International Mathematical Olympiad 1970
Bryan Thwaites (Leader) and Margaret Hayman (Deputy Leader), the British Team

This International Olympiad was the twelfth of the series and, as in recent years, invitations to participate had been received by certain countries of Western Europe. This year, Hungary was the host country, and all the arrangements were made at governmental level by the Hungarian Ministry of Cultural Affairs.

The organisation followed the pattern of previous years. Each participating country provided a “leader”, who was a member of the “jury” which was the sole arbiter of all things academic—the choice of questions, their marking and moderating, and the decision on prize-winners. A “deputy leader” looked after each country’s team of eight students, none of whom was yet at University.

For the British mathematicians, there are acute difficulties. All the Democratic Socialist countries of East Europe attach great importance to the Olympiad, and regard it as an operation from which much prestige can be gained. The same seems hardly less true for the Western European countries, with the one exception of Britain. It is not easy to see why the British Government is so reluctant to recognise the evident advantages to be derived from groups of young people of the very highest intelligence cohabitating and competing together over a period of 12 days or so.

Organisationally, the lack of official Government support creates considerable problems for the two leaders and the schoolboy participants. As Government acceptance is dependent on an assurance that adequate funds have been made available by private industry to the Committee for Mathematical Awards of the Mathematical Association, the official programme may not be made available to the teams beforehand, which makes the timetabling of the party largely a matter of guesswork. Thus, this year, the leaders were obliged to extrapolate from previous years to arrange the return journey, only to find on arrival in Hungary that the British delegation, to the amazement and disbelief of the other national delegations, would miss the final ceremonies and prizegiving.

The participation of a British delegation this year was in doubt until the very last moment, owing to the necessity of raising the cost—about £1000—from purely private sources, at very short notice.

To us, it is clear that, in the eyes of at least the communist countries, the British entry is not a serious one. Thus, when it comes to purely mathematical aspects, the questions submitted beforehand from Britain go unregarded; the British attitude to mathematics is barely heeded; and the final six questions are biased towards the kind of mathematics a typical clever Russian 17-year-old will be doing, and away from what a British sixth-former is likely to experience.

To be more precise, one might look at the six questions set this year, which follow this Report. If the Oxbridge scholarship examinations were to be used as a yardstick, one might opine that only one question—number 2—is of a kind which might be found as a scholarship question. Question 1 is too “trad” as a straight triangle rider; question 3 (a nice question) does not seem to lie within the typical orbit of sixth-form inequalities; question 4 deals with a kind of number theory which we seem to by-pass; question 5 is another inequality of (dare we say?) a rather uninspiring kind; and question 6 involves essentially an application of induction which would be unfamiliar to most English classrooms.

This year 14 countries took part, of which five were from Western Europe. From the total number of questions submitted earlier in the summer by each leader, it is the host country’s prerogative to choose some 12 problems to serve as a basis for discussion by the jury. To those accustomed to the difficulties of setting, for example, A-level questions, it will appear a formidable task for a committee of 14, speaking and thinking in 14 different languages with interpreters working in four languages, to arrive at six sensible, well-phrased, questions. Indeed, this process took two long days’ work.

As it turned out this year, the final stages of compiling each question seemed best done with an English draft text, and the English leader was able to make a substantial contribution at this stage, when it became clear that some other leaders would have preferred much more complicated and pseudo-rigorous formulations of the same questions.

When it came, finally, to approve the set of six perfected questions, the British leader abstained from voting. He felt that too much stress was being laid on the formulation of inequalities, and too little on the solution of any kind of equation or mathematical model; and that the overall balance in the mathematics was unreasonably, and unnecessarily, to the disadvantage of the British boys.

Against this was set the view that within that sphere of mathematics for which equalities can be formulated, the inequality is of greater interest and of more fundamental significance than the equality; that “solving equations” was not a very profitable intellectual exercise; that models outside the geometric, or those involving rates-of-change, were not deemed to be “within the syllabus”; and, finally, that if the boys were of the highest calibre, then the type of problem was irrelevant. (On these various points, it would be interesting to form some kind of British view for the benefit of future delegations).

The abstention caused a minor international incident, for it was at once construed as a veto. The jury then launched itself with never-flagging zeal into a seminar on the theory of democracy, during which the British leader expounded the principles of making up the rules of the game as you go along, with by no means a total lack of success. The abstention prevailed.

Thereafter, definitive versions of each question were agreed in the four “official” languages, each leader translated them into his own language—and then settled down to typing out 10 copies.

Another interesting comparison of national mathematical viewpoints arose when the number of marks for each question came to be decided. Eventually they were agreed as follows: 5, 7, 8, 6, 6, 8, totalling 40. The candidates were not told of these relative weights. No marking scheme was agreed within each question.

While all the preliminaries were going on (in a not very attractive holiday resort some 100 miles west of Budapest), the teams were settling elsewhere.

