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Journal title history:
➢ The first 32 issues, from Vol. 1, No. 1 (March 1975) to Vol. 4, No.2 (February 1978) were published under the name EUREKA.
➢ Issues from Vol. 4, No. 3 (March 1978) to Vol. 22, No.8 (December 1996) were published under the name Crux Mathematicorum.
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Mat he m
at ico ru m
ISSN 0705 - 0348
C R U X M A T H E M A T I C O P U M
Vol. io9 No. l
January 1984
Sponsored by Carleton-Qttawa Mathematics Association Mathematique d*Ottawa-Carleton
Public par ie College Algonquin, Ottawa
The assistance of the publisher and the support of the Canadian Ilather.atical Olympiad Committee, the Carleton University Department of Mathematics and Statistics, the University of Ottawa Department of Mathematics, and the endorsement of the Ottawa Valley Education Liaison Council are gratefully acknowledged.
* * ft ft ft -A- ft ft ft ft ft ft ft * & * ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft
* ft ft ft ft ft ft ft ft ft ft ft ft & ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft ft
CRUX J'ATHFtLATICORUX is a problem-solving journal at the senior secondary and university undergraduate levels for those who practise or teach mathematics. lis purpose is primarily educational, but it serves also those who read it for professional, cultural, or recreational reasons.
It is published monthly (except July and August.). The yearly subscription rate for ten issues is $2 2 in Canada, US$2 0 elsewhere. Back issues: $2 each. Bound volumes with index: Vols. 1C2 (combined) and each of Vols. 3-8, $17 in Canada and US$15 elsewhere. Cheques and money orders, payable to CRUX fiiATHEMATICORUM, should be jent to the managing editor.
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* * *
CONTENTS
A 1984 Farrago . . . . .
Why Newton and Leibniz had To Do It
The Puzzle Corner . . . . . . . .
The Olympiad Corner: 51 . . . . .
Problems - Probl&mes . . . . . . .
Solutions . . . . . . . . . . . .
A Magic Circle for 1984 . . . . .
Charles W. Trigg 2 . . . . . . . . 8
. . . . . . . . 8
. M.S. Klamkin 9
. . . . . . . . 18
. . . . . . . . 20
. Kama!a Kumari 34
- 1 -
- 2 -
A 1984 FARRAGO
CHARLES W. TRIGG
1 + 9 - 8 - ̂ 4 = 0 I
Although 4 divides 1984, no man should leap to the conclusion that he needs
to protect himself against the wiles of those of the opposite sex from 19 to 84.
They might feel that such an attitude goes against the grain. Rather, we would
like to ao to our farraqo for nourishment and even the possible yeasty stimulation
gained when either grain or thoughts ferment. Or, you may prefer the productivity
of controlled thouqht.
-1*9 + 8 + /4* = 1
Each digit of 1984 is a power; that is,
1 = l1, 9 = 32, 8 = 2 3s 4 = 2
2.
The sum of the bases is 8, and so is the sum of the exponents.
Also, 1984 = 123 + 162,
- 1 - 9 + 8 + 4^=2
: - / 9 " + 8 - 4 = 2 and 19 + 8 + vTT = 29, representing that non-leapable
date in every leap years 2/29.
l.(-g+8) + 4 = 3
1984 = 4-4-4*31, and 3 + 1 = 4 .
1984 = 26-3i = 1-32-2-31 = 4»2-1-31-2-4, two palindromic arrays.
The sum of the prime factors of 1984 is 43, the 13th odd prime.
1 + 9 + 8 + 4 = 2-11, and i + i = 2.
l.g.8-4 = 4-3-2-3-4.
1 - 9 + 8 + 4 = 4
The reverse of 1984 is 4891 = 67*73, the product of alternate primes; and
73 - 67 = 6, the smallest perfect number.
The overlapping blend of 1984 and its reverse is 19848919 a prime whose rank
in the sequence of primes is 147881, a prime whose rank is 13669, a prime whose
- 3 -
rank is 1614 9 and the litany must sadly end here. The other overlapping blend,
4891984 = i-2»8"305749 in which the only missing digit is 6,
(1 + 9)/(8A) = 5
1984 = i857024/9369 in which all ten digits are presente [Stewart Metchettes Journal of Recreational Mathematics, 13 : 1 (1980-81) 27. ]
1 + 9 - 8 + 4 = 6
?r = 3.1415926.. .1984,.., where the year ends in the 336ist decimal pi ace 9 and
31415926 = 1984 = 3361 (mod 9).
19 - 8 - 4 = 7
1 = -/9" + 8 - 49 -1 + /ST = 8/4, -1 - v^ + 8 = 4.
(1 + 9 - 8)°4 = 8
84 - 19 = 65 = 5*139 so 19 and 84 are terminal terms of arithmetic progressions
with common differences of 5 and 13. The longer ones with d = 59 contains four
primes and two squares. I t is
19 24 2£ 34 39 44 49* 54 5£ 64* 69 74 79_ 84.
In the six-term A.P., with d = 139 the units ' digi ts are d is t inc t , as are the tens'
dicrits. No o or 6 appears in any term of
1£ 3̂2~ 45 58 71 84,
but two of the terms are primes and one is a fifth power.
19 - 8 - /4~ = 9
1/1984 = 0.000504032..., wherein the nonzero digits are in descending order
of magnitude.
l-v /9"+8 + 4 = 10 1 I
1+ 9+ 8+ 4= 22= 4+ 8+ 9+ 1
19 + 98 + 84 + i+l = 242 = 14 + 48 + 89 + 91
198 + 984 + 841 + 419 = 2442 = 914 + 148 + 489 + 891
1984 + 9841 + 8419 + 4198 = 24442 = 8914 + 9148 + 1489 + 4891
19z
1982 984z
. n
1-9 + 8/1+ = 11
84z
8412 41*
4192 14*
9142 148*
89z +
4892 +
91z
8912
19842 + 98412 + 84192 + 41982 = 89142 + 91482 + 14892 + 48912
1 + 9 + 8 / 4 12
Two palindromic arrays for 1984:
i9eQ4 1»2«4®2*32«4«2*1S
I.90.4 = 9-3-2-l-9-9*3»l-3*9-9*l«2-3-9.
1*9 + 8/A" = 13
In a Schram sequence r'Journal of Recreational Mathematics, 14:2 (1981-82) 141],
each succeeding term is generated from its predecessor by taking the absolute
difference between the sum of the squares of the odd digits and the sum of the
squares of the even digits. For example:
198^ 16 35 34 7 49 65 11
Thus 1984 is a charm attached to a bracelet of nine terms9 includinq three squares,
one repunits and three primes. The sum of the loop members is 223, the 48th prime.
The sum of the loop members and the charm is 2207, the 329th prime.
1 + 9 + Q//Z - 14
In the reiterative operation wherein an integer and its reverse are addeds
NQ is the starting integer, and the kth versum, N«9 is the sum resulting from the
7cth addition. If #0 = 19849 then i7l9 a permutation of consecutive digits, is an
antipalindrome in that each digit differs from the corresponding digit of its
reverse. The first palindrome encountered in the sequence is #13.
N0 = 1984 6875 12661
4891 5786 16621
16222151 31344412
15122261 21444313
Ni = 6875 12661 29282 ,.. 31344412 52788725 = Nu
Other antipalindromes in this sequence are
Ns = 105149, Nw = 10474750, ff18 = 1059424960, and N2& = 102618981520.
- 5 -
1*9 + 8 - /4~ = 15
The recursive operation u = u u u u „ with each product reduced n+4 n n+1 n+2 n+3
modulo io, when applied to 1984 produces a 10-digit bracelet with a 2-digit charm attached, namely:
8 — 4
/ \
/ \ 1 - 9 - 8 2
\ / 4 6
\ / 6 — 2
The charm digits and those at the extremities of all but one diameter sum to 10.
1 + 9 + 8 - / 7 = 16
1984 is part of a 1560-digit additive bracelet wherein each element is the units1 digit of the sum of the four preceding digits, namely:
19842 37684 5302D 57462 91808 ... .
The complete bracelet is included in "A Diqital^ Bracelet for 1968", Journal of
Recreational Mathematics , 1 (April 1968) 108-111.
