 3)(n+ 4)

�:

Let

T (n) = (n+1)(n+2)(n+3)(n+4)S(n) =

nXk=0

�(�1)k

�n+ 4

k + 4

�(k+ 1)

�:

Now, for n � 1,

nXi=0

(�1)i�n

i

�= 0 (�)

since

(1� 1)n =

�n

0

���n

1

�+

�n

2

�+ : : :+ (�1)n

�n

n

�= 0:

270

Also

nXi=0

(�1)i�n

i

�i =

nXi=1

(�1)ii � n!

i! � (n� i)!+ (�1)0 � 0 � n!

0! � n!

=

nXi=1

(�1)in!

(i� 1)!(n� i)!

=

nXi=1

(�1)in

�n� 1

i� 1

�

= n

nXi=1

(�1)i�n� 1

i� 1

�

= �nnXi=1

(�1)i�1�n� 1

i� 1

�:

Substituting j = i� 1, (*) shows that

nXi=0

(�1)i�n

i

�i = �n

n�1Xj=0

(�1)j�n� 1

j

�= 0. (��)

Hence

T (n) =

nXk=0

(�1)k�n+ 4

k+ 4

�(k+ 1)

=

nXk=0

(�1)k+4�n+ 4

k + 4

�(k+ 1)

=

nXk=�4

(�1)k+4�n+ 4

k+ 4

�(k+ 1)

���3 + 2(n+ 4)�

�n+ 4

2

��:

Substituting j = k+ 4, gives

T (n) =

n+4Xj=0

(�1)j�n+ 4

j

�(j � 3)�

�2n+ 8� 3� (n+ 4)(n+ 3)

2

�

=

n+4Xj=0

(�1)j�n+ 4

j

�j

�3

n+4Xj=0

(�1)j�n+ 4

j

�� 1

2(4n+ 10� n2 � 7n� 12)

The �rst two terms are zero because of results (�) and (��) so

271

T (n) =n2 + 3n+ 2

2:

Then

S(n) =T (n)

(n+ 1)(n+ 2)(n+ 3)(n+ 4)

=n2 + 3n+ 2

2(n+ 1)(n+ 2)(n+ 3)(n+ 4)

=(n+ 1)(n+ 2)

2(n+ 1)(n+ 2)(n+ 3)(n+ 4)

=1

2(n+ 3)(n+ 4):

ThereforenX

k=0

(�1)k�n

k

�k3 + 9k2 + 26k + 24

=1

2(n+ 3)(n+ 4):

That completes the Corner for this issue. Send me Olympiad contests

and your nice solutions and generalizations for use in the Corner.

Congratulations!

The following contributors to CRUXwithMAYHEM distinguishedthem-

selves at the 38th IMO held in Argentina:

Name Country Award

Florian Herzig Austria Gold Medal

Sabin Cautis Canada Bronze Medal

Adrian Chan Canada Silver Medal

Byung Kyu Chun Canada Silver Medal

Mansur Boase United Kingdom Silver Medal

Toby Gee United Kingdom Bronze Medal

Carl Bosley USA Gold Medal

CRUX with MAYHEM o�ers hearty congratulations to these students.

We trust that their participation in solving problems in CRUXwithMAYHEM

was a contributory factor in their successes.

272

BOOK REVIEWS

Edited by ANDY LIU

The Puzzle Arcade, by Jerry Slocum,

published in 1996 by Klutz, 2121 Staunton Court, Palo Alto, CA 94306, USA.,

ISBN# 0-888075-851-4, coil bound, 48 pages, US$20 plus handling.

Quantum Quandaries, edited by Timothy Weber,

published in 1996 by the National Science Teachers' Association,

1840 Wilson Boulevard, Arlington, VA 22201-3000, USA.,

ISBN# 0-87355-136-2, softcover, 208 pages, US$7.95 plus handling.

Reviewed by Andy Liu, University of Alberta.

These two books are the most delightful additions to the literature of

popular mathematics recently, and should appeal to everyone.

Jerry Slocum, a retired executive in the aircraft industry, has a collec-

tion of mechanical puzzles numbering over 22000. The Puzzle Arcade fea-

tures a very small portion of this amazing trove. The book is profusely il-

lustrated with striking colours, and comes with several pockets containing

either complete puzzles or equipment needed for others. Apart from the

standard tangram-like puzzles, matchstick puzzles, topological entanglement

puzzles, optical illusions, mazes and word puzzles, there are many unlikely to

be familiar to the readers.

The readers are asked to put together two identical polyhedral pieces to

form a familiar object. This is perhaps the most annoying of the puzzles with

only two pieces. A three-piece puzzle features Sam Loyd's tantalizing ponies.

There is also a puzzle with four or �ve pieces. With four, the readers can form

the back of a playing card, but it takes all �ve to come up with the King of

Hearts. The list goes on and on, and the readers simply have to see the

book to relish it. Hints and answers are provided. The publisher has done a

fantastic job packing so much into a relatively compact o�ering. Purchased

separately, the puzzles will cost many times more than this bargain of a book!

Timothy Weber of the NSTA is the managing editor of the magazine

Quantum. It is the sister publication of the RussianmagazineKvant, a Russian

word which means quantum. About two-thirds of the material in Quantum

are translated from Kvant, the rest being original contributions in English.

About two-thirds of the material are mathematical, the rest being in physics.

It is published six times a year by Springer-Verlag New York Inc., P.O.Box

2485, Secaucas, NJ 07096-2485, USA.. Student subscription at US$18 per

volume is a steal.

One of the most popular columns inQuantum is theBrainteasers. There

are �ve in each issue, usually four in mathematics and one in physics. Many

are presented in whimsical style, and there are always the most delightful

273

cartoon illustrations. Quantum Quandaries, a pocket-sized book, contains

the �rst 100 Brainteasers and their solutions. Each Brainteaser is on one

page, with its solution overleaf for easy reference. Additional cartoon illus-

trations are provided by Sergey Ivanov. The only regret is that the book is

in black-and-white, whereas the original presentation in the magazine was

in full colour.

We conclude this review with a sample Brainteaser. Solve the alphametic

equation

USA+USSR=PEACE

where letters stand for di�erent digits if and only if they are di�erent.

What are the parametric equations, and the range of

the parameter, of the following curve?

6

-

274

A Fermat-Fibonacci Collaboration

K.R.S. Sastry

Fermat begins

Fermat the mischievous proposed the impossible: Find three squares in arith-

metic progression (AP ), the common di�erence (cd) being a square, [Dickson

[1] pp. 435]. Throughout this paper we deal with integer squares but, we

�nd that, what is impossible for squares is possible for triangles: 6, 21, 36

are triangular numbers in AP with cd of 15 also being triangular! Let us see

how we can generate triples of triangular numbers in AP whose cd is also

triangular. Curiously, the supplier is Fibonacci!

We recall that the nth triangular number is given by Tn =1

2n(n+ 1).

We also note that 8Tn +1 = (2n+1)2 is a perfect square, an odd square to

be exact. Simply retrace the preceding steps to see that the converse, every

odd square leads to a triangular number, holds. Furthermore, multiplying

each term of an AP by 8 and then adding 1 to them results in a new AP , but

the new cd is just eight times the old. Take an example:

AP : 2, 7, 12 results in the AP : 17, 57, 97 with cd's of 5 and 40: (1)

The upshot of all this is that the triangular numbers in AP can be made to

depend on the square numbers in AP .

An expression for squares in AP

Suppose the squares x2, y2, z2 are in AP . Then x2 + z2 = 2y2. We may

assume x, y, z are relatively prime in pairs and that x < y < z holds.

Hence x and z are both odd. Put x = p� q, z = p + q where p and q are

relatively prime integers having opposite parity. Then we have the famous

Pythagorean relation p2+q2 = y2 whose primitive solutions are well known:

p = m2 � n2, q = 2mn, y = m2 + n2. Here m and n are relatively prime

integers with opposite parity. This yields

x = jm2 � n2 � 2mnj; y = m2 + n2; z = m2 � n2 + 2mn: (2)

The cd: y2 � x2, equals 4mn(m2 � n2). This cannot be a square in view

of Fermat's proof of the fact mn(m2 � n2), the area of the (Pythagorean)

right-angled triangle with sidesm2�n2, 2mn,m2+n2 cannot be a square

[Dickson [1] pp. 615-616].

275

Triangular numbers in AP whose cd is triangular

Suppose the triangular numbers

Ta =1

2a(a+ 1);

Tb =1

2b(b+ 1);

Tc =1

2c(c+ 1);

are in AP . Then

8Ta + 1 = (2a+ 1)2 = A2;

8Tb + 1 = (2b+ 1)2 = B2;

8Tc + 1 = (2c+ 1)2 = C2;

are three odd squares in AP . Hence from the expressions in (2)

A = jm2 � n2 � 2mnj; B = m2 + n2; C = m2 � n2 + 2mn: (3)

The cd of the AP squares A2, B2, C2 is 4mn(m2 � n2). Therefore, from

the remark in (1), the triangular numbers Ta, Tb, Tc in AP have cd:

1

8(4mn(m2� n2)) =

1

2mn(m2� n2):

The very form of this cd suggests that we choose m2 � n2 = mn � 1. In

that case the cd:

1

2mn(m2� n2) =

1

2mn(mn� 1)

would be triangular. Thus we seek, in natural numbers, the solution of the

equation

m2 � n2 = mn� 1:

Fibonacci enters

Before solving the equation m2 � n2 = mn � 1, let us recall the

Fibonacci sequence

1; 1; 2; 3; 5; 8; 13;21; 34; � � �given recursively by F1 = F2 = 1, Fk = Fk�1+Fk�2, k > 2. It is also given

by Binet's formula

Fk =(1 +

p5)k + (1�p5)k

2kp5

; k = 1; 2; 3; � � � : (4)

276

Also, the Lucas sequence

1; 3; 4; 7; 11; 18; 29; � � �is likewise de�ned: L1 = 1, L2 = 3, Lk = Lk�1 + Lk�2, k > 2. We leave it

to the reader to deduce the following relations:

Lk = Fk�1 + Fk+1 =

1 +

p5

2

!k+

1�p5

2

!k: (5)

Now consider the solution of the equation m2 � n2 = mn� 1, put in

the form (2m� n)2� 5n2 = �4.

The �rst few trial solutions of the equation (2m�n)2�5n2 = �4 are

2m� n 1 4 11 � � �n 1 2 5 � � �m 1 3 8 � � �

It is already apparent that 2m � n = L2k�1, m = F2k, n = F2k�1,

k = 1, 2, 3, � � � . Direct calculations using (4) and (5) show that

L22k�1 � 5F 2

2k�1 = �4; k = 1; 2; 3; � � � :We leave the veri�cation to the reader.

Likewise the �rst few trial solutions of the equation

(2m� n)2� 5n2 = 4

are

2m� n 3 7 18 � � �n 1 3 8 � � �m 2 5 13 � � �

This time we observe that 2m� n = L2k, m = F2k+1, n = F2k and again

we leave the veri�cation that

L22k � 5F 22k = 4; k = 1; 2; � � �

to the reader.

A numerical example

Choose the solutionm = 8, n = 5. Then from (3), we have

2a+ 1 = A = jm2 � n2 � 2mnj = 41;

2b+ 1 = B = m2 + n2 = 89;

2c+ 1 = C = m2 � n2 + 2mn = 119:

277

This yields a = 20, b = 44, c = 59, Ta = 210, Tb = 990, Tc = 1770 in AP

whose cd, 780 =1

2(39)(40), is triangular too. The reader may have spotted

a like parity solution m = 5, n = 3. This makes a, b, c non-integral, since

A, B, C are even. This happens whenm = F6i�1, n = F6i�2, i = 1; 2; � � � ,a fact easily shown by induction.

Fermat concludes

The next question to settle would be: Are there three pentagonal numbers

in AP whose cd is pentagonal? The cases that have been settled are n = 3

and n = 4, instances of the general problem: For what values of n are there

three n-gonal numbers in AP whose cd is n-gonal too? It may be recalled

that the rth n-gonal number is

p(n; r) = (n� 2)r2

2� (n� 4)

r

2; n � 3; r = 1; 2; 3; � � � :

For n > 4, is this Fermat's newest last problem?

Reference

[1] L.E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY,

(1971), pp. 1-39, 435, 615-616.

2943 Yelepet

Dodballapur 561 203

Karnataka, India

278

THE SKOLIAD CORNERNo. 23

R.E. Woodrow

This number we give the problems of the 1995 Prince Edward Island

Mathematics Competition. Thanks go to Gordon MacDonald, University of

Prince Edward Island, organizer of the contest, for supplying us with the

problems, and the solutions which we will give next issue.

1995 P.E.I. Mathematics Competition

1. Find the area of the shaded region inside the circle in the following

�gure.

-�

6

?

1�1

1

�1

P

2. \I will be n years old in the year n2", said Bob in the year 1995.

How old is Bob?

3. Draw the set of points (x; y) in the plane which satisfy the equation

jxj + jx� yj = 4.

4. An autobiographical number is a natural number with ten digits

or less in which the �rst digit of the number (reading from left to right)

tells you how many zeros are in the number, the second digit tells you how

many 1's, the third digit tells you how many 2's, and so on. For example,

6; 210; 001; 000 is an autobiographical number. Find the smallest autobio-

graphical number and prove that it is the smallest.

279

5. A solid cube of radium is oating in deep space. Each edge of the

cube is exactly 1 kilometre in length. An astronaut is protected from its

radiation if she remains at least 1 kilometre from the nearest speck of radium.

Including the interior of the cube, what is the volume (in cubic kilometres) of

space that is forbidden to the astronaut? (You may assume that the volume

of a sphere of radius r is 4

3�r3 and the volume of a right circular cylinder of

radius r and height h is �r2h.)

6. Which is greater, 999! or 500999? (Where 999! denotes 999 factorial,

the product of all the natural numbers from 1 to 999 inclusive.) Explain your

reasoning.

Where do those errors come from?

Several readers wrote about the error in the solution of Problem 2(a)

of the Manitoba Contest 1995 [1995: 219].

2. (a) Find two numbers which di�er by 3 and whose squares di�er

by 63.

Corrections by Jamie Batuwantudawa, student, Fort Richmond Colle-

giate, Winnipeg, Manitoba; by Paul{Olivier Dehaye, Brussels, Belgium; and

by Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA.

The solution of the equations 6x+9 = 63 or 6x+9 = �63 is obvious:

x = 9 or x = �12, and not x = 7 or x = �8 as given.

Next we give solutions for the problems of the Twelfth W.J. Blundon

Contest, written February 22, 1995, [1997: 218{219].

THE TWELFTH W.J. BLUNDON CONTESTFebruary 22, 1995

1. (a) From a group of boys and girls, 15 girls leave. There are then left

two boys for each girl. After this, 45 boys leave. There are then 5 girls for

each boy. How many boys and how many girls were in the original group?

Solution. Call the number of boys b and the number of girls g. When

15 girls leave b = 2(g � 15). After 45 boys leave 5(b� 45) = g � 15. So

b = 2(5(b� 45)), b = 10b� 450, 9b = 450, b = 50; g � 15 = 50=2 = 25

and g = 40. Thus the total number of boys and girls was originally 90.

280

(b) A certain number of students can be accommodated in a hostel. If

two students share each room then two students will be left without a room.

If three students share each room then two rooms will be left over. How

many rooms are there?

Solution. Let s be the number of students and r the number of rooms.

At two students per room 2r = s � 2. Consider three per room, then

3(r � 2) = s. Now 2r = 3(r � 2) � 2 so r = 8. There are 8 rooms

(and 18 students).

2. How many pairs of positive integers (x; y) satisfy the equation

x

19+

y

95= 1?

Solution. Note that 95 = 19� 5. For x = 1; 2; : : : ; 18 we have

y

95= 1� x

19=

19� x

19=

95� 5x

95;

giving a solution. There are then 18 such pairs.

3. A book is to have 250 pages. How many times will the digit 2 be

used in numbering the book?

Solution. Two is used once in each decade in the one's place for a total

of 25 such occurrences. It is used 10 times in each hundred in the ten's place

for another 30 occurrences. And it occurs 51 times in the hundred's place,

for a grand total of 25 + 30 + 51 = 106 times.

4. Without using a calculator

(a) Show thatp7 +

p48 +

p7�p48 is a rational number.

Solution. Let x =p7 +

p48 +

p7�

p48. Then

x2 = 7 +p48 + 2

q7 +

p48

q7�

p48 + 7�

p48

= 14 + 2p49� 48 = 16

Thus x = 4, as x > 0.

(b) Determine the largest prime factor of 9919.

Solution. Considering the units digit we are led to try 7 as a factor,

yielding 9919 = 7� 1417. We are now led to try for a factor of 1417 ending

in 3 or 9 less than 38, (since 382 = 1444), and 1417 = 13� 109. Since 109

is prime, the answer is 109.

281

5. A circle is inscribed in a circular sector which is one sixth of a circle

of radius 1, and is tangent to the three sides of the sector as shown. Calculate

the radius of the inscribed circle.

Solution.

r r

r

Call the radius of the inscribed circle r. The central angle is 1

6�360� =

60�, so each small triangle is a 30�, 60�, 90� right triangle. The hypotenuse,

h, has length h = r sin 30� = r

2. So r + h = 1 gives 3

2r = 1, and r = 2

3.

6. Determine the units digit of the sum

2626 + 3333 + 4545:

Solution. The units digit is the same as for

626 + 333 + 545:

Now 6n � 6 mod 10 for n � 1 and

3n �

8>><>>:

3 n � 1 mod 4

9 n � 2 mod 4

7 n � 3 mod 4

1 n � 4 mod 4

; so 333 � 3 mod 10:

Finally 5n � 5 mod 10 for n � 1. Hence

2626 + 3333 + 4545 � 6 + 3 + 5 � 4 mod 10

and the last digit is 4.

7. Find all solutions (x; y) to the system of equations

x+ y+x

y= 19

x(x+ y)

y= 60:

282

Solution. Nowx

y=

60

x+ yso

x+ y +60

x+ y= 19; (x+ y)2� 19(x+ y) + 60 = 0

and x+ y = 4 or x+ y = 15.

If x + y = 4, x

y= 60, we get x = 60y, and 61y = 4, so y = 4

61and

x = 240

61.

If x+ y = 15, xy= 5 giving y = 3, x = 12.

The solutions for (x; y) are�240

61; 4

61

�and (12;3).

8. Find the number of di�erent divisors of 10800.

Solution. Now 10800 = 9�12�100 = 24�33�52. The factors must

be of the form 2a3b5c where 0 � a � 4, 0 � b � 3 and 0 � c � 2, so the

number of factors is 5� 4� 3 = 60.

9. Show that n4�n2 is divisible by 12 for any positive integer n > 1.

Solution. Now n4�n2 = (n� 1)n2(n+1). In any three consecutive

numbers, at least one is divisible by 3. Also if n is even, 4 divides n2, and if

n is odd, both n� 1 and n+ 1 are even so 4 divides n2 � 1. In either case

4 divides n4 � n2 and we are done.

10. Two clocks now indicate the correct time. One gains a second

every hour, and the other gains 3 seconds every 2 hours. In how many days

will both clocks again indicate the correct time?

Solution. There is a bit of a problem about whether the clock is a 12

hour or 24 hour clock. Let us assume a 12 hour clock in each case. The �rst

clock will gain 12 hours in 12�60�60 hours, or 1800 days, which is the �rst

time it will again tell the correct time. The second clock will gain 12 hours in

12� 60� 60 � �32

�hours, or 1200 days. At the end of 3600 days they will

both tell the correct time.

That completes the Corner for this number. I need suitable exam ma-

terials, and would welcome your suggestions for the evolution of this feature

of Crux with Mayhem.

283

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to the

Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,

PO Box 208283 Yale Station, New Have, CT 06520{8283 USA. The electronic

address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Playful Palindromes

What are palindromes? They are any sort of objects which are the same

when read backwards or forwards. An example of a palindromic word is

NOON. An example of a palindromic sentence is

ABLE WAS I ERE I SAW ELBA,

accredited to Napoleon. Here we concentrate on palindromic numbers such

as 12321, and henceforth the word palindrome will refer to palindromic num-

bers.

Quickly work out the following problems.

1. How many two digit palindromes are there?

2. How many three digit palindromes are there?

3. How many palindromes are there from 0 to 1998?

4. How many n-digit palindromes are there? (Consider the cases when n

is even and odd.

Palindromic Sums

Ask a friend to pick a positive integer. Tell this friend to reverse the

digits of the integer and add it to the original. If the sum is not a palindrome

then take the sum's reverse and add it to itself. Continue doing this until the

sum is a palindrome.

284

Examples

123 139

+ 321 + 931

444 1070

+ 0701

1771

It is clear that if all the digits are less than 5 then the process terminates

after the �rst step. The second example shows that a number with a digit

greater than 5 can also have a palindromic sum.

Now try these examples:

1. 1990 2. 89 3. 196

Answers

1. 68486 2. 8813200023188 3. No one has yet found the answer!

Unsolved Problem

Is it true that every integer has a palindromic sum (that is, the process

will eventually terminate)?

This is an example of where better algorithms and faster computers will

not help. Even if a super-fast computer with terabytes of memory were to

�nd the solution of 196, this leaves the problem open still as we do not know

that ALL integers have a palindromic sum.

Perhaps for the eager reader, an attempt should be made to classify as

much as possible those numbers which DO have palindromic sums. We have

already made one such class of numbers, namely those whose digits are all

less than 5. Are there any other classes of numbers that guarantee to give

palindromic sums?

Best of the Web

We present here some web sites which may be of interest to the math-

ematical web surfer:

http://www.math.princeton.edu/~kkedlaya/competitions.html

Kiran Kedlaya'sMath Competition Archive, with copies of old USAMO, IMO,

and Putnam papers, and many links to other related sites.

http://www.math.toronto.edu/mathnet/

285

U of T Mathnet. A collection of material for high school students and teach-

ers.

http://www.research.att.com/~njas/sequences/

Sloane's On-line Sequence Dictionary.

Thanks to Ben Chia, ex Cedarbrae C.I.

Decimal Expansion of Fractions

Cyrus Hsia

While programming on the computer, a couple of my friends, Peter

Plachta and Jacob Mouka noticed some interesting patterns. One such pat-

tern came about as follows: Stumped on a hard programming problem, they

tried examining their output. Through the complex mess of numbers, they

noticed that the decimal expansion of fractions, of the form 1

p, where p is a

prime number (other than 2 or 5) always had a repeating decimal pattern.

Furthermore, this repeating pattern appeared immediately after the deci-

mal point. For example, 1

3= 0:3. With many such examples generated on

the computer, they conjectured that this must be the case for all primes p,

p 6= 2; 5. In fact, their conjecture is correct. We will show this result and

some interesting generalizations. But �rst let us give some de�nitions so that

there will be no ambiguity in their meaning.

The decimal expansion of a fraction is of the form

0:a1a2 : : : anb1 : : : bm : : : b1 : : : bm : : : ;

where the sequence b1 : : : bm repeats inde�nitely. We write a bar over it to

indicate that it repeats, that is,

0:a1a2 : : : anb1 : : : bm:

De�nition 1.1. The repeating part in a decimal expansion is the number

(that is, the sequence of digits b1 : : : bm) under the bar. The astute reader

would notice that the bar could have been placed one decimal place over, as

0:a1a2 : : : anb1b2 : : : bmb1, or that a bar could have been placed over twice

the above sequence and still give the same value. So to make this de�nition

unambiguous, we will always mean the earliest and smallest such pattern

after the decimal point. Finally, we will say that fractions, like 1

2, that ter-

minate do not have a repeating part.

De�nition 1.2. With the above de�nition of the repeating part in mind, the

leading part is the number (that is, the sequence of digits a1 : : : an) immedi-

ately following the decimal point before the repeating part. Note that frac-

tions may not have a leading part, as in 1

3.

286

Finally, let us call a number good if it is relatively prime to 10, that

is it is not divisible by 2 or 5. With these de�nitions, let us show that the

conjecture above is true by �rst proving some simple results that we need.

We will assume from now on that we are looking at the decimal expansions

of 1

p, where p is a good prime.

Lemma 1.3. The decimal expansion of 1p, where p is a good prime, must have

a repeating part, that is, 1

p= 0:a1a2 : : : anb1 : : : bm, m � 1.

Proof. Left to the reader as an exercise.

Lemma 1.4. If 0:a1a2 : : : anb1 : : : bm is the decimal expansion of a fraction,

then an 6= bm.

Proof. Left to the reader as an exercise.

Lemma 1.5. For a decimal expansion,

0:a1 : : : anb1 : : : bm

=10n�1a1 + 10n�2a2 + � � �+ an

10n+

10m�1b1 + 10m�2b2 + � � �+ bm

10n(10m � 1);

which we writea1 : : : an

10n+

b1 : : : bm

10n(10m � 1)

in short form.

Proof. Left to the reader as an exercise.

Now with these lemmas we can prove the conjecture.

Conjecture 1.6. (Peter Plachta and Jacob Mouka)

The fraction 1

pin decimal notation consists only of a repeating part (that

is, it has no leading part) if p is a good prime.

Proof. Assume on the contrary that 1

p= 0:a1a2 : : : anb1 : : : bm, where n �

1, so that there is at least one digit in the leading part.

Case I:m = 0, that is, the decimal expansion terminates. Then

1

p= 0:a1 : : : an =

a1 : : : an

10n) 10n = p(a1 : : : an) ) p j 10n;

but p is good, so this is not possible.

Case II:m � 1. Then

1

p=a1 : : : an

10n+

b1 : : : bm

10n(10m � 1)

(by Lemma 1.5). Multiplying both sides by 10n(10m � 1)p, we get

10n(10m � 1) = p[(10m � 1)(a1 : : : an) + b1 : : : bm]

= p[10ma1 : : : an + b1 : : : bm � a1 : : : an]:

287

Since n � 1, 10 j 10n(10m � 1), so 10 divides the right hand side of

the equation. In particular, since m � 1,

10 j (b1 : : : bm � a1 : : : an) ) 10 j (bm � an):

Now 0 � an; bm � 9) an = bm, which by Lemma 1.4 is impossible.

Why does the fraction have to have a prime as its denominator? The

prime is what makes the particular types of fractions of Conjecture 1.6 have

no leading part. Then a natural question to ask is whether the converse

holds, that is, given a decimal expansion with only a repeating part, is the

denominator in its fractional representation a prime. To see that the converse

does not hold, look at the decimal expansion of 1

9, which is 0:1. It has only

a repeating part but the denominator is not prime.

It is not the fact that the denominator is a prime that makes Conjecture

1.6 true as the following theorem will show.

Theorem 1.7. For all positive integers N > 1, 1

Nin decimal notation has a

repeating part but no leading part if and only ifN is good.

First, we need a result that most of the readers will already be familiar

with. In fact, it has popped up a number of times in olympiad and pre-

olympiad type problems. Consider the numbers Mi consisting of i digits

made up entirely of 1's, for example, M1 = 1, M2 = 11, M3 = 111. Then

we have:

Lemma 1.8. For any good positive integer N , N divides some Mi.

Proof. Consider the N + 1 numbersM1, M2, : : : ,MN+1 modulo N . Then

by the Pigeonhole Principle, at least two of the Mi's are congruent, say

Mj �Mk (mod N), where j > k. Then we have

Mj �Mk (mod N) ) 11 : : : 1| {z }j

� 11 : : : 1| {z }k

(mod N)

) 11 : : : 1| {z }j�k

00 : : : 0| {z }k

� 0 (mod N)

) 11 : : : 1| {z }j�k

� 0 (mod N):

This can be done since neither 2 nor 5 divides N . This means that Mj�k is

divisible by N .

Proof of Theorem 1.7.

(Only if part) If 1

Nhas a repeating part but no leading part, then we have

1

N= 0:b1 : : : bm =

10m�1b1 + � � �+ bm

10m � 1

by Lemma 1.5. Cross-multiplying,we get 10m�1 = N(10m�1b1+� � �+bm).

Thus N is good.

288

(If part) If N is good, then by Lemma 1.8 there is an M = Mi divisible by

N . Let M = kN , where k is an integer, k � 1. We have then

1

N=

k

M=

k

11 : : : 1| {z }i

=9k

99 : : : 9| {z }i

=9k

10i � 1:

Hence,

1

N< 1 =) k < M =) 9k < 9M = 10i � 1 < 10i:

This means that 9k is at most an i-digit number; that is, 9k can be written

as b1 : : : bi, where some of the bi's can be 0. Thus

1

N=b1 : : : bi

10i � 1= 0:b1 : : : bi

as required.

Now it is time for the readers to roll up their sleeves and tackle some

problems.

Exercises

1.1 Prove Lemmas 1.3, 1.4, and 1.5.

1.2 (a) Show that any fraction of the form a

bin lowest terms, where a and

b are positive integers, a < b, also has no leading term if and only if b

is good.

(b) Show that (a) holds even when a � b and with the same conditions.

Thus a

b= a1a2 : : : an:b1 : : : bm if and only if b is good.

1.3 (Related exercise) Find the smallest positive integer n such that

n3 + 9n2 + 9n+ 7

1996

in decimal notation terminates (that is, it has no repeating part, just a

�nite leading part).

1.4 (Related exercise) Write a computer program to �nd the repeating part

in a fraction of the form 1

p, where p is a prime.

To �nish o�, fractions with denominators that have factors of 2 and 5 still

need to be considered. In fact, it turns out that there is a nice relationship

between the number of factors of 2 and 5 in the denominator and the number

of terms in the leading part of its decimal representation.

Lemma 2.1. IfN is of the form 2kM , where M is good, then 1

Nhas k terms

in its leading part, and it has a repeating part ifM > 1.

289

Proof. IfM = 1 then it is obvious that 1

2k= 5k

10khas k terms in the leading

part and no repeating part. IfM > 1, then we have

1

N=

1

2kM=

5k

10kM=

1

10k

5k

M

!:

By Exercise 1.2, we have 5k

M= a1 : : : an:b1 : : : bm. Dividing through by

10k and noting 1

Nis less than 1, we must have n < k. Now once we show

that an 6= bm, then we are �nished, as then

1

N=

1

10k

5k

M

!= 0: 0 : : : 0a1 : : : an| {z }

k terms

b1 : : : bm:

If an = bm then

5k�1

2M=

1

10

5k

M

!= a1 : : : an�1:anb1 : : : bm�1;

contradicting Exercise 1.2 that a fraction has a repeating part if and only if

the denominator is good.

Lemma 2.2. Similarly, if N is of the form 5kM , where M is good, then 1

N

has k terms in its leading part, and it has a repeating part ifM > 1.

Proof. Similar (in fact identical) to the proof of Lemma 2.1.

Now for the grand �nale...

Theorem 2.3. If N is of the form 2k5lM , where again M is good, then the

fraction 1

Nhasmax(k; l) terms in the leading part, and ifM > 1 then it also

has a repeating part.

Proof. Once again, this just takes from material shown before. If k or l

equals 0, then the result becomes Lemma 2.2 or 2.1. If k = l, then the result

is immediate, and in fact the leading part has k = l 0's. If k > l > 1, then we

have 1

2k5lM= 1

10l2k�lM. By Lemma 2.1, we have k � l terms in the leading

part of 1

2k�1M. Dividing by 10l just shifts the decimal point l places to the

left giving (k� l)+ l = k terms in the leading part (and similarly l terms for

l > k > 1).

More Exercises

2.1 (Exploration) In our long analysis, we have shown (not on purpose), that

the length of the leading part can be determined. However, the length

of the repeating part was never discussed. Determine any relationships

between the numbers in the fraction and the length of the repeating part

in its decimal representation.

2.2 (a) If we had 6 �ngers (3 on each hand?), then perhaps our system of

counting would be in base 6. In general, if the base was pq, where p

290

and q are distinct primes, then show an analogous result to the above

theorems.

(b) Do the same for any base B.

2.3 (More down to earth problem) Of the numbers 1995, 1996, : : : , 1999,

which number has the greatest number of terms in the leading part of

its reciprocal? With this, we hope you will be a leading part in this year!

Acknowledgements

Peter Plachta and JacobMouka, 3rd year students, University of Toronto

at Scarborough, for the conjecture.

An Ambivalent Sum

Naoki Sato

A conditionally convergent seriesPan is a series which converges, but

such that the seriesP janj diverges. One of the most famous examples of

such a series, which appears in textbooks everywhere, is the following:

1� 1

2+

1

3� 1

4+

1

5� 1

6+ � � � = ln2:

A basic, but unusual property of conditionally convergent series is that we

may rearrange the terms, so that the sum comes out to be any value we wish

(see Calculus, 3rd Ed., M. Spivak, pp. 476-8), so we must be careful when

manipulating the terms in determining the sum. For example, a question of

the 1954 Putnam asks to verify that

1 +1

3� 1

2+

1

5+

1

7� 1

4+

1

9+

1

11� 1

6+ � � � =

3

2ln2:

Same terms, di�erent sum. These results suggest a generalization, and the

form of the alternating harmonic series makes for an elegant analysis.

For positive integers r and s, let

S(r; s) = 1 +1

3+ � � �+ 1

2r � 1| {z }r terms

� 1

2� 1

4� � � � � 1

2s| {z }s terms

+1

2r + 1+

1

2r + 3+ � � �+ 1

4r� 1| {z }r terms

� 1

2s+ 2� 1

2s+ 4� � � � � 1

4s| {z }s terms

+ � � � :

291

A �rst step in summing S(r; s) will be looking at its partial sums, so

let Sn(r; s) be the sum of the �rst n(r+ s) terms of S(r; s). Assume for the

moment that r > s. Then,

Sn(r; s) =

�1 +

1

3+ � � �+ 1

2r � 1

���1

2+

1

4+ � � �+ 1

2s

�

+

�1

2r + 1+

1

2r+ 3+ � � �+ 1

4r � 1

�

��

1

2s+ 2+

1

2s+ 4+ � � �+ 1

4s

�+ � � �

+

�1

2(n� 1)r + 1+

1

2(n� 1)r+ 3+ � � �+ 1

2nr � 1

�

��

1

2(n� 1)s+ 2+

1

2(n� 1)s+ 4+ � � �+ 1

2ns

�

=

�1 +

1

3+ � � �+ 1

2r � 1� 1

2� 1

4� � � � � 1

2r

�

+1

2+

1

4+ � � �+ 1

2r� 1

2� 1

4� � � � � 1

2s

+

�1

2r + 1+

1

2r+ 3+ � � �+ 1

4r � 1

� 1

2r + 2� 1

2r+ 4� � � � � 1

4r

�

+1

2r + 2+

1

2r + 4+ � � �+ 1

4r� 1

2s+ 2� 1

2s+ 4

� � � � � 1

4s+ � � �

+

�1

2(n� 1)r + 1+

1

2(n� 1)r+ 3+ � � �+ 1

2nr � 1

� 1

2(n� 1)r+ 2� 1

2(n� 1)r + 4� � � � � 1

2nr

�

+1

2(n� 1)r+ 2+

1

2(n� 1)r + 4+ � � �+ 1

2nr

� 1

2(n� 1)s+ 2� 1

2(n� 1)s+ 4� � � � � 1

2ns

= A2nr +1

2ns+ 2+

1

2ns+ 4+

1

2ns+ 6+ � � �+ 1

2nr

= A2nr +1

2

�1

ns+ 1+

1

ns+ 2+

1

ns+ 3+ � � �+ 1

nr

�;

where

An = 1� 1

2+

1

3� 1

4+ � � �+ (�1)n+1

1

n:

292

We may estimate a more general sum by integrals. Since y = 1

xis

decreasing on (0;1), for 0 < d < a,

Za+md

a

1

xdx <

d

a+

d

a+ d+ � � �+ d

a+ (m� 1)d<

Za+(m�1)d

a�d

1

xdx

so that

1

dln

�a+md

a

�<

1

a+

1

a+ d+ � � �+ 1

a+ (m� 1)d

<1

dln

�a+ (m� 1)d

a� d

�

(if you are not sure why these inequalities are true, draw a graph).

So, plugging in a = ns+ 1, d = 1, m = n(r� s), we obtain

ln

�nr + 1

ns+ 1

�<

1

ns+ 1+

1

ns+ 2+ � � �+ 1

nr< ln

�r

s

�

=) limn!1

�1

ns+ 1+

1

ns+ 2+ � � �+ 1

nr

�= ln

�r

s

�:

Note that we could have also used the well-known result

limn!1

�1 +

1

2+

1

3+ � � �+ 1

n� lnn

�= ;

where is Euler's constant.

Finally, we have that

S(r; s) = limn!1

Sn(r; s) = ln2 +1

2ln

�r

s

�=

1

2ln

�4r

s

�:

The case r < s is argued similarly (the case r = s is obvious), and we

obtain the same value.

We make some observations:

(1) The value S(r; s) depends only on the ratio r=s, as can be predicted.

(2) The case r = 1, s = 4, produces the intriguing-looking formula

1� 1

2� 1

4� 1

6� 1

8+

1

3� 1

10� 1

12� 1

14� 1

16+ � � � = 0:

(3) The set fS(r; s)g = f12ln(4r

s) : r; s 2Z+g is dense in R.

293

(4) Going back to a result mentioned above, to rearrange the terms of a

conditionally convergent series so that it sums to a given value, say S,

we use the following algorithm: Assuming S � 0, we successively add

the �rst positive terms of the series until we go over S; then we add

the �rst negative terms of the series until we go under S, and so on

(if S < 0, we merely start with the negative terms). In fact, this is

the idea behind the proof. The nature of the conditionally convergent

series guarantees this algorithm produces a series with sum S.

It has been observed that if we try this with the alternating harmonic

series, the number of positive terms and negative terms taken each time

becomes constant. We can see this is because of (3); that is, any real

number can be approximated by a sum S(r; s).

J.I.R. McKnight Problems Contest 1978

The J.I.R. McKnight Problems Contest is a problem solving contest in

Scarborough, Ontario. It began in 1975 in honour of one of Scarborough's

�nest Mathematics teachers and continues today as a scholarship paper for

senior students. This paper, although tailored to students taking OAC (On-

tario Academic Credit) Mathematics courses, contains many problems that

are accessible to all high school students. With some ingenuity, most prob-

lems reveal some very beautiful results and we hope our readers will enjoy

this as well.

1. Verify that 13! = 1122962 � 798962.

2. Show that, for any real numbers x, y, and z,

(x+ y)z � (x2 + y2)

2+ z2:

3. Sum to k terms the series whose nth term is

n4 + 2n3 + n2 � 1

n2 + n:

4. Show that in order that the quadratic function

3x2 + 2pxy + 2y2 + 2ax� 4y + 1

may be resolved into factors linear in x and y, p must be a root of the

equation p2 + 4ap+ 2a2 + 6 = 0.

5. Find the sum of all 3{digit odd numbers that can be made using the

digits 1, 3, 5, 7, 8, 9, where no repeated digits are used in any number.

294

6. Find a cubic function f(x) with the following properties:

(a) It is a curve passing through (1; 1).

(b) The tangent to the curve at (1; 1) has slope 1.

(c) The sum of the roots of its corresponding equation is 1

5.

(d) The sum of the squares of the roots of its corresponding equation

is 121

25.

1984 Swedish Mathematics Olympiad

Qualifying Round

1. Solve the following system of equations (in 5 unknowns):

ab = 1;

bc = 2;

cd = 3;

de = 4;

ea = 6:

2. Show that if n is an odd natural number, then

n12 � n8 � n4 + 1

is divisible by 29.

3. Find all positive numbers x such that

x8�3x > x7:

4. ABC is an isosceles triangle with AB = AC. Choose a point D on

the side AB, and a point E on AC produced through C, such that

AD + AE = AB + AC. Show that DE > BC.

5. The numbers 1, 2, : : : , 9, are placed in a 3� 3 grid, so that no row, no

column and neither of the two major diagonals contains a sequence of

numbers ordered by size (either increasing or decreasing). Show that

the number in the central square must be odd.

6. A woman, who is not yet 100 years old, and one of her grandchildren

have the same birthday. For six years in a row, the woman's age was a

multiple of her grandchild's age. How old was the woman on the sixth

of these birthdays?

295

Final Round

1. Let A and B be two arbitrary points inside a circle C. Show that there

always exists a circle through A and B which lies completely inside C.

2. The squares in a 3 � 7 grid are coloured either blue or yellow. Con-

sider all rectangles of m � n squares in this grid, where 2 � m � 3,

2 � n � 7. Show that at least one of these rectangles has all four of its

corner squares the same colour.

3. Show that if a and b are positive numbers, then

�a+ 1

b+ 1

�b+1��a

b

�b:

4. Find positive integers p and q such that all the roots of the polynomial

(x2 � px+ q)(x2� qx+ p)

are positive integers.

5. Solve the system of equations

a3 � b3 � c3 = 3abc;

a2 = 2(a+ b+ c);

where a, b, and c are natural numbers.

6. The numbers a1, a2, : : : , a14 are positive integers. It is known that

14Xi=1

ai = 6558:

Show that the numbers a1, a2, : : : , a14 consist of the numbers 1, 2,

: : : , 7 each taken twice.

296

Swedish Mathematics Olympiad Solutions

1983 Final Round

1. The positive integers are added in groups as follows: 1, 2+3, 4+5+6,

7 + 8 + 9 + 10, and so on. What is the sum of the nth group?

Solution by Bob Prielipp, University of Wisconsin-Oshkosh, Oshkosh,

WI, USA.

Let Tk = k(k+1)=2 be the kth triangular number. Then the n numbers

in the nth group are Tn�1 + 1, Tn�1 + 2, : : : , Tn�1 + n = Tn. Using the

formula

S =n(a+ l)

2;

where S is the sum of a �nite arithmetic progression, a is the �rst term, l is

the �nal term, and n is the number of terms in the progression, the desired

sum is

n[(Tn�1 + 1+ Tn)]

2=n�(n�1)n

2+ 1+

n(n+1)

2

�2

=n3 + n

2:

Mayhem Problems

The Mayhem Problems editors are:

Richard Hoshino Mayhem High School Problems Editor,

Cyrus Hsia Mayhem Advanced Problems Editor,

Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only problems | the

next issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from the

previous issue be submitted by 1 October 1997, for publication in the issue

5 months ahead; that is, issue 2 of 1998. We also request that only students

submit solutions (see editorial [1997: 30]), but we will consider particularly

elegant or insightful solutions from others. Since this rule is only being im-

plemented now, you will see solutions from many people in the next few

months, as we clear out the old problems from Mayhem.

297

High School Problems

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,

Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>

H225. In Cruxmayhemland, stamps can only be bought in two de-

nominations, p and q cents, both of which are at least 31 cents. It is known

that if p and q are relatively prime, the largest value that cannot be created

by these two stamps is pq � p � q. For example, when p = 5 and q = 3,

one can a�x any postage that is higher than 15� 5� 3, or 7 cents, but not

7 cents itself. The governor of Cruxmayhemland tells you that 1997 is the

largest value that cannot be created by these stamps. Find all possible pairs

of positive integers (p; q) with p > q.

H226. In right-angled triangle ABC, with BC as hypotenuse, AB =

x and AC = y, where x and y are positive integers. Squares APQB,

BRSC, and CTUA are drawn externally on sides AB, BC, and CA re-

spectively. When QR, ST , and UP are joined, a hexagon is formed. Let K

be the area of hexagon PQRSTU .

(a) Prove that K cannot equal 1997. (HINT: Try to �nd a general formula

for K.)

(b) Prove that there is only one solution (x; y) with x > y so that

K = 1998.

H227. The numbers 2, 4, 8, 16, : : : , 2n are written on a chalkboard. A

student selects any two numbers a and b, erases them, and replaces them by

their average, namely (a+b)=2. She performs this operation n�1 times until

only one number is left. Let Sn and Tn denote the maximum and minimum

possible value of this �nal number, respectively. Determine a formula for

Sn and Tn in terms of n.

H228. Verify that the following three inequalities hold for positive

reals x, y, and z:

(i) x(x � y)(x� z) + y(y � x)(y � z) + z(z � x)(z � y) � 0 (this is

known as Schur's Inequality),

(ii) x4 + y4 + z4 + xyz(x+ y + z) � 2(x2y2 + y2z2 + z2x2),

(iii) 9xyz + 1 � 4(xy+ yz + zx), where x+ y + z = 1.

(Can you derive an ingenious method that allows you to solve the problem

without having to prove all three inequalities directly?)

298

Advanced Problems

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.

M1G 1C3 <hsia@math.toronto.edu>

A201. Consider an in�nite sequence of integers a1, a2, : : : , ak, : : :

with the property that every m consecutive numbers sum to x and every n

consecutive numbers sum to y. If x and y are relatively prime then show

that all numbers are equal.

A202. Let ABC be an equilateral triangle and � its incircle. IfD and

E are points on AB and AC, respectively, such that DE is tangent to �,

show thatAD

DB+AE

EC= 1:

(8th Iberoamerican Mathematical Olympiad, Mexico '93)

A203. Let Sn = 1+a+aa+ � � �+aa:::a , where the last term is a tower

of n� 1 a's. Find all positive integers a and n such that Sn = naSn�1

n .

A204. Given a quadrilateral ABCD as shown, with AD =p3,

AB + CD = 2AD, \A = 60� and \D = 120�, �nd the length of the

line segment from D to the midpoint of BC.

60�

120�

p3

A

B

C

D

Challenge Board Problems

Editor: Ravi Vakil, Department of Mathematics, Princeton University,

FineHall, Washington Road, Princeton, NJ 08544{1000USA <vakil@math.princeton.edu>

C73. Proposed by Matt Szczesny, 4th year, University of Toronto.

The sequence fang consists of positive reals, such that1Xn=1

an diverges.

Show that

1Xn=1

an

sndiverges, where sn is the nth partial sum, that is,

sn = a1 + a2 + � � �+ an.

299

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 812"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed to

the Editor-in-Chief, to arrive no later than 1 March 1998. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication.

Correction

Problem 2235 [1997: 168]was incorrectly attributed to Walther Janous,

Ursulinengymnasium, Innsbruck, Austria, when it was in fact proposed by

D.J. Smeenk, Zaltbommel, the Netherlands. The editor apologizes for this

error.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX,

either to the Editor-in-chief's departmental FAX machine in St. John's, New-

foundland, or to the Canadian Mathematical Society's FAX machine in Ot-

tawa, Ontario. While we understand the reasons for solvers wishing to use

this method, we have found many problems with it. The major one is that

hand-written material is frequently transmitted very badly, and at times is

almost impossible to read clearly. We have therefore adopted the policy that

we will no longer accept submissions sent by FAX. We will, however, con-

tinue to accept submissions sent by email or regular mail. We do encourage

email. Thank you for your cooperation.

300

2251. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

In the plane, you are given a circle (but not its centre), and points A, K, B,

D, C on it, so that arc AK = arc KB and arc BD = arc DC.

Construct, using only an unmarked straightedge, the mid-point of arc AC.

2252. Proposed by K.R.S. Sastry, Dodballapur, India.

Prove that the nine-point circle of a triangle trisects a median if and only

if the side lengths of the triangle are proportional to its median lengths in

some order.

2253. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle and Ib, Ic are the excentres of 4ABC relative to

sides CA, AB respectively.

Suppose that IbA2 + IbC

2 = BA2 +BC2 and that IcA2 + IcB

2 = CA2 +

CB2.

Prove that 4ABC is equilateral.

2254. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is an isosceles triangle with AB = AC. Let D be the point on

sideAC such that CD = 2AD. Let P be the point on the segment BD such

that \APC = 90�.

Prove that \ABP = \PCB.

2255. Proposed by Toshio Seimiya, Kawasaki, Japan.

Let P be an arbitrary interior point of an equilateral triangle ABC.

Prove that j\PAB � \PACj � j\PBC � \PCBj.2256. Proposed by Russell Euler and Jawad Sadek, Department of

Mathematics and Statistics,NorthwestMissouri State University,Maryville,

Missouri, USA.

If 0 < y < x � 1, prove thatln(x)� ln(y)

x� y> ln

�1

y

�.

2257. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

The diagonals AC and BD of a convex quadrilateral ABCD intersect

at the pointO. LetOK, OL,OM , ON , be the altitudes of triangles4ABO,4BCO, 4CDO, 4DAO, respectively.

Prove that if OK = OM and OL = ON , then ABCD is a parallelo-

gram.

301

2258. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

In a right-angled triangle ABC (with \C = 90�), D lies on the seg-

ment BC so that BD = ACp3. E lies on the segment AC and satis�es

AE = CDp3. Find the angle between AD and BE.

2259. Proposed by Paul Yiu, Florida Atlantic University, Boca Ra-

ton, Florida, USA.

Let X, Y , Z, be the projections of the incentre of 4ABC onto the

sidesBC,CA, AB respectively. LetX0, Y 0, Z0, be the points on the incircle

diametrically opposite to X, Y , Z, respectively. Show that the lines AX0,

BY 0, CZ0, are concurrent.

2260. Proposed by Vedula N. Murty, Andhra University, Visakhap-

atnam, India.

Let n be a positive integer and x > 0. Prove that

(1 + x)n+1 � (n+ 1)n+1

nnx:

2261. Proposed by Angel Dorito, Geld, Ontario.

Assuming that the limit exists, �nd

limN!1

1 +

2 + N+:::

1+:::

N + 1+:::

2+:::

!;

where every fraction in this expression has the form

a+ b+:::

c+:::

b+ c+:::

a+:::

for some cyclic permutation a, b, c of 1, 2, N .

[Proposer's comment: this problem was suggested by Problem 4 of

Round 21 of the International Mathematical Talent Search,Mathematics and

Informatics Quarterly, Vol. 6, No. 2, p. 113.]

2262. Proposed by Juan-Bosco Romero M�arquez, Universidad de

Valladolid, Valladolid, Spain.

Consider two triangles4ABC and4A0B0C0 such that \A � 90� and

\A0 � 90� and whose sides satisfy a > b � c and a0 > b0 � c0. Denote the

altitudes to sides a and a0 by ha and h0a.

Prove that (a)1

hah0a� 1

bb0+

1

cc0, (b)

1

hah0a� 1

bc0+

1

b0c.

302

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

2145. [1996: 170] Proposed by Robert Geretschl�ager, Bundesreal-

gymnasium, Graz, Austria.

Prove that

nYk=1

�ak + bk�1

� � nYk=1

�ak + bn�k

�for all a, b > 1.

Editor's composite solutionbased on the nearly identical solutions sub-

mitted by almost all the solvers.

Note that if n is odd, then both sides have the positive factor�n+ 1

2

�a + b

n�1

2 :

Hence the given inequality is equivalent to

bn=2cYk=1

�ka+ bk�1

� �(n� k+ 1)a+ bn�k

�

�bn=2cYk=1

�ka+ bn�k

� �(n� k + 1)a+ bk�1

�:

Therefore, it su�ces to show that, for all k = 1, 2, : : : , bn2c,

�ka+ bk�1

� �(n� k + 1)a+ bn�k

�� �

ka+ bn�k� �

(n� k + 1)a+ bk�1�: (1)

After expanding and cancelling equal terms, (1) becomes

(n� 2k + 1)abk�1 � (n� 2k+ 1)abn�k: (2)

We have n � 2k + 1 � n � 2�n

2

�+ 1 = 1 > 0, and bn�2k+1 � 0, which

implies bk�1 � bn�k. Thus, (2) follows immediately.

Solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;

THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece;

DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA;

F.J. FLANIGAN, San Jose State University, San Jose, California, USA;

RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Fer-

ris State University, Big Rapids, Michigan, USA; MITKO KUNCHEV, Baba

Tonka School of Mathematics, Rousse, Bulgaria; KEE-WAI LAU, Hong Kong;

DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; L. RICE, Univer-

sity of Toronto Schools, Toronto, Ontario; KRISTIAN SABO, student,

303

Osijek, Croatia; JOEL SCHLOSBERG, student, Hunter College High School,

New York NY, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY

SMITH,MountRoyal College, Calgary, Alberta; PANOS E. TSAOUSSOGLOU,

Athens, Greece; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo,

Ontario; and the proposer.

Several solvers noted that the given conditions can be relaxed to a > 0

(or even a � 1) and b � 1. This is obvious from the proof above.

2146. [1996: 171] Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with AB > AC, and the bisector of \A meets BC

at D. Let P be an interior point on the segment AD, and let Q and R be the

points of intersection of BP and CP with sides AC and AB respectively.

Prove that PB � PC > RB � QC > 0.

Solution by the proposer.

Let E be a point on AB such that AE = AC. Then we get

4AEP � 4ACP , so that \AEP = \ACP and PE = PC. Since

\BEP = 180� � \QCP > 180� � \ACB > \ABC > \EBP , we have

PB > PE = PC. Because AE = AC < AB, we obtain

\ABP < \AEP = \ACP . Since \BAP = \PAC, we have

\APR = \ACP + \PAC > \ABP + \BAP = \APQ.

Let F be a point on AB such that \APF = \APQ, then

\APF < \APR, so F is between A and R, and 4APF � 4APQ,so that AF = AQ, PF = PQ and \AFP = \AQP . Therefore,

AR > AF = AQ.

In addition, \ARP = \RPB + \RBP = \RPB + \ABQ and

\PFR = \PQC = \BQC = \BAC + \ABQ:

If \RPB = \BAC; \ARP = \PFR; so PR = PF = PQ;

if \RPB < \BAC; \ARP < \PFR; so PR > PF = PQ;

if \RPB > \BAC; \ARP > \PFR; so PR < PF = PQ:

We consider 2 cases.

Case I: \BAC � 90�: Because \RPB < \ARC < 90� � \BAC,

we have PR > PQ. Let G be a point on PR such that PG = PQ and

H a point on PB such that PH = PC. Then 4PGH � 4PQC; thus

we get GH = QC. As \BRG > \RAC � 90�, we get GB > RB, so

BH > GB �GH > RB �GH. Hence

PB � PC = BH > RB � QC:

304

Case II: \BAC < 90�: Then there exists a point O on AD such that

\BOC = 180� � \BAC.

(i) IfO coincides with P , we have \RPB = \BAC and so PR = PQ.

We take a point J on PB such that PJ = PC. Then 4PRJ � 4PQC,

thusRJ = QC: BecauseBJ > RB�RJ = RB�QC, we have PB�PC >

RB � QC.

(ii) If P is between A andO, then \BPC < \BOC = 180��\BAC,

so \RPB = 180� � \BPC > \BAC and from above, PR < PQ.

We take a point K on PQ such that PK = PR and a point M on PC

produced beyond C such that PM = PB(> PC). Let T be the intersec-

tion of KM and QC. As 4PKM � 4PRB, we get KM = RB and

\PKM = \PRB, so

\QKM = \ARP = \RPB + \RBP = \RPB + \ABQ

> \BAC + \ABQ = \BQC:

Therefore, \QKT > \KQT , thusKT < QT . Hence

RB � QC = KM � QC = (KT + TM)� (QT + TC)

= (KT �QT ) + (TM � TC) < TM � TC:

As TM � TC < CM = PB � PC, we obtain RB � QC < PB � PC.

(iii) If P is betweenO andD, then \BPC > \BOC = 180��\BAC,

so \RPB = 180� � \BPC < \BAC. Thus, from the above, PR > PQ.

We take pointsX and Y on PR and PB, respectively, such that PX = PQ

and PY = PC. Then 4PXY � 4PQC, so XY = QC.

As \BPC > \BOC = 180� � \BAC, we have \QPR + \RAQ >

180�, thus\ARP+\AQP < 180�: Since\AQP > \ARP , we get\ARP <

90�, so \BRP > 90�; hence RB < XB.

Therefore, Y B > XB � XY > RB � XY = RB � QC. Thus

PB � PC = Y B > RB �QC.

No other solutions were submitted.

2147. [1996: 171] Proposed by Hoe Teck Wee, Singapore.

Let S be the set of all positive integers x such that there exist positive

integers y andm satisfying x2 + 2m = y2.

(a) Characterize which positive integers are in S.

(b) Find all positive integers x so that both x and x+ 1 are in S.

305

Solution by Thomas Leong, Staten Island, NY, USA.

(a) We show that S = fx : x = 2r � 2s; r > s � 0; r; s 2 Zg.First suppose that x2 + 2m = y2 with x positive. Then 2m = y2 � x2 =

(y + x)(y � x), and so y + x = 2r and y � x = 2s where r > s � 1 (s 6= 0

since x and y are of the same parity). Thus 2x = 2r�2s, or x = 2r�1�2s�1

where r � 1 > s� 1 � 0. Conversely, if x = 2r � 2s with r > s � 0, then

with m = r + s+ 2 and y = 2r + 2s, we have x2 + 2m = y2.

(b) The desired set is fx : x = 2r � 2; r � 2 or x = 2r � 1; r � 1g.Now every odd number in S is of the form 2r � 1, r � 1. If x + 1 2 S is

odd, then x + 1 = 2r � 1 and x = 2r � 2 are both in S provided r � 2. If

x 2 S is odd, then x = 2r � 1 and x + 1 = 2r = 2r+1 � 2r are both in S

provided r � 1.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; THEODORE

CHRONIS, student, Aristotle University of Thessaloniki, Greece; CHARLES

R. DIMINNIE, Angelo State University, San Angelo, TX, USA; KEITH

EKBLAW, Walla Walla, Washington, USA; F.J. FLANIGAN, San Jose State

University, San Jose, California, USA; DAVIDHANKIN, Hunter CollegeCam-

pus Schools, New York, NY, USA; FLORIAN HERZIG, student, Perchtolds-

dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV

KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI

LAU, Hong Kong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA;

GIOVANNI MAZZARELLO, Florence, Italy and IAN JUNE GARCES, Ateneo

de Manila University, Manila, The Philippines; MICHAEL PARMENTER,

Memorial University of Newfoundland, St. John's, Newfoundland; HEINZ-

J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH, Mount Royal College,

Calgary, Alberta; LAWRENCE SOMER, The Catholic University of America,

Washington, DC; DAVID R. STONE, Georgia Southern University, States-

boro, Georgia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; KENNETH

M. WILKE, Topeka, Kansas, USA; and the proposer. There were 6 incorrect

or incomplete solutions.

KONE �CN �Y remarks the similarity between this problem and problem 13,

page 162 of Introduction to Number Theory by W.W. Adams and L.J.

Goldstein, which claims that the Diophantine equation x2 � y2 = mk is

solvable for x; y for any given m; k � 3. (In the present problem m = 2

and x and y are interchanged.)

Several solvers described S in terms of the binary expansions of its

members: namely any positive integer in S has a block of 1s followed by a

(possibly empty) block of 0s.

306

2148. [1996: 171] Proposed by Aram A. Yagubyants, Rostov na

Donu, Russia.

Suppose thatAD, BE andCF are the altitudes of triangleABC. Sup-

pose that L, M , N are points on BC, CA, AB, respectively, such that

BL = DC, CM = EA, AF = NB.

Prove that:

1. the perpendiculars to BC, CA, AB at L, M , N , respectively are con-

current;

2. the point of concurrency lies on the Euler line of triangle ABC.

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let H be the orthocentre and O the circumcentre. Re ect H in O to

get a point H0 on OH, and let L0, M 0, N 0 be the feet of the perpendicu-

lars from H0 onto BC, CA and AB, respectively. If X, Y , Z denote the

midpoints of BC, CA, AB, then HD k OX k H0L0 (and similarly for

the other sides). As HO = OH0 we have DX = XL0 and, together with

XC = XD + DC = BL0 + L0X = BX, yields that DC = BL0. Hence

from the de�nition of L and since L0 is also an interior point of BC, we have

L = L0 and analogously M = M 0 and N = N 0. The perpendiculars to the

sides of the triangle LH0;MH0, and NH0 intersect atH0, which lies on the

Euler lineHO of 4ABC, as we wanted to prove.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-

ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MITKO

KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria; L. RICE,

Woburn CI, Scarborough Ontario; TOSHIO SEIMIYA, Kawasaki, Japan;

SHAILESH SHIRALI, Rishi Valley School, India; D.J. SMEENK, Zaltbommel,

the Netherlands; and the proposer. One solution was incomplete.

Several readers reduced the result to the theorem, attributed some-

times to Steiner and sometimes to Carnot, that appeared recently in CRUX

with MAYHEM [1997: 122] in an alternative solution to problem 2120. Al-

though this approach leads to a nice solution here, it takes more work to set

up the machinery than to solve the problem without it.

2149. [1996: 171] Proposed by Juan-Bosco Romero M�arquez, Uni-

versidad de Valladolid, Valladolid, Spain.

Let ABCD be a convex quadrilateral andO is the point of the intersec-

tion of the diagonalsAC andBD. Let A0B0C0D0 be the quadrilateral whose

vertices, A0, B0, C0, D0, are the feet of the perpendiculars drawn from the

point O to the sides BC, CD, DA, AB, respectively.

Prove that ABCD is an inscribed (cyclic) quadrilateral if and only if

A0B0C0D0 is a circumscribing quadrilateral (A0B0, B0C0, C0D0, D0A0 are

tangents to a circle).

307

Solution by D.J. Smeenk, Zaltbommel, the Netherlands (with notation

modi�ed by the editor)

Assume ABCD is cyclic. We denote:

\DBA = \DCA = �; \ACB = \ADB = �;

\BDC = \BAC = ; \CAD = \CBD = �:

[Because of the right angles at C0 and D0,] AD0OC0 is cyclic, so that

\OD0C0 = \OAC0 = �. Likewise BD0OA0 is cyclic, so that \A0D0O =

\A0BO = �. Thus D0O is the bisector of \A0D0C0. Arguing in this manner

we conclude that the angle bisectors of A0B0C0D0 all pass through O and,

therefore, that this quadrilateral has an incircle.

For the converse assume that A0B0C0D0 has an incircle and obtain

A, B, C, D as the appropriate intersection points of the lines through A0

perpendicular to OA0, through B0 perpendicular to OB0, etc. We are to

show that ABCD is cyclic and O is the point where AC and BD intersect.

We denote:

\OA0D0 = \B0A0O = �; \OB0A0 = \C0B0O = �;

\OC0B0 = \D0C0O = ; \OD0C0 = \A0D0O = �:

with

�+ � + + � = �: (1)

AD0OC0 cyclic implies \OAC0 = \OD0C0 = �, and \D0AO =

\D0C0O = .

In a similar way,

\C0DO = � and \ODB0 = ;

\B0CO = � and \OCA0 = �;

\A0BO = � and \OBD0 = �:

Consider the quadrangle AOCD. We have

\OAD + \ADC + \DCO + \COA = 2�;

that is,

� + (� + ) + �+ \COA = 2�: (2)

Equations (1) and (2) imply \COA = �, so that O lies on AC. Arguing

analogously, O lies also on BD, so that O is the desired intersection point.

Finally, opposite angles have the sum

\BAD + \DCB = ( + �) + (�+ �) = �;

thus ABCD is cyclic.