That the entire British team and deputy leader reached Keszthely on Lake Balaton—after a 6.15 a.m. start, and the chaos of the railway station in Budapest, with an interpreter who didn’t know which train they were trying to catch—was most surprising. In Keszthely, they stayed in a boarding school for catering students where the accommodation was pleasant, and food was adequate but dull.

The first two days were left free, and the students made friends with members of other teams and swam in the lake, with the exception of Russia, East Germany and Hungary whose teams had long training sessions. After the papers were taken, the leaders at last broke out of their rigid isolation and were allowed to mix with the deputy leaders and the rest of the teams.

Once the scripts had been completed, furious activity broke forth. Each leader and deputy leader first marked their team’s papers, though without putting the marks on the scripts. Then a very complicated system of “co-ordination” took place, whereby each question would be read by its own small specialist team of young Hungarian mathematicians, whose job it was to ensure that the marking of that question over all 14 teams was fair.

The attitude to detailed explanation varied considerably between the coordinators, some being interested merely in the argument, and others insisting that even the most obvious statements were proved in detail. In the case of disagreement, the leader of the delegation from the country which set the question was the final arbiter. It seems that full marks were given for complete solutions, even if not by the best method, but marks were lost for incomplete argument even when the steps seemed trivial.

The jury then met and, after a relatively brief discussion about methods of marking and coordination—and of how far a marking scheme, and priorities of proof, should be decided by the jury before marking is started—settled to the very protracted task of deciding the marks for 1st, 2nd, and 3rd prizes.

Opinion divided between making the prizes reflect good work, and making them span at least one pupil from each country. Finally, after three hours of discussion, the decision was in accord with the second, rather than the first, principle:

1st prize 40–37 marks, 7 pupils;
2nd prize 36–30 marks, 11 pupils;
3rd prize 29–19 marks, 40 pupils.

Later, there was an equally long session on the relative merits of individual solutions for special prizes. A brief solution is not in itself sufficient, but one which can be generalised immediately, even if this has not been carried out, was deemed worthy of merit. How far a very good mathematical generalisation which depends on special knowledge, rather than elementary methods, was deserving of a prize was also discussed at some length.

Finally the following were agreed—an East German for question 3, a Hungarian for a generalisation of question 4, and another Hungarian, a Czech and a German for brief solutions of the questions, which are directly capable of some generalisation, even though this was not carried out, and two Hungarian students who found the equality in question 6.

Considering that G.B. is the only country which has no special training—even Sweden had a three-day training session before leaving for the competition—the British team did very well. If, however, we are to compete on equal terms then we must consider a method whereby the student can have some experience of the sort of questions that will be asked, and practice at answering such questions.

While the leader and deputy leader were mathematically engaged, the boys had more swimming, games and conversation, and the whole group had a pleasant reception given one evening by the local education department, and a boat trip on the Lake, which included a wine-tasting for the grown-ups.

In spite of having to leave early, the boys enjoyed their experience and gained a great deal, particularly from meetings with pupils from other countries. Only with the Mongolian team did they fail to communicate.

The following table lists the score by countries:

German Dem. Rep.221Sweden110
Great Britain180U.S.S.R.221

The numbers in each Class were I, 7 (British 1, German 1, Russians 2, Hungarians 3); II, 11; III, 40 (British 6). Bernard Silverman, City of London School, scored 39 out of 40.

The British team scored as follows by questions:


The average team score was 48.8 per cent. The British team average was 56.25 per cent.

Questions and Solutions

First Day (4 hours)

  1. M is a point on the side AB of the triangle ABC. r1, r2, r are the radii of the incircles of triangles AMC, BMC, ABC; and ρ1, ρ2, ρ are the radii of the ecircles of the same triangles corresponding to the sides AM, BM, AB. Prove that (r_1 / \rho_1).(r_2 / \rho_2) = (r / \rho).

    (Poland, 5 marks)


    One may first recall (or derive fairly simply) the standard triangle formulae
    r = 4R cos ½A cos ½B cos ½C
    and ρ = 4R sin ½A sin ½B cos ½C for ecircle opposite A.
    whence r / ho = cot ½A cot ½B.

    So denote CMA by D.


    (r_1 / ho_1).(r_2 / ho_2) = (cot ½A cot ½D) . (cot ½(\pi - D).cot ½B)
    = cot ½A.cot ½B = r / ho.


    (i) A “transformation geometry” solution would be welcomed.

    (ii) The interesting generalisation that ∏(rii) = r/ρ.

  2. Let a, b and n be integers greater than unity; and let the numbers a, b be the bases of two number systems.

    The numbers An, Bn have the same representation


    in the bases a, b, where xn ≠ 0 and xn−1 ≠0. The numbers obtained by deleting the first digit xn are An−1, Bn−1. Prove that a > b if and only if A_{n-1} / A_n < B_{n-1} / B_n.

    (Romania, 7 marks)


    Let p(t) = \sum_{i=0}^n x_i t^i and q(t) = \sum_{i=0}^{n-1} x_i t^i
    so that An/An−1 = p(a)/q(a).