19 - 8/4 = 17
Nonagonal numbers have the form N(n) = n(7n-5)/2. All nonagonal numbers of
the forms
#(1920 ± 768 + 20000k) and #(3795 ± 768 + 20000k), k = 0,1,2,,..,
terminate in 1984, The digits of #(3027) = 32061984 are distinct. Some that
start with 1984 are
#(7530) = 198434325, #(7531) = 198487036,
#(23810) = 1984146825, #(23815) = 1984980250.
l«9*(8/4) = 18 I I
The digits of 1984 can be distributed on the 12 mid-edges of a regular octahedron in two ways (shown on the Schlegel diagrams of Figure 1) so that each digit appears on the 4 edges issuing from each vertex.
Figure 1
In each distribution the orders around the vertices constitute the six cyclic
arrangements of four distinct elements, with the orders around opposite vertices
being opposite, Alsos the digits on the perimeters of the eight faces are the
combinations of the four digits taken three at a times with the same combination
appearing on opposite faces, but in reverse order.
1 + 9«(8/4) = 19
There is but one way to distribute the digits
of 1984 on the faces of a regular octahedron so that
different digits appear on the faces meeting at each
vertex. This distribution is shown on the Schl̂ egel
diagram of Figure 2, the isolated digit being the
one on the hidden face. The same digit appears on
opposite faces. The orders of digits around opposite
faces are opposite.
1 + 9 + 8 + A 20
Figure 2
41+5.446 = 1984 70
4454*4455 = 1984 25 70
44545*44546 = 1984 3015 70
445454*445455 = 1984 297115 70
1*9 + + 4 21
In the following determinants the 2-by»2 center is stationary, while the
198419841984 and its reverse rotate around the perimeter, and the value of each
is given in palindromic formB
- 7 -
1 9
4 1
8 4
9 1
4 1
8 1
9 4
1 4
4 8
1 1
9 4
8 4
8 9
4 1
1 4
9 8
8
9
8
4
9
9
8
8
9
9
8
1
1
9
8
4
L
1
9
8
8
4
1
9
1
4
8
9i
4|
8
9
1 1
|
= - 828
= -2*3®2«3»23
0
= 35e4s5s3 = -
= 2112
= 8*33*8
18
9
1
u |9
' 1
4
8
9
8
4
1
1
9
8
4
4
1
4
8
8
1
4
9
1
1
4
9
4
1
4
1
1 9
9 8
8 4
9 1 1
4 1 f
9 9
8 8
1 41
4 8 1
9 9
8 1
8 4|
8 91
9 1
8 4
9 81
= 3 ® 4 ° 7 ® 4 3
= 84*8
= 4*21B2®4
The circulant
9 + 8 8 4 = 22
1 9 8 4
4 1 9 8
8 4 1 9
9 8 4 1
- (1 + 9 + 8 + 4 )°296 .
-1 + (/9~)!*(8//4") = 23
On pages 134 and 138 of his History of Binary and Other Nondeoimal Numeration9
Tomash Publishers9 1981, Anton Glasser gives the expressions for 1984 in four
computer codess namely in:
Binary coded decimals 1984 = oooi 1001 1000 oioo;
Pure binary, 1984 = 11111000000:
Excess three, 1984 = oioo noo ion oni; and
Biquinary, 1984 = oioooio 1010000 1001000 onoooo.
(-1 + /9)*(8 + 4) = 24
. 8 -
The smallest power of 2 that ends in 198^ is 25k 9 and the next one is 255k .
Since, for all positive integers n%
2n+500 E 2n {md ^ ^
a consequence of the fact that 2 5 0 0- 1 = 9375 = 3*55 (mod 101*), it follows that
2 5 ^ s 255k 2lQ5k 2l55k
9 2 2 0 5 I +
all end in 1984, and they are the only powers of 2 that do.
19 + 8 - A = 25
And how was your year?
2404 Loring Street, San Diego? California 92109.
ft * ft
WHY NEWTON AND LEIBNIZ HAD TO DO IT
The following quotation is taken from N* Piskunovs Differential and Integral
Calculus (translated from the Russian by George Yankovsky), second edition9 Mir
Publishers,, Moscow, 1977, Vol, I, page 14. (The italics are in the text.)
"As we shall see throughout this course, the concept of a variable quantity
is the basic concept of differential and integral calculus. In 'Dialectics of
Mature^ Friedrich Engels wrote: 'The turning point in mathematics was Descartes8
variable magnitude. With that came motion and hence dialectics in mathematics,
and at once% too9 of necessity the differential and integral calculus.8"
Friedrich Engels gives compelling reasons why the calculus had to be in
vented (or discovered, depending on your philosophical point of view). But why
did it have to be done by Gottfried Wilhelm von Leibniz and Sir Isaac Newton?
Because Leibniz was Gottfriedrich and Newton was Engelish, * * *
THE PUZZLE CORNER
Answer to Puzzle No. 49 [1983: 311]:
314 138 198 or ^ 2 9 so E = 2.
62172 52302*
Answer to Puzzle No. 50 [1983: 311]
683 73
49859s with NIECE = 20959 and MAN = 182.
- 0 -
THE OLYMPIAD CORNER: 51
M.S. KLAMKIN
In recent yearss the mathematical preparation of students entering post-
secondary institutions in Canada and the United States has been steadily declining3 and the same thing can be said for the standards used in assessing this prepara
tion,, The elementary and secondary schools are equally to blame for this decline.
The parlous state of our precolleqe mathematical education has finally received
international exposure in a number of recent reports from the following prestig
ious groups:
National Commission on Excellence in Education;
National Science Board's Commission on Precolleae Education in Mathematics,
Science9 and Technology;
Twentieth-Century Fund Task Force on Federal Elementary and Secondary Educa
tion Policy;
National Task Force on Education for Economic Growth;
Carnegie Foundation for the Advancement of Teaching,
Preparation in mathematics and the physical sciences is much more demanding
almost everywhere abroad. As one example, I give below the 1981 matriculation
examination in mathematics for Grade 11 students in Hong Kong (for which I am
grateful to K.P. Shum9 The Chinese University of Hong Kong), It is worth noting
that, for a Hong Kong student to enter col lege, he or she must first complete
Grade 12 (2 years) and then take university examinations which include mathematics
at a still higher level. In contrasts the University of Alberta gives a diagnostic
(only) test to entering students enrolled in calculus. Considering the low level
of this test and the low grades achieved9 it is likely that the vast majority of
these students would be wiped out if they had to pass the following Hong Kong
Grade 11 test. The present-day economic situation and the competition for jobs
may soon result in students being better prepared in the secondary schools for
admittance to postsecondary institutions. If so9 it will not be a moment too
soon!
1981 HONG KONG GRADE 11 MATRICULATION EXAMINATION
Time: 3 hours
1, (a) Let P and Q be nonsingular square matrices of the same order. Simplify
(i) P G C G ' V 1 ) ,
(11) (Q'1P~1)PQ, (2 marks)
- 10
(b) Let M = PQ, where
and Q =
(1) Find a9 bs and o. *-i (11 marks) (ii) Find Q~ and P~ ; hence, or otherwise, find M~
(c) Using the result of (b)(ii), or otherwise, solve the following system of
linear equations:
2x + 2 = 7
Ux + y + 2z = 12
3?/ + z = -3, (3 marks)
28 (a) Prove by mathematical induction that, for any positive integer n%
l 3 + 23 + 33 + 980 + n3 = fn2(n+l)2e (6 marks)
(b) The set of positive integers is grouped as follows:
( 1 ) , ( 2 9 3 ) 9 (*+ s 5 9 6) 9 ( 7 9 8 9 9 , 1 0 ) 9 . . . ,
so that the nth group* G , consists of n consecutive positive integers.
(1) Show that the f i r s t term in Gk is * -.
( i i ) Calculate the sum of a l l the ter^s in G, .
