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257
THE ACADEMY CORNERNo. 26
Bruce ShawyerAll 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
Abstracts � R�esum�es
Canadian Undergraduate Mathematics Conference
1998 | Part 4
The Dirichlet-to-Neumann Map
Scott MacLachlan
University of British Columbia
There are two classic problems in potential theory - the Dirichlet and Neumannproblems. In most courses on partial di�erential equations the problems are treatedseparately; however there is a very elegant link between these problems that can beexplored via Fourier analysis. The Dirichlet-to-Neumannmap takes data for a Dirich-let problem and translates it into data for a Neumann problem without calculating asolution �rst. This map is useful in many areas, particularly in potential theory.
Syst �emes de racines
Maciej Mizerski
McGill University
Un des plus beaux r �esultats de la th �eorie des groupes et alg �ebres de Lie est laclassi�cation des alg �ebres de Lie semi-simples. Cette classi�cation se ram�ene �a l' �etudede la structure des syst �emes de racines. Dans ma pr �esentation, je vais introduireles syst �emes de racines et survoler leur classi�cation en utilisant les diagrammes deDynkin. Aussi je vais tenter de faire le lien avec les groupes et les alg �ebres de Lie.
Cardan's Formulas for Solving Cubic Equations Revisited
Afroze Naqvi
University of Regina
This presentation o�ers a very brief history of Algebra and then proceeds toderive Cardan's Formulas for solving cubic equations. These formulas will be used tosolve some cubic equations.
258
Mathematics of the Ancient Jews and Ancient Israel
Peter Papez
University of Calgary
As mathematicians we are well acquainted with the mathematics of the an-cient Babylonians, Greeks and Egyptians. Even Ancient Chinese mathematicians havegained great notoriety in recent years. But the mathematics of ancient Israel is cer-tainly noteworthy, if not impressive, and has not received the attention it deserves.The purpose of this discussion will be to introduce these impressive achievementsand discuss some problems and the solutions obtained by ancient Jewish scholars.In order to facilitate the discussion, a brief overview of ancient Jewish history willbe given. As well, the Talmud and Talmudic Law will be presented. The discussionwill encompass the very beginning construction of numerals and numeracy within theancient Jewish culture, proceed through the development of arithmetic, touch on im-portant notions concerning science and logic, and conclude with a discussion of somefairly advanced developments in sampling and statistics. Various problems and theirsolutions, as derived by the ancient Jews, will be used to illustrate these develop-ments and the entire discussion will be placed in a historical and Talmudic context.
Computing in the Quantum World
Christian Paquin
Universit �e de Montr �eal
The current computing model is based on the laws of classical physics. But theworld is not classical; it follows the laws of quantummechanics. A quantum computeris a model of computation based on quantummechanics. It has been proved that sucha model is more powerful than its classical counterpart, meaning that it can do thesame computations as a classical computer (in approximately the same time) but thereexist some problems for which the quantum computer is much faster. In this paper Iwill explain what is quantum information, why a quantum computer would be useful,what are the problems to build a quantum computer and how it will work.
Wavelet Compression on Fractal Tilings
Daniel Pich �e
University of Waterloo
Over the last decade, wavelets have become increasingly useful for studyingthe behaviour of functions, and for compression. Though much remains to be inves-tigated in this �eld, certain types of wavelets are fairly easy to construct, namely Haarwavelets. This paper ties together the theory of these wavelets with that of complexbases. An algorithm is proposed for doing wavelet analysis, with wavelets arising inthis fashion. This will enable further study of the properties of these wavelets.
259
Introduction to Representation Theory
Evelyne Robidoux
McGill University
This paper provides an introduction to representation theory, with emphasison representations of �nite groups. The background required will be only a bit ofgroup theory (that is, being familiar with the concepts of groups, homomorphisms,conjugacy classes) and linear algebra (vector spaces, linear transformations, innerproducts, direct sums). A classical reference for this material is the book by Serre,Linear Representations of Finite Groups, which I really recommend.
The E�ect of Impurities in One-Dimensional Antiferromagnets
Alistair Savage
University of British Columbia
A brief overview of a paper to be published shortly with Ian A�eck of theUniversity of British Columbia Department of Physics and Astronomy.
Authentication Codes Without Secrecy
Nelly Sim~oes
Simon Fraser University
Suppose Alice wants to say something to Bob and cares only about the authen-tication of her message. Authentication makes it possible for Bob to receive Alice'smessage and be certain it came from her. Basically an authentication code withoutsecrecy is a process in which a mathematical function transforms what Alice wants tosay to Bob, we call this the source state, into what is called an authentication tag andadds it to the source state to form the message. We also suppose that Alice and Bobmutually trust each other. Alice wants to use an unconditionally secure method ofauthentication. She does not want anyone to be able to modify her message, not evena person with lots of computer power. This is why Alice decides to use a combinator-ial method. We are going to explore an unconditionally secure way of authenticatingAlice's message.
Integer Triangles With a Side of Given Length
Jill Taylor
Mount Allison University
This presentation will give a glimpse into the history of Heronian triangles
(triangles with integer sides and integer area) and one special case of such triangleswhich I have researched this summer. However, the main focus will be a particularproblem involving Heronian triangles with a given perimeter. The solution of thisproblem requires both number theory and basic geometrical concepts.
260
Convergence and Transcendence in the Field of p{adic Numbers
Sarah Sumner
Queen's University
We will �rst explore convergence properties of series in Qp, and then studyinstances when the series
1Xn=0
anpn; an 2 Qp
is transcendental over Qp. We will give a general result describing a large class ofseries of this type which are transcendental over Qp. Our result unfortunately doesnot resolve the transcendence of
P1n=0 n! in Qp. However, our theorem applies to
show transcendence of1Xn=0
�nn!
where �n is a primitive n-th root of unity if (p; n) = 1 and is 1 otherwise.
Random Number Generation
Ren�ee Touzin
Universit �e de Montr �eal
Nowadays, in many scienti�c �elds, random number distributions are needed.Those distributions such as a binomial, a normal, an exponential are built from iiduniform (0; 1) distributions. But this distribution does not really exist. In fact, itconsists of a mathematical and deterministic algorithm that tries to behave stochas-tically.
The building of an e�cient generator requires a strong knowledge of mathe-matics and computer science. A priori, a generator must follow certain theoreticalcriteria. A posteriori, those same generators must pass many statistical tests to besure they look random even though they are deterministic. In this presentation, wewill talk about di�erent kinds of existing generators and the qualities of a good gen-erator. We will give examples of good and bad generators and �nally we will presenta simple implementation in C.
The Probabilistic Method: Proving Existence by Chance
Alexander Yong
University of Waterloo
The Probabilistic Method is based on a simple concept: in order to prove the
existence of some mathematical object, construct an appropriate probability space
and show that the object occurs with positive probability. We will investigate this
powerful proof technique via three examples.
261
THE OLYMPIAD CORNERNo. 199
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.
The �rst Olympiad we give this issue is a Danish contest. My thanks goto Ravi Vakil for collecting a copy and forwarding it to me when he was Cana-dian Team Leader to the International Mathematical Olympiad at Mumbai.
GEORG MOHR KONKURRENCEN I MATEMATIK1996
January 11, 9{13
Only writing and drawing materials are allowed.
1. \C in4ABC is a right angle and the legs BC and AC are both oflength 1. For an arbitrary point P on the leg BC, construct points Q, resp.R, on the hypotenuse, resp. on the other leg, such that PQ is parallel to ACand QR is parallel to BC. This divides the triangle into three parts.
Determine positions of the point P on BC such that the rectangularpart has greater area than each of the other two parts.
2. Determine all triples (x; y; z), satisfying
xy = z
xz = y
yz = x .
3. This year's idea for a gift is from \BabyMath", namely a series of9 coloured plastic �gures of decreasing sizes, alternating cube, sphere, cube,sphere, etc. Each �guremay be opened and the succeeding one may be placedinside, �tting exactly. The largest and the smallest �gures are both cubes.Determine the ratio between their side-lengths.
4. n is a positive integer. It is known that the last but one digit in thedecimal expression of n2 is 7. What is the last digit?
5. In a ballroom 7 gentlemen, A, B, C, D, E, F and G are sittingopposite 7 ladies a, b, c, d, e, f and g in arbitrary order. When the gentle-men walk across the dance oor to ask each of their ladies for a dance, one
262
observes that at least two gentlemen walk distances of equal length. Is thatalways the case?
The �gure shows an example. In this example Bb = Ee andDd = Cc.
A B C D E F G
f d b g c e a
s s s s s s s
s s s s s s s
The second two sets of problems come from the St. Petersburg CityMathematical Olympiad. Again my thanks go to Ravi Vakil, Canadian TeamLeader to the InternationalMathematical Olympiad at Mumbai for collectingthem for me.
ST. PETERSBURG CITY MATHEMATICAL OLYMPIADThird Round { February 25, 1996
11th Grade (Time: 4 hours)
1. Serge was solving the equation f(19x� 96=x) = 0 and found 11
di�erent solutions. Prove that if he tried hard he would be able to �nd atleast one more solution.
2. The numbers 1, 2, : : : , 2n are divided into two groups ofn numbers.Prove that pairwise sums of numbers in each group (sums of the form a+ a
included) have identical sets of remainders on division by 2n.
3. No three diagonals of a convex 1996{gon meet in one point. Provethat the number of the triangles lying in the interior of the 1996{gon andhaving sides on its diagonals is divisible by 11.
4. Points A0 and C0 are taken on the diagonal BD of a parallelogramABCD so that AA0kCC0. Point K lies on the segment A0C, and the lineAK meets the line C0C at the point L. A line parallel to BC is drawnthrough K, and a line parallel to BD is drawn through C. These two linesmeet at pointM . Prove that the points D, M , L are collinear.
5. Find all quadruplets of polynomials p1(x), p2(x), p3(x), p4(x)with real coe�cients possessing the following remarkable property: for allintegers x, y, z, t satisfying the condition xy � zt = 1, the equalityp1(x)p2(y)� p3(z)p4(t) = 1 holds.
6. In a convex pentagon ABCDE, AB = BC, \ABE + \DBC =
\EBD, and\AEB+\BDE = 180�. Prove that the orthocentre of triangleBDE lies on diagonal AC.
263
7. Two players play the following game on a 100�100 board. The �rstplayer marks some free square of the board, then the second puts a domino�gure (that is, a 2 � 1 rectangle) covering two free squares one of which ismarked. The �rst player wins if all the board is covered; otherwise the secondwins. Which of the players has a winning strategy?
Selective Round { March 10, 199611th Grade (Time: 5 hours)
1. It is known about real numbers a1, : : : , an+1; b1, : : : , bn that0 � bk � 1 (k = 1, : : : , n) and a1 � a2 � : : : � an+1 = 0. Prove theinequality:
nXk=1
akbk �[Pn
j=1 bj]+1Xk=1
ak .
2. Segments AE and CF of equal length are taken on the sides ABand BC of a triangle ABC. The circle going through the points B, C, Eand the circle going through the points A, B, F intersect at points B andD.Prove that the line BD is the bisector of angle ABC.
3. Prove that there are no positive integers a and b such that for alldi�erent primes p and q greater than 1000, the number ap+bq is also prime.
4. A Young tableau is a �gure obtained from an integral-sided rectangleby cutting out its cells covered by several integral-sided rectangles containingits right lower angle. We call a hook a part of the tableau consisting of somecell and all the cells lying either to the right of it (in the same row) or belowit (in the same column). A Young tableau of n cells is given. Let s be thenumbers of hooks containing exactly k cells. Prove that s(k+ s) � 2n.
5. In a triangle ABC the angle A is 60�. A point O is taken insidethe triangle such that \AOB = \BOC = 120�. A pointD is chosen on thehalf-line CO such that the triangle AOD is equilateral. The perpendicularbisector of the segment AO meets the line BC at point Q. Prove that theline OQ divides the segment BD into two equal parts.
6. There are 120 underground lines in a city; every station may bereached from each other with not more than 15 changes. We say that twostations are distant from each other if not less than 5 changes are neededto reach one of them from the other. What maximum number of pairwisedistant stations can be in this city?
7. a, b, c are integers. It is known that the polynomial x3 + ax2 +
bx + c has three di�erent pairwise coprime positive integral roots, and thepolynomial ax2+ bx+ c has a positive integral root. Prove that the numberjaj is composite.
8. Positive integers 1, 2, : : : , n2 are being arranged in the squares ofan n�n table. When the next number is put in a free square, the sum of the
264
numbers already arranged in the row and column containing this square iswritten on the blackboard. When the table is full, the sum of the numbers onthe blackboard is found. Show an example of this way of putting the numbersin squares such that the sum is minimum.
We next turn to readers' solutions to problems given earlier in the Cor-ner. First some catching up. Over the summer I had some time to sort and �lematerials, and discovered solutions to problems of the 1994 Balkan Olympiadmis�led with 1999 material. So we begin with solutions to the 1994 BalkanOlympiad [1997: 198].
2. [Greece] Show that the polynomial
x4 � 1994x3 + (1993 +m)x2 � 11x+m; m 2 Z
has at most one integral root. (Success rate: 39:66%)
Solutions by Mansur Boase, student, Cambridge, England.Consider
x4 � 1994x3 + (1993 +m)x2 � 11x+m . (�)
Suppose the given polynomial has two integral roots. Then neither can beodd for otherwise
(x4 + 1993x2)� (1994x3) +m(x2 + 1)� 11x
will be odd (as each of the terms in brackets is even) and hence non-zero.
Suppose x1 = 2r1a, r1 > 0 and a odd is a solution.
Considering the polynomial (mod 22r1), we then have thatm � 11x (mod 22r1). Hence m � 2r1(11a) (mod 22r1), and since a isodd, m must be of the form 2r1l1, l1 odd.
If x2 = 2r2b, b odd, is also a solution, then m = 2r2l2, l2 odd, so wemust have r1 = r2.
Thus, if there are two integral roots, both must be of the form 2rk,r > 1, k odd. The product of the two roots must be a multiple of 22r. Thequadratic which has these two roots as zeros is x2 + px + 22rq, where p, qare integers.
Now the given polynomial (*) can be factorized into two quadratics:
(x2 + px+ 22rq)(x2 + sx+ t) .
If s were not integral, then the coe�cient of x3 would not be integral in thequartic, and if t were not integral, the coe�cient of x2 would not be integralin the quartic. Thus s and tmust be integers andm = 22rqt. But the highest
265
power of 2 dividingm is 2r so r � 2r giving r = 0, a contradiction. Hencethe given quartic cannot have more than one integral root.
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%)
Solution by Mansur Boase, student, Cambridge, England.
Choose one of the people, A, and suppose A knows x1, x2, : : : , xr.Then by (i) no xi knows an xj for i 6= j. Therefore, by (ii), for each pairfxi; xjg there must exist an Xij who knows both xi and xj in order thatxi and xj have two common acquaintances A andXij. Now A cannot knowany of theXij. Thus by (ii) eachXij can have only two acquaintances amongx1, x2, : : : , xr, namely xi and xj , so all the Xij are distinct.