308

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and

MAR�IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid,

Spain (with two proofs of the \only if" portion); CHRISTOPHER J. BRADLEY,

CliftonCollege, Bristol,UK (\only if" portion); TOSHIO SEIMIYA,Kawasaki,

Japan; SHAILESH SHIRALI, Rishi Valley School, India; and the proposer.

FLORIAN HERZIG, student, Perchtoldsdorf, Austria commented that

the \only if" portion follows immediately from problem 3 in the 1990 Cana-

dian Mathematical Olympiad, CRUX [1990; 161, 198].

Seimiya added a warning that the parenthetical comment appended to

the statement of the problem actually leads to a di�erent result which turns

out to be not quite correct; if the lines A0B0; B0C0; C0D0; D0A0 are tangents

to a circle then it is not always true thatABCD is an inscribed quadrilateral.

For a counterexample, let \BAC+\BDC = �, and \BAC 6= �=2. If T is

the point where AB intersects CD, then T , A, O, D are concyclic, so that

by Simson's theorem, B0, C0, D0 are collinear. The incircle of 4A0B0D0is tangent to the lines A0B0, B0C0, C0D0, D0A0; yet A, B, C, D are not

concyclic.

2150. [1996: 171] Proposed by �Sefket Arslanagi �c, Berlin, Germany.

Find all real solutions of the equation

p1� x = 2x2 � 1 + 2x

p1� x2:

I. Solution by Kee-Wai Lau, Hong Kong.

By squaring both sides of the equation and simplifying we obtain

�x = 4x(2x2 � 1)p1� x2: (1)

Clearly x = 0 is not a solution. So by (1) we obtain

�1 = 4(2x2 � 1)p1� x2: (2)

Let y =p1� x2. Then (2) becomes

8y3 � 4y � 1 = 0

or (2y+ 1)(4y2� 2y� 1) = 0:

Since y is positive, y =1+

p5

4and x = �

s5�p5

8. For the negative

value of x, both 2x2 � 1 and 2xp1� x2 are negative and so cannot be a

solution. For 0 � x � 1 the continuous function

p1� x� (2x2 � 1) + 2x

p1� x2

309

is positive when x = 0 and negative when x = 1. Thus the function vanishes

at least once in the interval [0; 1]. We conclude that the only solution of the

equation is x =

s5�p5

8.

II. Solutionby David Doster, Choate Rosemary Hall, Wallingford, Con-

necticut, USA.

Squaring both sides of the given equation and simplifying the resulting

equation yields

4(1� 2x2)p1� x2 = 1:

Since x lies between �1 and 1, we may set x = sin �,

where ��=2 � � � �=2. Then we get 4(1 � 2 sin2 �)p1� sin2 � = 1,

or equivalently, 4(2 cos2 � � 1) cos � = 1. Now the equivalent cubic equa-

tion 8 cos3 � � 4 cos � � 1 = 0 can be rewritten as

(2 cos � + 1)(4 cos2 � � 2 cos � � 1) = 0:

Since cos � � 0, wemust have cos � =1+

p5

4. Thus sin � =

�

p10� 2

p5

4.

Since only the positive solution satis�es the original equation, we have

x =

p10� 2

p5

4. (�, incidentally, is �=5.)

Also solved by NIELS BEJLEGAARD, Stavanger, Norway;

CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; MIGUEL ANGEL

CABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student, Aris-

totle University of Thessaloniki, Greece; TIM CROSS, King Edward's School,

Birmingham, England; CHARLES R. DIMINNIE, Angelo State University,

San Angelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium,

Bamberg, Germany; RUSSELL EULER, NWMissouri State University, Mary-

ville, Missouri, USA; F.J. FLANIGAN, San Jose State University, San Jose,

California, USA; DAVID HANKIN, Hunter College Campus Schools, New

York, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;

RICHARD I. HESS, Rancho Palos Verdes, California, USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; AMIT KHETAN, Troy,

MI, USA; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan,

USA; MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bul-

garia; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA;

BEATRIZ MARGOLIS, Paris, France; J.A. MCCALLUM, Medicine Hat, Al-

berta; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; L. RICE, Uni-

versity of Toronto Schools, Toronto, Ontario; KRISTIAN SABO, student, Os-

ijek, Croatia; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on,

Spain; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGE

PROBLEM GROUP, Saratoga Springs, New York, USA; D.J. SMEENK, Zalt-

bommel, the Netherlands; DIGBY SMITH, Mount Royal College, Calgary,

Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Geor-

310

gia, USA; PANOS E. TSAOUSSOGLOU, Athens, Greece; NISHKA VIJAY, stu-

dent, Mount Allison University, Sackville, New Brunswick; EDWARD T.H.

WANG, Wilfrid Laurier University, Waterloo, Ontario; and the proposer.

There were 2 incorrect solutions.

Most solvers who used a trigonometric approach began immediately

with the statement that �1 � x � 1, and then used either a sine or co-

sine substitution. This assumes that we are restricting our attention to an

equation with real values, but we are told only that the variable x must be

real-valued. Only HANKIN explicitly showed that there were no real-valued

solutions to be found outside the above range.

2151. [1996: 217] Proposed by Toshio Seimiya, Kawasaki, Japan.

4ABC is a triangle with \B = 2\C. Let H be the foot of the per-

pendicular from A to BC, and let D be the point on the side BC where the

excircle touches BC. Prove that AC = 2(HD).

Solution by Crist �obal S �anchez{Rubio, I.B. Penyagolosa, Castell �on,

Spain.

Let s = (a + b + c)=2 be the semiperimeter of 4ABC; we have

b+CD = c+ BD = s, so that

DC = s� b = (a� b+ c)=2:

Let C0 be the image of A under the rotation about B through the angle

180� � \B. Because 4ABC0 is isosceles, from the condition \B = 2\C it

follows that \CAC0 is also isosceles so that HC = (a+ c)=2. Finally,

HD = HC �DC =a+ c

2� a� b+ c

2=

b

2=AC

2:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; GEORGI

DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of

Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gym-

nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,

Austria; RICHARD I.HESS, Rancho PalosVerdes, California,USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON,

Beaverton, OR, USA;MAR�IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo

Cano, Valladolid, Spain; LUIZ A. PONCE, Santos, Brazil; D.J. SMEENK, Zalt-

bommel, the Netherlands; and the proposer.

Many solvers used the fact that \B = 2\C is equivalent to the condi-

tion b2 = c(c+ a), which has appeared before in CRUX: [1976: 74], [1984:

278], and [1996: 265{267]. For a recent problem concerning integer{sided

triangles with \B = 2\C see the solution to problem 578 in The College

Math J. 28:3 (May, 1997) 233{235. Further references are given there.

311

2152. [1996: 217] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.

Let n � 2 and 0 � x1 � : : : � xn � �

2be such that

nXk=1

sinxk = 1.

Consider the set Sn of all sums x1 + : : :+ xn.

1. Show Sn is an interval.

2. Let ln be the length of Sn. What is limn!1

ln?

I. Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

1. First I prove that

n arcsin1

n� x1 + x2 + � � �+ xn �

�

2:

Since the graph of sinx is concave down for x 2 [0; �=2], the chord joining

the points (0; sin 0) and (�=2; sin�=2) lies below the graph. Hence [since

this chord has slope 2=�]

2x

�� sinx for all x 2 [0; �=2]

and we can deduce the right-hand side of the claim:

2

�(x1 + x2 + � � �+ xn) � sinx1 + sinx2 + � � �+ sinxn = 1:

The left-hand side follows immediately from Jensen's inequality, since

sinx is concave down for x 2 [0; �=2] and 0 � (x1 + � � �+ xn)=n < �=2:

1

n=

sinx1 + sinx2 + � � �+ sinxn

n� sin

�x1 + x2 + � � �+ xn

n

�:

Next I will show that the set Sn of all sums x1 + x2 + � � �+ xn is the

interval �n arcsin

1

n;�

2

�:

First let 0 � x � y � arcsinA be such that sinx+ siny = A, where

0 � A � 1 is �xed. I prove that x+ y is a continuous function f depending

on x only: for each value of x there is exactly one value of y such that all

conditions are ful�lled, since

siny = A� sinx 2 [0; A] if and only if y = arcsin(A� sinx):

Thus f(x) = x + arcsin(A � sinx). It is clear that f is continuous for

x 2 [0; arcsinA].

312

Now we consider the following process. Initially let x1 = � � � = xn =

arcsin(1=n) [so thatP

sinxi = 1]. In the �rst step keep x3; : : : ; xn con-

stant and vary x1; x2 such that sinx1 + sinx2 remains constant and x1 be-

comes zero. Thereby sinx1 + sinx2 + � � �+ sinxn remains constant (equal

to 1) throughout and x1 + x2 + � � �+ xn varies continuously because of the

previous result.

In the second step keep x1 = 0, x4; : : : ; xn constant and vary x2 and

x3 as before, with x2 ! 0. Again x1 + x2 + � � �+ xn varies continuously.

It is clear now how to vary the xi step by step such that in the end

x1 = x2 = � � � = xn�1 = 0 and as a consequence xn = �=2. As we see,

x1 + � � �+ xn has varied continuously from n arcsin(1=n) to �=2. Hence

Sn =

�n arcsin

1

n;�

2

�:

2. By l'Hopital's rule,

limn!1

ln = limn!1

��

2� n arcsin

1

n

�

=�

2� lim

x!0

arcsinx

x

=�

2� lim

x!0

1p1� x2

=�

2� 1:

II. Solution by Thomas C. Leong, The City College of City University of

New York, New York, NY, USA.

Equivalently, we consider the set

Y = fy = (y1; : : : ; yn) j 1 � y1 � � � � � yn � 0; y1+ � � �+ yn = 1g � Rn

and the image f(Y ) of Y under f(y) = arcsiny1 + � � � + arcsinyn. Note

that Sn = f(Y ).

1. Since Y is a connected subspace ofRn and f is a continuous function,

the image f(Y ) � R is also connected, and we know that the only connected

subspaces of Rare intervals. Thus Sn is an interval.

2. Since arcsinx is convex in [0; 1], we can use the majorization in-

equality. Since�1

n;1

n; : : : ;

1

n

�� (y1; y2; : : : ; yn) � (1;0; : : : ; 0)

[Editor's note: here x � y means that x is majorized by y; see for example

page 45 and then Theorem 108, page 89 of Hardy, Littlewood and P �olya's

Inequalities],

n arcsin1

n� arcsiny1 + � � �+ arcsinyn � arcsin 1 =

�

2;

313

with equality when (y1; : : : ; yn) is equal respectively to (1=n; : : : ; 1=n) and

(1; 0; : : : ; 0). Thus ln = �

2� n arcsin(1=n) and

limn!1

ln =�

2� lim

n!1

�n arcsin

1

n

�=�

2� lim

n!1

�arcsin(1=n)

1=n

�=�

2� 1:

Also solved by KEITH EKBLAW, Walla Walla, Washington, USA; and

RICHARD I. HESS, Rancho Palos Verdes, California, USA.

2153. [1996: 217] Proposed by �Sefket Arslanagi �c, Berlin, Germany.

Suppose that a, b, c 2 R. If, for all x 2 [�1; 1], jax2 + bx + cj � 1,

prove that

jcx2 + bx+ aj � 2:

Solution (virtually identical) by Florian Herzig, student, Perchtolds-

dorf, Austria, andWalther Janous, Ursulinengymnasium, Innsbruck, Austria.

Let f(x) = ax2+bx+c and g(x) = cx2+bx+a. Then by assumption

ja+ b+ cj = jf(1)j � 1, ja� b+ cj = jf(�1)j < 1 and jf(0)j � 1. Hence

jg(x)j =

����c(x2 � 1) + (a+ b+ c)1 + x

2+ (a� b+ c)

1� x

2

����� jcjjx2 � 1j+ ja+ b+ cj j1 + xj

2+ ja� b+ cj j1� xj

2

� jx2 � 1j+ j1 + xj2

+j1� xj

2

= 1� x2 +1+ x

2+

1� x

2

= 2� x2 � 2

for all x 2 [�1; 1].

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-

ladolid, Spain; THEODORE CHRONIS, student, Aristotle University of Thes-

saloniki, Greece; GEORGI DEMIZEV, Varna, Bulgaria, andMITKOKUNCHEV,

Baba Tonka School of Mathematics, Rousse, Bulgaria; F.J. FLANIGAN, San

Jose State University, San Jose, California, USA; DIGBY SMITH, Mount

Royal College, Calgary, Alberta; and the proposer. There was one incorrect

solution.

Several solvers pointed out that this is not a new problem. Herzig cited

the book International Mathematical Olympiad 1978{1985 (published by the

Mathematical Association of America, 1986). Chronis gave the reference Se-

lected Problems and Theorems in Elementary Mathematics | Arithmetic and

Algebra by D.O. Shklyarsky, N.N. Chentov, and I.M. Yaglom (exercise 304

on page 60). Demizev and Kunchev remarked that it can be found as prob-

lem #86 on page 35 of the book Mathematics Competitions by L. Davidov,

314

V.Petnob, I. Tonov, and V. Chukanov (So�a, 1977). Bellot Rosado men-

tioned that the same problem was proposed during the 2nd round of the

1996 Spanish Mathematical Olympiad.

Janous remarked that the problem can be restated as \Prove that if

p(x) is a real polynomial of degree 2 satisfying the condition that jp(x)j � 1

for all x 2 [�1; 1], then jx2p(1=x)j � 2 for all x 2 [�1; 1]." (Ed: x2p(1=x)

is to be written as a polynomial before substituting values of x.) He then

ventured the following conjecture:

Conjecture: If p(x) is a real polynomial with degree n such that

jp(x)j � 1 for all x 2 [�1; 1] then jxnp(1=x)j � 2n for all x 2 [�1; 1].

Can any reader prove or disprove this?

2154. [1996: 217] Proposed by K.R.S. Sastry, Dodballapur, India.

In a convex pentagon, the medians are concurrent. If the concurrence

point sections each median in the same ratio, �nd its numerical value. (A

median of a pentagon is the line segment between a vertex and the midpoint

of the third side from the vertex.)

C A0 D

B0

E

C0A

D0

B

E0

P

I Solution by Toshio Seimiya, Kawasaki, Japan.

Let the convex pentagon beABCDE, and letA0, B0, C0,D0, E0 be the

midpoints of CD, DE, EA, AB, BC, respectively. We assume that AA0,

BB0, CC0, DD0, EE0 are concurrent at P , and that

AP

PA0=

BP

PB0=

CP

PC0=

DP

PD0=

EP

PE0= �:

Then set D0E0 = 1. Since D0, E0 are midpoints of AB, BC, respectively,

we have AC = 2D0E0 = 2, and D0E0 k AC. Since

DP

PD0=

EP

PE0= �;

we have D0E0 k DE and DE=D0E0 = �, so that

DE = �D0E0 = �: (1)

315

Since AP=PA0 = CP=PC0 = �, we have AC=A0C0 = AP=PA0 = �, so

that

A0C0 =AC

�=

2

�: (2)

Since AC k D0E0 and D0E0 k DE, we get AC k DE, and since A0, C0 are

midpoints of CD, AE, respectively, we have

AC +DE = 2A0C0:

Therefore we have from (1) and (2)

2 + � =4

�;

from which we have �2 + 2�� 4 = 0. Thus we obtain � =p5� 1.

II Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-

tria.

(We will use the same notation as used by Seimiya above. Each bold-

face letter represents the vector to that point from the origin.) Then we have

D0 = 1

2(A+ B), E0 = 1

2(B+ C), etc. There must exist real numbers �; � > 0,

with �+ � = 1, such that

P = �A+ �A0 = �B+ �B0 = �C+ �C0 = �D+ �D0 = �E+ �E0:

Now

�D+ �D0 = �A+ �A0 () 2�D+ �(A + B) = 2�A+ �(C + D); (3)

�A+ �A0 = �B+ �B0 () 2�A+ �(C + D) = 2�B+ �(D + E): (4)

Equation (4) can be simpli�ed to

2�A+ �C = 2�B+ �E () 2�(B� A) = �(C � E);

which implies that AB k EC. Similarly,

BC k AD; CD k BE; DE k CA; EA k DB: (5)

Now equation (3) can be written as:

2�D+ �A � 2�A� �D = �C � �B;

(2�� �)(D � A) = �(C � B):

On the other hand from (5) we have

2�(C� B) = �(D � A):

Hence,

D� A =�

2�� �(C � B) =

2�

�(C� B);

316

whence we obtain

�

2�� �=

2�

�

4�2 � 2�� � �2 = 0

4�2 � 2�(1� �)� (1� �)2 = 0

5�2 = 1

� =

p5

5:

Thus the ratio we are considering has the numerical value

� : � = � : (1� �) = 1 : (p5� 1)

Also solved by SAM BAETHGE, Nordheim, Texas, USA; MIHAI CIPU,

Romanian Academy, Bucharest; MIGUELANGEL CABEZ �ONOCHOA, Logro ~no,

Spain; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba

Tonka SchoolofMathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{

Ludwig{Gymnasium, Bamberg, Germany; RICHARD I. HESS, Rancho Palos

Verdes, California, USA; D. KIPP JOHNSON, Valley Catholic High School,

Beaverton, Oregon; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the

proposer.

Seimiya also observes the following:

1. If four medians are concurrent at a point P , then the �fth median also

passes through P .

2. We assume that �ve medians are concurrent at a point P . If any three

medians are divided at P into the same ratio � : 1, then the other two

medians are divided at P into the same ratio � : 1.

3. This �gure is an image of a regular pentagon by an a�ne transforma-

tion.

2155. [1996: 218] Proposed by Christopher J. Bradley, Clifton Col-

lege, Bristol, UK.

Prove there is no solution of the equation

1

x2+

1

y8=

1

z2

in which y is odd and x; y; z are positive integers with highest common fac-

tor 1.

Find a solution in which y = 15, and x and z are also positive integers.

Solution by D. Kipp Johnson (modi�ed slightly by the editor), Valley

Catholic High School, Beaverton, Oregon, USA.

317

First, we �nd the general solution in N to the auxiliary equation

1

x2+

1

t2=

1

z2where t is odd. (1)

(1) can be written as z2(x2+ t2) = x2t2, which shows that x2+ t2 must be a

square. From well{known results, we then have x = k(2uv), t = k(u2�v2),where k 2 N is odd, and u, v are relatively prime integers of opposite parity

with u > v. Substituting into (1) and simplifying gives

z(u2 + v2) = 2kuv(u2� v2): (2)

Since u2 + v2 is odd, z = 2w for some w 2 N and (2) becomes

w(u2 + v2) = kuv(u2� v2): (3)

Since (u; v) = 1, (u2 + v2; uv) = 1. Furthermore, (u2 + v2; u2 � v2) = 1

since if p is a prime such that pju2 + v2, u2 � v2, then pj2u2, 2v2 implies

pj(2u2; 2v2). But (2u2; 2v2) = 2(u; v)2 = 2 and clearly p 6= 2 as u2 + v2 is

odd. This is a contradiction. Hence u2+ v2jk. Letting k = j(u2+ v2), then

(3) becomes

w = juv(u2� v2) where j 2 N is odd.

Therefore, wemay choose u, v subject to the stated conditions and arbitrarily

odd j 2 N to �nd the general solution of the original equation, which is given

by:

x = 2juv(u2+ v2); y4 = t = j(u2� v2)(u2+ v2); z = 2juv(u2� v2):

Since (x; y; z) = 1, we must have j = 1 and so y4 = u4 � v4, which is

impossible by Fermat's Last Theorem.

To �nd a solution in which y = 15, let

y4 = 154 = 34 � 54 = j(u� v)(u+ v)(u2 + v2): (4)

Since u�v, u+v, and u2+v2 are pairwise relatively prime, and the left side

of (4) contains only two distinct prime factors, u� v = 1 and (4) becomes

154 = j(2v+ 1)(2v2+ 2v+ 1): (5)

From (5) we see that 154 > 4v3 and so v < 24 resulting in 2v + 1 < 49.

Now, 2v+1 is odd and the only odd factors of 154 which are less than 49 are

3, 5, 9, 15, 25 and 45 with corresponding values of v = 1, 2, 4, 7, 12, and

22. A quick check shows that only in the case v = 1 is 2v2 + 2v + 1 also a

factor of 154. Thus u = 2 and (5) gives j = 33 � 53 = 3375 resulting in the

unique solution:

x = 67500; y = 15; z = 40500:

318

Also solved (both parts) by WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria, and the proposer. The second part of the problem was

also solved by GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV,

Baba Tonka School of Mathematics, Rousse, Bulgaria (jointly); HANS

ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Germany; SHAWN

GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS,

Rancho Palos Verdes, California, USA; MURRAY S. KLAMKIN, University of

Alberta, Edmonton, Alberta; and DIGBY SMITH, Mount Royal College, Cal-

gary, Alberta. All these solvers gave the same solution (67500;15; 40500).

However, most of their \proofs" erred in assuming that gcd(x; y; z) = 1

implies that x, y, z are pairwise relatively prime. Furthermore, Johnson

was the only one who gave a completely valid proof for the uniqueness of

the solution when y = 15, though Janous found this to be the case by using

DERIVE.

Both Hess and Klamkin considered the given equation without the re-

striction that y be odd and (x; y; z) = 1. Hess observed that an in�nite fam-

ily of solutions is given by x = 2mn(m2+n2)4(m2�n2)3, y = m4�n4, andz = 2mn(m2+n2)3(m2�n2)4 wherem, n 2 Nwithm > n while Klamkin

gave the family: x = 22 � 33 � 54 � n4, y = 15n, z = 22 � 34 � 53 � n4.

Both formulas yield the solution given above whenm = 2 and n = 1.

2156. [1996: 218] Proposed by Hoe Teck Wee, Singapore.

ABCD is a convex quadrilateral with perpendicular diagonalsAC and

BD. X and Y are points in the interior of sides BC and AD respectively

such thatBX

CX=BD

AC=DY

AY:

EvaluateBC �XYBX � AC :

Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.

Let Z be the point in the interior of side AB such that BZ : AZ =

BX : CX. Then it follows from

BZ

AZ=BX

CX=DY

AY

�=BD

AC

�

that XZ and AC as well as ZY and BD are parallel, which, considering

AC ? BD, implies that 4XY Z is a right triangle with legs XZ and Y Z.

ThusXZ

ZY=XZ

AC� ACBD

� BDZY

=BZ

AB� AZBZ

� ABAZ

= 1

[where XZ=AC = BZ=AB and BD=ZY = AB=AZ by similar triangles

while AC=BD = AZ=BZ by the de�nition of Z]. This means that4XY Z

319

is an equilateral right triangle, whence

XY : ZX =p2 : 1

and, consequently,

BC �XYBX � AC =

AC �XYZX �AC =

XY

ZX=p2:

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-

ladolid, Spain; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; GEORGI

DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of

Mathematics, Rousse, Bulgaria; FLORIAN HERZIG, student, Perchtoldsdorf,

Austria; RICHARD I.HESS, Rancho PalosVerdes, California,USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIO SEIMIYA, Ka-

wasaki, Japan; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK,

Zaltbommel, the Netherlands; and the proposer.

Bellot comments that facts about orthodiagonal quadrilaterals can be

found in Jordan Tabov, Simple properties of the orthodiagonal quadrilater-

als, Matematyka & Informatyka, 1:1 (1991) 1{5. Janous, using rectangular

coordinates, shows that if the angle from AC to BD were �, then

BC �XYBX � AC = 2 sin

�

2:

2157. [1996: 218] Proposed by �Sefket Arslanagi �c, Berlin, Germany.

Prove that 21997�1996� 1 is exactly divisible by 19972.

I. Solution by Theodore Chronis, student, Aristotle University of Thes-

saloniki, Greece.

From the Fermat-Euler theorem we have a'(m) � 1(mod m), for

every a;m such that gcd(a;m) = 1. We also have '(pk) = pk�1(p � 1),

when p is a prime number.

So '(19972) = 1997 � 1996 and 2'(19972) � 1(mod 1)9972.

II. Solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-

tria.

We have '(1997) = 1996 = 22 � 499.Furthermore, 2449 � 1585(mod 1)997 and 22�449 � �1(mod 1)997.

Hence 2 is of order 1996 modulo 1997; that is

� := 21996 � 1 � 0 (mod 1997):

On the other hand,

� � 672989 (mod 19972); that is, � 6� 0 (mod 19972):

320

But

� := 21997�1996� 1 = ((21996� 1) + 1)1997� 1 = (�+ 1)1997� 1

= 1 + 1997�+

�1997

2

��2 +

�1997

3

��3 + � � �+ �1997� 1

= 1997�+ 19973 � f;

where f is an integer. Thus � � 0(mod 1)9972, but � 6� 0(mod 1)9973.

Also solved by MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;

MIHAI CIPU, Romanian Academy, Bucharest, Romania; GEORGI DEMIZEV,

Varna, Bulgaria, andMITKOKUNCHEV, Baba Tonka School ofMathematics,

Rousse, Bulgaria; CHARLES R. DIMINNIE, Angelo State University, San An-

gelo, TX, USA; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg,

Germany; SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario;

FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,

Rancho Palos Verdes, California, USA; D. KIPP JOHNSON, Valley Catholic

High School, Beaverton, Oregon; MURRAY S. KLAMKIN, University of Al-

berta, Edmonton, Alberta; THOMAS LEONG, Staten Island, NY, USA; DAVID

E. MANES, SUNY at Oneonta, Oneonta, NY, USA; CAN ANH MINH, Uni-

versity of California, Berkeley, California; SOLEDAD ORTEGA and JAVIER

GUTI �ERREZ, students, Universidad de La Rioja, Logro ~no, Spain; YOLANDA

PELLEJERO, student, Universidad de La Rioja, Logro ~no, Spain; ROBERT

P. SEALY, Mount Allison University, Sackville, New Brunswick; HEINZ-

J�URGEN SEIFFERT, Berlin, Germany; and the proposer.

It is interesting to note that only Janous and Manes addressed the

question of \exact" divisibility; that is, they showed that no higher power of

1997 than 2 would work.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

321

THE ACADEMY CORNERNo. 13

Bruce Shawyer

All communications about this column should be sent to Bruce

Shawyer, Department of Mathematics and Statistics, Memorial University

of Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Mathematics Competitions

There are many mathematical competitions at many di�erent levels.

There is in fact a journal devoted to this. It is called Mathematics Compe-titions, and is the Journal of the World Federation of National Mathematics

Competitions. It is published by the Australian Mathematics Trust, and in-

formation on it can be obtained by contacting them at PO Box 1, Belconnen,

ACT 2616, Australia.

The journal covers all sorts of mathematical competitions, and articles

not only give problems and solutions, but also cover the philosophy of com-

petition and alternative ways of stimulating talented students.

The World Federation also makes two awards, the David Hilbert In-

ternational Award (which recognises contributions of mathematicians who

have played a signi�cant role over a number of years in the development

of mathematical challenges at the international level and which have been

a stimulus for mathematical learning), and the Paul Erd �os National Award

(which recognises contributions of mathematicians who have played a sig-

ni�cant role over a number of years in the development of mathematical

challenges at the national level and which have been a stimulus for the en-

richment of mathematics learning).

322

THE OLYMPIAD CORNERNo. 184

R.E. Woodrow

All communications about this column should be sent to Professor R.E.

Woodrow, Department of Mathematics and Statistics, University of Calgary,

Calgary, Alberta, Canada. T2N 1N4.

We begin this number with the problems of the Mock Test for the In-

ternational Mathematical Olympiad team of Hong Kong. My thanks go to

Richard Nowakowski, Canadian Team Leader, for collecting them at the 35th

IMO in Hong Kong.

INTERNATIONALMATHEMATICAL OLYMPIAD1994

Hong Kong Committee | Mock Test, Part ITime: 4.5 hours

1. In 4ABC, we have \C = 2\B. P is a point in the interior of

4ABC satisfying AP = AC and PB = PC. Show that AP trisects the

angle \A.

2. In a table-tennis tournament of 10 contestants, any two contestants

meet only once. We say that there is a winning triangle if the following

situation occurs: ith contestant defeated j th contestant, j th contestant defeated

kth contestant, and kth contestant defeated ith contestant. Let Wi and Li be

respectively the number of games won and lost by the ith contestant. Suppose

Li+Wj � 8 whenever the ith contestant beats the j th contestant. Prove that

there are exactly 40 winning triangles in this tournament.

3. Find all the non-negative integers x, y, and z satisfying that

7x + 1 = 3y + 5z :

Mock Test, Part IITime: 4.5 hours

1. Suppose that yz+ zx+ xy = 1 and x, y, and z � 0. Prove that

x(1� y2)(1� z2) + y(1� z2)(1� x2) + z(1� x2)(1� y2) �4p3

9:

323

2. A function f(n), de�ned on the natural numbers, satis�es:

f(n) = n� 12 if n > 2000; and f(n) = f(f(n+ 16)) if n � 2000:

(a) Find f(n).

(b) Find all solutions to f(n) = n.

3. Let m and n be positive integers where m has d digits in base

ten and d � n. Find the sum of all the digits (in base ten) of the product

(10n � 1)m.

As a second Olympiad set we give the problems of the Final Round of

the 45th Mathematical Olympiad written in April, 1994. My thanks go to

Marcin E. Kuczma, Warszawa, Poland; and Richard Nowakowski, Canadian

Team leader to the 35th IMO in Hong Kong, for collecting them.

45th MATHEMATICAL OLYMPIAD IN POLANDProblems of the Final Round | April 10{11, 1994

First Day | Time: 5 hours

1. Determine all triples of positive rational numbers (x; y; z) such that

x+ y + z, x�1 + y�1 + z�1 and xyz are integers.

2. In the plane there are given two parallel lines k and l, and a cir-

cle disjoint from k. From a point A on k draw the two tangents to the given

circle; they cut l at pointsB andC. Letm be the line throughA and the mid-

point of BC. Show that all the resultant lines m (corresponding to various

points A on k) have a point in common.

3. Let c � 1 be a �xed integer. To each subset A of the set f1, 2, : :: : , ng we assign a number w(A) from the set f1, 2, : : : , cg in such a way

that

w(A \ B) = min(w(A);w(B)) for A;B � f1; 2; : : : ; ng:

Suppose there are a(n) such assignments. Compute limn!1n

pa(n).

Second Day | Time: 5 hours

4. We have three bowls at our disposal, of capacities m litres, n litres

andm+ n litres, respectively; m and n are mutually coprime natural num-

bers. The two smaller bowls are empty, the largest bowl is �lled with water.

Let k be any integer with 1 � k � m + n � 1. Show that by pouring

water (from any one of those bowls into any other one, repeatedly, in an

unrestricted manner) we are able to measure out exactly k litres in the third

bowl.

324

5. Let A1; A2; : : : ; A8 be the vertices of a parallelepiped and let O be

its centre. Show that

4(OA21 + OA2

2 + � � �+ OA28) � (OA1 + OA2 + � � �+ OA8)

2:

6. Suppose that n distinct real numbers x1; x2; : : : ; xn (n � 4) satisfy

the conditions x1 + x2 + � � �+ xn = 0 and x21 + x22 + � � �+ x2n = 1. Prove

that one can choose four distinct numbers a, b, c, d from among the xi's in

such a way that

a+ b+ c+ nabc � x31 + x32 + � � �+ x3n � a+ b+ d+ nabd:

We now give three solutions to problems given in the March 1996 Cor-

ner as the Telecom 1993 Australian Mathematical Olympiad [1996: 58].

TELECOM 1993 AUSTRALIANMATHEMATICAL OLYMPIAD

Paper 1Tuesday, 9th February, 1993

(Time: 4 hours)

6. In the acute-angled triangle ABC, let D, E, F be the feet of alti-

tudes through A, B, C, respectively, andH the orthocentre. Prove that

AH

AD+BH

BE+CH

CF= 2:

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.

AH

AD+BH

BE+CH

CF= 3�

�HD

AD+HE

BE+HF

CF

�

= 3��[BHC]

[ABC]+

[CHA]

[ABC]+

[AHB]

[ABC]

�

= 3�[ABC]

[ABC]= 2:

7. Let n be a positive integer, a1; a2; : : : ; an positive real numbers

and s = a1 + a2 + � � �+ an. Prove that

nXi=1

ai

s� ai�

n

n� 1and

nXi=1

s� ai

ai� n(n� 1):

325

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

We give the solution by Boase.

nXi=1

s� ai

ai=

nXi=1

�s

ai� 1

�=

nXi=1

s

ai� n

nXi=1

ai

s= 1 and

nXi=1

s

ai

nXi=1

ai

s� (

X1)2 = n2;

by the Cauchy{Schwarz inequality.

ThusnXi=1

s

ai� n2:

Hence

nXi=1

s� ai

ai� n2 � n = n(n� 1).

To prove the �rst inequality, �rst note that

nXi=1

1

nXi=1

a2i �

nXi=1

ai

!2

= s2:

Hence

nXi=1

a2i �s2

n.

By the Cauchy{Schwarz inequality,

nXi=1

ai(s� ai)

nXi=1

ai

s� ai�

nXi=1

ai

!2

= s2:

Therefore

nXi=1

ai

s� ai�

s2Pni=1 ai(s� ai)

=s2

sPni=1 ai �

Pni=1 a

22

�s2

s2 � s2

n

=1

1� 1n

=n

n� 1:

So, both inequalities are proved.

8. [1996: 58] Telecom 1993 Australian Mathematical Olympiad.

The vertices of triangleABC in the xy{plane have integer coordinates,

and its sides do not contain any other points having integer coordinates. The

interior of ABC contains only one point, G, that has integer coordinates.

Prove that G is the centroid of ABC.

326

Solution by Mansur Boase, student, St. Paul's School, London, Eng-

land.

B C

A

B0C0

A0

r

G

By Pick's Theorem

[ABC] = 1 + 3

�1

2

�� 1 =

3

2;

[ABG] = 0 + 3

�1

2

�� 1 =

1

2;

[BCG] =1

2and

[CAG] =1

2:

Therefore[ABG]

[ABC]=

[BCG]

[ABC]=

[CAG]

[ABC]=

1

3:

HenceGA0

AA0=

GB0

BB0=

GC0

CC0=

1

3:

The unique point satisfying this above is well-known to be the centroid.

Next we give one solution from the JapanMathematical Olympiad 1993

given in the March 1996 Corner.

2. [1996: 58] Japan Mathematical Olympiad 1993.

Let d(n) be the largest odd number which divides a given number n.

Suppose that D(n) and T (n) are de�ned by

D(n) = d(1) + d(2) + � � �+ d(n);

T (n) = 1 + 2 + � � �+ n:

Prove that there exist in�nitely many positive numbers n such that

3D(n) = 2T (n).

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land, and Zun Shan and Edward T.H. Wang, Wilfrid Laurier University, Wa-

terloo, Ontario. We give the solution by Boase.

327

T (n) =n(n+ 1)

2:

Thus we need to prove that there are in�nitely many n for which

D(n) =n(n+ 1)

3so that 3D(n) = 2T (n) holds:

Consider

D(2n) = d(1) + d(3) + � � �+ d(2n � 1) + d(2) + d(4) + � � �+ d(2n)

= 1 + 3 + � � �+ (2n � 1) + d(1) + d(2) + � � �+ d(2n�1)

= 1 + 3 + � � �+ (2n � 1) +D(2n�1):

Now

1 + 3 + � � �+ (2n � 1) =2n(2n + 1)

2� 2

2n�1(2n�1 + 1)

2

= 2n�1(2n � 2n�1)

= 22n�2:

Thus D(2n) = D(2n�1) + 22n�2.

Now, D(21) = 2 and we shall prove by induction that D(2n) =22n + 2

3for

n � 0.

This holds for n = 0 and for n = 1. Suppose it holds for n = k.

Thus D(2k) =22k + 2

3.

Then

D(2k+1) = D(2k) + 22k =22k + 2

3+ 22k

=4(22k) + 2

3

=22k+2 + 2

3

and the result follows by induction. Now consider D(2n � 2).

D(2n � 2) = D(2n)� d(2n � 1)� d(2n)

=22n + 2

3� (2n � 1)� 1

=22n + 2

3� 2n

=22n � 3(2n) + 2

3=

(2n � 1)(2n � 2)

3:

328

ThusD(x) =x(x+ 1)

3for x = 2n�2, and there are in�nitely many such x.

Next we turn to comments and solutions from the readers to problems

from the April 1996 number of the Corner where we gave the selection test

for the Romanian Team to the 34th IMO as well as three contests for the

Romanian IMO Team [1996: 107{109].

SELECTION TESTS FOR THE ROMANIAN TEAM,34th IMO.

Part II | First Contest for IMO Team1st June, 1993

1. Find the greatest real number a such that

xpy2 + z2

+y

pz2 + x2

+zp

x2 + y2> a

is true for all positive real numbers x, y, z.

Solutions by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-

sity, Waterloo, Ontario.

We claim that a = 2. Let

f(x; y; z) =xp

y2 + z2+

ypz2 + x2

+zp

x2 + y2:

We show that f(x; y; z) > 2. Since f(x; y; z)! 2 as x! y and z ! 0, the

lower bound 2 is sharp. Without loss of generality, assume that x � y � z.

Since by the arithmetic{harmonic{mean inequality, we havepz2 + x2py2 + z2

+

py2 + z2

pz2 + x2

� 2;

it su�ces to show that

f(x; y; z) >

pz2 + x2py2 + z2

+

py2 + z2

pz2 + x2

or equivalently,

zpx2 + y2

>

pz2 + x2 � xpy2 + z2

+

py2 + z2 � ypz2 + x2

:

By simple algebra, this is easily seen to be equivalent to

zpy2 + z2(

pz2 + x2 + x)

+z

pz2 + x2(

py2 + z2 + y)

<1p

x2 + y2: (1)

329

Sincepy2 + z2 �

p2z2 =

p2 z,

pz2 + x2 > x and

p2 x �

px2 + y2,

we have

zpy2 + z2(

pz2 + x2 + x)

<z

p2 z(x+ x)

=1

2p2 x

�1

2px2 + y2

:

Thus to establish (1), it remains to show that

zpz2 + x2(

py2 + z2 + y)

<1

2px2 + y2

or equivalently

2zpy2 + z2 + y

<

sz2 + x2

x2 + y2:

Sincez2 + x2

x2 + y2= 1�

y2 � z2

x2 + y2;

which is an non-decreasing function of x, we have

z2 + x2

x2 + y2�

z2 + y2

2y2;

and thus it su�ces to show thatpz2 + y2p2 y

>2zp

y2 + z2 + y;

or equivalently

y2 + z2 + ypy2 + z2 > 2

p2 yz: (2)

Since y2 + z2 � 2yz, we have

y2 + z2 + ypy2 + z2 � 2yz+ y

p2z2

= (2 +p2)yz > 2

p2 yz

and thus (2) holds. This completes the proof.

2. Show that if x, y, z are positive integers such that x2 + y2 + z2 =

1993, then x+ y + z is not a perfect square.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-

tario. We give Wang's solution.

We show that the result holds for nonnegative integers x, y, and z.

Without loss of generality, we may assume that 0 � x � y � z. Then

3z2 � x2 + y2 + z2 = 1993

330

implies that

z2 � 665; z � 26:

On the other hand z2 � 1993 implies that z � 44 and thus 26 � z � 44.

Suppose that x + y + z = k2 for some nonnegative integer k. By the

Cauchy{Schwarz Inequality we have

k4 = (x+ y+ z)2 � (12 + 12 + 12)(x2 + y2 + z2) = 5979

and so k � b 4p5979c = 8. Since k2 � z � 26, k � 6. Furthermore, since

x2+y2+z2 is odd, it is easily seen that x+y+z must be odd, which implies

that k is odd. Thus k = 7 and we have x+ y+ z = 49.

Let z = 26 + d, where 0 � d � 18. Then

x+ y = 23� d) y � 23� d) x2 + y2 � 2y2

= 2(23� d)2 = 1058� 92d+ 2d2: (1)

On the other hand, from x2 + y2 + z2 = 1993 we get

x2 + y2 = 1993� z2 = 1993� (26 + d)2 = 1317� 52d� d2: (2)

From (1) and (2), we get

1317� 52d� d2 � 1058� 92d+ 2d2

or

3d2 � 40d � 259

which is clearly impossible since 3d2� 40d = d(3d� 40) � 18� 14 = 252.

This completes the proof.

Remark: It is a well-known (though by no means easy) result in clas-

sical number theory that a natural number n is the sum of three squares (of

nonnegative integers) if and only if n 6= 4l(8k + 7) where l and k are non-

negative integers. Since 1993 � 1(mod 8) it can be so expressed and thus

the condition given in the problem is not \vacuously" true. In fact, 1993 can

be so expressed in more than one way; for example,

1993 = 02 + 122 + 432

= 22 + 152+ � 422

= 22 + 302 + 332

= 112 + 242 + 362:

These representations also show that the conclusion of the problem is false

if we allow x, y, and z to be negative integers; e.g. if x = �2, y = �30,

z = 33 then x2 + y2 + z2 = 1993 and x + y + z = 12; and if x = �11,

y = 24, z = 36 then x2 + y2 + z2 = 1993 and x+ y+ z = 49 = 72.

331

4. Show that for any function f : P(f1;2; : : : ; ng) ! f1; 2; : : : ; ngthere exist two subsets, A and B, of the set f1; 2; : : : ; ng, such that A 6= B

and f(A) = f(B) = maxfi j i 2 A \Bg.

Comment by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Ontario.

The problem, as stated, is clearly incorrect since formaxfi : i 2 A\Bgto make sense, we must have A \ B 6= ;. For n = 1 clearly there are no

subsetsA andB withA 6= B andA\B 6= ;. A counterexample when n = 2

is provided by setting f(;) = f(f1g) = f(f2g) = 1 and f(f1;2g) = 2.

This counterexample stands ifmax is changed to min. The conclusion is still

incorrect ifA\B is changed to A[B. A counterexample would be f(;) = 2

and f(f1g) = f(f2g) = ff(1;2)g = 1.

Part III | Second Contest for IMO Team2nd June, 1993

3. Prove that for all integer numbers n, with n � 6, there exists an

n{point setM in the plane such that every point P inM has at least three

other points inM at unit distance to P .

Solution by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-

sity, Waterloo, Ontario.

The three diagrams displayed below illustrate the existence of such a

set. M1 is for alln = 3k,M2 is for alln = 3k+1 andM3 is for alln = 3k+2,

where k = 2; 3; 4; : : : . In each diagram, the solid lines connecting two points

all have unit length and the dotted lines, also of unit length, indicate how to

construct an (n + 3){point set with the described property from one with

n points.

s

s

s

ss

sc c

c

s

s

s

s s

s s

c c

c

s

s

s

ss

s

s

sc

c c

M1 M2 M3

(n = 6; 9; 12; : : : ) (n = 7; 10; 13; : : : ) (n = 8; 11; 14; : : : )

332

Part IV | Third Contest for IMO Team3rd June, 1993

1. The sequence of positive integers fxngn�1 is de�ned as follows:

x1 = 1, the next two terms are the even numbers 2 and 4, the next three

terms are the three odd numbers 5, 7, 9, the next four terms are the even

numbers 10, 12, 14, 16 and so on. Find a formula for xn.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; and by Zun Shan and Edward T.H. Wang, Wilfrid Laurier University,

Waterloo, Ontario. We give the solution of Shan and Wang.

We arrange the terms of the sequence fxng in a triangular array ac-

cording to the given rule so that the 1st level consists of a single 1 and for

all k � 2, the kth level consists of the k consecutive even (odd) integers that

follow the last odd (even) integer in the (k� 1)st level:

1

2 4

5 7 9

10 12 14 16

17 19 21 23 25

For any given n 2 N , let ln denote the unique integer such that�ln

2

�< n �

�ln + 1

2

�;

that is,

(ln � 1)ln

2< n �

ln(ln + 1)

2: (1)

Note that ln is simply the number of the level which xn is on. We claim that

xn = 2n� ln: (2)

When n = 1, clearly l1 = 1 and thus 2n� ln = 1 = x1. Suppose (2) holds

for some n � 1. There are two possible cases:

Case (i): If n + 1 ��ln+1

2

�; that is, if xn+1 and xn are on the same

level, then ln+1 = ln and hence

xn+1 = xn + 2 = 2n� ln + 2 = 2(n+ 1)� ln+1:

333

Case (ii): If n+ 1 >�ln+1

2

�then xn is the last number on the lthn level

and xn+1 is the �rst number on the lthn+1 level. Thus

ln+1 = ln + 1

and

xn+1 = xn + 1 = 2n� ln + 1 = 2(n+ 1)� ln+1:

Hence, by induction, (2) is established. From (1) we get

l2n � ln < 2n � (ln + 1)2 � (ln + 1):

Solving the inequalities, we easily obtain

ln <1 +

p1 + 8n

2� ln + 1:

Hence

ln =

&1 +

p1 + 8n

2

'� 1 (3)

where dxe denotes the least integer greater than or equal to x (that is, the

ceiling function). From (2) and (3) we conclude that

xn = 2n+ 1�

&1 +

p1 + 8n

2

':

That completes this number of the Corner. Sendme your nice solutions

as well as Olympiad contests.

334

BOOK REVIEWS

Edited by ANDY LIU

A Mathematical Mosaic, by Ravi Vakil,published by Brendan Kelly Publishing Inc., 1996,

2122 Highview Drive, Burlington, ON L7R3X4,

ISBN# 1-895997-04-6, softcover, 253+ pages, US$16.95 plus handling.

Reviewed by Jim Totten, University College of the Cariboo.

So, you have a group of students who have decided they want the extra

challenge of doing some mathematics competitions. You want a source of

problems which will pique the students' interest, and which also lead to fur-

ther exploration. The problem source should lend itself well to independent

work. The question is: where do you �nd the appropriate level enrichment

material? Many of us have already tried to answer this question and have a

collection of such problem books. Well, here is a book to be added to your

collection!

It is certainly a problem book, but it is much more than that. The au-

thor at one moment guides the reader through some very nice mathematical

developments, and throws out problems as they crop up in the development,

and in the next moment uses a problem as a starting point for some inter-

esting mathematical development.

With a few exceptions the problems in this book are not new, nor are the

solutions. They are, however, well organized, both by topic and by level of

mathematical maturity needed. Answers are NOT always provided; instead

there is often simply a solution strategy or hint given, and occasionally there

is simply a reference to some other source for a full-blown treatment. Even

when answers are provided, they are not tucked away at the end of the book,

but rather they are worked into another topic (usually later in the book, but

not always), where they become part of the development of another topic or

problem.

The author is a PhD candidate in pure mathematics at Harvard Univer-

sity (at the time the book was written). Being still very young, he knows how

to speak to today's teenagers. His sense of humour and general puckishness

is present throughout: just when you are lulled into some serious compu-

tation in probability, he deviously throws a trick question at you, that has

a totally non-obvious answer (non-obvious, that is, until you CAREFULLY

re-read the question).

Manymathematics books published today include short biographies on

famous mathematicians through history, especially those whose names come

up in the theory developed in the book. This book is no exception. But what

is unique about this book is the inclusion of Personal Pro�les of youngmath-

ematicians from several countries that he has met at International Mathe-

matical Olympiads (IMOs) in the past. The pro�les are quite diverse, which

335

means that most bright students could �nd one to identify with and to use as

a role model. The author and those he pro�les have taken a risk in doing this:

they have tried to predict some of the important mathematicians in the early

part of the next century. It should be interesting to follow their careers and

see if those predictions can come true, or if by placing them in the spotlight,

they �nd too much pressure to deal with.

The problems range from puzzles that elementary school children can do

to problems that provide training for Putnam candidates (toward the end of

the book). There are many cross-references and connections between seem-

ingly unrelated problems from di�erent areas of mathematics, connections

that most students would be unable to make. Many of these connections

are new to this reviewer. However, once made these connections are quite

clear.

As for his credentials, Ravi Vakil placed among the top �ve in the Put-

nam competition in all four of his undergraduate years at the University of

Toronto. Before that he won two gold medals and a silver medal in IMOs

and coached the Canadian team to the IMO from 1989 to 1995.

Tests for DivisibilitySome tests for divisibility are well known, such as the test for divisibil-

ity by 3: �nd the sum of the digits of the whole number n | if that sum is

divisible by 3, then the original whole number n is divisible by 3.

Many people know this test, but do not know why it works. So, why

does it work?

The basic principle is that the remainder obtained when 10k is divided

by 3, is 1. This is easy to check, since 10k = 3 � 3 � 10k�1 + 10k�1.This enables one to step down one power at a time. So, if d is a digit, the

\remainder" obtained when d� 10k is divided by 3 is d. (NB: this is not the

true remainder one gets when dividing by 3!)

If n = abc : : : def is a whole number, then we write it as

n = a� 10k + b� 10k�1 + c� 10k�2 + : : :+ d� 100 + e� 10 + f:

When we divide by 3, we get a \remainder" of a+ b+ c+ : : :+ d+ e+ f .

If this is divisible by 3, the the true remainder is 0, and so n is divisible

by 3.

Now, do you know, or can you construct, a test for divisibility by 7?

336

Packing Boxes with N{tetracubes

Andris CibulisRiga, Latvia

Introduction

With the popularity of the video game Tetris, most people are aware of

the �ve connected shapes formed by four unit squares joined edge to edge.

They are called the I{, L{, N{, O{ and T {tetrominoes, after the letter of the

alphabet whose shapes they resemble. They form a subclass of the polyomi-

noes, a favourite topic in research and recreational mathematics founded by

Solomon Golomb.

Here is a problem from his classic treatise, Polyominoes. Is it possible

to tile a rectangle with copies of a particular tetromino? Figure 1 shows that

the answer is a�rmative for four of the tetrominoes but negative for the

N{tetromino, which cannot even �ll up one side of a rectangle.

Figure 1

Getting o� the plane into space, we can join unit cubes face to face to

form polycubes. By adding unit thickness to the tetrominoes, we get �ve

tetracubes, but there are three others. They are shown in Figure 2, along

with the N{tetracube.

Is it possible to pack a rectangular block, or box, with copies of a par-

ticular tetracube? The answer is obviously a�rmative for the I{, L{, O{

and T {tetracubes, and it is easy to see that two copies of each of the three

tetracubes not derived from tetrominoes can pack a 2� 2� 2 box. Will the

N{tetracube be left out once again? Build as many copies of it as possible

and experiment with them.

If the k�m�n box can be packed with the N{tetracube, we call it an

N{box. Are there any such boxes? Certain types may be dismissed immedi-

ately.

337

���

���

���

�����

�����

PPPPP

PPPPP

PPPPP

PPP

PPP

���

���

���

�����

���

���

���

PPP

PPP

PPP

PPPPP

PPP

PPP

PPP

���

���

���

���

�����

�����

PPPPP

PPPPP

PPPPP

PPP

PPP

���

���

���

���

���

���

PPPPP

PPPPP

PPP

PPPPPPP

PPPPP

Figure 2

Observation 1.The k � m � n box cannot be an N{box if it satis�es at least one of the

following conditions:

(a) one of k;m and n is equal to 1;

(b) two of k;m and n are equal to 2;

(c) kmn is not divisible by 4.

It follows that the 2� 3 � 4 box is the smallest box which may be an

N{box. Figure 3 shows that this is in fact the case. The box is drawn in two

layers, and two dominoes with identical labels form a singleN{tetracube.

0

1

23

0

1

2

3

Figure 3

So there is life in this universe after all! The main problem is to �nd all

N{boxes.

N{cubes

If k = m = n, the k � m � n box is called the k{cube, and a cube

which can be packed by the N{tetracube is called an N{cube. We can easily

assemble the 12{cube with the 2�3�4N{box, which makes it anN{cube.

This is a special case of the following result.

338

Observation 2.

Suppose the k�m�n and `�m�n boxes are N{boxes. Let a, b and c be

any positive integers. Then the following are also N{boxes:

(a) (k+ `)�m� n;

(b) ak � bm� cn.

The 12{cube is not the smallest N{cube. By Observation 1, the �rst

candidate is k = 4. It turns out that this is indeed an N{cube. It can

be assembled from the 2 � 4 � 4 N{box, whose construction is shown in

Figure 4.

0

1 23

0

1 2

3

Figure 4

The next candidate, the 6{cube, is also anN{cube, but a packing is not

that easy to �nd. In Figure 5, we begin with a packing of a 2 � 6 � 6 box,

with a 1 � 2 � 4 box attached to it. To complete a packing of the 6{cube,

add a 2� 3� 4 N{box on top of the small box, ank it with two 2� 4� 4

N{boxes and �nally add two more 2� 3� 4 N{boxes.

0 1

2 3

4 5

0 1

2 3

4 5

6 7

8 9

6 7

8 9

Figure 5

Can we pack the 8{cube, the 10{cube, or others? It would appear that

as size increases, it is more likely that we would have an N{cube. How-

ever, it is time to stop considering one case at a time. We present a recur-

sive construction which expandsN{cubes into larger ones by adding certain

N{boxes.

Theorem 1.The k{cube is an N{cube if and only if k is an even integer greater than 2.

Proof:

That this condition is necessary follows from Observation 1. We now

339

prove that it is su�cient by establishing the fact that if the k{cube is an N{

cube, then so is the (k+ 4){cube. We can then start from either the 4{cube

or the 6{cube and assemble all others.

From the 2�3�4 and 2�4�4N{boxes, we can assemble all 4�m�nboxes for all even m;n � 4, via Observation 2. By attaching appropriate

N{boxes from this collection, we can enlarge the k{cube �rst to the

(k + 4) � k � k box, then the (k + 4) � (k + 4) � k box and �nally the

(k+ 4){cube. This completes the proof of Theorem 1.

Further Necessary Conditions

Observation 1 contains some trivial necessary conditions for a box to

be an N{box. We now prove two stronger results, one of which supersedes

(c) in Observation 1.

Lemma 1.The k�m� n box is not an N{box if at least two of k;m and n are odd.

Proof:We may assume that m and n are odd. Place the box so that the hor-

izontal cross-section is an m � n rectangle. Label the layers L1 to Lk from

bottom to top. Colour the unit cubes in checkerboard fashion, so that in any

two which share a common face, one is black and the other is white. We may

assume that the unit cubes at the bottom corners are black. It follows that

Li has one more black unit cube than white if i is odd, and one more white

unit cube than black if i is even.

Suppose to the contrary that we have a packing of the box. We will call

an N{tetracube vertical if it intersects three layers. Note that the intersec-

tion of a layer with any N{tetracube which is not vertical consists of two or

four unit cubes, with an equal number in black and white. The intersection

of a vertical N{tetracube with its middle layer consists of one unit cube of

each colour.

Since L1 has a surplus of one black unit cube, it must intersect `1 ver-

tical N{tetracubes in white and `1 + 1 vertical N{tetracubes in black, for

some non-negative integer `1. These N{tetracubes intersect L3 in `1 black

unit cubes and `1 + 1 white ones. Hence the remaining part of L3 has a

surplus of two black. They can only be packed with `3 vertical N{tetracubes

intersecting L3 in white, and `3+2 vertical N{tetracubes in black, for some

non-negative integer `3. However, the surplus in black unit cubes in L5 is

now three, and this surplus must continue to grow. Thus the k � m � n

box cannot be packed with the N{tetracube. This completes the proof of

Lemma 1.

Lemma 2.The k�m� n box is not an N{box if kmn is not divisible by 8.

340

Proof:

Suppose a k �m� n box is an N{box. In view of Lemma 1, we may

assume that at least two of k;m and n are even. Place the packed box so

that the horizontal cross-section is an m� n rectangle, and label the layers

L1 to Lk from bottom to top. De�ne vertical N{tetracubes as in Lemma 1

and denote by ti the total number of those which intersects Li; Li+1 and

Li+2; 1 � i � k� 2.

Since at least one of m and n is even, each layer has an even number

of unit cubes. It follows easily that each ti must be even, so that the total

number of vertical N{tetracubes is also even. The same conclusion can be

reached if we place the box in either of the other two non-equivalent orien-

tations. Hence the total number of N{tetracubes must be even, and kmn

must be divisible by 8. This completes the proof of Lemma 2.

N{Boxes of Height 2

We now consider 2 � m � n boxes. By Observation 1, m � 3 and

n � 3. By Lemma 2, mn is divisible by 4. We may assume that m is even.

First letm = 4. We already know that the 2� 4� 3 and 2� 4� 4 boxes are

N{boxes. Figure 6 shows that so is the 2� 4� 5 box. If the 2� 4� n box

is anN{box, then so is the 2� 4� (n+ 3) box by Observation 2. It follows

that the 2� 4� n box is an N{box for all n � 3.

0

1

23

0

1

2

3

Figure 6

Now letm = 6. Then n is even. We already know that the 2�6�4 box

is anN{box. However, the 2� 6� 6 box is not. Our proof consists of a long

case-analysis, and we omit the details. On the other hand, the 2 � 6 � 10

box is an N{box. In Figure 7, we begin with the packing of a 2� 3� 6 box

with a 2� 2� 3 box attached to it. We then build the mirror image of this

solid and complete the packing of the 2� 6� 10 box by adding a 2� 4� 3

N{box.

If the 2� 6�n box is an N{box, then so is the 2� 6� (n+4) box by

Observation 2. It follows that the 2� 6� n box is an N{box for n = 4 and

all even n � 8.

Theorem 2.The 2 � m � n box is an N{box if and only if m � 3; n � 3 and mn is

divisible by 4, except for the 2� 6� 6 box.

341

01

2

3

4 5

0

1

2

34 5

Figure 7

Proof:If m is divisible by 4, the result follows immediately from Observa-

tion 2. Let m = 4` + 2 for some positive integer `. We already know that

the 2� 10 � 6 box is an N{box. If the 2 � (4`+ 2)� n box is an N{box,

then so is the 2 � (4` + 6) � n box by Observation 2. This completes the

proof of Theorem 2.

The Main Result

Theorem 3.The k �m� n box, k � m � n, is an N{box if and only if it satis�es all of

the following conditions:

(a) k � 2;

(b) m � 3;

(c) at least two of k,m and n are even;

(d) kmn is divisible by 8;

(e) (k;m; n) 6= (2; 6; 6).

Proof:Necessity has already been established, and we deal with su�ciency.

We may assume that k � 3, since we have taken care of N{boxes of height

2. Consider all 3 � m � n boxes. By (c), both m and n must be even. By

(d), one of them is divisible by 4. All such boxes can be assembled from the

3� 2� 4 box.

Consider now the k �m � n box. We may assume that m and n are

even. If k is odd, then one of m and n is divisible by 4. Slice this box into

one 3 �m � n box and a number of 2 �m � n boxes. Since these are all

N{boxes, so is the k�m� n box.

Suppose k is even. Slice this box into a number of 2 �m � n boxes,

each of which is an N{box unless m = n = 6. The 4 � 6 � 6 box may be

assembled from the 4� 2� 3 box, and we already know that the 6{cube is

an N{cube. If the k� 6� 6 box is an N{box, then so is the (k+4)� 6� 6

box. This completes the proof of Theorem 3.

342

Research Projects

Problem 1.Try to prove that the 2� 6� 6 box is not an N{box. It is unlikely that any

elegant solution exists.

Problem 2.AnN{box which cannot be assembled from smallerN{boxes is called a prime

N{box. Find all prime N{boxes.

Problem 3.Prove or disprove that an N{box cannot be packed if we replace one of the

N{tetracubes by an O{tetracube.

Problem 4.For each of the other seven tetracubes, �nd all boxes which it can pack.

Bibliography

Bouwkamp, C. J. & Klarner, David, Packing a Box with Y {pentacubes, Jour-

nal of Recreational Mathematics, 3 (1970), 10-26.

G �obel, F. & Klarner, David, Packing Boxes with Congruent Figures, Indaga-

tiones Mathematicae, 31 (1969), 465-472.

Golomb, Solomon G., Polyominoes | Puzzles, Patterns, Problems & Pack-

ings, Princeton University Press, Princeton, 1994.

Klarner, David, A Search for N{pentacube Prime Boxes, Journal of Recre-

ational Mathematics, 12 (1979), 252-257.

Acknowledgement

This article has been published previously in a special edition of delta-k,

Mathematics for Gifted Students II, Vol. 33, 3 1996, a publication of the

Mathematics Council of the Alberta Teachers' Association and in AGATE,

Vol. 10, 1, 1996, the journal of the Gifted and Talented Education Council

of the Alberta Teachers' Association. It is reprinted with permission of the

author and delta-k.

343

THE SKOLIAD CORNERNo. 24

R.E. Woodrow

This number we give the problems of the 1995 Concours Math �ematique

du Qu �ebec. This contest comes to us from the organizers via the Canadian

Mathematical Society which gives partial support to the contest. My thanks

go to Th �er �ese Ouellet, secretary to the contest. The contest was written by

over 2000 students February 2, 1997 over a three hour period. This will also

test your French! We will, of course, accept solutions in either English or

French.

CONCOURS MATH �EMATIQUE DU QU �EBEC 1995February 2, 1995

Time: 3 hours

1. LA FRACTION SIMPLIFI �EE

Simpli�er la fraction1 358 024 701

1 851 851 865:

2. LA FORMULE MYST �ERE

Consid �erons les �equations suivantes

xy = p; x+ y = s; x1993 + y1993 = t; x1994 + y1994 = u:

En faisant appel aux lettres p, s, t, u mais pas aux lettres x, y, donnez une

formule pour la valeur

x1995 + y1995:

3. LA DIFF �ERENCE �ETONNANTE

Lorsque la circonf �erence d'un cercle est divis �ee en dix parties �egales,

les cordes qui joignent les points de division successifs forment un d �ecagone

r �egulier convexe. En joignant chaque point de division au troisi �eme suivant,

on obtient un d �ecagone r �egulier �etoil �e. Montrer que la di� �erence entre les

cot �es de ces d �ecagones est �egale au rayon du cercle.

344

4. LE TERRAIN EN FORME DE CERF-VOLANT

Abel Belgrillet est membre du Club des a �erocervidophiles (amateurs de

cerfs-volants) du Qu �ebec. Il dispose de quatre tron�cons de cloture rectilignes

AB, BC, CD, DA pour d �elimiter un terrain (en forme de cerf-volant, voir

�gure) sur lequel il s'adonnera �a son activit �e favorite cet �et �e. Sachant que les

tron�cons AB et DA mesurent 50 m chacun et que les tron�cons BC et DA

mesurent 120 m chacun, d �eterminer la distance entre les points A et C qui

maximisera l'aire du terrain.

A C

B

D

50

50

120

120

5. L'IN �EGALIT �E MODIFI �EE D'AMOTH DIEUFUTUR

(a) (2 points) L'in �egalit �e x2 + 2y2 � 3xy est-elle vraie pour tous les

entiers?

(b) (8 points) Montrer que l'in �egalit �e x2 +2y2 � 145xy est valide pour

tous r �eels x et y.