( i i i ) Hence, or otherwise, f ind the sum of al l the terms in the n groups
G\9 ( ? 2 J ^ 3 s • • • » £ , (10 marks)
3, (a) Solve the inequality loq2{ log3( log^x)} > oa (4 marks)
(b) Solve the equation 22x+3 + 7«2X - 1 = o. (5 marks)
(c) Solve the system of equations
Hog2^ - log^i/ = 4
l 1092(^-21/) = 5 . (7 marks)
4, (a) Let /(a?) = x2+px+q and ^(x) = x2-9x±rs where p$<?sr are constants;
and l e t a9B be the roots of fix) = 0 and a+2B* 2a+£ the roots of g(x) = o.
( i ) Find the value of p.
( i i ) When fix) is divided by x-29 the remainder is -10. Find the values
of q and r, (8 marks)
(b) Factorize L 1
i b
z3 h3 . Hence evaluate
2 8
3 27
5 125
(8 marks)
-11 -5, Let / (G) = 4cos2e - 2 s ine cos 0 + 2 s i n 2 e .
(a) Express f (G) In the form A + B cos (26+a)5 where A9 B, and a are
constants and o < a < TT/2. What are the values of these constants? (6 marks)
(b) Find the maximum and minimum values of / ( 8 ) . What are the general values
of e f o r which / ( e ) i s a maximum? (5 marks)
(c) Figure 1 shows the graph of y - /2 cos 26 fo r -TT < e < IT,
( i ) Sketch the graph of y = /2cos (20+~) on Figure 1 fo r -TT < e £ TT.
( i i ) Sketch the graph of y = f ( 0 ) on Figure l f o r -TT < e < TT8 (5 marks)
y
43
- 3
- 4
Figure 1
6, Given:
the l i n e (L)
the c i r c l e (C): x2 t yz = 59
i
2/ = war + a,
2 + „2 =
2 2 and the e l l i p s e (E): | - + *^
(a) I f the l i n e (L) touches the c i r c l e (C) , f i n d the r e l a t i o n between c and m,
(LI marks)
- 12 -(b) I f the l ine (L) touches the el l ipse (£) , f ind the relation between
a and m. (4 marks)
(c) If the l ine (£) touches both the c i rc le (C) and the e l l i p se (£ ) , de
termine the values of m. Hence find the equations of all the common tangents of
the c i r c l e (C) and the e l l i p se (# ) . (8 marks)
7S (a) Given:
A = ( l 9 2, 3S 49 5} and B = (3 5 49 5S 6} .
Let X = ix: x = 100a + loh + o9 where as2?sc e 4̂} s (In other words, X consists
of al l th ree -d ig i t numbers where each d ig i t is chosen from the se t A,)
Let I = {y\ y - 100a + 10b + e$ where a9h9c e B}.
Find the number of elements in each of the following se t s :
(1) x,
( 1 1 ) *.
(111) X n I ,
(1v) X u T9
(v) x - J (i.e., the set of elements in J but not in r). (8 marks)
(b) (1) Let P = (a, 3, es d9 e} s
Q = id, e9 f, g}, *
P = (a, b9 /, #9 h, i] .
Find Pu(6?nj?) and (Pu#)np.
(ii) Prove that, for any three sets A$ Bs and c9 A <=• C if and only if
Au(BnC) = UuB)n£. (8 marks)
8, (a) In a throw of a die, the probability of getting the number n is P(n).
It is known that
P ( D - Y2and p(2) = p(3) = P ( L° = p(5) = h
(1) What is the probability of getting an odd number in one throw of the
die?
(ii) Find P(6).
(iii) If the die is thrown twice, what is the probability that an odd num
ber is obtained in the first throw and an even number is obtained in the second
throw? (6 marks)
(b) The 115 employees of a firm are classified into three groups A , B , and c
according to their monthly salaries (see the table below).
Group
1 A B
C
Monthly Salary
below $1000
$1000 - $3000
above $3(300
Number of Employees ;
8
100 |
7 1
The distribution of the monthly salaries of the loo employees in Group B is
shown in the histogram in Figure 2.
Distribution of the Monthly Salaries of the 100 Employees in Group B
Number of Employees
15
0 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000
Monthly Salaries ($)
Figure 2
(i) Let a and o^ be the standard deviations of the monthly salaries of all B
the lis employees and the 100 employees in Group B9 respectively. Without calculat
e d the numerical values of a and a D 5 determine which standard deviation is greater. D
Explain briefly.
(ii) It is known that the mean and standard deviation of the monthly salaries
of the loo employees in Group B are $2100 and $4009 respectively.
(1) From June 1981 onwards9 the salaries of all the employees will be
increased by 10%, Calculate the mean and standard deviation of the monthly salaries
of the 100 employees in Group B in June 1981.
(2) If the firm has decided on a second salary revision giving an in
crease of $300 per month for every employee from September 1981 onwards, calculate
the mean and standard deviation of the monthly salaries of the 100 employees in
Group B in September 1981. (10 marks) END OF PAPER
- 14 .
Readers of this Corner are certainly aware of the International Mathematical Olympiads (I.M.O.). But many readers may not be aware that there are also international olympiads in physics and chemistry. For those student and teacher readers who are also interested in physics, I will give some information about the International Physics Olympiad (I.P.O.)- More information can be obtained from the journal Out-of-School Scientific and Technical Education , issue 18-19, obtainable for $1.25 US from I.C.C., Vekemanstraat 7i-73s 1120 Brussels, Belgium.
The f i rs t I.P.O. was organized in Warsaws Poland9 in 1967. The Rules and Regulations are given below. They are yery similar to those of the I.M.O.
1. Aims of the Competition. International Physics Olympiads are conducted for the purpose of widening
contacts in the field of Physics education at school level, for promotion of extra-curricula activities of talented students in Physics and exchange of experience gained through national Physics competitions (Olympiads). International Physics Olympiads help to establish friendly relations between young people of different countries and strengthen understanding and cooperation between nations. International Physics Olympiads help their participants to develop a creative approach to Physics studies and to master skills in application of the knowledge gained at school. They promote participants8 individual thinking and Physics abilities.
2. Organizer of the XII IPO. The Organizer of the IPO sends out official invitations not later than 20
January of the year of the Olympiad and provides equal conditions for all participating delegations.
3. Participants. Each participating country may send a delegation of five high school students
to the IPO, as well as students of professional training schools. At the beginning of the competition the age of the participants should not exceed 20 years.
4. Supervisors. The team of students is accompanied by two supervisors, one of whom is an
official representative of the delegation. At least one of the supervisors should be able to translate the problems from one of the working languages into the mother tongue of the delegations and to take part in evaluation of competition papers. On their arrival, the representatives of the delegations hand over to the Organizer the following documents!
a) list of participants with their names, dates of birth, home address, name of school and its address 1
b) list of supervisors with their names, occupation and home address.
5. Financial Arrangements* The country of origin covers the cost of transportation of a delegation to the
place of the Olympiad and back. The expenditures connected with the stay of delegations in the host country and the conducting of the competition are covered by the Organizer of the IPO.
6. Working Languages. The working languages of the Olympiad are Russian, English, French and German.
7. Schedule. The competition is conducted in two rounds. On the first day of competition
(first round) participants take part in solving theory problems, during the second day of competition (second round) they do experimental tasks. The participants
- 15 -
may use slide rules, electronic calculators and drawing accessories. Those things are supposed to be brought at choice from their own country,
8. International Jury. The representatives of delegations compose the International Jury. The Chair-
man of the Jury is appointed by the Organizer. When voting on issues concerning the competition each country as well as the Chairman has one vote. Decisions are taken but the decisive vote is that of the Chairman.
9. International Jury Responsibilities. The members of the Jury are supposed to: a) ensure that the present Rules and Regulations are strictly followed;
b) look after the accuracy of evaluation and equal application of evaluation criteria to all participants;
c) keep secret problems and tasks during the competition and restrain from offering help to the participants;
d) finalize the results of the Olympiad.
10. Competition Tasks. The problems and tasks for the competition are prepared by the Organizer.
They are taken from the following Physics topics; — Mechanics (including Oscillations and Wave Motion). — Thermodynamics and Molecular Physics. — Electricity and Magnetism. — Optics and Atomic Physics.
