The Yugoslavs were hosts to the Ninth International Mathematical Olympiad (IMO) in which thirteen countries took part. This was held in Cetinje, the former capital of the kingdom of Montenegro, from 3–12 July. The body responsible for the organization of the Olympiad was the Union of the Societies of Mathematicians, Physicists and Astronomers of Yugoslavia.
Mathematical Olympiads are mass problemsolving contests held each year, mainly in the Socialist countries of Europe. They have been fostered because they are regarded as of value in stimulating interest in, and raising standards among, second school students, and in helping to discover the highly talented.
The first IMO was held in Rumania in 1959, and up to this year participants were always from the Socialist countries. However, the Yugoslavs decided to invite other nations, and it was during a visit to Austria by the former Yugoslav Master of Education that an invitation to attend was made to Mr. A. C. Crosland, M.P., then Secretary of State for Education and Science. This was accepted, and the Guinness Awards, as organizers of the National Mathematical Contest and the British Mathematical Olympiad, was asked to be responsible for sending out the English team. Other nonCommunist countries taking part were: France, Sweden and Italy.
The English team was made up as follows: Head of the Delegation: Mr. R. C. Lyness, HMI; Deputy Head: Dr. N. A. Routledge, Maths. Master at Eton College. Organizer: Mr. Maurice Goldsmith.
Professor Walter Hayman, F.R.S., who with his wife sets and marks the questions for the British Olympiad, had a prior engagement in Canada. They had invited the following boys, who accepted: George CameronSmith, 18.0, King Edward VI School, Stafford; Anthony Laing Davies, 17.9, Robert Hill, 18.2, Malcolm John Williamson, 16.8, Manchester Grammar; Patrick Phair, 18.3, Michael Cullen, 18.3, Winchester; Simon Norton, 15.4, David Garland, 17.4, Eton.
The IME have no permanent staff, but the national host organization is responsible for all administration from year to year. The ‘sovereign’ body is the Jury which meets before and during the Olympiad. It is made up of the Heads and Deputies of all participating countries, although only one vote is allowed per country.
The Jury was comprised of the Heads of the twelve delegations, and discussed twelve questions which had been chosen by the host country from the bank of questions to which each country had previously supplied six. After much discussion six were chosen and marks assigned.
Some questions were found to be insufficiently original, some too easy, some too difficult and some outside the knowledge conventionally to be expected from students in all the participating countries. One question, number 6, was originally worded in a form involving a father leaving a hoard of gold coins to his sons in accordance with the terms of a will and the Jury changed the setting of this problem so as to accord better with modern socialist spirit.
The questions in the past have always included ‘planimetry’ and ‘terminals’ (traditional trigonometrical and geometrical constructions and proofs) and have not involved calculus, mechanics or more than easy coordinate geometry. Those students who had knowledge ‘outside the questions’ could use it, of course, to solve the problems.
When the questions had been answered, each Head and Assistant ‘marked’ the papers from his own country and submitted the marking for each question to a Yugoslavian coordinator who went through the marking and approved the final mark given.
The competitors must not have reached university, and the maximum age is 19 years 6 months. They sat for two papers on two successive days (4 and 5 July). Each paper contained three problems, all of which were to be done in 4 hours.
The problems presented this year were as follows:
ABCD is a parallelogram. AB = a, AD = 1, is the size of , and the three angles of triangle ABD are acute.
Prove that the four circles K_{A}, K_{B}, K_{C}, K_{D}, each of radius 1, whose centres are the vertices A, B, C, D, cover the parallelogram if and only if
(Czechoslovakia: 6 points)
One side, and only one side, of a tetrahedron is of length greater than 1. Show that its volume is equal to or less than .
(Poland: 7 points)
k, m and n are positive whole numbers, and m + k + 1 is a prime number greater than n + 1.
Write C_{s} for s(s + 1).
Prove that the product
(C_{m+1}C_{k})(C_{m+2}C_{k})....(C_{m+n}C_{k})
is divisible by the product C_{1}C_{2}C_{3}...C_{n}.
(England: 8 points)
The triangles A_{0}B_{0}C_{0} and A^{1}B^{1}C^{1} have all their angles acute. Describe how to construct one of the triangles ABC, similar to A^{1}B^{1}C^{1} and circumscribing A_{0}B_{0}C_{0} (so that A, B, C correspond to A^{1}, B^{1}, C^{1}, and AB passes through C_{0}, BC through A_{0}, and CA through B_{0}).
Among these triangles ABC, describe, and prove, how to construct the triangle with the maximum area.
(Italy: 6 points)
Consider the sequence (c_{n}):
c_{1} = a_{1} + a_{2}+ ... + a_{s}
c_{2} = a_{1}^{2} + a_{2}^{2} + ... + a_{s}^{2}
c_{n} = a_{1}^{n} + a_{2}^{n} + ... + a_{s}^{n}
where a_{1}, a_{2}, a_{3},
, a_{s} are real numbers, not all
equal to zero.
Being given that, among the numbers of the sequence (c_{n}), there is an infinity equal to zero, determine all the values of n for which c_{n} = 0.
(U.S.S.R.: 7 points)
In a sports meeting lasting n days there are m medals to be won.
On the first day, one medal and 1/7 of the remaining (m  1) medals are won.
On the second day, 2 medals and 1/7 of the remainder are won. And so on.