6. L' �ECHIQUIER COQUET

Trouver l'unique fa�con de colorier les 36 cases d'un �echiquier 6� 6 en

noir et blanc de sorte que chacune des cases soit voisines d'un nombre impair

de cases noires.

(Note: deux cases sont voisines si elles se touchent par un cot �e ou par

un coin). On ne demande pas de d �emontrer que la solution est unique.

7. LA FRACTION D'ANNE GRUJOTE

En base 10, 13= 0:333 : : : . �Ecrivons 0:�3 pour ces d �ecimles r �ep �et �ees.

Comment �ecrit-on 13dans une base b, o �u b est de la forme

(i) b = 3t (trois points),

(ii) b = 3t+ 1 (trois points),

(iii) b = 3t+ 2 (quatre points),

o �u t est un entier positif quelconque?

345

To �nish this number of the Skoliad Corner we give the o�cial solutions

to the 1995 P.E.I. Mathematics Competition given last issue [1997: 278].

My thanks go to Gordon MacDonald, University of Prince Edward Island for

forwarding the materials.

1995 P.E.I. Mathematics Competition

1. Find the area of the shaded region inside the circle in the following

�gure.

-�

6

?

S

O 1�1

1

�1

P

T1

T2

Solution. First determine the coordinates of the point P . Since P lies

on the line y = x, P = (a; a) for some value of a. Since P lies on the circle

x2 + y2 = 1, a2 + a2 = 1 and so a = �1p2. Hence P = (�1p

2; �1p

2). Partition

the shaded region as shown. Then the area of the quarter circle S is �4and

the area of the triangle T1 is 12(base)(height). The base is 1 and the height

is 1p2. (Turn the page upside down.) So the area of T1 is 1

2p2. The triangle

T2 has the same area so the total area of the shaded region is

�

4+

1p2:

2. \I will be n years old in the year n2", said Bob in the year 1995.

How old is Bob?

Solution. AssumingBob's age is an integer, Bobmust have a reasonable

chance to be alive in a year that is a perfect square. Since 442 = 1936,

452 = 2025 and 462 = 2116, Bob must be 45 years old in the year 2025.

Hence Bob is now 15 years old (or 14 years old if he hasn't had his birthday

yet this year).

3. Draw the set of points (x; y) in the plane which satisfy the equation

jxj + jx� yj = 4.

346

Solution. Consider the following four possible cases:

Case 1: When x � 0 and x � y. Then x+ x� y = 4 so 2x� y = 4.

Case 2: When x � 0 and x < y. Then x� (x� y) = 4 so y = 4.

Case 3: When x < 0 and x � y. Then �x+ x� y = 4 so y = �4.

Case 4: When x < 0 and x < y. Then �x� (x� y) = 4 so �2x+ y = 4.

Thus the set of points is made up of four line segments. When these

line segments are drawn we obtain the following parallelogram:

-

6

r

r

r

r

�4 �2

d

2 4x{axis

�4

�2

2

4

y{axis

4. An autobiographical number is a natural number with ten digits

or less in which the �rst digit of the number (reading from left to right)

tells you how many zeros are in the number, the second digit tells you how

many 1's, the third digit tells you how many 2's, and so on. For example,

6; 210; 001; 000 is an autobiographical number. Find the smallest autobio-

graphical number and prove that it is the smallest.

Solution. When you add the digits of an autobiographical number, you

count the total number of digits. (For example the digits of the above 10 digit

autobiographical number must sum to 10.) Using this fact and the process of

elimination we can �nd the smallest autobiographical number.

The only possible one-digit number whose digits sum to 1 is 1 and it is

not autobiographical so there are no one digit autobiographical numbers.

The only possible two digit numbers whose digits sum to two are (in

increasing order) 11 and 20 and they are not autobiographical so there are

no two digit autobiographical numbers.

The only possible three digit numbers whose digits sum to three are (in

increasing order) 102; 111; 120; 201;210 and 300 and they are not autobio-

graphical so there are no three digit autobiographical numbers.

The possible four digit numbers whose digits sum to four are (in in-

creasing order) 1003; 1012;1021; 1030; 1102; 1111;1120; 1201; : : : .

347

Checking these numbers we �nd that the �rst autobiographical number

in this list, and hence the smallest autobiographical number is 1201.

5. A solid cube of radium is oating in deep space. Each edge of the

cube is exactly 1 kilometre in length. An astronaut is protected from its

radiation if she remains at least 1 kilometre from the nearest speck of radium.

Including the interior of the cube, what is the volume (in cubic kilometres) of

space that is forbidden to the astronaut? (You may assume that the volume

of a sphere of radius r is 43�r3 and the volume of a right circular cylinder of

radius r and height h is �r2h.)

Solution. Along with the cube of radium, on each of the six faces of

the cube there will be a cube with sides of length 1 km of forbidden space.

Along each of the 12 edges of the cube of radium there will be a quarter of

a cylinder of radius 1 km and length 1 km of forbidden space, and at each of

the eight corners of the cube of radium there will be an eighth of a sphere of

radius 1 km of forbidden space.

Putting it all together, there are 7 cubes (with sides of length 1 km),

3 cylinders of radius 1 km and length 1 km, and 1 sphere of radius 1 km of

forbidden space. So the volume of forbidden space is:

7 + 3� +4

3� km

3= 7+

16

3� km

3:

6. Which is greater, 999! or 500999? (Where 999! denotes 999 factorial,

the product of all the natural numbers from 1 to 999 inclusive.) Explain your

reasoning.

Solution. Write

999! = 1 � 2 � 3 � 4 � � � 995 � 996 � 997 � 998 � 999

and successively pair o� numbers from the left with numbers from the right

so

999! = (1 � 999)(2 � 998)(3 � 997)(4 � 996) � � � (449 � 501)(500):

(Note that no number is paired with 500 and there are 499 such pairs.)

Now use the fact that (500�k)(500+k) = 5002�k2 < 5002 for each

value of k from 1 to 499 so

999! < (5002)(5002)(5002)(5002) � � � (5002)(500)

where 5002 is repeated 499 times. Hence

999! < (5002)499(500) = 5002(499)+1 = 500999;

so 500999 is greater.

348

MATHEMATICAL MAYHEMMathematical Mayhem began in 1988 as a Mathematical Journal for and byHigh School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to theMayhem Editor, Naoki Sato, Department of Mathematics, Yale University,PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronicaddress is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Invariants of Inscribed Regular n-gons

There are two unexpected (in their simultaneity) invariants of inscribed

regular n-gons.

Let P1P2 � � �Pn be a regular n-gon, and let P be a point on its circum-

circle. Then both

PP 21 + PP 2

2 + � � �+ PP 2n and PP 4

1 + PP 42 + � � �+ PP 4

n

are independent of the position of P .

Proof. We will show the result using complex numbers. Without loss

of generality, assume the radius of the circumcircle is 1, and that Pk is at the

complex number !k�1 in the complex plane, where ! = cis(2�n). Let z be

the complex number corresponding to P , so jzj = 1, and zz = jzj2 = 1.

Note

PP 2k+1 = jz � !kj2 = (z � !k)(z � !k)

= zz � !kz � !kz + !k!k = 2� !kz � !n�kz;

since !k =j!kj2!k

= 1!k

= !n

!k= !n�k. Hence,

PP 21 + PP 2

2 + � � �+ PP 2n =

n�1Xk=0

jz � !kj2 =

n�1Xk=0

(2� !kz � !n�kz)

= 2n�

n�1Xk=0

!k

!z �

n�1Xk=0

!n�k

!z

= 2n;

349

sincen�1Xk=0

!k =1� !n

1� != 0:

Therefore, the sum is indeed independent of z. Similarly,

PP 4k+1 = (2� !kz � !n�kz)2

= 4+ !2kz2 + !2(n�k)z2 � 4!kz � 4!n�kz + 2

= 6 + !2kz2 + !2(n�k)z2 � 4!kz � 4!n�kz;

soPP 4

1 + PP 42 + � � �+ PP 4

n

=

n�1Xk=0

(6 + !2kz2 + !2(n�k)z2 � 4!kz � 4!n�kz)

= 6n+

n�1Xk=0

!2k

!z2 +

n�1Xk=0

!2(n�k)

!z2

�4

n�1Xk=0

!k

!z � 4

n�1Xk=0

!n�k

!z = 6n:

Four Large Spheres and a Small Sphere

Wai Ling Yee

Problem: Given four identical spheres of radius R which are mutually

tangent, �nd the radius r of the sphere that will �t in the hole at the centre

of the tetrahedron that they form, in terms of R.

When I �rst tried this problem, I looked at several triangles and, after

many calculations, produced an answer. I came across this problem again in

chemistry when calculating the size of tetrahedral holes in ion lattices. Try

this:

Construct a cube so that the centres of the spheres are at alternate cor-

ners of the cube. The length of a face diagonal of the cube is 2R. Therefore,

the length of a body diagonal of the cube isp6R. Note that we can also

express the length of the body diagonal as 2(R+ r). Hence,

2(R+ r) =p6R implies that r =

p6� 2

2R:

Rider. This is the radius of the sphere that is externally tangent to all

four. What is the radius of the sphere that is internally tangent to all four?

350

IMO REPORT

Adrian Chan

Two weeks of training at St. Mary's University in Halifax in early July

kicked o� the month-long journey the 1997 Canadian IMO team would en-

dure. Four ights later, the team found itself in Mar del Plata, Argentina,

ready to participate in the 37th International Mathematical Olympiad.

This year's team members included: Adrian \Terrible Taco" Birka, Sabin

\Cursed Corner" Cautis, Adrian \Flash Flood" Chan, Jimmy \Fruits and Veg-

gies" Chui, Byung Kyu \Argentinian Polite" Chun, and Mihaela \My Dear-

est" Enachescu. We gave our team leader Dr. Richard \Richard" Nowakowski

and our deputy leader Naoki \Raspberry" Sato all they could handle. Spe-

cial thanks goes out to our coaches Dr. Bruce Shawyer and leader observer

Dr. Chris \Bob's Your Uncle" Small.

This year's contest was not as di�cult as last year's killer in India, but

still challenging. It was the largest IMO ever, with a record 82 countries

competing. Canada was well represented, bringing back 2 silver, 2 bronze,

and an honourable mention. Our team's scores were as follows:

CAN1 Adrian Birka 6

CAN2 Sabin Cautis 16 Bronze Medal

CAN3 Adrian Chan 25 Silver Medal

CAN4 Jimmy Chui 10 Honourable Mention

CAN5 Byung Kyu Chun 29 Silver Medal

CAN6 Mihaela Enachescu 21 Bronze Medal

As a team, Canada �nished 29th out of the 82 countries. Best of luck

to the graduating members of the team, Sabin and Byung, as they take their

math skills to the University of Waterloo next year. The remaining four mem-

bers of the team are eligible to return to Taiwan for next year's IMO. How-

ever, it seems that the team should focus more in the area of geometry!

Special thanks must also go out to Dr. Graham Wright of the Canadian

Mathematical Society for paying all our bills, and Professor Ed Barbeau of

the University of Toronto for conducting a year-long correspondence program

for all IMO hopefuls.

Visiting Argentina was a new experience for many of us, and was def-

initely a memorable one to all. The food was great, especially the beef and

the chocolate alfajores, and the IMO was very well-organized and run. Best

of luck to all IMO hopefuls who are working hard to be on the 1998 IMO

team, which will compete in Taipei, Taiwan.

351

Unique Forms

Naoki Sato

For the mathematician, there is something re-assuring about being able

to put an object into a unique form, or a canonical form, such as the prime

factorization of a positive integer, or a matrix in Jordan Canonical Form. Such

forms make working with these objects much easier and more tractable, as

well as allowing a way to get a handle on them. Here, we explore the idea

of the unique, canonical form.

1. Let p(x) = anxn + an�1x

n�1 + � � � + a1x + a0, and let c be a real

number. Show that there exist unique constants bn, bn�1, : : : , b1, b0,such that

p(x) = bn(x� c)n + bn�1(x� c)n�1 + � � �+ b1(x� c) + b0:

Can you �nd the bi explicitly?

What may seem like a lot of (linear) algebra to untangle, comes easily

undone with one idea: consider p(x+ c). This is a polynomial of degree at

most n, so there exist unique bi such that

p(x+ c) = bnxn + bn�1x

n�1 + � � �+ b1x+ b0;

which is equivalent to

p(x) = bn(x� c)n + bn�1(x� c)n�1 + � � �+ b1(x� c) + b0:

The expression may look familiar, because it is the nth approximant to a

power series around x = c, a central object of calculus. Suppose we take the

kth derivative of both sides. The \constant term" is then k!bk, so evaluating

both sides at x = c leads to

p(k)(c) = k!bk so that bk =p(k)(c)

k!:

A quick corollary: If p(x) is a polynomial, then (x � c)k j p(x) if andonly if p(c) = p0(c) = p00(c) = � � � = p(k�1)(c) = 0. (Both conditions are

equivalent to b0 = b1 = � � � = bk�1 = 0.)

2. Let N be a non-zero integer. Show that there exist unique positive

integers k and a1 < a2 < � � � < ak such that

N = (�2)ak + (�2)ak�1 + � � �+ (�2)a1:

(In other words, show that there is a unique representation of N in

base �2.)

352

We will use strong induction. The statement is true for N = 1 =

(�2)0 and for N = �1 = (�2)1 + (�2)0. Now, assume there is a positive

integer n such that the statement is true for N = 1; 2; : : : ; n and N =

�1;�2; : : : ;�n.

Let N = n + 1. If N is odd, then n is even, and n=(�2) is an inte-

ger between �1 and �n, so by the induction hypothesis, for some unique

distinct ai,

n=(�2) = (�2)ak + (�2)ak�1 + � � �+ (�2)a1;

which is equivalent to

N = n+ 1 = (�2)ak+1 + (�2)ak�1+1 + � � �+ (�2)a1+1 + (�2)0:

Note that the exponents are still distinct. The same trick works ifN is even,

and for N = �(n + 1). Subtracting o� 1 = (�2)0 if necessary, and then

dividing by �2, we obtain a form which is simultaneously seen to exist and

be unique, by the induction hypothesis. Hence, by strong induction, the

statement is proved for all N . Note that this also provides an algorithm for

�nding the exponents.

One may be tempted to isolate the higher exponents, but this tends

to get very messy. Instead, one can start with the lower exponents - after

seeing if (�2)0 is in the sum, they can then be picked o� inductively.

3. Let r be a positive rational number. Show that there exist unique pos-

itive integers k and a1 � a2 � � � � � ak such that

r =1

a1+

1

a1a2+

1

a1a2a3+ � � �+

1

a1a2a3 � � � ak:

Note that

r =1

a1+

1

a1a2+

1

a1a2a3+ � � �+

1

a1a2a3 � � � ak;

which is equivalent to

a1r � 1 =1

a2+

1

a2a3+ � � �+

1

a2a3 � � � ak:

This suggests �nding the right a1, reducing, and applying the same argument

to a1r � 1.

De�ne the sequences faig and frig as follows: set r1 = r, ai = d1=rie,ri+1 = airi � 1, i � 1, and we terminate when ri+1 = 0. For example, if

r1 = 3=7, then we obtain a1 = d7=3e = 3, r2 = 3 � 3=7� 1 = 2=7, a2 = 4,

r2 = 1=7, a3 = 7, and r3 = 0. And we see

3

7=

1

3+

1

3 � 4+

1

3 � 4 � 7:

353

We will show that in general, this is the only possible sequence which works.

Consider the choice of a1. Since a1r � 1, a1 must be at least d1=re.However, if a1 was set to d1=re+ 1, then

r =1

a1+

1

a1a2+

1

a1a2a3+ � � �+

1

a1a2a3 � � � ak

<1

a1+

1

a21+

1

a31+ � � �

=1

a1 � 1=

1

d1=re�

1

1=r= r;

contradiction. Hence, there is only one choice for a1, the one we have chosen

above. By the above reduction, there is only one choice for a2, a3, and so

on. Hence, we have uniqueness.

We also know terms will be generated, but how do we know the se-

quence always terminates? Let ri = pi=qi, as a reduced fraction. Then

ai = dqi=pie, so

ri+1 =pi

qi

�qi

pi

�� 1 =

pi

lqipi

m� qi

qi:

However,

pi

�qi

pi

�� qi < pi

�qi

pi+ 1

�� qi = pi:

Hence, the numerators of the ri are strictly decreasing. But as they are all

positive integers, the sequence must terminate at some point.

Finally, we must show the sequence faig is non-decreasing. We have

ai+1 =

2666

qi

pi

lqipi

m� qi

3777 �

qi

pi

lqipi

m� qi

>qi

pi>

�qi

pi

�� 1 = ai � 1:

Therefore, ai+1 � ai.

Rider. Does an analogous result hold for positive irrational numbers?

What is the sequence forp2? For e?

4. Let be a root of x3 � x � 1 = 0, and let p, q be polynomials with

rational coe�cients, such that q( ) 6= 0. Show that there exist unique

rationals a, b, and c such that

p( )

q( )= a 2 + b + c:

354

First, let us express p( )=q( )as a polynomial in . Note that x3�x�1

is irreducible over the rationals. Hence, there exist polynomials u(x) and

v(x) with rational coe�cients such that

u(x)q(x) + v(x)(x3� x� 1) = gcd(q(x); x3 � x� 1) = 1:

Then by substituting x = ,

u( )q( ) = 1 implies thatp( )

q( )= p( )u( ) = r( );

where r(x) = p(x)u(x), a polynomial with rational coe�cients.

Now, by the division algorithm for polynomials,

r(x) = (x3 � x� 1)w(x) + ax2 + bx+ c;

for some rationals a, b, and c (dividing r(x) by x3�x� 1 leaves, at most, a

quadratic remainder). Hence, r( ) = a 2 + b + c, which shows existence.

To see uniqueness, suppose r( ) = a0 2 + b0 + c0 implies that

(a � a0) 2 + (b � b0) + (c � c0) = 0; that is, is a root of the quadratic

(a � a0)x2 + (b � b0)x + (c � c0) = 0. But as indicated above, the cubic

x3 � x� 1 is irreducible; hence, all the coe�cients of the quadratic must be

zero, and a = a0, b = b0, c = c0, showing uniqueness.

5. Let a and b be distinct positive integers, and let d = gcd(a; b). Con-

sider the Diophantine equation ax + by = d. It is well-known that

all solutions are given by (x; y) = (u+ bt=d; v � at=d), where t is an

integer and (u; v) is some solution.

(i) Show that there exists a unique solution satisfying jxj � b=2d,

jyj � a=2d.

(ii) Show that the solution in (i) is given by the Euclidean Algorithm.

For example, for a = 4 and b = 7, we have

7 = 1 � 4 + 3;

4 = 1 � 3 + 1;

3 = 1 � 3;

so gcd(4,7) = 1, and

1 = 4� 3 = 4� (7� 4) = 2 � 4� 7:

Hence, the Euclidean Algorithm gives the solution (x; y) = (2;�1),

which satis�es jxj � 7=2 and jyj � 4=2,

We can assume d = 1, as a way of normalizing the equation, since

ax + by = d is equivalent to (a=d)x+ (b=d)y = 1. It is easy to check that

d may be factored back in.

355

(i) Uniqueness is immediate, since the solutions in x form an arithmetic pro-

gression with common di�erence b=d = b, and similarly for y, with di�erence

a=d = a.

If b = 1, then (x; y) = (0; 1) serves as a solution. If b = 2, then a is

odd, and (x; y) = (1; (1� a)=2) serves as a solution. Assume that b > 2.

Because the solutions in x form an arithmetic progression with di�er-

ence b, there exists a solution (x; y) with jxj � b=2, or �b=2 � x � b=2.

Then y = (1� ax)=b, so

y �1� a � b

2

b=

1

b�a

2> �

a

2;

and

y �1 + a � b

2

b=

1

b+a

2<a+ 1

2;

so y � a=2 implies that jyj � a=2.

(ii) We must qualify the statement somewhat | if either a or b is equal to 1,

then the Euclidean Algorithm does not explicitly provide a solution. We may

stipulate now that the solution is (x; y) = (1; 0) or (0,1), if a or b is 1,

respectively (they cannot both be 1, since they must be distinct).

We will use strong induction onmax(a; b). The statement is easily ver-

i�ed for max(a; b) � 2. Assume the statement is true for max(a; b) � n,

for some positive integer n. We will prove the statement for

max(a; b) = n+ 1. Assume b = n+ 1.

We seek solutions of ax + (n + 1)y = 1. By the division algorithm,

there exist unique non-negative integers q and r such that n+ 1 = aq + r,

0 � r � a� 1. Then re-arranging, the equation becomes ax+ (aq+ r)y =

a(x+ qy)+ ry = 1. Since a, r � n, by the induction hypothesis, there exist

solutions satisfying jx + qyj � r=2, jyj � a=2. By the triangle inequality,

jxj � r=2+ jqyj � r=2+qa=2 = (r+qa)=2 = (n+1)=2. Hence, by strong

induction, the statement is proved for all values of max(a; b).

1996 Balkan Mathematical Olympiad

1. Let O and G be the circumcentre and centroid of a triangle ABC re-

spectively. IfR is the circumradius and r is the inradius of ABC, show

that

OG �pR(R� 2r) :

2. Let p > 5 be a prime number and X = fp � n2jn 2 Z+; n2 < pg.Prove that X contains two distinct elements x, y, such that x 6= 1 and

x divides y.

356

3. Let ABCDE be a convex pentagon. Denote by M , N , P , Q, R the

midpoints of the segments AB, BC, CD, DE, EA respectively. If

the segments AP , BQ, CR, DN have a common point of intersection,

prove that this point also belongs to the segment EN .

4. Show that there exists a subset A of the set f1; 2; 3; : : : ; 21996 � 1ghaving the following properties:

(a) 1 2 A and 21996 � 1 2 A,(b) Every element ofAnf1g is the sum of two (not necessarily distinct)

elements of A, and

(c) The number of elements of A is at most 2012.

Mayhem Problems

The Mayhem Problems editors are:

Richard Hoshino Mayhem High School Problems Editor,

Cyrus Hsia Mayhem Advanced Problems Editor,

Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only solutions | the

next issue will feature only problems.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from the pre-

vious issue be submitted by 1 November 1997, for publication in the issue 5

months ahead; that is, issue 4 of 1998. We also request that only studentssubmit solutions (see editorial [1997: 30]), but we will consider particularly

elegant or insightful solutions from others. Since this rule is only being im-

plemented now, you will see solutions from many people in the next few

months, as we clear out the old problems from Mayhem.

357

High School Problems | Solutions

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,

Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>

H211. Proposed by Richard Hoshino, grade OAC, University of Toronto

Schools

Let p, q, and r be the roots of the cubic equation ax3�bx2+cx+d = 0,

where a, b, c, and d are real coe�cients.

Given that Arctan p + Arctan q + Arctan r = �4, where Arctan de-

notes the principal value, prove that a = b+ c+ d.

SolutionbyMiguel Carri �on �Alvarez, Universidad Compluteuse deMad-

rid, Spain.

Let � = Arctan p, � = Arctan q, and = Arctan r. Then we have

� + �+ = �4and

1 = tan(�+ �+ ) =tan(� + �) + tan

1� tan(� + �) tan =

tan �+tan �

1�tan � tan� + tan

1� tan �+tan �

1�tan � tan� tan

=tan � + tan�+ tan � tan � tan� tan

1� tan � tan�� tan � tan � tan � tan =

p+ q + r � pqr

1� pq � pr � qr:

Finally, a(x�p)(x�q)(x�r) = a[x3�(p+q+r)x2+(pq+pr+qr)x�pqr] =ax3 � bx2 + cx+ d. From the relationship between the coe�cients and the

roots of this cubic, we get p+ q+ r = ba, pq+ pr+ qr = c

a, and pqr = �d

a.

Using these relations in the above equation, we have

1 =

ba+ d

a

1� ca

) 1�c

a=b

a+d

a) a = b+ c+ d :

Also solved by BOB PRIELIPP, University of Wisconsin-Oshkosh, WI,

USA. Prielipp also points out that this problem is similar to problem E2299

(pp. 520-521of theMay 1972 issue of The American MathematicalMonthly.)

H212. Given two sequences of real numbers of length n, with the

property that for any pair of integers (i; j) with 1 � i < j � n, the ith term

is equal to the j th term in at least one of the two sequences. Prove that at

least one of the sequences has all its values the same.

SolutionbyMiguel Carri �on �Alvarez, Universidad Compluteuse deMad-

rid, Spain.

Let the two sequences be faig and fbig. Assume that faig does not

have all its values the same. Then, there is a pair (i; j) such that ai 6= aj ,

and hence bi = bj. Now, it is impossible that both ak = ai and ak = aj .

Hence, either ak 6= ai or ak 6= aj. In either case, bk = bi = bj. This is true

for all k, so the sequence of bi's has all its values the same.

358

H213. Two players begin with one counter each, initially on opposite

corners of an n� n chessboard. They take turns moving their counter to an

adjacent square. A player wins by being the �rst to reach the row opposite

from their initial starting row, or landing on the opponent's counter. Who

has the winning strategy? (Generalize to anm� n board.)

SolutionbyMiguel Carri �on �Alvarez, Universidad Compluteuse deMad-

rid, Spain.

On a square board, the �rst player always wins by crossing the board

to the opposite row. In n� 2 moves,

d

t

becomes d

t

?

�?

and the �rst player wins.

If there are more columns than rows, the same strategy works for the

�rst player.

If there are more rows than columns, one of the players wins by \cap-

turing" the opponent. If the parity of both dimensions of the board is the

same, both players start from squares of the same colour. In this case, the

�rst player always moves to the opposite colour and cannot capture the sec-

ond player. The second player wins by \closing in" on the �rst player and

capturing the �rst player. [Ed: Why is this always possible?] Now if the

parity of both dimensions of the board is opposite, the �rst player moves to-

wards the colour of the second player, and so can capture the second player

using the same strategy.

Advanced Problems | Solutions

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.

M1G 1C3 <hsia@math.toronto.edu>

A190. Find all positive integers n > 1 such that ab � �1(mod n)

implies that a+ b � 0(mod n).

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.

Let n =Qki=1 p

�ii where the pi are distinct primes. By the Chinese

Remainder Theorem, the given property holds for n if and only if it holds for

each p�ii ; that is, ab � �1(mod p�ii ), implies that a + b � 0(mod p�ii ).

Also, both a and b must be relatively prime with n.

359

Let Si be the set of positive integers both relatively prime with and less

than p�ii . Take x, u, and v 2 Si. Note that ux � vx mod p�ii is equivalent

to x(u� v) � 0(mod p�ii ), which, in turn, is equivalent to u = v, since x is

relatively prime with p�ii and u and v are less than p�ii . Hence, each element

of fx; 2x; : : : ; (pi�1)x; (pi+1)x; : : : ; (p�ii � 1)xg leaves a unique residuemodulo p�ii . Also note that the product of two numbers in Si is itself in Si.

Therefore, the set fx; 2x; : : : ; (pi � 1)x; (pi + 1)x; : : : ; (p�ii � 1)xg takenmodulo p�ii is a permutation of Si. Since �1 2 Si, for each x 2 Si, there

exists a unique y 2 Si such that xy � �1(mod p�ii ).

Case I: pi is odd.

Let a = 2. Then there exists a b such that 2b � �1(mod p�ii ).

Thus, 2 + b � 0(mod p�ii ), so that 4 � 1(mod p�ii ). Therefore

3 � 0(mod p�ii ), from which it follows that pi = 3, �i = 1. Therefore,

the only possible odd prime factor of n is 3, and only one such factor

may appear. By a simple check, n = 3 satis�es the given property.

Case II: pi = 2.

Let a = 3. Then there exists a b such that 3b � �1(mod 2�i). Thus

3+ b � 0(mod 2�i), so that 9 � 1(mod 2�i). Therefore 2�i j 8, fromwhich it follows that �i � 3. Thus n can have 0, 1, 2, or 3 factors of 2,

and again by a simple check, all satisfy the given property.

From the above two cases, the solutions for n are 2, 3, 4, 6, 8, 12, and 24.

This problem, which generalizes problem B-1 on the 1969 PutnamCom-

petition, is equivalent to �nding all n > 1, such that a relatively prime to n

implies that a2 � 1(mod n).

A191. Taken over all ordered partitions of n, show thatXk1+k2+���+km=n

k1k2 � � � km =

�m+ n� 1

2m� 1

�:

Solution.

The problem is particularly suitable for generating functions. As is con-

vention, let [xn]f(x) denote the coe�cient of xn in f(x). Consider each kibeing contributed from a factor of x + 2x2 + 3x3 + � � � . Then the number

we seek is:

[xn] (x+ 2x2 + � � � )(x+ 2x2 + � � � ) � � � (x+ 2x2 + � � � )| {z }m factors

= [xn]xm

(1� x)2m= [xn�m ]

1

(1� x)2m

= [xn�m]

��2m� 1

2m� 1

�+

�2m

2m� 1

�x+

�2m+ 1

2m� 1

�x2 + � � �

�

=

�m+ n� 1

2m� 1

�:

360

A192. Let ABC be a triangle, such that the Fermat point F lies in the

interior. Let u = AF , v = BF , w = CF . Derive the expression

u2 + v2 + w2 � uv � uw � vw =a2 + b2 + c2

2� 2

p3K:

Solution by Wai Ling Yee, University of Waterloo, Waterloo, Ontario.

Using the law of cosines on 4BCF ,4ACF , and4ABF :

a2 = v2 + w2 � 2vw cos 120� = u2 + v2 + vw; (1)

b2 = u2 + w2 � 2uw cos 120� = u2 + v2 + uw; (2)

c2 = u2 + v2 � 2uv cos 120� = u2 + v2 + uv: (3)

Adding the areas of4BCF , 4ACF , and4ABF :

K =1

2uv sin 120� +

1

2uw sin120� +

1

2vw sin120�

=

p3

4(uv+ vw+ uw): (4)

Therefore, u2 + v2 + w2 � uv � uw � vw = a2+b2+c2

2� 2

p3K, from

12[(1) + (2)+ (3)]� 2

p3(4).

Since u2 + v2+w2� uv� uw� vw � 0 for all real u, v,w, it follows

from this problem that a2 + b2 + c2 � 4p3K.

Challenge Board Problems | Solutions

Editor: Ravi Vakil, Department of Mathematics, Princeton Univer-

sity, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA

<vakil@math.princeton.edu>

C70. Prove that the group of automorphisms of the dodecahedron

is S5, the symmetric group on �ve letters, and that the rotation group of

the dodecahedron (the subgroup of automorphisms preserving orientation)

is A5.

Solution.

Assign a number to each of the dodecahedron's edges as shown in Fig-

ure 1. (Alternatively, think of it as painting each of the edges one of �ve

di�erent colours.) Notice that if an edge on a face F is numbered n, then

the edge from the opposite vertex in F (that isn't an edge of F ) has the same

number. Thus the entire numbering can be recovered from the numbering of

the edges of a simple face. Notice also that the edges around each face are

numbered 1 through 5 (in some order).

361

1

2

3 4

5

1

2

34

5

4

5

3

2

3

1

2

34

5

2 5

4 3

1 1

3 4

5 2

Figure 1.

By observation, if

�1 2 3 4 5

a b c d e

�is an element of A5, then there

is a unique face where the edges are numbered (clockwise) a-b-c-d-e. In

this way each element

�1 2 3 4 5

a b c d e

�of A5 induces a rotation of the

dodecahedron (by sending the face 1-2-3-4-5 to the face a-b-c-d-e, with edge

1 of the �rst face sent to edge a of the second). Conversely, each rotation

induces an element of A5. This set map is clearly a group homomorphism,

so the group of rotations of the dodecahedron is A5.

Essentially the same argument shows that the group of automorphisms

of the dodecahedron (including re ections) is S5.

Comments.

1. What are the rotation and automorphism groups of the icosahedron?

How about the other platonic solids?

2. The group A5 comes up often in higher mathematics. For example,

because of a simple property of A5, there are quintic polynomials with

integer co-e�cients whose roots can't be written in terms of radicals.

(This is not true of polynomials of degree less than 5. The proof involves

Galois theory.) Surprisingly, the description ofA5 as the automorphism

group of the dodecahedron comes up in a variety of contexts.

C72. A �nite group G acts on a �nite set X transitively. (In other

words, for any x; y 2 X, there is a g 2 G with g � x = y.) Prove that there

is an element of G whose action on X has no �xed points.

362

(The problem as stated is incorrect. We need to assume that X has

more than one element.)

Solution.

Fix any element x0 of X. The number of elements of G sending x0 to

y is independent of y 2 X: if y1 and y2 are any two elements of X, and

Si = fg 2 G j g � x0 = yig; i = 1; 2;

and z is an element of G sending y1 to y2 (z exists by the transitivity of

the group action), then zS1 = S2. In other words, the elements of S2 are

obtained from the elements of S1 by multiplication on the left by z. In par-

ticular, jS1j = jS2j.Let f(g;x) = 1 if g � x = x and 0 otherwise. Then by the previous

comment, for any �xed x 2 X,Pg2G f(g;x)

jGj=

1

jXj:

The average number of �xed points (over all elements of the group) is 1:

1

jGj

Xg2G

Xx2X

f(g;x)

!=

Xx2X

�Pg2G f(g;x)

jGj

�

=Xx2X

1

jXj

= 1:

One element of the group (the identity) has more than the average (as

jXj > 1), so there is an element with less than the average number (and

hence zero) �xed points.

Comments.

1. How does this relate to problem 1 on the 1987 IMO?:

Let pn(k) be the number of permutations of the set f1; : : : ; ng, n � 1,

which have exactly k �xed points. Prove that

nXk=0

kpn(k) = n!:

(Hint: Let X be the set f1; : : : ; ng andG the group of permutations on

X, acting on X in the natural way.)

2. Can the result be salvaged if the group action is not transitive?

363

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-

land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-

companied by a solution, together with references and other insights which

are likely to be of help to the editor. When a submission is submitted with-

out a solution, the proposer must include su�cient information on why a

solution is likely. An asterisk (?) after a number indicates that a problem

was submitted without a solution.

In particular, original problems are solicited. However, other inter-

esting problems may also be acceptable provided that they are not too well

known, and references are given as to their provenance. Ordinarily, if the

originator of a problem can be located, it should not be submitted without

the originator's permission.

To facilitate their consideration, please send your proposals and so-

lutions on signed and separate standard 812"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailed

to the Editor-in-Chief, to arrive no later than 1 April 1998. They may also

be sent by email to crux-editors@cms.math.ca. (It would be appreciated if

email proposals and solutions were written in LATEX). Graphics �les should

be in epic format, or encapsulated postscript. Solutions received after the

above date will also be considered if there is su�cient time before the date

of publication.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX,

either to the Editor-in-Chief's departmental FAXmachine in St. John's, New-

foundland, or to the Canadian Mathematical Society's FAX machine in Ot-

tawa, Ontario. While we understand the reasons for solvers wishing to use

this method, we have found many problems with it. The major one is that

hand-written material is frequently transmitted very badly, and at times is

almost impossible to read clearly. We have therefore adopted the policy that

we will no longer accept submissions sent by FAX. We will, however, con-

tinue to accept submissions sent by email or regular mail. We do encourage

email. Thank you for your cooperation.

364

2263. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle, and the internal bisectors of \B, \C, meet AC,

AB at D, E, respectively. Suppose that \BDE = 30�.

Characterize 4ABC.

2264. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a right angled triangle with the right angle at A. PointsD and

E are on sides AB and AC respectively, such that DEkBC. Points F and

G are the feet of the perpendiculars from D and E to BC respectively.

Let I, I1, I2, I3 be the incentres of4ABC,4ADE,4BDF ,4CEGrespectively. Let P be the point such that I2PkI1I3, and I3PkI1I2.

Prove that the segment IP is bisected by the line BC.

2265. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

Given triangle ABC, let ABX and ACY be two variable triangles

constructed outwardly on sidesAB and AC of4ABC, such that the angles

\XAB and \Y AC are �xed and \XBA + \Y CA = 180�. Prove that all

the linesXY pass through a common point.

2266. Proposed by Waldemar Pompe, student, University of War-

saw, Poland.

BCLK is the square constructed outwardly on side BC of an acute

triangle ABC. Let CD be the altitude of 4ABC (with D on AB), and let

H be the orthocentre of4ABC. If the lines AK and CD meet at P , show

thatHP

PD=AB

CD:

2267. Proposed by Clark Kimberling, University of Evansville, Evans-

ville, IN, USA and Peter Y�, Ball State University, Muncie, IN, USA.

In the plane of 4ABC, let F be the Fermat point and F 0 its isogonalconjugate.

Prove that the circles through F 0 centred at A, B and C meet pairwise

in the vertices of an equilateral triangle having centre F .

2268. Proposed by Juan-Bosco Romero M�arquez, Universidad de

Valladolid, Valladolid, Spain.

Let x, y be real. Find all solutions of the equation

2xy

x+ y+

sx2 + y2

2=pxy +

x+ y

2:

365

2269. Proposed by Crist �obal S �anchez{Rubio, I.B. Penyagolosa,

Castell �on, Spain.

Let OABC be a given parallelogram with \AOB = � 2 (0; �=2].

A. Prove that there is a square inscribable in OABC if and only if

sin�� cos� �OA

OB� sin�+ cos�

and

sin�� cos� �OB

OA� sin�+ cos�:

B. Let the area of the inscribed square be Ss and the area of the given

parallelogram be Sp. Prove that

2Ss = tan2 ��OA2 + OB2 � 2Sp

�:

2270. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Given 4ABC with sides a, b, c, a circle, centre P and radius � inter-

sects sidesBC, CA, AB in A1 and A2, B1 and B2, C1 and C2 respectively,

so thatA1A2

a=B1B2

b=C1C2

c= � � 0:

Determine the locus of P .

2271. Proposed by F.R. Baudert, Waterkloof Ridge, South Africa.

A municipality charges householders per month for electricity used ac-

cording to the following scale:

�rst 400 units | 4.5cj per unit;

next 1100 units | 6.1cj per unit;

thereafter | 5.9cj per unit.

IfE is the total amount owing (in dollars) for n units of electricity used,

�nd a closed form expression, E(n).

2272?. Proposed by no name on proposal { please identify!

Write r - s if there is an integer k satisfying r < k < s. Find, as a

function of n (n � 2), the least positive integer k satisfying

k

n-

k

n� 1-

k

n� 2- : : : -

k

2- k:

[The proposer has not seen a proof, and has veri�ed the conjectured

solutions for 2 � n � 600.]

366

2273. Proposed by Tim Cross, King Edward's School, Birmingham,

England.

Consider the sequence of positive integers: f1, 12, 123, 1 234, 12 345,: : : g, where the next term is constructed by lengthening the previous term at

its right-hand end by appending the next positive integer. Note that this next

integer occupies only one place, with \carrying" occurring as in addition: thus

the ninth and tenth terms of the sequence are 123 456789 and 1 234 567900

respectively.

Determine which terms of the sequence are divisible by 7.

2274. Proposed by V�aclav Kone �cn �y, Ferris State University, Big

Rapids, Michigan, USA.

A. Letm be a non-negative integer. Find a closed form for

nXk=1

mYj=0

(k+j).

B. Let m 2 f1; 2; 3; 4g. Find a closed form for

nXk=1

mYj=0

(k+ j)2.

C?. Let m and �j (j = 0; 1; : : : ;m) be non-negative integers. Prove or

disprove that

nXk=1

mYj=0

(k+ j)�j is divisible by

m+1Yj=0

(n+ j).

2275. Proposed by M. Perisastry, Vizianagaram, Andhra Pradesh,

India.

Let b > 0 and ba � ba for all a > 0. Prove that b = e.

367

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased to

consider for publication new solutions or new insights on past problems.

1940. [1994: 108; 1995: 107, 205; 1996: 321; 1997: 170] Proposed

by Ji Chen, Ningbo University, China.

Show that if x; y; z > 0,

(xy + yz+ zx)

�1

(x+ y)2+

1

(y+ z)2+

1

(z + x)2

��

9

4:

Comment by Vedula N. Murty, Andhra University, Visakhapatnam, In-

dia.

In response to Marcin Kuczma's comment [1997: 170], I present the

details of my proof of the assertion:

bc(b� c)2(2a2 � bc) + ca(c� a)2(2b2� ca) > 0

when

2a2 � bc < 0 < 2b2 � ca � 2c2 � ab;

where a, b, c are the side lengths of a triangle satisfying 0 < a � b � c.

We have

a+ b > c: (1)

So, 2�a2 + b2

�� (a+ b)2 > c(a+ b) implies that

bc� 2a2

2b2 � ca< 1; and

further that

bc

ac

bc� 2a2

2b2 � ca<

bc

ac: (2)

Thusc� a

c� b� 1 implies that

(c� a)2

(c� b)2�

c� a

c� b�

b

a; and further that

(c� a)2

(c� b)2�

bc

ac>

bc

ac

�bc� 2a2

2b2 � ca

�: So we have

bc(b� c)2(2a2 � bc) + ca(c� a)2(2b2� ca) > 0:

368

2158. [1996: 218] Proposed by P. Penning, Delft, the Netherlands.

Find the smallest integer in base eight for which the square root (also

in base eight) has 10 immediately following the `decimal' point.

In base ten, the answer would be 199, withp199 = 14:10673 : : : .

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let n be a positive integer with the given property. Letm be the integer

part ofpn and n = m2 + p. The base 8 representation of a real number x

is denoted by (x)8. By hypothesis,

m+1

8= m+ (0:10)8 �

pn < m+ (0:11)8 = m+

9

64;

which is equivalent to m2 + 14m+ 1

64� n < m2 + 9

32m+ 92

642:

It is easy to see that these inequalities can be replaced by the following

stronger ones by taking into account that the expressions on both sides can-

not be integers:

m2 +m+ 1

4� n � m2 +

9m

32:

[For example, any integer greater than or equal to m2 + m4+ 1

64must be

greater than or equal to m2 + m+1

4, since m is an integer.|Ed.] So

m+ 1

4� p �

9m

32:

The lower limit has to be less than the upper limit, hence m � 8. For

8 � m � 10 the lower limit is greater than 2 and the upper limit less than

3 [so no integer value of p exists]. If m = 11 then p = 3; therefore the

smallest m is 11 and so the smallest n ism2 + p = 124 = (174)8, wherep(174)8 = (13:10531 : : : )8:

Also solved by CHARLES ASHBACHER, Hiawatha, Iowa, USA;MIGUEL

ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; THEODORE CHRONIS, student,

Aristotle University of Thessaloniki, Greece; MIHAI CIPU, InstituteofMath-

ematics, Romanian Academy, Bucharest, Romania; GEORGI DEMIREV,

Varna, and MITKO KUNCHEV, Baba Tonka School of Mathematics, Rousse,

Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gymnasium, Bamberg, Ger-

many; ROBERT GERETSCHL �AGER, Bundesrealgymnasium, Graz, Austria;

SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD

I. HESS, Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursul-

inengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic

High School, Beaverton, Oregon, USA; HEINZ-J�URGEN SEIFFERT, Berlin,

Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

The curious fact that the smallest solution for base 8 is 124 = 112 + 3

and for base 10 is 199 = 142+3 (both written in base 10) is no coincidence!

Geretschl�ager, Godin, and the proposer consider other bases as well, and

369

show that the smallest (base 10) integer whose square root when written in

base b has 10 immediately following the \decimal" point is

8>><>>:

�b� 1

2

�2+ 1 if b is odd�

3b� 2

2

�2+ 3 if b is even:

Janous lists all (base 10) positive integers < 10000 which satisfy the

condition of the problem, and gets 144 such integers, starting 124, 229, 329,

: : : . From this list one can �nd, for example, that: the �rst occurrence of

a consecutive pair is 1860; 1861 (also pointed out by Engelhaupt); the last

integer not part of a consecutive pair appears to be 3616; the �rst consecutive

triple is 5644; 5645;5646; and so on. Can anyone �nd any general patterns

here for arbitrary bases?

Tsaoussoglou reports that the smallest integer whose square root in

base 10 has 100 immediately following the decimal point is 992+20 = 9821

(p9821 = 99:100958 : : : ), and that the corresponding answer for base 8 is

712 + 18 = 5059 (p5059 = (107:100657 : : : )8). What if we want 1000

following the decimal point? Or 1 followed by n zeros?

2159. [1996: 218] Proposed by Walther Janous, Ursulinengymnas-

ium, Innsbruck, Austria.

Let the sequence fxn; n � 1g be given by

xn =1

n(1 + t+ : : :+ tn�1)

where t > 0 is an arbitrary real number.

Show that for all k, l � 1, there exists an indexm = m(k; l) such that

xk � xl � xm.

Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton,

Oregon, USA.

Since xn = 1 for all n if t = 1, any indexm will su�ce. We distinguish

two other cases.

First, let t > 1. By the AM{GM inequality,

xn =1 + t+ � � �+ tn�1

n>

np1 � t � : : : � tn�1

=n

pt1+���+(n�1) =

n

ptn(n�1)=2

= t(n�1)=2:

This last quantity increases without bound as n ! 1, so there must be an

indexm for which xkxl � xm for given indices k; l.

370

Second, let 0 < t < 1. Then

xn =1 + t+ � � �+ tn�1

n<

1 + 1 + � � �+ 1

n= 1;

so xkxl < xk and xkxl < xl, and we may choose the indexm to equal either

k or l.

Also solved by THEODORE CHRONIS, student, Aristotle University

of Thessaloniki, Greece; MIHAI CIPU, Institute of Mathematics, Romanian

Academy, Bucharest, Romania; GEORGI DEMIREV, Varna, and MITKO

KUNCHEV, Baba Tonka School ofMathematics, Rousse, Bulgaria; FLORIAN

HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos

Verdes, California, USA; MURRAY S. KLAMKIN, University of Alberta, Ed-

monton, Alberta; THOMAS C. LEONG, City College of City University of New

York, New York, USA; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and the

proposer.

Some solvers noted that since x1 = 1 for all t,m = 1 will work for all

k and l whenever t < 1. Cipu and the proposer found that m = k + l � 1

works when t > 1.

2160. [1996: 218] Proposed by Toshio Seimiya, Kawasaki, Japan.

4ABC is a triangle with\A < 90�. LetP be an interior point ofABC

such that \BAP = \ACP and \CAP = \ABP . Let M and N be the

incentres of4ABP and4ACP respectively, and letR1 be the circumradius

of 4AMN . Prove that

1

R1

=1

AB+

1

AC+

1

AP:

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let a; b; c denote the sides BC; CA;AB and �; �; the angles of the

triangle. Then

\APB = 180��\ABP�\BAP = 180��\ABP��+\CAP = 180���

and by the cosine law in4APB

c2 = AP 2 +BP 2 + 2AP � BP cos�:

4ABP and4CAP are by hypothesis similar, whence BP = c�APb

and

c2 =AP 2

b2(b2 + c2 + 2bc cos�):

By setting x2 = b2 + c2 + 2bc cos�

AP =bc

xand BP =

c � APb

=c2

x:

371

Let X be the point where the incircle of 4APB touches AP . Then

PX =AP + BP � AB

2=

c � (b+ c� x)

2x

=c � ((b+ c)2 � x2)

2x � (b+ c+ x)

=2bc2 sin2 �

2

x � (b+ c+ x)

and since \XPM = 12\APB = 90� � �

2we get

PM =PX

sin �2

=2bc2 sin �

2

x � (b+ c+ x):

Since 4NPM and4CPA are similar,MN = b�PMAP

, and by using the fact

that \MAN = �2, we have

1

R1

=2 sin �

2

MN

=2 � bc

xsin �

2

b � 2bc2 sin �

2

x�(b+c+x)

=b+ c+ x

bc

=1

AB+

1

AC+

1

AP:

Also solved by MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain;

RICHARD I.HESS, Rancho PalosVerdes, California, USA;WALTHER JANOUS,

Ursulinengymnasium, Innsbruck, Austria; MAR�IA ASCENSI �ON L �OPEZ

CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; D.J. SMEENK, Zalt-

bommel, the Netherlands; and the proposer.

2161. [1996: 219] Proposed by Juan-Bosco Romero M�arquez, Uni-

versidad de Valladolid, Valladolid, Spain.

Evaluate1Xn=1

1

(2n� 1)(3n� 1):

Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA

(modi�ed slightly be the editor).

Let S denote the given summation. We show that

S = 2 ln2�3

2ln3 +

p3�

6� 0:64527561:

372

Using partial fractions, we have

S =

1Xn=1

�2

2n� 1�

3

3n� 1

�

=

1Xn=1

�2

2n� 1�

2

2n

�+

1Xn=1

�1

n�

3

3n� 1

�

= 2

1Xn=1

(�1)n�11

n� 3

1Xn=1

�1

3n� 1�

1

3n

�

= 2 ln2� 3

1Xn=1

�1

3n� 1�

1

3n

�: (1)

Since the sequence of functions ffn(x)g, where fn(x) = x3n+1 � x3n+2, is

uniformly convergent on [0; 1],

1Xn=1

�1

3n� 1�

1

3n

�=

�1

2�

1

3

�+

�1

5�

1

6

�+

�1

8�

1

9

�+ � � �

=

Z 1

0

(x� x2 + x4 � x5 + x7 � x8 + � � � ) dx

=

Z 1

0

x(1� x)(1 + x3 + x6 + � � � )dx

=

Z 1

0

x(1� x)

1� x3dx =

Z 1

0

x dx

1 + x+ x2

=

Z 1

0

x+ 12

1 + x+ x2dx �

1

2

Z 1

0

1�x+ 1

2

�2+ 3

4

dx

=

�1

2ln(1 + x+ x2)�

1

2�2p3tan�1

�2x+ 1p3

������1

0

=1

2ln3�

p3

3

�tan�1

p3� tan�1

1p3

�

=1

2ln3�

p3

3��

6: (2)

Substituting (2) into (1), we �nd S = 2 ln2� 32ln 3 +

p3�6

, as claimed.

Also solved by THEODORE CHRONIS, student, Aristotle University

of Thessaloniki, Greece; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO

KUNCHEV, Baba Tonka School of Mathematics, Rousse, Bulgaria (jointly);

RICHARD I.HESS, Rancho PalosVerdes, California, USA;WALTHER JANOUS,

Ursulinengymnasium, Innsbruck, Austria; D. KIPP JOHNSON, Valley Catholic

High School, Beaverton, OR, USA; MURRAY S. KLAMKIN, University of Al-

berta, Edmonton, Alberta; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; and

the proposer. Two incorrect solutions were also received.

373

About half of the solvers used the non{elementary approach of consid-

ering the Euler psi-function (x) =�0(x)

�(x)where �(x) denotes the gamma

function. Both Hess and Sei�ert pointed out that the evaluation of the more

general sumP1n=1

1(n+p)(n+f)

(p > �1, f > �1, p 6= f) can be found in

Tables of Integrals, Series, and Products by I.S. Gradsteyn and I.M. Ryzhik

(Academic Press, 1994, 5th edition). Speci�cally, the following formulas can

be found on p. 952 and p. 954, respectively:

(x) = � �1Xn=1

�1

n� 1 + x�

1

n

�; x > 0;

�1

2

�= � � 2 ln2;

�2

3

�= � +

p3�

6�

3

2ln3:

(Here, denotes Euler's constant.)

Since S =P1n=1

�1

n� 1

2

� 1

n� 1

3

�, one obtains easily from the above

formulas that S = �23

��

�12

�= 2 ln2� 3

2ln3 +

p3�6

as in the solution

above.

2162. [1996: 219] Proposed by D.J. Smeenk, Zaltbommel, the Neth-

erlands.

In4ABC, the Cevian lines AD, BE, and CF concur at P . [XY Z] is

the area of4XY Z. Show that

[DEF ]

2[ABC]=PD

PA�PE

PB�PF

PC

I. Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain

(slightly shortened by the editor).

It is a standard exercise (for example, #13.5.6 in H.S.M. Coxeter, In-

troduction to Geometry) to show that

[DEF ]

[ABC]=

��� + 1

(�+ 1)(�+ 1)(� + 1)

with � =AF

FB, � =

BD

DC, � =

CE

EA.

Since in our case, according to Ceva's Theorem, ��� = 1, all we have

to prove is that

PA

PD�PB

PE�PC

PF= (�+ 1)(�+ 1)(� + 1):

374

But it is true because

PA

PD=

[PCA]

[PDC];

PB

PE=

[PAB]

[PEA];

PC

PF=

[PBC]

[PFB];

�+ 1 =AF

FB+ 1 =

[PAF ]

[PFB]+ 1 =

[PAB]

[PFB];

�+ 1 =BD

DC+ 1 =

[PBD]

[PDC]+ 1 =

[PBC]

[PDC];

and

� + 1 =CE

EA+ 1 =

[PCE]

[PEA]+ 1 =

[PCA]

[PEA]:

II. Comment by Murray S. Klamkin, University of Alberta, Edmonton,

Alberta.

The corresponding more general problem for a simplex is given in [2]

and also with a slightly modi�ed proof in [3]. It is shown that if V0, V1, : : : ,

Vn denote the n+1 vertices of a simplex S inn{dimensional Euclidean space

and if V 00 , V

01 , : : : , V

0n denote the n + 1 vertices of an inscribed simplex S0

such that the cevians ViV0i are concurrent at a point P within S, then

VOL S0

VOL S=

n�0�1 : : : �n

(1� �0)(1� �1) : : : (1� �n):

Here the barycentric representation for P is P = �0V0+�1V1+ : : :+�nVnwhere �0 + �1 + : : :+ �n = 1 and �i � 0. It is also shown that the volume

of S0 is a maximum if P is the centroid.

Also PV 0=PV = �i=(1� �i). For this and other metric properties of

concurrent cevians of a simplex see [1987: 274{275].

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-

ladolid, Spain; MIGUEL ANGEL CABEZ �ON OCHOA, Logro ~no, Spain; GEORGI

DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba Tonka School of

Mathematics, Rousse, Bulgaria; HANS ENGELHAUPT, Franz{Ludwig{Gym-

nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,

Austria; RICHARD I.HESS, Rancho PalosVerdes, California,USA; WALTHER

JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S. KLAMKIN,

University of Alberta, Edmonton, Alberta; TOSHIO SEIMIYA, Kawasaki,

Japan; and the proposer.

Janous refers to [4, p. 342, item 1.3] for a closely related identity.

Bellot quotes Victor Th �ebault who reported an 1881 reference for the

exercise that began our featured solution. This exercise is often attributed

375

to Routh (1891). For those who read Bulgarian, he also provides a reference

concerning interesting properties of a cevian triangle DEF : Hristo Lesov,

Projections of noteworthy points of the triangle following the respective ce-

vians (in Bulgarian). Matematyka & Informatyka, 5 (1994) 42{49. He fur-

ther mentions [1, p. 78] where a problem of Langendonck is solved: Given

a4ABC, �nd the probability of selecting a point P inside the triangle such

that it is possible to form a triangle whose sides equal the respective dis-

tances from P to the sides of the 4ABC. The probability turns out to be

[DEF ]=[ABC].

[1] Heinrich D�orrie, Mathematische Miniaturen, F. Hirt, Breslau, 1943.

[2] M.S. Klamkin, A volume inequality for simplexes, Univ. Beograd Publ.

Elektrotehn. Fak. Ser. Mat. Fiz. No. 357{380 (1971) 3{4.

[3] M.S. Klamkin, International Mathematical Olympiads 1978{1985,

Math. Assoc. Amer., Washington, D.C., 1986, pp. 87{88.

[4] D.S. Mitrinovi�c et. al., Recent Advances in Geometric Inequalities,

Kluwer Academic Publishers, 1989.

2163. [1996: 219] Proposed by Theodore Chronis, student, Aristotle

University of Thessaloniki, Greece.

Prove that if n;m 2 N and n � m2 � 16, then 2n � nm.

I. Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let x =pn � m � 4. Then we �rst prove that x2 � 2x. This inequality is

equivalent tox

lnx�

2

ln2:

De�ne f(x) as x=(lnx). Then

f 0(x) =lnx� 1

(lnx)2

which is positive for all x � 4. Moreover f(x) is continuous and di�er-

entiable in the interval [4;1) and f(4) = f(2). Hence x2 � 2x, and so,

nm � (x2)x � (2x)x = 2n.

II. Solution by D. Kipp Johnson, Valley Catholic High School, Beaver-

ton, Oregon.

We �rst show that the inequality is true for anym � 4with the smallest

allowable n, namely n = m2, and then we induct on n.

For the �rst step we letm � 4 and n = m2 and consider the following

equivalent statements:

2m2

� (m2)m () 2m2

� 22m log2m

() m2 � 2m log2m

() m=2 � log2m:

376

We establish this last statement by noticing that, for m � 4, we have

1

m� 1� 21=2 � 1

1 +1

m� 1� 21=2

log2

�1 +

1

m� 1

��

1

2

log2m� log2(m� 1) �1

2:

This permits us to write the followingm� 4 inequalities:

log2 5� log2 4 �1

2

log2 6� log2 5 �1

2

.

.

.

log2(m)� log2(m� 1) �1

2;

which, when added, telescope to produce the desired log2m � m=2.

For the second step we induct on n. We shall assume that for some

n � m2 � 16 we have 2n � nm. Multiplying each side by 2 we obtain

2n+1 � 2 � nm. It will su�ce to show that 2 � nm � (n + 1)m, which is

equivalent to 2 � (1 + 1=n)m. The following sequence of inequalities does

the trick:

2 �pe �

s�1 +

1

m2

�m2

��1 +

1

m2

�m��1 +

1

n

�m;

and the proof is �nished.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,

Bosnia and Herzegovina; MIHAI CIPU, Romanian Academy, Bucharest, Ro-

mania; GEORGI DEMIZEV, Varna, Bulgaria, and MITKO KUNCHEV, Baba

Tonka School of Mathematics, Rousse, Bulgaria; CHARLES R. DIMINNIE,

Angelo State University, San Angelo, TX, USA; RUSSELL EULER and JAWAD

SADEK, NWMissouri State University,Maryville,Missouri; SHAWNGODIN,

St. Joseph Scollard Hall, North Bay, Ontario; FABIANMARTIN HERCE, stu-

dent, Universidad de La Rioja, Logro ~no, Spain; RICHARD I. HESS, Ran-

cho Palos Verdes, California, USA; JOE HOWARD, New Mexico Highlands

University, Las Vegas, NM, USA; WALTHER JANOUS, Ursulinengymnasium,

Innsbruck, Austria; MURRAY S. KLAMKIN, University of Alberta, Edmon-

ton, Alberta; ROBERT P. SEALY, Mount Allison University, Sackville, New

Brunswick; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; DIGBY SMITH,

377

Mount Royal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Lau-

rier University, Waterloo, Ontario; and the proposer. There was 1 incom-

plete solution.

Hess actually proves a somewhat stronger result by replacing the con-

dition n � m2 � 16 by the 2 conditions n � 16 and n � m2.

2164. [1996: 273] Proposed by Toshio Seimiya, Kawasaki, Japan.

Let D be a point on the side BC of triangle ABC, and let E and F be

the incentres of triangles ABD and ACD respectively. Suppose that B, C,

E, F are concyclic. Prove that

AD +BD

AD+ CD=AB

AC:

Solution by the Con Amore Problem Group, the Royal Danish School

of Educational Studies and Informatics, Copenhagen, Denmark.

The intersection points of the line EF with AB;AC, and AD will be

called G;H and K, respectively, and we put \B = 2y, and \C = 2z, so

\GBE = \EBD = y, and \DCF = \FCH = z. Since B;C;E; F are

concyclic, \GEB = \FCD = z, and \CFH = \EBD = y; and so

\AGH = y + z = \AHG

giving

AG = AH: (1)

Since AE bisects \A in 4GAK, and AF bisects \A in 4KAH, we have

[using (1)]: KEEG

= KAAG

= KAAH

= KFFH

, whence KEEG

+ 1 = KFFH

+ 1, or

KG

EG=

KH

FH: (2)

In4ABD let the inradius, the altitude from D, and the area be r1, h1 and

T1, respectively; and in 4ACD let the corresponding quantities be r2; h2and T2. In 4AKG, and 4AKH let the altitude from K be k1 and k2,

respectively. Then h1h2

= k1k2, whence

h1

k1=h2

k2(3)

and [using (2)]

k1

r1=KG

EG=KH

FH=k2

r2; (4)

and further, (3) - (4) imply

h1

r1=h2

r2: (5)

378

Now

h1 � AB = 2T1 = r1 � (AB + BD +DA);

so

h1

r1=AB +BD +DA

AB= 1+

BD +DA

AB; (6)

and similarly,

h2

r2=AC + CD +DA

AC= 1 +

CD+DA

AC; (7)

From (5) - (7) it easily follows that

AD +BD

AD+ CD=AB

AC:

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-

ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;

RICHARD I. HESS, Rancho Palos Verdes, California, USA; CRIST �OBAL

S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain; D.J. SMEENK, Zalt-

bommel, the Netherlands; and the proposer.

Bellot Rosado notes that this problem is a special case of problem 1206

[H. Demir, Mathematics Magazine, 58, no. 1, January 1985; solution by

V. D. Mascioni, Mathematics Magazine, 59, no. 1, February 1986]

2165. [1996: 273] Proposed by Hoe Teck Wee, student, Hwa Chong

Junior College, Singapore.

Given a triangle ABC, prove that there exists a unique pair of points

P and Q such that the triangles ABC, PQC and PBQ are directly similar;

that is, \ABC = \PQC = \PBQ and \BAC = \QPC = \BPQ,and the three similar triangles have the same orientation. Find a Euclidean

construction for the points P and Q.

Solution by the Con Amore Problem Group, the Royal Danish School

of Educational Studies and Informatics, Copenhagen, Denmark.

We identify the Euclidean plane with the complex plane, and points

with complex numbers. We may assume without loss of generality that

B = 0: (1)

Suppose such a P and Q exist. Since 4ABC � 4PQC, there is a

point (complex number) X such that

P = XA; (2)

and

Q = XC; (3)

379

and besides:

P � Q

C �Q=A

C: (4)

Moreover, if (2) - (4) hold, then P; Q are as desired. Now (2) - (4) imply that

XA�XC

C �XC=A

C;

or

XA �XC = A�XA;

or

X =A

2A� C; (5)

and then, by (2) - (3):

P =A2

2A�C(6)

and

Q =AC

2A� C: (7)

On the other hand, if X;P;Q satisfy (5) - (7) then (2) - (4) hold. This proves

the existence and uniqueness of P andQ with the desired properties. To �nd

a Euclidean construction for P and Q, de�ne

R = 2A� C;

and note that

A =R+ C

2;

that is, R is the image of C by re ection in A. Also note that

Q

C=A

R;

that is, triangles ABR and QBC are similar, and similarly oriented. Finally

A+ Q

2=

1

2

�AC

2A� C+ A

�=

A2

2A� C= P

so P is the midpoint of AQ. All this gives us the following construction:

(A) Re ect C in A to get R.