11. Evaluation. The criteria for the evaluation are set by the Organizer. The highest score
for solving theory problems is 10 points, for experimental tasks—20 points. Checking of the competition papers is done by the members of a working group appointed by the Organizer. The representatives of delegations are given an opportunity to look through their participants' papers and make remarks to the members of the working group if there are any. *
12. Prizes. Final results of the participants are decided at the International Jury meeting
on the basis of the evaluation marks. The guidelines for the prize distribution are as follows: 1st Degree Diploma is awarded to those with more than 90% of that score; 2nd Degree Diploma is awarded to participants having 78% to 90% of the highest score; and 3rd Degree Diploma—to those having 65% to 77% of it. Participants with scores of 50% to 64% get an Encouragement Diploma. Those having scores less than 50% are given a certificate stating their participation in the Olympiad. There is no team winner in the IPO.
23. Organizerf s Responsibilities. In addition to abovementioned responsibilities, the Organizer of the IPO should:
a) guarantee the course of the competition in accordance with the present Rules and Regulations;
b) work out the schedule for the competition and other events to be organized durincr the Olympiad;
c) ascertain on arrival that the participants meet the requirements of the Olympiad;
d) provide the translation of problems, tasks and cues into the working languages;
e) provide interpreters for delegations and the International Jury meetings;
f) ensure carrying out of the experimental tasks in accordance with the safety rules;
g) organize closing of the IPO and presenting the diplomas and prizes to the participants of the Olympiad.
- 16 -
I now give as a sample a set of three theory problems and one experimental
problem in physics. They are taken from the reference given above, where solu
tions and a scoring scheme can also be found. The Russian journal Kvant and
the Hungarian journal KozSpiskolai Matematikai Lapok both have in every issue
sets of practice problems (in English) in both mathematics and physics.
1, A test tube of mass M is placed in vacuum. A barrier of mass m whose
thickness is negligible divides the volume of the test tube in two equal
parts. The closed part of the test tube contains a mole of ideal monatomic gas of
molar mass p at a temperature 2\ The barrier is released ands moving without
friction, leaves the test tube. Then the gas flows out.
What is the final velocity of the test tube if it is at rest at the moment
when the barrier starts its motion? Neglect the momentum of the gas until the
barrier leaves the test tube and the heat exchange between the gas and both the
test tube and the barrier. The change of the temperature after the barrier has
left the test tube is neglected too. Gravitational pull is not taken into consid
eration as well.
2 a An electric bulb of resistance i?0 = 2.0 U operating at a nominal voltage
UQ = u.5 V is fed from an accumulator with an E.M.F. E = 6.0 V9 its in
ternal resistance is negligible*
(i) Let the nominal voltage be applied tq, the bulb through a slide-wire
rheostat connected potentiometrically. What is the rheostat resistance and to
what maximum current should the rheostat stand so that the efficiency of the
system should be not less than no = o»6?
(ii) What is the maximum possible efficiency for the given light bulb (at
its nominal voltage) and accumulators and how are they to be connected through a
suitably chosen rheostat so that the efficiency n should reach its maximum value?
3, A radiowave receiver of the radioastronomical observatory is placed on
the sea coast at a height h = 2 m above sea level. When a radiostar radiating
electromagnetic waves of wavelength X = 21 cm rises above the horizon the receiver
registers alternating maxima and minima* The receiver registers only those elec
tromagnetic waves whose electric vector E vibrates parallel to the water surface.
The registered signal is proportional to E2. Define:
(i) The angles of altitude of the star above the horizon along the celestial
sphere at which maxima and minima are registered (in general form).
(ii) Will the signal in the receiver be increased or decreased immediately
when the radiostar rises? Explain why.
- 17 -
(iii) The ratio of sianals in the first maximum avd the next minimum. When
the electromagnetic wave is reflected by the water surface, the ratio of amplitudes
of the registered electric field of the reflected (E _ ) and incident {E. ) refl mc
waves is described by the relation
r e f l _ n - COS <j> E. ~ n + COS <j> 5
mc
where w is the refractive index and <£ is the incident angle of the electromagnetic wave. For air-water boundary at X = 21 cm, n = 9„
(iv) Will the ratio of the signals received at adjacent maxima and minima be increased or decreased with the rising of the star above the horizon?
Assume the sea surface to be smooth while solving the problem.
L\t Experimental Task.
You have at your disposal an elastic rubber cord suspended vertically on a stand. The initial length of the cord is J0 = 150 mm9 i ts initial mass together with a pan is 5 g. A set of loads with masses from 10 to 100 gs a stopwatch, a measurement ruler9 a drawing curve and plotting paper are also at your disposal,, The acceleration due to gravity is assumed to be 10 m/s2.
1. Load consecutively the elastic*cord with masses from 15 to 105 g. Tabulate and plot Graphically in an appropriate scale the experimental dependence of the extension AJ on the force of strain F.
2. Usinq the results of measurements performed in accordance to Point i , calculate and write in a table the values of the volume of the cord loaded with masses from 35 to 95 g. Perform the calculations consecutively for each two adjacent loading values within the given interval. Write the formulae used in the calculations. Express by formulae your assumption about the dependence of the volume on the loading. For the Young's modulus the tabulated value E = 2®io6 N/m2 is accepted. While discussing the results you should bear in mind that at given loadings Hooka's law for rubber holds approximately and the deviations from this law can reach up to io%.
3. Measure the rubber cord volumes using a stopwatch and loading the cord with a mass of 60 g9 Write the formulae you have used.
instruction. Do not extend the rubber cord without need and do not let i t stay loaded longer than necessary for the measurements. Do not extend the rubber cord to such extent that the amplitudes of its vibration to be larger than 15-20 mm.
Editor's Note* All communications about this column should be sent to Professor M.S. Klamkin, Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada T6G 2G1.
- l e -
p p n E L E M S - - P P O B L F f E S
Problem proposals and solutions should be sent to the editor, whose address appears on the front page of this issue. Proposals should, whenever possible, be accompanied by a solution, references, and other insights which are likely to be of help to the editor. An asterisk (*) after a number indicates a problem submitted without a solution.
Original problems are particularly sought. But other interesting problems may also be acceptable provided they are not too well known and references are given as to their provenance. Ordinarily, if the originator of a problem can be located, it should not be submitted by somebody else without his permission.
To facilitate their consideration, your solutions, typewritten or neatly handwritten on signed, separate sheets, should preferably be mailed to the editor before June 1, 1984, although solutions received after that date will also be considered until the time when a solution is published.
872 8 [1983: 241] (Corrected) Proposed by George Tsintsifas, Tliessaloniki^
Greece.
Let T be a given trianale ABC with sides a$b9o and circumradius R9 and let P
be any point in the plane of T. It is known (M.S. Klamkin* "Triangle inequalities
from the triangle inequality"9 Elemente der Mathematik, Vol. 34 (1979), No. 3)
that there exists a triangle T0 = TQ(P)$ possibly deaenerate§ with sides <z°PAs
&«PB9 and <?«PC. Find the locus of all the points P for which
PA«PB«PC < R'FQ ,
where i?0 = FQ(P) is the circunradius of TQ . When does equality occur?
901B Proposed by Charles W. Trigg3 San Diegos California,
In
BOXER = HITS,
the X doubles as a m u l t i p l i c a t i o n s ign . Find (a) the fewest HITS, and (b) the
most HITS9 tha t the BOXER can de l i ve r .
902, Proposed by J. Chris Fishers University Libre de Bmtxelles3 Belgique.
(a) For any po in t P on a side of a given t r i a n g l e 9 define Q to be tha t
point on the t r i a n g l e f o r which PQ bisects the area. What is the locus of the
midpoint o f PQ?
(b) Like the curve in par t ( a ) , the locus of the midpoints of the perimeter-
b isec t ing chords of a t r i a n g l e (see Crux 674 [1982: 256]) has an o r i en ta t i on tha t
is opposite to that of the given t r i a n g l e . Is th i s a general p r inc ip le? More
prec ise ly 9 given a t r i a n g l e and a fami ly of chords j o i n i ng Pit) to Q( t ) 9 where
( i ) Pit) and Q(t) move counterclockwise about the t r i ang le as t increases and
( i i ) Pit) * Q(t) f o r any t , does the midpoint of PQ always trace a curve tha t i s
clockwise-oriented?
