INTERNATIONAL MATHEMATICAL
OLYMPIAD 1967

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 problem-solving 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 non-Communist 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 Cameron-Smith, 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 co-ordinate 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 co-ordinator 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:

1st Day

ABCD is a parallelogram. AB = a, AD = 1, $\alpha$ is the size of $\angle DAB$ , 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 $a <= cos \alpha + \sqrt{3} sin \alpha$

(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 C₁C₂C₃...C_n.

(England: 8 points)

2nd Day

The triangles A₀B₀C₀ and A¹B¹C¹ have all their angles acute. Describe how to construct one of the triangles ABC, similar to A¹B¹C¹ and circumscribing A₀B₀C₀ (so that A, B, C correspond to A¹, B¹, C¹, and AB passes through C₀, BC through A₀, and CA through B₀).

Among these triangles ABC, describe, and prove, how to construct the triangle with the maximum area.

(Italy: 6 points)
Consider the sequence (c_n):
c₁ = a₁ + a₂+ ... + a_s
c₂ = a₁² + a₂² + ... + a_s²
c_n = a₁ⁿ + a₂ⁿ + ... + a_sⁿ
where a₁, a₂, a₃, , 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 ninety-nine 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
France¹	4	10	9	6	12	—	—	—	131
Hungary	34	31	38	33	23	26	38	28	251
Italy²	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.

International Mathematical Olympiad 1967

Solutions to Problems [see SCIENCE TEACHER, October 1967, p. 30]

The solutions, prepared by Dr. N. A. Routledge, are as follows:

Let O₁ 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.

$\therefore$ If , the four circles cover the parallelogram.

Equally,

[diagram for question 1]

if O₂ is the circumcentre of BCD,
it is clear that R = O₂C < O₁C (since O₁ and O₂ are images in BD).

$\therefore$ If the four circles cover the parallelogram, they cover O₁, and so one of O₁A, O₁B, O₁D, O₁C is . $\therefore$ .

Hence is a N.S.C. for the parallelogram to be covered.

Now $2R = BD / sin \alpha$ (well known)
$\therefore$ $4R^2 sin^2 \alpha = BD^2 = 1 + a^2 - 2a cos \alpha$
$\therefore$ $4(1 - R^2) sin^2 \alpha = 3 sin^2 \alpha - (a - cos \alpha)^2$

Now $cos \alpha = AN < a$
$\therefore$ $a - cos \alpha <= \sqrt{3} sin \alpha$ is also a N.S.C., or $a <= cos \alpha + \sqrt{3} sin \alpha$ .

Let us assume that all sides save CD are , and that CD is unrestricted.

[diagram for question 2]

For any tetrahedron it is clear that we increase the volume if we change CD so that $ABC \perp ABD$ (since Vol. = × ABD × height of C above ABD).

$\therefore$ for any tetrahedron, .
Consider ABD:

[diagram for question 2]

Since , D lies in the common region of the two circles, radius 1, centres A and B. Hence .

Let AB = 2x. Then $h_2 <= \sqrt{1 - x^2}$ . Similarly $h_1 <= \sqrt{1 - x^2}$ .

$\therefore$ Vol. . Now if ,
, for .

$\therefore$
for .

$\therefore$ Volume .

Note that C_p - C_q	=	p(p+ 1) - q(q + 1)
	=	p² - q² + p - q
	=	(p - q)(p + q + 1)

$\therefore$ $\prod_{r=1}^n (C_{m+r} - C_k) =$
$\prod_{r=1}^n (m+r-k) \prod_{r=1}^n (m+r+k+1)$
$= \prod_{r=1}^n(m+r-k) \prod_{r=0}^n(m+r+k+1)/(m+k+1)$ ,
, say.

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).

$\therefore$ 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)!

$\therefore$ x is divisible by their product, n!(n + 1)!(m + k + 1).

$\therefore$ $\prod_{r=1}^n (C_{m+r} - C_k)$ is divisible n!(n + 1)!
$= \prod_{r=1}^n C_r$ .

Lemma. In the figure, if AB varies so that it

[diagram for question 4]

passes through X, and A and B are on the ‘outer’ segments of the circles, AB is longest when it is $\perp$ to XY.

For $\angle YAB$ and $\angle YBA$ are fixed, so $\triangle ABY$ is of fixed shape and thus it is enough to maximize AY.

For this AY is a diameter, and so $\angle AXY = 90°$ . (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 A₀B₀C₀, 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 B₀C₀ 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.)

Since not all a_i = 0, C_2n > 0.

$\therefore$ C_2n+1 = 0 for $\infty$ many n. (A).

If, among the a_i, two are equal and opposite, replace them both by 0. Since x²ⁿ⁺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.
$\therefore$ C_n = 0 for all odd n.
or (ii) some , but there are no equal and opposite pairs.

Case (ii) is impossible, for if M = one of the a_i of maximum modulus, then, since

as $n --> \infty$
if |a_i| < |M|, then as $n --> \infty$ , where k () is the number of a_i = M. Thus for all sufficiently large n, conflicting with (A), which still holds for the transformed (a_i).

Let a_r = number of medals still not awarded. at start of rth day.
$\therefore$ a₁ = m.
and $a_r - a_{r+1} = r + (1/7)(a_r - r)$ () where a_n+1 = 0.
$\therefore$ 6a_r - 7a_r+1 = 6r.

A little rough work leads us to
6(a_r + 6r - 42) = 7(a_r+1 + 6(r + 1) - 42)
$\therefore$ $a_r + 6r - 42 = (6/7)^{r-1} (a_1 + 6 . 1 - 42)$
().

Since a₁ = m, a_n+1 = 0, then

$\therefore$ 7ⁿ(n - 6) = 6^n-1 (m - 36) (i)
$\therefore$ 6^n-1 | n - 6. But for n = 2, ..., 5, no power of 6 divides n - 6.

But 6^n-1 > n - 1 for (easily proved, by induction)
$\therefore$ 6^n-1 > n - 6, > 0 for n > 6 and so 6^n-1 does not divide n - 6.

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:

Day	Number of medals awarded that day	Number of medals then left
1	1+5	30
2	2+4	24
3	3+3	18
4	4+2	12
5	5+1	6
6	6	0

Solutions reproduced from The Science Teacher volume 11 number 4 (March 1968) pages 14 and 15.

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INTERNATIONAL MATHEMATICAL OLYMPIAD 1967

1st Day

2nd Day

International Mathematical Olympiad 1967

Solutions to Problems [see SCIENCE TEACHER, October 1967, p. 30]

INTERNATIONAL MATHEMATICAL
OLYMPIAD 1967