(B) Construct Q such that A and Q are on opposite sides of the line

BC, \CBQ = \RBA, and \QCB = \ARB.

380

(C) Construct P as the midpoint of AQ.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,

Rancho Palos Verdes, California, USA; GOTTFRIED PERZ, Pestalozzigymna-

sium, Graz, Austria (two solutions); TOSHIO SEIMIYA, Kawasaki, Japan;

and the proposer.

2166. [1996: 273] Proposed by K.R.S. Sastry, Dodballapur, India.

In a right-angled triangle, establish the existence of a unique interior

point with the property that the line through the point perpendicular to any

side cuts o� a triangle of the same area.

Solution by Toshio Seimiya, Kawasaki, Japan.

Let the right-angled triangle be ABC, with right angle at A. We may

assume without loss of generality that AB � AC. P is an interior point of

4ABC. The line through P perpendicular to BC meets BC;AC at D;E

respectively (ifAB = AC, thenE = A); the line throughP perpendicular to

CAmeets BC;CA at F;G respectively, and the line through P perpendicu-

lar toAB meets BC;AB atH;K respectively. We assume that the triangles

BHK;CDE;CFG have the same area. Since these triangles are similar,

they must be congruent. Since 4CDE � 4CGF , we have CD = CG, so

that 4CDP � 4CGP . Thus we have \PCD = \PCG. Therefore, CP is

the bisector of \C.

Let M be the intersection of AP with BC (if AB = AC; M = D).

Since 4BHK � 4FCG, we have BH = FC, so that

BF = HC: (1)

Since PFkAB and PHkAC, we get

BF : BM = AP : AM = CH : CM (2)

From (1) and (2) we have BM = CM . Therefore, AM is the median of

4ABC. Thus P must be the intersection of the median AM with the bi-

sector of \C. It is interesting to note that the bisector of \B does not pass

through P if AB 6= AC.

Conversely, let P be the intersection of the median AM with the bi-

sector of \C. Draw the linesDE;FG;HK through P perpendicular to BC,

CA, AB respectively, as shown in the �gure. Since \PCD = \PCG, we

have 4PCD � 4PCG, so that CD = CG. As 4CFG is 4CED we

get 4CFG � CED. Since PFkAB and PHkAC, we have BF : BM =

AP : AM = CH : CM . As BM = CM , we get BF = CH, so that

BH = FC. Because 4KBH is 4GFC, we have 4KBH � 4GFC.

Since 4CDE;4CGF;4HKB are congruent, they have the same area.

A number of solvers used a coordinate approach. If A = (0; 0),

B = (0; c), C = (b;0) and BC = a, then P =�

b2

2b+a; bc2b+a

�.

381

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,

Spain; SAMBAETHGE, Nordheim, Texas, USA; CHRISTOPHER J. BRADLEY,

Clifton College, Bristol, UK; CON AMORE PROBLEM GROUP, Royal Dan-

ish School of Educational Studies, Copenhagen, Denmark; C. DIXON, Royal

Grammar School, Newcastle upon Tyne, England; DAVID DOSTER, Choate

Rosemary Hall, Wallingford, Connecticut,USA; HANS ENGELHAUPT, Franz{

Ludwig{Gymnasium, Bamberg, Germany; FLORIAN HERZIG, student, Per-

chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,

USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAV

KONE �CN �Y, Ferris State University, BigRapids, Michigan,USA; D.J. SMEENK,

Zaltbommel, the Netherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece;

and the proposer. There were two incorrect solutions.

The proposer comments: \I am unable to establish the existence of such

a point in a general triangle. If we replace \perpendicular" by \parallel" then

the problem is easy and yields the unique point centroid of a triangle."

2167. [1996: 274] Proposed by �Sefket Arslanagi �c, Berlin, Germany.

Prove, without the aid of the di�erential calculus, the inequality, that

in a right triangle

a2(b+ c) + b2(a+ c)

abc� 2 +

p2;

where a and b are the legs and c the hypotenuse of the triangle.

Solution by Mihai Cipu, Institute of Mathematics, Romania Academy,

Bucharest, Romania, and CICMA, Concordia University, Montreal Quebec.

Since a2(b+c)+b2(a+c) = c(a2+b2)+ab(a+b) the given inequality

is equivalent to

c(a� b)2 � (p2c� a� b)ab =

ab(a� b)2p2c+ a+ b

:

In an isosceles triangle this relation holds trivially, while in the general

case it is equivalent to (p2c+a+b)c � ab, a simple consequence of c2 � ab.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,

UK; CON AMORE PROBLEM GROUP, Royal Danish School of Educational

Studies, Copenhagen, Denmark; HANS ENGELHAUPT, Franz{Ludwig{Gym-

nasium, Bamberg, Germany; FLORIAN HERZIG, student, Perchtoldsdorf,

Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; JOHN

G. HEUVER, Grande Prairie CompositeHigh School, Grande Prairie, Alberta;

WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;

MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV

KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAI

LAU, Hong Kong; VEDULA N. MURTY, Andhra University, Visakhapatnam,

India; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB

382

PRIELIPP and JOHN OMAN, University of Wisconsin{Oshkosh, Wisconsin,

USA; JUAN-BOSCO ROMERO M�ARQUEZ, Universidad de Valladolid, Val-

ladolid, Spain; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; TOSHIO

SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands;

PANOS E. TSAOUSSOGLOU, Athens, Greece; EDWARD T.H. WANG, Wilfrid

Laurier University, Waterloo, Ontario; and the proposer. There were four

incorrect solutions.

From the proof above it is obvious that equality holds if and only if

a = b. Both Kone�cn �y and Wang pointed out that a weaker version of this

problem with 2 +p2 replaced by � appeared as problem No. 238 in the

January, 1983 issue of the College Mathematics Journal. Wang remarked

that in the published solution (CMJ 15 (4), 1984; 352{353)Walter Blumberg

and Marshall Fraser independently proved the more general result that

a2(b+ c) + b2(a+ c)

abc� 2 + csc(C=2)

where a, b, and c are the sides of an arbitrary triangle with C being the

largest angle. The present problem is the special case when C = �=2.

Klamkin gave a proof for this more general result. Romero M�arquez asked

whether the stronger inequality

a2(b+ c) + b2(a+ c)

abc�

p2c

a+ b� c

holds. The triangle with a = 3, b = 4, and c = 5 provides a simple coun-

terexample. Janous remarked that the given inequality could be considered

as the special case (when n = 1) of the problem of �nding the minimum

value of

Tn =a2n(bn + cn) + b2n(an + cn)

(abc)n

where a, b, c are the legs and hypotenuse of a right triangle and n > 0 is

a real number. He proposed the conjecture that Tn � 2 + 2(2�n)=2 for all

n � n0 where n0 is the solution of the equation 2(n+4)=2 + n = 2.

2168?. [1996: 274]Proposed by Jan Ciach, Ostrowiec �Swi�etokrzyski,Poland.

Let P be a point inside a regular tetrahedron ABCD, with circumra-

dius R and let R1; R2; R3; R4 denote the distances of P from vertices of the

tetrahedron. Prove or disprove that

R1R2R3R4 �4

3R4;

and that the maximum value of R1R2R3R4 is attained.

383

Solution by Murray S. Klamkin, University of Alberta, Edmonton, Al-

berta. [Ed: First, we give Klamkin's proof for a triangle, which was not

published as a solution to 2073? [1995: 277; 1996: 282]]

Let P be given by the vector ~P = x1 ~A1 + x2 ~A2 + x3 ~A3, where

x1 + x2 + x3 = 1 and x1, x2, x3 � 0 (barycentric coordinates). Then

PA21 =

���~P � ~A1

���2=

���x2 � ~A2 � ~A1

�+ x3

�~A3 � ~A1

����2= 3

�x22 + x2x3 + x23

�;

etc. We have taken R = 1, so that the side of the triangle��� ~A2 � ~A1

��� = p3.

The given inequality now takes the form

�x22 + x2x3 + x23

� �x23 + x3x1 + x21

� �x21 + x1x2 + x22

��

3

64;

or in homogeneous form

64�x22 + x2x3 + x23

� �x23 + x3x1 + x21

� �x21 + x1x2 + x22

�� 3 (x1 + x2 + x3)

6(1)

where (only) x1, x2, x3 � 0.

We assume without loss of generality that x1 � x2 � x3 � 0 so that

we can let x3 = c, x2 = b + c and x1 = a + b + c with a, b, c � 0 and

a+ b+ c > 0. On substituting back into (1), we get

3a6 + 36a5b+ 54a5c+ 116a4b2 + 348a4bc+ 213a4c2 + 160a3b3

+816a3b2c+ 1128a3bc2 + 468a3c3 + 80a2b4 + 864a2b3c

+2232a2b2c2 + 2232a2bc3 + 765a2c4 + 480ab4c+ 2208ab3c2

+3888ab2c3 + 3060abc4 + 918ac5 + 192b5c+ 1104b4c2

+2592b3c3 + 3060b2c4 + 1836bc5 + 459c6 � 0

since all terms on the left side are non-negative (this expansion was con-

�rmed usingMathematica [Ed: and usingDERIVE]). There is equality only ifa = c = 0, so that if x1 +x2 +x3 = 1, x1 = x2 = 0:5 and x3 = 0, then the

point P must be a mid-point of a side.

For a regular tetrahedron, we proceed as before: let P be given by the

vector ~P = x1 ~A1 + x2 ~A2 + x3 ~A3 + x4 ~A4, where x1 + x2 + x3 + x4 = 1

and x1, x2, x3, x4 � 0. Then

PA1 =���~P � ~A1

���=

���x2 � ~A2 � ~A1

�+ x3

�~A3 � ~A1

�+ x4

�~A4 � ~A1

���� ;

384

so that

PA21 = a2

�x22 + x23 + x24 + x3x4 + x4x2 + x2x3

�;

etc., where a is an edge of the tetrahedron. So if we take R = 1, we have

a2 = 83. the desired inequality then becomes (in homogeneous form)

9 (x1 + x2 + x3 + x4)8 � 28F1F2F3F4; (2)

where

F1 = x22 + x23 + x24 + x3x4 + x4x2 + x2x3;

and the other Fi's are obtained by cyclic interchange of the indices 1, 2, 3, 4.

Assuming without loss of generality that x1 � x2 � x3 � x4, we

take x4 = d, x3 = d + c, x2 = d + c + b, x1 = d + c + b + a, where

a, b, c, d � 0 and x1 > 0. On substituting back into (2) and expanding

out (using Mathematica), we get a polynomial [Ed: we have omitted this

polynomial, since it would cover two whole pages of CRUX with MAYHEM]

which is non-negative since all the terms are non-negative. There is equality

only if a = c = d = 0. Hence x1 = x2 = 0:5 and so P must be the

mid-point of an edge.

I also conjecture for the correponding result for a regular n{dimensional

simplex:

the product of the distances from a point within or on a regular simplex

to the vertices is a maximum only when the point is a mid-point of an edge.

Analogously to inequality (2), the conjecture is that

3n�1 (x0 + x1 + : : :+ xn)2n+2 � 22n+2F0F1 : : : Fn

(x0, x1, : : : , xn � 0), where Fi is the complete symmetric homogeneous

polynomial of degree 2 with unit coe�cients of all the variables x0, x1, : : : ,

xn, except xi, and there is equality only if two of the variables are equal and

the rest are zero.

Also solved by CON AMORE PROBLEM GROUP, Royal Danish School

of Educational Studies, Copenhagen, Denmark; and RICHARD I. HESS, Ran-

cho Palos Verdes, California, USA.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

385

THE ACADEMY CORNERNo. 14

Bruce Shawyer

All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Christopher Small writes:

I notice that in the Hints { 2 section of the Bernoulli Trials,the statement that the hint does not work for question 9 has beeninterpolated.

It seems to me that the statement is correct as I originally gaveit to you. A counterexample is easily found with a = c. Forexample, a = c = 1 and b = 1=2 is a counterexample.

This month, we present solutions to some of the problems in a univer-

sity entrance scholarship examination paper from the 1940's, which appeared

in the April 1997 issue of CRUX with MAYHEM.

1. Find all the square roots of

1� x+p22x� 15� 8x2:

Solution.

First we note that the only way to generate the given expression as a

square, is to square pax+ b+

pcx + d:

There are, of course, two square roots, being � this quantity. So we have

�pax+ b+

pcx + d

�2= (a+ c)x+ (b+ d) + 2

pacx2 + (ad+ bc)x+ bd

= �x+ 1+ 2

r�2x2 +

11

2x� 15

4:

386

Since this is an identity, we equate coe�cients, yielding

a+ c = �1; (1)

b+ d = 1; (2)

ac = �2; (3)

ad+ bc = 112; (4)

bd = �15

4: (5)

Solving (1) and (2) for c and d respectively, and substituting in (3) and (5)

gives

a2 + a� 2 = (a+ 2)(a� 1) = 0;

b2 � b� 154

= (b� 52)(b+ 3

2) = 0:

so that a = �2; 1 and b = 52;�3

2, giving the corresponding values of

c = 1;�2 and d = �32; 52.

We must now substitute these into (4), and this leads to the solution that the

square roots of the given expression are

��q

�2x+ 52+

qx� 3

2

�:

2. Find all the solutions of the equations:

x+ y + z = 2;

x2 + y2 + z2 = 14;

xyz = �6:

Solution.First we recall the expressions involving the roots of a cubic:

(P �x)(P � y)(P � z) = P 3� (x+ y+ z)P 2+ (xy+ yz+ zx)P �xyz:

We also note that

(x+ y+ z)2 � (x2 + y2 + z2)

2= xy + yz + zx:

So, from two of the given three expressions, we get

xy + yz+ zx = �5:

Thus, the solution of the given equations is the set of roots of the cubic

equation

P 3 � 2P 2 � 5P + 6 = 0:

387

It is easy to check that P = 1 is a root: so it is easy to factor into

(P � 1)(P + 2)(P � 3) = 0;

giving that the other two roots are P = �2 and P = 3.

Thus, the solution set of the given equations is any permutation of 1;�2; 3.

3. Suppose that n is a positive integer and that Ck is the coe�cient of xk in

the expansion of (1 + x)n. Show that

nXk=0

(k+ 1)C2k =

(n+ 2) (2n� 1)!

n! (n� 1)!:

Solution.

Consider

nX

k=0

(k+ 1)C2k

!xn. Since Cn�k = Ck, we can write this as

nXk=0

�(k+ 1)Ckx

k� �Cn�kx

n�k�:

This is the term in xk in the product nX

k=0

(k+ 1)Ckxk

! nX

k=0

Ckxk

!:

The right member of this product is (1 + x)n. There are several ways to

determine the left member: for example,

x(1 + x)n =

nXk=0

Ckxk+1;

so that, on di�erentiating, we have that

(1 + x)n + nx(1 + x)n�1 =

nXk=0

(k+ 1)Ckxk:

Thus, the product above is

(1 + x)n�1(1 + (n+ 1)x)� (1 + x)n = (1 + (n+ 1)x)(1 + x)2n�1:

The coe�cient of xn in this product is

(2n� 1)!

(n� 1)!n!(n+ 1) +

(2n� 1)!

n! (n� 1)!=

(n+ 2) (2n� 1)!

n! (n� 1)!:

388

THE OLYMPIAD CORNERNo. 185

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

To begin this number, we give the ten problems of the Seventh Irish

Mathematical Olympiad written May 7, 1994. My thanks go to Richard

Nowakowski for collecting the problems when he was Canadian Team leader

at the IMO in Hong Kong.

IRISH MATHEMATICAL OLYMPIAD 1994Second Paper | 7 May 1994

Time: 3 hours

1. Let x, y be positive integers with y > 3 and

x2 + y4 = 2�(x� 6)2 + (y+ 1)2

�:

Prove that x2 + y4 = 1994.

2. Let A, B, C be three collinear points with B between A and C.

Equilateral triangles ABD, BCE, CAF are constructed with D, E on one

side of the line AC and F on the opposite side. Prove that the centroids

of the triangles are the vertices of an equilateral triangle. Prove that the

centroid of this triangle lies on the line AC.

3. Determine with proof all real polynomials f(x) satisfying the equa-

tion

f(x2) = f(x)f(x� 1):

4. Consider the set of m� n matrices with every entry either 0 or 1.

Determine the number of such matrices with the property that the number

of \1"s in each row and in each column is even.

5. Let f(n) be de�ned on the set of positive integers by the rules

f(1) = 2 and f(n+ 1) = (f(n))2 � f(n)+ 1; n = 1; 2; 3; : : : :

Prove that for all integers n > 1

1� 1

22n�1

<1

f(1)+

1

f(2)+ � � �+ 1

f(n)< 1� 1

22n:

389

6. A sequence xn is de�ned by the rules

x1 = 2

and

nxn = 2(2n� 1)xn�1; n = 2; 3; : : : :

Prove that xn is an integer for every positive integer n.

7. Let p, q, r be distinct real numbers which satisfy the equations

q = p(4� p)

r = q(4� q)

p = r(4� r):

Find all possible values of p+ q+ r.

8. Prove that for every integer n > 1

n((n+ 1)2=n � 1) <

nXi=1

2k+ 1

i2< n

�1� n�2=(n�1)

�+ 4:

9. Let w, a, b, c be distinct real numbers with the property that there

exist real numbers x, y, z for which the following equations hold:

x+ y+ z = 1

xa2 + yb2 + zc2 = w2

xa3 + yb3 + zc3 = w3

xa4 + yb4 + zc4 = w4:

Express w in terms of a, b, c.

10. If a square is partitioned into n convex polygons, determine the

maximum number of edges present in the resulting �gure.

[You may �nd it helpful to use a theorem of Euler which states that if a

polygon is partitioned into n polygons, then v � e + n = 1 where v is the

number of vertices and e is the number of edges in the resulting �gure].

Next we turn to the \o�cial" results of the 38th IMOwhich was written

in Mar del Plato, Argentina, July 24 and 25, 1997. My source this year was

the contest WEB site. I hope that I have made no serious errors in compiling

the results and transcribing names.

This year a total of 460 students from 82 countries took part. This is

somewhat up from last year. Sixty-�ve countries sent teams of 6 (the number

invited to participate in recent years). But there were 13 teams of smaller

size, 3 of �vemembers, 2 of size four, 7 of size three, and 1 with two members.

390

The contest is o�cially an individual competition and the six problems

were assigned equal weights of seven marks each (the same as the last 16

IMO for a maximum possible individual score of 42 and a total possible of

252 for a national team of six students). For comparison see the last 16 IMO

reports in [1981: 220], [1982: 223], [1983: 205], [1984: 249], [1985: 202],

[1986: 169], [1987: 207], [1988: 193], [1989: 193], [1990: 193], [1991:

257], [1992: 263], [1993: 256], [1994: 243], [1995: 267] and [1996: 301].

There were 4 perfect scores. The jury awarded �rst prize (Gold) to the

thirty-nine students who scored 35 or more. Second (Silver) prizes went to

the seventy students with scores from 25 to 34, and third (Bronze) prizes

went to the one hundred and twenty-two students with scores from 15 to

24. Any student who did not receive a medal, but who scored full marks

on at least one problem, was awarded honourable mention. This year there

were seventy-eight honourable mentions awarded. The median score on the

examination was 14.

Congratulations to the Gold Medalists, and especially to Ciprian

Manolescu who scored a perfect paper last year as well.

Name Country Score

Eftekhari, Eaman Iran 42

Manolescu, Ciprian Romania 42

Bosley, Carleton United States of America 42

Do Quoc Anh Vietnam 42

Frenkel, P �eter Hungary 41

Pap, Gyula Hungary 41

Ivanov, Ivan Bulgaria 40

Terpai, Tam�as Hungary 40

Maruoka, Tetsuyuki Japan 40

Curtis, Nathan United States of America 40

Najnudel, Joseph France 39

Dourov, Nikolai Russia 39

Rullg�ard, Hans Sweden 39

Gyrya, Pavlo Ukraine 39

Lam, Thomas Australia 38

Han, Jia Rui China 38

Ni, Yi China 38

Zou, Jin China 38

Woo, Jee Chul Republic of Korea 38

Hornet, Stefan Laurentiu Romania 38

Summers, Bennet United Kingdom 38

Tchalkov, Rayko Bulgaria 37

An, Jin Peng China 37

Sun, Xiao Ming China 37

Salmasian, Hadi Iran 37

Leptchinski, Mikhail Russia 37

Tcherepanov, Evgueni Russia 37

391

Name Country Score

Herzig, Florian Austria 36

Lippner, G �abor Hungary 36

Bahramgiri, Mohsen Iran 36

Battulga, Ulziibat Mongolia 36

Farrar, Stephen Australia 35

Zheng, Chang Jin China 35

Podbrdsk �y, Pavel Czech Republic 35

Holschbach, Armin Germany 35

Bayati, Mohsen Iran 35

Merry, Bruce Republic of South Africa 35

Grechuk, Bogdan Ukraine 35

Kabluchko, Zajhar Ukraine 35

Next we give the problems from this year's IMO Competition. Solu-

tions to these problems, along with those of the 1997 USA Mathematical

Olympiad will appear in a booklet entitled Mathematical Olympiads 1997which may be obtained for a small charge from: Dr. W.E. Mientka, Executive

Director, MAA Committee on HS Contests, 917 Oldfather Hall, University of

Nebraska, Lincoln, Nebraska, 68588, USA.

38th INTERNATIONALMATHEMATICAL OLYMPIADJuly 24{25, 1997 (Mar del Plata, Argentina)

First Day | Time: 4.5 hours

1. In the plane the points with integer coordinates are the vertices of

unit squares. The squares are coloured alternately black and white (as on a

chessboard).

For any pair of positive integers m and n, consider a right-angled tri-

angle whose vertices have integer coordinates and whose legs, of lengthsm

and n, lie along edges of the squares.

Let S1 be the total area of the black part of the triangle and S2 be the

total area of the white part. Let

f(m;n) = jS1 � S2j :

(a) Calculate f(m;n) for all positive integersm and n which are either

both even or both odd.

(b) Prove that f(m;n) � 12maxfm;ng for allm and n.

(c) Show that there is no constant C such that f(m;n) < C for all m

and n.

2. Angle A is the smallest in the triangle ABC.

The points B and C divide the circumcircle of the triangle into two

arcs. Let U be an interior point of the arc between B and C which does not

contain A.

392

The perpendicular bisectors of AB and AC meet the line AU at V

and W , respectively. The lines BV and CW meet at T .

Show that

AU = TB + TC:

3. Let x1; x2; : : : ; xn be real numbers satisfying the conditions:

jx1 + x2 + � � �+ xnj = 1

and

jxij �n+ 1

2for i = 1; 2; : : : ; n:

Show that there exists a permutation y1; y2; : : : ; yn of x1; x2; : : : ; xn such

that

jy1 + 2y2 + � � �+ nynj �n+ 1

2:

.

Second Day | Time: 4.5 hours

4. An n � n matrix (square array) whose entries come from the set

S = f1; 2; : : : ; 2n�1g, is called a silver matrix if, for each i = 1; : : : ; n, the

ith row and the ith column together contain all elements of S. Show that

(a) there is no silver matrix for n = 1997;

(b) silver matrices exist for in�nitely many values of n.

5. Find all pairs (a; b) of integers a � 1, b � 1 that satisfy the equation

a(b2) = ba:

6. For each positive integer n, let f(n) denote the number of ways of

representing n as a sum of powers of 2 with nonnegative integer exponents.

Representations which di�er only in the ordering of their summands

are considered to be the same. For instance, f(4) = 4, because the number

4 can be represented in the following four ways:

4; 2 + 2; 2 + 1 + 1; 1 + 1 + 1 + 1:

Prove that, for any integer n � 3:

2n2=4 < f(2n)< 2n

2=2:

As the IMO is o�cially an individual event, the compilation of team

scores is uno�cial, if inevitable. These totals and the prize awards are given

in the following table.

393

Rank Country Score Gold Silver Bronze Total

1. China 223 6 { { 6

2. Hungary 219 4 2 { 6

3. Iran 217 4 2 { 6

4.{5. Russia 202 3 2 1 6

4.{5. United States of America 202 2 4 { 6

6. Ukraine 195 3 3 { 6

7.{8. Bulgaria 191 2 3 1 6

7.{8. Romania 191 2 3 1 6

9. Australia 187 2 3 1 6

10. Vietnam 183 1 5 { 6

11. Republic of Korea 164 1 4 1 6

12. Japan 163 1 3 1 5

13. Germany 161 1 3 2 6

14. Republic of China (Taiwan) 148 { 4 2 6

15. India 146 { 3 3 6

16. United Kingdom 144 1 2 2 5

17. Belarus 140 { 2 4 6

18. Czech Republic 139 1 2 2 5

19. Sweden 128 1 3 { 4

20.{21. Poland 125 { 2 2 4

20.{21. Yugoslavia 125 { 2 3 5

22.{23. Israel 124 { 1 5 6

22.{23. Latvia 124 { 1 4 5

24. Croatia 121 { 1 4 5

25. Turkey 119 { 1 4 5

26. Brazil 117 { 1 4 5

27. Colombia 112 { { 6 6

28. Georgia 109 { 1 3 4

29. Canada 107 { 2 2 4

30.{31. Hong Kong 106 { { 5 5

30.{31. Mongolia 106 1 { 3 4

32.{33. France 105 1 { 1 2

32.{33. Mexico 105 { 1 3 4

34.{35. Armenia 97 { { 3 3

34.{35. Finland 97 { - 4 4

36. Slovakia 96 { 1 2 3

37.{38. Argentina 94 { { 3 3

37.{38. The Netherlands 94 { 2 { 2

39. Republic of South Africa 93 1 { 2 3

40. Cuba 91 { 1 2 3

41.{42. Belgium 88 { { 3 3

41.{42. Singapore 88 { { 4 4

43. Austria 86 1 { 1 2

44. Norway 79 { { 3 3

45. Greece 75 { 1 { 1

394

Rank Country Score Gold Silver Bronze Total

46.{47. Kazakhstan 73 { { 1 1

46.{47. Former Yugoslav Republic of

Macedonia

73 { { 3 3

48.{49. Italy 71 { { 1 1

48.{49. New Zealand 71 { { 2 2

50. Slovenia 70 { { 2 2

51. Lithuania 67 { 1 1 2

52. Thailand 66 { { 1 1

53.{54. Estonia 64 { { 2 2

53.{54. Peru 64 { { 2 2

55. Azerbaijan 56 { { 1 1

56. Macao 55 { { { {

57.{59. Denmark 53 { { 1 1

57.{59. Moldova (team of 3) 53 { { 2 2

57.{59. Switzerland (team of 5) 53 { { 2 2

60.{61. Iceland 48 { 1 { 1

60.{61. Morocco 48 { { { {

62. Bosnia and Herzegovina

(team of 5)

45 { { 1 1

63. Indonesia 44 { { { {

64. Spain 39 { { { {

65. Trinidad and Tobago 30 { { { {

66. Chile 28 { { { {

67. Uzbekistan (team of 3) 23 { { { {

68. Ireland 21 { { { {

69.{70. Malaysia 19 { { { {

69.{70. Uruguay 19 { { { {

71.{72. Albania (team of 3) 15 { { { {

71.{72. Portugal (team of 5) 15 { { { {

73. Philippines (team of 2) 14 { { { {

74. Bolivia (team of 3) 13 { { { {

75. Kyrgyztan (team of 3) 11 { { { {

76.{78. Kuwait (team of 4) 8 { { { {

76.{78. Paraguay 8 { { { {

76.{78. Puerto Rico 8 { { { {

79. Guatemala 7 { { { {

80. Cyprus (team of 3) 5 { { { {

81. Venezuela (team of 3) 4 { { { {

82. Algeria (team of 4) 3 { { { {

395

This year the Canadian Team slid to 29th place from 16th last year and

19th the previous year. The Team members were:

Byung Kyu Chun 29 Silver

Adrian Chan 25 Silver

Mihaela Enachescu 21 Bronze

Sabin Cautis 16 Bronze

Jimmy Chui 10 Honourable Mention

Adrian Birka 6

The Canadian Team Leader was Richard Nowakowski, of Dalhousie

University, and the Deputy Team Leader was Naoki Sato, currently a stu-

dent at the University of Toronto and former Canadian Silver medalist.

The Chinese Team placed �rst this year. Its members were:

Jia Rui Han 38 Gold

Yi Ni 38 Gold

Jin Zou 38 Gold

Jin Peng An 37 Gold

Xiao Ming Sun 37 Gold

Chang Jin Zheng 35 Gold

Congratulations to the Chinese Team!!

Now we turn to solutions to problems of the Czechoslovakia Mathe-

matical Olympiad 1993 [1996: 109].

CZECHOSLOVAK MATHEMATICAL OLYMPIAD 1993Final Round

1. Find all natural numbers n for which 7n� 1 is a multiple of 6n� 1.

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; and by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, On-tario.

No such n exists. Suppose n is a natural number for which

6n � 1 j 7n � 1. Since 6n � 1 mod 5 we have 5 j 7n � 1. From 7 � 2;

72 � �1, 73 � 3 and 74 � 1 mod 5, we see that 7n � 1 � 0 mod 5 if

and only if 4 divides n. In particular, n must be even, which implies, since

62 � 1 mod 7 that 6n � 1 mod 7; that is, 7 j 6n � 1. However, obviously

7 6 j 7n � 1, which is a contradiction.

2. A 19� 19 table contains integers so that any two of them lying on

neighbouring �elds di�er at most by 2. Find the greatest possible number

of mutually di�erent integers in such a table. (Two �elds of the table are

considered neighbouring if they have a common side.)

396

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

We shall �rst prove that the value 71 for the number of di�erent integers

is attainable. We can assume, without loss of generality, that the number in

the top left hand square is 0. We can then form the following array:

0 1 3 5 � � � 31 33 34

2 3 5 7 � � � 33 35 36

4 5 7 9 � � � 35 37 38

6 7 9 11 � � � 37 39 40

.

.

....

.

.

....

.

.

....

.

.

.

32 33 35 37 63 65 66

34 35 37 39 65 67 68

35 37 39 41 67 69 70

The di�erence in the values of opposite corners of the array cannot exceed

twice the number of moves to get from one to the other passing through

adjacent squares. This equals

2� 36 = 72:

Thus there can be at most 73 di�erent values since every square in the array

can be reached from the top left square in at most 36 moves.

We need to prove that there is no array with 72 or 73 di�erent numbers.

The value 73 can only occur if the di�erence between every pair of ad-

jacent squares is exactly 2, for otherwise one of the shortest paths from op-

posite corners can pass through the pair of adjacent squares whose value is

not 2, giving a contradiction.

If the number of distinct values is 72, then the bottom right hand square

must have value 71, so in any shortest path, there is a pair of adjacent squares

with di�erence only 1.

To have 72 values, all the even and odd numbers less than or equal to

71 must be in the table.

There is a (possibly broken) dividing line separating the odd and even

numbers, all numbers on the side of 71 being odd.

If there is a gap on the line, then there is a shortest path from 71 to 0

without crossing the line, which is impossible.

Considering 1 and 71 we see that the shortest distance must be 35,

so that one of the squares adjacent to 0 must contain 1. The broken line

separating even and odd entries cannot have the form

397

or

for otherwise these would be a shortest path from 0 to 71 with more than

one di�erence of 1. It follows that the even entries are con�ned to the �rst

row or the �rst column, contradicting that there are 36 of them.

4. A sequence fang1n=1 of natural numbers is de�ned recursively by

a1 = 2 and an+1 = the sum of 10th powers of the digits of an, for all n � 1.

Decide whether some numbers can appear twice in the sequence fang1n=1.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

Let an have dn digits, so dn � 1 + log10 an.

Then an+1 � dn910 � (1 + log10 an)9

10 < an for an su�ciently large, asx

1+logx!1 as x!1. Thus an+1 < an if an

1+log10 an> 910.

Let K be the �rst natural number for which K

1+log10K> 910.

Then if ai � K, the sequence decreases until there is a term < K.

Hence the sequence has in�nitely many terms < K, and there must be

a term repeated (in�nitely often).

5. Find all functions f : Z!Zsuch that f(�1) = f(1) and

f(x) + f(y) = f(x+ 2xy) + f(y� 2xy)

for all integers x, y.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

f(1) + f(y) = f(1 + 2y) + f(�y) (1)

f(y)+ f(�1) = f(�y)+ f(�1+ 2y) (2)

f(1) = f(�1) and f(y) = f(y), so equating (1) and (2)

f(1 + 2y) + f(�y) = f(�y)+ f(�1+ 2y):

Hence f(2y+ 1) = f(2y� 1) for all integers y. (?)

Also f(y) + f(1) = f(3y) + f(1� 2y) and since f(1) = f(1� 2y)

(making use of (?)) f(y) = f(3y).

Also, equating this with equation (1)

f(3y)+ f(1� 2y) = f(1 + 2y) + f(�y);

398

and since by (?) f(1� 2y) = f(1 + 2y)

f(3y) = f(�y):

Thus f(y) = f(3y) = f(�y) and f(y) = f(�y) for all y.If y is odd, then f(y) = f(y� 2xy) so f(x) = f(x+ 2xy).

Thus f(2a) = f(2a(1 + 2y)).

Thus an oddmultiple of a number and that number give the same value.

Therefore if n = 2ka, a odd, then f(n) = f(2k). Therefore all func-

tions must be such that f(2ka) = f(2k), k � 0 where a is odd and f(2k)

for each k can take any value as can f(0).

These functions always satisfy the conditions since f(�1) = f(1) and

f(x+ 2xy) + f(y� 2xy) = f(x(1 + 2y)) + f(y(1� 2x))

= f(x) + f(y):

For the remainder of this number of the Corner we turn to readers'

solutions to 10 of the problems of the Baltic Way | 92 Contest given in the

May 1996 number of the Corner [1996: 157{159].

MATHEMATICAL TEAM CONTEST\BALTIC WAY | 92"

Vilnius, 1992 | November 5{8

1. Let p, q be two consecutive odd prime numbers. Prove that p + q

is a product of at least 3 positive integers > 1 (not necessarily di�erent).

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; byMansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Wind-sor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario.

Since p and q are odd, p+ q is even and has a factor of 2. The corre-

sponding factor 12(p+q) lies strictly between p and q and is not prime since p

and q are consecutive prime numbers. It must therefore have two factors at

least. Hence p+ q is a product of at least 3 positive integers greater than 1.

2. Denote by d(n) the number of all positive divisors of a positive

integer n (including 1 and n). Prove that there are in�nitely many n such

that nd(n)

is an integer.

399

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Wind-sor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, Ontario. We give Bradley's solution.

Let p be a prime. Then pp�1 has factors 1; p; p2; : : : ; pp�1, which are

p in number. Hence, for such n, pp�1=d(pp�1) = pp�2 which is integral.

Hence there are in�nitely many n such that n=d(n) is an integer.

3. Find an in�nite non-constant arithmetic progression of positive in-

tegers such that each term is neither a sum of two squares, nor a sum of two

cubes (of positive integers).

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, Universityof Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give Godin's solution.

Looking at squares mod 4 we see

x x2 mod 4

0 0

1 1

2 0

3 1

thus if x � 3 mod 4, it is impossible to express x as a sum of two squares.

Similarly, looking at cubes mod 7 we see

x x3 mod 7

0 0

1 1

2 1

3 6

4 1

5 6

6 6

and again if x � 3 mod 7 it is impossible to express x as a sum of 2 cubes.

Thus the sequence s0 = 3, sn+1 = sn + 28 has all sn � 3 mod 4 and

sn � 3 mod 7 so all the terms are not the sum of 2 squares or cubes.

4. Is it possible to draw a hexagon with vertices in the knots of an in-

teger lattice so that the squares of the lengths of the sides are six consecutive

positive integers?

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, Clifton College, Bristol, UK; and by Shawn

400

Godin, St. Joseph Scollard Hall, North Bay, Ontario. We give Bradley's so-lution.

No. It is not possible.

Six consecutive positive integers must contain three odd numbers and

three even numbers. To form them as the sums of squares must involve

an odd number of odd squares. The sum of all the numbers whose squares

are made is therefore odd. But to move from (0; 0; 0) to (0; 0; 0) involves

a total number which is even (going out and coming back to put it loosely).

[A pentagon, on the other hand, can be made, for example,

(0; 0; 0)! (0;1; 1)! (1; 2; 2)! (1;2; 4)! (1; 1; 2)! (0;0; 0)

involving sides whose squares are 2; 3; 4; 5; 6 | with two odd numbers.]

5. Given that a2 + b2 + (a + b)2 = c2 + d2 + (c + d)2, prove that

a4 + b4 + (a+ b)4 = c4 + d4 + (c+ d)4.

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; byMansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University ofWindsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; andby Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Wegive Wang's solution.

This follows immediately from the fact that

x4 + y4 + (x+ y)4 = 2�x4 + 2x3y + 3x2y2 + 2xy3 + y4

�= 2

�x2 + xy + y2

�2=

1

2

�x2 + y2 + (x+ y)2

�2for all x, y.

6. Prove that the product of the 99 numbers k3�1

k3+1, k = 2; 3; : : : ; 100,

is greater than 23.

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; byMansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, University ofWindsor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; andby Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. Wegive Wang's solution and comment.

Let P (n) =Qn

k=2k3�1

k3+1where n � 2. We show that in general

P (n) =2(n2 + n+ 1)

3n(n+ 1)

from which it follows that P (n) > 23. Since

(k+ 1)2 � (k+ 1) + 1 = k2 + k + 1;

401

we have

P (n) =

nYk=2

(k� 1)(k2 + k + 1)

(k+ 1)(k2� k + 1)

=

Qn�2k=0 (k+ 1)Qn

k=2(k+ 1)�Qnk=2(k

2 + k + 1)Qn�1k=1 k

2 + k + 1

=1 � 2

n(n+ 1)� n

2 + n+ 1

3=

2(n2 + n+ 1)

3n(n+ 1):

Remark: It is easy to see that the sequence fP (n)g, n = 2; 3; : : : is

strictly decreasing and from the result established above we see that

limn!1P (n) = 23. This result is well known and can be found in, for ex-

ample, Theory and Application of In�nite Series, by K. Knopp (Ex. 85(2) on

p. 28).

7. Let a =1992p1992. Which number is greater:

a

qq

q

a

a

a

9>>>>=>>>>;1992 or 1992?

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Michael Selby, Universityof Windsor, Windsor, Ontario; and by Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario. We give Godin's solution.

Let f(x) = ax. Clearly p > q if and only if f(p) > f(q); that

is, f is strictly increasing. Similarly if fn(x) is de�ned by f1(x) = f(x)

and fn+1(x) = f(fn(x)), so fn(x) = f f : : : f| {z }n

(x) then p > q if and only

if fn(p) > fn(q) for all n.

Now 1992 > a.

Thus f1992(1992) > f1992(a). But f(1992) = 1992, so f1992(1992) =

1992. Now f1992(a) is the expression in question so 1992 is the larger.

Editor's Note. Selby points out that if xk = fk(a) then limk!1 xk = L

where a = L1=L or L = 1992.

8. Find all integers satisfying the equation

2x � (4� x) = 2x+ 4:

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; byMansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,

402

St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Wind-sor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and byEdward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We giveTsaoussoglou's solution.

2x =2(x+ 2)

4� x> 0; so � 2 < x < 4:

Checking

x = �11

26= 2(�1) + 4

4� (�1)=

2

5

x = 0 1 =2(0 + 2)

4� 0is a solution

x = 1 2 =2(1 + 2)

4� 1is a solution

x = 2 4 =2(2 + 2)

4� 2is a solution

x = 3 8 =2(3 + 2)

4� 3is not a solution

9. A polynomial f(x) = x3 + ax2 + bx + c is such that b < 0 and

ab = 9c. Prove that the polynomial has three di�erent real roots.

Solutions by Christopher J. Bradley, Clifton College, Bristol, UK; byBob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; and byMichael Selby, University of Windsor, Windsor, Ontario. We give Selby'ssolution.

Suppose a = 0. Then c = 0 and f(x) = x3 + bx. Therefore there are

three roots: x = 0, x =p�b, x = �p�b. If a > 0, then c = ab

9< 0, since

b < 0. Now f(0) = c < 0 and since limx!1 f(x) =1, there is some r1 in

(0;1) such that f(r1) = 0.

Also f(�a) = �a3 + a3 � ab+ c = �8ab

9> 0.

Since f(0) < 0 and f(�a) > 0, there is a root r2 in the interval

(�a; 0). Finally, since f(x) ! �1 as x ! �1 and f(�a) > 0 there

must be a root r3 in (�1;�a).If a < 0, c = ab

9> 0. Hence f(0) = c > 0 while f(x) ! �1 as

x! �1. Therefore we have a root t1 in (�1; 0).

f(�a) = �a3+a3�ab+ c = �89ab < 0. Hence there must be a root

t2 in (0;�a). Finally f(x) ! 1 as x ! 1. Since f(�a) < 0 there need

be a root t3 in (�a;1).

In all cases we have three distinct roots.

403

10. Find all fourth degree polynomials p(x) such that the following

four conditions are satis�ed:

(i) p(x) = p(�x), for all x,(ii) p(x) � 0, for all x,

(iii) p(0) = 1,

(iv) p(x) has exactly two local minimum points x1 and x2 such that

jx1 � x2j = 2.

Solutions by Christopher J. Bradley, Clifton College, Bristol, UK; byShawn Godin, St. Joseph Scollard Hall, North Bay, Ontario; and by MichaelSelby, University of Windsor, Windsor, Ontario. We give Selby's solution.

Condition (i) implies p(x) is even.

p(x) = ax4 + bx3 + cx2 + dx+ e;

p(�x) = ax4 � bx3 + cx2 � dx+ e:

Hence we have 2bx3 + 2dx = 0 for all x. Thus b = d = 0.

Also p(0) = 1. Therefore p(x) has the form

p(x) = ax4 + cx2 + 1:

Now p0(x) = 4ax3 + 2cx. The critical points are x = 0, x2 = �c

2a. Hence

we must have �c

2a> 0.

p00(x) = 12ax2+2c. Since we want exactly two local minima, we must

have at x2 = �c2a

, p0(x) = 12a��c2a

�+ 2c = �4c > 0. Therefore c < 0 and

a > 0.

Further we want p(x) � 0. Since p(x) ! 1 as jxj ! 1 then since

p(0) = 1, we need at x2 = �c2a

, p(x) = a c2

4a2+c

��c2a

�+1 � 0. So �c2

4a+1 � 0

or 4a � c2. Also x1 =

q�c2a

, x2 = �q�c2a

, jx1 � x2j = 2 )����q�c2a

���� = 1 or

�c = 2a. Since 4a � c2, we have 4a � 4a2 or a(1� a) � 0.

Therefore since a > 0, 1� a � 0 and we have 0 < a � 1, c = �2a.

The polynomials are of the form ax4 � 2ax2 + 1 where 0 < a � 1.

That completes the Corner for this issue. Send me your nice solutions

and contest materials.

404

BOOK REVIEWS

Edited by ANDY LIU

144 Problems of the Austrian-Polish Mathematics Competition 1978{1993,compiled by Marcin Emil Kuczma,

University of Warsaw,

published by the Academic Distribution Center, 1993,

1216 Walker Road, Freeland, Maryland 21053, USA,

ISBN# 0-9640959-0-4, softcover, 157+ pages, US$20 plus handling.

40th Polish Mathematics Olympiad 1989/90,edited by Marcin Emil Kuczma,

University of Warsaw,

published by Science, Culture, Technology Publishing,

AMK Central Post O�ce, Box 0581, Singapore 915603,

no ISBN#, softcover, 49+ pages, US$10 plus handling.

Reviewed by Andy Liu, University of Alberta.

Both books under review are by Marcin Kuczma, one of the world's

leading problem proposers. In IMO95 in Canada, two of his problemsmade it

and a third nearly did. He is responsible for a substantial number of problems

in these two books, which come with the highest recommendation. While

originality can never be guaranteed since great minds tend to think alike

across space and time, the author strived to come up with something new

and exciting, and has largely succeeded.

The �rst book contains all the problems of the Austrian-Polish Math-

ematics Competition since its inception in 1978, up to 1993. It is one of

the world's oldest regional mathematics competitions, attesting to the close

friendship of the mathematical communities in these two nations. Apart

from six problems given over two days in the IMO format, there is a team

competition in which the six members collaborate to solve three problems in

three hours. These students are usually those ranked immediately below the

IMO team members. While the APMC may be considered a consolation for

a graduating high school student, it is a wonderful experience and excellent

preparation for those who have further aspirations in making the IMO team

in future years.

The second book is somewhat of an anomaly in publishing, in that it

contains only the problems of one competition in one year. There are 24

problems in all, consisting of 12 from the First Round of the 1989/90 Polish

Mathematics Olympiad, and 6 from each of the Second and the Third Rounds.

Discounts are o�ered for orders of multiple copies. Contact the publishers

for details.

405

The Introduction in each of these two books provides valuable infor-

mation about the respective competitions, but of course the main value lies

in the problems. We give below a sample from each book. Both should be

on the shelf of anyone serious about mathematics competitions.

Problem 5, Austrian-Polish Mathematics Competition, 1985.

We are given a certain number of identical sets of weights; each set consists of

four di�erent weights expressed by natural numbers (of weight units). Using

these weights we are able to weigh out every integer mass up to 1985 (inclu-

sive). How many ways are there to compose such a set of weights given that

the joint mass of all weights is the least possible?

Problem 21, Third Round, Polish Mathematical Olympiad, 1989/90.

The edges of a cube are numbered 1 through 12.

(a) Show that for every such numbering, there exist at least eight triples

of integers (i; j; k), with 1 � i < j < k � 12, such that the edges

labelled i, j and k are consecutive sides of a polygonal line.

(b) Give an example of a numbering for which a ninth triple with these

properties does not exist.

406

MORE UNITARY DIVISOR PROBLEMS

K.R.S. Sastry

A (positive) integral divisor d of a (natural) number n is called a unitary di-

visor of n if and only if d andn

dare relatively prime; that is

�d;n

d

�= 1.

For example, 9 is a unitary divisor of 18 because

�9;

18

9

�= (9; 2) = 1.

But, 3 is not a unitary divisor of 18 because�3; 18

3

�= (3; 6) = 3 6= 1. This

novel de�nition of divisibility produces interesting analogues and contrasts

with the results of ordinary divisibility. See [2, 3, 4]. In this paper we con-

sider the solutions of two more new unitary divisor problems.

BACKGROUND MATERIAL: Let n = p�1

1 p�2

2 � � � p�kk denote the prime de-

composition of n. Let d�(n) denote the number of unitary divisors of n.

Then, see [2, 3, 4], we have

��(1) = 1; d�(n) = 2k for n > 1: (1)

The Euler � function counts the number �(n) of natural numbers that

are less than and relatively prime to n. Then, see [1], we have

�(1) = 1; �(n) =

kYi=1

(pi � 1)p�i�1

i for n > 1: (2)

The �rst problem asks us to determine the positive integers n for which

d�(n) = �(n). The second asks us to establish a curious result concerning

the number of unitary divisors of each unitary divisor of n.

THE FIRST PROBLEM: Determine the set S = fn : d�(n) = �(n)g.Solution: In view of equations (1) and (2), we have to solve the equation

(p1 � 1)p�1�11 (p2 � 1)p

�2�12 � � � (pk � 1)p

�k�1k = 2k: (3)

The right hand side of equation (3) is a power of 2, so each factor on its left

hand side is to be necessarily a power of 2. Two cases arise because of the

even prime 2.

(I) p1 = 2 and pi are distinct odd primes for 2 � i � k. Then �2 � 1 =

�3�1 = � � � = �k�1 = 0 and p2�1 = 2�2 ; p3�1 = 2�3 ; � � � ; pk�1 = 2�k .

Thus pi = 2�i + 1 are distinct odd primes for

2 � i � k: (4)

(II) pi are all odd primes for 1 � i � k. Then as in the preceding, �i = 1

and pi = 2�i + 1 are distinct odd primes for

1 � i � k: (5)

407

(A) First we note the obvious solution n = 1 that follows from

d�(1) = 1 and �(1) = 1.

(B) k = 1. This implies that n is formed from a single prime.

(I) k = 1, p1 = 2. So n = 2�1. This yields 2�1�1 = 21; �1 = 2. Hence

n = 22 = 4. It is easily veri�ed that d�(4) = �(4) = 2.

(II) k = 1, p1 an odd prime. This yields (p1 � 1) = 21; p1 = 3. Hence

n = 3. Again d�(3) = �(3) = 2.

(C) k = 2. This implies that n is formed from powers of two distinct primes

p1; p2.

(I) k = 2, p1 = 2 and p2 an odd prime. Hence n = 2�1p2. This yields

2�1�12�2 = 22, (�1 � 1) + �2 = 2. This equation has two solutions:

(C1) �1 = 1; �2 = 2 =) n = 2�1�2�2 + 1

�= 10;

(C2) �1 = 2; �2 = 1 =) n = 2�1�2�2 + 1

�= 12:

(II) k = 2, p1, p2 are both odd primes. Hence n = p1p2, 2�12�2 = 22,

�1 + �2 = 2. Now p1 and p2 are distinct odd primes so �1 + �2 = 2 are

distinct natural numbers. Hence �1+�2 > 2 and there is no solution in this

case.

(D) k = 3. In this case, n is the product of powers of three distinct primes.

(I) k = 3, p1 = 2, and p2, p3 are distinct odd primes. Hence

n = 2�1p2p3, 2�1�12�22�3 = 23, (�1 � 1) + �2 + �3 = 3. Since �1 is

at least 1 and only one of �2 and �3 can equal 1, we have the single solution

�1 = 1, �2 = 1, �3 = 2. This gives n = 2�1�2�1 + 1

� �2�2 + 1

�= 30.

(II) k = 3, p1, p2, p3 are distinct odd primes. This leads to the equation

�1 + �2 + �3 = 3 that has no solution in natural numbers, as explained

above.

(E) k > 3, n = p�1

1 p�2

2 � � � p�kk .

(I) k > 3, p1 = 2 yields (�1 � 1) + �2 + �3 + � � �+ �k = k, which has no

solution for the reason explained above in (D).

(II) k > 3 yields �1 + �2 + � � � + �k = k, and this equation too does not

have a solution.

Combining (A), � � � , (E) we have determined S = f1; 3; 4; 10; 12; 30g.The veri�cation that if n 2 S then d�(n) = �(n) is left to the reader.

THE SECOND PROBLEM: The fascinating property of the natural numbers

that 13 + 23 + � � � + n3 = (1 + 2 + � � �+ n)2motivates us to look for a

collection of integers with the similar property. For instance the collection

of the integers 1, 2, 2, 2, 4, 4, 4, 8 exhibit that property:

13+23+23+23+43+43+43+83 = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)2:

408

How to generate collections of such integers exhibiting the above property?

Here is an algorithm:

(A1) Start with any integer n. Say n = 30.

(A2) Write down the unitary divisors of n. The unitary divisors of 30 are

1, 2, 3, 5, 6, 10, 15, 30.

(A3) Nowwrite down the number of unitary divisors of each unitary divisor

of n. In view of (1) each of these integers will be a power of 2. In the present

case we have respectively

1; 2; 2; 2; 4; 4; 4; 8:

Lo and behold! We have generated a collection of integers having the men-

tioned property. Intrigued? Try this on n = 24 or n = 420 or any other

value. In general we have the following result:

Let a1, a2, � � � , a� denote the unitary divisors of an integer n. Let

d�(ai) denote the number of unitary divisors of ai. Then

�Xi=1

[d�(ai)]3=

"�X

i=1

d�(ai)

#2:

Proof: Let n = p�1

1 p�2

2 � � � p�kk denote the prime decomposition of n. Then

its 2k = � unitary divisors are:

a1; a2; � � � ; a� = 1;

p�1

1 ; p�2

2 ; � � � ; p�kk ;

p�1

1 p�2

2 ; p�1

1 p�3

3 ; � � � ; p�k�1

k�1 p�kk ;

.

.

.

p�1

1 p�2

2 � � � p�kk :

Observe that the unitary divisors have been put together, for convenience,

in collections containing�k

0

�;

�k

1

�;

�k

2

�; � � � ;

�k

k

�

terms respectively. Here�k

r

�=

k(k�1)���(k�r+1)

r!denotes the binomial coe�-

cient for 0 � r � k. We now use (1) to write down d�(ai) for 1 � i � 2k:

1; 2; 2; � � � ; 2; 22; 22; � � � ; 22; 23; 23; � � � ; 23; � � � ; 2k:

Observe that these collections contain, respectively,�k

0

�term equalling 1:�

k

1

�terms equalling 2;

�k

2

�terms equalling 22; � � � ; �k

k

�term equalling 2k.

HenceX[��(ai)]

3= 13 +

�23 + 23 + � � �

�k

1

�terms

�

+

�26 + 26 + � � �

�k

2

�terms

�+ � � �+ �

23k�:

409

This is nothing but the expansion of

�1 + 23

�k= 9k = 32k:

Also

X��(ai) = 1 +

�2 + 2 + � � �

�k

1

�terms

�

+

�22 + 22 + � � �

�k

2

�terms

�+ � � �+ �

2k�

= (1 + 2)k = 3k:

It is easy to see that the assertion follows.

Finally we invite the reader to state and prove an analogous result for

the ordinary (positive integral) divisors of (positive) integers n. For other

problems on cubes of natural numbers see [5, 6].

References

[1] L.E. Dickson, History of the Theory of Numbers, Vol. I, Chelsea Publi-

cations, (1971), pp. 113{114.

[2] R.T. Hanson and L.G. Swanson, Unitary Divisors, Mathematics Maga-zine, 52 (September 1979), pp. 217{222.

[3] K.R.S. Sastry, Unitary Divisor Problems, CruxMathematicorum, 22 (Feb-

ruary 1996), pp. 4{9.

[4] K.R.S. Sastry, Unitary Divisor Analogues, Mathematics and ComputerEducation, 30 (Fall 1996), pp. 300{311.

[5] K.R.S. Sastry, Fermat-Pell: An Observation and an Application, Mathe-matics and Computer Education, 28 (Winter 1994), pp. 24{35.

[6] K.R.S. Sastry, Cubes of Natural Numbers in Arithmetic Progression, CruxMathematicorum, 18, (1992), pp. 161{164.

K.R.S. Sastry

Jeevan Sandhya

Doddakalsandra Post

Raghuvanahalli

Bangalore 560 062, India

410

THE SKOLIAD CORNERNo. 25

R.E. Woodrow

This number we give the problems of the 1997 Alberta High School

Mathematics Prize Exam, second round, written last February by students

invited on the basis of the �rst round results from November. My thanks

go to Ted Lewis, Chair of the Alberta High School Mathematics Prize Exam

Board for forwarding the contest, which is partially supported by the Cana-

dian Mathematical Society.

ALBERTA HIGH SCHOOLMATHEMATICS COMPETITION

February 11, 1997Second Round

1. Find all real numbers x satisfying jx� 7j > jx + 2j+ jx� 2j.Remark. Note that jaj is called the absolute value of the real number a.

It has the same numerical value as a but is never negative. For example,

j3:5j = 3:5, while j � 2j = 2. Of course, j0j = 0.

2. Two lines b and c form a 60� angle at the point A, and B1 is a

point on b. From B1, draw a line perpendicular to the line b meeting the

line c at the point C1. From C1 draw a line perpendicular to c meeting the

line b at B2. Continue in this way obtaining points C2, B3, C3, and so on.

These points are the vertices of right trianglesAB1C1; AB2C2; AB3C3; : : : .

If area (AB1C1) = 1, �nd

area (AB1C1)+ area (AB2C2)+ area (AB3C3)+ � � �+ area (AB1997C1997):

3. A and B are two points on the diameter MN of a semicircle.

C, D, E and F are points on the semicircle such that \CAM = \EAN =

\DBM = \FBN . Prove that CE = DF .

4. (a) Suppose that p is an odd prime number and a and b are positive

integers such that p4 divides a2 + b2 and p4 also divides a(a + b)2. Prove

that p4 also divides a(a+ b).

(b) Suppose that p is an odd prime number and a and b are positive

integers such that p5 divides a2+ b2 and p5 also divides a(a+ b)2. Show by

an example that p5 does not necessarily divide a(a+ b).

5. The picture shows seven houses represented by the dots, connected

by six roads represented by the lines. Each road is exactly 1 kilometre long.

You live in the house marked B. For each positive integer n, how many

411

ways are there for you to run n kilometres if you start at B and you never

run along only part of a road and turn around between houses? You have to

use the roads, but you may use any road more than once, and you do not

have to �nish at B. For example, if n = 4, then three of the possibilities are:

B to C to F to G to F ; B to A to B to C to B; and B to C to B to A to B.

s s s

s s

s s

F G

D E

A B C

Next we give the solutions to the 1995 Concours Math �ematique du

Qu �ebec which comes to us from Th �er �ese Ouellet, secretary to the contest.

We give the solutions in French, the language of the competition.

CONCOURS MATH �EMATIQUE DU QU �EBEC 1995February 2, 1995

Time: 3 hours

1. LA FRACTION �A SIMPLIFIER

Simpli�ez la fraction

1 358 024 701

1 851 851 865:

Solution.

1 358 024 701

1 851 851 865=

11� 123 456 791

15� 123 456 791=

11

15:

2. LA FORMULE MYST �ERE

Consid �erons les �equations suivantes

xy = p; x+ y = s; x1993 + y1993 = t; x1994 + y1994 = u:

En faisant appel aux lettres p, s, t, u mais pas aux lettres x, y, donnez une

formule pour la valeur

x1995 + y1995:

412

Solution. On a successivement

su = (x+ y)(x1994+ y1994)

= x1995 + xy1994 + x1994y + y1995

= x1995 + y1995 + xy � (x1993 + y1993)

= x1995 + y1995 + pt:

D'o �u l'on tire

x1995 + y1995 = su� pt:

3. LA DIFF �ERENCE �ETONNANTE

Lorsque la circonf �erence d'un cercle est divis �ee en dix parties �egales,

les cordes qui joignent les points de division successifs forment un d �ecagone

r �egulier convexe. En joignant chaque point de division au troisi �eme suivant,

on obtient un d �ecagone r �egulier �etoil �e. Montrer que la di� �erence entre les

cot �es de ces d �ecagones est �egale au rayon du cercle.

Solution. On consid �ere le cercle divis �e en dix parties �egales. On joint lespoints CD et DE, par des segments �egaux aux cot �es du d �ecagone convexe,

les points BE et CF , par des segments �egaux aux cot �es du d �ecagone �etoil �e,

et les diam �etres AF et BG. Alors CD kBE kAF et DE kCF kBG. En

vertu des propri �et �es des parall �elogrammes, on a

BE � CD = BE �HE

= BH = OF = r:

s

ss

s

s

s

s

s

s

sA

B

C

D E

F

GO

H

r

4. LE TERRAIN EN FORME DE CERF-VOLANT

Abel Belgrillet est membre du Club des a �erocervidophiles (amateurs de

cerfs-volants) du Qu �ebec. Il dispose de quatre tron�cons de cloture rectilignes

AB, BC, CD, DA pour d �elimiter un terrain (en forme de cerf-volant, voir

�gure) sur lequel il s'adonnera �a son activit �e favorite cet �et �e. Sachant que les

413

tron�cons AB et DA mesurent 50 m chacun et que les tron�cons BC et CD

mesurent 120 m chacun, d �eterminez la distance entre les points A et C qui

maximisera l'aire du terrain.

Solution. D �ecoupons le terrain selon AC:

l'aire de

50 120

50 120

A C

D

B

= l'aire de

50 120

120 50

A C

D

B

Mais, l'aire d'unparall �elogramme de cot �es donn �es estmaximale lorsque

ce parall �elogramme est un rectangle (car dans ce cas, la hauteur sera maxi-

male). On doit donc avoir que l'angle bB = 90�. Le triangle ABC est donc

rectangle en B. Ainsi

AC =

qAB

2+ BC

2=p2500 + 14400 = 130:

5. L'IN �EGALIT �E MODIFI �EE D'AMOTH DIEUFUTUR

(a) (2 points) L'in �egalit �e x2 + 2y2 � 3xy est-elle vraie pour tous les

entiers?

(b) (8 points) Montrez que l'in �egalit �e x2 +2y2 � 145xy est valide pour

tous r �eels x et y.

Solution. (a) Non. Voici quelques couples qui fournissent un contre-

exemple:

x y

3 2

4 3

5 3

5 4

6 4

(b) 1 �ere solution. Consid �erons l'in �egalit �e (x� �y)2 � 0, qui est vraie

pour tout couple de r �eels x, y. Elle est �equivalente �a x2 + �2y2 � 2�xy.

En posant � =p2, on trouve x2+2y2 � 2

p2xy = (2; 828 : : : )xy � 14

5xy.

On peut aussi montrer que 2p2 � 14

5en �elevant au carr �e de chaque cot �e.

2i �eme solution. Montrer que x2 + 2y2 � 145xy revient �a d �emontrer

l'in �egalit �e 5x2 + 10y2� 14xy � 0. Or,

5x2 + 10y2� 14xy = x2 � 2xy + y2 + 4x2 � 12xy+ 9y2

= (x� y)2 + (2x� 3y)2

� 0

414

la derni �ere in �egalit �e r �esultant du fait qu'une somme de carr �es est toujours

sup �erieure ou �egale �a z �ero.

6. L' �ECHIQUIER COQUET

Trouvez l'unique fa�con de colorier les 36 cases d'un �echiquier 6� 6 en

noir et blanc de sorte que chacune des cases soit voisine d'un nombre impair

de cases noires.

(Note: deux cases sont voisine si elles se touchent par un cote-ou par

un coin). On ne demande pas de d �emontrer que la solution est unique.

Solution. Ayant une solution, chacune des quatre symm�etries possi-

bles (les deux diagonales, la verticale et l'horizontale) engendre une nou-

velle solution. Or, puisque la solution est unique, elle doit n �ecessairement

etre sym�etrique selon ces quatre axes de sym�etries. Il s'ensuit que les cases

num�erot �ees 1 ci-dessous sont toutes de la meme couleur. De meme pour les

cases num�erot �ees 2, num�erot �ees 3; : : : ; num�erot �ees 6.