- 19 -
9 0 3 , Proposed by Stanley Rabinowitz3 Digital Equipment Corp,; Nashua^ New
Hampshire,
Let ABC be an acute-angled t r i anq le w i th circumcenter 0 and orthocenter H,
(a) Prove that an e l l i p s e wi th foc i 0 and H can be inscr ibed in the t r i a n g l e .
(b) Show how to constructs wi th straiahtedae and compass, the points L9M9N
where th i s e l l i p s e is tanqent to the sides BC3CA9AB9 respect ive ly s of the t r i a n g l e .
(c) Prove tha t AL9BM9CN are concurrent.
904 a Proposed by George Tsintsifass Thessalonikis Greece.
Let M be any po in t in the plane of a given t r i ang le ABC* The cevians
AM,BM,CM in te rsec t the l ines BC9CA9AB in A !S B \ C \ respect ive ly . Find the locus
of the points M such tha t
rMCB'l + rMAC'] + [MBA1] = rMC*Bi + ^MA'Cl - [MB'A],
where the square brackets denote the sianed area o f a t r i a n o l e .
9 0 5 , Proposed by J.T. Groenman3 A.vnhems The Netherlands.
Let ABC be a t r i a n q l e that i s not r ight -ang led a t B or C. Let D be
the foo t of the perpendicular from A upon BC9 and l e t M and N be the f ee t of the
perpendiculars from D upon AB and ACS respec t i ve ly .
(a) Prove t h a t , i f /A = 90°9 then /BMC = /BNC. (This problem is given
wi thout proof in M.N. Aref and Wi l l iam Wernick, Problems and Solutions in Euclidean
Geometrys Dover, New Yorks 1968, p. 95, Ex. 8.}
(b) Prove or disprove the converse of Dart ( a ) . J.
906', Proposed by E.J. Eckert and P.P. Vestergaard, Institute of Electronic
Systems3 Aalborg University Centres Denmark.
Let P denote the set of ra t iona l points on the u n i t c i r c l e c9 that i s , the
set of a l l points Cr / t 9 s/t) where r*s9t are in tegers , t > o9 and r 2 +s 2 = t2 . I t
is known tha t P i s dense in c. Let T be the subset of P fo r which t is prime.
Is T dense i n C?
907 , Proposed by Kenneth S. Williams^ Carleton University3 Ottawa.
The four consecutive pos i t i ve integers 769 77 s 78s 79 are such tha t
no one of them is expressible as the sum of two squares. Prove that there are
i n f i n i t e l y many such quadruples of consecutive in tegers .
908 . Proposed by M.S. Klamkin3 University of Alberta.
Determine the maximum value of
p = sinaA . sinBB . sinYCs
where A9B9C are the anales of a t r i ang le and a s3 sy are given pos i t i ve numbers.
- 2C -
QOQ t Proposed by Stan Wagon, Smith Colleges Northampton, Massachusetts.
For which positive inteqers n is i t true that, whenever an integer's
decimal expansion contains only zeros and ones, with exactly n ones, then the
inteqer is not a perfect square?
910, Proposed by 0. Bottema, Delft, The Netherlands.
Determine the locus of the centers of the conies through the incenter
and the three excenters of a given trianale. * & *
S O L U T I O N S
No problem is ever permanently closed. The editor will always be pleased to consider for publication new solutions or new insights on past problems.
777 , r1982: 246] Proposed by 0. Bottema, Delft, The Netherlands; and J.T.
Groenman, Amhem, The Netherlands,
Let Q = ABCD be a convex quadrilateral with sides AB = a, BC = b9 CD = a,
DA = d9 and area LQl. The following theorem is well known: If Q has both a cir-
cumcircle and an incircle5 then LQl = Sabed.
Prove or disprove the following converse: If Q has a circumcircle and rQl = v/abod9 then there exists a circle, tangent to the four lines AB, BC, CD,
and DA.
Solution by George Tsintsifas, ThessaloniJli, Greece (revised and extended
by the editor).
We will say that the convex quadrilateral Q is cyclic if it has a circumcircle^
and that it is circumscribed? le if it has an incircle or an excircle. Consider the
following three theorems:
THEOREM 1. If Q is cyclic and circvmscribable, then LQl = /abed,
THEOREM 2. If LQl = /abed and Q is circvmscribable, then Q is cyclic.
THEOREM 3. If LQl = /abed and Q is cyclic, then Q is circumscribable.
The well-known theorem mentioned in the proposal is a weakened form of Theorem l,
and Theorem 3 is a restatement of the proposed problem, If Q is a nonsquare rec
tangle, then the hypothesis of Theorem 3 is true and its conclusion false. We will
prove these three theorems with the additional assumption, for Theorem 3 only,
that Q is not a nonsquare rectangle. We will need the following lemmas, in which
s - (a+b+c+d)/2.
LEMMA 1. Q is cyclic if and only if B+D = IT.
LEMMA 2. Q has an incircle if and only if s-b-d = 0.
LEMMA 3. If Q does not have a pair of opposite sides parallel (that is, if
- 21 -
its sides produced form a complete quadrilateral)3 then Q has an excircle if and
only if (s~a-d) (s-e-d) - 0,
LEMMA 4. For any QJ
tQl2 = li(s-a) - abed co%1^,
where the product is cyclic over a3bscsd,
LEMMA 5, For all numbers asb3c3ds
H(s-a) = s (s-a-d) (s-b-d) (s-e-d) + abed.
Lemmas l and 4 are classical results. The conditions of Lemmas 2 and 3 are
equivalent to a+a = b+d$ and b+e = a+d or a+b = e+ds respectively. These condi
tions are aivens with numerous historical references, in Sauve ClL The identity
of Lemma 5 is easily verified, but we do not recall seeing it before in the liter-
ature* Our proofs of the three theorems will be based on Lemmas 1,2,3 and on the
formula
LQl2 = s (s-a-d) (s-b-d) (s-e-d) + abed - abed cos2 ~^, (1)
valid for any Q$ which is an immediate consequence of Lemmas 4 and 5.
Proof of Theorem 1. It follows from the hypothesis that
R+n cos-y- = 0 and {s-a-d <<?-:- j- -'•) - 3,
and (l) reduces to LQl2 = abed. Proof of Theorem 2. I t fo l lows from the hypothesis that
LQl2 = abed and (s-a-dHs-b-dHs-e-d) = 0.
Thus ( l ) i s equivalent to cos(B+D)/2 = o, so B+D = TT and c is c y c l i c .
Proof of Theorem 3. I t fo l lows from the hypothesis tha t W 2 = abed and
cos(B+D)/2 = os so ( l ) i s equivalent to
(s-a-d)(s-b-d)(s-e-d) = 0.
If s-b-d - os then Q has an incircle. Suppose (s-a-d)(s-c-d) - o. If Q does not
have a pair of opposite sides parallel, then it has an excircle. If it has a pair
of opposite sides parallel, then the other two opposite sides are also parallel,
so 0 is a rectangle, hence a square9 and it has an incircle. In any case, Q is
circumscribable provided it is not a nonsquare rectangle.
Also solved by ELWYN ADAMS, Gainesville, Florida; LEON BANKOFF, Los Angeles, California; HXPPOLYTE CHARLES, Waterloo, Quebec? JORDI DOU, Barcelona, Spain; GALI SALV7TORE, Perkins, Quebec? KESIRAJU SATYANARAYANA, Gagan Mahal Colony,
- 22 -
Hyderabad, I n d i a ; D . J . SMEENK, Zaltboiranel, The N e t h e r l a n d s ; and t h e p r o p o s e r s (two s o l u t i o n s ) .
Editor's comment.
Three of the above solvers proved to be faint-hearted: they simply gave the
nonsquare rectangle counterexample (a technical knockout of the problem) and then
prudently ret ired from the f ie ld without achieving honour.
Theorem 2 appeared as Problem B-6 in the t h i r t y - f i r s t William Lowell Putnam
Mathematical Competition held on December 59 1970. A solution was published in r 2 ]
with the following comment: "This elegant solution was presented by Robert Oliver.
Pe was the only student to receive a perfect score on this problem." Readers
w i l l f ind i t interesting to compare Oliver's solution with the one given here.