Any person who is not A, nor an acquaintance of A must by (ii) be anXij. Thus the total number of people must be
�r
2
�+ r + 1.
Now r > 2 and n > 4.
If r = 3; then n = 7 (A) ,If r = 4; then n = 11 (B) ,If r = 5; then n = 16 (C) .
Let us label the people 1; 2; : : : ; n.
Case (A): Without loss of generality suppose 1 knows 2, 3 and 4 and that 5knows 2 and 3, 6 knows 2 and 4 and 7 knows 3 and 4.
Now 5 must have three acquaintances, so he must know one of 6 and7. But he has common acquaintances with both 6 and 7, contradicting (i).
Case (B): Without loss of generality, suppose 1 knows 2, 3, 4 and 5 and 6,7, 8, 9, 10, 11 know pairs f2; 3g, f2; 4g, f2; 5g, f3;4g, f3; 5g and f4; 5grespectively.
Then 6 cannot know 7, 8, 9 or 10 as he has common acquaintances witheach of them. So 6 can only know 2, 3 and 11, while he must know r = 4
people, a contradiction.
Case (C): We claim that n = 16 is the smallest number of people requiredin such a set, by noting that cases (A) and (B) fail and that the followingacquaintance table satis�es (i) and (ii):
266
Person Acquaintances1 2 3 4 5 6
2 1 7 8 9 10
3 1 7 11 12 13
4 1 8 11 14 15
5 1 9 12 14 16
6 1 10 13 15 16
7 2 3 14 15 16
8 2 4 12 13 16
9 2 5 11 13 15
10 2 6 11 12 14
11 3 4 9 10 16
12 3 5 8 10 15
13 3 6 8 9 14
14 4 5 7 10 13
15 4 6 7 9 12
16 5 6 7 8 11
Now we turn to solutions to problems of the Fourth Grade of the 38th
Mathematics Competition of the Republic of Slovenia [1998: 132].
1. Prove that there does not exist a function f : Z ! Z, for whichf(f(x)) = x+ 1 for every x 2 Z.
Solutions by Mohammed Aassila, CRM, Universit �e de Montr �eal,
Montr �eal, Qu �ebec; by Pierre Bornsztein, Courdimanche, France; by Masoud
Kamgarpour, Carson Graham Secondary School, North Vancouver, BC; by
Pavlos Maragoudakis, Pireas, Greece; and by Edward T.H. Wang, Wilfrid
Laurier University,Waterloo, Ontario. We give the solutionbyMaragoudakis.
Suppose that there is such a function. Then f(f(f(x))) = f(x) + 1.Since f(f(x)) = x+ 1, we get f(x+ 1) = f(x) + 1.
By induction f(x + n) = f(x) + n for every n 2 N. Alsof(x) = f(x� n+ n) = f(x� n) + n so
f(x� n) = f(x)� n for every n 2 N .
Finally f(x+ y) = f(x) + y for x; y 2 Z.
For x = 0, f(y) = f(0) + y. For y = f(0), f(f(0)) = f(0) + f(0).
But f(f(0)) = 1; thus 2f(0) = 1, a contradiction.
2. Put a natural number in every empty �eld of the table so that youget an arithmetic sequence in every row and every column.
267
0
74
103
186
Solutions by Pierre Bornsztein, Courdimanche, France; by Masoud
Kamgarpour, Carson Graham Secondary School, North Vancouver, BC;
and by Pavlos Maragoudakis, Pireas, Greece. We give the solution of
Kamgarpour.
Let us say the numbers adjacent to 0 are a0 and a1, with a0 in the rowand a1 the column. We know that in an arithmetic sequence every term isthe arithmetic mean of the term before and after. Therefore, we can putnumbers in the chart as follows:
3a1 74
2a1 a1 � a0 + 103 186
a11
2(a1 + 103) 103 1
2(309� a1) 206� a1
0 a0 2a0 3a0 4a0
Now
1
2(186 + 4a0) = 206� a1 ===) 93 + 2a0 = 206� a1
===) 2a0 + a1 = 113 , (1)
and
74 + 1
2(a1 + 103) = 2(103+ a1 � a0) ===) 3a1 � 4a0 = �161 . (2)
Solving (1) and (2) gives a0 = 50 and a1 = 13, so we can easily put numbersin every �eld as shown:
52 82 112 142 172
39 74 109 144 179
26 66 106 146 186
13 58 103 148 193
0 50 100 150 200
268
3. Prove that every number of the sequence
49 , 4489 , 444889 , 44448889 , : : :
is a perfect square (in every number there are n fours, n � 1 eights and anine).
Solutions by Pavlos Maragoudakis, Pireas, Greece; by Bob Prielipp,
University ofWisconsin{Oshkosh,Wisconsin,USA; and by Edward T.H.Wang,
Wilfrid Laurier University, Waterloo, Ontario. We give Prielipp's solution.
Let 43829 denote the number 444889 and 627 denote the number 667.
We shall show that
(6n�17)2 = 4n8n�19
for each positive integer n.
Because (6n�17)2 =�6(10n�1)
9+ 1
�2, it su�ces to establish that
�6(10n � 1)
9+ 1
�2
= 4n8n�19 for each positive integer n . (�)
Let n be an arbitrary positive integer. Then
�6(10n � 1)
9+ 1
�2=
�6 � 10n + 3
9
�2
=(2 � 10n + 1)2
9
=4 � 102n + 4 � 10n + 1
9
=40n�140n�11
9= 4n8n�19 .
4. Let Q be the mid-point of the side AB of an inscribed quadrilat-eral ABCD and S the intersection of its diagonals. Denote by P and R
the orthogonal projections of S on AD and BC respectively. Prove thatjPQj = jQRj.
269
Solutions by Pierre Bornsztein, Courdimanche, France; and by Toshio
Seimiya, Kawasaki, Japan. We give the solution of Seimiya.
A
B
C D
EF
P
Q
R
S
b
b
Let E be the point symmetric to A with respect to P , and let F be thepoint symmetric to B with respect to R. Then we have
SE = SA; \SEA = \SAE ,
and SF = SB; \SFB = \SBF .
As A, B, C, D are concyclic we get \CAD = \CBD . Thus,
\SEA = \SAE = \SBF = \SFB .
Consequently we have \ASE = \BSF . Thus we get
\BSE = \BSA+ \ASE = \BSA+ \BSF = \FSA .
Since SB = SF and SE = SA, we have
4SEB � 4SAF . (SAS)
Thus we get EB = AF . Since P , Q, R are mid-points of AE, AB, BFrespectively, we have
PQ =1
2EB and QR =
1
2AF .
Therefore we have PQ = QR .
That completes the Corner for this issue. Send me your nice solutionsas well as contest materials.
270
BOOK REVIEWS
ALAN LAW
In Polya's Footsteps, by Ross Honsberger,published by the Mathematical Association of America, 1997,ISBN # 0-88385-326-4, softcover, 328+ pages, $28.95.Reviewed byMurray S. Klamkin.
This is another in a series of books by the author on problems and solutionstaken from various national and international olympiad competitions and very manyof which have appeared previously in the Olympiad Corner of CruxMathematicorum.Also as the author notes, the solutions are his own unless otherwise acknowledged.Otherwise how would it look if everything was copied from elsewhere. I �nd thesesolutions of the author to be a mixed bag, especially in light of the following twoquotations,
A good proof is one that makes us wiser. Yu. I. Manin
Proofs really aren't there to convince you something is true | they're there toshow you why it is true. Andrew Gleason
Some of his solutions are a welcome addition, but there are also quite a num-ber for which the previously published ones are much better and certainly not asoverblown.
I now give some speci�c comments on some of the problems and their solutions.
� page 14. Given any sequence of r digits, is there a perfect square k2 with thesedigits immediately preceding the last digit of k2? The nice solution of the im-possibility is ascribed to Andy Liu, but not from the University of Calgary. As arelated aside, it could be mentioned that in problem 4621, School Science and
Mathematics, it is shown that the given sequence of digits can be the �rst r ormiddle r digits of an in�nite number of squares k2.
� page 26. Here it is shown in an elongated fashion that
nYi=1
�1 +
1
ai
�� (n+ 1)
n,
where
nXi=1
a1 = 1 and ai � 0. This is a well-known inequality appearing in
many places. For a more elegant proof, we can use H�older's Inequality to im-
mediately obtain the known inequality
nYi=1
�1 +
1
ai
� 1
n
� 1 +1
npa1a2 � � � an
.
We can then �nish o� using the AM{GM Inequality
1
n
nXi=1
ai
!n� a1a2 � � �an.
This proof not only leads to learning about the important H�older's Inequality,but also leads to generalizations. For example, let
Pai = A and
Pbi = B.
ThennYi=1
n
r1 +
1
ai+
1
bi� 1 +
n
A+
n
B.
� page 60. Not noted in this nice geometry problem, is that PQRS also has theproperty of having the least perimeter for quadrilaterals inscribed in ABCD.
271
� page 76. Here the author generalizes a Chinese problem by determining theremainder when F (xn+1) is divided by F (x) where F (x) = 1 + x + � � �+xn�1 + xn, and again gives an elongated solution. At the end he noted: \Analternative solution is given in [1992: 102]; additional comment appears in[1994: 46]." Left out is that the additional comment gives a wider generaliza-tion and a solution which is simpler and more compact.
� page 93. Here we are to determine the area of a triangle ABC given threeconcurrent cevians AD;BE and CF intersecting at P , where AD = 20,BP = 6 = PE, CF = 9 and PF = 3. The solution here is almost fourpages long and uses a guess in solving an irrational equation since it is knownthat the solution is an integer. For a much more compact solution with moreunderstanding, consider the following. Let [G] denote the area of a �gure
G. Then immediately, [APB]
[ABC] =3
3+9 = 14 and [APC]
[ABC] =6
6+6 = 12 , so that
[BPC]
[ABC]= 1
4and AP = 15. Now lettinq \BPC = � � �, \CPA = � � �
and \APB = � � , we have 2[BPC] = 6 � 9 sin� = 12 [ABC],
2[CPA] = 9 � 15 sin� = [ABC], and 2[APB] = 15 � 6 sin = 12[ABC].
All that is left is to determine one of the angles �, � and , whose sum is �,and so are angles of some triangle with sides proportional to 15, 12 and 9, re-spectively. Since this is a right triangle, � = �
2and [ABC] = 2 � 6 � 9 = 108.
Note that even if the latter was not a right triangle, we still can determine theangles.
� page 96. Here we have an elementary but long solution to �nd the smallestpositive integer n which makesmn � 1 divisible by 21989 no matter what oddinteger greater than 1 might be substituted for m. For another way of show-ing that n = 21987 is the smallest n, we use the known extension of Euler'sTheorem a�(n) � 1 (mod n), where (a; n) = 1, to the � function. Here,
�(2�) = �(2�) if � = 0, 1, 2;
�(2�) = 1
2�(2�) if � > 2;
�(p�) = �(p
�) if p is an odd prime;
and �(2�p�11 p
�21 � � � p�nn ) is the least common multiple of �(2�), �(p�11 ),
�(p�22 ), : : : , �(p�nn ), where 2, p1, p2, : : : , pn are di�erent primes. Then there
is no exponent � less than �(n) for which the congruence a� � 1 (mod n) issatis�ed for every integer a relatively prime ton. Hence �(21989) = 1
2�(21989)
= 21987. See R.D. Carmichael, The Theory of Numbers Wiley, NY 1914.
� page 106. Here the problem is to determine the acute angle A of a triangle if itis given that vertexA lies on the perpendicular bisector joining the circumcentreO and the orthocentre H. A simpler solution follows almost immediately byusing vectors. Letting A, B and C be vectors from O to the vertices A, B andC, respectively. We have H = A + B + C so that jAj = jB + Cj. Squaringthe latter equation, we get R2 = R2 + R2 + 2B � C = 4R2 � a2. Hencea = R
p3 = 2R sinA so that A = 60�. Note that many metric properties
of a triangle can be determined easily using this vector representation and thefollowing ones for the centroid G = A+B+C
3 and the incentre I = aA+bB+cCa+b+c .
For example, OH2 = (A + B + C)2 = A2 + B2 + C2 + 2B � C + 2C � A+
2A �B = 9R2 � a2� b2� c2, so that we have a simple proof of the inequality9R2 � a2 + b2 + c2, or equivalently, 9
4� sin2 A + sin2 B + sin2C. It also
272
follows immediately that O;H and G are collinear with OH = 3OG. As anexercise for the reader, show that OI2 = R2 � 2Rr.
� page 118. Here AB, CD and EF are chords of a circle that are concurrentat K and are inclined to each other at 60� angles. One has to show thatKA + KD + KE = KB + KC + KF . Again the referred to solutionis neater and shorter. Another easy solution is to use the polar equation ofthe circle with origin atK.
� page 122. Here E is a point on a diameter AC of a circle centre O and one hasto determine the chord BD through E which yields the quadrilateral ABCDof the greatest area. Here the author claims that BD should be perpendic-ular to AC. However, this is correct only if OA � OE
p2. For a com-
plete and simpler solution, let OA = r, OE = a and \BEC = �. Then
BE = �a cos �+pr2 � a2 sin2 � and ED = a cos � +
pr2 � a2 sin2 �, so
that [ABCD] = 2r sin �pr2 � a2 sin2 �. Letting t = sin2 �, we wish to max-
imize F � t(r2 � a2t). Now F is increasing from t = 0 to r22a2
. Consequently,
if r � ap2, F is maximized for t = r2
2a2, and if r > a
p2, F is maximized
for t = 1.
� page 133. Here we have to maximize abcd where a, b, c and d are integers witha � b � c � d and a + b + c + d = 30. Again a previous solution in CRUX
is simpler and more general and points out the use of the widely applicableMajorization Inequality.
� page 134. Here we are given tangents to a circle K from an external point Pmeeting K at A and B. We want to determine the position of C on the mi-nor arc AB such that the tangent at C cuts from the �gure a triangle PQRof maximum area. The author reduces the problem to minimizing the tangentchord QCR or, in his terms, tan�+ tan(90� � �� �) where 2� is the given
angle APB. Then using calculus, he obtains the minimizing angle � as 90���2 ,
which places C at the mid-point of arc AB which is to be expected intuitively.He also justi�es the minimum by examining the second derivative. My crit-icism here is one should eschew calculus methods in these problems. Sincetan�+tan(90�����) = cos�
sin(�+�) cos�= 2 cos �
sin�+sin(�+2�), we have a more
elementary and simpler solution for the minimizing angle.
� page 152. Here we have to show that if for every point P inside a convexquadrilateral ABCD, the sum of the perpendiculars to the four sides (or theirextensions if necessary) is constant, then ABCD is a parallelogram. The solu-tion given in approximately three pages is attributed to me. However, I don'tquite recognize it: my solution in CRUX was considerably shorter.