1 2 3 3 2 1

2 4 5 5 4 2

3 5 6 6 5 3

2 4 5 5 4 2

1 2 3 3 2 1

Pour que les cases 1 touchent �a un nombre impair de cases noires, les

cases 4 doivent n �ecessairement etre noires, puisque chacune d'elles est voi-

sine de deux cases 2 et d'une case 4. De plus, les cases 4 �etant voisines d'une

case 1, d'une case 6, et de deux cases 2, 3 et 5, il est clair que les cases 1

et 6 doivent etre de couleurs di� �erentes. De plus, les cases 4 �etant noires,

les cases 6 sont n �ecessairement blanches, puisqu'elles sont voisines chacune

d'une case 4, de deux cases 5 et de trois cases 6. Les cases 1 sont donc noires,

puisque nous avons d �ej �a �etabli qu'elles sont de couleurs di� �erentes que les

cases 6. En consid �erant maintenant les cases 3, et ce qu'on sait d �ej �a, on

conclut que les cases 2 et 3 doivent etre de meme couleur. Maintenant, en

consid �erant les cases 2, sachant que 1 et 4 sont noires et que 2 et 3 sont de

meme couleur, on d �eduit que 5 doit etre noir. En consid �erant la position 5,

maintenant, on conclut �nalement que les cases en 2 et en 3 sont blanches,

ce qui nous donne �nalement:

415

7. LA FRACTION D'ANNE GRUJOTE

En base 10, 13= 0:333 : : : . �Ecrivons 0:�3 pour ces d �ecimales r �ep �et �ees.

Comment �ecrit-on 13dans une base b, o �u b est de la forme

(i) b = 3t (trois points),

(ii) b = 3t+ 1 (trois points),

(iii) b = 3t+ 2 (quatre points),

o �u t est un entier positif quelconque?

Solution. (i) Notons 13= t

3t, d'ou 1

3= 0:t en base 3t.

(ii) Supposons que 13= 0:d1d2d3 : : : . Alors 3t+1

3= d1:d2d3 : : : et d1

est �egal �a la partie enti �ere de 3t+13

. Donc d1 = t.

Soustrait d1 = t de 3t+1

3. On obtient

3t+ 1

3� t =

3t+ 1

3� 3t

3=

1

3= 0:d2d3 : : : :

Le proc �ed �e donne t = d2 = d3 = : : : . Donc 13= 0:t.

(iii) Supposons encore 1

3= 0:d1d2d3 : : : . Alors 3t+2

3= d1:d2d3 : : : :

La partie enti �ere de 3t+2

2est t. Donc d1 = t.

3t+ 2

3� 3t

3=

2

3= 0:d2d3 : : : :

On multiplie encore par 3t + 2. On obtient d2:d3d4 : : : =2(3t+2)

3= 6t+4

3,

dont la partie enti �ere est 2t+1. Donc d2 = 2t+1. On soustrait d2 = 2t+1

et on obtient

6t+ 4

3� 3(2t+ 1)

3=

(6t+ 4)� (6t+ 3)

3

=1

3= 0:d3d4 : : : :

On voit que t = d1 = d3 = d5 = : : : et 2t+1 = d2 = d4 = d6 = : : : : Donc

1

3= 0:t (2t+ 1):

416

V �eri�cation

0:t (2t+ 1) =(3t+ 2)t+ (2t+ 1)

(3t+ 2)2

�1 +

1

(3t+ 2)2+

1

(3t+ 2)4+ � � �

�

=3t2 + 4t+ 1

(3t+ 2)2

(1

1� 1(3t+2)2

)

=3t2 + 4t+ 1

(3t+ 2)2

�(3t+ 2)2

(3t+ 2)2 � 1

�

=3t2 + 4t+ 1

9t2 + 12t+ 3

=1

3:

That completes the Skoliad Corner for this issue. Please send me suit-

able contest materials at this level, as well as your comments and suggestions

for future columns.

417

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to the

Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,

PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic

address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

Factorial Fanaticism

If you've seen one too many math contests, then the following factorial

problem will be old hat: \How many factors of 2 are there in 100!?" For

those who don't recognize \!" as a mathematical symbol, here is its meaning:

n! is the product of all the integers from n down to 1, or

n! = n� (n� 1)� (n� 2)� � � � � 3� 2� 1:

For example, 5! = 5� 4� 3� 2� 1 = 120.

So how many factors of 2 are there in 100!? Well, there are1002

= 50 even numbers giving 50 factors of 2. But every multiple of 4 gives

an additional factor of 2. There are 1004

= 25 of them. Continuing, each

multiple of 8 gives another factor of 2, and there are b1008c = 12 of them,

and so on for multiples of 16, 32, and 64. This is enough, as the next power

of 2, that is 128, is greater than 100, so there are no multiples of it which are

less than 100. The total number of factors of 2 is then

50 + 25 + 12 + 6 + 3 + 1 = 97:

[Ed.: Coincidence?]

Now try and �nd how many factors of 5 there are 100!. Find a general

formula for �nding the number of primes p in n! for given integers p and n.

A similar problem to the one above is to calculate the number of con-

secutive zeros at the end of 100!. It is equal to the number of factors of 5

418

in 100! (Why?). Once you have braved this question, you can easily �gure

out a method for �nding the number of factors of m in n! for given integers

m and n.

In fact, it has been shown (and is an excellent exercise for the reader to

show) that the number of factors of a prime number p in n! is

n� sn

p� 1;

where sn is the sum of the digits of n when expressed in base p.

A natural question to ask, after knowing all the zeros at the end of n!,

is \What is the last non-zero digit in n!?" One way, perhaps not the most

elegant, is to write n! in the form p�1

1 p�2

2 � � � p�rr , where the pi are all the

distinct primes dividing n. The �1, : : : , �r, are calculated by the method

above. Since we don't care about the trailing zeros, say k of them, remove a

factor of 10k and then take the quotient modulo 10. For those who are not

familiar with modular arithmetic, all this does is �nd the units digits.

For example, 6! = 720 = 24 � 32 � 5. Removing a factor of 10 to get

rid of the trailing zeros gives 72 = 23 � 32. Then 23 � 32 is congruent to

2 modulo 10, that is, the units digit is 2. Thus, the last non-zero digit of 6!

is 2.

Now for some more interesting questions.

1. How many digits does n! have?

2. What is the initial digit of n!, or more generally its rth digit?

3. How many digits are zeros in n!? How many are 1's? Likewise for the

other digits.

419

Solving the Quartic

Cyrus Hsia

Most senior high school students are familiar with the general solution

to the quadratic equation ax2 + bx+ c = 0. Either it is derived or given as

a mere fact in class, and the notorious solution is

x =�b�pb2 � 4ac

2a:

It is rarely the case, however, that the solution to the general cubic

equation ax3+ bx2+ cx+ d = 0 is ever derived much less even mentioned.

Here we do not tread this path, but refer the reader to [1] and [2] for its

treatment. Instead, we go one step further and give the process of solving

the general quartic (fourth degree) equation ax4 + bx3 + cx2 + dx+ e = 0.

Readers are encouraged to read [1] and [2] to learn or refresh their

memories on solving the cubic equation. This is because we wish to reduce

the quartic equation into that of a cubic equation. The process we give here

is a variant of the solution credited to Lodovico Ferrari [3].

Motivations

Readers who are familiar with solving cubic equations know that the

�rst step is to reduce the general cubic into the depressed cubic equation

x3 + px+ q = 0. This may be of some value in the quartic case as well.

Also, if the quartic equation could be written as a di�erence of squares,

that is (x2 + p)2� (qx+ r)2 = 0, then solving the original quartic becomes

an exercise in solving two quadratic equations | namely, solving

x2 + p+ qx+ r = 0 and x2 + p� qx� r = 0:

To this end, we try to convert the general quartic equation into this

form somehow. Supposing we have reduced ax4 + bx3 + cx2 + dx+ e = 0

into the form x4+ux2 + vx+w = 0. (How do we do this? See the method

below.) Then we compare coe�cients in

x4 + ux2 + vx+w = (x2 + p)2 � (qx+ r)2

= x4 + (2p� q2)x2 + (�2qr)x+ (p2 � r2)

to solve for p, q, and r in terms of the known u, v, andw. (How is this done?

See the method below.) Voil �a! The solution falls apart.

Method for Solving the General Quartic Equation

� Step 1: Reduce ax4 + bx3 + cx2 + dx + e = 0 to the form

x4 + ux2 + vx + w = 0. To do this, simply divide through by a.

(We must have a 6= 0; else the equation would no longer be a quartic.)

Next, make the substitution x = t� b4a

to get t4 + ut2 + vt+ w = 0.

420

� Step 2: Convert the reduced quartic from Step 1 to a di�erence of

squares as described above. To do this, we need to solve for p, q, and r

in terms of the known u, v, and w. By comparing coe�cients, we have

three equations in three unknowns:

2p� q2 = u

�2qr = v

p2 � r2 = w

Solving for p, we have v2 = 4q2r2 = 4(2p�u)(p2�w). This is a cubic

equation in p, which can be solved.

� Step 3: Take the di�erence of squares found in Step 2 and solve for the

two quadratic equations to get the four solutions of the quartic equation

t4+ut2+vt+w = 0. Remember, these are not the desired solutions.

Make the substitution x = t � b4a

for each of the four solutions to

get the four solutions of the original quartic equation. (Why didn't we

multiply by a?)

Afterword

This seems quite tedious, and is thus one reason why it is not taught in

classrooms. However, note how simple solving the quartic is if a black-box

was given to automatically solve a cubic. The quartic is solved by using this

black-box once and solving an additional two quadratic equations.

The best way to learn a technique is to use it. Here is a problem where

the reader is encouraged to apply the above algorithm. Although there are

more elegant ways to solve this, by using Ferrari's method the reader will

gain a greater understanding of solving polynomial equations.

Solve the equation

x2 +x2

(x+ 1)2= 3:

(1992 CMO Problem 4)

We end with a note that polynomial equations of degree �ve or higher

cannot be solved, so readers can be relieved that there are not longer processes

which reduce polynomials one degree at a time to reach a solution.

References

[1 ] Hlousek, Daniela and Lew Dion, \Cardano's Little Problem", Mathe-matical Mayhem, Volume 6, Issue 3, March/April 1994, pp. 15-18.

[2 ] Dunham, William, Journey Through Genius, Penguin Books, New York,

1990, pp. 142-149.

[3 ] Dunham, William, Journey Through Genius, Penguin Books, New York,

1990, p. 151.

421

Pizzas and Large Numbers

Shawn Godin

The following is an excerpt from a television commercial for a certainunnamed pizza chain:

Customer: So what's this new deal?

Pizza Chef: Two pizzas.

Customer: [Towards four-year-old Math Whiz] Two pizzas. Write that down.

Pizza Chef: And on the two pizzas choose any toppings - up to �ve [from the

list of 11 toppings].

Older Boy: Do you...

Pizza Chef: ...have to pick the same toppings on each pizza? No!

Math Whiz: Then the possibilities are endless.

Customer: What do you mean? Five plus �ve are ten.

Math Whiz: Actually, there are 1 048 576 possibilities.

Customer: Ten was a ballpark �gure.

Old Man: You got that right.

[At this point, the camera fades to a picture of hot pizzas; the announcer's voice

and a very small Roman who says \Pizza! Pizza!"]

This simple advertisement caused an uproar in the mathematics com-

munity ([1], [2], [3], [4], [5]). The question is, should a four year old be

trusted with higher mathematics? To answer the question, we should �rst

do a little analysis.

Before jumping into the problem, let us state some assumptions:

(i) Each pizza may have up to �ve toppings on it (given).

(ii) Both pizzas do not have to have the same toppings (given).

(iii) A pizza may have no toppings (although a boring choice, it seems rea-

sonable since most pizza places include a price for \sauce and cheese"

as the basic \no item" pizza).

(iv) The order that the items are put on the pizza does not matter. So a

pizza with ham and bacon is the same as a pizza with bacon and ham (it

seems obvious, but this decision will a�ect our solution).

(v) The order that you order the pizzas, or the order that they are put in

the box, or handed to you, does not matter. That is, ordering a pizza

with ham and a second pizza with mushrooms is the same as ordering a

pizza with mushrooms and another with ham (again, it seems obvious,

but this decision will a�ect our solution).

422

(vi) Double items are not allowed (although it is allowed in most pizza

places, we will consider this simpler problem �rst then extend it).

Now o� to the solution. First, we will look at the number of ways that

one pizza can be made. Our pizza may have zero, one, two, three, four, or

�ve toppings, so the total number of pizzas available to us is:�11

5

�+

�11

4

�+ � � �+

�11

0

�= 1024:

Thus, if we are to choose two pizzas we have two possibilities:

Case 1: Both pizzas are the same. In this case, there are 1024 ways to

pick the �rst pizza and, having picked the �rst one, we are locked in for our

second one (it must be the same as the �rst) so there are 1024 ways to pick

two pizzas if they must be the same.

Case 2: Both pizzas are di�erent. In this case, there are 1024 pizzas to

choose from and I want to pick two, so the number of ways to do this is:�1024

2

�= 523 776:

So the total number of ways to pick the two pizzas is:

1024 + 523 776 = 524 800;

which does not agree with the four year old. Now, as luck would have it, if we

disregard our assumption (v), the number of ways to create pizzas in Case 2

is P (1024;2) = 1 047 552, giving a total of 1024 + 1 047 552 = 1 048 576,

which is the four year old's claim.

It would seem that the child regards the order that the pizzas are or-

dered in as important. Indeed, Jean Sherrod of Little Caesars Enterprises

explained in a letter to Mathematics Teacher ([3]) that that is indeed what

they assumed. If they are assuming this, that is disregarding our assumption

(v), would not it also make sense to regard the order that the toppings were

asked for as important? How many pizzas would be possible if we rewrote

assumption (iv) so that order was important?

Let us pursue the truth even further. As is well known to pizza con-

noisseurs, and as is admitted by Sherrod ([3]), a pizza may have a multiple

selection of an item. Double pepperoni would count as two choices, both

of the same topping. Weber and Weber ([4]) consider the multiple topping

question with an interesting technique.

Suppose our 11 toppings are as follows: pepperoni (P), bacon (B),

sausage (S), ham (H), anchovies (A), mushrooms (M), green peppers (GP),

tomatoes (T), green olives (GO), black olives (BO), and hot peppers (HP).

When you order a pizza, the order is processed on an order form that lists all

423

possible toppings separated by vertical lines. The order form may look like

this:

P j B j S j H j A j M j GP j T j GO j BO j HP

and the person taking the order just marks an X over the symbol for

each topping you order. If you wanted double pepperoni, two X's would be

placed in the P slot, counting as two toppings.

If we allow multiple toppings, we can calculate the total number of

pizzas possible by considering the total number of ways that the order form

can be �lled out. For example, if you ordered your own favourite pizza, the

\Cardiac Conspiracy", which has double bacon, double pepperoni, and ham,

the order form (if we just look at the lines and X's) would look like this:

XX j XX j j X j j j j j j jSimilarly, if you ordered a \Trip To The Sea" pizza (quintuple anchovies),

the order form would look like this:

j j j j XXXXX j j j j j jThus, each 5 item pizza will be represented by an arrangement of 5 X's

and 10 lines. The number of ways to arrange these 15 symbols is:

15!

5!10!= 3003:

This is in fact the binomial coe�cient C(15;5), or C(15; 10). This re-

sult makes more sense if we think of the problem as picking 5 of the 15

positions to mark with an X (or alternately, pick 10 of the 15 spaces to mark

with a line).

To account for all possible pizza choices, we need to consider pizzas with

zero, one, two, three, four, or �ve toppings. If we do this we get:

15!

10!5!+

14!

14!4!+ � � �+ 10!

10!0!= 4368:

So, the number of choices for two pizzas is then C(4368; 2) + 4368 =

9 541 896. So now we have a lot of numbers to deal with. In our �rst

calculation, we saw that there are 1024 possible pizzas allowing no multiple

toppings. In our latest calculation, we saw that there are a total of 4368

pizzas if multiple toppings are allowed. Thus, since the original 1024 has

been accounted for in the 4368, it would seem that 4368 � 1024 = 3344

pizzas have multiple toppings. Our next problem to consider is to classify

these 3344 pizzas.

Consider the case where a double topping is allowed and is used. Then

a �ve item pizza could be made up of a double item and three singles, or two

double items and a single. The method of Weber and Weber for enumerating

pizza choices here becomes too cumbersome to work with. For example, if

I marked a double item with a D and a single with an X, then arranging 10

424

lines, a D and 3 X's to get our �rst case does not take into account X's beside

the D (more than a double) or even two or three X's together (which means

we recount later cases). So to solve the problem, we will have to go back to

the fundamental counting theorem.

First, we will consider the pizza made up of a double item and three sin-

gles. Our �rst task is to select the item that will be double. This can be done

inC(11; 1) = 11ways. Next, pick the three singles from the other remaining

toppings. This can be done in C(10;3) = 120 ways. So the total number

of pizzas with a double item and three singles is C(11;1)C(10;3) = 1320.

Continuing in this manner, we can account for all pizzas with at least one

double topping and no topping is allowed more than twice (can you identify

each term with the case it represents?):�11

1

��10

3

�+

�11

2

��9

1

�+

�11

1

��10

2

�+

�11

2

�+

�11

1

��10

1

�+

�11

1

�= 2486:

Now we have accounted for 1024 + 2486 = 3510 of the total 4368

possible single pizzas. At this point, the reader should have enough ammu-

nition to attack the remaining problem. The following questions are still

unanswered.

1. How many pizzas are possible if triple (quadruple, quintuple) toppings

are allowed? (pick your cases carefully!)

2. How many choices of two pizzas do you have if you are allowed to pick

an item at most twice (three, four, �ve times)?

3. Using local pizza company choices, how many pizzas (or pairs of pizzas)

are possible from each pizza place?

4. Who in town makes the best pizza?

Have fun trying to solve these and other problems of your creation.

Mathematics is a demanding endeavour, so if you feel hungry you may want

to pick up the phone...

References

1. N. Kildonan, problem 1946, Crux Mathematicorum, 20:5 (1994) 137.

2. M. Woltermann, \Pizza Pizza", Mathematics Teacher, 87:6 (1994) 389.

3. J. Sherrod, \Little Caesars Responds",Mathematics Teacher, 87:6 (1994)474.

4. G. Weber and G. Weber, \Pizza Combinatorics", The College Mathe-matics Journal, 26:2 (1995) 141-143.

5. S. Godin, Solution to problem 1946, Crux Mathematicorum, 21:4 (1995)

137-139.

425

Minima and Maxima of Trigonometric Expressions

Nicholae Gusita

Problem 1. Find the minimum and maximum values of

y1 = sinx+ cosx;

for 0� � x < 360�.

Solution. We write

y1 = sinx+ sin(90� � x) = 2 sin45� cos (x� 45�) =p2 cos (x� 45�) :

Then the minimum of �p2 occurs when x � 45� = 180�, so that

x = 225�, and the maximum ofp2 occurs when x � 45� = 0, so that

x = 45�.

Problem 2. Find the minimum and maximum values of

y2 = 3 sinx+ 4 cosx;

for 0� � x < 360�.

Solution. If we put � = tan�1�43

�, then

y2 = 3 (sinx+ tan� cosx) = 3

�sinx+

sin�

cos�cosx

�=

3 sin(x+ �)

cos�:

But

cos2 � =1

1 + tan2 �=

9

25;

so that

cos� =3

5;

since 0� < � < 90�. Then y2 = 5 sin (x+ �). It is clear now that the

minimum of �5 occurs when x+ � = 270�, so that x = 270� � �, and the

maximum of 5 occurs when x+ � = 90�, so that x = 90� � �.

Problem 3. Find the minimum and maximum values of

y3 = a sinx+ b cosx;

for 0� � x < 360�, a, b 2 Rnf0g.Solution. Let � = tan�1

�ba

�, �90� < � < 90�. Assume a > 0. Then

y3 = a (sinx+ tan� cosx) = a

�sinx+

sin�

cos�cosx

�=a sin (x+ �)

cos�:

426

But as in Problem 2,

cos2� =1

1 + tan2 �=

a2

a2 + b2;

so that

cos� =ap

a2 + b2:

Therefore, y3 =pa2 + b2 sin(x+ �).

The minimum of �pa2 + b2 occurs when x + � = 270�, so that x =

270� � �, and the maximum ofpa2 + b2 occurs when x+ � = 90�, so that

x = 90� � �.

If a < 0, then using similar reasoning, we �nd the minimum occurs

when x = 90� � �, and maximum when x = 270� � �.

Problem 4. Given two positive angles a and b whose sum is 150�, prove

that the maximum value of sina+ cos b isp3.

Solution. We have

y4 = sina+ cos b = sina+ sin (90� � b)

= 2 sin

�a� b

2+ 45�

�cos

�a+ b

2� 45�

�

= 2 sin (a� 30�) cos 30� =p3 sin(a� 30�) :

The maximum ofp3 occurs when a�30� = 90�, so that a = 120�, b = 30�.

Problem 5. Find the minimum and maximum values of

y5 = a sin2 x+ b sinx cosx+ c cos2 x;

0 � x < 360�, a, b, c 2 R.Solution. Since sin

2x = 1

2(1� cos 2x), cos2 x = 1

2(1 + cos 2x), and

sinx cosx = 12sin2x,

y5 =a

2(1� cos 2x) +

b

2sin2x+

c

2(1 + cos 2x)

=a+ c

2+

1

2[b sin2x+ (c� a) cos 2x] :

By problem (3), the minimum and maximum values are

a+ c�qb2 + (a� c)

2

2and

a+ c+

qb2 + (a� c)

2

2

respectively.

427

J.I.R. McKnight Problems Contest 1979

1. Sand is being poured on the ground forming a pile in the shape of a cone

whose altitude is 23of the radius of its base. The sand is falling at the

rate of 20 L/s. How fast is the altitude of the pile increasing when it is

42 cm high?

2. Solve:

tan�1�x� 1

x+ 1

�+ tan�1

�2x� 1

2x+ 1

�= tan�1

�23

36

�:

3. A cylinder is inscribed in a cone, the altitude of the cylinder being equal

to the radius of the base of the cone. Find the measure of angle �, if

the ratio of the total surface area of the cylinder to the area of the base

of the cone is 3 : 2.

�

-�R

6

?

R

4. Solve the equation jxj+ jx� 1j+ jx� 2j = 4.

5. Solve for x and y.

2x+y = 6y

3x = 3 � 2y+1

6. A �xed circle of radius 3 has its centre at (3;0). A second circle has its

centre at the origin, its radius is approaching zero. A line joins the point

of intersection of the second circle and the y-axis to the intersection of

the two circles. Find the limit of the point of intersection of this line

with the x-axis.

428

Mayhem Problems

The Mayhem Problems editors are:

Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only problems | the

next issue will feature only solutions.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from this

issue be submitted by 1 December 1997, for publication in the issue 5 months

ahead; that is, issue 4 of 1998. We also request that only students submit

solutions (see editorial [1997: 30]), but we will consider particularly elegant

or insightful solutions from others. Since this rule is only being implemented

now, you will see solutions from many people in the next few months, as we

clear out the old problems from Mayhem.

High School Problems

Editor: Richard Hoshino, 17 Norman Ross Drive, Markham, Ontario,

Canada. L3S 3E8 <rhoshino@undergrad.math.uwaterloo.ca>

H229. Here's a simple way to remember how many books there are

in the Bible. Remember that there are x books in the Old Testament, where

x is a two-digit integer. Then multiply the digits of x to get a new integer y,

which is the number of books in the New Testament. Adding x and y, you

end up with 66, the number of books in the Bible. What are x and y?

H230. Dick and Cy stand on opposite corners (on the squares) of a 4

by 4 chessboard. Dick is telling too many bad jokes, so Cy decides to chase

after him. They take turns moving one square at a time, either vertically or

horizontally on the board. To catch Dick, Cy must land on the square Dick is

on. Prove that:

(i) If Dick moves �rst, Cy can eventually catch Dick.

(ii) If Cy moves �rst, Cy can never catch Dick.

(Can you generalize this to a 2m� 2n chessboard?)

H231. Let O be the centre of the unit square ABCD. Pick any point

P inside the square other than O. The circumcircle of PAB meets the cir-

cumcircle of PCD at points P and Q. The circumcircle of PAD meets the

circumcircle of PBC at points P and R. Show that QR = 2 � OP .

429

H232. Lucy and Anna play a game where they try to form a ten-digit

number. Lucy begins by writing any digit other than zero in the �rst place,

then Anna selects a di�erent digit and writes it down in the second place,

and they take turns, adding one digit at a time to the number. In each turn,

the digit selected must be di�erent from all previous digits chosen, and the

number formed by the �rst n digits must be divisible by n. For example, 3,

2, 1 can be the �rst three moves of a game, since 3 is divisible by 1, 32 is

divisible by 2 and 321 is divisible by 3. If a player cannot make a legitimate

move, she loses. If the game lasts ten moves, a draw is declared.

(i) Show that the game can end up in a draw.

(ii) Show that Lucy has a winning strategy and describe it.

Advanced Problems

Editor: Cyrus Hsia, 21 Van Allan Road, Scarborough, Ontario, Canada.

M1G 1C3 <hsia@math.toronto.edu>

A205. Find all functions f : R! Rsuch that f(x) � x and f(x+y)�f(x) + f(y) for all x, y 2 R.

A206. Let n be a power of 2. Prove that from any set of 2n�1 positive

integers, one can choose a subset ofn integers such that their sum is divisible

by n.

A207. Given triangle ABC, let A0, B0, and C0 be points on sidesBC,

CA, and AB respectively such that 4A0B0C0 � 4ABC. Find the locus of

the orthocentre of all such triangles A0B0C0.

A208. Let p be an odd prime, and let Sk be the sum of the products

of the elements f1; 2; : : : ; p � 1g taken k at a time. For example, if p = 5,

then S3 = 1 � 2 � 3 + 1 � 2 � 4 + 1 � 3 � 4 + 2 � 3 � 4 = 50. Show that p j Sk for

all 2 � k � p� 2.

Challenge Board Problems

Editor: Ravi Vakil, Department of Mathematics, Princeton Univer-

sity, Fine Hall, Washington Road, Princeton, NJ 08544{1000 USA

<vakil@math.princeton.edu>

C74. Prove that the k-dimensional volume of a parallelopiped in Rn

spanned by the vectors ~v1, : : : , ~vk is the determinant of the k � k matrix

fvi � vjgi;j.

430

PROBLEMS

Problem proposals and solutions should be sent to Bruce Shawyer, De-partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.

In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.

To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81

2"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailedto the Editor-in-Chief, to arrive no later than 1 May 1998. They may alsobe sent by email to crux-editors@cms.math.ca. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX, either

to the Editor-in-Chief's departmental FAX machine in St. John's, Newfoundland, or

to the Canadian Mathematical Society's FAX machine in Ottawa, Ontario. While we

understand the reasons for solvers wishing to use this method, we have found many

problems with it. The major one is that hand-written material is frequently transmit-

ted very badly, and at times is almost impossible to read clearly. We have therefore

adopted the policy that we will no longer accept submissions sent by FAX. We will,

however, continue to accept submissions sent by email or regular mail. We do en-

courage email. Thank you for your cooperation.

2276. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Quadrilateral ABCD is cyclic with circumcircle �(0; R).

Show that the nine-point (Feuerbach) circles of 4BCD, 4CDA,

4DAB and4ABC have a point in common, and characterize that point.

431

2277. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.

For n � 1, de�ne

un =

�1

(1; n);

2

(2; n); : : : ;

n� 1

(n� 1; n);

n

(n; n)

�;

where the square brackets [ ] and the parentheses ( ) denote the least com-

mon multiple and greatest common divisor respectively.

For what values of n does the identity un = (n� 1)un�1 hold?

2278. Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel, Alcorc �on,Madrid, Spain.

Determine the value of an, which is the number of ordered n{tuples

(k2; k3; : : : ; kn; kn+1) of non-negative integers such that

2k2 + 3k3 + : : :+ nkn + (n+ 1)kn+1 = n+ 1:

2279. Proposed byWalther Janous, Ursulinengymnasium, Innsbruck,Austria.

With the usual notation for a triangle, prove that

Xcyclic

sin3A cosB cosC =sr

4R4

�2R2 � s2 + (2R+ r)2

�:

2280. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle with incentre I. Let D be the second intersection of

AI with the circumcircle of4ABC. Let X, Y be the feet of the perpendic-

ulars from I to BD, CD respectively.

Suppose that IX + IY = 12AD. Find \BAC.

2281. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC is a triangle, and D is a point on the side BC produced beyond

C, such that AC = CD. Let P be the second intersection of the circumcircle

of4ACD with the circle on diameter BC. Let E be the intersection of BP

with AC, and let F be the intersection of CP with AB.

Prove that D, E, F , are collinear.

2282. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

A line, `, intersects the sides BC, CA, AB, of 4ABC at D, E, F

respectively such that D is the mid-point of EF .

Determine the minimum value of jEF j and express its length as ele-

ments of 4ABC.

432

2283. Proposed by Waldemar Pompe, student, University of War-saw, Poland.

You are given triangle ABC with \C = 60�. Suppose that E is an

interior point of line segment AC such that CE < BC. Suppose that D is

an interior point of line segment BC such that

AE

BD=BC

CE� 1 :

Suppose that AD and BE intersect in P , and the circumcircles of AEP and

BDP intersect in P and Q. Prove that QE k BC.

2284. Proposed by Toshio Seimiya, Kawasaki, Japan.

ABCD is a rhombus with \A = 60�. Suppose that E, F , are points

on the sides AB, AD, respectively, and that CE, CF , meet BC at P , Q

respectively. Suppose that BE2 +DF 2 = EF 2.

Prove that BP 2 +DQ2 = PQ2.

2285. Proposed by Richard I. Hess, Rancho Palos Verdes, California,USA.

An isosceles right triangle can be 100% covered by two congruent tiles.

Design a connected tile so that two of them maximally cover a non-

isosceles right triangle. (The two tiles must be identical in size and shape

and may be turned over so that one is the mirror image of the other. They

must not overlap each other or the border of the triangle.)

What coverage is achieved for a 30{60{90 right triangle?

2286.Proposed by Zun Shan and Edward T.H. Wang, Wilfrid Laurier Univer-

sity, Waterloo, Ontario.It is well known and easy to show that the product of any four consec-

utive positive integers plus one, is always a perfect square. It is also easy to

show that the product of any two consecutive positive integers plus one is

never a perfect square. It is possible that the product of three consecutive

integers plus one is a perfect square. For example:

2� 3� 4 + 1 = 52 and 4� 5� 6 + 1 = 112:

(a) Find the next largest natural number n such that n(n+ 1)(n+ 2) + 1

is a perfect square.

(b)? Are there any other examples?

433

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2090. [1995: 307] Proposed by Peter Iv �ady, Budapest, Hungary.For 0 < x < �=2 prove that

�sinx

x

�2<�2 � x2

�2 + x2:

Essentially the same solution was sent in by Murray S. Klamkin, Uni-versity of Alberta, Edmonton, Alberta; V�aclav Kone �cn �y, Ferris State Univer-sity, Big Rapids, Michigan, USA; and Heinz-J �urgen Sei�ert, Berlin, Germany.

For 0 < x < �2, we have

�sinx

x

�2=

1Yn=1

�1� x2

(n�)2

�2<

�1� x2

�2

�2<

�2 � x2

�2 + x2:

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,Bosnia and Herzegovina; THEODORE CHRONIS, student, Aristotle Univer-sity of Thessaloniki, Greece; LUIS V. DIEULEFAIT, IMPA, Rio de Janeiro,Brazil; PETER HURTHIG, Columbia College, Burnaby, BC; WALTHERJANOUS, Ursulinengymnasium, Innsbruck, Austria; KEE-WAI LAU, HongKong; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; CHRISWILDHAGEN, Rotterdam, the Netherlands; and the proposer. Two incom-plete solutions were received.

Most solvers proved a better inequality, and some commented that the in-terval could be extended to (��; �).Klamkin and Manes both point out the complimentary inequality (which isnot as easy to prove):

1� t2

1 + t2� sin�t

�tfor all real t.

See American Mathematical Monthly, 76 (1969), pp. 1153-1154, R. Redhef-fer, problem 5642.

Janous gave the extension to: what is the value of � that gives the best in-

equality of the type �sinx

x

�2<

�2 � x2

�2 + x2

which is valid for all x 2 (0; �=2)? We leave this as a challenge to our otherreaders!

434

2169. [1996: 274] Proposed by D.J. Smeenk, Zaltbommel, the Neth-erlands.

AB is a �xed diameter of circle �1(0;R). P is an arbitrary point of its

circumference. Q is the projection onto AB of P . Circle �2(P1PQ) inter-

sects �1 at C and D. CD intersects PQ at E. F is the midpoint of AQ.

FG ? CD, where G 2 CD. Show that:

1. EP = EQ = EG,

2. A, G and P are collinear.

Solution by Toshio Seimiya, Kawasaki, Japan.

1. Let H be the second intersection of PQ with �1. Since PH ? AB,

AB is the perpendicular bisector of PH, so thatPQ = QH and

\PAQ = \HAQ. Since PC = PQ = PD, we get \PHC =

\PDC = \PCD, so that 4PHC � 4PCE, from which we have

PH : PC = PC : PE:

Thus we have PH � PE = PC2 = PQ2. As PH = 2PQ, we have

2PQ � PE = PQ2, so that 2PE = PQ; thus, PE = EQ.

Since FG ? CD and PQ ? AB, we have \GFQ = \PEC. As

4PHC � 4PCE we get \PEC = \PCH = \PAH = 2\PAQ.

Thus \GFQ = 2\PAQ. Since F;E are midpoints of AQ;PQ, we

get FEjjAP , so that \PAQ = \EFQ. Thus \GFQ = 2\EFQ, so

that \GFE = \EFQ. Hence we have 4GFE � 4QFE, so that

EG = EQ. Therefore EP = EQ = EG.

2. Since 4GFE � 4QFE we have FG = FQ = AF , so that \GAF =12\GFQ = \EFQ. Thus we have AGjjFE. Since AP jjFE;A;G and

P are collinear.

Also solved by Francisco BellotRosado, I.B. Emilio Ferrari, Valladolid, Spain;FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; GOTTFRIED PERZ, Pestalozzigymnasium, Graz,Austria; CRIST �OBAL S �ANCHEZ{RUBIO, I.B. Penyagolosa, Castell �on, Spain;and the proposer.

435

2170. [1996: 274] Proposed by Tim Cross, King Edward's School,Birmingham, England.

Find, with justi�cation, the positive integer which comes next in the

sequence 1411; 4463;4464; 1412; 4466; 4467;1413; 4469; : : : .

[Ed.: the answer is NOT 4470.]

Editor's summary based on the solutions and comments submitted bythe solvers whose names appear below.

Most solvers felt (and the editors agree) that the answer could be any-

thing, as stated in the following comment by Murray Klamkin:

Any number can be the next term! A set of numbers is a se-quence mathematically if and only if a rule of formation is given.The given set of numbers is not a sequence and so the problem ismeaningless. Given any �nite set of n numbers, one can always�nd an in�nite number of formulae which agree with the givenn terms, and such that the (n+1)th term is completely arbitrary.

However, some solvers did come up with interesting formulae or reasonings

to \justify" the answer that they gave. Some examples:

I. (Diminnie) Let gn = n� 3�n

3

�, and let

xn = 31487 + bn=3c2� (�1)g(n)

+ 12

n�1 + (�1)g(n)

�g(n) + 76

�(�1)g(n)� 1

�o:

Then x1, x2, : : : , x8 agree with the given terms, and x9 = 4470.

In general, if k is any number and if we de�ne

yn = xn + bn=9c(k� 4470);

then y1, y2, : : : , y8 agree with the given terms, and y9 = k.

II. (Hurthig and the proposer) Squaring each of the given terms reveals

199021; 19918369; 19927296; 1993744;

19945156; 19954089; 1996569; 19971961;

and the given numbers an (n = 1, 2, : : : , 8), are the least positive integers

whose squares begin with the digits 1990, 1991, : : : , 1997. That is, an is

the smallest positive integer k such that k2 begins with the same digits as

1989 + n.

This leads to a9 = 447.

III. (Bradley) The given numbers are the integer parts of the square roots of

1991� 103; 1992� 104; 1993� 104; 1994� 103;

1995� 104; 1996� 104; 1997� 103; 1998� 104:

436

Hence the next term would bejp

1999� 104k, or 4471.

IV. (Hess) Let f(n) =jnp10� 1

3

k. The the given numbers are

n; f(n)+ 2; f(n+ 1); n+ 1; f(n+ 1) + 2; f(n+ 2); n+ 2; f(n+ 2) + 2;

with n = 1411. Thus the next term is f(n+ 3) = f(1414) = 4471.

Other submitted answers include 4479 (Kone �cn �y) and 44610 (Ortega

and Guti �errez).

Solved by HAYO AHLBURG, Benidorm, Spain; CHRISTOPHER J.BRADLEY, Clifton College, Bristol, UK; CHARLES R. DIMINNIE, AngeloState University, San Angelo, TX, USA; RICHARD I. HESS, Rancho PalosVerdes, California, USA; PETER HURTHIG, Columbia College, Burnaby, BC;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; SOLEDAD ORTEGA andJAVIER GUTI �ERREZ, students, University of La Roija, Logro ~no, Spain; andthe proposer.

2171. [1996: 274] Proposed by Juan-Bosco Romero M�arquez, Uni-versidad de Valladolid, Valladolid, Spain.

Let P be an arbitrary point taken on an ellipse with foci F1 and F2,

and directrices d1, d2, respectively. Draw the straight line through P which

is parallel to the major axis of the ellipse. This line intersects d1 and d2 at

points M and N , respectively. Let P 0 be the point where MF2 intersects

NF1.

Prove that the quadrilateral PF1P0F2 is cyclic.

Does the result also hold in the case of a hyperbola?

Solution by Richard I. Hess, Rancho Palos Verdes, California, USA,modi�ed by the editor.

I: The Ellipse: x2

a2+ y2

b2= 1.

Let F1, F2 be�a�; 0�,��a

�; 0�, respectively, where � > 1, and

b =

p�2�1

�a. Let P be (a sin�; b cos �) and M be (�a; b cos �).

By symmetry, P 0 is on the y{axis. Let P 0 be (0;�d).

437

-

6

qq

p

q

P 0

q

Cq qq

F1F2

MN P

d1d2

Thusd

a=�=

b cos �

�a� a=�;

so that d =b cos �

�2 � 1.

Choose C to be the point (0; �) such that CF1 = CF2 = CP 0. Thus

(� + d)2 = �2 +a2

�2;

so that2�b cos �

�2 � 1+

b2 cos2 �

(�2 � 1)2=

a2

�2=

b2

�2 � 1;

giving

� =b

2 cos �� b cos �

2(�2� 1):

We now show that CP = CP 0. If this were true, we would have

(� � b cos �)2 + a2 = sin2 � =

�� +

b cos �

�2 � 1

�2;

or

�2b� cos � + b2 cos2 � +�2b2

�2 � 1sin2 � =

2b� cos �

�2 � 1+

b2 cos2 �

(�2 � 1)2;

or

�b2 + b2 cos2 �

�2 � 1+ b2 cos2 � +

b2�2

�2 � 1=

b2

�2 � 1+�2b2 cos2 �

�2 � 1;

or�2

�2 � 1� 1 =

1

�2 � 1;

438

which is clearly true.

Thus C is the centre of the cyclic quadrilateral PF1P0F2.

II: The Hyperbola: x2

a2� y2

b2= 1.

Let F1, F2 be�a�; 0�,��a

�; 0�, respectively, where 0 < � < 1, and

b =

p1��2

�a. Let P be (a sinh�; b cosh �) and M be (�a; b cosh �).

By symmetry, P 0 is on the y{axis. Let P 0 be (0; d).

Choose C to be the point (0; �) such that CF1 = CF2 = CP 0.

-

6

qC

qq

F1F2

qqMN

qP 0

q

P

Follow the procedure in case I to show that C is the centre of the cyclic

quadrilateral PF1P0F2.

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; TOSHIOSEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, the Netherlands;and the proposer (for the ellipse only).

Smeenk notes that it is easy to verify that PP 0 is a normal to the el-lipse. Bellot Rosado refers to E.A. Maxwell's Elementary Coordinate Geom-

etry, Oxford University Press, 1952, which contains the two following relatedproblems:

1. P , Q are two points on an ellipse with foci S, S0, such that PQ is

perpendicular to SS0.

Prove that PS, QS, PS0, QS0 touch a circle, and identify its centre.

2. Tangents TP , TQ are drawn to an ellipse with foci S, S0. A line through

S parallel to TQ meets S0T in U , and a line through S0 parallel to TP

meets ST in V .

Prove that S, S0, U , V lie on a circle.

439

2172. [1996: 274] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Let x; y; z � 0 with x + y + z = 1. For �xed real numbers a and b,

determine the maximum c = c(a; b) such that a+ bxyz � c(yz+ zx+xy).

SolutionbyMihai Cipu, Institute ofMathematics, Romanian Academy,Bucharest, Romania.

The answer is

c(a; b) = min

�4a; 3a+

b

9

�;

provided that a � 0.

For x = y = 1=2, z = 0 one obtains c � 4a, while by substituting x =

y = z = 1=3 it follows that c � 3a+b=9. Thus c(a; b) � min(4a; 3a+b=9).

Now to �nish the proof we shall show that for all real numbers a � 0 and b,

a+ bxyz � min

�4a; 3a+

b

9

�� (xy + yz + zx) (1)

for all x; y; z � 0, x+ y+ z = 1.

To this end we shall use the fact that for any such triple x; y; z there

exists a Euclidean triangle whose sides have lengths 1� x; 1� y;1� z. The

triangle is degenerate if xyz = 0. Let us denote by r, resp. R, the radius of

the incircle, resp. circumcircle, of the associated triangle. Using well-known

formulae and the hypothesis, one easily �nds

xyz = r2 and xy + yz + zx = r2 + 4Rr: (2)

Here r and R have non-negative values subject to the restrictions

16Rr� 5r2 � 1 and R � 2r. [Editor's note. The hypothesis x+ y+ z = 1

means that the associated triangle has semiperimeter s = 1. Thus (2) follows

from the known identities

(s� a1)(s� a2)(s� a3) = r2s andX

(s� a1)(s� a2) = r2+4Rr

which hold for any triangle with sides a1; a2; a3 | see for example equations

(15) and (16), page 54 of Mitrinovi�c, Pe�cari�c and Volenec, Recent Advancesin Geometric Inequalities. And the restriction 16Rr � 5r2 � 1 is just the

known inequality 16Rr�5r2 � s2; see (3.6) on page 166 of Recent Advances,or item 5.8 of Bottema et al., Geometric Inequalities.]

We note that for x = y = 0, z = 1 one gets a � 0 [else there is no

solution for c]. Using this fact, in the case b � 9a we have

min(4a; 3a+ b=9) = 4a

and

a+ br2 � a(1 + 9r2) � a(4r2 + 16Rr);

440

so that (1) holds. In the opposite case b � 9a we get

a(1� 3r2 � 12Rr) � a(4Rr � 8r2) � (4Rr � 8r2)b=9

[since 4Rr � 8r2 = 4r(R� 2r) � 0], or equivalently

a+ br2 � (3a+ b=9)(r2 + 4Rr);

which is (1) again.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,Bosnia and Herzegovina; FLORIAN HERZIG, student, Perchtoldsdorf, Aus-tria; RICHARD I. HESS, Rancho Palos Verdes, California, USA; MURRAYS. KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; and the proposer. Oneincorrect solution was sent in.

Most solvers noted that a solution exists only if a � 0.

2173. [1996: 275, 1997: 169]ProposedbyWalther Janous, Ursulinen-gymnasium, Innsbruck, Austria.

Let n � 2 and x1; : : : ; xn > 0 with x1 + : : :+ xn = 1.

Consider the terms

ln =

nXk=1

(1 + xk)

s1� xk

xk

and

rn = Cn

nYk=1

1 + xkp1� xk

where

Cn = (pn� 1)n+1(

pn)n=(n+ 1)n�1:

1. Show l2 � r2.

2. Prove or disprove: ln � rn for n � 3.

I. Solution to Part 1 by Richard I. Hess, Rancho Palos Verdes, Califor-nia, USA.

For n = 2, we get C2 = 2=3 and [since x1 + x2 = 1]

l2 = (1 + x1)

rx2

x1+ (1 + x2)

rx1

x2=

x1 + x2 + 2x1x2px1x2

;

r2 =2

3

�1 + x1p

x2

��1 + x2p

x1

�:

441

Thus

px1x2(r2 � l2) =

2

3(1 + x1 + x2 + x1x2)� x1 � x2 � 2x1x2

=1

3(1� 4x1x2)

=1

3

�(x1 + x2)

2 � 4x1x2�

=1

3(x1 � x2)

2 � 0:

Therefore l2 � r2 with equality if and only if x1 = x2 = 1=2.

II. Partial solution to Part 2 by the proposer.We show that ln � rn in the case n = 3 (which indeed was the starting

point for the whole problem).

Putting x1 = x, x2 = y, x3 = z, the desired inequality l3 � r3 reads

(1 + x)

r1� x

x+ (1 + y)

s1� y

y+ (1 + z)

r1� z

z

� 3p3

4� (1 + x)(1 + y)(1 + z)p

1� xp1� y

p1� z

(1)

where x; y; z 2 (0; 1) such that x+ y+ z = 1.

We now recall the di�cult Crux problem 2029 of Jun-huaHuang, solved

by Kee-Wai Lau on [1996: 129]:

wbwc + wcwa +wawb � 3Fp3; (2)

where wa; wb; wc; F are the angle bisectors and the area of a triangle. We

claim that this inequality is equivalent to inequality (1). Indeed, let us apply

the transformation a = y + z, b = z + x, c = x + y where

x; y; z > 0, converting any triangle inequality into an algebraic inequality

valid for positive numbers. Then it's not di�cult to see that x + y + z = s

(the semiperimeter of the triangle), whence x = s � a, y = s � b and

z = s � c. Furthermore, due to homogeneity we may and do put s = 1,

whence F =ps(s� a)(s� b)(s� c) =

pxyz. Also, by the known formula

wc =2ab cos(C=2)

a+ b=

2ab

a+ b

ss(s� c)

ab=

2

a+ b

ps(s� c)ab

(for example, see [1995: 321]) we get (using x+ y + z = 1)

wc =2

x+ y+ z + z

pz(y+ z)(z+ x) =

2

1 + z

pz(1� x)(1� y);

442

and similarly for wa and wb. Hence (2) is equivalent to

Xcyclic

4 �pxyp(1� y)(1� x)(1� z)

(1 + x)(1 + y)� 3

p3pxyz;

which is equivalent to (1). Since (2) is true, so is (1).

Forn � 4 I do not have any idea of how to settle whether the inequality

ln � rn is true. It may be interesting and useful to see a purely algebraic

proof of inequality (1).

Part 1 also solved by �SEFKET ARSLANAGI �C, University of Sarajevo,Sarajevo, Bosnia and Herzegovina; and the proposer. The proposer didPart 2 only in the case n = 3 (given above). One other reader sent in asolution to Part 2 which the editor considers to be faulty. Readers are in-vited to try �nishing o� this problem completely, or even just the specialcase n = 4.

2174. [1996: 275] Proposed by Theodore Chronis, student, AristotleUniversity of Thessaloniki, Greece.

Let A be an n� n matrix. Prove that if An+1 = 0 then An = 0.

Solution by John C. Tripp, Southeast Missouri State University, CapeGirardeau, Missouri.

We consider A as a linear transformation on an n-dimensional vector

space. We assume that An+1 = 0. Let x be any element of the vector space.

The set of vectors

V = fx; Ax;A2x; : : : ; Anxghas n + 1 elements, so it is linearly dependent. Let k be the smallest non-

negative integer such that Akx is a linear combination of the other vectors

in V . We have

Akx = c1Ak+1x+ c2A

k+2x+ c3Ak+3x+ � � �+ cn�kA

nx;

for some scalars c1; c2; c3; : : : ; cn�k, and

Anx = An�kAkx

= An�k(c1Ak+1x+ c2A

k+2x+ c3Ak+3x+ � � �+ cn�kA

nx)

= c1An+1x+ c2A

n+2x+ c3An+3x+ � � �+ cn�kA

2n�k)

= An+1(c1x+ c2Ax+ c3A2x+ � � �+ cn�kA

n�k�1x) = 0:

Since x was arbitrary, we have An = 0.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; MIHAI CIPU, Romanian Academy, Bucharest, Romania; CON AMOREPROBLEMGROUP, Royal Danish Schoolof Educational Studies, Copenhagen,Denmark; LUZ M. DeALBA, Drake University, Des Moines, Iowa; F.J.

443

FLANIGAN, San Jose State University, San Jose, California, USA; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; KEE-WAI LAU, Hong Kong; WALDEMAR POMPE, student,University of Warsaw, Poland; E. RAPPOS, University of Cambridge, Eng-land; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; SKIDMORE COLLEGEPROBLEMGROUP, Saratoga Springs, New York, USA; DIGBY SMITH,MountRoyal College, Calgary, Alberta; EDWARD T.H. WANG, Wilfrid Laurier Uni-versity, Waterloo, Ontario; and the proposer.

Most solvers used the Cayley-Hamilton Theorem. Wang commentsthat this problem is a special case of a more general and well-known resultwhich states that \ifA is an n�n complex matrix such thatAk = 0 for somek � 1 (that is, A is nilpotent), then An = 0". Indeed, some solvers provedthis more general result.

2175. [1996: 275] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

The fraction1

6can be represented as a di�erence in the following ways:

1

6=

1

2� 1

3;

1

6=

1

3� 1

6;

1

6=

1

4� 1

12;

1

6=

1

5� 1

30:

In how many ways can the fraction1

2175be expressed in the form

1

2175=

1

x� 1

y;

where x and y are positive integers?

Solution by D. Kipp Johnson, Valley Catholic High School, Beaverton,Oregon, USA.

Notice that1

2175=

1

x� 1

y

so that

x =2175y

y+ 2175= 2175� 21752

y + 2175:

Thus x will be an integer if and only if y + 2175 is a factor of 21752, and x

will be positive whenever y is, since then 21752=(y + 2175) < 2175, and

y will be positive whenever y + 2175 > 2175, so we seek factors of 21752

which exceed 2175. But 21752 = 32 � 54 �292 has (2+ 1)(4+1)(2+1) = 45

positive factors, one of which is its square root, 2175. Since the factors of

21752 come in pairs whose product is 21752, exactly half of the other 44

factors exceed 2175, giving 22 solutions in positive integers. The smallest is

x = 300; y = 348. This immediately generalizes to the solution in positive

444

integers of 1=n = 1=x� 1=y. Since � (n2) (the number of divisors of n2) is

odd for a perfect square, there will be (� (n2)�1)=2 solutions to the equation

1=n = 1=x� 1=y.

Also solved by HAYO AHLBURG, Benidorm, Spain; SAM BAETHGE,Nordheim, Texas, USA; THEODORE CHRONIS, student, Aristotle Universityof Thessaloniki, Greece; TIM CROSS, King Edward's School, Birmingham,England; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX,USA; KEITH EKBLAW,Walla Walla, Washington,USA; HANS ENGELHAUPT,Franz{Ludwig{Gymnasium, Bamberg, Germany; F.J. FLANIGAN, San JoseState University, San Jose, California, USA; FLORIAN HERZIG, student, Per-chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; KEE-WAILAU, Hong Kong; J.A. MCCALLUM, Medicine Hat, Alberta; ROBERT P.SEALY, Mount Allison University, Sackville, New Brunswick; DAVID R.STONE, Georgia Southern University, Statesboro, Georgia, USA; PANOS E.TSAOUSSOGLOU, Athens, Greece; LAMARR WIDMER, MessiahCollege, Grantham, PA, USA; KENNETH M. WILKE, Topeka, Kansas, USA;and the proposer. There were 4 incorrect solutions.

Flanigan refers the interested reader to problem #10501 of the Ameri-

can Mathematical Monthly, volume 103, number 2, February 1996, page 171.

2176. [1996: 275] Proposed by �Sefket Arslanagi�c, University of Sara-jevo, Sarajevo, Bosnia and Herzegovina.

Prove that

n

vuut nYk=1

(ak + bk) � n

vuut nYk=1

ak + n

vuut nYk=1

bk

where a1; a2; : : : ; an > 0 and n 2 N.Solution by Sai C. Kwok, San Diego, CA, USA.

Using the arithmetic-geometric mean inequality, we have

nY

k=1

ak

ak + bk

! 1n

� 1

n

nXk=1

ak

ak + bk

and

nY

k=1

bk

ak + bk

! 1n

� 1

n

nXk=1

bk

ak + bk

The result follows by adding the above two inequalities.

445

Also solved by THEODORE CHRONIS, student, Aristotle Universityof Thessaloniki, Greece; yMIHAI CIPU, Romanian Academy, Bucharest,Romania, and Concordia University, Montreal, Quebec; ROBERTGERETSCHL �AGER, Bundesrealgymnasium, Graz, Austria; FLORIANHERZIG,student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes,California, USA; JOHN G. HEUVER, Grande Prairie Composite High School,Grande Prairie, Alberta; yWALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids,Michigan, USA; MITKO KUNCHEV, Baba Tonka School of Mathematics,Rousse, Bulgaria; yCANANHMINH, student,University of CaliforniaBerke-ley, Berkeley, CA, USA; ySOLEDAD ORTEGA and JAVIER GUTI �ERREZ,students, University of La Rioja, Logro ~no, Spain; WALDEMAR POMPE, stu-dent, University of Warsaw, Poland; JUAN-BOSCO ROMERO M�ARQUEZ,Universidad de Valladolid, Valladolid, Spain; yHEINZ-J�URGEN SEIFFERT,Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; GEORGETSAPAKIDIS, Agrinio, Greece; and the yproposer. A proof for the case n = 4

was submitted by V.N. Murty. ( The symbol y before a solver's name indi-cates that the solver's solutionwas virtually the same as the one highlightedabove.)

Clearly, the condition that b1; b2; : : : ; bn > 0 was inadvertantly leftout from the original statement. All solvers assumed, explicitly or implicitly,that this was the case. However, only Hess gave a simple example to showthat the inequality need not be true without the aforementioned condition:take n = 2, a1 = a2 = 1 and b1 = b2 = �1.

Janous pointed out that equality holds if and only if the vectors(a1; a2; : : : ; an) and (b1; b2; : : : ; bn) are proportional.

Klamkin commented that the given inequality is an immediate specialcase of Jensen's generalization of H �older's Inequality, and referred readersto D.S.Mitrinovi�c et. al., Recent Advances in Geometric Inequalities, KluwerAcademic Publishers, 1989, pp. 50{54.

Kone �cn �y remarked that it is known that if A and B are n�n positivesemi-de�nite Hermitian matrices, then

npdet(A+B) � n

pdetA+

npdetB

(see, for example, Inequalities: Theory of Majorization and Its Application

by Albert W. Marshall and Ingram Olkin, Academic Press Inc., 1979, p. 475).If we let A and B be the n� n diagonal matrices with diagonal entries ak'sand bk's respectively (i = 1, 2, : : : , n), then the proposed inequality followsimmediately.

Sei�ert pointed out that the proof given above can be found on p. 178of the book Classical and New Inequalities in Analysis, Kluwer AcademicPublishers, 1992.

446

2177. [1996: 317] Proposed by Toshio Seimiya, Kawasaki, Japan.ABCD is a convex quadrilateral, with P the intersection of its diag-

onals and M the mid-point of AD. MP meets BC at E. Suppose that

BE : EC = (AB)2 : (CD)2. Characterize quadrilateral ABCD.

Solution by Gottfried Perz, Pestalozzigymnasium, Graz, Austria.

P

A B

C

D

M

E�

�

�

d(P; CD)

�

d(P; AB)

Let

� = \MPA = \EPC;

� = \BPE = \DPM:

Then, applying the Sine Rule

to the triangles 4APM and

4DPM , we get

sin� =AM � sin\AMP

AP;

sin� =DM � sin(180� � \AMP )

DP=AM � sin\AMP

DP;

whence

sin �

sin�=DP

AP: (3)

Applying the law of sines to the triangles4CPE, and4BPE, we get

sin � =CE � sin\CEP

CP;

sin� =BE � sin(180�� \CEP )

BP=BE � sin\CEP

BP;

whence

sin�

sin�=CE � BPBE � CP =

CD2 �BPAB2 �CP : (4)

From (1), (2) follows that quadrilateral ABCD has the desired property if

and only if

AB2

CD2=

AP � BPCP �DP =

[ABP ]

[CDP ]=

AB � d(P; AB)

CD � d(P; CD)

(where [XY Z] denotes the area of 4XY Z and d(U; V W ) the distance of

U from V W ); that is, if

AB : CD = d(P;AB) : d(P;CD):

447

This implies that ABCD is characterized by the fact that in the triangles

4ABP and 4CDP , having \APB = \CPD in common (ABCD is con-

vex!), the ratio of the side opposite to P and the altitude passing through P

is the same, which means that4ABP and4CDP are directly or inversely

similar.

In the �rst case, we have \BAP = \DCP ; that is, ABCD is a

trapezoid with parallel sides AB and CD. In the second case, we have

\BAP ( = \BAC) = \PDC ( = \BDC), whence ABCD is an inscribed

quadrilateral (A and D are at the same side of BC).

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; SABIN CAUTIS, student, Earl Haig Secondary School, North York, On-tario; CON AMORE PROBLEM GROUP, Royal Danish School of EducationalStudies, Copenhagen, Denmark; RICHARD I. HESS, Rancho Palos Verdes,California, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;WALDEMAR POMPE, student, University of Warsaw, Poland; PANOS E.TSAOUSSOGLOU, Athens, Greece; and the proposer. Four incomplete orincorrect solutions were received.

2178. [1996: 318] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

If A;B;C are the angles of a triangle, prove that

sinA sinB sinC � 8�sin3A cosB cosC + sin3B cosC cosA

+sin3C cosA cosB�

� 3p3�cos2A+ cos2B + cos2C

�:

Solution by Florian Herzig, student, Perchtoldsdorf, Austria.I shall prove a stronger version with 6 sinA sinB sinC of the left side.

We use the following known identities and inequalities:

sin 2A+ sin2B + sin2C = 4 sinA sinB sinC (1)

cos2A+ cos2B + cos2C = 1� 2 cosA cosB cosC (2)

sinA sinB sinC � 3p3

8(3)

cosA cosB cosC � 1

8(4)

where A, B, C are the angles of a triangle. Therefore

sin3A cosB cosC + sin3B cosC cosA+ sin3C cosA cosB

= sinA(1� cos2A) cosB cosC + sinB(1� cos2B) cosC cosA

+ sinC(1� cos2C) cosA cosB

448

= � cosA cosB cosC(sinA cosA+ sinB cosB + sinC cosC)

+ sinA cosB cosC + cosA(sinB cosC + sinC cosB)

= �1

2cosA cosB cosC(sin2A+ sin2B + sin2C)

+ sinA(cosB cosC + cosA)

= �2 cosA cosB cosC sinA sinB sinC + sinA sinB sinC

= sinA sinB sinC(1� 2 cosA cosB cosC)

= sinA sinB sinC(cos2A+ cos2B + cos2C):

Combining (2) and (4) yields

cos2A+ cos2B + cos2 � 3

4:

By using this inequality and (3) we get

6 sinA sinB sinC � 8(sin3A cosB cosC + sin3B cosC cosA

+ sin3C cosA cosB)

= 8 sinA sinB sinC(cos2A+ cos2B + cos2C)

� 3p3(cos2A+ cos2B + cos2C)

as we wanted to show. Equality holds as in (3) and (4) if and only if

A = B = C = 60�.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,Bosnia and Herzegovina; RICHARD I. HESS, Rancho Palos Verdes, Cali-fornia, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;MURRAY S. KLAMKIN,University of Alberta, Edmonton,Alberta; MICHAELLAMBROU, University of Crete, Crete, Greece; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the pro-poser.

Janous, in his solution, used a new identity, which is presented as aproblem 2279 in this issue.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

449

THE ACADEMY CORNERNo. 15

Bruce Shawyer

All communications about this column should be sent to BruceShawyer, Department of Mathematics and Statistics, Memorial Universityof Newfoundland, St. John's, Newfoundland, Canada. A1C 5S7

Memorial University Undergraduate

Mathematics Competition

September 25, 1997

Answer as many questions as you can. Complete solutions carry more credit

than scattered comments about many problems.

1. Determine whether or not the following system has any real solutions.

If so, state how many real solutions exist.

x+1

x= y; y +

1

y= z; z +

1

z= x:

2. The surface area of a closed cylinder is twice the volume. Determine

the radius and height of the cylinder given that the radius and height

are both integers.

3. Prove that

1 +1

4+

1

9+ : : :+

1

n2< 2:

4. Describe the set of points (x; y) in the plane for which

sin(x+ y) = sinx+ siny:

5. In a parallelogram ABCD, the bisector of angle ABC intersects AD

at P . If PD = 5, BP = 6 and CP = 6, �nd AB.

6. Show that, where k+ n � m,

nXi=0

�n

i

��m

k+ i

�=

�m+ n

n+ k

�:

Send me your nice solutions!

450

THE OLYMPIAD CORNERNo. 186

R.E. Woodrow

All communications about this column should be sent to Professor R.E.Woodrow, Department of Mathematics and Statistics, University of Calgary,Calgary, Alberta, Canada. T2N 1N4.

This number we give the 24 problems proposed to the jury, but not

selected for the 37th International Mathematical Olympiad in July 1996 at

Mumbai, India. My thanks go to Ravi Vakil, Canadian Team Leader to the

37th IMO for collecting this and other contest material and forwarding it to

me.

PROBLEMS PROPOSED TO THE JURYBUT NOT USED AT THE

37th INTERNATIONALMATHEMATICAL OLYMPIAD

July 1996 | Mumbai, India

1. Let a, b and c be positive real numbers such that abc = 1. Prove

thatab

a5 + b5 + ab+

bc

b5 + c5 + bc+

ca

c5 + a5 + ca� 1:

When does equality hold?

2. Let a1 � a2 � � � � � an be real numbers such that for all integers

k > 0,

ak1 + ak2 + � � �+ akn� 0:

Let p = maxfja1j; : : : ; janjg. Prove that p = a1 and that

(x� a1)(x� a2) � � � (x� an) � xn � an1

for all x > a1.

3. Let a > 2 be given, and de�ne recursively:

a0 = 1; a1 = a; an+1 =

a2n

a2n�1

� 2

!an:

Show that for all integers k > 0, we have

1

a0+

1

a1+

1

a2+ � � �+ 1

ak<

1

2(2 + a�

pa2 � 4):

451

4. Let a1; a2; : : : ; an be non-negative real numbers, not all zero.

(a) Prove that xn�a1xn�1�� � ��an�1x�an = 0 has precisely one positive

real root.

(b) Let A =Pn

j=1 aj , and B =Pn

j=1 jaj and let R be the positive real root

of the equation in (a). Prove that AA � RB.