    It suffices to prove that p(t)/q(t) is monotonically increasing.

    Which follows from p(t)/q(t) = 1 + (x_n t^n / q(t)) = 1 + x_n [ \sum_{i=0}^{n-1} (x_i / t^{n-i}) ]^{-1}.

  3. The real numbers a0, a1, ....., an, ..... satisfy

    1 = a0a1 ≤ ..... ≤ an ≤ .....    (1)

    The numbers b1, ....., bn, ..... are then defined by

    b_n = \sum_{k=1}^n ( 1 - (a_{k-1} / a_k) ) (1 / \sqrt{a_k}).

    (i) Prove that 0 ≤ bn < 2 for all n.

    (ii) Also, given c such that 0 ≤ c < 2, prove that there exist numbers a0, a1, ..... satisfying the property (1) and for which bn > c for infinitely many n.

    (Sweden, 8 marks)


    (i) A simple calculus proof follows, though it would not be admitted by most other European countries.

    First, bn ≥ 0 is trivial from the definitions.


    b_n = \sum_{k=1}^n a_k^{-3/2}(a_k - a_{k-1}) < \int_1^{a_k} x^{-3/2} dx = -2[a_k^{-1/2} - 1] < 2.

    (ii) Choose x > 0. c < x(x + 1) < 2, and put ak = x−2k.

    Then b_n = \sum_{k=1}^n (1 - x^2)x^k = x(1 + x)(1 - x^n) which increases monotonically to the limit x(1 + x) as n → ∞.


    This limit can clearly be made to lie as close to 2 as required. What therefore is wrong with the following argument produced by one of the candidates?

    The maximum value of bn will be obtained by ensuring that the ai’s give the maximum value of (bi −bi−1) for each bi−1. Thus, regarding ai−1 as fixed,

    (d(b_i - b_{i-1}) / da_i( = (d / da_i) [ a_i^{-1/2} - a_{i-1} a_i^{-3/2} ] = -½a_i^{-3/2} + (3/2)a_{i-1}a_i^{-5/2}

    so that (bi − bi−1) is maximised for ai = 3ai−1.

    Thus, with a0 = 1, ai = 3i, and max b_n = \sum_{i=1}^n (1 - (1/3)) (1 / 3^{i/2}) = (2/3) . (1 / \sqrt{3}) ((1 - (1 / 3^{ni/2})) / (1 - (1 / \sqrt{3}))) < {2\over 3}{1\over \sqrt{3}-1} < 1 !

Second Day (4 hours)

  1. Find the set of all positive integers n with the following property:

    “Given the set { n, n + 1, n + 2, n + 3, n + 4, n + 5 }, it is possible to partition it into two sets such that the product of the members of one set equals the product of the members of the other set.”

    (Czechoslovakia, 6 marks)


    If a prime p divides any of the six numbers, it must divide at least two of them for the property to hold. Thus p ≤ 5. Furthermore 5 divides only n, n + 5. Thus the only prime division of n + 1, n + 2, n + 3, n + 4, are 2 and 3, but of these 4 numbers two are odd, and so must be powers of 3, which is impossible since their difference is 2. Hence the answer is the empty set.

  2. In the tetrahedron ABCD, DB is perpendicular to DC, and the foot of the perpendicular from D to the plane ABC is the orthocentre of the triangle ABC.

    Prove that (AB + BC + CA)2 ≤ 6(DA2 + DB2 + DC2). For what tetrahedra does the equality hold?

    (Bulgaria, 6 marks)


    First it may be proved by a variety of methods that all three angles at D are right-angles.

    Second, 6(DA2 + DB2 + DC2) − (AB + BC + CA)2
    = 3(AB2 + BC2 + CA2) − (AB + BC + CA)2
    = (AB − BC)2 + (BC − CA)2 + (CA − AB)2
    ≥ 0.

    The equality holds only if AB = BC = CA.

  3. 100 coplanar points are given, no three of which are collinear.

    Prove that, of all the triangles which can be drawn with these points as vertices, not more than 70 per cent are acute-angled.

    (Russia, 8 marks)


    For a quadrilateral, at least one angle ≥ 90°, so at least 1 out of the 4C3 = 4 triangles is obtuse.

    With additional, fifth, point, there are 4C2 extra triangles (making 10 in all) and there are 4C3 = 4 extra quadrilaterals which give at least ½ . 4 = 2 extra obtuse triangles, to make at least 3 obtuse triangles all told.

    Given 100 points, every five points therefore contribute at least 3/10 = 30 per cent obtuse triangles, and so in all not more than 70 per cent of triangles is acute-angled.

Report reproduced from Science Teacher volume 14 number 1 (October 1970) pages 1–6.

I have been unable to locate the copyright holder of Science Teacher; if you believe you own the copyright, please let me know.

Note an erratum notice by Geoff Smith to the solution to problem 1.

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