15th International Mathematical Olympiad
MOSCOW, JULY 1973—A Personal Report by F. J. Budden

A few weeks before the start of the Olympiad I was asked to act as leader of the British delegation in place of Mr. R. C. Lyness, who had had to withdraw for medical reasons. I had the benefit of knowing the ropes, having been with the 13th IMO in Czechoslovakia in 1971. But the advantages of familiarity with the procedures in preparation and organisation, and of personal acquaintance with many of the delegates from other countries were somewhat offset by the fact that I was suddenly precipitated into the battle with little opportunity of preparation for it.

Moreover, for various reasons it was found necessary for the leader to travel in the same flight to Moscow as the team and deputy leader. This meant that I was unable to participate in the extremely important work of selection of questions. The 15 leaders of the remaining countries, forming the IMO “jury” had deliberated continuously over a period of two days, Friday and Saturday, 6 and 7 July, and by the time the British team arrived, the final selection of six questions had been settled. Whether the British leader’s presence at these meetings would have influenced the choice of questions will never be known. Certainly he would have had much to say, and meetings would have gone on longer!

The participating countries were:

Although I thought I “knew the ropes”, there were several departures from the procedure of two years previously in CS. Then the students had been accommodated in hostels entirely separately from the adults, and so had been subject to only a minimum of adult supervision (a fact which had troubled more than one delegate, including myself). In Moscow the arrangements were different, for each of the deputy leaders lived in the same hotel as the students. This not only ensured proper supervision, but it also meant that the members of the team were able to feel in continuous contact with their leaders instead of this contact being intermittent, and the British deputy leader quickly established a firm rapport with the team which contributed greatly to their morale.

The most notable contrast between 1971 and 1973 was the consequence of the IMO being held in a capital city. In Zilina, 1971, the whole competition was compact and intimate—persons and places were always within easy reach. In Moscow, by contrast, the family feeling was largely lost because personnel were so widely dispersed: students and deputies in the University Hotel, leaders in the Ukraine Hotel some five miles nearer the heart of the metropolis; while the rather ordinary school in which the Olympiad was held was itself several miles from each of these hotels.

These long lines of communication meant that one was continuously dependent upon private transport—buses and cars laid on by the organisers, or else upon the public buses or underground trains. Moreover, within the mammoth Ukraine Hotel the leaders were widely dispersed throughout the building. The British delegate in room 1707 (the hotel has 29 floors) would visit other delegates in remote rooms, and on one occasion a visit to room 776 to collect a batch of papers and straight back to 1707 took no less than 22 minutes, most of this time being spent waiting for lifts (the sole means of vertical translation). So large was the restaurant, so long were the hours during which a particular meal was available, and so great was the time spent waiting to be served, that it was unlikely one would meet one’s colleagues at a meal except by assignation; the result was less opportunity for fraternisation than had been the case in CS.

Again, in the provincial town of Zilina, the IMO had been a major event, marked with TV coverage, a special postal franking stamp, and flags flying. In Moscow, the event was submerged, and seemed to receive little mention, especially as the Moscow Film Festival was in prospect! There was less ceremonial occasions (dinners, parties), and less organised trips (in CS we were within easy reach of country of great scenic beauty). On the other hand, the actual business of the competition seemed to be got through with much more despatch, and never did meetings go on till the late hours as had occurred two or three times in CS. A strong factor here was that, whereas in CS proceedings had been translated in Russian, English, German, and sometimes French, as well as Czechoslovakian, here in Moscow only English and Russian seemed to be considered necessary. Indeed, in general conversation it was gratifying to note that English seemed to be the most favoured “master-language” with which almost all delegates were familiar, to varying degrees. It was rather surprising to find that in Moscow no attempt was made to mobilise any photocopying apparatus, which could have greatly reduced the amount of manual work in the preparation of the question papers and in the dissemination of results and marks.

On 7 July, the British party arrived one hour late at Moscow airport owing to an industrial dispute at Heathrow. This delay had the happy result of causing us to miss the experience of flying through a violent electric storm which had occurred at Moscow at the scheduled arrival time. Customs and passports were formalities, only straightforward questions being asked, and it was not long before the parties were on the way to their hotels, having been met by Russian IMO representatives at the airport.

I had a few hours in which to acquaint myself with the six selected questions ready for a meeting on Sunday morning, the purpose of which was to agree on the formulation of the questions in their final terms. After much discussion, master versions were produced in Russian and English, and these were precisely translated by leaders into their own respective languages.