� page 158. Here one has equilateral triangles drawn outwardly on the sides ofa triangle ABC, so that the three new vertices are D, E and F . Given D, Eand F , one has to construct ABC. Here I have no criticisms of the solution,especially since the author �rst reviews some basic ideas of translations and ro-tations in the plane and then uses these to get a nice solution. I just like to pointout that a solution using complex numbers, although not particularly elegant,is very direct. Let (A; B;C) = (z1; z2; z3) and (D;E; F ) = (w1; w2; w3).Then 2w1 = z2+z3+i
p3(z2�z3). The other two equations follow by a cyclic
change in the subscripts. After some simple algebra, we �nd
2z1 = w1+w21+i
p3
2�w3
1�ip3
2, so that z1 is easy to construct given w1; w2
273
andw3. We can then construct z2 and z3. It also follows, by summing the threeequations, that the centroids of ABC and DEF coincide.
� page 184. Here we have a problem on numbering an in�nite checkerboard withan approximate three page solution and with no attribution. Andy Liu informedme of a prior source of this problem. It occurs with an approximately one pagesolution in A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Prob-
lems with Elementary Solutions II, Holden-Day, San Francisco, 1967, p.129.
� page 203. Here we have to �nd the minimum value of the function f(x) =pa2 + x2 +
p(b� x)2 + c2, where a, b and c are positive numbers. This is
a classic minimum distance problem and is treated in many places. The au-thor repeats the classic re ection solution given in CRUX and also refers to analternate solution given there but does not indicate that this solution gives im-mediate generalizations via Minkowski's Inequality which is another useful toolfor competition contestants: p
pap + xp + yp+ p
pbp + (c� x)p + (d� y)p �
pp(a + b)p + cp + dp where p > 1. If 1 > p > 0, the inequality is reversed.
� page 216. Here we have to show there are in�nitely many non-zero solutionsto the Diophantine equation x2 + y5 = z3. The author, after a page of work,comes up with the solution (x; y; z) = (215a+10; 26a+4; 210a+7). More generalequations of this type have appeared as problems very many times in the last100 years with more general solutions. For example, consider the equation xr+ys + zt = wu where r, s, t and u are given positive integers with u relativelyprime to rst. We now just let x = a(ar +bs+ct)mst; y = b(ar +bs+ct)mtr
and z = c(ar + bs + ct)mrs so that xr + ys + zt = (ar + bs + ct)mrst+1.On setting w = (ar + bs + ct)n, we have to ensure that there are integers mand n such that mrst + 1 = nu. Since u is relatively prime to rst, there isan in�nite set of such pairs of positive integers m and n. A much more di�cultproblem is to �nd relatively prime solutions.
� page 247. Here one has to show that (1 + x1)(1 + x2) � � � (1 + xn) �1+S+ S2
2!+ � � �+ Sn
n!, where the x's are positive numbers and S =
Pxi. In
addition to his solution, the author refers to an alternate solution in CRUX in1989. It should be pointed out that the following is a known stronger inequal-
ity: (1+x1x)(1+x2x) � � � (1+xnx) � 1+xS+ x2S2
2!+� � �+ xnSn
n!(x > 0).
Here the � sign indicates majorization; that is, the coe�cient of xr on the leftis less than or equal to the coe�cient of xr on the right for r = 1, 2, : : : , n.See [1982: 296].
Aside from my criticisms above, the book is a nice collection of problems andsome essays. In a review [1998: 78] of the author's previous book in the same vein,From Erd }os to Kiev, Bill Sands was somewhat critical of the random order of theproblems. While that is true here also, I would not be too critical of this providedthe solutions were given in order; that is, having all the geometry solutions together,the algebra solutions together, and so on. For when one is writing a mathematicscompetition, the order of problems is essentially at random. But when one is learningto solve problems on one's own, it will be more e�ective to have solutions to likeproblems linked together. For a student competitor using this book, I strongly advisethat she or he also look at the corresponding solutions in CRUX with MAYHEM.
Finally, I do not think the book's title, \In P �olya's Footsteps", is appropriate.
274
APROPOS BELL AND STIRLING NUMBERS
Antal E. Fekete
Introduction
In 1877 Dobi ~nski stated [1] that there exist integers qn such that
0n
0!+
1n
1!+
2n
2!+
3n
3!+ � � � = qn
�1
0!+
1
1!+
1
2!+
1
3!+ � � �
�= qne ,
and he calculated their values for n = 1 through 5 (Table 1). Indeed, qnare the Bell numbers, so named in honour of the American mathematicianEric Temple Bell (1883-1960), who was among the �rst to popularize thesenumbers; see [2] and [3], where further references can be found. It may beshown that qn is just the sum of Stirling numbers of the second kind:
qn =
�n
1
�+
�n
2
�+ � � � +
�n
n
�.
Since�n
k
is the number of k{member quotient sets of an n{set, the Bell
number qn is the number of all quotient sets of an n{set. It can also becalculated via recursion in terms of the Stirling numbers of the �rst kind; thefollowing formula is due to G.T. Williams [5]:�
n
n
�qn �
�n
n� 1
�qn�1 +� � � � + (�1)n�1
�n
1
�q1 = 1 ,
where�n
k
�are the coe�cients of the polynomial of degreenwith roots 0 , 1 , 2 ,
: : : , (n� 1):�n
n
�xn�
�n
n� 1
�xn�1+� � � �+(�1)n�1
�n
1
�x = x(x�1)(x�2) � � � (x�n+1) .
R�enyi numbers
Here we show that there exist integers rn such that
0n
0!�
1n
1!+
2n
2!�
3n
3!+� � � � = rn
�1
0!�
1
1!+
1
2!�
1
3!+� � � �
�=
rn
e.
We calculate their values for n = 1 through 4 by bringing to the numeratorthe polynomial with coe�cients
�n
k
�, and splitting it into linear factors which
we can then cancel.
r1 = �1: X0
(�1)nn
n!=X1
(�1)n1
(n� 1)!= �
1
e.
Copyright c 1999 CanadianMathematical Society
275
r2 = 0:
X0
(�1)nn2
n!=
X0
(�1)nn2 � n
n!�
1
e=X0
(�1)n(n� 1)n
n!�
1
e
=X2
(�1)n1
(n� 2)!�
1
e=
1
e�
1
e= 0 .
r3 = 1:
X0
(�1)nn3
n!=
X0
(�1)nn3 � 3n2 + 2n
n!+
0
e+
2
e
=X0
(�1)n(n� 2)(n� 1)n
n!+
2
e
=X2
(�1)n1
(n� 3)!+
2
e= �
1
e+
2
e=
1
e.
r4 = 1:
X0
(�1)nn4
n!=
X0
(�1)nn4 � 6n3 + 11n2 � 6n
n!+
6
e�
6
e
=X0
(�1)n(n� 3)(n� 2)(n� 1)n
n!
=X4
(�1)n1
(n� 4)!=
1
e.
The recursion formula for rn in terms of Stirling numbers of the �rst kind�n
n
�rn �
�n
n� 1
�rn�1 +� � � � + (�1)n�1
�n
1
�r1 = (�1)n
shows that rn is an integer for all n (Table 1). To prove it we write theleft{hand side as
eX0
(�1)k�n
n
�kn �
�n
n�1�kn�1 +� � � � + (�1)n�1
�n
1
�k
k!
= eX0
(�1)k(k� n+ 1) � � � (k� 1)k
k!= e
Xn
(�1)k
(k� n)!= (�1)n .
We call rn R�enyi numbers in honour of the Hungarian mathematician Alfr �edR �enyi (1921-1970) who �rst studied them [4]. They can be expressed as thealternating sum of the Stirling numbers of the second kind:
rn = ��n
1
�+
�n
2
��+ � � � + (�1)n
�n
n
�.
276
Related numbers
We now introduce related numbers exhibitingproperties similar to thoseof Bell and R �enyi numbers. There exist integers an, bn such that
0n
0!+
2n
2!+
4n
4!+ � � � = an cosh(1) + bn sinh(1) ;
and1n
1!+
3n
3!+
5n
5!+ � � � = an sinh(1) + bn cosh(1)
which can be calculated via recursion in terms of the Stirling numbers of the�rst kind:�n
n
�an �
�n
n� 1
�an�1 +� � � � + (�1)n�1
�n
1
�a1 =
�1 if n is even0 if n is odd�
n
n
�bn �
�n
n� 1
�bn�1 +� � � � + (�1)n�1
�n
1
�b1 =
�0 if n is even1 if n is odd
Furthermore, there exist integers cn, dn such that
0k
0!�
2k
2!+
4k
4!�+ � � � = ck cos(1)� dk sin(1) ;
and1k
1!�
3k
3!+
5k
5!�+ � � � = ck sin(1) + dk cos(1) ,
which can be calculated via recursion in terms of the Stirling numbers of the�rst kind:
�n
n
�cn �
�n
n� 1
�cn�1 +� � � � + (�1)n�1
�n
1
�c1 =
8<:
1 if n = 4k
�1 if n = 4k + 2
0 if n is odd
�n
n
�dn �
�n
n� 1
�dn�1 +� � � � + (�1)n�1
�n
1
�d1 =
8<:
0 if n is even1 if n = 4k+ 1
�1 if n = 4k + 3
Table 1: Bell, R �enyi and related numbers
n 0 1 2 3 4 5 6 7 8 9 10
qn 1 1 2 5 15 52 203 877 4140 21147 115975
rn 1 �1 0 1 1 �2 �9 �9 50 267 413
an 1 0 1 3 8 25 97 434 2095 10707 58194
bn 0 1 1 2 7 27 106 443 2045 10440 57781
cn 1 0 �1 �3 �6 �5 33 266 1309 4905 11516
dn 0 1 1 0 �5 �23 �74 �161 57 3466 27361
These related numbers are useful in calculating the number of quotient sets
277
of an n{set with an even or odd number of members:
an =qn + rn
2=
�n
2
�+
�n
4
�+
�n
6
�+ � � � ;
and bn =qn � rn
2=
�n
1
�+
�n
3
�+
�n
5
�+ � � � .
We also have
cn = ��n
2
�+
�n
4
���n
6
�+� � � � ;
and dn =
�n
1
���n
3
�+
�n
5
��+ � � � .
Extended Bell numbers
There exist integers qkn such that
kn
0!+
(k+ 1)n
1!+
(k+ 2)n
2!+
(k+ 3)n
3!+ � � � = e qk;n+1 .
In particular, q1n = qn; for this reason we call qkn the extended Bell num-bers. For each �xed k we have a recursion formula in terms of the Stirlingnumbers of the �rst kind
qkn =
(�1)k�1��k
1
�qn �
�k
2
�qn+1 +
�k
3
�qn+2 �+ � � � + (�1)k�1
�k
k
�qn+k�1
�,
providing the �rst of three methods to calculate the extended Bell numbers(Table 2).
A second method is via the recursion
qkn = kqk;n�1 + qk+1;n�1 .
For a �xed n, the extended Bell numbers q0;n , q1;n , q2;n , : : : are inarithmetic progression of order n � 1. Therefore qkn � Qn�1(k + 1) is apolynomial of degree n� 1 in the variable k. We have the recursion
Qn(k) � (k� 1)Qn�1(k) + Qn�1(k+ 1)
enabling us to calculate these polynomials:
Q0(k) � 1 (constant)Q1(k) � k
Q2(k) � k2 + 1
Q3(k) � k3 + 3k+ 1
Q4(k) � k4 + 6k2 + 4k + 4
Q5(k) � k5 + 10k3 + 10k2 + 20k+ 11
278
Q6(k) � k6 + 15k4 + 20k3 + 60k2 + 66k+ 41
Q7(k) � k7 + 21k5 + 35k4 + 140k3 + 231k2 + 287k+ 162
Q8(k) � k8 + 28k6 + 56k5 + 280k4 + 616k3 + 1148k2 + 1296k+ 715
. . . . . . . . . . . . . . .
A thirdmethod of calculating qkn is furnishedby the di�erence equation�nqk1 = qk�1;n that may be verbalized as follows. In Table 2, the k{th row
can be obtained by calculating the �rst member of each of the higher order
di�erence sequences of the (k+ 1){st row. This property is useful not onlyin calculating the k{th row from the (k + 1){st, quickly and e�ciently, butalso the other way around.
Table 2: Extended Bell Numbers
knn 1 2 3 4 5 6 7 8 n
� � � � � � � � � ��7 1 �6 37 �233 1492 �9685 63581 �421356 ��6 1 �5 26 �139 759 �4214 23711 �134873 ��5 1 �4 17 �75 340 �1573 7393 �35178 ��4 1 �3 10 �35 127 �472 1787 �6855 ��3 1 �2 5 �13 36 �101 293 �848 ��2 1 �1 2 �3 7 �10 31 �21 ��1 1 0 1 1 4 11 41 162 �0 1 1 2 5 15 52 203 877 �1 1 2 5 15 52 203 877 4140 �2 1 3 10 37 151 674 3263 17007 �3 1 4 17 77 372 1915 10481 60814 �4 1 5 26 141 799 4736 29371 190497 �5 1 6 37 235 1540 10427 73013 529032 �6 1 7 50 365 2727 20878 163967 1322035 �7 1 8 65 537 4516 38699 338233 3017562 �k � � � � � � � � qkn
For example, suppose that row 3 is given and we wish to �nd the next, row 4.We enter row 3 as the slanting row indicated by the boxed numbers below,and calculate entries in successive slanting rows as the sum of the adjacenttwo entries in the previous slanting row.
1 5 26 141 799 4736 : : :
4 21 115 658 3937 : : :
17 94 543 3279 : : :
77 449 2736 : : :
372 2287 : : :
1915 : : :
10481 : : :
279
Then row 4 appears as the top line of the calculation. It is clear that, giventhe initial condition qn, we can reconstruct the entire table for qkn as theunique solution to the di�erence equation.
Problems
In passingwe mention some further results, proposed here as problemsthat the reader may wish to solve using various ideas presented above.
(1) Show that
12
1!+
32
3!+
52
5!+ � � � =
22
2!+
42
4!+
62
6!+ � � � .
Are there exponents other than n = 2 for which the equality holds?
(2) Show that
13
1!�
23
2!+
33
3!�+ � � � =
14
1!�
24
2!+
34
3!�+ � � � .
Are there pairs of exponents other than 3, 4 for which an equality ofthis type holds?
(3) Show that
1n
1!�
3n
3!+
5n
5!�+ � � � = �n
�1
1!�
1
3!+
1
5!�+ � � �
�
and0n
0!�
2n
2!+
4n
4!�+ � � � = �n
�1
0!�
1
2!+
1
4!�+ � � �
�
simultaneously hold for n = 3. Find all other n having the same prop-erty. If follows that�
0n
0!�
2n
2!+
4n
4!�+ � � �
�2+
�1n
1!�
3n
3!+
5n
5!�+ � � �
�2
is an integer for n = 3. Clearly, this is also true for n = 1. Find allother n having the same property.
(4) Clearly,�0n
0!+
2n
2!+
4n
4!+ � � �
�2��1n
1!+
3n
3!+
5n
5!+ � � �
�2
is an integer for n = 1. Find all other n having the same property.
(5) Show that there are integers rkn such that
kn
0!�
(k+ 1)n
1!+
(k+ 2)n
2!�
(k+ 3)n
3!+� � � � =
rk;n+1
e.
In particular r1n = rn, in consequence of which we may call rkn theextended R �enyi numbers.