REFERENCES
1. L§o SauvS, ,!0n Circumscribable Quadrilaterals", this journal, 2 (1976)
63-67.
2. J.H. McKay9 "The William Lowell Putnam Mathematical Competition", American
Mathematical Monthly, 78 (1971) 763-770.
A * *
779 i [1982: 246] Proposed by E, ¥»estelmans University College^ London^
England,
Suppose A - X + il, where X and Y are real square matrices. If A is in
vertible., show that
A ' 1 = IC^+I 2)- 1
if and only if X and I commute.
If A is singular, can X2 + Y2 be invertible? If A is invertible9 can X2 = Y2 = o
(the zero matrix)?
Solution by the proposer.
If A is invertible, so are I and
AA = (X-tiY)(X~iY) = X2 + I2 + i(YX-XY). (1)
Suppose XY = YX. Then x2+r2 is invertible by (Dand AA(X1+Y2)~1 = i% from which
follows
A"1 = I(X2+Y2)'1. (2)
Conversely9 suppose Z2+72 is invertible and (2) holds. Then
j = AA'1 = AAiX2^2)"1 = I + i(YX-XY)(X2+Y2)~1>
and XY = YX follows.
21 if
Here 4 is singular and X2+Y2 = 21 is invertible, so the answer to the first
question is yes.
If
* = ( s o1) «•< * = ( ; ; ) . « - * = (<° o1) • Here ,4 is invertible and X2 - I2 = o5 and the answer to the second question is
also yes.
Also solved by KENT D. BOKLAN, student, Massachusetts Institute of Technology; and partially solved by MARCO A. ETTRICK, New York Technical College.
A 3. A
780, [1982: 247] Proposed by Stanley Rabinowitz3 Digital Equipment Corp. s
Merrimack; New Hampshire,
Prove that one can take a walk on Pascal's triangle? stepping from one element
only to one of its nearest neighbors, in such a way that each element ( ) gets
stepped on exactly {m) times.
Solution adapted from those submitted by Friend E. Kierstead3 Jr.s Cuyahoga
Falls3 Ohio; and the proposer. •
Let P be a finite Pascal trianole of m+i rows, so that the last row contains m
the coefficients of (a+bf1. The walker will not work up much of a sweat if m = o9
so we assume that m > l. We will show by induction that there exists a walk on p
starting at ( ) in which each number (J.), i = o,i,...,j, j = o§i,.B.s??7s gets
stepped on exactly (̂ ) times, with the walker going at all times from a number to
one of its nearest neighbors. If after his walk is completed the walker were to
head directly back to ( ), then the extended walk would constitute an Eulerian
circuit in a connected directed qraph whose nodes are the numbers in p and whose r m
directed arcs are the segments of the walk.
Before setting up the induction we explain some notation. We will denote by
r. the row containing the coefficients of (a-tb )d. The midnodes of r. are the two
nodes closest to. the center if j is odd or the two nodes flanking the central node
if j is even. We will let w denote a walk on P startinq at (°) in which each 777 777 *" U
node {i ) gets stepped on exactly {i ) times, with the walker going at all times from a node to one of the nearest nodes9 and endina at a midnode of r if 7?? is odd or
m at the central node of r if m is even. Whenever such a walk w exists, we will
77? 777
. 24 -
denote by r the node of r last stepped ons and by N A a midnode of r „ nearest ^ 777 771+1 7 7 7 + 1
to N . 77?
flow for the induction. I t is clear that there exists a walk W\ on px (there
are in fact two). Suppose there exists a walk wm on P for some m > l . For
£ = o , i , . . . 9TT, between the consecutive nodes (^t1) and [m.+*) of r , we place 2(77) X % ' H + l' 777+1 r H'
directed arcs, half in one direction and half in the other, and then replace one
of the arcs ending at N by an arc from N to N A. If the walker, having 7 7 7 + 1 777 7 7 7 + 1
completed walk w , now follows the arc from N to N „, heads awav from the center m* m 7 7 7 + 1 s
to the end of r , and then weaves back and forth along the directed arcs, going
as far as he can each time before reversing direction, then, when he has traversed
all the arcs, node (m* ) will have been stepped on exactly once; for i = 1,2,... ,777+1,
node (' 1 ) will have been stepped on exactly 771 \ [771 \ /777+l\
times; and the walk will end either at the central node or at a midnode of r „, 7 7 7 + 1
dependino on the parity of m\ so a walk w A w i l l have been accomplished. The 7 7 7 + 1
induction is complete. "k & &
781, -1982: 276] "Proposed by Alan Wayne^ Holiday3 Florida,
The Philosophic Father
Your brother's first to bathe, my son, So do not look so W R Y .
You'll have to use the T U B in turn, There3s room for only O N E .
There's very little room, I guess. Aye, therefs the R U B! But try.
Use soap on cloth to R U B . Don't spurn My counsel? answer "Y E S".
In time, you111 find it's true-Much later, by the B Y E —
As ashes resting in an U R N , There will be room for T W O .
Regard the preceding three patterns of three capitalized words as interdepen
dent arithmetic additions in the decimal system. Restore the digits.
Solution by the proposer.
From the outside columns of the three additions, none of W, T, 0, R, Y, B9 U,
E9 or N can be zero, so S = o9 Then B = 59 so E is odd and Y is even. This
implies that 2U+1 = E. Thus
(U9 E) = (1, 3), (3, 7), or (4, 9).
2 r
If (U9 E) = (3, 7), then Y = 2S R = i, N = 49 and 0 = l = R. If (U, E) = (49 g),
then Y = 4 = U. Hence (U, E) = (l, 3), and there follow immediately Y = 89 R = 4,
N = 65 0 = 95 W = 2, and T = 7.
The unique solution is
248 415 583 715 415 146 9639 830* 729*
Also solved by MEIR FEDER, Haifa, Israel? J.A.H. HUNTER, Toronto, Ontario? ALLAN WM. JOHNSON JR., Washington, D.C.; J.A. McCALLUM, Medicine Hat, Alberta? ADAM MILLING, Grade 11 student, A.B. Lucas Secondary School, London, Ontario? STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hampshire? RAM REKHA TIWARI, Radhaur, Bihar, India? CHARLES W. TRIGG, San Diego, California? KENNETH M. WILKE, Topeka, Kansas? and ANNELIESE Z1MMERMANN, Bonn, West Germany.
782, r1982: 276] Proposed by H.S.M. Coxeter, University of Toronto,
(a) Sketch the plane cubic curve qiven by the parametric equations
x = a(B-y)2, y = B(y-a)2§ z = y(a-B)2
s a f-B + y = 09
where (x$y$z) are barycentr ic (or a r e a l , or t r i l i n e a r ) coordinatess re fer red to an
equ i la te ra l t r i a n q l e . In what respect do i t s asymptotes behave d i f f e r e n t l y from
those of a hyperbola?
(b) El iminate the parameters a,B9y so as to obtain a s ing le equation
x 3+z/ 3+s3 + a(x2y+x2z+y2z+y2x+z2x+z2y) + bxyz - 0
f o r ce r ta in numbers a and b,
(c) What equation does the curve have in terms of polar coordinates?
Solution by G,P. Henderson^ Campbellarofts Ontario.
Let 0 be the centre and u the inradius of the equ i la te ra l t r i a n g l e . We re
present any po in t P of the plane by i t s trilinear coordinates (x9y9z) referred to
the equ i l a te ra l t r i a n g l e , tha t i s * xsy9z are the d i rec ted perpendicular distances
from P to the sides of the t r i ang le 9 the pos i t i ve d i r ec t i on fo r each coordinate
being from 0 to the corresponding s ide . Thus the coordinates of 0 are (u9u,u)
and the coordinates of any po in t P sa t i s f y
x + y + z = 3w.
In the sequels a l l sums w i l l be cyclic over x9y9z or a ^ y .
Set
lyz - v, xyz - w9 EBy = B, aBy = C. (1)
- 26 -
Then9 since la = o,
Za2 = -7B% Za3 = 3C\ Z(£2y+3y2) = -3C.