5. Let P (x) be the real polynomial, P (x) = ax3+bx3+cx+d. Prove

that if jP (x)j � 1 for all x such that jxj � 1, then

jaj + jbj+ jcj+ jdj � 7:

6. Let n be an even positive integer. Prove that there exists a positive

integer k such that

k = f(x)(x+ 1)n + g(x)(xn + 1)

for some polynomials f(x), g(x) having integer coe�cients. If k0 denotes

the least such k, determine k0 as a function of n.

7. Let f be a function from the set of real numbers R into itself such

that for all x 2 R, we have jf(x)j � 1 and

f

�x+

13

42

�+ f(x) = f

�x+

1

6

�+ f

�x+

1

7

�:

Prove that f is a periodic function (that is, there exists a non-zero real num-

ber c, such that f(x+ c) = f(x) for all x 2 R).8. Let the sequence a(n), n = 1; 2; 3; : : : ; be generated as follows:

a(1) = 0, and for n > 1,

a(n) = a(bn=2c) + (�1)n(n+1)=2:

(Here btc is the greatest integer less than or equal to t.)

(a) Determine the maximum and minimum value of a(n) over n � 1996 and

�nd all n � 1996 for which these extreme values are attained.

(b) How many terms a(n), n � 1996, are equal to 0?

9. Let triangle ABC have orthocentre H, and let P be a point on its

circumcircle, distinct from A, B, C. Let E be the foot of the altitudeBH, let

PAQB and PARC be parallelograms, and let AQ meet HR in X. Prove

that EX is parallel to AP .

10. Let ABC be an acute-angled triangle with jBCj > jCAj, andlet O be the circumcentre, H its orthocentre, and F the foot of its altitude

CH. Let the perpendicular to OF at F meet the side CA at P . Prove that

\FHP = \BAC.

452

11. Let ABC be equilateral, and let P be a point in its interior.

Let the lines AP , BP , CP meet the sides BC, CA, AB in the points

A1, B1, C1 respectively. Prove that

A1B1 � B1C1 � C1A1 � A1B � B1C �C1A:

12. Let the sides of two rectangles be fa; bg and fc; dg respectively,

with a < c � d < b and ab < cd. Prove that the �rst rectangle can be placed

within the second one if and only if

(b2� a2)2 � (bc� ad)2 + (bd� ac)2:

13. Let ABC be an acute-angled triangle with circumcentre O and

circumradius R. Let AO meet the circle BOC again in A0, let BO meet the

circle COA again in B0 and let CO meet the circle AOB again in C0. Provethat

OA0 � OB0 � OC0 � 8R3:

When does equality hold?

14. Let ABCD be a convex quadrilateral, and let RA, RB , RC , RD

denote the circumradii of the triangles DAB, ABC, BCD, CDA respec-

tively. Prove that RA+RC > RB+RD if and only if \A+\C > \B+\D.

15. On the plane are given a point O and a polygonF (not necessarily

convex). Let P denote the perimeter of F , D the sum of the distances from

O to the vertices of F , and H the sum of the distances from O to the lines

containing the sides of F. Prove that D2 �H2 � P 2=4.

16. Four integers are marked on a circle. On each step we simulta-

neously replace each number by the di�erence between this number and the

next number on the circle, moving in a clockwise direction; that is, the num-

bers

a, b, c, d are replaced by a � b, b � c, c � d, d � a. Is it possible after

1996 such steps to have numbers a, b, c, d such that the numbers jbc� adj,jac � bdj, jab� cdj are primes?

17. A �nite sequence of integers a0; a1; : : : ; an is called quadratic iffor each i in the set f1; 2; : : : ; ng we have the equality jai � ai�1j = i2.

(a) Prove that for any two integers b and c, there exists a natural number n

and a quadratic sequence with a0 = b and an = c.

(b) Find the smallest natural number n for which there exists a quadratic

sequence with a0 = 0 and an = 1996.

18. Find all positive integers a and b for which�a2

b

�+

�b2

a

�=

�a2 + b2

ab

�+ ab;

453

where, as usual, btc refers to the greatest integer which is less than or equal

to t.

19. Let N0 refer to the set of non-negative integers. Find a bijective

function f from N0 into N0 such that for allm;n 2 N0,

f(3mn+m+ n) = 4f(m)f(n)+ f(m) + f(n):

20. A square (n�1)� (n�1) is divided into (n�1)2 unit squares in

the usual manner. Each of the n2 vertices of these squares is to be coloured

red or blue. Find the number of di�erent colourings such that each unit

square has exactly two red vertices. (Two colouring schemes are regarded as

di�erent if at least one vertex is coloured di�erently in the two schemes.)

21. Let k,m, n be integers such that 1 < n � m�1 � k. Determine

the maximum size of a subset S of the set f1; 2; 3 : : : ; k � 1; kg such that

no n distinct elements of S add up to m.

22. Determine whether or not there exist two disjoint in�nite sets

A and B of points in the plane satisfying the following conditions:

(a) No three points inA[B are collinear, and the distance between any two

points in A[ B is at least 1.

(b) There is a point ofA in any triangle whose vertices are in B, and there is

a point of B in any triangle whose vertices are in A.

23. A �nite number of beans are placed on an in�nite row of squares.

A sequence of moves is performed as follows: at each stage a square contain-

ing more than one bean is chosen. Two beans are taken from this square;

one of them is placed on the square immediately to the left while the other

is placed on the square immediately to the right of the chosen square. The se-

quence terminates if at some point there is at most one bean on each square.

Given some initial con�guration, show that any legal sequence of moves will

terminate after the same number of steps and with the same �nal con�gura-

tion.

24. Let U be a �nite set and f , g be bijective functions from U onto

itself. Let

S = fw 2 U : f(f(w)) = g(g(w))gand

T = fw 2 U : f(g(w)) = g(f(w))g;and suppose that U = S [ T . Prove that, for m 2 U , f(w) 2 S if and only

if g(w) 2 S.

As always we welcome your nice original solutions which di�er from

the o�cial solutions provided by the proposers and the selection committee.

454

As an example of an Olympiad which may not be as widely circulated,

and for which you may not have already seen solutions, we give the four

problems of the 4th Class for the Croatian National Mathematical Competi-

tion of May 13, 1994 and the three problems of the Croatian Mathematical

Olympiad of May 14, 1994.

My thanks go to Richard Nowakowski, Canadian Team Leader at the

35th IMO in Istanbul for collecting these problems.

CROATIAN NATIONALMATHEMATICAL COMPETITION

Fourth ClassMay 13, 1994

1. One member of an in�nite arithmetic sequence in the set of natural

numbers is a perfect square. Show that there are in�nitely many members

of this sequence having this property.

2. For a complex number z let w = f(z) =2

3� z.

(a) Determine the set fw : z = 2+ iy; y 2 Rg in the complex plane.

(b) Show that the function w can be written in the form

w � 1

w � 2= �

z � 1

z � 2:

(c) Let z0 =12and the sequence fzng be de�ned recursively by

zn =2

3� zn�1; n � 1:

Using the property (b) calculate the limit of the sequence fzng.3. Determine all polynomials P (x) with real coe�cients such that for

some n 2 N we have xP (x� n) = (x� 1)P (x), 8 x 2 R.4. In the plane �ve points P1, P2, P3, P4, P5 are chosen having integer

coordinates. Show that there is at least one pair (Pi; Pj), for i 6= j such that

the line PiPj contains a point Q, with integer coordinates, and is strictly

between Pi and Pj.

.

Additional Competition for the OlympiadMay 14, 1994

1. Find all ordered triples (a; b; c) of real numbers such that for every

three integers x; y; z the following identity holds:

jax+ by+ czj+ jbx+ cy + azj+ jcx+ ay + bzj = jxj+ jyj+ jzj:

455

2. Construct a triangle ABC if the lengths jAOj, jAU j and radius r

of the incircle are given, where O is the orthocentre and U the centre of the

incircle.

3. Let P be the set of all lines of the plane M . Does there exist a

function f : P !M having the following properties:

(a) the function f is an injection:

(b) f(p) 2 p, 8 p 2 P ?

That should provide some problems for your puzzling pleasure over the

next couple of months. Now we return to readers' solutions to problems

featured in earlier numbers of the Corner.First, an apology. Somehow, in shifting my �les around we misplaced

solutions by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain, to two

problems that we discussed in the October number of the Corner. His name

should be added as a solver of problems 6 and 7 of the Telecom 1993 Aus-

tralian Mathematical Olympiad in the solutions given [1997: 324{325].

Last number we gave solutions by the readers to the �rst ten problems

of the \Baltic Way | 92" contest given in the May 1996 number [1996: 157{

159].

MATHEMATICAL TEAM CONTEST\BALTIC WAY | 92"

Vilnius, 1992 | November 5{8

11. Let Q+ denote the set of positive rational numbers. Show that

there exists one and only one function f : Q+ ! Q+ satisfying the following

conditions:

(i) If 0 < q < 12then f(q) = 1 + f

�q

1�2q

�.

(ii) If 1 < q � 2 then f(q) = 1 + f(q� 1).

(iii) f(q) � f(1q) = 1 for all q 2 Q+.

Solution by Michael Selby, University of Windsor, Windsor, Ontario.By a change of variable eq = 1

1�2q , we have from (i),

f

� eq1 + 2eq

�= 1+ f(eq); (0 < eq <1); or f

1

1eq+ 2

!= 1 + f

11eq

!:

Calling t = 1eqand using (iii) we have

1

f(t+ 2)= 1 +

1

f(t); 0 < t <1; t 2 Q+: (1)

456

Then

1

f(t+ 4)=

1

f(t+ 2+ 2)= 1 +

1

f(t+ 2)= 1 + 1 +

1

f(t)

= 2 +1

f(t):

Hence, we can evaluate f(t+ 2k), k � 0, k an integer, if we know f(t).

Observe that condition (ii) can be rewritten as f(1 + t) = 1 + f(t),

t 2 Q+, 0 < t � 1.

We can now evaluate f(2k+ 1 + q) as follows: Since

1

f(2 + q)= 1 +

1

f(q); we have

1

f(2 + 1 + q)= 1 +

1

f(1 + q):

If 0 < q � 1, then1

f(3 + q)= 1 +

1

1 + f(q). Hence f(3 + q), 0 < q � 1

can be evaluated if f(q) is known. Once f(3 + q) is known, we obtain

1

f(5 + q)=

1

f(2 + 3 + q)= 1 +

1

f(3 + q);

and1

f(2k+ 1+ q)+ 1 +

1

f(2k� 1 + q); 1 � q > 0:

Therefore, we can now evaluate

f(2k+ q); f(2k+ 1 + q) 0 < q � 1; (2)

for all k � 0, k an integer, if we know f(q).

Furthermore, we can evaluate f(n), n � 1.

First f(1) = 1 since putting q = 1 in (iii) gives (f(1))2 = 1. Now

f(2) = 1 + f(1) = 2 from (ii). We follow recursively, f(3):

1

f(3)= 1 +

1

f(1)= 2

and1

f(2k+ 1)= 1 +

1

f(2k� 1):

Similarly1

f(2k+ 2)= 1 +

1

f(2k)and f(2) = 2:

Thus any such function is uniquely de�ned on the integers.

Finally, we can evaluate the function at any q from the values on the

positive integers. Let q = a

b, where (a; b) = 1.

457

Write a = bq1+ r1 where q1 is a non-negative integer, and 0 � r1 < b

is an integer. If r1 = 0, f(q) = f(q1) which is determined.

If a � r1 < b, we apply f(ab) = f(q1 +

r1

b). This is determined if the

value of f(r1b) is known using (2). Now 0 < r1

b< 1. We now compute f( b

r1).

b = r1q2+r2, r2 < r1. Continuing, since 0 � rk+1 < rk, rj = 0 for some j,

and we will have an expression for which f is evaluated at an integer. Hence

f exists and is uniquely determined.

12. Let N denote the set of positive integers. Let ' : N ! N be a

bijective function and assume that there exists a �nite limit

limn!1

'(n)

n= L:

What are the possible values of L?

Solution by Michael Selby, University of Windsor, Windsor, Ontario.We claim L must be 1.

Considermaxf'(1); : : : ; '(n)g = jn. We note that jn � n, since ' is

one-to-one. Let in 2 f1; 2; : : : ; ng be such that '(in) = jn. Then

'(in)

in� 1:

Since

limn!1

'(n)

n= L; lim

n!1'(in)

in= L � 1: (1)

Now consider Sn = fn 2 N : '(n) � ng. Sn must be in�nite. First Sn 6= ;for if Sn = ; then '(k) > k for all k and there is no k0 with '(k0) = 1.

Suppose Sn is �nite, with k the largest value in the set. Then '(n) > n

for n � k + 1. Consider f1; 2; : : : ; k + 1g. Since '(n) > k + 1 for

n � k + 1, the only integers which can be pre-images of f1; 2; : : : ; k + 1gare f1; 2; : : : ; kg. This is not possible, since ' is one-to-one and onto.

Therefore Sn = fn 2 N : '(n) � ng is in�nite. Choose a sequence,

nk 2 Sn with nk !1. We now have limk!1

'(nk)

nk= L. However

'(nk)

nk� 1:

Thus

L � 1: (2)

From (1) and (2), L = 1.

458

13. Prove that for any positive x1; x2; : : : ; xn, y1; y2; : : : ; yn the in-

equalitynXi=1

1

xiyi� 4n2Pn

i=1(xi + yi)2

holds.

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; by Christopher J. Bradley, Clifton College, Bristol, UK; byMichael Selby, University of Windsor, Windsor, Ontario; by Bob Prielipp,University ofWisconsin{Oshkosh,Wisconsin,USA; by PanosE. Tsaoussoglou,Athens, Greece; and by Edward T.H. Wang, Wilfrid Laurier University, Wa-terloo, Ontario. We feature Bradley's solution.

Now,1

xy� 4

(x+ y)2

since (x+ y)2 � 4xy, as (x� y)2 � 0. So

nXi=1

1

xiyi�

nXi=1

4

(xi + yi)2(�)

Lemma. (a1+a2+ � � �+an)(a2a3 � � � an+a1a3a4 � � � an+a1a2a4 � � �an + � � �+ a1a2 � � � an�1) � n2a1a2 � � � an.

This follows from separate applications of the AM{GM inequality to

the two terms on the left. It follows that

a1 + a2 + � � �+ an � n2

1a1

+ 1a2

+ � � �+ 1an

:

Now put

ai =1

(xi + yi)2; i = 1; : : : ; n

and then

1

(x1 + y1)2+

1

(x2 + y2)2+ � � �+ 1

(xn + yn)2� n2Pn

i=1(xi + yi)2:

Combining this with (�) showsnXi=1

1

xiyi� 4n2Pn

i=1(xi + yi)2:

14. There is a �nite number of towns in a country. They are connected

by one direction roads. It is known that, for any two towns, one of them can

be reached from the other one. Prove that there is a town such that all the

remaining towns can be reached from it.

459

Solutions by Mansur Boase, student, St. Paul's School, London, Eng-land; and by Christopher J. Bradley, Clifton College, Bristol, UK. We give thesolution by Boase.

We prove the result by induction on the number, n, of towns. If n � 2

the result is immediate.

Label the towns A1; A2; : : : ; Ak. We shall prove that if the statement

holds for all n < k, then it is also true for n = k, so by induction it will be

true for all n.

We can split up the towns excluding A1 into two sets M and N , M

containing those towns which can be reached from A1 and N those which

cannot be reached from A1.

Thus, every town inN can reach A1, and there is no route from a town

inM to a town in N .

If N is empty, then A1 is the desired town.

If this is not the case, then, since for any two towns in N , one of them

can be reached from the other, and there is no route from outsideN into N ,

the routes in question must pass through towns in N .

By the induction hypothesis, since jN j < k, there is a town inN which

can reach all other towns in N . It can also reach A1, and thus all towns

inM . Therefore, this is the town which can reach all the other towns in the

country, and the result is proved.

15. Noah has 8 species of animals to �t into 4 cages of the ark. He

plans to put species in each cage. It turns out that, for each species, there

are at most 3 other species with which it cannot share the accommodation.

Prove that there is a way to assign the animals to their cages so that each

species shares with compatible species.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

Let the animals be vertices of a graph. Join two animals by an edge if

they are compatible. Now we have a graph with 8 vertices, and each vertex

is joined to at least 4 others. So, by Dirac's theorem on Hamiltonian cycles,

there must be a Hamiltonian cycle, and if we take consecutive pairs of animals

in this cycle, we can put them in the same cage, and we have the required

solution.

17. Quadrangle ABCD is inscribed in a circle with radius 1 in such

a way that one diagonal, AC, is a diameter of the circle, while the other

diagonal, BD, is as long as AB. The diagonals intersect in P . It is known

that the length of PC is 25. How long is the side CD?

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; and by Christopher J. Bradley, CliftonCollege, Bristol, UK.We give the solution of Arslanagi�c.

460

CD

OB

P

A

The triangle ABD is isosceles because AB = BD. Let O be the cen-

tre of the circumcircle. Then BO ? AD. Because CD ? AD (AC is a

diameter), we get CDkBO; that is,4PCD � 4POB, and it follows that

CD

OB=PC

PO; that is

CD =OB � PCPO

=1 � 2

535

=2

3:

18. Show that in a non-obtuse triangle the perimeter of the triangle

is always greater than two times the diameter of the circumcircle.

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; by Mansur Boase, student, St. Paul's School, London,England; by Christopher J. Bradley, Clifton College, Bristol, UK; by BobPrielipp, University of Wisconsin{Oshkosh, Wisconsin, USA; and by EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario. We give Prielipp'ssolution.

In this solutionR; r; s will denote the circumradius, inradius and semi-

perimeter of a triangle. We shall show that in a non-obtuse triangle the

perimeter is always greater than or equal to 2(2R) + 2r.

Lemma. If A is an angle of triangle ABC, then cosA is a root of the

equation

4R2t3 � 4R(R+ r)t2 + (s2 + r2 � 4R2)t+ (2R+ r)2 � s2 = 0 (�)Proof. Since a = 2R sinA, and s� a = r cot(A

2),

s = a+ (s� a) = 2R sinA+ r cot

�A

2

�

= 2Rp(1� cosA)(1 + cosA) + r

s1 + cosA

1� cosA:

461

Thus

s2 = 4R2(1� cosA)(1 + cosA) + 4Rr(1 + cosA) + r21 + cosA

1� cosA

so

4R2(1� cosA)2(1 + cosA) + 4Rr(1 + cosA)(1� cosA)

+r2(1 + cosA)� s2(1� cosA) = 0:

Hence

rR2 cos3A� 4R(R+ r) cos2A+ (s2 + r2 � 4R2) cosA

+(2R+ r)2 � s2 = 0

making cosA a root of the equation (�).Corollary 1. IfA, B, andC are the angles of triangleABC, then cosA,

cosB, and cosC are the roots of the equation (�).Corollary 2. If A, B, and C are the angles of triangle ABC, then

cosA cosB cosC =s2 � (2R+ r)2

4R2:

Corollary 3. If A is the largest angle of triangle ABC, then

s > 2R+ r if A < 90�

s = 2R+ r if A = 90�

and s < 2R+ r if A > 90�:

Corollary 4. In a non-obtuse triangle the perimeter of the triangle is

always greater than or equal to 2(2R) + 2r.

Proof. If the triangle is an acute triangle, then s > 2R + r, and

2s > 2(2R) + 2r. If the triangle is a right triangle, then s = 2R + r.

Thus 2s = 2(2R) + 2r.

Corollary 5. In a non-obtuse triangle the perimeter of the triangle is

always greater than twice the diameter of the circumcircle.

19. Let C be a circle in the plane. Let C1 and C2 be nonintersecting

circles touching C internally at points A and B respectively. Let t be a com-

mon tangent of C1 and C2, touching them at points D and E respectively,

such that both C1 and C2 are on the same side of t. Let F be the point of

intersection of AD and BE. Show that F lies on C.

462

Solution by Christopher J. Bradley, Clifton College, Bristol, UK.Let SA be the tangent to C1 and C at A and TB be the tangent to C2

and C at B, as shown.

C

t

A

S

F

E

B

GH

D

C1

C2

�

90�� �

90�� '

'

�

'

O

Let \SAD = � and \TBE = '. Let O be the centre of C. AO meets

C1 again at G and since it is a common radius, AG is a diameter of C1.

BO meets C2 again at H and BH is likewise a diameter of C2.

We have \DAG = 90�� � and \DGA = �. By the alternate segment

theorem \GDE = 90� � � and since \ADG = 90� (angle in a semicir-

cle) it follows that \FDE = �. Similarly \FED = ' and so \DFE =

180� � � � '. Also \EBH = 90� � '.

Considering the angles of the (re-entrant) quadrilateral FAOB we have

re ex \AOB = 360�� (90�� �)� (90��')� (180�� ��') = 2�+2'.

So \AOB = 360� � 2� � 2' = 2\DFE. But O is the centre of circle C,

and AB is an arc of C, so F lies on C. (Converse of the angle at the centre

= twice angle at circumference).

20. Let a � b � c be the sides of a right triangle, and let 2p be its

perimeter. Show that p(p�c) = (p�a)(p�b) = S (the area of the triangle).

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; byMansur Boase, student, St. Paul's School, London, Eng-land; by Christopher J. Bradley, CliftonCollege, Bristol, UK; by ShawnGodin,St. Joseph Scollard Hall, North Bay, Ontario; by Bob Prielipp, University ofWisconsin{Oshkosh, Wisconsin, USA; by Michael Selby, University of Wind-sor, Windsor, Ontario; by Panos E. Tsaoussoglou, Athens, Greece; and byEdward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

Since the triangle is a right triangle we have c2 = a2 + b2, p = a+b+c2

,

and S = ab

2.

463

Then

p(p� c) =a+ b+ c

2

�a+ b+ c

2� c

�=

(a+ b)2� c2

4

=a2 + b2 + 2ab� c2

4=ab

2= S;

and

(p� a)(p� b) =

�a+ b+ c

2� a

��a+ b+ c

2� b

�

=c+ b� a

2

c� b+ a

2

=c2 � (b� a)2

4=c2 � (b2 + a2) + 2ab

4=ab

2= S;

as required.

We conclude this number of the Corner with solutions to some of the

problems of the 8th Iberoamerican Mathematical Olympiad, September 14{

15, 1993 (Mexico) which we gave last year [1996: 159{160].

8th IBEROAMERICAN MATHEMATICAL OLYMPIADSeptember 14{15, 1993 (Mexico)

1. (Argentina) Let x1 < x2 < � � � < xi < xi+1 < � � � be all the

palindromic natural numbers, and for each i, let yi = xi+1�xi. How many

distinct prime numbers belong to the set fy1; y2; y3; : : : g ?Solutions by Mansur Boase, student, St. Paul's School, London, Eng-

land; and by Shawn Godin, St. Joseph Scollard Hall, North Bay, Ontario. Wegive Boase's solution.

The �rst few palindromic numbers are

1; 2; 3; : : : ; 9; 11; 22;33; : : : ; 99; 101; 111; : : : :

Now 11� 9 = 2 and 22� 11 = 11.

We shall show that these are the only two prime values which a yi term

can take.

If xi and xi+1 have di�erent numbers of digits, then xi will be of the

form 99 : : : 9 and xi+1 of the form 10 : : : 01, so yi = xk+1 � xi = 2.

We can consider, without any loss of generality only yi where xi has

more than two digits since yi can only be prime if yi = 2 or 11 for xi with

one or two digits from the above list of the �rst xi.

If xi and xi+1 end in the same digit, then 10 divides yi, so yi cannot

be prime. If xi and xi+1 end in di�erent digits, say r and s, then s = r+1,

464

xi is of the form r999 : : : 9r, and xi+1 is of the form (r + 1)0 : : : 0(r + 1).

Then yi = xi+1 � xi = (r + 1) � (10 � r) = 11. Thus only two distinct

primes belong to the set fy1; y2; : : : g.2. (Mexico) Show that for any convex polygon of unit area, there exists

a parallelogram of area 2 which contains the polygon.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

We shall show more generally that there exists a rectangle of area 2

containing the polygon. The result is obviously true for a triangle. To prove

this, construct a rectangle on the longest side of the triangle and circumscrib-

ing the triangle. If the area of the triangle is 1, then the rectangle will have

area 2.

B C

A

If the polygon has more than three vertices, then choose the two ver-

tices of the polygon which are furthest apart. Call them B and C. Draw

perpendiculars to the line BC at B and at C to give lines l1 and l2, respec-

tively.

C

B

l2

l1

D

E

All the vertices of the polygon must lie between these two lines. (Otherwise

there would be two vertices further apart than jBCj.)Now consider the smallest rectangle which circumscribes the polygon

and with one pair of opposite sides lying on l1 and l2. Suppose this polygon

touches the polygon again at D and E.

C

B

E

D

U S

T

R

Let the vertices of the rectangle be R, S, T and U with D on UR and

E on ST . Then it is easy to see that

[RUCB] = 2[BCD]

465

and

[BSTC] = 2[BEC]

sinceBCkRU andBCkST . [RSUT ] = 2[CDBE] � 2(area of polygon)= 2

since the polygon is convex. We can �nd an even larger rectangle of area 2

containing the polygon.

4. (Spain) Let ABC be an equilateral triangle, and � its incircle. If

D and E are points of the sides AB and AC, respectively, such that DE is

tangent to �, show thatAD

DB+AE

EC= 1:

Solutions by �Sefket Arslanagi�c, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; and by Mansur Boase, student, St. Paul's School, London,England. We give Arslanagi�c's solution.

A B

C

�

Ip

D

E

a

aa

Let AB = AC = BC = a and BD = p; that is, AD = a � p, and

CE = q; that is, AE = a� q. The circle � is inscribed in the quadrilateral

and we get

ED + BC = BD + CE

or

ED+ a = p+ q

or

ED = p+ q � a: (�)By the law of cosines for the triangle ADE, it follows that

ED2 = AE2 +AD2 � 2AE � AD cos 60�;

so, from (�)(p+ q � a)2 = (a� q)2 + (a� p)2 � (a� q)(a� p)

and from this we obtain

a =3pq

p+ q:

Now, we have

AD = a� p =p(2q� p)

p+ q

466

and

AE = a� q =q(2p� q)

p+ q;

that is,

AD

DB+AE

EC=p

p

(2q� p)

p+ q+q

q

(2p� q)

p+ q=p+ q

p+ q= 1;

as required.

6. (Argentina) Two non-negative integer numbers, a and b, are \cu-

ates" (friends in Mexican) if the decimal expression of a+b is formed only by

0's and 1's. Let A and B be two in�nite sets of non-negative integers, such

that B is the set of all the numbers which are \cuates" of all the elements of

A, and A is the set of all the numbers which are \cuates" of all the elements

of B. Show that in one of the sets A or B there are in�nitely many pairs of

numbers x, y such that x� y = 1.

Solution by Mansur Boase, student, St. Paul's School, London, Eng-land.

Suppose an integer of A ends with the digit `r'. Then all integers in B

must have a last digit the same as for 10 � r or 11 � r in order that they

are all \cuates" of A. If B contains elements with last digits the same as for

10� r and 11� r, then every element of Amust end in the last digit r to be

\cuates" with integers ofB of both last digits. Thus either setA or set B has

all integers ending in the same digit. Without loss, assume that all elements

of A end in `r'.

Now consider an element of B which is a \cuate" of all the integers of

A. Let us say it is of type (i) if it ends with the last digit 10�r and of type (ii)

if it ends with the last digit of 11� r. If we change the last digit we obtain

another number which is a \cuate" of all the elements of A, and hence in

B. The di�erence between these pairs is 1. It follows therefore that there

are equal numbers of each type in B, and as B is in�nite, there are in�nitely

many pairs x; y in B such that x� y = 1.

That completes the Corner for this month. Send me your Olympiad

contest materials and your nice solutions to problems in the Corner.

467

BOOK REVIEWS

Edited by ANDY LIU

Learn from the Masters! edited by Frank Swetz, John Fauvel, Otto Bekken,

Bengt Johansson, Victor Katz, published by the MAA, Classroom Resource

Materials Series, 1995, ISBN# 0-88385-703-0, softcover, 312+ pages, $23.00.

Reviewed byMaria de Losada, Bogot �a, Colombia.

A rich and varied collection of thoughts directed primarily toward the

use of the history of mathematics for e�ective and enriched teaching (and

learning), these are papers given at the \Kristiansand (Norway) Conference"

of 1988. The areas and topics cover a broad range corresponding to di�erent

tastes and personal interests, divided in sections on school and higher math-

ematics. Frank Swetz's Using Problems from the History of Mathematics in

Classroom Instruction is a superb example of history transcending anecdotal

information and grasping the relationship between problem solving and the

gradual construction of meaning (especially clear in the section Illustratingthe growth of mathematical pro�ciency), an essential component of each in-

dividual student's coming to grips with mathematical concepts. The choice

of problems and the orientation involving the speci�c ways in which they can

be used to enrich instruction is excellent. Other papers that focus on math-

ematical thinking explore algorithms and analogies, modelling and heuristic

reasoning, as well as Man-Keung Siu'sMathematical Thinking and History of

Mathematics.

In the latter section, amidst the very �ne selection o�ered, the article

by Israel Kleiner The Teaching of Abstract Algebra: An Historical Perspective

stands out. Kleiner describes his approach as genetic, but notes that he tries

\to show how attempts to solve the problems give rise to the abstract theory.

This is, of course, the historical sequence of events." He further addresses

the question of how \history provides the opportunity to raise a number of

general issues about the nature of mathematics".

It is most unfortunate that the random sprinkling of photographs of

notable mathematicians throughout the book should place that of Emmy

Noether opposite the title In Hilbert's Shadow.

De�nitely of interest to those who lean toward using history as a re-

source for enriching their teaching and the mathematical thinking of their

students, as well as to those whose interest in historical subjects is just be-

ginning to be awakened.

468

Dissecting Rectangular Strips Into

Dominoes

Frank Chenstudent, D. S. McKenzie Junior High School, Edmonton

Kenneth Nearey and Anton Tchernyistudents, Grandview Heights Junior High School, Edmonton

The First Problem

A domino is de�ned to be a 1� 2 or 2� 1 rectangle. The �rst one is said to

be horizontal and the second vertical. In our Mathematics Club, we learned

to count the number gn of di�erent ways of dissecting a 3 � 2n strip into

dominoes. The sequence fgng satis�es the recurrence relation

gn = 4gn�1 � gn�2; (1)

for all n � 2, with initial conditions g0 = 1 and, as shown in Figure 1,

g1 = 3.

Figure 1

The generating functionG(x) for the sequence is de�ned to be the for-

mal power series

g0 + g1x+ g2x2 + � � � :

It is easy to deduce from (1) that

G(x) =1� x

1� 4x+ x2: (2)

In examining the dissections of the 3� 2n strip, we observed that they

fall into two kinds. A dissection of the �rst kind can be divided by a vertical

line into two substrips without splitting any dominoes. Such a line is called

a fault line. A dissection of the second kind, called a fault-free dissection,has no fault lines. Figure 2 shows an example of each, using 3� 4 strips.

469

Figure 2

Let fn be the number of fault-free dissections of the 3�2n strip. From

Figure 1, f1 = 3. For alln � 2, a fault-free dissection cannot start with three

horizontal dominoes. It must start o� as shown in Figure 3, and continue by

adding horizontal dominoes except for a �nal vertical one.

Figure 3

It follows that fn = 2 for all n � 2, and the sequence satis�es a trivial

recurrence relation

fn = fn�1

for all n � 3, with initial conditions f0 = 1; f1 = 3 and f2 = 2. Let

F (x) = f0 + f1x+ f2x2 + � � � be the generating function for the sequence.

Then F (x) = �1+x+2(1+x+x2+ � � � ) = �1+x+ 21�x . This simpli�es

to

F (x) =1 + 2x� x2

1� x: (3)

Having solved the simpler problem of counting fault-free dissections of

the 3 � 2n strip, we make use of our result to �nd an alternative solution

to the general problem of �nding all dissections of this strip. They can be

classi�ed according to where the �rst fault line is. This is taken to be the

right end of the strip if the dissection is fault-free. Then the strip is divided

into a 3 � 2k substrip on the left and a 3 � 2(n� k) substrip on the right,

where 1 � k � n.

Since the �rst substrip is dissected without any fault lines, it can be

done in fk ways. The second substrip can be dissected in gn�k ways as we

do not care whether there are any more fault lines. Hence

gn = f1gn�1 + f2gn�2 + � � �+ fng0: (4)

470

From the values of fn, gn = 3gn�1 + 2gn�2 + 2gn�3 + � � � + 2g0. If we

subtract from this expression gn�1 = 3gn�2 + 2gn�3 + � � � + 2g0, we have

gn � gn�1 = 3gn�1 � gn�2, which is equivalent to (1).

We now derive (2) in another way. It follows from (4) that for all n � 1,

2gn = f0gn + f1gn�1 + f2gn�2 + � � �+ fng0: (5)

Multiplying F (x) and G(x) yields

F (x)G(x) = (f0 + f1x+ f2x2 + � � � )(g0 + g1x+ g2x

2 + � � � )= f0g0 + (f0g1 + f1g0)x+ (f0g2 + f1g1 + f2g0)x

2 + � � � :

In view of (5), this becomes F (x)G(x) = g0+2g1x+2g2x2+� � � = 2G(x)�1

so that

G(x) =1

2� F (x): (6)

Substituting (3) into (6) yields (2).

The Second Problem

Let gn be the number of ways of dissecting a 4 � n strip into dominoes.

Then g0 = 1; g1 = 1 and, as shown in Figure 4, g2 = 5. It is not hard to

verify that g3 = 11. We wish to determine the in�nite sequence fgng via

recurrence relations and generating functions.

Figure 4

Let fn be the number of fault-free dissections of the 4 � n strip. We

have f0 = 0; f1 = 1 and from Figure 4, f2 = 4. For odd n � 3, the

only fault-free dissections are the extensions of the second and third ones

in Figure 4, with horizontal dominoes except for a �nal vertical one. Hence

fn = 2. For even n � 4; fn = 3 since we can also include similar extensions

of the fourth dissection in Figure 4.

As in the solution of the First Problem, we have

gn = f1gn�1 + f2gn�2 + � � �+ fng0:

471

This leads to

gn = gn�1 + 5gn�2 + gn�3 � gn�4 (7)

for all n � 4, with initial conditions g0 = 1; g1 = 1; g2 = 5 and g3 = 11.

Also,

F (x) = 1 + x+ 4x2 + 2x3 + 3x4 + 2x5 + 3x6 + � � �= �2� x+ 2x2

+3(1 + x2 + x4 + x6 + � � � ) + 2x(1 + x2 + x4 + � � � )= �2� x+ x2 +

3 + 2x

1� x2

=1+ x+ 3x2 + x3 � x4

1� x2: (8)

Substituting (8) into (6), which is still valid here, we have

G(x) =1� x2

1� x� 5x2 � x3 + x4: (9)

We now give an alternative solution to the Second Problem, along the

line of the solution to the First Problem we learned at the Mathematics Club.

We classify the dissections of the 4�n strip into �ve types according to how

they start. These correspond to those in Figure 4 if we ignore the vertical

dominoes in the second column. Call these Types A, B, C, D and E, and let

their numbers be an; bn; cn; dn and en, respectively.

By symmetry, we have bn = cn so that

gn = an + 2bn + dn + en (10)

for all n � 1. In a Type A dissection, we are left with a 4 � (n� 2) strip,

which can be dissected in gn�2 ways. Hence

an = gn�2 (11)

for all n � 3, with a1 = 0 and a2 = 1.

In a Type B dissection, if we complete the second column with a vertical

domino, the remaining 4� (n� 2) strip can be dissected in gn�2 ways. Theonly alternative is to �ll the second column with two horizontal dominoes.

The remaining part can be dissected in bn�1 ways, so that

bn = gn�2 + bn�1 (12)

for all n � 3, with b1 = 0 and b2 = 1.

In a Type D dissection, the situation is similar except that if we �ll

the second column with two horizontal dominoes, we must then also �ll the

472

third column with two more horizontal dominoes. The remaining part can be

dissected in dn�2 ways, so that

dn = gn�2 + dn�2 (13)

for all n � 3, with d1 = 0 and d2 = 1.

Finally, in a Type E dissection, after �lling the �rst column with two

vertical dominoes, we are left with a 4� (n�1) strip which can be dissected

in gn�1 ways. Hence

en = gn�1 (14)

for all n � 2, with e1 = 1. Now (7) follows from (10), (11), (12), (13) and

(14) since

gn = an + 2bn + dn + en

= gn�2 + (2gn�2 + 2bn�1) + gn�2 + dn�2 + gn�1= gn�1 + 4gn�2 + 2(gn�3 + bn�2) + dn�2= gn�1 + 4gn�2 + 2gn�3 + 2bn�2 + dn�2= gn�1 + 4gn�2 + 2gn�3 + gn�2 � an�2 � en�2= gn�1 + 5gn�2 + gn�3 � gn�4:

Using (7), (1� x� 5x2� x3 + x4)G(x) simpli�es to 1� x2. Hence (9) also

follows.

Supplementary Problem

Let fn denote the number of fault-free dissections of the 4�n strip. Find a

recurrence relation for the sequence ffng with initial conditions.

Acknowledgement

This article has been published previously in a special edition of delta-k,Mathematics for Gifted Students II, Vol. 33, 3 1996, a publication of the

Mathematics Council of the Alberta Teachers' Association and in AGATE, Vol.10, 1, 1996, the journal of the Gifted and Talented Education Council of the

Alberta Teachers' Association. It is reprinted with permission of the authors

and delta-k.

473

THE SKOLIAD CORNERNo. 26

R.E. Woodrow

This numberwe give the 30 problems of the KangourouDesMath �emati-

ques, �Epreuve Europ �eenne, which was given Friday March 22, 1996 to about

500,000 students in 16 European countries, and in 8 African countries, with-

out counting French schools around the world. My thanks go to Ravi Vakil,

Canadian Team leader to the 37th IMO in Mumbai, India, who collected a

great deal of contest material and forwarded it to me. My copy is in French,

and we give it in that language. The contestants are given 75 minutes, and

no calculators are allowed.

KANGOUROU DES MATH �EMATIQUESMarch 22, 1996Time: 75 minutes

1. Les repr �esentants de 12 pays ont choisi pour vous ces 30 questions.Chaque question a �et �e discut �ee 10 minutes. Quelle a �et �e la dur �ee totale de

la discussion? (3 points)

A. 360 min B. 300 min C. 120 min D. 52 min E. 40 min.

2. Dans la �gure ci-contre, l'aire de la r �egion laiss �ee en blanc est

6 cm2. Quelle est l'aire de la r �egion grise? (3 points)

A. 3 cm2 B. 4 cm2 C. 6 cm2 D. 9 cm2 E. 12 cm2.

3. Quel est le plus grand nombre? (3 points)

A. 1� 9� 9� 6 B. 19� 9� 6 C. 1� 99� 6

D. 1� 9� 96 E. 19� 96.

4. En utilisant une et une seule fois chacun des chi�res 1; 2; 3 et 4, je

peux �ecrire di� �erent nombres. Je peux �ecrire par exemple 3241. Quelle est

le di� �erence entre le plus grand et le plus petit nombre ainsi fabriqu �es?

(3 points)

A. 2203 B. 2889 C. 3003 D. 3087 E. 3333.

474

5. Un cercle et un rectangle s'aimaient d'amour tendre, \Mais mal-

heureusement, dit le cercle, que nous croissions ou r �etr �ecissions, nous ne

pourrons jamais avoir plus de n points communs!" �A votre avis, combien

vaut n? (3 points)

A. 2 B. 4 C. 5 D. 6 E. 8.

6. Le nombre 110

+ 1100

+ 11000

est �egale �a: (3 points)

A. 31110

B. 31000

C. 1111000

D. 1111110

E. 3111

.

7. La �gure ci-contre repr �esente un grand carr �e d'aire 1 m2. Une diag-

onale est partag �ee en trois segments de meme longueur. Le segment m �edian

est une diagonale du petit carr �e gris. Quelle est l'aire de ce petit carr �e?

(3 points)

A. 110m2 B. 1

9m2 C. 1

6m2 D. 1

4m2 E. 1

3m2.

8. La salle d'un th �eatre comporte 26 reng �ees de 24 places chacune.

Toutes les places sont num�erot �ees, en commen�cant par le premier rang. Dans

quelle rang �ee se trouve le si �ege num�erot �e 375? (3 points)

A. 12 �eme B. 13 �eme C. 14 �eme D. 15 �eme E. 16 �eme.

9. Parmi les phrases ci-dessous, quelles sont les phrases vraies?(3 points)

(1)La somme de deux nombres n �egatifs est toujours n �egative.

(2)La somme d'une nombre positif et d'un nombre n �egatif est toujours pos-

itive.

(3)La somme d'une nombre n �egatif et de deux nombres positifs est toujours

positive.

A. aucune B. la (1) seule C. le (1) et la (2)

D. la (2) et la (3) E. toutes les trois.

10. a est un nombre de deux chi�res, b est le nombre obtenu en

�ecrivant deux fois, cote �a cote, les deux chi�res de a. Quel est le quotient de

b par a? (3 points)

A. 10 B. 11 C. 99 D. 100 E. 101.

475

11. Un triangle �equilat �eral et un hexagone sont inscrits dans un meme

cercle. Si l'on divise l'aire de l'hexagone par l'aire du triangle, quel est le

quotient obtenu? (4 points)

A. 1:5 B. 2 C. 3 D. 4 E. �.

12. Un kangourou a dans sa poche 3 chaussettes blanches, 2 chaus-

settes noires et 5 chaussettes grises. Sans regarder, il veut en prendre une

paire. Quel nombre minimum de chaussettes lui faut-il sortir pour etre sur

qu'il en a bien deux de la meme couleur? (4 points)

A. 2 B. 3 C. 4 D. 7 E. 10.

13. �A la fete foraine, une �llette a achet �e cinq �echettes. Chaque fois

qu'elle touche la cible, elle a deux �echettes gratis. Elle a lanc �e en tout 17

�echettes. Combien de fois a-t-elle touch �e la cible. (4 points)

A. 4 B. 6 C. 7 D. 12 E. 17.

14. Les bissectrices de deux angles d'un triangle font entre elles un

angle de 110�. Combien vaut le troisi �eme angle de ce triangle? (4 points)

110�

?

A. 30� B. 40� C. 45� D. 55� E. 70�.

15. Une vieille montre retarde de 8 minutes par vingt-quatre heures.

De combien deminutes dois-je l'avancer ce soir �a 22 heures si j'ai absolument

besoin qu'elle me donne l'heure exacte demain matin �a 7 heures? (4 points)

A. 1 min 40 s B. 2 min 20 s C. 3 min D. 4 min 30 s E. 6 min.

476

16. La somme, en degr �es, des angles marqu �es sur la �gure ci-contre

est �egale �a: (4 points)

r

r

r

A. 120 B. 150 C. 180 D. 270 E. 360.

17. Un bidon plein de lait p �ese 34 kg. Le meme bidon quand il est �a

moiti �e plein p �ese 17; 5 kg. Combien p �ese le bidon vide? (4 points)

A. 1 kg B. 0; 5 kg C. 1; 5 kg D. 2 kg

E. il n'y a pas assez de donn �ees.

18. Le cot �es d'un triangle mesurent 8 cm, 15 cm et 17 cm. Quelle est

son aire? (4 points)

A. 40 cm2 B. 60 cm2 C. 68 cm2 D. 80 cm2

E. on ne peut pas la calculer.

19. Entre 6 heures ce matin et 18 heures ce soir, combien de fois les

deux aiguilles de ma montre feront-elles un angle droit? (4 points)

A. 2 B. 6 C. 12 D. 22 E. 24.

20. Simone a un gros tas de dalles triangulaires. Toutes ces dalles ont

une forme identique: ce sont des triangles �equilat �eraux de 1 dm de cot �e. De

combien de dalles Simone aura-t-elle besoin pour daller un grand triangle

�equilat �eral de 2 m�etres de cote? (4 points)

A. 200 B. 300 C. 400 D. 600 E. 800.

21. En d �ecoupant un coin d'un cube en bois, on a obtenu le solide

ci-contre. Maintenant on d �ecoupe de la meme fa�con les sept autres coins du

cube. On a alors un solide qui a quatorze faces (les faces triangulaires ne se

touchent pas et ne se recoupent pas). Quel est le nombre s de sommets et le

nombre a d'aretes du solide obtenu? (5 points)

A. s = 24; a = 36 B. s = 36; a = 24

C. s = 10; a = 15 D. s = 24; a = 32

E. s = 36; a = 18.

477

22. On compte le nombre de points d'intersection de quatre droites

distinctes. Quel est le nombre qu l'on est sur de ne pas trouver? (5 points)

A. 0 B. 2 C. 3 D. 5 E. 6.

23. Combien y a-t-il de triangles, dont les cot �es ont pour mesures (en

centim �etres) des nombres entiers, et dont le p �erim �etre est �egal �a 15 cm?

(5 points)

A. 1 B. 5 C. 7 D. 19 E. 45.

24. Marine et Claire se partagent un cone glac �e en le coupant �a mi-

hauteur. Marine en a plus que Claire! (5 points)

12

12

A. 1 fois et demie plus B. 2 fois plus C. 3 fois plus

D. 7 fois plus E. 8 fois plus.

25. Nous sommes sur une ligne de m�etro circulaire. Vingt-quatre

trains s'y d �eplacent dans la meme direction, �a intervalles r �eguliers et roulant

tous �a la meme vitesse. Demain, on doit rajouter des trains a�n de dimin-

uer de 20% les intervalles entre deux trains. Combien y aura-t-il de trains

suppl �ementaires demain sur la ligne? (5 points)

A. 2 B. 3 C. 5 D. 6 E. 12.

26. Dans la �gure ci-contre, (AB) est parall �ele �a (CD). De plus

AD �DC �CB et AB = AC. Combien vaut l'angle D? (5 points)

A B

CD?

A. 108� B. 120� C. 130� D. 150� E. on ne peut pas la savoir.

478

27. Charles a attribu �e �a tous ses livres un code de trois lettres, en util-

isant, l'ordre alphab �etique: AAA;AAB; AAC; : : : ; AAZ; ABA;ABB : : :

Charles a 2203 livres. Quel est le dernier code utilis �e par Charles quand il a

cod �e toute sa collection? (5 points)

A. CFS B. CHT C. DGS D. DFT E. DGU .

28. Cinq personnes sont assises autour d'une table ronde. Chacune

a�rme �a son tour: \Mes deux voisins, de droite et de gauche, sont des

menteurs". On sait que les menteurs mentent toujours et que quelqu'un qui

n'est pas un menteur dit toujours la v �erit �e. De plus tout le monde connait

la v �erit �e en ce qui concerne ses deux voisins. Combien y a-t-il de menteurs

�a cette table? (5 points)

A. 2 B. 3 C. 4 D. 5 E. on ne peut pas la savoir.

29. On a coup �e quatre droles de parts suivant les diagonales d'un drolede gateau plat �a quatre cot �es. J'ai mang �e une part. Mes amis m �econtents

ont pes �e les trois restantes et ont trouv �e 120 g. 200 g. et 300 g. Combien

pesait la part que j'ai mang �ee? (5 points)

200

300120

?

A. 120 g B. 180 g C. 280 g D. 330 g E. 500 g.

30. Dans la suite de chi�res 122333444455555 : : : , chaque entier est

�ecrit autant de fois que sa valeur. Quel est le 1996 �eme chi�re �ecrit?

(5 points)

A. 0 B. 3 C. 4 D. 5 E. 6.

479

Last issue we gave the problems of the Second Round of the Alberta

High School Mathematics Competition, of February 11, 1997. The solutions

we give were taken from the contest web site

http://www.math.ualberta.ca/~ahsmc/sample.htm

where more information about the contest and the solutions may be found.

The solutions are selected from contestants' work. My thanks to E. Lewis,

University of Alberta, Chair of the contest, for supplying us with materials.

ALBERTA HIGH SCHOOLMATHEMATICS COMPETITION

February 11, 1997Second Round

1. Find all real numbers x satisfying jx� 7j > jx + 2j+ jx� 2j.Remark. Note that jaj is called the absolute value of the real number a.

It has the same numerical value as a but is never negative. For example,

j3:5j = 3:5 while j � 2j = 2. Of course, j0j = 0.

Solution by Laura Harms, Lorne Jenken High School, Barrhead, Al-berta.

If x is in (�1;�2), the inequality becomes 7�x > (�2�x)+(2�x)which simpli�es to x > �7. Hence all x in (�7;�2) satisfy the inequality.

If x is in [�2; 2], then jx� 2j+ jx+ 2j = 4, while jx� 7j is never lessthan 5, so all x in [�2; 2] satisfy the inequality.

If x is in (2; 7), then the inequality becomes 7� x > (x� 2)+ (x+2)

which simpli�es to x < 73. Hence all x in (2; 7

3) satisfy the inequality.

If x is in [7;1), then x � 7 < x � 2 < (x � 2) + (x + 2), and the

inequality is not satis�ed.

In summary, x satis�es the inequality if and only if �7 < x < 7=3.

2. Two lines b and c form a 60� angle at the point A, and B1 is a

point on b. From B1, draw a line perpendicular to the line b meeting the

line c at the point C1. From C1 draw a line perpendicular to c meeting the

line b at B2. Continue in this way obtaining points C2, B3, C3, and so on.

These points are the vertices of right trianglesAB1C1; AB2C2; AB3C3; : : : .

If area (AB1C1) = 1, �nd

area (AB1C1)+ area (AB2C2)+ area (AB3C3)+� � �+area (AB1997C1997):

Solution by Margaret Tong, James Fowler High School, Calgary, Al-berta.

Clearly, trianglesABnCn are similar to each other. In a (30�; 60�; 90�)triangle, the hypotenuse is twice as long as the shorter leg. Let AB1 = x.

Then AC1 = 2x and AB2 = 4x. It follows that area(ABnCn) =

16 � area(ABn�1Cn�1), so that the desired total area is given by

T = 1 + 16 + 162 + � � � + 161996. Multiplying this by 16, we have 16T =

480

16+ 162+ 163 + � � �+161997. Subtraction yields 15T = 161997� 1 so that

T = ( 1

15)(161997� 1).

3. A and B are two points on the diameter MN of a semicircle. C,

D, E and F are points on the semicircle such that \CAM = \EAN =

\DBM = \FBN . Prove that CE = DF .

Solution by Byung-Kyu Chun, Harry Ainlay High School, Edmonton,Alberta.

Complete the circle. Extend EA to cut it at C0, and extend DB to

cut it at F 0. By symmetry, AC = AC0 so that m(AC0C) = m(ACC0).Similarly,m(BF 0F ) = m(BFF 0). Now m(EC0C) = 180� �m(CAC0) =180��2m(CAM) = 180��2m(FBN) = 180��m(FBF 0) = m(DF 0F ).Since the arcsCE andDF subtend equal angles at the circle, they have equal

measure. It follows that the chords CE and DF are equal.

M NBA

F

F 0

E

D

D0

C

C0

4. (a) Suppose that p is an odd prime number and a and b are positive

integers such that p4 divides a2 + b2 and p4 also divides a(a + b)2. Prove

that p4 also divides a(a+ b).

(b) Suppose that p is an odd prime number and a and b are positive

integers such that p5 divides a2+ b2 and p5 also divides a(a+ b)2. Show by

an example that p5 does not necessarily divide a(a+ b).

Solution to part (a) by Byung-Kyu Chun, Harry Ainlay High School,Edmonton, Alberta.

Note that a(a+b)2 = a(a2+b2)+2a2b. Since p4 divides both a(a+b)2

and a2+b2, it must also divide 2a2b. Since p is an odd prime, p4 divides a2b.

Suppose p2 does not divide a. Then the only powers of p that can possibly

divide a2 are p or p2. Since p4 divides a2b, it follows that p2 must divide b.

Hence p4 divides b2. However, this contradicts p4 dividing a2 + b2 but not

a2. It follows that we must have p2 dividing a. Then p4 divides a2 so that it

also divides b2. Hence p2 divides b, and it also divides a+ b. It follows that

481

p4 divides a(a+ b).

Solution topart (b) given by Byung-KyuChun,Harry AinlayHigh School,Edmonton, Alberta; and by Jason Ding, ArchbishopMacDonald High School,Edmonton, Alberta.

We look for a, b and p such that p5 divides a2+ b2, p2 divides a and b,

but p3 does not divide a+ b. Setting a = p2x and b = p2y, these conditions

become p divides x2 + y2 and p does not divide x+ y. We can pick x = 2,

y = 1 and p = 5. This gives a = 50 and b = 25 as an example

5. The picture shows seven houses represented by the dots, connected

by six roads represented by the lines. Each road is exactly 1 kilometre long.

You live in the house marked B. For each positive integer n, how many

ways are there for you to run n kilometres if you start at B and you never

run along only part of a road and turn around between houses? You have to

use the roads, but you may use any road more than once, and you do not

have to �nish at B. For example, if n = 4, then three of the possibilities are:

B to C to F to G to F ; B to A to B to C to B; and B to C to B to A to B.

r r r

r r

r r

F G

D E

A B C

Solution by Byung-Kyu Chun, Harry Ainlay High School, Edmonton,Alberta.

The number of ways to travel n = 1 kilometre is 2, the ways being B

to A and B to C. For n = 2, we have 4 ways, B via A to B, B via C to

B, B via C to D and B via C to F . Note that the answer is the same had

we started at D or F . We guess that the number of ways for any n is exactly

2n. We now prove this by induction, in an unusual manner. We consider

separately the cases n = 2k and n = 2k+ 1.

In the even case, the result is true for k = 0. Suppose that the number

of ways for n = 2(k � 1) is exactly 22(k � 1). At this point, we must be at

one of B, D, or F . As pointed out before, there are 4 ways to go another 2

kilometres. Hence, for n = 2k, the number of ways is 4 � 22(k�1) = 22k.

For n = 2k+1, we use the established fact that for n = 2k, the number

of ways is exactly 22k. Again, after 2k kilometres, we must be at one of B,

D, or F . In each case, there are two ways to go the extra kilometre, bringing

the total to 2 � 22(k�1) = 22k+1 for n = 2k+ 1.

That completes the Skoliad Corner for this number. Please send me

contest materials for use in the Skoliad as well as any comments or sugges-

tions about what you would like to see featured here.

482

MATHEMATICAL MAYHEM

Mathematical Mayhem began in 1988 as a Mathematical Journal for and by

High School and University Students. It continues, with the same emphasis,

as an integral part of Crux Mathematicorum with Mathematical Mayhem.

All material intended for inclusion in this section should be sent to the

Mayhem Editor, Naoki Sato, Department of Mathematics, Yale University,

PO Box 208283 Yale Station, New Haven, CT 06520{8283 USA. The electronic

address is still

mayhem@math.toronto.edu

The Assistant Mayhem Editor is Cyrus Hsia (University of Toronto).

The rest of the sta� consists of Richard Hoshino (University of Waterloo), Wai

Ling Yee (University of Waterloo), and Adrian Chan (Upper Canada College).

Shreds and Slices

A Note on Convexity

A function f is convex on I (I an interval) if

f(�x+ (1� �)y)� �f(x) + (1� �)f(y) 8 x; y 2 I; � 2 [0; 1];

and J{convex on I(see [1]) if

f

�x+ y

2

�� f(x) + f(y)

28 x; y 2 I:

Similarly, f is concave on I if

f (�x+ (1� �)y)� �f(x) + (1� �)f(y) 8 x; y 2 I; � 2 [0; 1];

and J{concave is de�ned similarly. Several sources claim that J{convexity is

su�cient for convexity (see [2]), but we intend to make this more precise.

First, f convex clearly implies that f is J{convex. We will prove that:

(i) If f is convex on an open interval I, then f is continuous on I, and

(ii) If f is J{convex and continuous on an interval I, then f is convex on I.

Proof. (i) Let a 2 I, � > 0. We will show that on some interval

around a, f(x) < f(a) + � and f(x) > f(a)� �.

Choose any b 2 I. Assume b > a. Then the graph of f(x) on [a; b]

lies under the chord joining (a; f(a)) and (b; f(b)) (see Figure 1), which in

turn lies under the line y = f(a) + � on some interval to the right of a,

including a. Applying a similar argument when b < a, we �nd an interval

around a on which f(x) < f(a) + �.

483

q

q

a b

y = f(x)

(b; f(b))

(a; f(a))y = f(a) + �

Figure 1.

q

q

c a b

y = f(x)

(a; f(a))

(c; f(c))

y = f(a) � �

Figure 2.

Now, if f(x) � f(a) for all x 2 I, then we are done, so assume

f(b) < f(a) for some b 2 I. Assume b > a. Then f(x) > f(a) for all

x < a by convexity, which certainly implies f(x) > f(a)� �. Take some

c < a, and consider the line joining (c; f(c)) and (a; f(a)). Then the graph

of f(x) to the right of a lies above this line, which in turn lies above the line

y = f(a)� � on some interval to the right of a, including a. The case b < a

is similar.

(ii) Since f is J{convex, it is clear that

f(�x+ (1� �)y) � �x+ (1� �)f(y)

for � = 0, 1, 12, 14, 34, and by an induction argument, for any dyadic rational

between 0 and 1; that is, a fraction of the form m

2n. Since the dyadic rationals

are dense in [0; 1], we can �nd a sequence which converges to any given real �

in [0; 1]. By taking a limit along this sequence, the identity is shown to be

true for all � 2 [0; 1].

Remark. In (i), I must be open, as seen in the example

f(x) =

�0 if 0 < x < 1

1 if x = 0 or x = 1:

Then f is convex, but not continuous.

Hence, (ii) allows for an easy way to check convexity, which is useful

for setting up Jensen's inequality, without resorting to the second derivative

test if calculus does not appeal to you. Also, the corresponding results hold

for f concave.

Example 1. Show that sinx is concave on [0; �].

Proof.

sinx+ siny = 2 sin

�x+ y

2

�cos

�x� y

2

�� 2 sin

�x+ y

2

�;

so that

484

sin

�x+ y

2

�� sinx+ siny

2:

Example 2. Show thata

x+ bis convex on (�b;1), where a > 0.

Proof. By two applications of AM-GM,

ax+y

2+ b

� ap(x+ b)(y+ b)

� 1

2

�a

x+ b+

a

y + b

�:

Example 3. Let k be a positive integer. Show that xk is concave on [0;1).

Proof. We show by induction that

�x+ y

2

�k� xk + yk

2

for all x, y � 0. The result is certainly true for k = 1. Assume it holds for

some k. Assume without loss of generality that x � y. Then

(xk � yk)(x� y) = xk+1 � xky� xyk + yk+1 � 0;

so that xky+ xyk � xk+1 + yk+1. Then,

�x+ y

2

�k+1

=

�x+ y

2

�k �x+ y

2

�

� xk + yk

2

!�x+ y

2

�

=xk+1 + xky + xyk + yk+1

4

� 2(xk+1 + yk+1)

4=

xk+1 + yk+1

2:

References

1. E. Lozansky and C. Rousseau,Winning Solutions, Springer-Verlag, NewYork, 1996.

2. D. S. Mitrinovi�c, Analytic Inequalities, Springer-Verlag, Berlin, 1970.

485

The Equation of the Tangent to the nth Circle

Krishna Srinivasan

Let n circles of radius r be tangent to each other in a row, such that

the centre of each lies on the x-axis, and the �rst circle passes through the

origin. Let l be the tangent of the nth circle passing through the origin, as

shown in the diagram. What is the equation of this tangent l?

: : :

O

y

x

O1 O2 On

M `

�r r r

Figure 1.

Let On be the centre of the nth circle, M the point of tangency, and � the

angle formed between ` and the x-axis. Then the slope of ` is tan�. Also,

\OMOn = 90� (since the radius is perpendicular to the tangent). There-

fore, the value of � is sin�1(MOn

OOn). Since the radius is r, MOn = r and

OOn = 2nr�r. Consequently, the slope of the tangent is tan(sin�1( 12n�1)).

Let us derive tan(sin�1 x), for 0 � x < 1. Let � = sin�1 x. Then

tan(sin�1 x) = tan � =sin�

cos �=

sin �p1� sin2 �

=xp

1� x2:

Substituting x = 12n�1 , we �nd the value of the slope is

1p(2n� 1)2 � 1

=1

2pn2 � n

:

And �nally, since the tangent passes through the origin, the equation of ` is

y =x

2pn2 � n

: (1)

It is interesting to note that the radius r does not appear in (1). This

shows that the line y = x=2p2 (substituting n = 2 into (1)), for example,

is always tangent to the second circle, regardless of the radius. [Ed: This

makes sense geometrically. Why?]

Coordinatizing the plane, as we have done, can make a problem simpler,

as the following problems show:

486

Problem: Circles P and Q are tangent; each has radius 1. PQ extended

meets the circles at A and B, and AC and BTC are tangents to circle P , as

shown in the diagram. Compute AC.

(1993 ARML, Team Questions)

q q

q

BQ P

A

C

T

q q

q

Figure 2.

Solution: The slope of BC is 1=2p2, by (1). Hence,

AC = AB=2p2 =

p2.

Problem: Three equal circles are tangent as shown. The line AD is

drawn from point A on the left circle, and tangent to the circle at the right

at point D. How long is the chord BC of the circle in the middle?

(1996-1997 Scarborough Mathematics League)

AB

C

D

q

q

q

q

-� 10

Figure 3.

Solution: Establish a coordinate system, as in Figure 1. As found above, the

equation of the tangent is y = x=2p6 and the equation of the second circle

is (x � 15)2 + y2 = 25. Solving for these two equations gives the points

of intersections, which are B(72�8p6

5; 6p6�45

) and C(72+8p6

5; 6p6+4

5). With

this information, the length of the chord BC can be calculated to 8 units.

[Ed: Knowing the slope of AD leads to a nice Euclidean solution.

By secant theorem, AB � AC = 10 � 20 = 200 and by similar triangles,

AB + BC=2 = 6p6. Now �nd BC.]

487

Combinatorial Games

Adrian Chan

Introduction

De�nition: One type of combinatorial game is a two-person game such that:

(i) Players alternate removing counters from a �nite collection according

to a set of rules, and

(ii) The last player to remove a counter wins.

The classic combinatorial game of this sort is Bouton's Nim or Nim.

The game consists of some number of piles with some number of counters in

each. A player, on his turn, may remove any number of counters from any

one pile.

Example: Say there are three piles of 1, 3, and 5 counters. The game could

proceed as follows:

(1; 3; 5)1! (1;3; 2)

2! (1;1; 2)1! (1; 1; 0)

2! (1; 0; 0)1! (0; 0; 0),

and player 1 wins!