280
(6) Show that there are integers akn , bkn such that
(2k)n
0!+
(2k+ 2)n
2!+
(2k+ 4)n
4!+ � � �
= ak;n+1 cosh(1) + bk;n+1 sinh(1)
and (2k+ 1)n
1!+
(2k+ 3)n
3!+
(2k+ 5)n
5!+ � � �
= ak;n+1 sinh(1) + bk;n+1 cosh(1) .
In particular, a1n = an , b1n = bn.
(7) Show that there are integers ckn , dkn such that
(2k)n
0!�
(2k+ 2)n
2!+
(2k+ 4)n
4!�+ � � �
= ck;n+1 cos(1)� dk;n+1 sin(1)
and (2k+ 1)n
1!�
(2k+ 3)n
3!+
(2k+ 5)n
5!�+ � � �
= ck;n+1 sin(1) + dk;n+1 cos(1) .
In particular, c1n = cn , d1n = dn.
(8) For a �xed k, write a recursion formula in terms of the Stirling numbersof the �rst kind for each of rkn , akn, etc.
(9) Show that, for n �xed, each of the sequences of numbers rkn , akn,etc., is in arithmetic progression of order n� 1. Write polynomials ofdegree n, Rn(k), An(k), etc., de�ned such that rkn � Rn�1(k + 1),akn � An�1(k+ 1), etc.
(10) Write a recursion formula for each of the polynomials Rn(k), An(k).Write the polynomials in explicit form up to n = 8.
(11) Write a recursion formula for each of rkn , akn, etc.
(12) Find and tabulate the values of each of rkn , akn, etc.
(13) Recall that the di�erence equation �nqk1 = qk�1;n uniquely deter-mines the extended Bell numbers qkn under the initial conditionq1n = qn. Write a di�erence equation for the extended R �enyi num-bers rkn under the initial condition r1n = rn. Do the same for each ofakn, bkn, etc.
(14) Let n be �xed; continue the sequence of extended Bell numbers qkn fornegative k (there are at least three ways of doing that) in order to justifyTable 2 for k = �1, �2, �3, : : : . Do the same for each of rkn, akn,etc.
281
(15) Show that for each �xed k we have
qk+1;n+1 = qk1 +
�n
1
�qk2 +
�n
2
�qk3 + � � � +
�n
n
�qk;n+1
and
qk+1;n�1 = qk1��n
1
�qk2+
�n
2
�qk3�+ � � � +(�1)n�1
�n
n
�qk;n+1 .
(16) Show that for each �xed k andm we have
qk+m;n+1 = qk1mn+
�n
1
�qk2m
n�1+
�n
2
�qk3m
n�2+ � � �+�n
n
�qk;n+1 , ;
in particular,
qm;n+1 = q0mn +
�n
1
�q1m
n�1 +
�n
2
�q2m
n�2 + � � � +�n
n
�qn .
(17) Show that�k
1
�q1n +
�k
2
�q2n + � � � +
�k
k
�qkn = qn+k�1 .
(18) Show that
qn � qn�1 +� � � � + (�1)n+1q1 = qn�1(0) = q�1;n+2 .
References
1. G. Dobi ~nski, Summierung Der ReihePnm=n! f �urm = 1, 2, 3, 4, 5, : : : Arch.
f �ur Math. und Physik, vol. 61 (1877), pp. 333-6.
2. Eric Temple Bell, Exponential numbers, Am. Math. Monthly, vol. 41, no. 7(1934), pp. 411-419.
3. Gian-Carlo Rota, The Number of Partitions of a Set, Am. Math. Monthly,vol. 71 (1964), pp. 498-504.
4. Alfr �ed R �enyi, New methods and results in Combinatory Analysis (in Hungar-ian), Magyar Tudom�anyos Akad �emia III Oszt �alya K�ozlem�enyei, vol. 16 (1966),pp. 77-105.
5. G.T. Williams, Numbers generated by the function eex�1, Am.Math. Monthly,
vol. 53 (1945), pp. 323-7.
6. D.E. Knuth, TwoNotes on Notation, Am.Math.Monthly, vol. 99 (1992), pp. 403-22.
7. A.E. Fekete, Apropos Two Notes on Notation, Am. Math. Monthly, vol. 101(1994), pp. 771-8.
Memorial University of NewfoundlandSt.John's, CANADA A1C 5S7e-mail: [email protected]
282
THE SKOLIAD CORNERNo. 39
R.E. Woodrow
This issue we give another example of a team competition with the prob-lems of the 1998 Florida Mathematics Olympiad, written May 14, 1998. Thecontest was organized by Florida Atlantic University. My thanks go to JohnGrant McLoughlin, Memorial University of Newfoundland for sending methe problems.
FLORIDA MATHEMATICS OLYMPIADTEAM COMPETITION
May 14, 1998
1. Find all integers x, if any, such that 9 < x < 15 and the sequence
1 , 2 , 6 , 7 , 9 , x , 15 , 18 , 20
does not have three terms in arithmetic progression. If there are no suchintegers, write \NONE."
2. A sequence a1, a2, a3; : : : is said to satisfy a linear recurrence rela-
tion of order two if and only if there are numbers p and q such that, for allpositive integers n,
an+2 = pan+1 + qan .
Find the next two terms of the sequence
2 , 5 , 14 , 41 , : : : ,
assuming that this sequence satis�es a linear recurrence relation of ordertwo.
3. Seven tests are given and on each test no ties are possible. Eachperson who is the top scorer on at least one of the tests or who is in thetop six on at least four of these tests is given an award, but each person canreceive at most one award. Find the maximum number of people who couldbe given awards, if 100 students take these tests.
4. Some primes can be written as a sum of two squares. We have, forexample, that
5 = 12 + 22; 13 = 22 + 32; 17 = 12 + 42 ,
29 = 22 + 52; 37 = 12 + 62; and 41 = 42 + 52 .
283
The odd primes less than 108 are listed below; the ones that can be writtenas a sum of two squares are boxed in.
3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 ,
37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 ,
73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 .
The primes that can be written as a sum of two squares follow a simple pat-tern. See if you can correctly �nd this pattern. If you can, use this patternto determine which of the primes between 1000 and 1050 can be written asa sum of two squares; there are �ve of them. The primes between 1000 and1050 are
1009 , 1013 , 1019 , 1021 , 1031 , 1033 , 1039 , 1049 .
No credit unless the correct �ve primes are listed.
5. The sides of a triangle are 4, 13, and 15. Find the radius of theinscribed circle.
6. In Athenian criminal proceedings, ordinary citizens presented thecharges, and the 500-man juries voted twice: �rst on guilt or innocence, andthen (if the verdict was guilty) on the penalty. In 399BCE, Socrates (c469{399)was charged with dishonouring the gods and corrupting the youth of Athens.He was found guilty; the penalty was death. According to I.F. Stone's calcu-lations on how the jurors voted:
(i) There were no abstentions;
(ii) There were 80more votes for the death penalty than there were forthe guilty verdict;
(iii) The sum of the number of votes for an innocent verdict and thenumber of votes against the death penalty equalled the number of votes infavour of the death penalty.
a) How many of the 500 jurors voted for an innocent verdict?
b) How many of the 500 jurors voted in favour of the death penalty?
7. Find all x such that 0 � x � � and
tan3 x� 1 +1
cos2 x� 3 cot
��
2� x
�= 3 .
Your answer should be in radian measure.
Last issue we gave the problems of the Newfoundland and LabradorTeachers' Association Mathematics League. Here are the answers.
284
NLTA MATH LEAGUEGAME 1 | 1998{99
1. Find a two-digit number that equals twice the product of its digits.
Solution. Denote the number by ab; We get 10a + b = 2a � b. Try-ing a = 1; 2; : : : ; 9, the only integral solution is with a = 3, b = 6 and36 = 2 � 3 � 6.
2. The degree measures of the interior angles of a triangle are A, B, Cwhere A � B � C. If A, B, and C are multiples of 15, how many possiblevalues of (A;B;C) exist?
Solution. Let A = 15m, B = 15n, and C = 15p. Then 15m+ 15n+
15p = 180 so m + n + p = 12 and m � n � p. Since m, n � 1 we havem+ n � 2 and p � 10. Also 3p � 12 so p � 4. For p �xed, 4 � p � 10 wehave m+ n = 12� p, or m = 12� p� n. This leads to the solutions
p 4 5 5 6 6 6 7 7 8 8 9 10
n 4 4 5 3 4 5 3 4 2 3 2 1
m 4 3 2 3 2 1 2 1 2 1 1 1
There are twelve solutions.
3. Place an operation (+;�;�;�) in each square so that the expres-sion using 1, 2, 3, � � � , 9 equals 100.
1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 = 100 .
You may also freely place brackets before/after any digits in the expression.Note that the squares must be �lled in with operational symbols only.
Solution. Here is one solution:
(((1 + 2 + 3 + 4 + 5)� 6)� 7) + 8 + 9 = 100 .
How many solutions are there?
4. A, B and C are points on a line that is parallel to another linecontaining pointsD andE, as shown. Point F does not lie on either of theselines.
r r r
r r
r
A B C
F
D E
How many distinct triangles can be formed such that all three of itsvertices are chosen from A, B, C, D, E, and F ?
285
Solution. Any choice of three of the six vertices determines a triangleexcept when they lie on a line; that is, except for the one choice fA;B;Cg.The total is thus
�6
3
�� 1 = 20� 1 = 19.
5. Michael, Jane and Bert enjoyed a picnic lunch. The three of themwere to contribute an equal amount of money toward the cost of the food.Michael spent twice as much money as Jane did buying food for lunch. Bertdid not spend any money on food. Instead, Bert brought $6 which exactlycovered his share. How much (in dollars) of Bert's contribution should begiven to Michael?
Solution. Bert brought $6, which exactly covered his share, so the totalcost of food is 3�$6 = $18. Now Michael spent twice as much as Jane so hespent $12 and she spent $6. The total amount of $6 brought by Bert shouldgo to Michael.
6. Two semicircles of radius 3 are inscribed in a semicircle of radius 6.A circle of radius R is tangent to all three semicircles, as shown. Find R.
3 3
33
RR
R
h
r
r
r r r
r r
Solution. Join the centres of the two smaller semicircles and the centre ofthe circle. This forms an isosceles triangle with equal sides 3+R and base 6units. Call the altitude of this triangle h. The altitude extends to a radius ofthe large semicircle, so h+R = 6. By Pythagoras, h2 + 32 = (R+ 3)2, so
(6� R)2 + 32 = (R+ 3)2 ,
36� 12R+ R2 + 9 = R2 + 6R+ 9 ,
36 = 18R ,
2 = R .
The radius of the small circle is 2.
7. If 5A = 3 and 9B = 125, �nd the value of AB.
Solution. Now 5A = 3, so 52A = 32 = 9 and
52AB = (52A)B = 9B = 125 = 53 ,
so 2AB = 3 and AB = 3
2.
286
8. The legs of a right angled triangle are 10 and 24 cm respectively.
Let A = the length (cm) of the hypotenuse,B = the perimeter (cm) of the triangle,C = the area (cm2) of the triangle.
Determine the lowest common multiple of A, B, and C.
Solution. Then
A =p102 + 242 = 26
B = 10 + 24 + 26 = 60
C = 1
2� 10 � 24 = 120
lcm(26;60; 120) = 3� 8� 5� 13 = 1560.
9. A lattice point is a point (x; y) such that both x and y are integers.For example, (2;�1) is a lattice point, whereas,
�3; 1
2
�and
��1
3; 23
�are not.
How many lattice points lie inside the circle de�ned by x2 + y2 = 20? (DoNOT count lattice points that lie on the circumference of the circle.)
Solution. Since 42 < 20 < 52 we have that �4 � x � 4. For a �xedx in the range we must have �
p20� x2 < y <
p20� x2 for integer x, y
solutions corresponding to interior points of the circle.
x �4 �3 �2 �1 0p20� x2 2
p11
p16
p19
p20
y �1 , 0 , 1 �3 , : : : , 3 �3 , : : : , 3 �4 , : : : , 4 �4 , : : : , 4
Lattice pts. 2 � 3 = 6 2 � 7 = 14 2 � 7 = 14 2 � 9 = 18 9
The total number is then 6 + 14 + 14 + 18 + 9 = 61.
10. The quadratic equation x2 + bx+ c = 0 has roots r1 and r2 thathave a sum which equals 3 times their product. Suppose that (r1 + 5) and(r2+5) are the roots of another quadratic equation x2+ ex+ f = 0. Giventhat the ratio of e : f = 1 : 23, determine the values of b and c in the originalquadratic equation.
Solution. Now r1 + r2 = �b and r1r2 = c, so as r1 + r2 = 3r1r2 weget 3c = �b. Similarly we have
r1 + r2 + 10 = �e ,(r1 + 5)(r2 + 5) = f .
Thus
10� b = �eand b� 10 = e ,
r1r2 + 5(r1 + r2) + 25 = f ,
c� 5b+ 25 = f .
287
From ef= 1
23, 23(b� 10) = c� 5b+ 25. Using b = �3c
23(�3c� 10) = c+ 15c+ 25 ,
so that � 69c� 230 = 16c+ 25 ,
�255 = 85c ,
�3 = c ,
and 9 = b .
The quadratic is x2 + 9x� 3 = 0.
RELAY
R1. Operations � and � are de�ned as follows:
A �B =AB +BA
A+ Band A � B =
AB � BA
A� B.
Simplify N = (3 � 2) � (3 � 2). Write the value of N in Box #1 of the relayanswer sheet.
Solution.
3 � 2 =32 + 23
3 + 2=
9 + 8
5=
17
5,
3 � 2 =32 � 23
3� 2=
9� 8
1= 1 ,
N = (3 � 2) � (3 � 2) =
�17
5� 1�=
�17
5
�1+ (1)17=5
17
5+ 1
= 1 .
R2. A square has a perimeter of P cm and an area of Q sq.cm. Giventhat 3NP = 2Q, determine the value of P . Write the value of P in Box #2of the relay answer sheet.
Solution. From R1,N = 1, so 3P = 2Q. Also the side length is s = P4,
so Q = (P4)2 = P 2
16. So 3P = 2 � P 2
16or 24P = P 2. This gives P = 0 or
P = 24. We use P = 24, assuming the square is not degenerate.
R3. List all two-digit numbers that have digits whose product is P .Call the sum of these two-digit numbers S. Write the value of S in Box #3of the relay answer sheet.
Solution. P = 1 � 24 = 2 � 12 = 4 � 6 = 8 � 3. The two-digit numbersare 46, 64, 38 and 83; so
S = 231 .
288
R4. How many integers between 6 and 24 share no common factorswith S that are greater than 1?
Solution. Now 231 = 11� 21 = 11� 7� 3. The numbers between 6
and 24 that share no common factor with 231 are
8 , 10 , 13 , 16 , 17 , 19 , 20 , 23 .
There are 8 of them.
TIE-BREAKER
Find the maximum value of
f(x) = 14�px2 � 6x+ 25 .