For various reasons, not perhaps unconnected with the speed of meals service, the buses carrying the 125 participants together with the deputies did not arrive until half an hour later than the scheduled commencement of the opening ceremony. This was presided over by Professor Markuschevich and was quite brief. The teams soon found themselves being marshalled into eight ordinary classrooms, no classroom containing two from the same country. The papers were distributed, and candidates had a brief opportunity to ask for clarification, as a result of which some minor misprints came to light. The four-hour session lasted till almost 2.30 p.m., after which they all had to be fed. Because of this there was some delay in the boat trip along the River Moscow, which had been arranged for the combined parties.

That evening the leaders with their deputies were able to begin the marking of the papers. So far as a marking scheme was concerned, all that had been arranged at the morning’s meeting was that the points for the six questions should be 6+6+8+6+6+8 = 40. Subdivision of marks within questions was left to the discretion of individual teams, as it was thought that any discrepancies between teams could be ironed out during the process of co-ordination which followed. Marking of paper I continued the next morning, while the teams were writing paper II. It may be interesting to record that the whole process of marking plus co-ordination occupied the British leaders a total of 22 hours. This might well have been kept below 20 hours, if only the authorities had insisted on candidates using one side of the paper!

Co-ordination of the questions was carried out by a team of Russian university mathematicians. Two or three would be assigned to each of the six questions, and they would scrutinise the answers of their particular question for each of the 125 candidates, with the assistance of interpreters where necessary. The object was, of course, to secure uniformity of marking. In this difficult task, the co-ordinators deserve the highest praise for their penetrating grasp of the complex arguments in many languages, for their perception in detecting subtle flaws, and for their patient determination to see fair play.

By the morning of Thursday, 12 July, complete results of some countries were beginning to emerge, but the full set of results were not known till the beginning of the final meeting on Friday. In the 13th IMO, national aggregates had been computed, displayed and discussed, whereas it was noteworthy on the present occasion that this “team” aspect was played down, so that the truly individual nature of the Olympiad was properly recognised.

The chief business of the meeting on Friday morning was to decide upon the award of prizes, the main decision being to fix the boundaries of the first, second and third class prizes. This proved to be fairly straightforward, and the final decision was as follows:

Evidently, it was determined that golds and silvers should not be obtained too cheaply.

Goto (who came in the GB team at the last minute as first reserve) distinguished himself with a first prize (35 points), obtaining fifth place in the competition; Beasley (25) narrowly missed a second prize, and was closely followed by Treitel (22) and Herzmark (21). Pritchard (19) and Scholl (18) were also awarded third prizes.

A happy result of the demarcation of the marks for prizes was that Cuba received its first prize ever. Prompt action by the Cuban leader, via the Cuban Embassy, resulted in front page treatment in the Cuban newspapers the following day, where their successful candidate was given a status little short of that of a national hero.

The performance by the British team may be accounted very satisfactory: their most praiseworthy effort was in the way they responded to Q.6, with a total of 25 points for that question from the whole team. This was second only to USSR (30); the next best was France with 16. Had aggregates been computed, GB would have been placed fifth in the league. Another observation of interest is the remarkable recovery of France, hitherto always near the bottom of the international table, now advanced to sixth place; this was more remarkable since their team was apparently selected from schools confined to the Paris area.

A sprinkling of prizes for solutions of special merit had been awarded in the past. Co-ordinators had noted specially meritorious solutions during their work, and GB had proposed three: Herzmark for his solution of Q.1, Beasley for Q.3, and Pritchard for Q.5. (These we plan to publish in the next issue of “Science Teacher”, February 1974.) It was considered that to qualify for a special prize, most, if not all, of the following requirements should be fulfilled: the solution should be elegant, original (unique), flawless, brief, and the result of the question should, if possible, be generalised by the candidate. Unfortunately, two or three solvers had shared Herzmark’s solution for Q.1, so this was disqualified under the first criterion; Beasley’s for Q.3 was much liked by members of the jury, but it had to be withdrawn under the third criterion, since Beasley had omitted to distinguish maximum from minimum. In the end, no special prizes were awarded to any candidate. Possibly, the restrictions were too severe.