As
3w = Zx = Z a ( £ - y ) 2 = Z($2y+3y2) - 6a6y = -9(7,
we have
u = -3C. ( 2 )
Similarly s
V = B 3 + 27C2 and w = ~<?(4£3 + 27C 2) . (3)
Eliminating £ and C from (2) and (3), we obtain
27w3 - 12uv + 9h) = 0, (4)
that is9
(Za;)3 - k(lx)(Zyz) + 9 ^ 2 = 09
which reduces to
Ix3 - Z{y1z^-yz1) + 3xyz = 0 . (5)
This has the form required in part (b), with a = -1 and 2) = 3.
Usinq Zx = 3^s (5) can be written in the form
(3x - ^u)(3y ~ ^u)(3z - 4w) = - u 3 , (6)
This is a necessary condition for 09z/9s) to be on the locus. We w i l l show that
i t is also a suff ic ient condition, so we can take (6) to be the equation of the
locus. I t follows that
4w ^u ^u x = 1 - * = T- z = T
are asymptotes of the locus.
Conversely9 l e t x3y9z satisfy (6) and Zx - 3u« We have to show that there exist a,3sy such that Za = o and x = a(3-y)2
9 etc* We claim that, for a suitable choice of cube roots (to be specified below), satisfactory values of a989y are
a = fe~:""W3, 6 = v*y~- 4w/3, y = 3fe - W 3 . (7)
For any choice of cube roots in (7)9 we have
Za 3 = Zx - ^u - -u (8)
and, from (6), a3B3y3 = -(w/3)39 We f i r s t choose cube roots a09$o9Yo in (?) such
that
- 27 -
a03oYo = - 3 . (9)
Then, from (8) and (9)9 Za03 - 3a03oYo = 09 or
(a0+6o+Yo)^0+ojeo+aj2To)^o+a)23o+a)Y0) = 0S
where cu is a primitive cube root of unity. So, for one of the choices
(a93,y) = (a093o»Yo) o r ^a0 .̂ Bo sa)2y0 ) or (a0 ,a)230 *̂ YO )»
we have a+3+Y = 0 and a3y = -w/3. Final ly s
x = a3 + ™ = a3 - 4CX3Y = a (a2 - 43y) = a{(3+y)2 ~ 48y} r a (3 -y ) 2s
with similar expressions for y and z. For real ^9z/ ss s we can choose real cube
roots in (7) except for the point 0(usu,u), To obtain this point we can take for
a s39y the three cube roots of -w/3.
For our system of polar coordinates ( r , 9 ) , we take the pole to be the point 0
and the i n i t i a l ray paral le l to the l ine x = o. Then
3: = w + r s i n S , y ~ u + v sin (0+-^-), z - u + r sin (e+-r-).
To get the polar equation of the locus/we eliminate x,y9z from these and (4) .
Since x9y,z are the roots of
t3 - 3ut2 + vt - w - 0,
r s i n e , r sin (e+2ir/3), r sin (e+^Tr/3) are the roots of
it-m)3 - 3u(t+u)2 + v(t-i-u) - w = 09
that i s ,
t 3 + {v-3u2)t + U3 - ^ = 0,
where we have used (4) to eliminate w. The sum of the products of the roots taken
two a t a time is
v - 3u2 = r 2 (s in 6 sin (e+~0 + sin 6 sin (8+^-) + sin (e+^) sin (6+^)}
= ^ { c o s f - - cos (20+^L) + c o s ^ - cos ( 2 6 ^ ) + c o s f - cos 26}
4 " (10)
Similarly9
M3 . " i U E i s f n a e .
- 28 -
Eliminatina v from the last two equations, we obtain the polar equation
r2(r sin 36 - u) = o. (11)
This is a necessary condition for (r,6) to be on the locus, and we will show that
the condition is also sufficient. It will then follow that the locus consists
of the isolated double point (acnode) r = o and the curve
v sin 39 = ua (12)
We have already seen that r = o is on the locus, so9 to prove the sufficiency of ( l i ) , we assume that r = u esc 38 and f ind a93,y such that
a + 3 + T = 09
a(8~y)2 = w(i + sin 6 esc 36),
8(y-a)2 = u(l + sin (6+^) esc 36), (13)
y(a-6)2 = w( i + sin (6+-^) esc 36).
From (1), a,B9y are the roots of t 3 + Bt - C = o. But from (2) , (3) , and (10),
c = - | and B3 - - ^ 2 c s c 2 3 6 ,
The eauation for a9B9y is therefore
t3 - ^J^3Q • t \ |
Using Cardan's method to solve th is , we f ind
3 °-
. . ? — esc 36 * s ine ,
~ c s c 36 • sin (6+^ ) ,
: ^ c s 7 3 6 ^ s i n ( 8 + f ) ,
and substitution in (13) shows that this actually is a solution.
Using (11) and the fact that x = 4u/3, y = 4w/3, s = W 3 are asymptotes,
we can sketch the curve (see f igure) . The asymptotes are inflectional, which
means that the two branches that approach an asymptote are on the same side of i t
in both directions, whereas for a hyperbola they are on opposite sides.
Also solved by W.J. BLUNDON, Memorial University of Newfoundland; O. BOTTEMA, Delft, The Netherlands? W.L. EDGE,-University of Edinburgh, Scotland (parts (a) and (b) only); J.T. GROENMAN, Arnhem, The Netherlands (parts (a) and (b) only); KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; and the proposer.
Edi tor?s commen t.
The u in the polar equation (12) is a scale factor that can be chosen at will.
Bottema took the circumradius of trie reference triangle to be unity and obtained
v sin 36 = I. The proposer did the same but oriented the polar axis differently
and obtained r cos 38 = - \ . Some of the other solvers obtained far more complicated
expressions for the polar equation. The proposer noted that the curve is drawn
and described on page 32 of Ha68 Forder's Geometry (Hutchinson, London, 1950)9 and
he discussed the curve (among other thinos) in a recent paper entitled "The Lehmus
Inequality" (submitted to Aequationes Mathematicae).
The solution we have featured was the only one which could lay any claim to
completeness because the solver took the trouble to verify that his equations, in
both the trilinear and polar cases, constituted necessary and sufficient conditions
for a point to lie on the locus.
- 30 -
783, £1982: 277] Proposed by R.C. Lynesss Southwold3 Suffolk, England.
Let n be a fixed natural number. We are interested in finding an in
finite sequence (vo9vi9v29...) of strictly increasing positive integers, and a
finite sequence (u0 ,^i,... 9u ) of nonzero integers such that, for all integers
777 > H$
2 2 UnV + U \V „ + . . . + U2V = WnU2 + U -\V2 „ + . . . + U V2 . ( 1 )
u m A 7 7 7 - 1 n m-n u 777 l m-1 n m-n
(a) Prove that (1) holds i f
and
coefficient of xT in {i-x)n
v = coefficient of xr in (1-x) n \
(b)" Find other sequences (u ) and (i> ) for which (1) holds.
Solution to part (a) by Gali Salvatove3 Perkinss Quebec.
The proposed sequences are
u = (-iff), v = 0,1,..S9n
and
t; = ( n w), r = 0.1.2,... 2? 2?
We note that (u ) is a f i n i t e sequence of nonzero integers ands since i t is easily
verif ied that v > v , that (v ) is a s t r i c t l y increasing in f in i te sequence of
positive inteaers, as required by the proposal, Our proof w i l l be based on the
following ident i ty , given by Bruce in r i ] , which is valid for a l l integers m > n:
xn ° 2n 1 2n n 2n
or, since u2 = u2 , r n~r
tW.\2 0 fJ7i+n\ o /i7i+n-l\ or wi \ ,~N xnl 0 v 2n ' l v 2n ' nK2n
We will apply to (3), in two different ways, the difference operator A,
defined for any sequence (w ) by
Aw =w -w , A w = A(A w ) , fc = 1 , 2 , 3 , . . . . 777 777+I 777 777 m
This operator is known to have the following property (see9 e.g., Ferrar [2]):
AW = UnW + U-iW . + . . . + W W . 777 u 777+n x m+n-1 n m
- 31 -From th i s property fo l lows
A (n) ="o( „ ) + " i ( n ) + . . . + *„(„)
u w iV T P - 1 n v 777-n ;
= UQV2 + U iV2 „ + . . . + U V2 (4) u w l 7 7 7 - 1 n m-n K J
On the other hand, A is a linear operator and
A(p = (%+1) - (*) = ( I , ) .