Solution.
f(x) = 14�px2 � 6x+ 25
= 14px2 � 6x+ 9� 9 + 25
= 14�p(x� 3)2 + 16 .
For f(x) to be maximum we want (x�3)2+16 to be minimum. This occurswhen x = 3, and
f(3) = 14�p16 = 14� 4 = 10 .
That completes the Skoliad Corner for this number. Sendme your com-ments, suggestions, and especially suitable material for use in the Corner.
Announcement
The second volume in the ATOM series has just been published. Algebra |
Intermediate Methods by Bruce Shawyer. Contents include:
Mathematical Induction, Series, Binomial Coe�cients,Solutions of Polynomial Equations, and Vectors and Matrices.
For more information, contact the CMS O�ce | address on outside backcover.
289
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
Mathematical Mayhem, Department of Mathematics, University of Toronto,
100 St. George St., Toronto, Ontario, Canada. M5S 3G3. The electronicaddress is still
The Assistant Mayhem Editor is Cyrus Hsia (University of Western On-tario). The rest of the sta� consists of Adrian Chan (Upper Canada College),Jimmy Chui (University of Toronto), David Savitt (Harvard University) andWai Ling Yee (University of Waterloo).
Shreds and Slices
Another Combinatorial Proof
Dear Cyrus:In the most recent issue of Crux Mathematicorum with Mathematical
Mayhem, Dave Arthur's two-step combinatorial proof of the identity
(n� r)
�n+ r� 1
r
��n
r
�= n
�n+ r � 1
2r
��2r
r
�(1)
was presented, and a one-step combinatorial proof was sought. I have sucha proof, but before I present the polished �nal version, let me explain howI was led to it.
Arthur's solution proceeded in two steps. He counted by two di�erentmethods:
(I) the number of ways of choosing two distinct sets A and B, each with relements, from a set with n+ r � 1 elements, and
(II) the number of ways of choosing two distinct sets C and D, where Chas one element and D has r elements, from a set with n elements.
It occurred to me that both (I) and (II) were ways to �nish o� the selec-tion of three sets A, B, and C, with r, r, and 1 elements respectively, froma set with n+ r elements. We could �rst choose the small set C, and thenchoose sets A and B from the remaining n+ r� 1 elements as in (I), or wecould choose one of the big sets A, and then choose sets C and B = D from
290
the remaining n elements as in (II). Counting both these methods yields theidentity
(n+ r)
��n+ r� 1
2r
��2r
r
��=
�n+ r
r
���n
r
�(n� r)
�: (2)
We can easily see that this identity is equivalent to (1) by multiplying bothsides of (2) by n and using the identity
n
�n+ r
r
�= (n+ r)
�n+ r � 1
r
�(3)
on the right-hand side.
There are two �sthetic problems with this: �rst, we derived the soughtidentity (1) with an extra factor of n + r on both sides; second, we usedthe identity (3) which, though elementary, can be thought of as needing acombinatorial proof of its own (choosing disjoint subsets A andW of sizes rand 1 from a set with n + r elements). The following proof of (1) is free ofthese defects.
Proof of (1). Suppose we have a circular arrangement of n + r boxes,and we have r amber balls, r blue balls, 1 cyan ball, and 1 white ball. Wewant to arrange these balls in the boxes (arrangements that di�er only by arotation being regarded as the same), such that two balls are not allowed tobe in the same box, except that the white ball may be in a box with a blue orcyan ball. We claim that both sides of (1) count the number of ways of doingthis. The two counting methods are:
1. First place the white ball in a box; since rotated arrangements are con-sidered the same, this can be done in only one way, but we have nowused up our freedom to rotate the boxes. Next, place the r amber ballsinto r of the remaining n + r � 1 empty boxes, which can be done in�n+r�1
r
�ways. Then, place the r blue balls into r of the n boxes that
do not contain an amber ball (recall that a blue ball can coexist with awhite ball), which can be done in
�n
r
�ways. Finally, place the cyan ball
in one of the n� r boxes not containing an amber or blue ball (again,the white ball is allowed if it doesn't also contain a blue ball), whichcan be done in n� r ways. The total number of such arrangements isthus �
n+ r� 1
r
��n
r
�(n� r) : (4)
2. This time, place the cyan ball �rst; as before, there is only one way todo this up to rotation. Next, select 2r of the remaining n+r�1 emptyboxes (which we will momentarily �ll with the 2r amber and blue balls),which can be done in
�n+r�1
2r
�ways. Then, place the r amber balls into
r of these selected 2r boxes, and place the r blue balls into the other r
291
selected boxes; this can be done in�2r
r
�ways. Finally, place the white
ball into any of the n boxes that does not contain an amber ball, whichcan be done in n ways. The total number of such arrangements is thus�
n+ r � 1
2r
��2r
r
�n : (5)
Since (4) and (5) are the left- and right-hand sides of (1), respectively,this completes the proof. �
We remark that we could rephrase what we are counting as follows: wewant to select three disjoint subsets A, B, and C, with r, r, and 1 elementsrespectively, from a set S with n+ r elements, and also label one elementW of S that is not in A, except that selections that di�er only by a cyclicpermutation of S are to be considered equal.
Yours sincerely,
Greg MartinDepartment of Mathematics, University of Toronto
Mayhem Problems
The Mayhem Problems editors are:
Adrian Chan Mayhem High School Problems Editor,Donny Cheung Mayhem Advanced Problems Editor,David Savitt 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 | thenext issue will feature only solutions.
We warmly welcome proposals for problems and solutions. With theschedule of eight issues per year, we request that solutions from this issuebe submitted in time for issue 6 of 2000.
High School Problems
Editor: Adrian Chan, 229 Old Yonge Street, Toronto, Ontario, Canada.M2P 1R5 <[email protected]>
We correct problem H253, which �rst appeared in Issue 3.
H253. Find all real solutions to the equationp3x2 � 12x+ 52 +
p2x2 � 12x+ 162 =
p�x2 + 6x+ 280 .
292
H257. Find all integers n such that n2�11n+63 is a perfect square.
H258. Solve in integers for x and y:
6(x! + 3) = y2 + 5 .
H259. Proposed by Alexandre Tritchtchenko, student, Carleton Uni-
versity, Ottawa, Ontario.Solve for x:
2m�n sin(2nx)m�nYi=1
cos(2m�ix) = 1 .
H260. Proposed byMohammed Aassila, CRM,Universit �e deMontr �eal,
Montr �eal, Qu �ebec.Let x1, x2, : : : , xm be real numbers. Prove that
X1�i<j�n
cos(xi � xj) � �n
2.
Advanced Problems
Editor: Donny Cheung, c/o Conrad Grebel College, University of Wa-terloo, Waterloo, Ontario, Canada. N2L 3G6 <[email protected]>
A233. Proposed by Naoki Sato, Mayhem Editor.In C81, we de�ned the following sequence: a0 = 0, a1 = 1, and
an+1 = 4an � an�1 for n = 1, 2, : : : . This sequence exhibits the followingcurious property: For n � 1, if we set (a; b; c) = (an�1; 2an; an+1), thenab + 1, ac + 1, and bc + 1 are always perfect squares. For example, forn = 3, (a; b; c) = (a2; 2a3; a4) = (4; 30; 56), and indeed, 4 � 30 + 1 = 112,4 � 56 + 1 = 152, and 30 � 56 + 1 = 412. Show that this property holds.Generalize, using the sequence de�ned by a0 = 0, a1 = 1, and an+1 =
Nan� an�1, and the triples (a; b; c) = (an�1; (N � 2)an; an+1), where Nis an arbitrary integer.
A234. In triangle ABC, AC2 is the arithmetic mean of BC2 andAB2. Show that cot2B � cotA cotC. (Note: cot � = cos �= sin�.)
(1997 Baltic Way)
A235. ProposedbyMohammed Aassila, CRM, Universit �e deMontr �eal,
Montr �eal, Qu �ebec.The convex polygon A1A2 � � �An is inscribed in a circle of radius R.
Let A be some point on this circumcircle, di�erent from the vertices. Set
293
ai = AAi, and let bi denote the distance from A to the line AiAi+1, i = 1,2, : : : , n, where An+1 = A1. Prove that
a21
b1+a22
b2+ � � �+
a2n
bn� 2nR .
A236. ProposedbyMohammed Aassila, CRM, Universit �e deMontr �eal,
Montr �eal, Qu �ebec.
For all positive integers n and positive reals x, prove the inequality
�2n
1
�x+ 1
+
�2n
3
�x+ 3
+ � � �+�
2n
2n�1�
x+ 2n� 1<
�2n
0
�x
+
�2n
2
�x+ 2
+ � � �+�2n
2n
�x+ 2n
.
Challenge Board Problems
Editor: David Savitt, Department of Mathematics, Harvard University,1 Oxford Street, Cambridge, MA, USA 02138 <[email protected]>
C87. Proposed by Mark Krusemeyer, Carleton College.
Find an example of three continuous functions f(x), g(x), and h(x)
from R to R with the property that exactly �ve of the six composite functionsf(g(h(x))),f(h(g(x))),g(f(h(x))), g(h(f(x))),h(f(g(x))), andh(g(f(x)))are the same function and the sixth function is di�erent.
C88.
(a) Let A be an n�n matrix whose entries are all either +1 or �1. Provethat j detAj � nn=2.
(b) It is conjectured that for there to exist an n�nmatrix A whose entriesare all either +1 or �1 and such that jdetAj = nn=2, it is necessaryand su�cient that n = 1, n = 2, or n is divisible by 4. Prove that thiscondition is necessary.
(c) Can you construct such a matrix, for n equal to a power of 2? Forn = 12?
294
Problem of the Month
Jimmy Chui, student, University of Toronto
Problem. Let a, b, and c be positive real numbers. Prove that
aabbcc � (abc)(a+b+c)=3 .
(1995 CMO, Problem 2)
Solution I. Taking log of both sides, we see that we must prove
a log a+ b log b+ c log c �a+ b+ c
3logabc .
Let f(x) = x logx. Then f 0(x) = logx + 1 and f 00(x) = 1=x. We can seethat f 00(x) > 0 for x > 0. So, f is convex for x > 0. By Jensen's Inequality,
f(a) + f(b) + f(c)
3� f
�a+ b+ c
3
�
===)a loga+ b log b+ c log c
3�
a+ b+ c
3log
�a+ b+ c
3
�.
By the AM-GM Inequality, a+b+c3
� 3pabc . Thus, we have
a log a+ b log b+ c log c � (a+ b+ c) log
�a+ b+ c
3
�
� (a+ b+ c) log(3pabc)
=a+ b+ c
3logabc .
Solution II. Recall Chebychev's Inequality: Let x1, x2, : : : , xn and y1,y2, : : : , yn be two real sequences, either both increasing or both decreasing.Then
x1y1 + x2y2 + � � �+ xnyn
n�
x1 + x2 + � � �+ xn
n�y1 + y2 + � � �+ yn
n.
(In other words, the average of the products is greater than or equal to theproduct of the averages.)
Without loss of generality, let a � b � c. Then loga � log b � log c.By Chebychev's Inequality,
a loga+ b log b+ c log c
3�
a+ b+ c
3�loga+ log b+ log c
3,
which implies that a log a+ b log b+ c log c � a+b+c3
log abc .
295
Solution III. Recall the Weighted AM-GM-HM Inequality: Let x1, x2,: : : , xn be positive real numbers, and let w1, w2, : : : , wn be non-negativereal numbers which sum to 1. Then
w1x1 + w2x2 + � � �+ wnxn � xw1
1 xw2
2 � � �xwn
n �1
w1
x2+ w2
x2+ � � �+ wn
xn
.
Take x1 = a, x2 = b, x3 = c, w1 = a=(a+ b+ c), w2 = b=(a+ b+ c),and w3 = c=(a + b + c). Then using the GM-HM portion of the aboveinequality, we obtain
aa=(a+b+c)bb=(a+b+c)cc=(a+b+c) = (aabbcc)1=(a+b+c)
�1
1
a+b+c+ 1
a+b+c+ 1
a+b+c
=a+ b+ c
3.
By the AM-GM Inequality,
aabbcc ��a+ b+ c
3
�a+b+c� (abc)(a+b+c)=3 .
J.I.R. McKnight Problems Contest 1986
Solutions
3. Prove that the sum of the squares of the �rst n even natural numbersexceeds the sum of the squares of the �rst n odd natural numbers byn(2n+1). Hence, or otherwise, �nd the sum of the squares of the �rstn odd natural numbers.
Partial solution by Luyun Zhong-Qido, Columbia International College,
Hamilton, Ontario, Canada.The sum of the �rst n positive integers is given by
12 + 22 + � � �+ n2 =n(n+ 1)(2n+ 1)
6. (1)
Therefore, 12 + 22 + � � �+ (2n)2 =2n(2n+ 1)(4n+ 1)
6, (2)
and 22 + 42 + � � �+ (2n)2 =4n(n+ 1)(2n+ 1)
6. (3)
296
From (2) and (3),
12 + 32 + � � �+ (2n� 1)2 =2n(2n+ 1)(4n+ 1)
6�
4n(n+ 1)(2n+ 1)
6
=2n(2n+ 1)[(4n+ 1)� 2(n+ 1)]
6
=2n(2n� 1)(2n+ 1)
6. (4)
From (3) and (4),
22 + 42 + � � �+ (2n)2 � [(12 + 32 + � � �+ (2n� 1)2]
=4n(n+ 1)(2n+ 1)
6�
2n(2n� 1)(2n+ 1)
6
=2n(2n+ 1)[2(n+ 1)� (2n� 1)]
6
=3 � 2n(2n+ 1)
6= n(2n+ 1) .
4. (b) Prove that in any acute triangle ABC,
cotA+ cotB + cotC =a2 + b2 + c2
4K,
where K is the area of triangle ABC.
Solution by Luyun Zhong-Qido, Columbia International College, Hamil-
ton, Ontario, Canada.We have the following relations in a triangle:
2R =a
sinA=
b
sinB=
c
sinC, (1)
K =1
2ab sinC =
1
2ac sinB =
1
2bc sinA . (2)
From (1) and (2),
K =1
2bc sinA =
1
2
�a sinB
sinA
��a sinC
sinA
�sinA = a2
sinB sinC
2 sinA,
so a2 =2K sinA
sinB sinC.
Likewise, b2 =2K sinB
sinA sinC, c2 =
2K sinC
sinA sinB.
We note that
cotA+ cotB =cosA
sinA+
cosB
sinB=
cosA sinB + cosB sinA
sinA sinB
=sin(A+B)
sinA sinB=
sinC
sinA sinB.
297
Therefore,
a2 + b2 + c2
4K=
2K sinAsinB sinC
+ 2K sinBsinA sinC
+ 2K sinCsinA sinB
4K
=2K(sin2A+ sin2B + sin2C)
4K sinA sinB sinC
=1
2
�sinA
sinB sinC+
sinB
sinA sinC+
sinC
sinA sinB
�
=1
2(cotB + cotC + cotA+ cotC + cotA+ cotB)
= cotA+ cotB + cotC .
J.I.R. McKnight Problems Contest 1989
PART A
1. A curve has equation y = x3 � 3x. A tangent is drawn to the curve atits relative minimum point. This tangent line also intersects the curveat P . Find the equation of the normal to the curve at P .
