Olympiad 1969
F. R. Watson and L. Beeson

The XIth International Mathematical Olympiad was held in Bucharest in July, 1969, following acceptance of an invitation from the Romanian Ministry of Education by the Department of Education and Science. The travel expenses of the British party were met by the Guinness Awards, which again managed all the arrangements.

We were the guests of the Romanian Ministry of Education which provided accommodation, pocket money, and sightseeing trips on a very generous scale, as will be seen from Mr. Beeson’s account.

The British team consisted of: F. R. Watson, University of Keele, Institute of Education (Head of delegation), L. Beeson, Bishop Otter College of Education, Chichester (Deputy Head), and a team of eight boys selected on the basis of the 5th B.M.O. held in May, 1969 (see Science Teacher, October, 1969):—D. J. Aldous (Exeter School), P. M. Bennett (Royal Grammar School, Newcastle), A. J. McIsaac (Charterhouse), S. P. Norton (Eton College), J. F. Segal (Dulwich College), P. D. Smith (Manchester Grammar School), A. G. Trangmar (Dulwich College), N. S. Wedd (Kelly College, Tavistock).

Fourteen countries took part, and Austria sent an observer. The results by countries are given below, though it should be noted that direct comparisons are misleading: some national teams are chosen with great care and train seriously for the Olympiad—in one country the eight contestants were hastily nominated by eight convenient schools only a few days before departure. Further, familiarity with the content of the questions varies widely from country to country.

CountryTotalNo. in each class
E. Germany24044

The three British prizewinners were, in order, Norton, Aldous, and Smith. In addition 13 prizes for solutions of special elegance were awarded, and three of these went to Norton (question 2), and to Aldous and Wedd (question 5).

The 112 candidates (7 of them girls), after a few days sightseeing and settling down in Bucharest, sat the two 4-hour papers in the Nicolae Balcescu High School on the mornings of July 10 and 11. Before this, the international jury, consisting of the 14 heads of delegations, had been engaged in selecting the questions; parallel definitive versions were then prepared in four working languages (English, French, German and Russian), which each head then translated, where necessary, and typed out for his own team. Photocopying facilities would have been very welcome here.

The Jury meetings were held at Snagov, some 30 km. north of Bucharest, in a lakeside camp provided for young people aspiring to represent Romania in rowing, canoeing and athletic events. This isolation from the rest of the delegations—to ensure secrecy—meant that we rapidly got to know one another through working, eating and relaxing together (bathing in the lake was very pleasant). Some of our Romanian interpreters, and some members of the Jury, were highly multilingual, but even those of us with little linguistic prowess soon became involved in animated, if ungrammatical, conversation. The cordial atmosphere which resulted undoubtedly helped us in our work and contributed greatly to the enjoyment of our stay.

After the contest, the two leaders of each delegation marked the papers of their own team, their work being co-ordinated by six Romanian interpreters, assigned one to each question, to ensure comparability of marking. (The Romanian marking of each question was co-ordinated by the leaders from its country of origin). The final meetings of the Jury, now in the Balcescu High School, were to determine the standards for the award of 1st, 2nd and 3rd and special prizes, and to discuss arrangements for the XII Olympiad to be held in Hungary.

When these were concluded, we had one or two days free for sightseeing in Bucharest before we were rejoined by the rest of the delegation for the final ceremony. Here, prizes and certificates of participation were presented by the President of the Romanian Mathematical Society, Academician Professor G. C. Moisil, who had earlier presided with great courtesy and perceptiveness at several of our Jury meetings.

The Ceremony, which included an address by the Deputy Minister of Education, was extensively reported on T.V. and radio, and in the press, and it was apparent that the Romanians, in common with other East European countries, regard education in mathematics as of considerable importance.

Certainly, this Olympiad seemed to be fulfilling one of its aims—that of stimulating interest in mathematics. The other effects are perhaps equally important—the opportunity for pupils and leaders to meet their counterparts in other countries, to compare experiences, to widen their horizons, both mathematically and personally, to visit a country with a very different social and political system, to form friendships and exchange addresses. I feel that our team performed very creditably—and that we are right to regard these occasions, not too intensely, as more of a holiday than an ordeal. The memories that remain—the pleasant relations with leaders of other delegations, and the warm hospitality of our Romanian hosts—a most enjoyable visit.

Mr. Beeson writes:

The team of British boys and I were met late on July 7 at Bucharest airport by the secretary of the Mathematical Association of Romania, together with our interpreter. From this moment on we received every possible help from the Mathematical Association and from the Romanian Ministry of Education, who were our joint hosts. Our “base” in Bucharest was a students’ hostel at the University, with a good canteen next door.