A Simple Game: One Pile Nim with Restriction

Let us look at a simple one pile game of Nim, with the restriction that

a player may only remove one or two counters at a time. Consider the game

where there are 7 counters to play with, denoted Nim(7; 1; 2). First, try

playing with a friend. You'll probably see a strategy for one of the players.

The strategy can be developed by looking at the game in a di�erent way.

Construct a directed graph, where the vertices represent the number

of counters, and the directed edges represent possible moves. Note that

the graph has no cycles since positions cannot repeat. Hence, a game is

represented by a path from the initial vertex (7 counters) to the terminal

vertex (0 counters).

- - - -PPPPPq

PPPPPq�����1

�����1R R R� �

7 5 4 2 1

6 3 0

Figure 1.

488

Safe or Unsafe?

A vertex is safe if a player moving to that vertex has a winning strategy,

and unsafe otherwise. We can label each vertex of our directed graph as safe

or unsafe according to the following instructions:

(i) The terminal position is safe,

(ii) If all moves from a vertex lead to an unsafe vertex, then that vertex is

safe, and

(iii) If there is a move from a vertex to a safe vertex, then that vertex is

unsafe.

We can relabel our directed graph of Nim(7; 1; 2) to get:

- - - -PPPPPq

PPPPPq�����1

�����1R R R� �

U U U U U

S S S

Figure 2.

The Winning Strategy

1. A player starts on an unsafe vertex, and can move to a safe vertex (iii).

2. The other player is forced to move to an unsafe vertex (ii).

3. The original player can again move to a safe vertex (iii). In fact, the

original player can alwaysmove to a safe vertex, and after some number

of moves, he will land on the terminal vertex.

Since the terminal position is a safe vertex, the original player will win!

Exercise: Who should win a similar Nim game with a pile of 11 counters?

With 12 counters?

Grundy-Values for Games

For any combinatorial game, we can label each vertex of the directed

graph with a non-negative integer instead of the labels safe and unsafe. This

number, the Grundy-Value, is found by the following process:

(i) The terminal vertex is labelled 0.

(ii) For every other vertex, consider the set of the labels of the vertices it

points to. The label of the vertex is then the smallest non-negative

integer not appearing in this set.

489

Example: Our game of Nim(7; 1; 2) with Grundy-Values appears as

- - - -PPPPPq

PPPPPq�����1

�����1R R R� �

1 2 1 2 1

0 0 0

Figure 3.

Is it a coincidence that the safe positions coincide with those vertices

assigned a Grundy-Value of 0? Let us investigate. Denote the sets of ver-

tices with safe positions, unsafe positions, Grundy-Values of 0, and positive

Grundy-Values as S, U , Z, and N respectively.

Proposition. S = Z and U = N .

Proof by Induction. The terminal position is in both S and Z by de�nition.

Assume that there exists some subset of vertices P in which the proposition

is true; that is, for any vertex in P , it is in S if and only if it is in Z, and it is

in U if and only if it is in N .

Let us look at vertices that are not members of P . There exists a vertex

not in P such that all of its emanating edges lead to a vertex in P (to the

reader: why?). Let this vertex be A, and consider all vertices that A leads

to.

If A leads to a vertex in P that is in S (and so in Z), then A is unsafe

so A is in U , and A leads to a vertex with Grundy-Value 0, so A itself must

have positive Grundy-Value, and A is inN .

Otherwise, A leads to vertices in P that are only in U (and so inN ), so

A is safe and A is in S, and every vertex A leads to has a positive Grundy-

Value, so A itself has Grundy-Value 0 and A is in Z.

Hence, the statement is true for P and A, a larger set of vertices. We

can repeat the argument until we have included all vertices.

The Original Game of Nim

First consider a one-pile game of Nim, with no restrictions on how

many counters are taken. Obviously, the �rst person can take all the counters

and win. If we compute the Grundy-Values for such a game, the Grundy-

Value of a pile of n counters is n.

Now consider a game of Nim with two piles of counters. Say there are

6 in one and 7 in the other, denoted Nim(6; 7). We can look at the game

as a grid instead of a directed graph: You start with a marker at (6; 7) and

a move is a translation going left or down, but not both. Whoever lands on

(0; 0) wins. (See Figure 4.)

Exercise: Fill the grid with Grundy-Values. Which positions are safe (win-

ning)? Who will win, the �rst player or the second player?

490

(6; 7)

(0;0)

Figure 4.

Finding the Strategy: Nim Sum

De�nition: TheNim Sum of a combinatorial game is the binary addition of the

Grundy-Values of each independent game. However, there is the additional

condition that there is no carrying. For example, the nim sum of 12 and 12is 02, and the nim sum of 11102 and 1012 is 10112.

Example: Let us look at the game Nim(4; 5; 6; 7). Notice that the game is the

same as the simultaneous games of Nim(4), Nim(5), Nim(6), and Nim(7),

where the player can choose which game to play in and make a move in.

Grundy-Value Binary Representation

Nim(4) 4 1 0 0

Nim(5) 5 1 0 1

Nim(6) 6 1 1 0

Nim(7) 7 1 1 1

Nim Sum 0 0 0

If the Nim Sum is 0, then the position is balanced. Otherwise, it is

unbalanced.

Problem: Prove that all balanced positions are safe positions. Hint: Look at

the de�nition of safe and unsafe positions.

Exercise: Analyze Nim(n; 1; 2; 3) and Nim(n; 1; 3; 4). What are the Grundy-

Values? Does the �rst or second player win?

With this concept of the Nim Sum, youwill be able to �nd the strategy in

any combinatorical game and determine if you will win or lose! The concept

of Nim Sum is a very e�ective tool not only in these games, but also other

areas of mathematics.

491

Mayhem Book Reviews

Donny Cheung

The Art of Problem Solving: A Resource for the Mathematics Teacher, Editedby Alfred S. Posamentier, published by Corwin Press Inc., 2455 Teller Road,

Thousand Oaks CA, 91320-2218, 1996, ISBN 0-8039-6362-9, softcover, 465

pages.

The subtitle does not quite do this book justice. This wonderful col-

lection of twenty independent contributions has brought together a diverse

spectrum of professionals to cover a wide range of topics. The result is a

problem solving resource which is both entertaining and informative.

Having no singular theme other than problem solving in general, this

book is ideal as an exposition of the many di�erent aspects of problem solv-

ing. Unlike other books on problem solving, this book is not singularly inter-

ested in documenting individual problem solving techniques, teaching meth-

ods, or historical anecdotes. Rather, a good mixture has been obtained, and

this is the strength of the book.

The book begins with a chapter titled \Strategies for Problem Explo-

ration", which individually explores 13 powerful problem solving strategies.

But also contained within the book are a chapter promoting the use of prob-

lem solving as a teaching paradigm, a chapter investigating the reasons that

many students make the errors that they do, a chapter on cooperative learn-

ing, a chapter dedicated to the pigeonhole principle, and a chapter exploring

mathematically gifted students from a psychological viewpoint.

For the most part, this book is written in a conversational tone, and is

very readable. There are problems scattered throughout the book, and there

is a handy system of boxed and circled numerals which make it easy to �nd

speci�c problems from a certain topic of mathematics, or speci�c problems

which highlight a certain problem solving technique.

Perhaps this is not the ideal book for those looking for hundreds of

problems to solve, but for anyone who is interested in problem solving in

itself, and things related to problem solving, I highly recommend this book.

Students! Get Ready for the Mathematics for SAT I, by Alfred S. Posamen-

tier, published by Corwin Press Inc., 1996, ISBN 0-8039-6415-3, softcover,

206 pages.

Teachers! Prepare Your Students for the Mathematics for SAT I, by Alfred

S. Posamentier and Stephen Krulik, published by Corwin Press Inc., 1996,

ISBN 0-8039-6416-1, softcover, 116 pages.

492

Each year, college-bound students around the world write the Scholas-

tic Assessment Test, whose scores are used in college admissions, mostly in

the United States. So for these students, preparing for this test is usually

considered a very important task both in the classroom and in homes.

To this end, there have been a great number of books which are aimed

at helping students prepare for the SAT. The two books reviewed here are

designed to help a student prepare for the Mathematics portion of the SAT,

and to help a teacher prepare his or her students.

Without an e�ort on the part of the student, no book can help prepare

anyone for the SAT. However, a well-designed book can certainly be helpful.

This set of books is organized and well laid out. The book for students

includes a very useful summary of basic mathematical facts, and includes

both sample problems with in-depth solutions and timed sample tests for

practice. Throughout the sample problems are also good tips for answering

di�erent types of questions that are asked on the SAT.

The book for teachers details ten important problem solving strategies,

with excellent illustrative problems and in-depth analyses of the problems.

It also discusses some variousmore non-routine types of problems, and some

more technical points about the test, such as e�cient calculator use. And, it

includes, as an appendix, the summary of mathematical facts that appears in

the student's book.

On the whole, I think that these books are worth looking into for any-

one who is going to write the SAT, or has students who are going to.

493

J.I.R. McKnight Problems Contest 1980

1. Sum to k terms the series whose nth term is

n4 + 3n3 + 2n2 � 1

n2 + n:

2. Find all the functions of the form f(x) = a+bxb+x

, where a and b are

constants, f(2) = 2f(5), and f(0)+ 3f(�2) = 0.

3. Solve for x and y:

2x � 3y = 6

2x+1

3y�1= 5

4. Perpendicular tangents are drawn to the circles represented by

x2 + y2 = a2 and x2 + y2 = b2. Find the equation of the locus of

the point of intersection of these tangents, and name the locus.

ab

5. Prove that

sinx+ siny+ sin z � sin(x+ y+ z)

cosx+ cos y+ cos z + cos(x+ y + z)

= tan

�x+ y

2

�tan

�y + z

2

�tan

�z + x

2

�:

6. A square sheet of tin, a inches on a side, is to be used to make an open-

top box by cutting a small square of tin from each corner and bending

up the sides. How large a square should be cut from each corner for the

box to have as large a volume as possible?

494

Mayhem Problems

The Mayhem Problems editors are:

Richard Hoshino Mayhem High School Problems Editor,Cyrus Hsia Mayhem Advanced Problems Editor,Ravi Vakil Mayhem Challenge Board Problems Editor.

Note that all correspondence should be sent to the appropriate editor |

see the relevant section. In this issue, you will �nd only solutions | the

next issue will feature only problems.

We warmly welcome proposals for problems and solutions. With the

new schedule of eight issues per year, we request that solutions from this

issue be submitted by 1 February 1998, for publication in the issue 5 months

ahead; that is, issue 4 of 1998. We also request that only students submit

solutions (see editorial [1997: 30]), but we will consider particularly elegant

or insightful solutions from others. Since this rule is only being implemented

now, you will see solutions from many people in the next few months, as we

clear out the old problems from Mayhem.

High School Solutions

H214. Show that 3(a+ b+ c)� 8 � abc+ c for all positive integers

a, b, c � 2.

SolutionbyMiguel Carri �on �Alvarez, Universidad Complutense deMad-rid, Spain.

Let a = x+ 2, b = y + 2, and c = z + 2, so x, y, z � 0. Then

3(6 + x+ y + z)� 8 � (2 + x)(2 + y)(2 + z) + 2 + z

implies that

18 + 3(x + y + z)� 8 � 8 + 4(x + y + z) + 2(xy + yz + xz) + xyz + 2 + z ;

which implies that

10 + 3(x + y + z) � 10 + 4(x+ y + z) + 2(xy + yz + xz) + xyz + z ;

which, in turn, imples that

0 � x+ y + 2z + 2(xy + yz + xz) + xyz ;

and the last inequality is trivially true.

H215. Consider a square EQUI. Let L, A, and T be points on the

interior such thatEQA and IEL are isoceles triangles with basesEQ and IE

respectively. Show that UT = IL ifEAL andALT are equilateral triangles.

495

Solution.

I

E

U

Q

L

A

T

�

�

We have the following facts: EA = AQ and EL = LI, since EAQ and ELI

are isosceles triangles. Also, AL = LI and AT = TL = AL, since EAL

and ALT are equilateral triangles.

To show that UT = IL, we will show that 4ALI is congruent to

4ATU .

Let � = \AEQ. Then \LEI = �, since 4AEQ �= 4LEI. Hence,

90� = \IEQ = 2�+ \LEA = 2�+ 60�, so � = 15�.

Then \ILA = 360� � \ELA� \ELI = 360� � 150� � 60� = 150�.Therefore, 4ALI �= 4ELI, and AI = EI = IU so \AIU = \EIU �\EIA = 90� � 30� = 60�. Therefore, 4AIU is equilateral, so AI = AU .

Finally, 4ALI �= 4ATU since AL = AT , AI = AU , and \LAI =

60� � \IAT = \TAU . Hence, UT = IL as required.

H216. Triangulate a square such that all edges of the triangles on the

interior of the square are equal in length and this length is smaller than the

length of the side of the square. (Hint: See the previous problem H215.)

Solution.

Consider the �gure in the solution of problemH215. ExtendEA through

A to meet QU at R and extend EL through L to meet UI at S as shown.

I

E

U

Q

L

A

T

S

R

Now ER is the diameter of the circumcircle of4ERQ since \EQR =

90�. Further, Amust be the centre of the circle since EA = AQ and A is on

the diameter ER. Thus, AR = RE. Likewise, LS = LE in4ESI.

496

Note that4ERS is equilateral and SR = 2AL. Therefore, ST = TR

and this value is equal to all the other lengths in the interior shown in the

solution of problem H215.

Rider. Are there other possible triangulations of the square with all thelengths in the interior equal?

Advanced Solutions

A191. Taken over all ordered partitions of n, show that

Xk1+k2+���+km=n

k1k2 � � � km =

�m+ n� 1

2m� 1

�:

Additional Solution by Waldemar Pompe.

Consider n soccer players. Let A(n;m) be the number of ways they

can be divided into m teams (blocks) such that in each team there is a goal-

keeper indicated. A partition which has k1, k2, : : : , km players in the blocks

gives k1k2 � � � km ways of choosing goal-keepers. Adding these products over

all partitions gives A(n;m).

On the other hand, A(n;m) can be found as follows: Place m+ n� 1

players in a row and choose 2m� 1 of them. Starting with the �rst of these

2m�1 players, we make every alternate player a goal-keeper (resulting inm

of them), and the remainingm� 1 are discarded, creatingm� 1 gaps in our

row of players. There are thenm blocks, with a goal-keeper in each. It can be

veri�ed that there is a one to one correspondence between such formations

and the formations above. Hence, the equality in question holds.

Challenge Board Solutions

We begin with a couple of corrections. J. Chris Fisher of the University

of Regina has kindly pointed out that the statement of problem C70 (and

hence the solution last issue) is wrong. The problem was as follows:

C70. Prove that the group of automorphisms of the dodecahedron

is S5, the symmetric group on �ve letters, and that the rotation group of

the dodecahedron (the subgroup of automorphisms preserving orientation)

is A5.

In fact, the group of automorphisms of the dodecahedron is A5 � C2,

where C2 is the two-element group. The published proof showed that the

rotation group wasA5, and then incorrectly described the rest of the problem

as similar.

497

In fact, the isomorphism between the automorphism group of the do-

decahedron and A5 � C2 can be described explicitly: given a dodecahedron

with edges numbered as described in the problem, any automorphism in-

duces a permutation of the numbers which can be checked to be even (and

hence an element ofA5). Also, the map to C2 sends an automorphism to the

identity if it is a rotation, and to the other element if it is not (that is, if it is

a re ection). The reverse map is similar: there is one rotation that permutes

the 5 types of edge-labels in any even way, and one re ection.

Chris suggests Coxeter's Introduction to Geometry (especially p. 273)

as a reference. One of the warning ags which tipped Chris o� was that the

erroneous proof implied that S5 is an isometry group of Euclidean three-

space. Coxeter lists the possible �nite isometry groups, and S5 does not

appear (and indeed S5 doesn't turn up as an isometry group until Euclidean

four-space).

(Also, the �gure accompanying the solution contained a typo: the \3"

on the central line of symmetry should be a \1".)

1

2

3 4

5

1

2

34

5

4

5

1

2

3

1

2

34

5

2 5

4 3

1 1

3 4

5 2

Thank you, Chris, for pointing out the error and explaining how to rec-

tify it! And if any other reader is suspicious of anything written in these

pages, please let us know.

C71. Let L1, L2, L3, L4 be four general lines in the plane. Let pij be

the intersection of lines Li and Lj. Prove that the circumcircles of the four

triangles p12p23p31, p23p34p42, p34p41p13, p41p12p24 are concurrent.

Solution.For convenience, let C123, C234, C341, and C412 be the circles. By

symmetry, it su�ces to show that the �rst three are concurrent.

This proof is slightly diagram-dependent, but it is short and sweet and

geometric. Label the three angles as in the �gure. By the exterior angle

theorem, c = a + b. Let Q be the intersection of C234 with C341, other

than p34. Then

498

r r r

r

r

r

r

C134

C234

P34

P13

P12

P24

P23

P14

Q

L1L2L3

L4

\p13Qp23 = \p23Qp34 � \p13Qp34

= c� b

= a

= \p13p12p23:

Thus Q is on the circle C123 as well.

Very rough sketch of another solution.

This solution is intended to intrigue and entice the reader, rather than

rigorously solve the problem. All statements made here can be made rigor-

ous.

This problem originally came to our attention because of a connection

to a sophisticated result:

Lemma. Let C be a degree d curve in the plane. Let A andB be degree

e curves that each meet C at de distinct points (say a1, : : : , ade and b1, : : : ,

bde respectively). Suppose that ai = bi for i = 1, : : : , de � 1. Then, if

d � 3, ade = bde; that is, the last points are also equal.

Amazingly, this result fails if d is 1 or 2. The proof relies on the fact

that there is no rational parametrization of curves of degree greater than 2,

while a line such as x = 0 has the parametrization (0; t) and a conic such as

the circle x2 + y2 = 1 has the parametrization�1� t2

1 + t2;

2t

1 + t2

�:

(An even more sophisticated result: these deep facts are related to the fact

that the diophantine equations x + y = z and x2 + y2 = z2 have lots

of solutions in integers, while if n > 2, xn + yn = zn has only a �nite

number of solutions even if x, y, and z are allowed to be quite general, for

example of the form r + sp2 where r and s are integers. This result, stated

appropriately, is a consequence of Siegel's Theorem, and relates in an obvious

way to Fermat's Last Theorem.)

499

Even the application of the lemma to this problem has some subtleties.

As before, it su�ces to show that C123, C234, and C341 are concurrent. Let

Q be the intersection of C234 and C341 (other than p34) as before, and let

P be the intersection of C123 and C234 (other than p23). Let C be the cubic

curve that is the union of C123 and L4, A the cubic that is the union of C234

and L1, and B the cubic that is the union of C341 and L2. Then C and A

intersect at the six points fp12; : : : ; p34g, and at P . They also intersect at

two points at in�nity (10; i0) and (1

0; �i0) (whatever that means!). The cubics

C and B intersect at the same nine points (including the two strange ones!),

except the P is replaced by Q. By the lemma, P = Q, and we are done.

Of course, a lot of further explanation is required to even make sense

of all this, but such explanation is beyond the scope of Mayhem.

Comments.

1. This problem is a �rst of an in�nite sequence of theorems, called the Cli�ord Theorems.

Call the intersection of two lines in general position their Cli�ord point. For three lines

in general position, there are three Cli�ord points of pairs of lines; call the circle through

the three points the Cli�ord circle of the three lines. Then, according to this problem,

given four lines in general position, the four Cli�ord circles (of the four triples of lines) are

concurrent; call this point the Cli�ord point of the four lines. In general, if n is odd, given

n general lines, then the n Cli�ord points of all (n � 1)-subsets of the n general lines

are concyclic, and the resulting circle is called the Cli�ord circle of the n general lines.

Similarly, if n is even, given n general lines, then the n Cli�ord circles of all (n � 1)-

subsets of the n general lines are concurrent, and the resulting point is called the Cli�ord

point of the n general lines. The theorems implicit in these de�nitions are the Cli�ord

theorems.

This theorem is discussed in Liang-shin Hahn's book Complex Numbers and Geometry

(published by the Mathematical Association of America). This is a wonderful, beautiful

book, and possibly the best place to learn about how complex numbers can be used to

make Euclidean plane geometry simple. Hahn proves all the Cli�ord theorems (in Section

2.3) using a simple lemma, which he proves using complex numbers, but which can also

be proved with ordinary Euclidean geometry:

Lemma. Suppose there are four circles C1, C2, C3, and C4 in a plane. Let C1 and C2intersect at z1 and w1, C2 and C3 intersect at z2 and w2, C3 and C4 intersect at z3and w3, C4 and C1 intersect at z4 and w4. Then the points z1, z2, z3, z4 are concyclic

if and only if w1, w2, w3, w4 are concyclic.

If anyone has a slick proof of the Cli�ord theorems (or even the next case), we would be

interested in seeing it.

2. What needs to be done to make the �rst (geometric) proof rigorous?

3. Can you make sense of the weird points at in�nity (10; i0) and (1

0;�i

0) described in the

sketch of the second solution? In your way of making sense of them, can you see why

they lie on every circle?

4. Without understanding the intricacies of the second sketch, can you use the sketch's

lemma, with a little hand waving, to prove Hahn's lemma above? Can you use it to

prove any other well-known results in Euclidean geometry? (Pappus' theorem seems like

a good possibility.) If so, we would love to hear from you!

500

PROBLEMSProblem proposals and solutions should be sent to Bruce Shawyer, De-

partment ofMathematics and Statistics,Memorial University of Newfound-land, St. John's, Newfoundland, Canada. A1C 5S7. Proposals should be ac-companied by a solution, together with references and other insights whichare likely to be of help to the editor. When a submission is submitted with-out a solution, the proposer must include su�cient information on why asolution is likely. An asterisk (?) after a number indicates that a problemwas submitted without a solution.

In particular, original problems are solicited. However, other inter-esting problems may also be acceptable provided that they are not too wellknown, and references are given as to their provenance. Ordinarily, if theoriginator of a problem can be located, it should not be submitted withoutthe originator's permission.

To facilitate their consideration, please send your proposals and so-lutions on signed and separate standard 81

2"�11" or A4 sheets of paper.

These may be typewritten or neatly hand-written, and should be mailedto the Editor-in-Chief, to arrive no later than 1 June 1998. They may alsobe sent by email to crux-editors@cms.math.ca. (It would be appreciated ifemail proposals and solutions were written in LATEX). Graphics �les shouldbe in epic format, or encapsulated postscript. Solutions received after theabove date will also be considered if there is su�cient time before the dateof publication.

Where to send your solutions and proposals

There has been an increase in the number of solutions and proposals

sent to the CanadianMathematical Society's Head O�ce in Ottawa, Ontario.

Please note the instructions above and send them directly to the Editor-in-

Chief.

Solutions submitted by FAX

There has been an increase in the number of solutions sent in by FAX,

either to the Editor-in-Chief's departmental FAXmachine in St. John's, New-

foundland, or to the Canadian Mathematical Society's FAX machine in Ot-

tawa, Ontario. While we understand the reasons for solvers wishing to use

this method, we have found many problems with it. The major one is that

hand-written material is frequently transmitted very badly, and at times is

almost impossible to read clearly. We have therefore adopted the policy that

we will no longer accept submissions sent by FAX. We will, however, con-

tinue to accept submissions sent by email or regular mail. We do encourage

email. Thank you for your cooperation.

501

2287. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

Let G denote the point of intersection of the medians, and I denote

the point of intersection of the internal angle bisectors of a triangle. Using

only an unmarked straightedge, construct H, the point of intersection of the

altitudes.

2288. Proposed by Victor Oxman, University of Haifa, Haifa, Israel.

In the plane are a circle (without centre) and �ve points A, B, C, D,

E, on it such that arc AB = arc BC and arc CD = arc DE. Using only

an unmarked straightedge, construct the mid-point of arc AE.

2289?. Proposed by Clark Kimberling, Evansville, IN, USA.Use any sequence, fckg, of 0's and 1's to de�ne a repetition-resistant

sequence s = fskg inductively as follows:

1. s1 = c1, s2 = 1� s1;

2. for n � 2, let

L = maxfi � 1 : (sm�i+2; : : : ; sm; sm+1)

= (sn�i+2; : : : ; sn; 0) for somem < ng;

L0

= maxfi � 1 : (sm�i+2; : : : ; sm; sm+1)

= (sn�i+2; : : : ; sn; 1) for somem < ng:

(so thatL is themaximal length of the tail-sequence of (s1; s2; : : : ; sn; 0)

that already occurs in (s1; s2; : : : ; sn), and similarly for L0), and

sn+1 =

8<:

0 if L < L0;1 if L > L0;cn if L = L0:�

For example, if ci = 0 for all i, then

s = (0; 1; 0; 0; 0; 1; 1; 0; 1; 0; 1; 1; 1; 0; 0; 1; 0; 0;

1; 1; 1; 1; 0; 1; 1; 0; 0; 0; 0; 0; 1; 0; 1; 0; : : : )�

Prove or disprove that s contains every binary word.

2290. Proposed by Panos E. Tsaoussoglou, Athens, Greece.For x; y; z � 0, prove that�

(x+ y)(y+ z)(z+ x)�2 � xyz(2x+ y + z)(2y+ z + x)(2z+ x+ y):

2291. Proposed by K.R.S. Sastry, Dodballapur, India.Let a, b, c denote the side lengths of a Pythagorean triangle. Suppose

that each side length is the sum of two positive integer squares. Prove that

360jabc.

502

2292. Proposed by K.R.S. Sastry, Dodballapur, India.

A convex quadrilateral Q has integer values for its angles, measured in

degrees, and the size of one angle is equal to the product of the sizes of the

other three.

Show that Q is either a parallelogram or an isosceles trapezium.

2293. Proposed by Claus Mazanti Sorensen, student, Aarhus Uni-versity, Aarhus, Denmark.

A sequence, fxng, of positive integers has the properties:

1. for all n > 1, we have xn�1 < nxn;

2. for arbitrarily large n, we have x1x2 : : : xn�1 < nxn;

3. there are only �nitely many n dividing x1x2 : : : xn�1.

Prove that

1Xk=1

(�1)kxkk!

is irrational.

2294. Proposed by Zun Shan and Edward T.H. Wang, Wilfrid LaurierUniversity, Waterloo, Ontario.

For the annual Sino-Japanese \Go" tournament, each country sends a

team of seven players, Ci's and Ji's, respectively. All players of each country

are of di�erent ranks (strengths), so that

C1 < C2 < : : : < C7 and J1 < J2 < : : : < J7:

Each match is determined by one game only, with no tie. The winner then

takes on the next higher ranked player of the opponent country. The tourna-

ment continues until all the seven players of one country are eliminated, and

the other country is then declared the winner. (For those who are not fa-

miliar with the ancient Chinese \Chess" game of \Go", a better and perhaps

more descriptive translation would be \the surrounding chess".)

(a) What is the total number of possible sequences of outcomes if each

country sends in n players?

(b)? What is the answer to the question in part (a) if there are three countries

participating with n players each, and the rule of the tournament is

modi�ed as follows:

The �rst match is between the weakest players of two countries

(determined by lot), and the winner of each match then plays the

weakest player of the third country who has not been eliminated (if

there are any left). The tournament continues until all the players

of two countries are eliminated.

503

2295. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Find three positive integers a, b, c, in arithmetic progression (with pos-

itive common di�erence), such that a+b, b+c, c+a, are all perfect squares.

2296. Proposed by Vedula N. Murty, Andhra University, Visakhap-atnam, India.

Show that sin2�x

2>

2x2

1 + x2for 0 < x < 1.

Hence or otherwise, deduce that � <sin�x

x(1� x)< 4 for 0 < x < 1.

2297. Proposed by Bill Sands, University of Calgary, Calgary, Al-berta.

Given is a circle of radius 1, centred at the origin. Starting from the

point P0 = (�1;�1), draw an in�nite polygonal path P0P1P2P3 : : : going

counterclockwise around the circle, where each PiPi+1 is a line segment tan-

gent to the circle at a point Qi, such that jPiQij = 2jQiPi+1j. Does this

path intersect the line y = x other than at the point (�1; 1)?2298. Proposed by Bill Sands, University of Calgary, Calgary, Al-

berta.The \Tickle Me" Feather Company ships its feathers in boxes which

cannot contain more than 1 kg of feathers each. The company has on hand

a number of assorted feathers, each of which weighs at most one gram, and

whose total weight is 1000001=1001 kg.

Show that the company can ship all the feathers using only 1000 boxes.

2299. Proposed by Walther Janous, Ursulinengymnasium, Inns-bruck, Austria.

Let x, y, z > 0 be real numbers such that x+ y + z = 1. Show that

Ycyclic

�(1� y)(1� z)

x

�(1�y)(1�z)=x� 256

81:

Determine the cases of equality.

2300. Proposed by Christopher J. Bradley, Clifton College, Bristol,UK.

Suppose thatABC is a triangle with circumradiusR. The circle passing

through A and touching BC at its mid-point has radius R1. De�ne R2 and

R3 similarly. Prove that

R21 + R2

2 +R23 � 27

16R2:

504

SOLUTIONS

No problem is ever permanently closed. The editor is always pleased toconsider for publication new solutions or new insights on past problems.

2179?. [1996: 318] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

For real numbers a � �1, we consider the sequence

F (a) :=

8<:�1 +

1

n

�pn(n+a)

; n � 1

9=; :

Determine the sets D, respectively I, of all a, such that F (a) strictly de-

creases, respectively increases.

Solution by Michael Lambrou, University of Crete, Crete, Greece.We show that the sequence F (a) is strictly increasing if and only if

�1 � a � 1

and strictly decreasing if and only if

a >4(log 1:5)2 � (log 2)2

(log 2)2 � 2(log 1:5)2� 1:168188898;

where all logarithms here and below are natural logarithms.

We approach this as follows. For a �xed a, the sequence F (a) is strictly

decreasing if and only if for all integers n � 1 we have

�1 +

1

n+ 1

�p(n+1)(n+1+a)

<

�1 +

1

n

�pn(n+a)

:

Taking logarithms and squaring, this is equivalent (since all quantities in-

volved are positive) to

(n+ 1)(n+ 1 + a)

�log

n+ 2

n+ 1

�2< n(n+ a)

�log

n+ 1

n

�2;

and so

a

"(n+ 1)

�log

n+ 2

n+ 1

�2� n

�log

n+ 1

n

�2#

< n2�log

n+ 1

n

�2� (n+ 1)2

�log

n+ 2

n+ 1

�2:

The quantity in square brackets on the left is negative, because the function

f1(x) = x

�log

x+ 1

x

�2(1)

505

has derivative

f 01(x) =

�log

x+ 1

x

��log

x+ 1

x� 2

x+ 1

�

<

�log

x+ 1

x

��1

x� 2

x+ 1

�< 0 for x 2 [1;1)

[the �rst inequality follows because log(1 + z) < z for all z > 0] so f1decreases for x 2 [1;1). Thus the sequence F (a) is strictly decreasing if

and only if

a >n2�log n+1

n

�2 � (n+ 1)2�log n+2

n+1

�2(n+ 1)

�log n+2

n+1

�2� n

�log n+1

n

�2 (2)

for all integers n � 1.

Next we show that the right-hand side of (2) is a decreasing function

of n. It is more convenient to study the right-hand side as a function of a

continuous variable x on [1;1). Thus we are to show

(x+ 1)2�log x+2

x+1

�2� (x+ 2)2

�log x+3

x+2

�2(x+ 2)

�log x+3

x+2

�2� (x+ 1)

�log x+2

x+1

�2

<x2�log x+1

x

�2 � (x+ 1)2�log x+2

x+1

�2(x+ 1)

�log x+2

x+1

�2� x

�log x+1

x

�2 :

We work on a more general case of an inequality, for appropriate func-

tions f : [1;1)! R, of the form

(x+ 1)f(x+ 1)� (x+ 2)f(x+ 2)

f(x+ 2)� f(x+ 1)<

xf(x)� (x+ 1)f(x+ 1)

f(x+ 1)� f(x): (3)

Here putting f equal to the function f1 in (1) gives the desired inequality.

We shall need the following easy lemma.

Lemma. If f : [b;1) ! R is a strictly monotonic, twice di�eren-

tiable function with f(x) > 0 for all x in its domain, and further satis�es

f(x)f 00(x) < 2 (f 0(x))2 there, then inequality (3) holds.

Proof. Indeed, observe that the function g(x) = 1=f(x) satis�es

g00(x) =2 (f 0(x))2 � f(x)f 00(x)

(f(x))3

> 0 ;

506

so g is strictly convex. Applying convexity to x, x+ 2 and 12(x+ (x+2)) =

x+ 1 we see that

1

f(x+ 1)<

1

2

�1

f(x)+

1

f(x+ 2)

�;

that is,

2f(x)f(x+ 2) <�f(x) + f(x+ 2)

�f(x+ 1): (4)

If f is strictly monotonic, then the quantity�f(x+ 1)� f(x)

� �f(x+ 2)� f(x+ 1)

�is strictly positive. Multiplying (3) by this quantity and simplifying, we get

(4), which is therefore equivalent to (3). 2

Returning to the proof, we will show that the function f1 given by (1),

which we have already shown is strictly monotonic, satis�es the rest of the

conditions of the lemma. Clearly f1(x) > 0 on [1;1), and we need to show

f1f001 < 2(f 01)

2. Here

f 01(x) =

�log

x+ 1

x

�2� 2

x+ 1log

x+ 1

x

and

f 001 (x) =2

x(x+ 1)2� 2

x(x+ 1)2log

x+ 1

x:

Thus upon cancellation we need to show

1� logx+ 1

x<

�(x+ 1) log

x+ 1

x� 2

�2(5)

on [1;1).

This unfortunately is a little tedious, as the sides of (5) are almost equal

for large x, so we will delay its proof until the end. Assuming (5) and return-

ing to inequality (2), if we call its right-hand side an, then we have shown

that (an) is strictly decreasing. Hence the condition a > an for all integers

n � 1 is equivalent to

a > a1 =4(log1:5)2 � (log 2)2

(log 2)2 � 2(log 1:5)2� 1:168188898:

Conclusion: F (a) is strictly decreasing if and only ifa > a1 = 1:16818889 : : : .

Similarly, F (a) is strictly increasing if and only if a < an for all inte-

gers n � 1. But as (an) decreases and is bounded below (by zero), this is

equivalent to a � liman. We show that liman = 1. This can be done by

507

l'Hopital's Rule taking x!1 on the continuous analogue of an; or, arguing

asymptotically, we have�log

n+ 1

n

�2=

�log

�1 +

1

n

��2=

�1

n� 1

2n2+

1

3n3+ O(1=n4)

�2

=1

n2

�1� 1

n+

11

12n2+O(1=n3)

�

and hence

an =

�1� 1

n+ 11

12n2+ O(1=n3)

�� �1� 1n+1

+ 1112(n+1)2

+ O(1=n3)�

�1

n+1� 1

(n+1)2+ O(1=n3)

�� � 1

n� 1

n2+ O(1=n3)

�=

�1(n+1)2

+ O(1=n3)

�1(n+1)2

+ O(1=n3)�! 1;

as claimed.

To complete the proof we must show (5). Set w = 1x+1

for x � 1 so

that w � 1=2 and x

x+1= 1�w. Now (5) becomes

1 + log(1�w) <

�1

wlog(1� w) + 2

�2;

and hence we have to show on 0 < w � 1=2 that

[log(1� w)]2 + (4w� w2) log(1� w) + 3w2 > 0: (6)

Now we need the following estimates of log(1� w): for 0 < w � 1=2

we have

�w� w2

2� w3

3� w4

4� w5

5� 13w6

42

< log(1� w) < �w � w2

2� w3

3� w4

4� w5

5� w6

6: (7)

The second inequality is simply a truncation of the Taylor series of log(1�w).The �rst simply says

13w6

42>

w6

6+w7

7+w8

8+ � � � ;

this is true enough, for 0 < w � 1=2 implies 0 < w

1�w � 1, and we have

w6

6+w7

7+w8

8+ � � � <

w6

6+w7

7(1 + w+ w2 + � � � )

=w6

6+w7

7

�1

1� w

�� w6

6+w6

7=

13w6

42:

508

Substituting inequalities (7) into the left-hand side of (6), we get that

it is enough to show for 0 < w � 1=2 that

�w+

w2

2+w3

3+w4

4+w5

5+w6

6

�2

+(4w�w2)

��w� w2

2� w3

3� w4

4� w5

5� 13w6

42

�+3w2 > 0:

The lengthy but routine calculation gives

1

12w4 +

1

6w5 +

19

90w6 � 71

210w7 + (positive terms) > 0:

But this last inequality is certainly true since, for 0 < w � 1=2, we have

1

12w4 � 71

210w7 � 1

12w4 � 71

210w4

�1

2

�3=

69w4

1680> 0:

[Editor's note: some minor adjustments seemed to be needed to

Lambrou's asymptotic proof that liman = 1, and also to one of his coef-

�cients toward the end of the solution. These have been made in the above

writeup.]

Also solved by CON AMORE PROBLEM GROUP, The Royal DanishSchool of Educational Studies, Copenhagen, Denmark; RICHARD I. HESS,Rancho Palos Verdes, California, USA; and V �ACLAV KONE �CN �Y, Ferris StateUniversity, Big Rapids, Michigan, USA. Two other readers sent in incorrector incomplete solutions.

A couple of readers mentioned that F (a) is eventually decreasing (forlarge enough n) whenever a > 1, as can be seen from the above solution.

One reader pointed out the similar problem 442 in the College Math-

ematics Journal, solution in Vol. 23 (1992) pp. 71{72, in which the exponentis n+ a instead of

pn(n+ a).

2180. [1996: 318] Proposed by Juan-Bosco Romero M�arquez, Uni-versidad de Valladolid, Valladolid, Spain.

Prove that if a > 0; x > y > z > 0; n � 0 (natural), then

1. ax(yz)n(y� z) + ay(xz)n(z � x) + az(xy)n(x� y) � 0;

2. ax coshx(y � z) + ay cosh y(z� x) + az cosh z(x� y) � 0.

Solutionby Joe Howard, New MexicoHighlands University, Las Vegas,NM, USA.

A characterization for a function f to be convex is that, for x > y >

z > 0,

(y� z)f(x) + (z � x)f(y)+ (x� y)f(z) � 0:

509

[See, for example, D.S. Mitrinovi�c, Analytic Inequalities, Springer, Berlin,1970, p. 16.]

Since result 1 is equivalent to

(y� z)ax

xn+ (z � x)

ay

yn+ (x� y)

az

zn� 0;

it is su�cient to show that f(t) =at

tnis a convex function of t for n � 0.

The case n = 0 is easy, so suppose that n > 0, and without loss of

generality, suppose that a = e. Then

f 00(t) =et�t2 � 2t+ n(n+ 1)

�tn+2

:

The discriminant of the quadratic equation t2 � 2t + n(n + 1) is

�4(n2 + n� 1) < 0. Hence f 00(t) > 0, and so f is convex.

For result 2, we must show that f(t) = at cosh t is convex. Again,

assume without loss of generality, that a = e. Thus

f 00(t) = 2et (cosh t+ sinh t) = 2e2t > 0:

Thus f is convex, and the inequality follows.

Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sara-jevo, Bosnia and Herzegovina; THEODORE CHRONIS, student, AristotleUniversity of Thessaloniki, Greece; FLORIAN HERZIG, student, Perchtolds-dorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; MURRAY S.KLAMKIN, University of Alberta, Edmonton, Alberta (�rst part only);MICHAEL LAMBROU, University of Crete, Crete; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; and the pro-poser.

Klamkin read the second part as cosh((x(y � z)), etc., which, on re- ection, is a reasonable interpretation, and gives a trivial result. We wonderhow many other readers did this?

2181. [1996: 318] Proposed by �Sefket Arslanagi �c, Berlin, Germany.Prove that the product of eight consecutive positive integers cannot be

the fourth power of any positive integer.

Solutionby Joe Howard, New MexicoHighlands University, Las Vegas,NM, USA. [Slightly modi�ed by the editor.]

Without loss of generality, let

P = (n� 3)(n� 2)(n� 1)n(n+ 1)(n+ 2)(n+ 3)(n+ 4):

510

Then

P = n8 + 4n7 � 14n6� 56n5 + 49n4 + 196n3 � 36n2� 144n:

If P is a 4th power, then it would be of the form

F =�n2 + an+ b

�4;

where a, b are constants. But then b = 0 since P has a zero constant term.

Thus F = n4(n+ a)4, and clearly a 6= 0.

But this implies that the coe�cients of n3, n2 and n in F are all zero. How-

ever, P has these coe�cients non-zero. Hence P cannot be of the form F .

It has been pointed out by many readers that this problem has ap-peared before. Most readers referred to theAmerican Mathematical Monthly,

1936, p. 310 for the solution to #3703 (posed by Victor Th �ebault in 1934,p. 522). Another reference was made to Honsberger's monograph Mathe-

matical Morsels, where it appears on p. 156 as \A Perfect 4th Power". Sev-eral readers also made reference to the general problem of proving that theproduct of (two or more) consecutive integers is never a square, which wasestablished in 1975 by Erd }os and Selfridge [1]. Because the solutionby Cautiswas quite di�erent from any of these published solutions, we have decidedto publish it here. The interested reader is directed to these other sourcesfor a di�erent solution.

Comments and/or solutions were submitted also by FRANCISCOBELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain; CHRISTOPHERJ. BRADLEY, Clifton College, Bristol, UK; ADAM BROWN, Scarborough, On-tario; GORAN CONAR, student, Vara�zdin, Croatia; THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece; F.J.FLANIGAN, San Jose State University, San Jose, California, USA; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon; MURRAY S.KLAMKIN, University of Alberta, Edmonton, Alberta; V �ACLAV KONE �CN �Y,Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU,University of Crete, Greece; MICHAEL PARMENTER, Memorial Universityof Newfoundland, St. John's, Newfoundland; HARRY SEDINGER, St. Bon-aventure University, St. Bonaventure, NY, USA; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands; DIGBYSMITH, Mount Royal College, Calgary, Alberta; DAVID R. STONE, GeorgiaSouthern University, Statesboro, Georgia, USA; EDWARD T.H. WANG, Wil-frid Laurier University, Waterloo, Ontario; KENNETH M. WILKE, Topeka,Kansas, USA; PAUL YIU, Florida Atlantic University, Boca Raton, Florida,USA; and the proposer.

Sei�ert remarks that A. Guibert proved the result in 1862. This isstated by L.E. Dickson in his History of the Theory of Numbers, Vol. II,

1952, pp. 679-680.

511

Janous notes the following deep theorem by Erd }os and Selfridge:

The product of two or more consecutive positive integers is never a

power of a positive integer; that is, the Diophantine equation

(n+ 1)(n+ 2) : : : (n+ k) = xk

has no integer solutions with k; l � 2 and n � 0.

Reference

1. P. Erd }os and J.L. Selfridge, The product of consecutive integers is nevera power, Illinois J. Math. 19 (1975), 292{301.

2182. [1996: 318] Proposed by Robert Geretschl�ager, Bundesreal-gymnasium, Graz, Austria.

Many CRUX readers are familiar with the card game \Crazy Eights", of which

there are many variations. We de�ne the game of \Solo Crazy Eights" in the

following manner:

We are given a standard deck of 52 cards, and are dealt k of these at

random, 1 � k � 52. We then attempt to arrange these k cards according to

three rules:

1. Any card can be chosen as the �rst card of a sequence;

2. A card can be succeeded by any card of the same suit, or the same num-

ber, or by any eight;

3. Anytime in the sequence that an eight appears, any suit can be \called",

and the succeeding card must be either of the called suit, or another

eight. (This means that, in e�ect, any card can follow an eight).

The game is won if all dealt cards can be ordered into a sequence according

to rules 1{3. If no such sequence is possible, the game is lost.

What is the largest value of k for which it is possible to lose the game?

Solution by Michael Lambrou, University of Crete, Greece.We show that if 34 or fewer cards are dealt, then it is possible to lose the

game, but for any 35 cards or more, the game always ends successfully. Thus

k = 34.

For example, the 34 or fewer cards may be chosen as follows; (a) ace

of hearts, (b) no eights, (c) any subset of the 33 cards consisting of any suit

which is not a heart and any rank which is neither an ace nor an eight. In this

situation the ace cannot be linked to any other card (as there are no other

aces nor hearts nor eights); thus we have a losing hand.

Let us now show that for 35 cards or more, the game can end success-

fully. Note that if eights were present, we could exchange them with any

undealt cards, arrange the cards, and re-exchange the eights (if there are in-

su�cient undealt cards, we could simply place the eights at the end). Thus

512

it is su�cient to show that we can obtain success on 35 or more cards which

do not include eights.

The algorithm below is based on two simple observations:

1. Any set of ranks each of which appears 3 times among the dealt cards

(that is, is represented by 3 suits) can be arranged in a sequence follow-

ing the rules of the game. This is so because any two such ranks have at

least two suits in common so, running through all suits of a �xed rank,

we can link it to the next rank utilising a common suit which we can go

through similarly leaving last a suit that links it with a further rank in

the set. (Clearly a common suit between the second and the third rank,

which is di�erent from the suit that linked the �rst to the second rank,

always exists). This can be repeated until we exhaust the set.

2. Any rank which appears twice among the dealt cards can be linked to

a rank which appears three times, since at least one suit is common to

both.

Let Ai, (i = 0; 1; 2; 3; 4), denote the set of ranks that appear i times

(that is, represented with i suits) among the dealt cards, and let ni = jAij,the number of elements in Ai. Since we are excluding eights, we have

n0 + n1 + n2 + n3 + n4 = 12

and we also have

n1 + 2n2 + 3n3 + 4n4 � 35:

We will show that n1+n2 � n4+1 and that if n2 = 0 this can be improved

to n1 � n4. We argue by contradiction. Suppose that n4 + 1 < n1 + n2.

Then

35 � n1 + 2n2 + 3n3 + 4n4

= n1 + 2n2 + 3(12� n0 � n1 � n2 � n4) + 4n4

= 36� 3n0 � 2n1 � n2 + n4

� 36� 2n1 � n2 + n4

< 36� 2n1 � n2 + (n1 + n2 � 1)

= 35� n1;

which is impossible since n1 � 0. If further we have n2 = 0, and we assume

that n4 < n1 we would get

35 � n1 + 2n2 + 3n3 + 4n4

= n1 + 0+ 3(12� n0 � n1 � 0� n4) + 4n4

= 36� 3n0 � 2n1 + n4

� 36� 2n1 + n4

< 36� 2n1 + n1

= 36� n1;

513

which forces n1 < 1, but then 0 � n4 < n1 is impossible.

We are now in position to describe an algorithm arranging all 35 cards

according to rules 1{3. Suppose �rst that n2 = 0. Then n1 � n4 and we may

write A1 = fb1; b2; : : : ; bn1g and A4 = fc1; c2; : : : ; cn4g. We start with a

card, say of rank b1, in A1. We link progressively this card with each card of

A1 having the same suit. After this suit has been exhausted within A1, we

link with a card, say of rank c1, of A4 of the same suit (as the elements of

A4 have all four suits dealt, this is always possible). We then run through

all four cards of rank c1, in some order ending with a suit for which there

still exist cards inA1. (This is always possible unlessA1 is exhausted.) Then

exhaust the cards of A1 with the same suit and jump back to A4 (this is

always possible as n1 � n4). Repeat this process untilA1 is exhausted, then

go through the rest of A4 ending with a suit that is the same as some suit

for some rank in A3. Finally jump to A3 and complete the process, which is

possible from observation 1 above.

On the other hand, if n2 6= 0, and thus n1 + n2 � n4 + 1, we modify

the algorithm from above as follows. After we exhaust A1 and return to A4,

we then go toA2, select a rank, exhaust the (two) suits of this rank, and jump

back to A4. Continue back and forth between A4 and A2, running each rank

through all its suits, being sure to order the suits for each rank from A4 so

that it matches a suit for a rank still remaining in A2. IfA2 is exhausted �rst

we complete the rest of A4 and jump to A3 as described for the �rst case. If

A4 is exhausted �rst we go from A2 to A3 which can be done by observation

2 above, as long as we have taken care to end up for the last rank of A2 with

a suit which matches the suit for at least one rank from A3. (This may take

a little bit of planning for the ordering of the last rank of A4 as well!)

Also solved by the proposer. There were two incomplete solutions.

Geretschl�ager also asks about a generalization to replace 52 = 4 �12+4

by n � k + j. He observes that for n = 4 (that is, 4k + j), an analogousargument to the one presented would yield a maximum number of 3k � 2.He then asks for general conditions such that the resultingmaximum numberis (n�1)(k�1)+1, or to �nd outwhat other numbers (if any) could turn up.

2183. [1996: 319] Proposed by V�aclav Kone �cn �y, Ferris State Univer-sity, Big Rapids, Michigan, USA.

Suppose that A, B, C are the angles of a triangle and that k, l, m � 1.

Show that:

0 < sinkA sinlB sinm B

� kkllmmSS

2

�(Sk2 + P )�

k

2

��(Sl2 + P )�

l

2

��(Sm2 + P )�

m

2

�;

where S = k + l +m and P = klm.

514

Editor's and Proposer's comments.This problem has already been posed; see 908 [1984: 19] Proposed by

Murray S. Klamkin, University of Alberta, Edmonton, Alberta.

Determine the maximum value of

P � sin�A � sin� B � sin C;where A, B, C are the angles of a triangle and �, �, are given positive

numbers.

A solution by Walther Janous, Ursulinengymnasium, Innsbruck, Aus-

tria, is given in [1985: 93]. He sent in a solution this time pointing out that

he had done so before!

Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;RICHARD I. HESS, Rancho Palos Verdes, California, USA; MICHAELLAMBROU, University of Crete, Crete, Greece; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; and the proposer.

2184. [1996: 319] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.

Let n be a positive integer and let an denote the sum

bn=2cXk=0

(�1)k�n� k

k

�:

Prove that the sequence fan : n � 0g is periodic.Composite solution from Michael Lambrou, University of Crete, Crete,

Greece and Kee-Wai Lau, Hong Kong.Since �

n� k

k

�=

�n� 1� k

k

�+

�n� 1� k

k � 1

�;

we have

an =

bn=2cXk=0

(�1)k�n� 1� k

k

�+

bn=2cXk=1

(�1)k�n� 1� k

k� 1

�

=

bn=2cXk=0

(�1)k�n� 1� k

k

��

b(n�1)=2cXk=0

(�1)k�n� 2� k

k

�

=

b(n�1)=2cXk=0

(�1)k�n� 1� k

k

��

b(n�2)=2cXk=0

(�1)k�n� 2� k

k

�

= an�1 � an�2:

515

It follows that an+3 = an+2� an+1 = (an+1 � an)� an+1 = �an, and so

an+6 = �an+3 = an, showing that the sequence fan : n � 0g is periodicwith period 6.

In fact, the sequence takes consecutively the values 1; 1; 0;�1;�1; 0,inde�nitely.

Also solved by PAUL BRACKEN, CRM,Universit �e deMontr �eal, Qu �ebec;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; DAVID DOSTER,Choate Rosemary Hall, Wallingford, Connecticut, USA; F.J. FLANIGAN, SanJose State University, San Jose, California, USA; FLORIAN HERZIG, stu-dent, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, Cal-ifornia, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;D. KIPP JOHNSON, Beaverton, Oregon, USA; V �ACLAV KONE �CN �Y, FerrisState University, Big Rapids, Michigan, USA; HEINZ-J�URGEN SEIFFERT,Berlin, Germany; and the proposer.

Bradley and Hess commented that the corresponding sum without thefactor (�1)k would yield the sequence of Fibonacci numbers. [Ed.: This canbe found in many elementary books on Combinatorics, for example, Exer-cise 21 on page 87 of Basic Techniques of Combinatorial Theory by DanielI.A. Cohen.] Flanigan pointed out that this problem appears, with technicaldi�erences in the de�nition of an, in the book Concrete Mathematics by Gra-ham, Knuth and Patashuik, (1989), 177{179. Lau pointed out that with therecurrence relation an = an�1 � an�2 and the initial values a0 = a1 = 1,one can prove easily, by induction, that

an = cos

�n�

3

�+

1p3sin

�n�

3

�; n = 0; 1; 2; : : : :

This fact was also derived by the proposer.

2185. [1996: 319] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

Notice that

22 + 42 + 62 + 82 + 102 = 4 � 5 + 5 � 6 + 6 � 7 + 7 � 8 + 8 � 9;

that is, the sum of the �rst n (in this case 5) even positive squares is equal

to the sum of some n consecutive products of consecutive pairs of positive

integers.

Find another value of n for which this happens.

(NOTE: this problem was suggested by a �nal exam that I marked recently.)

I. Solution by Tim Cross, King Edward's School, Birmingham, England.We require to �nd positive integers n; k for which

22+42+ � � �+(2n)2 = k(k+1)+(k+1)(k+2)+ � � �+(k+n�1)(k+n);

516

which is equivalent to

4

nXr=1

r2 =

nXr=1

(k+ r� 1)(k+ r) = k(k� 1)

nXr=1

1+ (2k� 1)

nXr=1

r+

nXr=1

r2;

3 � n6(n+ 1)(2n+ 1) = k(k� 1)n+ (2k� 1)

n

2(n+ 1);

1

2(n+ 1)(2n+ 1) +

1

2(n+ 1) = k(k� 1) + k(n+ 1);

and �nally

k2 + nk� (n+ 1)2 = 0: (1)

We thus look for positive integer solutions

k =�n+

pn2 + 4(n+ 1)2

2

and we require the discriminant� = 5n2+8n+4 to be a perfect square, say

� = �2 for some positive integer �. This condition leads to a Pell equation

(5n+ 4)2 � 5�2 = �4.Examining the more general form x2 � 5y2 = �4, we �nd solutions

(x; y) = (1; 1); (4; 2); (11;5); (29;13); : : : :

The solution x = 5n+ 4 = 29 gives n = 5, the example given.

Pell equations have solution-pairs which satisfy similar second-order

recurrence relations. In this case, xk and yk both satisfy

uk = 3uk�1 � uk�2; k � 3; (2)

with (x1; y1) = (1;1) and (x2; y2) = (4; 2). (Notice that the sequence

fykg = 1; 2; 5; 13; : : : is that of alternate Fibonacci numbers.)

If we take the sequence fxkg = 1; 4; 11; 29; : : : we see that terms are

alternately � 1 mod 5 and� 4 mod 5. Since we need x � 4 mod 5, we put

vk = x2k and can then derive from (2) the sequence

vk = 7vk�1 � vk�2; k � 3; with v1 = 4 and v2 = 29:

[Editor's note. For example, from (2) we get

x2k = 3x2k�1 � x2k�2 = 3(3x2k�2 � x2k�3)� x2k�2= 7x2k�2 � (3x2k�3 � x2k�2) = 7x2k�2 � x2k�4

and the recurrence for the v's follows.] Then, since nk = (vk� 4)=5, we can

deduce the sequence fnkg de�ned by

nk = 7nk�1 � nk�2 + 4; k � 3; with n1 = 0 and n2 = 5:

517

This gives the sequence

fnkg = 0 (trivially); 5; 39; 272; 1869; 12815; : : :

of suitable values of n.

II. Solution by John Oman and Bob Prielipp, University of Wisconsin{Oshkosh, Wisconsin, USA.

[Editor's note: Oman and Prielipp �rst derived equation (1), which

however they wrote in the form

n2 � (k� 2)n� (k2 � 1) = 0: (3)

Here and below, their notation has been changed to agree with Solution I.]

Considering equation (3) as a quadratic in n, a necessary condition for

it to have integer solutions is for the discriminant k(5k � 4) to be a per-

fect square. Thus 5k � 4 = x2 and k = y2 for some positive integers x

and y. [Editor's note. Since gcd(k;5k � 4) = 1; 2 or 4, the only alterna-

tive is 5k � 4 = 2x2 and k = 2y2, which implies 5y2 � 2 = x2 and thus

x2 � 3 mod 5, impossible. Note that we thus get the same equation

x2 = 5y2 � 4 as in Solution I.]

The following sequences for km andnm solve (3) and provide additional

solutions to the problem:

km = y2m

and nm =km � 2 +

pkm(5km � 4)

2=y2m� 2 + xmym

2;

where

xm = (2 +p5)

3 +

p5

2

!m+ (2�

p5)

3�p

5

2

!m

and

ym =

5 + 2

p5

5

! 3 +

p5

2

!m+

5� 2

p5

5

! 3�p

5

2

!m:

The �rst few values generated by these formulas are

m x y k n

0 4 2 4 5

1 11 5 25 39

2 29 13 169 272

3 76 34 1156 1869

[Editor's note. Oman and Prielipp then noted that

xm = L2m+3 and ym = F2m+3;

518

the (2m+3)rd Lucas and Fibonacci numbers, respectively (where, as usual,

F1 = F2 = 1 and L1 = 1, L2 = 3, both sequences then generated by the

familiar Fibonacci recurrence). This follows because

3�p5

2=

1�p

5

2

!2

and 2�p5 =

1�p5

2

!3

;

and so

xm =

1 +

p5

2

!2m+3

+

1�p

5

2

!2m+3

= L2m+3

and

ym =1p5

1 +

p5

2

!2m+3

� 1p5

1�p

5

2

!2m+3

= F2m+3:

Thus km = F 22m+3 and (after some manipulations)nm = F2m+3F2m+4�1.]

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; ADAM BROWN, Scarborough, Ontario; THEODORE CHRONIS, student,Aristotle University of Thessaloniki, Greece; CHARLES R. DIMINNIE, An-gelo State University, San Angelo, TX, USA; KEITH EKBLAW, Walla Walla,Washington, USA; JEFFREY K. FLOYD, Newnan, Georgia, USA; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; V �ACLAVKONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA; MICHAELLAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong;J.A. MCCALLUM,Medicine Hat, Alberta; ROBERT P. SEALY, Mount AllisonUniversity, Sackville, New Brunswick; HEINZ-J�URGEN SEIFFERT, Berlin,Germany; DIGBY SMITH, Mount Royal College, Calgary, Alberta; DAVIDR. STONE, Georgia SouthernUniversity, Statesboro, Georgia, USA; EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, Ontario; KENNETH M.WILKE, Topeka, Kansas, USA; and the proposer. One incorrect solution wassent in.

Besides Oman and Prielipp, and (to a smaller extent) Cross, no otherreaders seem to have noticed the presence of the Fibonacci numbers in thesolution to this problem. One more occurrence, which Oman and Prielippdon't mention, is that the largest number k + n on the right-hand side ofthe given equation is F 2

2m+4, which �ts nicely with the fact that the smallest

number on the right-hand side is k = F 22m+3. In fact, only Brown com-

mented on the fact that these numbers are squares! So for example, theequation arising from the next-smallest solution n = 39 is

22 + 42 + � � �+ 782 = 25 � 26 + 26 � 27 + � � �+ 63 � 64;

519

where 25 = F 25 and 64 = F 2

6 (and 78 = 2F5F6�2, as follows from SolutionII). In general, the required equation reads

22 + 42 + � � �+ (2F2t�1F2t � 2)2

= F 22t�1(F

22t�1 + 1) + (F 2

2t�1 + 1)(F 22t�1 + 2) + � � �+ (F 2

2t � 1)F 22t

for any integer t � 2.

2186. [1996: 319] Proposed by Vedula N. Murty, Andhra University,Visakhapatnam, India.

Let a, b, c respectively denote the lengths of the sides BC, CA, AB of

triangle ABC. Let G denote the centroid, let I denote the incentre, let R

denote the circumradius, r denote the inradius, and let s denote the semi-

perimeter.

Prove that

GI2 =1

9(a+ b+ c)

�(a� b)(a� c)(b+ c� a)

+ (b� c)(b� a)(c+ a� b) + (c� a)(c� b)(a+ b� c)�:

Deduce the (known) result

GI2 =1

9

�s2 + 5r2 � 16Rr

�:

Solution by Kee-Wai Lau, Hong Kong.

Let A be the origin, AB = u and AC = v. It is well known that

AG =1

3(u+ v) and AI =

b

a+ b+ cu +

c

a+ b+ cv:

Hence

GI =

�b

a+ b+ c� 1

3

�u +

�c

a+ b+ c� 1

3

�v:

520

Since u � u = c2; v � v = b2 and u � v = bc cosA = 12(b2 + c2 � a2), so

GI2 =

�b

a+ b+ c� 1

3

�2c2 +

�c

a+ b+ c� 1

3

�2b2

+

�b

a+ b+ c� 1

3

��c

a+ b+ c� 1

3

�(b2 + c2 � a2)

=1

9(a+ b+ c)2

�(a+ c� 2b)2c2 + (a+ b� 2c)2b2

+ (a+ c� 2b)(a+ b� 2c)(b2 + c2 � a2)�

=1

9(a+ b+ c)

�� a3 � b3 � c3 + 2a2b+ 2ab2

+ 2b2c+ 2bc2 + 2c2a+ 2ca2 � 9abc�

=1

9(a+ b+ c)

�(a� b)(a� c)(b+ c� a)

+ (b� c)(b� a)(c+ a� b) + (c� a)(c� b)(a+ b+ c)�

as required. Since

s =a+ b+ c

2; r =

s(s� a)(s� b)(s� c)

sand R =

abc

4rs;

we have

1

9

�s2+ 5r

2 � 16Rr�

=1

9

�(a + b+ c)2

4+

5(b + c� a)(a + c� b)(a+ b� c)

4(a + b+ c)�

8abc

a + b+ c

�

=1

9(a+ b+ c)

�(a+ b+ c)3 + 5(b+ c� a)(a + c� b)(a + b� c)� 32abc

4

�

=1

9(a+ b+ c)

�� a

3 � b3 � c

3+ 2a

2b+ 2ab

2

+ 2b2c+ 2bc

2+ 2c

2a + 2ca

2 � 9abc�

= GI2;

as required.

Bellot Rosado notes that a variation of this problem was proposed by

Cezar Co�snit� �a, solved by T.C. Esty, with the second form ofGI given by D.L.

MacKay [Problem E415, American Mathematical Monthly (1940), solution

p. 712].

Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Val-ladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK;FLORIAN HERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS,

521

Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; MICHAEL LAMBROU,University of Crete, Crete,Greece; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA;PANOS E. TSAOUSSOGLOU, Athens, Greece; and the proposer.

2187. [1996: 320] Proposed by Syd Bulman-Fleming and EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.

It is easy to show that the maximum number of bishops that can be placed

on an 8� 8 chessboard, so that no two of them attack each other, is 14.

(a) Prove or disprove that in any con�guration of 14 non-attacking bishops,

all the bishops must be on the boundary of the board.

(b) Describe all of the con�gurations with 14 non-attacking bishops.

Solution by Kee-Wai Lau, Hong Kong.