2. (a) In a triangle whose sides are 5, 12, and 13, �nd the length of thebisector of the larger acute angle.
(b) This large rectangle has been cut into eleven squares of varioussizes. The smallest square has an area of 81 cm2. Find the dimen-sions of the large rectangle.
3. Solve the following system of equations for x and y, where x, y 2 R:
x3 � y3 = 35 , xy2 � yx2 = 30 .
298
4. The combined volume of two cubes with integral sides is equal to thecombined length of all their edges. Find the dimensions of all cubessatisfying these conditions.
5. A car at an intersection is heading west at 24 m/s. Simultaneously, asecond car, 84 m north of the �rst car, is travelling directly south at10 m/s. After 2 seconds:
(a) Find the rate of change of the distance between the 2 cars.
(b) Using vector methods, determine the velocity of the �rst car rela-tive to the second.
(c) Explain fully why the magnitude of the vector in (b) is not neces-sarily the same as the rate of change in (a).
PART B
1. Prove that
sin 1� + sin3� + sin5� + � � �+ sin97� + sin99� =sin2 50�
sin1�.
2. Determine the nth term and the sum of n terms for the series
3 + 5 + 10 + 18 + 29 + � � � .
3. A circle has equation x2+y2 = 1. Lines with slope 1
2and�1
2are drawn
through C(0; 0) to form a sector of the circle. Find the dimensionsof the rectangle of maximal area which can be inscribed in this sectorgiven that two sides of the rectangle are parallel to the y{axis and therectangle is entirely below the x-axis.
4. In triangle ABC, angles A, B, and C are in the ratio 4 : 2 : 1. Provethat the sides of the triangle are related by the equality 1
a+ 1
b= 1
c.
5. Prove that 7 +
p37
2
!n+
7�
p37
2
!n� 1
is divisible by 3 for all n 2 N.
299
Derangements and Stirling Numbers
Naoki Satostudent, Yale University
In this article, we introduce two important classes of combinatorialnumbers which appear frequently: derangements and Stirling numbers. Wealso hope to emphasize the importance and usefulness of basic counting prin-ciples.
We can think of a permutation on n objects as a 1-1 function � fromthe set f1, 2, : : : , ng to itself. For example, for n = 3, the map � given by�(1) = 1, �(2) = 3, and �(3) = 2 is a permutation on 3 objects, namelythe elements of f1; 2; 3g; a permutation essentially re-arranges the elements.Note that there are n! di�erent permutations on n objects. Then, a derange-ment is a permutation � which has no �xed points; that is, �(i) 6= i for alli = 1, 2, : : : , n. Alternatively, if we think of a permutation on n objects asa distribution of n letters to n corresponding envelopes, then a derangementis a permutation where no letter is inserted into the correct correspondingenvelope. Let Dn denote the number of derangements on n objects. Thenatural question to ask is, what is the formula for Dn?
Problems.
1. Write down all permutations on 2, 3, and 4 elements. How many ofthese are derangements? Is there a systematic way of writing down allpermutations on n elements?
2. What is the total number of �xed points over all permutations on n
elements? You should be able to guess the answer from small cases.
We immediately see thatD1 = 0 andD2 = 1 (by convention, D0 = 0).Assume that n � 2; we will derive a recurrence relation for theDn, by divid-ing all derangements on n elements into two categories, and then countingthe number in each category. Let � be a derangement on the n elements1, 2, : : : , n. Let k = �(1), so k 6= 1 since � is a derangement. There aretwo cases: �(k) = 1 or �(k) 6= 1. Let An be the number of derangementsfor which �(k) = 1, and let Bn be the number of derangements for which�(k) 6= 1.
If �(k) = 1, then � swaps the elements 1 and k, leavingwhat � does tothe remaining elements 2, 3, : : : , k�1, k+1, : : : , n to be considered. Since� re-arranges these elements, it too acts as a derangement on these n � 2
elements, of which there are Dn�2. Going back, there are n � 1 possiblevalues for k, as 1 is omitted, so An = (n� 1)Dn�2.
Copyright c 1999 CanadianMathematical Society
300
If �(k) 6= 1, then let � be the permutation which swaps 1 and k, andleaves everything else �xed. Recall that for maps f and g, f � g denotes thecomposition of the two maps; that is, (f � g)(x) = f(g(x)). In this case, iff and g are permutations, then f � g is the permutation which arises fromapplying g, then applying f . Consider the permutation � � �. We see that(� � �)(1) = �(�(1)) = �(k) 6= 1, (� � �)(k) = �(�(k)) = �(1) = k,and for all other elements i, (� � �)(i) = �(�(i)) = �(i) 6= i, since �is a derangement. Thus, � � � is a permutation which �xes the element k,and which deranges all n� 1 others (this is why we composed � with �), ofwhich there are Dn�1. Again, there are n� 1 possible values of k (again, 1is omitted), so Bn = (n� 1)Dn�1. Therefore,
Dn = An +Bn
= (n� 1)Dn�2 + (n� 1)Dn�1
= (n� 1)(Dn�1 +Dn�2) .
Problems.
3. Show that in general, for permutations � and �, � � � 6= � � �. Canyou determine when � � � = � � �?
4. For a positive integer k, let �k denote the permutation � composedwith itself k times; that is,
�k = � � � � � � � � �| {z }k
.
Prove that for any permutation �, there exists a positive integer k suchthat �k = 1. Moreover, if � is a permutation on n elements, then�n! = 1. Here, 1 stands for the identity permutation, the permutationwhich takes every element to itself.
5. Classify all permutations � on n elements such that � � � = �2 = 1.
6. Prove that for any permutation �, there exist unique permutations �and such that � � � = � � = 1. Problem 5 shows that � = ispossible; is this true in general? Hint: Apply Problem 4!
The next few terms in the sequence are thenD3 = 2,D4 = 9,D5 = 44,etc. We can use this recurrence to derive an explicit formula for Dn. By therelation,
Dn � nDn�1 = �Dn�1 + (n� 1)Dn�2
= � (Dn�1 � (n� 1)Dn�2)
= (�1)2 (Dn�2 � (n� 2)Dn�3)
= � � �= (�1)n�2(D2 � 2D1) = (�1)n .
301
Hence,
Dn
n!=
Dn�1
(n� 1)!+
(�1)n
n!
=Dn�2
(n� 2)!+
(�1)n�1
(n� 1)!+
(�1)n
n!= � � �=
1
0!�
1
1!+
1
2!� � � �+ (�1)n
1
n!
===) Dn = n!
�1
0!�
1
1!+
1
2!� � � �+ (�1)n
1
n!
�.
Remember this expression; as mentioned above, it comes up a lot! Wenote in passing that
1
e=
1Xk=0
(�1)k1
k!=
1
0!�
1
1!+
1
2!� � � � ,
so that
Dn �n!
e
for large n. We also note that it is possible to derive the formula for Dn
using the Principle of Inclusion-Exclusion. We �nish o� derangements withtwo quick problems.
Problem. Let n be a positive integer. Show that
nXk=0
�n
k
�Dk = n! .
Solution. It looks as if we might have to plug away at some algebra,but a combinatorial approach is much more natural and painless. Note thatthe RHS, n!, is simply the number of permutation on n objects, and this isa cue. What is the number of permutations which derange k elements, oralternatively, which �x n� k elements? We �rst choose n�k elements, andderange the rest, which there are Dk ways of doing, for a total of�
n
n� k
�Dk =
�n
k
�Dk .
The result follows from summing over k (since every permutation derangesk elements for some k).
In general, ifx0, x1, x2, : : : and y0, y1, y2, : : : are sequences satisfying
nXk=0
�n
k
�xk = yn ,
302
then it is possible to recover fxng from fyng, via
xn =
nXk=0
(�1)n�k�n
k
�yk .
Indeed, for yn = n!, we obtain that xn = Dn, since
nXk=0
(�1)n�k�n
k
�k! =
nXk=0
(�1)n�kn!
(n� k)!
= n!
�1
0!�
1
1!+
1
2!� � � �+ (�1)n
1
n!
�= Dn .
Problem. Let n be a positive integer. Prove that
nXk=0
k
�n
k
�Dk = (n� 1) � n! .
Solution. Left as an exercise for the reader. (Follow the same type ofreasoning as the previous problem.) Hint: Combinatorially, what does theLHS represent?
We now draw our attention to the Stirling numbers, in particular thoseof the second kind (as opposed to the �rst kind for those in suspense). Thesearise in the following situation: Suppose that we have n distinguishable ballsto distribute among k indistinguishable boxes, and no box can be empty.How many such distributions are there? First, consider the case where theboxes are distinguishable.
If the boxes were allowed to be empty, there would be kn such dis-tributions; assign a box to each ball. Following the Principle of Inclusion-Exclusion, we subtract the number of distributions with at least 1 box empty,which is �
k
k � 1
�(k� 1)n ,
since we must choose out of k � 1 boxes for each ball. We then add thenumber of distributions with at least 2 boxes empty, which is�
k
k � 2
�(k� 2)n ,
and so on. Hence, the total number of distributions is
nXi=0
(�1)i�k
i
�(k� i)n .
303
Then, for the case where the boxes are indistinguishable, we may \removethe labels" of the boxes, and divide by a factor of k!, to obtain the numberof distributions of the original problem:
S(n; k) :=1
k!
nXi=0
(�1)i�k
i
�(k� i)n .
These are the Stirling numbers of the second kind. We list the �rst few here:
n S(n; k); k = 1; 2; : : : ; n
1 1
2 1 1
3 1 3 1
4 1 7 6 1
5 1 15 25 10 1
An important property of these Stirling numbers, one that may be no-ticeable in the table, is the following:
S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .
This rule easily allows us to generate further rows in the table. We stressthat although we derived a formula for S(n; k) above, it is in general bet-ter to consider the combinatorial signi�cance of this number. So, we give acombinatorial proof here, and leave an algebraic proof for the reader.
Assume that the balls are labelled 1 through n (recall that they are dis-tinguishable). Consider balln. Either it is in a box all by itself, or with others.For the �rst case, the number of distributions is simplyS(n � 1; k � 1), as we can add one box containing ball n to a distributionof n� 1 balls among k � 1 boxes. For the second case, temporarily removeball n. This leaves n � 1 balls among k non-empty boxes, of which thereare S(n� 1; k) distributions. We can then add ball n to any of the k boxes,giving rise to k distinct distributions. Hence,
S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .
Problem. Show that
S(n; k) =X
1a12a2 � � � kak ,
where the sum is taken over all�n�1k�1
�decompositions of n� k into k non-
negative integers a1, a2, : : : , ak: a1+a2+� � �+ak = n�k. (A decompositionof a non-negative integer N into m parts is a sequence of m non-negativeintegers, the sum of which is N . So, 3 has 4 di�erent decompositions into 2parts: 3 = 0 + 3 = 1 + 2 = 2 + 1 = 3 + 0.)
Solution. The table above makes this virtually obvious, but we will eshout the details. Draw in arrows in the table, joining entries in consecutive
304
rows. An arrow from S(n � 1; k � 1) has weight 1 and an arrow fromS(n � 1; k) has weight k. By virtue of the identity proved above, anentry in the table is equal to the sum of the weights of the paths leadingto it, where the weight of a path is simply the product of the weights of thearrows in it (think of this as a tweaked Pascal's Triangle).
There are�n�1k�1
�paths from entry S(1;1) to S(n; k), and the weight of a
path is precisely the term in the given sum. The result follows from summingover all paths, which clearly gives the given expression.
Problem. Show that
S(n+ 1; k) =
nXi=0
�n+ 1
i
�S(i; k � 1) .
Solution. In a distribution of n+1 balls among k boxes, select one boxas \blue". Let i be the number of balls in the blue box, so 1 � i � n + 1.From the n+1 balls, we may choose i to be in the blue box, leavingn�i+1
to be distributed among k� 1 boxes. There are�n+ 1
i
�S(n� i+ 1; k� 1) =
�n+ 1
n� i+ 1
�S(n� i+ 1; k� 1)
such distributions in this case. The result follows from summing over i.
Problems.
1. Let n be a positive integer. Prove that
nXk=0
k
�n
k
�Dk = (n� 1) � n! .
2. Here is an examplewhich illustrates the Principle of Inclusion-Exclusion:Let A, B, and C be subsets of a set S. Let A denote the complementof A. Prove that
jA\B\Cj = jSj�jAj�jBj�jCj+jA\Bj+jA\Cj+jB\Cj�jA\B\Cj .
For example, let S be the set of all distributions of n distinguishableballs among 3 distinguishable boxes, letA be the subset of distributionswith box 1 empty, and so on. ThenA is the subset of distributions withbox 1 containing at least one ball, and so on, so A\B\C is the subsetof distributions with all three boxes containing at least one ball. Nowdetermine what the formula above turns into. This is a useful formula,because the terms such as jAj and jA \ Bj are easy to compute. Thisformula also generalizes to any number of subsets.
3. By using the derived formula for S(n; k), show that
S(n; k) = S(n� 1; k� 1) + kS(n� 1; k) .
305
4. Let n and k be positive integers, with k < n. Verify that there are�n�1k�1
�decompositions of n� k into k parts. If we restrict the parts to
be positive integers, then how many decompositions are there of n intotal?
5. De�ne a sequence of polynomials fpn(x)g as follows: p1(x) = p(x)
is given, and pn+1(x) = xp0n(x). Find a closed formula for pn(x), interms of p(x) and n.
6. (a) Let n be a positive integer. Prove that the following is an identityin x:
xn =
nXk=0
k!S(n;k)
�x
k
�.
(b) Prove that
1n + 2n + � � �+mn =
nXk=0
k!S(n;k)
�m+ 1
k+ 1
�.
Naoki SatoDepartment of Mathematics
Yale UniversityNew HavenConnecticut
306
PROBLEMS
Problem proposals and solutions should be sent to Bruce Shawyer, Departmentof Mathematics and Statistics, Memorial University of Newfoundland, St. John's,
Newfoundland, Canada. A1C 5S7. Proposals should be accompanied by a solution,
together with references and other insights which are likely to be of help to the edi-tor. When a submission is submitted without a solution, the proposer must include
su�cient information on why a solution is likely. An asterisk (?) after a numberindicates that a problem was submitted without a solution.
In particular, original problems are solicited. However, other interesting prob-
lems may also be acceptable provided that they are not too well known, and refer-
ences are given as to their provenance. Ordinarily, if the originator of a problem canbe located, it should not be submitted without the originator's permission.
To facilitate their consideration, please send your proposals and solutions
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 January 2000. They may also be sent by email to
[email protected]. (It would be appreciated if email proposals and solu-tions 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. Please note that we do not acceptsubmissions sent by FAX.
2446. Correction: Proposed by Catherine Shevlin, Wallsend upon
Tyne, England.A sequence of integers, fang with a1 > 0, is de�ned by
an+1 =
8>><>>:
an2
if a(n) � 0 (mod 4),3an + 1 if a(n) � 1 (mod 4),2an � 1 if a(n) � 2 (mod 4),an+1
4if a(n) � 3 (mod 4).