We had two or three excursions before the contest, seeing both old and new parts of the capital and some of the country nearby. We had a discussion partly in French, with a lady inspector of education—we tried to explain the English system of education and argued “State versus Public” with her and among ourselves. We got to know some of the other 14 teams and tried to relax before the contest paper.

The opening ceremony was impressive, with television coverage and journalists in attendance. As soon as the papers began, the rain started, and other delegations called it ‘English weather’. On both days, the British team seemed well able to sum up how well they had tackled the questions. This meant that they were happy on the first day, and generally rather miserable on the second. Our interpreter arranged visits to lakes near the city for swimming, and helped to raise our spirits.

While we two leaders stayed in Bucharest to mark the contest papers, the teams left in three heavily laden coaches for a week’s trip round the north-east of the country. I was able, through the kindness of our hosts, to join the boys after three days. We were obviously taken on a most interesting and varied route, so it is worth recording the principal places—other visitors to the country might want to follow the same route. The buses went through Bacau, to some spectacular country round Bicaz, visited monasteries in Moldavia at Agapia, Sucevita, Moldovita and Voronet. We stayed near the border with the U.S.S.R. at Suceava for two nights, saw superb alpine pastures covered in wild flowers and snow-capped mountains. We stayed generally in student hostels, or, as at Tirgu Mures and at Brasov, in boarding schools.

Our party was rather more than 150 in all. Great care was taken of this valuable group of young people—we had a doctor and nurse travelling with us, for instance. This proved fortunate, since one of the Swedish team had to have an operation for appendicitis. He was left in hospital, with the Swedish interpreter staying in the town near him. Our coach drivers were most skilled at coping with well-engineered, but sometimes rough, roads. The insurance liability sometimes daunted me.

Often our middle meal of the day (not served at mid-day!) was provided in school canteens. They were a little different from many school dinners, however, often starting with brandy and usually having four courses! Perhaps our best day was the last of our trip, when we were given a splendid welcome at a holiday camp for children good at mathematics and science. We were told such camps were run for those good at other subjects also, and that they did no work at the camps—it was simply a reward for previous efforts. Even the shyest of our team left with half-a-dozen addresses of Romanian girls.

We returned to Bucharest for the closing ceremony. The Romanians had rearranged this just to suit the British team, who had booked their flight on the original closing day. A television interview with Simon Norton and the other two First Prize winners was broadcast that evening, as part of the coverage of the closing ceremony.

We all enjoyed the privilege of going. We were sorry not to be able to communicate with all the delegations—the Mongolians, for instance, spoke virtually nothing but their own language. We learnt how far mathematics is international, and how sometimes very clever boys feel depressed at their inability compared with even cleverer boys. We were reminded that something more than mathematical ability is needed in personal relationships. Often I, at least, found myself thinking of the way that we in Britain might stage such an Olympiad, and what we should want our guests to see and do. Should we lose much mathematically if each team had to be four boys and four girls? We should undoubtedly gain socially. Certainly, if we do ever act as hosts, we shall find it very difficult to match the kindness and generosity of our Romanian friends.


Paper I (4 hours)

  1. Prove that there are infinitely many natural numbers a with the following property: the number z = n4 + a is not prime for any natural number n.

    (E. Germany, 5 marks)

  2. Let a1, a2, ... an be real constants, x a real variable and

    f(x) = cos(a_1 + x) + cos(a_2 + x) / 2 + ... + ... + cos(a_n + x) / 2^(n-1)

    Prove that, if f(x1) = f(x2) = 0, then x_1 - x_2 = m\pi, where m is an integer.

    (Hungary, 7 marks)

  3. For each value of k = 1, 2, 3, 4, 5 find the necessary and sufficient conditions on the number a > 0 for there to exist a tetrahedron with k edges of length a and the remaining 6 - k edges of length 1.

    (Poland, 7 marks)

Paper II (4 hours)

  1. A semi-circular arc \gamma is drawn on AB as diameter. C is a point of \gamma distinct from A and B, and D is the orthogonal projection of C on AB. We consider three circles \gamma_1, \gamma_2, \gamma_3 which have AB as a common tangent. Of these \gamma_1 is the circle which is inscribed in the triangle ABC and \gamma_2, \gamma_3 are both tangential to the line-segment CD and to \gamma. Prove that \gamma_1, \gamma_2, \gamma_3 have a second tangent in common.

    (Holland, 6 marks)

  2. Given n > 4 points in a plane such that no three are collinear, prove that one can find at least (n - 3 \choose 2) convex quadrilaterals whose vertices are four of the given points.