(a) We prove the result.

Clearly we need consider only the black bishops. The diagram below

shows the black squares of a chessboard rearranged so that the bishops will

now move vertically or horizontally. [The diagram is labelled the same as

the standard labelling of the squares of a chessboard, with the rows (ranks)

numbered 1 to 8 and the columns (�les) labelled a to h. | Ed.]

h8

f8 g7 h6

d8 e7 f6 g5 h4

b8 c7 d6 e5 f4 g3 h2

a7 b6 c5 d4 e3 f2 g1

a5 b4 c3 d2 e1

a3 b2 c1

a1

In any column and any row there is at most one bishop. [Thus there

are at most 7 black bishops and similarly at most 7 white bishops, for the

given maximum of 14. Moreover, to attain this maximum, there must be a

black bishop in each column in the above diagram. | Ed.]

Denote the bishop in the kth column by Bk. If B1 is on a7 then B7

must be on h2, and if B1 is on b8 then B7 must be on g1. Next, B2 must be

on a5 or d8 and correspondingly B6 must be on h4 or e1. Next B3 must be

on a3 or f8 and correspondingly B5 must be on h6 or c1. Finally B4 must

be on a1 or h8. This shows that the non-attacking bishops must be on the

boundary of the board.

(b) From part (a) we see that there are 24 = 16 ways to locate the black

bishops. Similarly there are 16 ways to locate the white bishops. Thus there

are 256 con�gurations with 14 non-attacking bishops.

522

Also solved (usually the same way) by SHAWN GODIN, St. JosephScollard Hall, North Bay, Ontario; RICHARD I. HESS, Rancho Palos Verdes,California, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece;and the proposers. Part (a) was also solved by FLORIAN HERZIG, student,Perchtoldsdorf, Austria.

2188. [1996: 320] Proposed by Victor Oxman, University of Haifa,Haifa, Israel.

Suppose that a, b, c are the sides of a triangle with semi-perimeter s

and area 4. Prove that1

a+

1

b+

1

c<

s

4 :

I. Solution Paul Yiu, Florida Atlantic University, Boca Raton, Florida,USA.

Let C be the angle opposite to side c of the triangle. Since

4 = 1

2ab sinC � 1

2ab, we have 1

a� b

24 : Here, equality holds if and only if

the angle opposite to c is a right angle.

Similarly, 1b� c

24 , and 1c� a

24 . Since equality cannot hold simulta-

neously, we have1

a+

1

b+

1

c<b+ c+ a

24 =s

4 :

II. Solutionby Theodore Chronis, student, Aristotle University of Thes-saloniki, Greece.

I will prove the stronger inequality:

1

a+

1

b+

1

c�p3

2� s4 :

Let a = x+ y; b = y+ z; c = z + x, where x; y; z > 0. Then�1

a+

1

b+

1

c

�2=

�1

x+ y+

1

y + z+

1

z + x

�2

��

1

2pxy

+1

2pyz

+1

2pzx

�2

=1

4

�1

pxy

+1

pyz

+1pzx

�2

� 3

4

�1

xy+

1

yz+

1

zx

�(by Cauchy-Schwarz)

Now s

4 = x+y+zp(x+y+z)xyz

=q

x+y+z

xyz=q

1xy

+ 1yz

+ 1zx.

Thus 1a+ 1

b+ 1

c�

p32� s

4 : The equality holds only when a = b = c.

523

Also solved by HAYO AHLBURG, Benidorm, Spain; MIGUELAMENGUAL COVAS,Cala Figuera,Mallorca, Spain; CLAUDIOARCONCHER,Jundia��, Brazil; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo,Bosnia and Herzegovina; FRANCISCO BELLOT ROSADO, I.B. Emilio Fer-rari, Valladolid, Spain; CHRISTOPHER J. BRADLEY, Clifton College, Bris-tol, UK; ADAM BROWN, Scarborough, Ontario; SABIN CAUTIS, student,Earl Haig Secondary School, North York, Ontario; GORAN CONAR, student,Gymnasium Vara�zdin, Vara�zdin, Croatia; FLORIAN HERZIG, student, Per-chtoldsdorf, Austria; RICHARD I. HESS, Rancho Palos Verdes, California,USA; JOHN G. HEUVER, Grande Prairie Composite High School, GrandePrairie, Alberta; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Aus-tria; MURRAY S. KLAMKIN, University of Alberta, Edmonton, Alberta;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;STEFAN and ALEXANDER LAMBROU, students, Crete, Greece; MICHAELLAMBROU, University of Crete, Crete, Greece; CAN AN+H MINH, Univer-sity of California, Berkeley, California; GOTTFRIED PERZ, Pestalozzigymna-sium, Graz, Austria (2 solutions); BOB PRIELIPP, University of Wisconsin{Oshkosh, Wisconsin, USA; JUAN-BOSCO ROMERO M �ARQUEZ, Universidadde Valladolid, Valladolid, Spain; HEINZ-J�URGEN SEIFFERT, Berlin, Ger-many; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel, theNetherlands; PANOS E. TSAOUSSOGLOU, Athens, Greece; GEORGETSAPAKIDIS, Agrinio, Greece; ADRI�AN UBIS MATI�INEZ, student, I.B.Sagasta, Logro ~no, Spain; MELETIS VASILIOU, Elefsis, Greece; and the pro-poser.

2189. [1996: 361] Proposed by Toshio Seimiya, Kawasaki, Japan.

The incircle of a triangle ABC touches BC at D. Let P and Q be

variable points on sidesAB and AC respectively such that PQ is tangent to

the incircle. Prove that the area of triangle DPQ is a constant multiple of

BP �CQ.Solution by Florian Herzig, student, Perchtoldsdorf, Austria.

Let E; F; T be the points of contact of the incircle with CA;AB and

PQ; de�ne x := AP , y := AQ as well as z := PQ. Then

x+ y + z = x+ y + PT +QT

= (x+ PF ) + (y +QE)

= AF + AE = 2(s� a) (1)

where s is the semi-perimeter of4ABC. Applying the cosine rule to4APQyields

z2 = x2 + y2� 2xy cosA = x2 + y2 � 2xy � b2 + c2 � a2

2bc:

524

From equation (1) we get

4(s�a)2�4(s�a)(x+y)+x2+2xy+y2 = x2+y2�xy � b2 + c2 � a2

bc

which is equivalent to

xy ��b2 + 2bc+ c2 � a2

bc

�+ 4(s� a)2 = 4(s� a)(x+ y)

xy�(b+ c)2 � a2

�+ 4bc(s� a)2 = 4bc(s� a)(x+ y)

4xys(s� a) + 4bc(s� a)2 = 4bc(s� a)(x+ y)

xys+ bc(s� a) = bc(x+ y): (2)

Now letR be the circumradius of4ABC and use the sine law [in the second

equality] and (2) [in the third] to get

[DPQ] = [ABC]� AP � AQ � sinA2

� BP � BD � sinB2

�CD �CQ � sinC2

=1

4R

�abc� xya� (c� x)(s� b)b� (b� y)(s� c)c

�=

1

4R

habc� axy � bc(s� b)� bc(s� c) + bx(s� b)

+ cy(s� c) +�xys+ bc(s� a)� bc(x+ y)

�i=

1

4R

�abc� axy � abc+ bx(s� b� c) + cy(s� b� c)

+ xys+ bc(s� a)�

=1

4R

�xy(s� a)� bx(s� a)� cy(s� a) + bc(s� a)

�

=(s� a)

4R(c� x)(b� y) =

(s� a)

4R� BP � CQ:

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, Valladolid, Spain;CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol,UK; RICHARD I.HESS,Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State University, BigRapids, Michigan, USA; D.J. SMEENK, Zaltbommel, the Netherlands; andthe proposer. Several solvers mentioned that this problem generalizes theproposer's earlier 1862 [1993: 203], [1994: 172-173].

525

2190. [1996: 361] Proposed by Walther Janous, Ursulinengymnas-ium, Innsbruck, Austria.

Determine the range of

sin2A

A+

sin2B

B+

sin2C

C

where A, B, C are the angles of a triangle.

Solution by Kee-Wai Lau, Hong Kong.Denote the function of the problem by f(A;B;C). We show that

0 < f(A;B;C) � 27

4�:

The �rst inequality follows from the de�nitions. By considering the degen-

erate triangle A = �, B = 0, C = 0, we see that it is also sharp. We now

prove the second inequality.

For 0 < x � �

2, let g(x) = sin2 x

x. We have

dg

dx=

sinx(2x cosx� sinx)

x2and

d2g

dx2=

h(x)

x3; where h(x) = 1� cos(2x) + 2x2 cos(2x)� 2x sin(2x) :

Since dh

dx= �4x2 sin(2x) � 0 and h(0) = 0, we have h(x) � 0. It follows

that d2g

dx2� 0, so that g is concave. Hence, if4ABC is acute angled, then

f(A;B;C) � 3g

�A+B + C

3

�= 3g

��

3

�=

27

4�:

Equality holds if and only if A = B = C = �

3.

Now, suppose that one of the angles of 4ABC is obtuse. We may

assume that \A > �

2, \B + \C < �

2.

Hence f(A;B;C) <2

�+

sin2B

B+

sin2C

C� 2

�+ 2

sin2�B+C2

�B+C

2

[by the concavity of g(x) together with sin2AA

< 1A< 2

�for A > 2

�].

It is easy to show that tanx � 2x for 0 � x � �

4. Hence

dg(x)

dx� 0

for 0 � x � �

4. It follows that

f(A;B;C) <2

�+ 2

sin2��

4

��

4

=6

�<

27

4�:

Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;RICHARD I. HESS, Rancho Palos Verdes, California, USA; and the proposer.

526

Three unsatisfactory solutions were received. Two of these dependedon an inequality that was passed o� as \known". Such claims must be ac-companied by a reference. Otherwise, how are the editors supposed to knowthat the claim is true? More importantly, the purpose of the solution shouldbe to explain why the result is true; consequently any reference should beaccessible to CRUX with MAYHEM readers.

A key step in the third rejected submission was claimed to be \ob-vious". The only thing obvious to this editor was that such claims do notbelong in mathematical arguments.

2191. [1996: 361] Proposed by �Sefket Arslanagi�c, University of Sara-jevo, Sarajevo, Bosnia and Herzegovina.

Find all positive integers n, that satisfy the inequality

1

3< sin

��

n

�<

1

2p2:

Editor's composite solutionbased on the ones submitted by the solverswhose names appear below.

Since sin� = 0, n = 1 is clearly not a solution.

Since sin��

n

�is decreasing for n � 2 and since

sin

��

8

�=

1

2

q2�

p2 >

1

2p2

we have sin��

n

�> 1

2p2for 2 � n � 8.

On the other hand, since sinx < x for x > 0, we have, for all n � 10,

sin��

n

�< �

n� �

10< 1

3.

We now show that

1

3< sin

��

9

�<

1

2p2

(1)

and son = 9 is the only solution. [Ed: Though (1) can be veri�ed numerically

by using a calculator, as many solvers did, the proposer's original intention

was an \analytic" proof like the one presented below.]

Let � = �

9and let r = sin�. Then from

p3

2= sin

��

3

�= sin(3�) = 3 sin� � 4 sin3 � = 3r � 4r3;

we see that r is a positive root of the polynomial f(x) = 4x3 � 3x+p32.

527

Note that

f(�1) = �1 +p3

2< 0; f(0) =

p3

2> 0;

f

�1

3

�= �23

27+

p3

2> 0; f

�1

2p2

�=

�54p2+

p3

2< 0;

and f(1) = 1 +

p3

2> 0:

Since f is a continuous function, we conclude that it has three real

roots, one in each of the three intervals: (�1; 0),�13; 1

2p2

�and

�1

2p2; 1�.

But sin��

9

�< �

9� 0:349 < 1

2p2, and so (1) follows.

Solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH andROGER ZARNOWSKI (jointly)Angelo State University, San Angelo, TX, USA;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece;F.J. FLANIGAN, San Jose State University, San Jose, California, USA; SHAWNGODIN, St. Joseph Scollard Hall, North Bay, Ontario; RICHARD I. HESS,Rancho Palos Verdes, California, USA; WALTHER JANOUS, Ursulinengym-nasium, Innsbruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA;V �ACLAV KONE �CN �Y, Ferris State University, Big Rapids, Michigan, USA;MICHAEL LAMBROU, University of Crete, Crete, Greece; HEINZ-J�URGENSEIFFERT, Berlin, Germany; D.J. SMEENK, Zaltbommel, the Netherlands;DIEGO SOT�ES and JAVIER GUTI�ERREZ, students, University of Rioja,Logro ~no, Spain; and DAVID R. STONE, Georgia Southern University, States-boro, Georgia, USA.

There were also nine incomplete or (partially) incorrect solutions.Though these solutions all give the correct answer n = 9, they contain vari-ous errors. Most of these errors pertain to the analysis of the three roots off(x) which is actually more subtle than it appears to be. Some solvers erro-neously claimed that one of the roots is greater than one. UsingMAPLE, we�nd easily that the three roots are: �0:9848077, 0:3420201, and 0:6427876.Only one (joint) solver stated that the three roots are, in fact, cos

�19�18

�,

sin��

9

�, and sin

�7�9

�. However, they made the mysterious and wrong state-

ment that sin�7�9

�< 1

3.

Some other solvers, after checking that f�13

�and f

�1

2p2

�have oppo-

site signs, jump to the conclusion that sin��

9

�must be the root in the interval�

13; 1

2p2

�. Clearly, to make this argument valid, one has to show that the

other positive root is not in this interval.

528

2192. [1996: 362] Proposed by Theodore Chronis, student, AristotleUniversity of Thessaloniki, Greece.

Let fang be a sequence de�ned as follows:

an+1 + an�1 =

�a2

a1

�an; n � 1:

Show that if

����a2a1���� � 2, then

����ana1���� � n.

Solution by Can Anh Minh, University of California, Berkeley.By the triangle inequality we have

jan+1j+ jan�1j � jan+1 + an�1j =����a2a1 an

���� =����a2a1

���� janj � 2janj

Thus we have jan+1j � janj � janj � jan�1j, and therefore

janj � jan�1j � jan�1j � jan�2j � � � � � ja2j � ja1j � ja1j

Thus we have

janj � ja1j =nX

k=2

(jakj � jak�1j) � (n� 1)ja1j

It follows that janj � nja1j, or equivalently����ana1���� � n:

Also solved by GERALD ALLEN, CHARLES DIMINNIE, TREY SMITH,and ROGER ZARNOWSKI (jointly), Angelo State University, San Angelo,Texas; CHETAN T. BALWE, Pune, India; CHRISTOPHER J. BRADLEY, CliftonCollege, Bristol, UK; SABIN CAUTIS, student, Earl Haig Secondary School,North York, Ontario; DAVID DOSTER, Choate Rosemary Hall, Wallingford,Connecticut, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria;RICHARD I. HESS, Rancho Palos Verdes, California, USA; CYRUS HSIA, stu-dent, University of Toronto, Toronto, Ontario; WALTHER JANOUS, Ursul-inengymnasium, Innsbruck, Austria; V �ACLAV KONE �CN �Y, Ferris State Uni-versity, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University ofCrete, Crete, Greece; KEE-WAI LAU, Hong Kong; DIGBY SMITH, MountRoyal College, Calgary, Alberta; and the proposer.

529

2193. [1996: 362] Proposed by Luis V. Dieulefait, IMPA, Rio deJaneiro, Brazil.

(a) Prove that every positive integer is the di�erence of two relatively prime

composite positive integers.

(b) Prove that there exists a positive integer n0 such that every positive in-

teger greater than n0 is the sum of two relatively prime composite positive

integers.

Solution to (a) byMichael Lambrou, University of Crete, Crete, Greece,modi�ed slightly by the editor.

Note that for all natural numbers k, we have:

2k + 1 = (k+ 1)2 � k2 (1)

and

2k = (2k+ 1)(8k+ 1)� (4k+ 1)2 (2)

Since gcd(k; k+ 1) = 1, (1) gives a required representation for all odd inte-

gers greater than 3. But 1 = 9� 8 and 3 = 25� 22, and so a required rep-

resentation exists for all odd natural numbers. On the other hand, straight-

forward computations show that

�(8k+ 4)(2k+ 1)(8k+ 1) + (8k+ 5)(4k+ 1)2 = 1

and so

gcd�(2k+ 1)(8k+ 1); (4k+ 1)2

�= 1:

Thus (2) gives a required representation for all even natural numbers.

Solution to (b) by the proposer, modi�ed by the editor.Let �(n)denote Euler's totient function and let �(n)denote the prime-

counting function. [Ed: that is, �(n) is the number of primes p such that

p � n.] It is known (see Hardy and Wright, An Introduction to the Theoryof Numbers) that

liminfn!1

�(n)n

log logn

= e� >1

2;

where is Euler's constant. Hence, for su�ciently large n, we have

�(n) >2

2 log logn: (1)

On the other hand, by Tschebyche�'s theorem, we have

�(n) <6

5

n

logn(2)

530

if n is large enough. From (1) and (2), we obtain

limn!1

�(n)

�(n)> lim

n!15 logn

12 log logn= 1;

which implies, for large enough n, that

�(n) � 2��(n) + 1

�: (3)

Note that, if n = a + b, then (a; n) = 1 is equivalent to (b;n) = 1, and to

(a; b) = 1.

Consider the �(n) ordered decompositions of n : n = k + (n � k),

where 1 � k � n, such that gcd(k;n) = 1. If we strike out those pairs in

which the �rst or second summand is a prime or 1, then we are deleting at

most 2(�(n) + 1) pairs, and so from (3), we conclude that there is at least

one decomposition n = a+ b where a and b are relatively prime composite

natural numbers, and our proof is complete.

Solved (both parts) by RICHARD I. HESS, Rancho Palos Verdes, Cal-ifornia, USA; MICHAEL LAMBROU, University of Crete, Crete, Greece; andthe proposer. Part (a) only was solved by FLORIAN HERZIG, student, Per-chtoldsdorf, Austria; and WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria.

Regarding (a), Janous actually obtained a stronger result by showingthat there are, in fact, in�nitely many required representations. Regard-ing (b), Hess remarked that a computer search seems to indicate thatn0 � 210. Using arguments similar to those given in the proof above,Lambrou proved a stronger result, namely:

For any given integer k � 1, there exists a positive integer mk such

that every integer greater thanmk can be written as the sum of two relatively

prime composite natural numbers in at leastmk di�erent ways.

It is not di�cult to see that this result also follows easily from the proposer'sproof presented above.

2194. [1996: 362] Proposed by Christopher J. Bradley, Clifton Col-lege, Bristol, UK.

Prove or disprove that it is possible to �nd a triangle ABC and a transver-

sal NML with N lying between A and B, M lying between A and C, and

L lying on BC produced, such that BC, CA, AB, NB, MC, NM , ML,

and CL are all of integer length, andNMCB is a cyclic inscriptable quadri-

lateral.

Editor's comment. There is a terminology problem here. It appears (from

his solution) that the proposer had intended the word inscriptable to mean that a

circle can be inscribed in the quadrangle NMCB. Although several of the solvers

531

assumed this to be the case, all reference books that I consulted disagree. The Oxford

English Dictionary and Nathan Altshiller Court (in his 1925 College Geometry) both

say that an inscriptible quadrangle (note the spelling) is one that can be inscribed

in a circle; today one more commonly calls such quadrangles (that can be inscribed

in circles) cyclic, concyclic, or inscribable | take your choice. On the other hand,

one must keep in mind that English (the language in which, for example, in ammable

means ammable and unravel means ravel) is so unpredictable that the proposer

might well have references to back up his terminology. For a quadrangle that con-

tains an inscribed circle, one would say circumscribing or avoid the issue and just

say \a quadrangle with an inscribed circle". The matter really becomes muddy here:

Court calls the latter quadrangle circumscriptible, while the OED agrees in one place

(under inscriptible), but contradicts itself in another, where a quadrangle is said to

be circumscribable if it \may be circumscribed by a circle". The moral of this story:

be circumspect (or maybe, circumspectable). As it turns out, our featured solutions

all produce cyclic quadrangles having inscribed circles!

I. Solution by Michael Lambrou, University of Crete, Crete, Greece(somewhat edited).

We show that any triangle ABC with rational sides and \A 6= \C

has a transversal LMN such that NMCB satis�es both circle conditions,

and has all relevant lengths rational. Hence, multiplying by an appropriate

number, we can obtain a similar triangle with the required properties.

Let a triangle ABC with rational sides be given. By renaming, we may

assume that B, which may be acute, right or obtuse, is larger than C. Note

that B cannot equal C, as the transversal | in order to produce a cyclic

quadrangle | would then be parallel to BC.

Consider as transversal LMN such that

\AMN = B (making NMCB cyclic), and (4)

LM is tangent to the incircle of triangle ABC (5)

(forcingNMCB to have an inscribed circle). We must show that the lengths

NB, MC, NM , ML and CL are all rational.

We have that the triangles ANM and ACB are similar. Let

� = AM=AB be the similarity ratio. We use the fact that the sides of

triangle ABC are rational to show that � 2 Q.

�(BC + AB + AC)

= MN + AM +AN

= MN + (AC �MC) + (AB �NB)

= [(MN + BC)� (NB +MC)] + AB + AC �BC

= 0 +AB + AC � BC 2 Q :

But BC + AB +AC 2 Q, so that � 2 Q as well.

532

Next, byMenelaus'sTheorem applied to triangleABC with transversal

LMN , it follows that

BL

CL� CMAM

� ANBN

= 1 2 Q ;

so that BL=CL 2 Q. Since BC=CL = (BL� CL)=CL, we have LC 2 Qand BL 2 Q.

Similarly, byMenelaus's Theorem applied to triangleMLC with trans-

versal AB, we see that LN 2 Q, and the desired result follows.

II. Families of solutionsbyMichael Lambrou, University of Crete, Crete,Greece and (independently) by Richard I. Hess, Rancho Palos Verdes, Cali-fornia, USA.

Let AMN and LBN be congruent right triangles with right angles at

M and B. Denote the legs by p ( = AM = LB), q ( = MN = BN), and

the hypotenuse by r ( = NL = NA =pp2 + q2). The solution is achieved

by choosing p, q, r so that p divides r + q. Hess provided the examples in

column 2, and Lambrou, column 3.

Hess Lambrou

p = AM = LB 2k+ 1 �2,

q = MN = BN 2k(k+ 1) ��,

r = NL = NA (k+ 1)2 + k2 ��, where

�2 + �2 = �2,

r + q = AB = LM (2k+ 1)2 (�+ �)�,

q(q+ r)=p =MC

= BC

2k(k+ 1)(2k+ 1) (�+ �)�,

r(q + r)=p = AC

= LC

(2k+ 1)��(k+ 1)2 + k2

� (�+ �)�

= �2 + �2 + ��.

In each case, NMCB has a circumscribed circle because of the right

angles at B and M , while it has an inscribed circle because of its symmetry

(so that NM + CB = NB + CM ).

Lambrou also sent in a family of asymmetric solutions for positive

integers m > n:

AB = nm2(m� n)(n2+ 1);

AC = n2m(m� n)(m2 + 1);

BC = mn(m� n)(m+ n)(mn� 1); and

AN =1

mnAC; AM =

1

mnAB:

We leave it to the reader to verify the details. (Remember to check the var-

ious triangle inequalities.)

Also solved by FLORIAN HERZIG, student, Perchtoldsdorf, Austria;D.J. SMEENK, Zaltbommel, the Netherlands; and the proposer.

533

2195. [1996: 362] Proposed by Bill Sands, University of Calgary,Calgary, Alberta.

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random,

one after the other, and with the balls replaced after each draw.

What is the probability that the three-element sequence obtained has the

properties that the smallest element is odd and that only the smallest ele-

ment, if any, is repeated?

For example, the sequences 453 and 383 are acceptable, while the sequences

327 and 388 are not.

(NOTE: this problem was suggested by a �nal exam that I marked recently.)

I. Solution by David Hankin, Hunter College Campus Schools, NewYork, NY, USA.

Let uk be the number of acceptable sequences chosen from balls num-

bered 1 to 2k, and let ak be the number of these acceptable sequences that

contain 1. We �rst show that, for k � 1,

uk � ak = uk�1;

where u0 = 0. Note that uk � ak is the number of acceptable sequences

that do not contain 1. The number of these sequences is the same as the

number of acceptable sequences that can be chosen from balls numbered 3

to 2k. Clearly, this is equal to uk�1.

To �nd ak, note that sequences that contain 1 must have one, two or

three 1's. There is one sequence with three 1's. For sequences with two 1's,

there are 2k�1 ways to choose the third element and 3 arrangements of the

three elements; so there are 3(2k� 1) such sequences. Similarly, there are

6�2k�12

�= 6(2k� 1)(k� 1) sequences with one 1. Thus

ak = 1+3(2k�1)+6(2k�1)(k�1) = 12k2�12k+4 = 4[k3� (k�1)3]:

Now (since u0 = 0)

un =

nXk=1

(uk � uk�1) =nX

k=1

ak = 4

nXk=1

[k3 � (k� 1)3] = 4n3:

Thus the requested probability is

4n3

(2n)3=

1

2:

II. Solution by Michael Lambrou, University of Crete, Crete, Greece.We show that there is a one-to-one onto pairing that pairs each favour-

able triplet abc (which for convenience we will denote (a; b; c)) with an un-

favourable one. Once this is done, the independence of events shows that

534

the sought probability is 1=2, as the favourable events exactly match the un-

favourable ones.

We denote the smallest element among (a; b; c) by 2s � 1, where

1 � s � n. The pairing will be described for the case when 2s � 1 oc-

curs in the �rst position (and is perhaps repeated in one or both of the other

positions). The other two possibilities are dealt with in a similar fashion,

by cyclic change of order. Note that some care must be taken not to double

count the triplets that belong to more than one situation.

Here is how we do our pairing.

� For �xed s and p with 2s � 1 < p � 2n� 1, map, in any one-to-one

onto fashion, the 2n� 2s triplets of the form

(2s� 1; p; q) where 2s� 1 < q � 2n and q 6= p

to the 2n� 2s triplets of the form

(2s; p+ 1; v) where 2s < v � 2n;

� for 2s� 1 < q � 2n� 1, map (2s� 1; 2s� 1; q) to (2s;2s; q+ 1);

� for 2s� 1 < q � 2n� 1, map (2s� 1; 2n; q) to (2s� 1; q; q);

� map (2s� 1; 2s� 1; 2n) to (2n; 2n;2s� 1); and �nally

� map (2s� 1; 2s� 1; 2s� 1) to (2s; 2s; 2s).

It is easy to see that this map is not ambiguous. It is also easy to check

that all favourable and all unfavourable triplets have been taken into account

once and once only. This completes the argument.

Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol,UK; ROBERT GERETSCHL �AGER, Bundesrealgymnasium, Graz, Austria;SHAWN GODIN, St. Joseph Scollard Hall, North Bay, Ontario; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; D. KIPP JOHNSON, Beaverton, Oregon, USA; W. MOSER,McGill University, Montr �eal, Qu �ebec; ROBERT P. SEALY, Mount AllisonUniversity, Sackville, New Brunswick; D. N. SHETH, Sir Parashuram Col-lege, Pune, India; DIGBY SMITH, Mount Royal College, Calgary, Alberta;and the proposer. Two incorrect solutions were sent in.

Lambrou was the only solver to �nd a \combinatorial" proof that theprobability is exactly 1=2. He also sent in two other solutions, one of whichis similar to solution I. Bradley's solution is also similar to solution I.

535

2196. [1996: 362] Proposed by Juan-Bosco Romero M�arquez, Uni-versidad de Valladolid, Valladolid, Spain.

Find all solutions of the diophantine equation

2(x+ y) + xy = x2 + y2;

with x > 0, y > 0.

Solution by Sam Baethge, Nordheim, Texas, USA.Let y = rx with r rational. The given equation becomes a quadratic

in r:

r2x2 � r(x2 + 2x) + (x2 � 2x) = 0

Then r =�x+ 2�p�3x2 + 12x+ 4

�=2x. The discriminant can be writ-

ten as 16 � 3(x � 2)2 and must be the square of an integer. Since x is a

positive integer the only solutions are (x; r) = (4;1), (4; 12), or (2; 2). This

produces (x; y) = (4; 4), (4; 2), or (2; 4).

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca,Spain; MANSUR BOASE, student, St. Paul's School, London, England;CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; SABIN CAUTIS,student, Earl Haig Secondary School, North York, Ontario; THEODORECHRONIS, student, Aristotle University of Thessaloniki, Greece; GORANCONAR, student, Vara�zdin, Croatia; PAUL-OLIVIER DEHAYE, Brussels, Bel-gium; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut,USA; C. FESTRAETS-HAMOIR, Brussels, Belgium; F.J. FLANIGAN, San JoseState University, San Jose, California, USA; ROBERT GERETSCHL �AGER, Bun-desrealgymnasium, Graz, Austria; SHAWN GODIN, St. Joseph Scollard Hall,North Bay, Ontario; DAVID HANKIN, Hunter College Campus Schools, NewYork, NY, USA; FLORIAN HERZIG, student, Perchtoldsdorf, Austria; CYRUSHSIA, student, University of Toronto, Toronto, Ontario; IGNOTUS, Villeta,Colombia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;D. KIPP JOHNSON, Beaverton, Oregon, USA; KEE-WAI LAU, Hong Kong;J.A. MCCALLUM, Medicine Hat, Alberta; CAN ANH MINH, University ofCalifornia, Berkeley; MICHAEL PARMENTER, Memorial University of New-foundland, St. John's, Newfoundland; GOTTFRIED PERZ, Pestalozzigymna-sium, Graz, Austria; BOB PRIELIPP, University of Wisconsin{Oshkosh, Wis-consin, USA; ROBERT P. SEALY, Mount Allison University, Sackville, NewBrunswick; HEINZ-J�URGEN SEIFFERT, Berlin, Germany; D.J. SMEENK, Zalt-bommel, the Netherlands; DIGBY SMITH, Mount Royal College, Calgary,Alberta; DAVID R. STONE, Georgia Southern University, Statesboro, Geor-gia, USA; ADRI�AN UBISMATI�INEZ, Logro ~no, Spain; DAVID C. VELLA, Skid-more College, Saratoga Springs, New York; EDWARD T.H. WANG, WilfridLaurier University,Waterloo, Ontario; KENNETHM.WILKE, Topeka, Kansas,USA; and the proposer. There were seven incorrect solutionsand one incom-plete solution.

Vella (in one of his two submitted solutions)used a change of variablesto turn the given equation into the equation of an ellipse with centre on the

536

x{axis and major axis aligned with the x{axis, and then used a geometricapproach to solve the problem. The proposer also suggests a generalization:

Solve the diophantine equation:

z(x+ y) + xy = x2 + y2

2197. [1996: 363] Proposed by Joaqu��n G �omez Rey, IES Luis Bu ~nuel,Alcorc �on, Madrid, Spain.

Let n be a positive integer. Evaluate the sum:

1Xk=n

�2k

k

�(k+ 1)22k+1

:

Solution by David Doster, Choate Rosemary Hall, Wallingford, Con-necticut, USA.

The value of the sum is1

4n

�2n

n

�. This follows directly from

1Xk=0

�2k

k

�(k+ 1)22k+1

= 1 (1)

and

n�1Xk=0

�2k

k

�(k+ 1)22k+1

= 1� 1

4n

�2n

n

�: (2)

To prove (1), note that it is easy to show that

�2k

k

�(k+ 1)22k+1

= �� 1

2

k+ 1

�(�1)k+1

�they both simplify to

1 � 3 � 5 � � � (2k� 1)

2k+1(k+ 1)!

�. Therefore

1Xk=0

�2k

k

�(k+ 1)22k+1

= �1Xk=0

(�1)k+1

� 12

k + 1

�= �

1Xk=1

(�1)k�12

k

�

= � �1 +

1Xk=0

(�1)k�12

k

�!= 1�

p1 + (�1) = 1:

537

To prove (2), we use induction. After checking at n = 1, we use the inductive

assumption to get

nXk=0

�2k

k

�(k+ 1)22k+1

= 1� 1

4n

�2n

n

�+

�2n

n

�(n+ 1)22n+1

= 1� 1

4n

�2n

n

��1� 1

2(n+ 1)

�= 1� 1

4n

�2n

n

�2n+ 1

2(n+ 1)

= 1� 1

4n+1

�2n

n

�(2n+ 1)(2n+ 2)

(n+ 1)2= 1� 1

4n+1

�2n+ 2

n+ 1

�;

which proves the identity.

Also solved by PAUL BRACKEN, Universit �e de Montr �eal, Montr �eal,Qu �ebec; CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK; FLORIANHERZIG, student, Perchtoldsdorf, Austria; RICHARD I. HESS, Rancho PalosVerdes, California, USA; WALTHER JANOUS, Ursulinengymnasium, Inns-bruck, Austria; MICHAEL LAMBROU, University of Crete, Crete, Greece;HEINZ-J�URGEN SEIFFERT, Berlin, Germany; STANWAGON,Macalester Col-lege, St. Paul, Minnesota, USA; and the proposer.

Janous �nds the sum by considering the generating function of the se-

quence fCkg =n

1k+1

�2k

k

�oof Catalan numbers and in doing so, obtains, as

a by-product, the recurrence relation

Cn =1

n+ 1

4n � 2

n�1Xk=0

Ck � 4n�1�k!:

He believes that this is a new identity, and wonders whether there is a purelycombinatorial proof and/or interpretation of it.

Wagon obtained the sum by usingMATHEMATICA, and remarked thatmore sophisticated symbolic algebra software ( work of Petkovsek, Wilf andZeilberger) can provide certi�cates that the closed-form formula obtained isactually valid. He also commented that: \Thus sums such as this are really

best left to computers, : : : ". This editor is interested in knowing how manyof our readers would agree with him.

Crux MathematicorumFounding Editors / R �edacteurs-fondateurs: L �eopold Sauv �e & Frederick G.B. Maskell

Editors emeriti / R �edacteur-emeriti: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical MayhemFounding Editors / R �edacteurs-fondateurs: Patrick Surry & Ravi Vakil

Editors emeriti / R �edacteurs-emeriti: Philip Jong, Je� Higham,

J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia

538

YEAR END FINALEAgain, a year has own by! This year saw the merger of CRUX and MAYHEM

come to fruition. The logistical problems were greater than I expected, and we were

slow at getting the �rst few issues to press. But now, it seems to ow smoothly.

Thanks to all for their patience and assistance.

The online version of CRUX with MAYHEM continues to attract attention.

Thanks are due to LOKI JORGENSON, NATHALIE SINCLAIR, and the rest of the

team at SFU who are responsible for this.

There are many people that I wish to thank most sincerely for particular con-

tributions. First and foremost is BILL SANDS. Bill is of such value to me and to the

continuance of CRUXwithMAYHEM. Aswell, I thankmost sincerely, CATHY BAKER,

ROLAND EDDY, CHRIS FISHER, BILL SANDS, JIM TOTTEN, and EDWARD WANG,

for their regular yeoman service in assessing which solutions should be highlighted;

DENIS HANSON, D. FARENICK, C. FISHER, A. LIU, R. MCINTOSH, J. MACLAREN,

D. RUOFF, M. TSATSOMEROS, H. WESTON, for ensuring that we have quality arti-

cles; ANDY LIU, ROBERT GERETSCHL �AGER, MURRAY S. KLAMKIN, MAR�IA FALK

de LOSADA, J �OSZEF PELIK �AN, GOTTFRIED PERZ, JIM TOTTEN, for ensuring that

we have quality book reviews, ROBERT WOODROW (and JOANNE LONGWORTH),

who carries the heavy load of two corners, one somewhat new and the other of long

standing, and RICHARD GUY for sage advice whenever necessary. The editors of the

MAYHEM section, NAOKI SATO, CYRUS HSIA, ADRIANCHAN, RICHARDHOSHINO,

RAVI VAKIL and WAI LING YEE, all do a sterling job. I also thank two of our reg-

ulars who assist the editorial board with proof reading; THEODORE CHRONIS and

WALDEMARPOMPE. The quality of these people are vital parts of what makes CRUX

with MAYHEM what it is. Thank you one and all.

As well, I would like to give special thanks to our retiring Associate Editor,

CLAYTON HALFYARD, for keeping me from printing too many typographical

errors; andmy colleagues, PETERBOOTH, RICHARDCHARRON, EDGARGOODAIRE,

ERIC JESPERS, MIKE PARMENTER, DONALD RIDEOUT, NABIL SHALABY, in the

Department of Mathematics and Statistics atMemorial University for their occasional

sage advice. I have also been helped by some Memorial University students, DON

HENDER, COLIN HILLIER, PAUL MARSHALL, SHANNON SULLIVAN, as well as a

WISE Summer student, ALYSON FORD. The staff of the Department of Mathemat-

ics and Statistics at Memorial University deserve special mention for their excellent

work and support: ROS ENGLISH, MENIE FRENCH, WANDA HEATH, and LEONCE

MORRISSEY; aswell as the computer and networking expertise of RANDYBOUZANE.

Also the assistance of ELLEN WILSON at Mount Allison University is much appre-

ciated. Not to mention GRAHAM WRIGHT, Managing Editor, would be a travesty.

Graham has kept somuch on the right track. He is a pleasure to workwith. The CMS's

TEX Editor, MICHAEL DOOB has been very helpful in ensuring that the printed mas-

ter copies are up to the standard required for the U of T Press who continue to print a

�ne product. Finally, I would like to express real and heartfelt thanks to the Heads of

my Department, BRUCE WATSON and HERBERT GASKILL, and to ALAN LAW, Dean

of Science of Memorial University, and WILLIE DAVIDSON, Acting Dean of Science

of Memorial University, without whose support and understanding, I would not be

able to do the job of Editor-in-Chief.

Last but not least, I send my thanks to you, the readers of CRUX. Without you,

CRUX would not be what it is. Keep those contributions and letters coming in. I do

enjoy knowing you all.

539

INDEX TO VOLUME 23, 1997

Crux Articles

A Probabilistic Approach to Determinants with Integer Entries

Theodore Chronis .................................................................... 23

Folding the Regular Heptagon

Robert Geretschl �ager ................................................................. 81

Heronian Triangles with Associated Inradii in Arithmetic Progression

Paul Yiu ............................................................................. 146

Folding the Regular Nonagon

Robert Geretschl �ager ............................................................... 210

A Fermat-Fibonacci Collaboration

K.R.S. Sastry ........................................................................ 274

Packing Boxes with N-tetracubes

Andris Cibulis ....................................................................... 336

More Unitary Divisor Problems

K.R.S. Sastry ........................................................................ 407

Dissecting Rectangular Strips Into Dominoes

Frank Chen, Kenneth Nearey and Anton Tchernyi ................................. 468

The Academy Corner Bruce Shawyer

February No. 8 .......................................................................... 3

March No. 9 ............................................................................ 65

April No. 10 ........................................................................... 129

May No. 11 Bernoulli Trials 1997 Christopher Small ................................. 193

September No. 12 Bernoulli Trials 1997 { Hints and Answers Christopher Small .... 257

October No. 13 ....................................................................... 321

November No. 14 ..................................................................... 385

December No. 15 ..................................................................... 449

The Olympiad Corner R.E. Woodrow

February No. 179 ........................................................................ 6

March No. 180 ......................................................................... 66

April No. 181 .......................................................................... 131

May No. 182 .......................................................................... 196

September No. 183 ................................................................... 260

October No. 184 ...................................................................... 322

November No. 185 .................................................................... 388

December No. 186 .................................................................... 450

The Skoliad Corner R.E. Woodrow

February No. 19 ........................................................................ 25

March No. 20 ........................................................................... 89

April No. 21 ........................................................................... 150

May No. 22 ........................................................................... 218

September No. 23 ..................................................................... 278

October No. 24 ....................................................................... 343

November No. 25 ..................................................................... 411

December No.26 ...................................................................... 473

Miscellaneous

Welcome / Bienvenue ................................................................... 1

Editorial ................................................................................ 30

In memoriam | Dr. Leon Banko� Clayton Dodge ................................... 145

540

Book Reviews Andy Liu

Shaking Hands in Corner Brook and other Math Problems

by Peter Booth, Bruce Shawyer and John Grant McLoughlin (Editors)

Reviewed by Robert Geretschl �ager and Gottfried Perz .............................. 20

Leningrad Mathematical Olympiads 1987{1991

by Dmitry Fomin and Alexey Kirichenko

Reviewed by J �oszef Pelik �an .......................................................... 78

The Lighter Side of Mathematics

by Richard K. Guy and Robert E. Woodrow (Editors)

Reviewed by Murray S. Klamkin ................................................... 143

Cien Problemas de Matem �aticas:

Combinatoria, �Algebra, Geometr��a, One Hundred Mathematics Problems:

Combinatorics, Algebra, Geometry

by Francisco Bellot Rosado and Mar��a Ascensi �on L �opez Chamorro

Reviewed by Mar��a Falk de Losada ................................................. 209

The Puzzle Arcade

by Jerry Slocum

Quantum Quandaries

by Timothy Weber (Editor)

Reviewed by Andy Liu .............................................................. 272

A Mathematical Mozaic

by Ravi Vakil

Reviewed by Jim Totten ............................................................ 334

144 Problems of the Austrian-Polish Mathematics Competition 1978{1993

by Marcin Emil Kuczma

40th Polish Mathematics Olympiad 1989/90

by Marcin Emil Kuczma

Reviewed by Andy Liu .............................................................. 405

Learn from the Masters!

edited by Frank Swetz, John Fauvel, Otto Bekken, Bengt Johansson and Victor Katz

Reviewed by Maria de Losada ..................................................... 467

Problems

February 2137, 2201{2212 ............................................................. 45

March 2214{2225 ..................................................................... 109

April 2173, 2226{2237 ................................................................. 166

May 2238{2250 ....................................................................... 242

September 2251{2262 ................................................................. 299

October 2265{2275 .................................................................... 363

November 2276{2286 ................................................................. 431

December 2297{2300 .................................................................. 500

Solutions

February 2101{2112 .................................................................... 49

March 2113{2124 ..................................................................... 112

April 1940, 2125{2128, 2130{2132, 2134{2140 ....................................... 170

May 1940, 2133, 2141{2144 ........................................................... 246

September 2145{2157 ................................................................. 302

October 1940, 2158{2168 ............................................................. 367

November 2090, 2169{2178 ........................................................... 434

December 2179{2197 ................................................................. 504

541

Mathematical Mayhem

February ................................................................................ 30

March .................................................................................. 94

April ................................................................................... 152

May ................................................................................... 224

September ............................................................................. 283

October ................................................................................ 348

November ............................................................................. 418

December ............................................................................. 482

Mayhem Articles

A Journey to the Pole | Part I

Miguel Carri �on �Alvarez ............................................................... 36

Matrix Exponentials | An Introduction

Donny Cheung ....................................................................... 94

A Journey to the Pole | Part II

Miguel Carri �on �Alvarez .............................................................. 152

A Pattern in Permutations

John Linnell ......................................................................... 158

What is the next term?

Naoki Sato .......................................................................... 226

Euler's and DeMoivre's Theorem

Soroosh Yazdani ..................................................................... 231

Decimal Expansion of Fractions

Cyrus Hsia .......................................................................... 285

An Ambivalent Sum

Naoki Sato .......................................................................... 290

Unique Forms

Naoki Sato .......................................................................... 351

Solving the Quartic

Cyrus Hsia .......................................................................... 420

Pizzas and Large Numbers

Shawn Godin ........................................................................ 422

Minima and Maxima of Trigonometric Expressions

Nicholae Gusita ..................................................................... 426

The Equation of the Tangent to the nth Circle

Krishna Srinivasan ................................................................. 485

Combinatorial Games

Adrian Chan ........................................................................ 487

Shreds and Slices

February ................................................................................ 30

May ................................................................................... 224

September ............................................................................. 283

October ................................................................................ 348

November ............................................................................. 482

Positive Matrices and Positive Eigenvalues ............................................ 31

Newton's Relations .................................................................... 32

Mathematically Correct Sayings ....................................................... 34

Fibonacci Residues .................................................................... 224

Playful Palindromes ................................................................... 283

The Best of the WEB .................................................................. 284

Invariants of Inscribed Regular n-gons ............................................... 348

Four Large Spheres and a Small Sphere .............................................. 349

Factorial Fanaticism ................................................................... 418

A Note on Convexity .................................................................. 482

542

Mayhem Problems

February ................................................................................ 42

March ................................................................................. 100

April ................................................................................... 163

May ................................................................................... 235

September ............................................................................. 296

October ................................................................................ 356

November ............................................................................. 429

December ............................................................................. 494

High School Problems

February: H217, H218, H219, H220 ................................................... 43

April: H221, H222, H223, H224 ...................................................... 164

September: H225, H226, H227, H228 ................................................ 297

November: H229, H230, H231, H232 ................................................ 429

High School Solutions

March: H205, H206, H207 ............................................................ 101

May: H208, H209, H210 .............................................................. 236

October: H211, H212, H213 .......................................................... 357

December: H214, H215, H216 ........................................................ 494

Advanced Problems

February: A193, A194, A195, A196 ..................................................... 44

April: A197, A198, A199, A200 ........................................................ 165

September: A201, A202, A203, A204 ................................................. 298

November: A205, A206, A207, A208 .................................................. 430

Advanced Solutions

March: A185, A186, A187 ............................................................. 102

May: A184, A188, A189 ............................................................... 239

October: A190, A191, A192 ........................................................... 358

December: A191 ...................................................................... 496

Challenge Board Problems

February: C70, C71, C72 ............................................................... 44

April: C70, C71, C72 .................................................................. 165

September: C73 ....................................................................... 298

November: C74 ....................................................................... 430

Challenge Board Solutions

March: C64, C86, C69 ................................................................. 106

October: C70, C72 ..................................................................... 360

December: C70, C71 .................................................................. 496

Contests

J.I.R. McKnight Problems Contest 1978 .............................................. 293

1984 Swedish Mathematics Olympiad ................................................ 294

1983 Swedish Mathematics Olympiad Solutions ..................................... 296

1996 Balkan Mathematical Olympiad ................................................. 355

J.I.R. McKnight Problems Contest 1979 .............................................. 428

J.I.R. McKnight Problems Contest 1980 .............................................. 493

Miscellaneous

Contest Dates .......................................................................... 35

IMO Report Richard Hoshino .......................................................... 41

IMO CORRESPONDENCE PROGRAM E.J. Barbeau ................................. 161

IMO Report Adrian Chan ............................................................. 350

Book Reviews Donny Cheung ......................................................... 491

543

Proposers, solvers and commentators in the PROBLEMS and SOLUTIONS

sections for 1997 are:

Hayo Ahlburg: 2170, 2175, 2188

Gerald Allen: 2191, 2192

Miguel Amengual Covas: 2102, 2104, 2112, 2114, 2151, 2162,

2166, 2188, 2189, 2196

Claudio Arconcher: 2114, 2120, 2124, 2142, 2188

Federico Ardila: 2102, 2103, 2106, 2107, 2108, 2111, 2112, 2113,

2114, 2115, 2117, 2119, 2120, 2121, 2124

�Sefket Arslanagi �c: 2090, 2103, 2113, 2114, 2117, 2121, 2124,

2128, 2132, 2133, 2137, 2150, 2153, 2157, 2163, 2167, 2172, 2173, 2176,

2178, 2180, 2181, 2180, 2181, 2188, 2191

Charles Ashbacher: 2118, 2119, 2158

Sam Baethge: 2114, 2118, 2122, 2124, 2125, 2126, 2143, 2144,

2154, 2166, 2175, 2196

Chetan T. Balwe: 2192

Niels Bejlegaard: 2128, 2130, 2136, 2137, 2142, 2150

Francisco Bellot Rosado: 2102, 2103, 2114, 2128, 2130, 2137,

2148, 2149, 2153, 2156, 2162, 2164, 2169, 2171, 2181, 2186, 2188, 2189

Mansur Boase: 2111, 2112, 2115, 2118, 2119, 2122, 2196

Carl Bosley: 2102, 2102, 2105, 2111, 2114, 2115, 2124

Paul Bracken: 2184, 2197

Christopher J. Bradley: 2102, 2103, 2105, 2107, 2108, 2109,

2112, 2114, 2117, 2118, 2120, 2121, 2122, 2124, 2125, 2126, 2127, 2128,

2130, 2133, 2137, 2140, 2141, 2142, 2143, 2144, 2145, 2147, 2148, 2149,

2150, 2155, 2164, 2165, 2166, 2167, 2170, 2171, 2174, 2175, 2177, 2178,

2181, 2184, 2185, 2186, 2188, 2189, 2191, 2192, 2194, 2195, 2196, 2197

Jeremy T. Bradley:

Adam Brown: 2181, 2195, 2188

Sydney Bulman{Fleming: 2123, 2187

Miguel Angel Cabez �on Ochoa: 2102, 2112, 2114, 2114, 2144,

2147, 2150, 2151, 2154, 2156, 2157, 2158, 2160, 2162

Joseph Callaghan: 2126, 2128

Sabin Cautis: 2177, 2181, 2188, 2192, 2196

Krzysztof Chelmi �nski: 2121

Ji Chen: 1940, 2101

Han Ping Davin Chor: 2102, 2103, 2108, 2113, 2125, 2126,

2127, 2137

Theodore Chronis: 2090, 2102, 2105, 2108, 2112, 2113, 2140,

2143, 2144, 2145, 2147, 2150, 2153, 2157, 2158, 2159, 2161, 2163, 2174,

2175, 2176, 2180, 2181, 2185, 2188, 2191, 2192, 2196

Jan Ciach: 2168

Mihai Cipu: 2154, 2157, 2158, 2159, 2163, 2167, 2172, 2174, 2176

Goran Conar: 2181, 2188, 2196

Tim Cross: 2103, 2108, 2112, 2114, 2118, 2121, 2122, 2124, 2125,

2126, 2143, 2144, 2150, 2170, 2175, 2185

Luz M. DeAlba: 2174

Paul-Olivier Dehaye: 2196

Georgi Demizev: 2151, 2153, 2154, 2155, 2156, 2157, 2158, 2159,

2161, 2162, 2163

Luis V. Dieulefait: 2090, 2115, 2193

Charles R. Diminnie: 2111, 2118, 2122, 2124, 2125, 2126, 2132,

2140, 2143, 2144, 2147, 2150, 2157, 2163, 2170, 2175, 2185, 2191, 2192

C. Dixon: 2166

David Doster: 2102, 2103, 2124, 2132, 2140, 2145, 2150, 2166,

2184, 2192, 2196, 2197

Jordi Dou: 2110, 2114, 2117, 2124

Keith Ekblaw: 2111, 2147, 2152, 2175, 2185

Hans Engelhaupt: 2102, 2103, 2107, 2114, 2118, 2119, 2122,

2124, 2125, 2126, 2131, 2133, 2137, 2140, 2143, 2144, 2150, 2151, 2154,

2155, 2157, 2158, 2162, 2166, 2167, 2174, 2175

Russell Euler: 2150, 2163

Noel Evans: 2122

C. Festraets-Hamoir: 2196

F.J. Flanigan: 2104, 2132, 2140, 2145, 2147, 2150, 2153, 2174,

2175, 2181, 2184, 2191, 2196

Je�rey K. Floyd: 2112, 2125, 2185

Ian June L. Garces: 2125, 2126, 2143, 2147

Toby Gee: 2115

Robert Geretschl�ager: 2145, 2158, 2176, 2182, 2195, 2196

Kurt Girstmair: 2129

Shawn Godin: 2111, 2112, 2118, 2122, 2123, 2125, 2126, 2140,

2143, 2144, 2155, 2157, 2158, 2163, 2187, 2191, 2195, 2196

Joaqu��n G �omez Rey: 2135, 2184, 2197

Javier Guti �errez: 2157, 2170, 2176, 2191

David Hankin: 2114, 2118, 2143, 2144, 2147, 2150, 2195, 2196

G.P. Henderson: 2129, 2136

Fabian Martin Herce: 2163

Florian Herzig: 2101, 2102, 2103, 2105, 2107, 2108, 2112, 2113,

2114, 2118, 2119, 2121, 2122, 2124, 2125, 2127, 2130, 2131, 2132, 2133,

2135, 2137, 2140, 2143, 2144, 2147, 2148, 2150, 2151, 2152, 2153, 2156,

2157, 2158, 2159, 2160, 2162, 2163, 2165, 2166, 2167, 2169, 2172, 2174,

2175, 2176, 2178, 2180, 2181, 2183, 2184, 2185, 2186, 2187, 2188, 2189,

2190, 2192, 2193, 2194, 2195, 2196, 2197

Richard I. Hess: 2109, 2111, 2112, 2113, 2114, 2115, 2118, 2119,

2122, 2123, 2124, 2125, 2126, 2128, 2129, 2130, 2131, 2132, 2133, 2135,

2143, 2144, 2145, 2147, 2150, 2151, 2152, 2154, 2155, 2156, 2157, 2158,

2159, 2160, 2161, 2162, 2163, 2164, 2165, 2166, 2167, 2168, 2169, 2170,

2171, 2172, 2173, 2174, 2175, 2176, 2177, 2178, 2179, 2180, 2181, 2183,

2184, 2185, 2186, 2187, 2188, 2189, 2190, 2191, 2192, 2193, 2194, 2195,

2197

John G. Heuver: 2102, 2107, 2120, 2133, 2167, 2176, 2188

Joe Howard: 2163, 2180, 2181

Cyrus Hsia: 2102, 2107, 2111, 2112, 2192, 2196

Peter Hurthig: 2090, 2114, 2116, 2123, 2124, 2126, 2170

Ignotus: 2196

Peter Iv �ady: 2090

Walther Janous: 2090, 2101, 2102, 2103, 2104, 2105, 2106, 2107,

2108, 2112, 2113, 2114, 2115, 2116, 2118, 2120, 2121, 2122, 2123, 2124,

2125, 2126, 2130, 2132, 2133, 2135, 2137, 2140, 2141, 2143, 2144, 2145,

2147, 2148, 2150, 2151, 2152, 2153, 2154, 2155, 2156, 2157, 2158, 2159,

2160, 2161, 2162, 2163, 2166, 2167, 2169, 2170, 2171, 2172, 2173, 2174,

2175, 2176, 2178, 2179, 2180, 2181, 2183, 2184, 2185, 2186, 2188, 2189,

2190, 2191, 2192, 2193, 2195, 2196, 2197

D. Kipp Johnson: 2143, 2144, 2151, 2154, 2155, 2157, 2158, 2159,

2161, 2163, 2175, 2181, 2184, 2185, 2191, 2195, 2196

Michael Josephy: 2115, 2123

Yang Kechang: 2106, 2116

Amit Khetan: 2144, 2150

Murray S. Klamkin: 2105, 2108, 2112, 2113, 2116, 2155, 2157,

2159, 2161, 2162, 2162, 2163, 2167, 2168, 2170, 2172, 2178, 2180, 2181,

2183, 2188

V�aclav Kone �cn �y: 2101, 2102, 2106, 2107, 2109, 2112, 2114, 2116,

2117, 2118, 2124, 2129, 2133, 2137, 2140, 2143, 2144, 2145, 2147, 2150,

2166, 2167, 2170, 2172, 2175, 2176, 2179, 2181, 2183, 2184, 2185, 2188,

2189, 2191, 2192

Marcin E. Kuczma: 1940, 2113, 2120

Mitko Kunchev: 2102, 2103, 2113, 2114, 2124, 2130, 2145, 2148,

2150, 2151, 2153, 2154, 2155, 2156, 2157, 2158, 2159, 2161, 2162, 2163,

2176

Sai Chong Kwok: 2176

Alexander Lambrou: 2188

Michael Lambrou: 2177, 2178, 2179, 2180, 2181, 2183, 2184,

2185, 2186, 2187, 2188, 2191, 2192, 2193, 2194, 2195, 2197

Stefan Lambrou: 2188

Luke Lamothe: 2112

Kee-Wai Lau: 1940, 2090, 2101, 2103, 2106, 2108, 2109, 2113,

2116, 2117, 2123, 2124, 2132, 2145, 2147, 2150, 2167, 2174, 2175, 2184,

2185, 2186, 2187, 2190, 2192, 2196

Thomas Leong: 2134, 2147, 2152, 2157, 2159

544

Kathleen E. Lewis: 2112, 2122, 2125, 2126

Mar��a Ascensi �on L �opez Chamorro: 2103, 2114, 2149,

2151, 2160

B. M???y: 2143, 2144

David E.Manes: 2090, 2111, 2112, 2113, 2118, 2119, 2122, 2123,

2125, 2126, 2131, 2132, 2143, 2144, 2145, 2147, 2150, 2157, 2161

Beatriz Margolis: 2140, 2150

Giovanni Mazzarello: 2125, 2126, 2143, 2147

Tara McCabe: 2122

J.A. McCallum: 2150, 2175, 2185, 2186

John Grant McLoughlin: 2112, 2122, 2125

Vasiliou Meletis: 2103

Norvald Midttun: 2132

Can Anh Minh: 2157, 2176, 2188, 2192, 2196

W. Moser: 2195

Vedula N. Murty: 1940, 2108, 2113, 2167, 2186, 2188

John Oman: 2167, 2185

Soledad Ortega: 2157, 2170, 2176

Victor Oxman: 2091, 2109, 2128, 2142, 2188 Michael Par-

menter: 2123, 2147, 2181, 2196

Yolanda Pellejer: 2157

P. Penning: 2102, 2103, 2107, 2109, 2112, 2114, 2118, 2119, 2120,

2122, 2123, 2124, 2125, 2126, 2127, 2130, 2133, 2136, 2137, 2158

Gottfried Perz: 2102, 2123, 2125, 2126, 2137, 2143, 2150, 2156,

2165, 2167, 2169, 2177, 2188, 2196

Carl Pomerance: 2134

Waldemar Pompe: 2102, 2103, 2114, 2117, 2127, 2134, 2139,

2177

Luiz A. Ponce: 2151

Bob Prielipp: 2167, 2195, 2186, 2188, 2196

Cory Pye: 2112, 2118, 2122, 2125

Stanley Rabinowitz: 2129

E. Rappos: 2174

L. Rice: 2140, 2145, 2148, 2150

Juan-Bosco Romero M�arquez: 2108, 2149, 2161, 2167,

2171, 2176, 2180, 2188, 2196

Kristian Sabo: 2145, 2150

Jawad Sadek: 2163

Mar��a Mercedes S�anchez Benito: 2064, 2065, 2067, 2068,

2069, 2071, 2077

Crist �obal S �anchez{Rubio: 2150, 2151, 2164, 2169

Bill Sands: 2125, 2126, 2185, 2195

K.R.S. Sastry: 2104, 2133, 2140, 2154, 2166

Joel Schlosberg: 2107, 2108, 2112, 2115, 2118, 2119, 2122, 2125,

2134, 2143, 2145

Robert P. Sealy: 2104, 2111, 2112, 2123, 2125, 2126, 2132, 2157,

2163, 2175, 2185, 2195, 2196

Harry Sedinger: 2140, 2181

Heinz-J�urgen Sei�ert: 2101, 2102, 2105, 2108, 2112, 2113,

2118, 2122, 2125, 2128, 2132, 2135, 2140, 2143, 2144, 2145, 2147, 2150,

2154, 2156, 2157, 2158, 2159, 2161, 2163, 2167, 2174, 2176, 2178, 2180,

2181, 2183, 2184, 2185, 2188, 2191, 2196

Toshio Seimiya: 2102, 2103, 2114, 2117, 2120, 2124, 2127, 2128,

2130, 2133, 2137, 2141, 2142, 2146, 2148, 2149, 2151, 2154, 2156, 2160,

2162, 2164, 2165, 2166, 2167, 2169, 2171, 2177, 2188, 2189

Zun Shan: 2104

D. N. Sheth: 2195

Catherine Shevlin: 2124

Shailesh Shirali, Rishi Valley School, India: 2117, 2148,

2149

D.J. Smeenk: 2102, 2103, 2107, 2114, 2117, 2120, 2124, 2127,

2130, 2133, 2137, 2141, 2148, 2149, 2150, 2151, 2156, 2160, 2162, 2164,

2166, 2167, 2169, 2171, 2181, 2188, 2189, 2191, 2194, 2196

Digby Smith: 2112, 2116, 2118, 2140, 2143, 2144, 2145, 2147,

2150, 2153, 2155, 2163, 2174, 2181, 2185, 2192, 2195, 2196

Trey Smith: 2191, 2192

Lawrence Somer: 2147

Diego Sot �es: 2191

David R. Stone: 2118, 2119, 2122, 2123, 2125, 2126, 2132, 2134,

2143, 2144, 2147, 2150, 2175, 2181, 2185, 2192, 2195, 2196

John C. Tripp: 2174

Panos E. Tsaoussoglou: 2102, 2105, 2108, 2113, 2116, 2117,

2118, 2122, 2124, 2143, 2144, 2145, 2147, 2150, 2158, 2166, 2167, 2175,

2176, 2177, 2178, 2180, 2186, 2188

George Tsapakidis: 2102, 2107, 2117, 2176, 2188

Adri �an Ubis Mati�inez: 2188, 2196

Meletis Vasiliou: 2102, 2107, 2114, 2117, 2120, 2124, 2188

David C. Vella: 2196

Nishka Vijay: 2150

Edward T.H. Wang: 2123, 2125, 2132, 2145, 2150, 2163, 2167,

2174, 2181, 2185, 2187, 2196

Hoe Teck Wee: 2101, 2103, 2105, 2111, 2119, 2131, 2147, 2156,

2165

Chris Wildhagen: 2090, 2102, 2111, 2112, 2113

Kenneth M. Wilke: 2125, 2143, 2147, 2175, 2181, 2196

Lamarr Widmer: 2175

Aram A. Yagubyants: 2137, 2148

Paul Yiu: 2118, 2130, 2137, 2181, 2188

Roger Zarnowski: 2191, 2192

Con Amore Problem Group: 2164, 2165, 2166, 2167, 2168,

2174, 2177, 2179

Skidmore College Problem Group: 2140, 2150, 2174

Proposers, solvers and commentators in the MAYHEM PROBLEMS and

SOLUTIONS sections for 1997 are:

Miguel Carri �on �Alvarez: H205, H206, H207, H211, H212,

H213, H214, A184, A188, C69

Donny Cheung: A184

J. Chris Fisher: C70

Nicolas Guay: A184

Waldemar Pompe: A191

Bob Prielipp: H207, H208, H211

Naoki Sato: C64

Matt Szczesney: C73

Vin de Silva: C68

Ravi Vakil: A184

Sam Vandervelde: C68

Edward T.H. Wang: A184, A185, A189

Eric Wepsic: C68

Samuel Wong: H205, H207

Wai Ling Yee: H209, A195, A197, A188, A189, A190, A192