Prove that there is an integer m such that am = 1.
(Compare OQ.117 in OCTOGON, vol 5, No. 2, p. 108.)
2451. Proposed by Michael Lambrou, University of Crete, Crete,
Greece.Construct an in�nite sequence, fAng, of in�nite subsets of N with the
following properties:
(a) the intersection of any two distinct sets An and Am is a singleton;
(b) the singleton in (a) is a di�erent one if at least one of the distinct setsAn, Am, is changed (so the new pair is again distinct);
(c) every natural number is the intersection of (exactly) one pair of distinctsets as in (a).
307
2452. Proposed by Antal E. Fekete, Memorial University of New-
foundland, St. John's, Newfoundland.Establish the following equalities:
(a)
1Xn=1
(2n+ 1)2
(2n+ 1)!=
1Xn=1
(2n+ 2)2
(2n+ 2)!.
(b)1Xn=1
(�1)n(n+ 1)3
(n+ 1)!=
1Xn=1
(�1)n(n+ 1)4
(n+ 1)!.
(c)
1Xn=1
(�1)n(n+ 1)6
(n+ 1)!=
1Xn=1
(�1)n(n+ 1)7
(n+ 1)!.
2453. Proposed by Antal E. Fekete, Memorial University of New-
foundland, St. John's, Newfoundland.Establish the following equalities:
(a)
1Xn=1
(�1)n(2n+ 1)3
(2n+ 1)!= �3
1Xn=1
(�1)n1
(2n+ 1)!.
(b)
1Xn=1
(�1)n(2n)3
(2n)!= �3
1Xn=1
(�1)n1
(n+ 1)!.
(c)
1Xn=1
(�1)n(2n+ 1)2
(2n+ 1)!
!2
+
1Xn=1
(�1)n(2n)2
(2n)!
!2
= 9 .
2454. Proposed by Gerry Leversha, St. Paul's School, London, Eng-
land.
Three circles intersect each other orthogonally at pairs of points A andA0, B and B0, and C and C0. Prove that the circumcircles of 4ABC and4AB0C0 touch at A.
2455. Proposed by Gerry Leversha, St. Paul's School, London, Eng-
land.
Three equal circles, centred at A, B and C intersect at a common pointP . The other intersection points are L (not on circle centre A), M (noton circle centre B), and N (not on circle centre C). Suppose that Q is thecentroid of 4LMN , that R is the centroid of 4ABC, and that S is thecircumcentre of4LMN .
(a) Show that P , Q, R and S are collinear.
(b) Establish how they are distributed on the line.
2456. Proposed by Gerry Leversha, St. Paul's School, London, Eng-
land.
Two circles intersect orthogonally at P . A third circle touches them atQ andR. LetX be any point on this third circle. Prove that the circumcirclesof 4XPQ and4XPR intersect at 45�.
308
2457. Proposed by Gerry Leversha, St. Paul's School, London, Eng-
land.
In quadrilateral ABCD, we have \A + \B = 2� < 180�, andBC = AD. Construct isosceles triangles DCI, ACJ and DBK, where I,J andK are on the other side of CD from A, such that \ICD = \IDC =
\JAC = \JCA = \KDB = \KBD = �.
(a) Show that I, J andK are collinear.
(b) Establish how they are distributed on the line.
2458. Proposed by Nikolaos Dergiades, Thessaloniki, Greece.Let ABCD be a quadrilateral inscribed in the circle centre O, radius
R, and letE be the point of intersection of the diagonalsAC andBD. Let Pbe any point on the line segment OE and letK, L,M , N be the projectionsof P on AB, BC, CD, DA respectively.
Prove that the linesKL, MN , AC are either parallel or concurrent.
2459. Proposed by Vedula N. Murty, Visakhapatnam, India, modi-
�ed by the editors.Let P be a point on the curve whose equation is y = x2. Suppose that
the normal to the curve at P meets the curve again at Q. Determine theminimal length of the line segment PQ.
2460. Proposed by V�aclav Kone �cn �y, Ferris State University, Big
Rapids, Michigan, USA.
Let y(x) =3
qx+
px2 � 1 +
3
qx�
px2 � 1 for 0 � x � 1.
(a) Show that y(x) is real valued.
(b) Find an in�nite sequence fxng1n=0 such that y (xn) can be expressed interms of square roots only.
2461. Proposed byMohammed Aassila, CRM,Universit �e deMontr �eal,
Montr �eal, Qu �ebec.Suppose that x0, x1, : : : , xn are integers which satisfy x0 > x1 >
: : : > xn. Let
F (x) =
nXk=0
akxn�k , ak 2 R , a0 = 1 .
Prove that at least one of the numbers jF (xk)j, (k = 0, 1, : : : , n) is
greater thann!
2n.
2462. Proposed by Vedula N. Murty, Visakhapatnam, India.If the angles, A, B, C of4ABC satisfy
cosA sinA
2= sin
B
2sin
C
2,
prove that 4ABC is isosceles.
309
SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to
consider for publication new solutions or new insights on past problems.
2324. [1998: 109, 1999: 50] Proposed by Joaqu��n G �omez Rey, IES
Luis Bu ~nuel, Alcorc �on, Madrid, Spain.
Find the exact value of1Xn=1
1
un, where un is given by the recurrence
un = n! +
�n� 1
n
�un�1 ;
with the initial condition u1 = 2.
Solution by Charles R. Diminnie, Angelo State University, San Angelo,
TX, USA; Walther Janous, Ursulinengymnasium, Innsbruck, Austria; and
Edward T.H. Wang, Wilfrid Laurier University, Waterloo, Ontario.
We �rst show that un = n! + (n� 1)!. This is true for n = 1, sinceu1 = 2. Suppose that un = n! + (n� 1)! for some n � 1. Then
un+1 = (n+ 1)! +n
n+ 1(n! + (n� 1)!)
= (n+ 1)! +n
n+ 1(n� 1)!(n+ 1)
= (n+ 1)! + n! ,
completing the induction.
Hence, for allm � 1, we have
mXn=1
1
un=
mXn=1
1
n! + (n� 1)!=
mXn=1
1
(n� 1)!(n+ 1)
=
mXn=1
n+ 1� 1
(n+ 1)!=
mXn=1
�1
n!�
1
(n+ 1)!
�= 1�
1
(m+ 1)!,
from which it follows that
1Xn=1
1
un= lim
m!1
mXn=1
1
un= 1.
Also solved by the proposer.
2339. [1998: 234] Proposed by Toshio Seimiya, Kawasaki, Japan.
A rhombus ABCD has incircle �, and � touches AB at T . A tangentto � meets sides AB, AD at P , S respectively, and the line PS meets BC,CD at Q, R respectively. Prove that
310
(a)1
PQ+
1
RS=
1
BT,
and
(b)1
PS�
1
QR=
1
AT:
Solution by Michael Lambrou, University of Crete, Crete, Greece.
If PS meets � at U we show, slightly generalizing (and correcting) thesituation, that, depending on the position of U , we have
1
PQ�
1
RS= �
1
BT,
1
PS�
1
QR= �
1
AT,
with an appropriate choice of � each time.
Using as coordinate axes the two (perpendicular) diagonals, meeting atO say, we may assume that A, B, C, D have coordinates A(a; 0), B(0; b),C(�a; 0), D(0;�b) respectively, where a, b > 0: The radius of � is OTwhich, being perpendicular to AB is the altitude of OAB. Writing OT = h
we have h � AB = OA � OB = ab, and the coordinates of U are of the form(h cos �; h sin �). As PS ? OU , the equation of PS is clearly
y sin� = �x cos � + h .
Line AB is x
a+ y
b= 1, so the coordinates of P , being on AB and PS, are
easily seen to be
(xP ; yP ) =1
b sin � � a cos �
�a(b sin� � h); b(h� a cos �)
�. (1)
Similarly (or quicker, by replacing a by �a) we �nd the coordinates of Q as
(xQ; yQ) =1
b sin� + a cos �
�� a(b sin� � h); b(h+ a cos �)
�.
(2)
Assume now that the position of U is such that PS cuts, say, AB andBC internally (the rest of the cases follow by a trivial adaptation of ourargument here) and so the coordinates of P are both positive [and Q has anegative x{coordinate and a positive y{coordinate]. By (1) we haveb sin� > h > a cos �. Thus
PQ =
q(xP � xQ)2 + (yP � yQ)2 =
2ab(b sin� � h)
(b sin�)2 � (a cos �)2.
Similarly
RS =2ab(b sin� + h)
(b sin�)2 � (a cos �)2,
311
and so
1
RS�
1
PQ=
(b sin �)2 � (a cos �)2
2ab
�1
b sin � + h�
1
b sin � � h
�
= �(b sin�)2 � (a cos �)2
2ab�
2h
(b sin �)2 � h2.
Writing h2 = a2b2
(AB)2= a2b2
a2+b2, this simpli�es to� (a2+b2)h
ab3which equals� 1
BT
as BT � AB = OB2 = b2, as required.
The proof of 1
PS� 1
QR= � 1
ATis similar and routine.
Also solved by CHRISTOPHER J. BRADLEY, Clifton College, Bristol, UK.
Bradley and Lambrou were the only readers who recognized that the problem was incor-
rect as stated. There were eleven partial solutions.
2346. [1998: 236] Proposed by Juan-Bosco Romero M�arquez, Uni-
versidad de Valladolid, Valladolid, Spain.
The angles of4ABC satisfy A > B � C. Suppose that H is the footof the perpendicular from A to BC, that D is the foot of the perpendicularfrom H to AB, that E is the foot of the perpendicular from H to AC, thatP is the foot of the perpendicular from D to BC, and that Q is the foot ofthe perpendicular from E to AB.
Prove that A is acute, right or obtuse according as AH �DP �EQ ispositive, zero or negative.
Solution by Nikolaos Dergiades, Thessaloniki, Greece.
As noted by the proposer, this is a generalization of a problem he pro-posed in College Mathematics Journal 28, no. 2, March 1997, 145-6.
Let S = AH �DP � EQ. Then, since \DHA = \PDH = \B, wehave
DP = DH cosB = AH cos2B .
Similarly EQ = AH cos2C, so,
S = AH � AH cos2B � AH cos2C = AH(1� cos2B � cos2C)
= AH(sin2B � cos2C) = AH
�1� cos 2B
2�
1 + cos 2C
2
�= �AH cos(B + C) cos(B �C) ,
or S = AH cosA cos(B � C), where cos(B �C) > 0, and hence
A is acute if S > 0 , A is right if S = 0 , A is obtuse if S < 0 .
Also solved by FRANCISCO BELLOT ROSADO, I.B. Emilio Ferrari, and MAR �IA
ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; CHRISTOPHER
312
J. BRADLEY, Clifton College, Bristol, UK; GORAN CONAR, student, Gymnasium Vara�zdin,
Vara�zdin, Croatia; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; VJEKOSLAV
KOVA �C, student, University of Zagreb, Croatia; MICHAEL LAMBROU, University of Crete,
Crete, Greece; GERRY LEVERSHA, St. Paul's School, London, England; TOSHIO SEIMIYA, Ka-
wasaki, Japan; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the
proposer.
2348. [1998: 236] Proposed by D.J. Smeenk, Zaltbommel, the Neth-
erlands.Without the use of trigonometrical formulae, prove that
sin (54�) = 1
2+ sin (18�) :
I. Solution by Richard I. Hess, Rancho Palos Verdes, California, USA.
From Figure 1 below where we assume that AC = 1 we have BC = sin54�,AM = MC = MB = BN = 1
2, \MBN = 36�, \MBQ = \QBN =
18�. ThusMQ = 1
2sin 18�.
Now \CMN = 36� which implies CN = MN = 2MQ = sin18�.Since BC = BN + CN we obtain sin54� = 1
2+ sin18�.
A
BC N
M
Q
54�
54�
72�
72�
36�
36�
18�
18�
Figure 1.
II. Solution by Nikolaos Dergiades, Thessaloniki, Greece.On the circle centred at O with radius r having a diameter AD we take thepoints B and C such that arcAB = arcBC = 36� (see Figure 2 below).Then arcCD = 108�, so that
AB = BC = 2r sin 18� , CD = 2r sin54� .
Since \BOA = \CDA, we see that BO k CD. Choosing E on CD suchthat BE k AD, we see that the quadrilateral BEDO is a rhombus; thusED = r. Since \CEB = \CDA = 36�, we have \CBA = 144�,\OBA = \OBC = 72�, and \EBO = 36�. It then follows that\CBE = 36� and CE = CB = 2r sin 18�. Thus
2r sin 54� = CD = CE + ED = 2r sin 18� + r ,
or sin54� =1
2+ sin18� .
313
AO
D
B
C
E
Figure 2.
III. Solution by Jeremy Young, student, Nottingham High School, Eng-
land.
Set x = sin18� and set y = sin54�. First consider the triangle inFigure 3 below, where we set AC = 1. Then BC = x and CD = 1. ThusAB =
p1� x2 and AD =
p1� x2 + (1 + x)2 =
p2(1 + x). Then
sin54� = y =1+ xp2(1 + x)
=) 2y2� 1 = x .
A
B C D
1p1� x2
x 1
p2(1 + x)
18�!
36�
108�
72�
36�
E
F G H
K
L
1
p1� y2
y 1�y
1
18�
18�54
�
72�
72�
36�
72�
18�
Figure 3. Figure 4.
In Figure 4 above where EG = 1, we have FG = y, EK = 1,GH = 1� y, and EF =
p1� y2 = KH. Then
LG = x =1
2
p(1� y)2 + 1� y2 =
1
2
p2� 2y .
which yields 1� 2x2 = y. Adding this to the previous result (2y2 � 1 = x)gives
2(y+ x)(y� x) = y + x ,
2(y� x) = 1 , since x, y > 0 =) y + x 6= 0,
y = x+1
2.
Also solved by SAM BAETHGE, Nordheim, Texas, USA; FRANCISCO BELLOT ROSADO,
I.B. Emilio Ferrari, and MAR �IA ASCENSI �ON L �OPEZ CHAMORRO, I.B. Leopoldo Cano, Val-
ladolid, Spain; M. BENITO and E. FERN �ANDEZ, Logro ~no, Spain; CHRISTOPHER J. BRADLEY,
314
Clifton College, Bristol, UK; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX,
USA; DAVID DOSTER, Choate Rosemary Hall, Wallingford, Connecticut, USA; DIANE and ROY
DOWLING, University ofManitoba, Winnipeg, Manitoba; RICHARDEDEN, student, Ateneo de
Manila University, Philippines; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria;
ANGEL JOVAL ROQUET, Spain; GEOFFREY A. KANDALL, Hamden, CT; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA; VJEKOSLAV KOVA �C, student, University of
Zagreb, Croatia; MICHAEL LAMBROU, University of Crete, Crete, Greece; KEE-WAI LAU, Hong
Kong; GERRY LEVERSHA, St. Paul's School, London, England; VEDULA N. MURTY, Dover,
PA, USA; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; BOB PRIELIPP, University
of Wisconsin{Oshkosh, Wisconsin, USA; TOSHIO SEIMIYA, Kawasaki, Japan; J. SUCK, Essen,
Germany; PANOS E. TSAOUSSOGLOU, Athens, Greece; PAUL YIU, Florida Atlantic University,
Boca Raton, Florida, USA; and the proposer.