On the nth day (the last) exactly n medals are won.
How many days did the meeting last, and what was the total number of medals?
(Hungary: 8 points)
About half the contestants were awarded prizes. England gained one First Prize, two Second Prizes and four Third Prizes, and our total of marks (231), was beaten only by Russia (275), Hungary (251) and East Germany (257), who make elaborate arrangements for preparing for the Olympiad, with special schools, training camps, etc. A young English contestant (from Eton College) gained one of the three “certificates of special elegance”.
There were ninetynine competitors, of whom only one, from Bulgaria, was a girl.
The Yugoslav Minister of Education was present at the opening and at the award ceremony. He expressed his pleasure at the interest in mathematics and expressed the goodwill that arises from this international fraternization. This was most apparent in the visits to the beautiful coast of Jugoslavia and places in the impressive mountains.
The 1968 IMO is to be held in Moscow.
The table below gives a consolidated score obtained by all countries participating.
1  2  3  4  5  6  7  8  

Bulgaria  7  14  24  20  20  11  21  42  159 
Czechoslovakia  16  27  30  24  9  13  11  29  159 
East Germany  42  30  23  42  39  35  13  33  257 
England  24  28  18  36  34  28  41  22  231 
France^{1}  4  10  9  6  12  —  —  —  131 
Hungary  34  31  38  33  23  26  38  28  251 
Italy^{2}  20  35  19  7  26  3  —  —  147 
Mongolia  10  11  26  5  6  9  7  13  87 
Poland  22  8  12  7  18  5  9  20  101 
Romania  34  42  22  17  29  19  28  23  214 
U.S.S.R  37  35  32  42  27  28  39  25  275 
Sweden  16  10  15  28  9  20  14  23  135 
Jugoslavia  13  18  11  18  26  22  6  22  136 
Notes: ^{1} took part on second day only; ^{2} had only six competitors.
Report reproduced from The Science Teacher volume 11 number 1 (October 1967) pages 30 and 31.
The solutions, prepared by Dr. N. A. Routledge, are as follows:
Q1 
Let O_{1} be the circumcentre of ABD, and R the circumradius. It is clear that 3 circles of radius R, centres A, B and D, cover the triangle. If , the four circles cover the parallelogram. Equally, if O_{2} is the circumcentre of
BCD, If the four circles cover the parallelogram, they cover O_{1}, and so one of O_{1}A, O_{1}B, O_{1}D, O_{1}C is . . Hence is a N.S.C. for the parallelogram to be covered. Now (well known) Now  
Q2 
Let us assume that all sides save CD are , and that CD is unrestricted. For any tetrahedron it is clear that we increase the volume if we change CD so that (since Vol. = × ABD × height of C above ABD). for any tetrahedron, . Since , D lies in the common region of the two circles, radius 1, centres A and B. Hence . Let AB = 2x. Then . Similarly . Vol. . Now if , Volume .  
Q3 
Now it is well known that a sequence of n (or n + 1) consecutive whole numbers (positive or negative or mixed) is divisible by n! (or (n + 1)!) (this follows by considering binomial coefficients). x is divisible by n!(n + 1)! But x is clearly divisible by (m + k + 1), which, being prime and > n + 1, has no common factor with n!(n + 1)! x is divisible by their product, n!(n + 1)!(m + k + 1). is divisible n!(n + 1)!  
Q4 
Lemma. In the figure, if AB varies so that it passes through X, and A and B are on the ‘outer’ segments of the circles, AB is longest when it is to XY. For and are fixed, so is of fixed shape and thus it is enough to maximize AY. For this AY is a diameter, and so . (Note that, if one of the ‘outer’ arcs is less than a semicircle, this cannot be so.) For the main problem it is easy to see that A and B (and C, in fact) lie on arcs of circles on the sides of A_{0}B_{0}C_{0}, and since ABC is always of the same shape, to maximize its area we need only maximize AB, and using the lemma seems to lead directly to a solution of the problem. However, not all points on the arc on B_{0}C_{0} are possible positions for A (in some cases) (and similarly for B), and it is necessary to show that our construction for a maximum triangle leads to permissible positions for A and B. (This is left as an exercise for the reader.)  
Q5 
Since not all a_{i} = 0, C_{2n} > 0. C_{2n+1} = 0 for many n. (A). If, among the a_{i}, two are equal and opposite, replace them both by 0. Since x^{2n+1} + (x)^{2n+1} = 0, the property (A) still holds. Do this as often as possible. Finally, either (i) all a_{i} = 0, in which case,
originally, the a were capable of being grouped
into equal and opposite pairs. Case (ii) is impossible, for if M = one of the a_{i} of maximum modulus, then, since as  
Q6 
Let a_{r} = number of medals still not awarded. at
start of rth day. A little rough work leads us to Since a_{1} = m, a_{n+1} = 0,
then But 6^{n1} > n  1 for (easily proved, by
induction) This leaves only n = 6, and so m = 36 from (i). We have shown that there can be only one solution. But we have still to show that m = 36, n = 6 corresponds to a realisable situation (it might be that, although (i) is satisfied, we get fractional numbers of medals.) This follows from the table:

Solutions reproduced from The Science Teacher volume 11 number 4 (March 1968) pages 14 and 15.
I have been unable to locate the copyright holder of The Science Teacher; if you believe you own the copyright, please let me know.
Return to IMO Register home page