So if we apply it once to (3), we obtain
A(„) ="0A(2„) + «iM ^ ) + ... + ^ A ( 2 j
and i f we apply i t repeatedly , f o r a to ta l of n t imes, the f i n a l r esu l t is
A W / 7 7 7 \ 2 2(m+n\ 9f^+n~^\ 2(m\ v n ' 0 v n ' lx n ' nKn/
0 77? l v m - 1 nK m-n
- u2v + u2v „ + . . . •+ u2v . (5) 0 777 1 7 7 7 - 1 n 7 7 7 - n
Finally, the desired result (l) follows (4) and (5).
Par t (a) was a l so solved by the proposer .
Editor's comment.
In [ l ] , Bruce gives e x p l i c i t l y only the special cases of i d e n t i t y (2) f o r
n - 1 ,2 ,3 ,4 , w i th a p l a u s i b i l i t y argument f o r the general case. He adds tha t
"the general r esu l t i s t rue and i s eas i l y proved by induc t i on " . In sp i te of the
fac t tha t the above so lu t ion hangs e n t i r e l y on the v a l i d i t y of ( 2 ) , our solver
did not give a proof of i t , presumably because he (or she: is Gali a man's name
or a woman's?) f e l t i t was "easy". Our proposer, in his proof of par t ( a ) , also
used ( 2 ) , but he proved i t only f o r the special case n = 3, and then used i t to
estab l ish ( i ) 9 but s t i l l only f o r n - 3, He added that "using the methods above
the resu l ts can be eas i l y genera l ised" . Easy is as easy does, and Bruce doesn' t9
our proposer doesn ' t , and Gali doesn ' t . I f none of these three worthies saw f i t
to give a proof o f ( 2 ) , we suspect i t may be because i t i s n ' t a l l that easy.
Easy or no t , a formal proof of (2) is required to complete par t ( a ) . The ed i t o r
- 32 -
will Gladly reopen the problem when such a proof is available.
Another question left danqling is that raised by part (b) . Are the se
quences (w ) and (v^ of part (a) the only ones that satisfy (D? A characteri
zation of all such sequences would be the last word in rounding off this interest
ing problem.
REFERENCES
1. Ian Bruce, "Binomial Identities", The Mathematical Gazette , 65 (December
1981) 282-285.
2. W.L. Ferrar, Higher Algebra, Oxford University Press, 19439 p. 29.
7o^t, C1982: C77] Proposed by Stanley Rabinowitzs digital Equipment Corp.s
Merrimack^ New Hampshire.
Let F (aj/b.^9 i = i s 2 f ,«,* s/??9 be the Farey sequence of order n , that i s ,
the ascending sequence of irreducible fractions between o and 1 whose denominators do not exceed n. (For example,
F5 f£ M 1 2 1 3 2 3 {1* 59 49 3 9 59 2 s 59 3 s 4 s 1 1 )
5s l]>
with m = l l e ) P^ove that, i f P0 = (0,0*) and P. = (a.,2?.), £ = 1,2,...,;??, are
la t t ice points in a Cartesian coordinate plane, then PQPi...Pm is a simple
polygon of area O - D / 2 .
Solution I ,
The sequence ia./b.) is strictly increas
ing. So, as a point P traverses the polygon
P. PQPI P in the sense of increasing i % m
(the figure shows the polygon for the special
case n = 5), the incl inat ion of the segment
P0P stays at TT/2 as P goes from P0 to P l9 i t
decreases s t r i c t l y from ir/2 to TT/M- as P goes from Pi to P , and i t stays at TT/M- as P goes
m from P back to Pn. From this it follows that
m
(i) no two sides of the polygon can cross,
so the polygon ts simple;
(ii) the area of the polygon is the sum of
the areas A. of the trianales PnP.P. „ for = 1,2,...^1.
Now, for each i9 A. ^Vul'^itl' since b.a. -a.b. „ = 1 [i, i ^+l ^ ̂ +1 Theorem 28]. Hence the area of the polygon is J(w-i).
- 33 -
Solution II. rAs in solution I ] , the polygon is simple.
Following Hardy & Wright r i , page 29], we call a l a t t i c e point P(x3y) visible
(from P0) if there is no l a t t i c e point on P0P between P0 and P8 In order that
P(x,y) be v i s i b l e s i t is necessary and suff ic ient that x/y be irreducible* Since
each a./b. is i r reducible s i t follows that all points Pi s P?, .* „SP are visible*
For each i - 1 ,2 , . . . .w-l , the parallelogram with adjacent sides P0P» and
P0P» . , whose area is b .a . j-a.b. . = 1, has no l a t t i c e point in i t s in te r ior u z+l % ^ + l ^ ^ + l K
f"i, Theorem 34].
From the l a s t two paragraphs, i t follows that the polygon has only the 777+1
l a t t i c e points P 0 § P l 9 . e 8 9 P on i t s boundary and no l a t t i c e point in i t s i n t e r i o r .
I t now follows from Pick's Theorem (see? e . g . , [2]) that the area of the polygon
is
0 + J 0/7+1) - 1 = i(777-l).
Solutions were received from the COPS of Ottawa? HOWARD EVES? University of Maine at Orono; FRIEND H, K1ERSTEAD, JR., Cuyahoga Falls, Ohio; LEROY Fa MEYERS, The Ohio State University; KENNETH M. WILKE, Topeka, Kansas; and the proposer.
Edit or fs comment.
Of the six solutions received* three were more or less equivalent to solution
I9 and three to solution II. Solution I is shorter, simpler, and, in this editor's
opinion, esthetically more satisfying.
REFERENCES
1. G.H. Hardy & E.M. Wright, An Introduction to the Theory of Numbers,
Fourth ed i t ion , Oxford University Press, i960.
2. A. Liu, "A Direct Proof of Pick's Theorem", Crux Mathematicorum, M- (1978)
242-24*4. J. A J-
785, C1982: 277] Proposed by Ja Chris Fisher^ University of Regina.
Suppose a closed different!able curve has exactly one tangent line par
allel to every direction. More precisely, suppose that the curve has parametriza-
tion ^(9): [O92TT] + R2 for which
(i) dv/dQ - (r(9)cos9, r(9)sin9) for some continuous real-valued'function
r(6)9 and
(ii) 7(9) = 7(0+7T).
Prove that the curve fs described in the clockwise sense as 0 runs from 0 to TT,
- y\ -Solution by the proposer.
Let t?(0): ro927r] -»» c be the complex-valued function corresponding to 7(0) ,
and suppose r (6 ) has the Fourier ser ies
r (6 ) = Z P^e
Since r (0 ) is r e a l , p , = p ,̂ for al l k. Furthermore, no term of the ser ies for
v(B) has a constant f i r s t der ivat ive; thus p = o, and so px = o as well . As
can be verified by d i f fe ren t ia t ing , th is means that
id ! v tQk i(k+l)Q pk i(l-k)B ;(0) =fife^ + i j (SJLe \~2 k+1 F F
Since v(B) = V(0+TT) , only even powers of e* appear , and v(B) s i m p l i f i e s to
/ £ n 1 r (9k i2k% pk i(2-2k)Bs
V = 2 ^ I now show that the area of the region bounded by v(B) is negative, which implies
that v(B) is described in the clockwise sense, This area equals
/2ir
v 1 f2l*f
Qk i2kB" h t)(2-2fc)ew- -i2kB i(2k-2)B^ Im / vvldB = - Im \ -? J {^e + jI^e )(Pfee + pke )dB
2TT K\ = k\J^ 7r (se+2^ ) = ^irnr^T < °-
A MAGIC CIRCLE FOR 1984
May the year represented by the magic
constant (obtainable in 32 different ways)
of the adjoining magic c i r c l e , which
contains 16 d i s t i n c t integers from 487
to 494 and 498 to 505, be full of cheer,
pleasure, and happiness for Crux and
i t s readers.
KAMALA KUMARI Daughter of SHREE RAM REKHA TIWARI Radhaur, Bihar, India