2351. [1998: 302] Proposed by Paul Yiu, Florida Atlantic University,
Boca Raton, Florida, USA.
A triangle with integer sides is called Heronian if its area is an integer.
Does there exist a Heronian triangle whose sides are the arithmetic,geometric and harmonic means of two positive integers?
Solution by Michael Lambrou, University of Crete, Crete, Greece.
We show, slightly generalizing the given situation, that no triangle withsides a,
pac, c, with a, c 2 N can have integral area. Indeed, if the area
� 2 N we would have by Heron's formula:
�2 =1
16(a+
pac+ c)(a�
pac + c)(a+
pac� c)(�a+
pac+ c)
=1
16
�4a2c2 � (a2 + c2 � ac)2
�.
If (a; c) = t so that a = pt, c = qt for some p, q 2 N with (p; q) = 1 weobtain
4�
t2=p4p2q2� (p2 + q2� pq)2 .
But the left hand side is rational; so 4p2q2 � (p2 + q2 � pq)2 must be aninteger perfect square, say T 2. But this is impossible because p2 + q2 � pq
is odd (as p, q are not both even) and so
T 2 = 4p2q2 � (p2 + q2 � pq)2 � 0� 1 � 3 (mod 4) ,
giving a contradiction, as no square is congruent to 3 modulo 4. This com-pletes the proof that � =2 N.
Also solved by DUANE BROLINE, Eatsern Illinois University, Charleston, Illinois.
USA; THEODORE CHRONIS, student, Aristotle University of Thessaloniki, Greece; RICHARD
I. HESS, Rancho Palos Verdes, California, USA; JEREMY YOUNG, student, Nottingham High
School, Nottingham, UK; and the proposer.
315
2352. [1998: 302] Proposed by Christopher J. Bradley, Clifton Col-
lege, Bristol, UK.
Determine the shape of4ABC if
cosA cosB cos(A�B) + cosB cosC cos(B �C)
+ cosC cosA cos(C � A) + 2cosA cosB cosC = 1 .
Solution by Nikolaos Dergiades, Thessaloniki, Greece.Since cosA cosB � sinA sinB = cos(A + B) = � cosC, we have
(cosA cosB + cosC)2=�1� cos2A
� �1� cos2B
�, or
cos2A+ cos2B + cos2C + 2 cosA cosB cosC = 1 . (1)
Also,
cos(A+ B) cos(A�B) = cos2A cos2B � sin2A sin2B
= cos2A cos2B ��1� cos2A
� �1� cos2B
�= cos2A+ cos2B � 1 . (2)
Using (2), we have
cosA cosB cos(A�B) = 1
2(cos(A� B) + cos(A+ B)) cos(A� B)
= 1
2
�cos2(A�B) + cos2A+ cos2B � 1
�= 1
2
�cos2A+ cos2B � sin2(A�B)
�.
Hence, the given equality is equivalent to
cos2A+ cos2B + cos2C + 2 cosA cosB cosC
�1
2
�sin2(A� B) + sin2(B � C) + sin2(C � A)
�= 1 ,
which, in view of (1), becomes
sin2(A� B) + sin2(B � C) + sin2(C � A) = 0 .
Hence, A = B = C, which means that 4ABC is equilateral.
Also solved by MICHEL BATAILLE, Rouen, France; THEODORE CHRONIS, student,
Aristotle University of Thessaloniki, Greece; C. FESTRAETS-HAMOIR, Brussels, Belgium;
WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; JUN-HUA HUANG, the Mid-
dle School Attached To Hunan Normal University, Changsha, China; MICHAEL LAMBROU,
University of Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; GERRY LEVERSHA, St. Paul's
School, London, England; VEDULA N. MURTY, Visakhapatnam, India; HEINZ-J �URGEN
SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan; D.J. SMEENK, Zaltbommel,
the Netherlands; ARAMTANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA,
USA; PARAYIOU THEOKLITOS, Limassol, Cyprus; PANOS E. TSAOUSSOGLOU, Athens, Greece;
PAUL YIU, Florida Atlantic University, Boca Raton, Florida, USA; JEREMY YOUNG, student,
Nottingham High School, Nottingham, UK; and the proposer. There was also one incorrect
solution.
316
Most solvers showed that the given equality is equivalent to eitherP
cos2(A�B) = 3,
or toP
cos 2(A � B) = 3, both of which are clearly equivalent toP
sin2(A � B) = 0,
obtained in the solution above.
Four solvers derived the equalityQcos(A � B) = 1, from which the conclusion also
follows immediately. The solution given above is self-contained, and does not use any known
identity (for example,P
cos(2A) = �1�4QcosA, which was used by a few solvers) beyond
the elementary formula transforming product into sum. Lambrou obtained the slightly stronger
result that �XcosA cosB cos(A�B)
�+ 2 cosA cosB cosC � 1 ,
with equality holding if and only if4ABC is equilateral.
2353. [1998: 302] Proposed by Christopher J. Bradley, Clifton Col-
lege, Bristol, UK.
Determine the shape of4ABC if
sinA sinB sin(A�B) + sinB sinC sin(B � C)
+ sinC sinA sin(C � A) = 0 .
Solutionby Gerry Leversha, St. Paul's School, London, England (slightlymodi�ed by the editor).
Since
sinA sinB sin(A� B)
= sin2A sinB cosB � sin2B sinA cosA
= 1
4((1� cos(2A)) sin(2B)� (1� cos(2B)) sin(2A))
= 1
4((sin(2B)� sin(2A)) + sin(2(A� B))) ,
the given equality is equivalent toXsin(2(A� B)) = 0 . (1)
Using elementary formulae transforming sums into products, we have
sin(2(A�B)) + sin(2(B� C)) = 2 sin(A�C) cos(A+ C � 2B) ,
and thus
sin(2(A� B)) + sin(2(B � C)) + sin(2(C � A))
= 2 sin(A� C) cos(A+ C � 2B) + 2 sin(C � A) cos(C �A)
= 2 sin(C �A) (cos(C �A)� cos(A+ C � 2B))
= �4 sin(A� B) sin(B � C) sin(C � A) . (2)
From (1) and (2), we see that the given equality is equivalent to
sin(A�B) sin(B �C) sin(C � A) = 0 ,
317
which holds if and only if at least two of A, B, C are equal to each other.Therefore, the given equality holds if and only if4ABC is isosceles.
Also solved by MICHEL BATAILLE, Rouen, France; THEODORE CHRONIS, student,
Aristotle University of Thessaloniki, Greece; NIKOLAOS DERGIADES, Thessaloniki, Greece;
C. FESTRAETS-HAMOIR, Brussels, Belgium; PETER HURTHIG, Columbia College, Burnaby,
BC; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; JUN-HUA HUANG, the
Middle School Attached To Hunan Normal University, Changsha, China; V �ACLAV KONE �CN �Y,
Ferris State University, Big Rapids, Michigan, USA; MICHAEL LAMBROU, University of
Crete, Crete, Greece; KEE-WAI LAU, Hong Kong; VEDULA N. MURTY, Visakhapatnam,
India; HEINZ-J �URGEN SEIFFERT, Berlin, Germany; TOSHIO SEIMIYA, Kawasaki, Japan;
ARAM TANGBOONDOUANGJIT, Carnegie Mellon University, Pittsburgh, PA, USA; PARAYIOU
THEOKLITOS, Limassol, Cyprus; PAUL YIU, Florida Atlantic University, Boca Raton, Florida,
USA; JEREMY YOUNG, student, Nottingham High School, Nottingham, UK; and the proposer.
There were also two incorrect solutions.
The solution given above is interesting since it shows that the conclusion A = B or
B = C or C = A does not depend on the assumption that A+B +C = �; that is, that A,
B and C are the three angles of a triangle.
2354. [1998: 302] Proposed by Herbert G �ulicher, Westfalische
Wilhelms-Universit�at, M�unster, Germany.In triangle P1P2P3, the line joining Pi�1Pi+1 meets a line �j at the
point Si;j (i; j = 1; 2; 3, all indices taken modulo 3), such that all the pointsSi;j , Pk are distinct, and di�erent from the vertices of the triangle.
1. Prove that if all the points Si;i [note the correction] are non-collinear,then any two of the following conditions imply the third condition:
(a)P1S3;1
S3;2P2
�P2S1;2
S1;3P3
�P3S2;3
S2;1P1
= �1 ;
(b)S1;2S1;1
S1;1S1;3�S2;3S2;2
S2;2S2;1�S3;1S3;3
S3;3S3;2= 1 ;
(c) �1, �2, �3 are either concurrent or parallel.
2. Prove further that (a) and (b) are equivalent if the Si;i are collinear.
Here, AB denotes the signed length of the directed line segment [AB].
Solution by G �unter Pickert, Giessen, Germany.
De�ne Si = Sii and Qi = �iTSi�1Si+1. If �ijjSi�1Si+1 (so that Qi
is at in�nity), then replaceSi+1Qi
QiSi�1by �1.
(a) From Menelaus' Theorem applied to 4Si+1Si�1Pi and the line �i(1 = 1, 2, 3),
Si+1Qi
QiSi�1�Si�1Si�1;i
Si�1;iPi�PiSi+1;i
Si+1;iSi+1
= �1 . (1)
318
De�ne
a =
3Yi=1
PiSi�1;i
Si�1;i+1Pi+1
, b =
3Yi=1
Si;i+1Si
SiSi;i�1, c =
3Yi=1
Si+1Qi
QiSi�1.
The given conditions (a), (b), (c) say (respectively) that a = �1,b = 1, and c = 1. [The value c = 1 is just Ceva's Theorem applied to4S1S2S3, and extended to allow the possibility that one or two of theQi are at in�nity.] From (1) we get
�1 = c �QSi�1Si�1;iQSi+1;iSi+1
�QPiSi+1;iQSi�1;iPi
= c � b � a�1 ,
so that �a = b � c .Two of c, b, �a are equal to 1 if and only if two of (a), (b), (c) hold, inwhich case the third product is 1 and the third condition holds too.
(b) If the Si are collinear thenQi = Si and c = 1; therefore (a) is equivalentto (b).
Also solved by JUN-HUA HUANG, the Middle School Attached To Hunan Normal Uni-
versity, Changsha, China; and the proposer.
Huang's solution is much the same as our featured solution. G �ulicher based his on his
problem 64 in Math. Semesterber. 40 (1992) 91{92.
2355. [1998: 303] Proposed by G.P. Henderson, Garden Hill, Camp-
bellcroft, Ontario.
For j = 1, 2, : : : , m, let Aj be non-collinear points with Aj 6= Aj+1.Translate every even-numbered point by an equal amount to get new pointsA02, A04, : : : , and consider the sequence Bj , where B2i = A02i andB2i�1 = A2i�1. The last member of the new sequence is either Am+1 orA0m+1 according asm is even or odd.
Find a necessary and su�cient condition for the length of the pathB1B2B3 : : : Bm to be greater than the length of the path A1A2A3 : : : Am
for all such non-zero translations.
CRUX 1985 [1994: 250; 1995: 280] provides an example of such a con-�guration. There, m = 2n, the Ai are the vertices of a regular 2n{gon andA2n+1 = A1.
Solution by the proposer.
(Editor's note: In the original submission the paths above were givenas B1B2 � � �Bm+1 and A1A2 � � �Am+1. Since this makes the upper boundson summations simpler we have opted to present the solution with theseendpoints, although logically it makes no di�erence.)
319
We use the same letter for a pointP and a vector�!P . Let the translation
in the problem be�!X . Then
�!B2k =
�!A02k =
�!A2k +
�!X , k = 1, 2, : : : , b(m+ 1)=2c ,
and the lengths of the new segments are����!B2 ��!B1
��� =
����!A02 ��!A1
��� =
����!X +�!A2 �
�!A1
��� ,����!B3 ��!B2
��� =����!A3 �
�!A02
��� =����!X �
��!A3 �
�!A2
���� ,� � � .
The lengths of the paths are
L0 =
mXj=1
�����!Bj+1 ��!Bj
��� =
mXj=1
����!X � (�1)j���!Aj+1 �
�!Aj
����and L =
mXj=1
�����!Aj+1 ��!Aj
��� .Now L0 is to be a minimum at
�!X =
�!0 . If we form the partial derivatives
of L0 with respect to the components of�!X and set
�!X =
�!0 , we get
mXj=1
(�1)j���!Aj+1 �
�!Aj
������!Aj+1 �
�!Aj
��� =�!0 . (1)
The minimum of a sum like L0 does not always occur at a point where thederivatives are zero. However, in this case we will prove that (1) actually isthe required condition.
That is, the sum of unit vectors along the odd segments�!A2 �
�!A1,�!
A4 ��!A3, : : : , is equal to the sum of unit vectors along the even segments.
The lengths of the segments are arbitrary provided they are greater than zero.It is only their directions that matter. In the case of the regular 2n{gon, both
sums are�!0 because each consists of unit vectors parallel to the sides of a
regular n{gon.
Set dj =�����!Aj+1 �
�!Aj
��� > 0 and�!Cj = (�1)j
���!Aj+1 �
�!Aj
�=dj ,
a unit vector parallel to the jth segment. Equation (1) becomes
mXj=1
�!Cj =
�!0 . (2)
The lengths of the paths are L0 =
mXj=1
����!X � dj�!Cj
��� and L =
mXj=1
dj .
320
To prove the necessity of (2), assume L0 � L for all�!X .
Set yj =����!X � dj
�!Cj
��� � dj . ThenmXj=1
yj � 0 . Squaring both sides of
����!X � dj�!Cj
��� = dj + yj and dividing by dj , we get
�!X2 =dj � 2
�!X �
�!Cj = 2yj + y2j=dj
�!X2
mXj=1
(1=dj)� 2�!X �
mXj=1
�!Cj = 2
mXj=1
yj +
mXj=1
(y2j=dj) � 0 .
When we set�!X =
mXj=1
�!Cj
� mXj=1
(1=dj) , this becomes
0@ mXj=1
�!Cj
1A
2
� 0 ,
and we have (2).
Given (2) we are to prove that L0 > L for all�!X 6=
�!0 . We have����!X � dj
�!Cj
��� =����!X � dj
�!Cj
��� ����!Cj��� � �����!X � dj�!Cj
���!Cj
��� ; (3)
L0 �mXj=1
�����!X � dj�!Cj
���!Cj
��� �������mXj=1
��!X � dj
�!Cj
���!Cj
������=
�������!X �
mXj=1
�!Cj �
mXj=1
dj
������ = L .
If L0 = L for some�!X 6=
�!0 , there is equality in (3) for all j. Then�!
X � dj�!Cj = cj
�!Cj , where cj 6= �dj ; thus
�!Cj =
�!X =(cj + dj) , and the
points are collinear.
All of the above is valid in a Euclidean space of any number of dimen-sions.
There were no other solutions submitted.
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