    (Mongolia, 7 marks)

  3. Given x1 > 0, x2 > 0, x1y1 - z12 > 0, and x2y2 - z22 > 0, prove that:

    8/((x_1 + x_2) (y_1 + y_2) - (z_1 + z_2)^2) <= 1/(x_1 y_1 - z_1^2) + 1/(x_2 y_2 - z_2^2)

    Give necessary and sufficient conditions for equality.

    (U.S.S.R., 8 marks)


  1. z = n4 + a may be considered as the difference of two squares by writing n^4 + a = n^4 + 2n^2\sqrt{a} + a - 2n^2\sqrt{a}
    Then \sqrt{a} = 2k^2, i.e. a = 4k4 gives z = (n^2 + 2k^2)^2 - 4n^2k^2 = ((n + k)^2 + k^2)((n - k)^2 + k^2). For k > 1 each factor exceeds 1, so z is not prime. Thus if a is of the form 4k4, z is not prime for any n. Evidently there are an infinity of such a.

  2. f(x) \equiv A cos x - B sin x where A = \sum_{k=1}^n (cos a_k / 2^{k-1}), B = \sum_{k=1}^n (sin a_k / 2^{k-1})

    f(x1) = f(x2) = 0 gives { A cos x1 - B sin x1 = 0
    A cos x2 - B sin x2 = 0

    If one of A, B is non-zero, it follows that sin(x2 - x1) = 0 and x_2 - x_1 = m\pi, m integer.

    Thus we must show that A, B are not both zero. Suppose the contrary.

    Then cos a_1 + \sum_{k=2}^n (cos a_k / 2^{k-1}) = 0 = sin a_1 + \sum_{k=2}^n (sin a_k / 2^{k-1}) whence 1 + \sum_{k=2}^n (\cos(a_1 - a_k) / 2^{k-1}) = 0. But L.H.S. >= 1 - \sum_{k=2}^n (1 / 2^{k-1}) |cos(a_1 - a_k)| >= (1 / 2^{n-1}) > 0 contradiction.

  3. [Diagram for Question 3]

    By writing 1/b we can reduce the cases k = 4, 5 to the cases k = 2, 1. Thus we need only consider k = 1, 2, 3.

    k = 1

    Let AB have length a, M be the mid-point of CD. The \triangle in equality in \triangle AMB gives a < \sqrt{3} as a necessary condition. It is also sufficient since for a < \sqrt{3} we can construct \triangleAMB and then ABCD.

    k = 2

    Case (i) Edges of length a are opposite—say AB, CD. \triangleAMB gives a < \sqrt{2} as the necessary (and sufficient) condition.

    Case (ii) Edges of length a are adjacent—say AB, AC. Radius of circum circle of \triangleABC is less than 1, which leads to \sqrt{2-\sqrt{3}} < a < \sqrt{2+\sqrt{3}}. Since ½ < \sqrt{2-\sqrt{3}} this condition is sufficient for the existence of \triangleABC and so of ABCD. Also \sqrt{2} > \sqrt{2-\sqrt{3}}, thus for k = 2 we have a < \sqrt{2+\sqrt{3}} as the necessary and sufficient condition.

    k = 3

    Case (i) The edges a concurrent—say AB, AC, AD
    Then RABC < a is necessary and sufficient—i.e. a > 1/\sqrt{3}

    Case (ii) The edges a coplanar. This leads analogously to a < \sqrt{3}

    Case (iii) The edges a are two opposite edges and one other. By considering the circumscribing parallelepiped (4 of whose faces are rectangles) of the tetrahedron, it can be shown that (\sqrt{5}-1)/\sqrt{2} < a < (\sqrt{5}+1)/\sqrt{2} is necessary and sufficient.

    (However, as the ranges in (i) and (ii) overlap it is not necessary to consider case (iii)).

    Thus for k = 3 there is no restriction on a.

    For k = 4 we have a > 1/\sqrt{2+\sqrt{3}} = \sqrt{2-\sqrt{3}}, and for k = 5, a > 1/\sqrt{3}.

  4. [Diagram for Question 4]

    Let the centres be Mi, their projections on AB, Ni, the radii ri, i = 1, 2, 3.

    Let AD = p, DB = q.

    We require to show that M1 M2 M3 are collinear.

    In \triangle O M2 N2, (½c - r2)2 = r22 + (q + r2 - ½c)2
    (r2 + q)2 = qc = a2.
    i.e. r2 + q = a   Similarly r3 + p = b
    BN1 = s - b. N1 N3 = s - b - q + r3 = s - c and so N1 N3 = N1 N3 by symmetry.

    Thus we must prove (r_2 + r_3)/2 = r_1.

    But (r_2 + r_3)/2 = (a + b - p - q)/2 = s - c = r, since \hat{C} = 90°

    Hence M1 is the mid-point of M2 M3 and the three circles have a second common tangent

  5. We can choose 3 of the points A, B, C so that all remaining points lie within the angle A\hat{B}C. Let X, Y be any other point distinct from A, B, C.

    Then if XY meets {BC } externally,
    the quadrilateral {XYBC } is convex and at least one

    of these must hold. Thus there are at least as many convex quadrilaterals as pairs X, Y—i.e. at least n-3C2.

    Note: 8 students, 2 of them British, proved a much stronger result, replacing n-3C2 by {}_n C_5 / (_{n-4}).

    The following is the proof given by N. S. Wedd.

    (i) Any set of 5 points contains at least one c.q.

    (ii) There are nC5 sets of 5 points, each containing at least one c.q.—though some sets may have c.q.’s in common with other sets of 5.

    (iii) Each c.q. can belong to at most (n - 4) such sets of 5 i.e. can be counted at most (n - 4) times.

    Hence there are at least {}_n C_5 / (_{n-4}) c.q. [which he then proves >= {}_{n-3}C_2 for n >= 5].

  6. Only 7 correct solutions were received for this difficult problem; one of these, due to S. P. Norton, is given here:—

    xi > 0, x_iy_i > z_i^2 >= 0 ==> yi > 0 and \exists ki, wi

    such that xi = wiki } wi, ki > 0.
    yi = wi/ki

    We require to prove:

    8 / ((w_1k_1 + w_2k_2) (w_1/k_1 + w_2/k_2) - (z_1 + z_2)^2) = 8/D <= 1 / (w_1^2 - z_1^2) + 1 / (w_2^2 - z_2^2) for wi > 0, wi2 > zi2, k1 > 0, (and so wi > /zi).

    Now (w_1k_1 + w_2k_2)(w_1/k_1 + w_2/k_2) - (w_1 + w_2)^2 = (w_1w_2/k_1k_2)(k_1 - k_2)^2 >= 0 with equality iff k1 = k2

    \therefore D >= (w_1 + w_2)^2 - (z_1 + z_2)^2 >= (z/_1/ + /z_2/)^2 - (z_1 + z_2)^2 >= 0.

    and so L.H.S. <= 8 / ((w_1 + w_2)^2 - (z_1 + z_2)^2) which must now be proved <= 1 / (w_1^2 - z_2^2) + 1 / (w_2^2 - z_2^2)

    Now ai > 0, bi > 0 gives (a_1 b_1 - a_2 b_2)^2 + b_1b_2(a_1 - a_2)^2 + a_1a_2(b_1 - b_2)^2 >= 0 with equality iff a1 = a2, b1 = b2.

    i.e. (a_1b_1 + a_2b_2)(a_1 + a_2)(b_1 + b_2) >= 8a_1, a2 b1 b2

    whence 1/a_1b_1 + 1/a_2b_2 >= 8/(a_1 + a_2)(b_1 + b_2)

    Writing wi + zi = ai, wi - zi = bi we have the desired result

    For equality we must havek1 = k2,a1 = a2,b1 = b2
    i.e.k1 = k2,w1 = w2,z1 = z2
    i.e.x1 = x2,y1 = y2,z1 = z2
    and these are clearly sufficient.

Notes on the Questions

The students produced solutions of considerable variety. There was admiration for the way our team used calculus or complex variable methods in Q2, for example, and there were interesting solutions of Q6 using the minimum of the quadratic form P1(t) = x1t2 + 2z1t + y1.

Q6 was found hard by everyone (though 3 Hungarians produced correct solutions), and Q4 (our worst question) was also answered badly. Q2, 3 and 5 were generally found easier; we did particularly well on Q3.

It was evident from the results that the questions conformed more closely to the syllabuses taught in some countries than in others (the Swedes, for example, could attempt Q4 only by co-ordinate methods); this factor, and the careful selection and preparation of I.M.O. entrants which are traditional in some East European countries, make comparisons between countries rather suspect.

However, it should be stated that the choice of questions, within the overall broad framework, was determined by a majority vote of the International Jury. It is necessary, too, to exclude considerable areas of our customary Sixth Form work, since, for example, analysis and mechanics are not commonly taught in European schools.

There seemed to be some support among members of the Jury for questions—such as Q5—involving “perspicacité” rather than background knowledge, and perhaps some increase in the proportion of these would be helpful.

Report reproduced from New Science Teacher volume 13 number 2 (December 1969) pages 59–63.

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