On 14 July, all participant teams went for an excursion, the first occasion on which they had been taken outside the Moscow perimeter. The closing ceremony was in the morning of Sunday, 15 July. Besides the presentation of the 65 prizes, there were speeches of congratulation and valediction, and the proceedings finished with a speech by Professor Bausch of East Germany, in which he invited representatives of the nations to compete in the 16th IMO which was to be held in Erfurt in DDR in 1974. Meanwhile, each delegation had been issued with a preliminary outline of the 1974 programme, from which it was clear that this particular IMO would be very well organised. It is understood unofficially that it is likely that the USA may be participating for the first time.

Later that day, David Goto was interviewed through an interpreter on Russian TV. Unfortunately, the festivities described below prevented the transmission being seen by the rest of the company.

The final event was a dinner for all in the University Hotel. At least, it started by being a dinner, but seemingly inexhaustible supplies of champagne and vodka ensured the speedy evaporation of decorum; the top table was regaled by successive teams displaying their vocal talents in impromptu contributions; music and dancing, including a not-to-be-forgotten performance of “Black Eyes” by one of the co-ordinators, and other convivial proceedings continued until the early hours.

Throughout the period of the Olympiad, there was a great deal of fraternisation between the students of the various teams. Several members of the British team organised an international bridge competition; addresses, souvenirs and invitations were exchanged, and it appeared that the IMO succeeded socially, as well as in its primary purpose. The British leaders found the society of their colleagues most agreeable. From many of them they received gifts, and these they were in the happy position of being able to reciprocate thanks to financial support by the British Awards Committee. Gifts were also presented to some of the Russians whose courtesy, hospitality and charm contributed so much to the success of the occasion.

Total Points by Country (maximum 320)

*(5 entrants only)

The Questions

1. O is a point lying in a straight line l. OP₁, OP₂, ..., OP_n are unit vectors with all P_i lying in a plane containing the line l and all of them on the same side of l. Prove that, if n is odd, then where |OM| denotes the length of the vector OM.

   (CS)

2. Find out whether or not there exists a finite set M of non-coplanar points in three-dimensional space such that, for any two points there exist two other points such that the straight lines AB and CD are parallel and do not coincide.

   (PL)

3. Find the minimal value of a² + b², where a and b are real numbers such that the equation x⁴ + ax³ + bx² + ax + 1 = 0 has at least one real root.

   (S)

4. A soldier has to check whether there are any mines in a region shaped like an equilateral triangle, including its boundary. The range of his detector is one half the altitude of the triangle. Starting from one vertex, which path should he choose to minimise the distance he walks to check the entire region?

   (YU)

5. A non-empty set G of non-constant functions f of the form f(x) = ax + b, where a, b real, with , and x real, has the following properties:

   1. If , then (where ), i.e. the set is closed under function composition.
   2. If , then the inverse function, where f^-1(x) = (x-b)/a for f(x) = ax + b.
   3. For each there exists an such that f(x_f) = x_f.

   Prove that there exists a real number k such that f(k) = k for all .

   (PL)

6. Let a₁, ..., a_n be n given positive numbers, and let q be a fixed real number, such that 0 < q < 1. Find n real numbers b₁, b₂, ..., b_n such that

   1. a_k < b_k for all k from 1 to n.
   2. for all k from 1 to n - 1.
   3. 

   (S)

Report reproduced from Science Teacher volume 17 number 2 (December 1973) pages 8, 10 and 11.

15th International Mathematical Olympiad

Solutions of Questions (see Science Teacher, December 1973)

1. (Based on solution by G. Herzmark)

   Suppose OP_m+1 is the middle of the 2m + 1 vectors, and makes angle with OX. Then we have OP_r (r = 1, 2, ..., m) within angle XOP_m+1 and OP_s (s = m + 2, m + 3, ..., 2m + 1) within the obtuse angle YOP_m+1. Let , .

   Resolute of resultant in direction .

   But .

   And .

   Hence the resolute along .

   Finally the resultant, being greater than its resolute, must also , equality occurring only when (r = 1, 2, ..., m) and (m + 2, ..., 2m + 1).

   

2. The British team produced five complete solutions, all correct and all essentially different. One general solution is a 3-D lattice of points forming a block a units × b units × c units (), containing (a + 1)(b + 1)(c + 1) points, and this requires that a, b, c have an H.C.F. > 1. The simplest solution (a = b = c = 2 with 27 points) was widely found, some solvers noticing that the middle point may be omitted giving 26 points. In the general case, all the internal points may be omitted so that only the points on the surface need to be taken.

   But not a single one of the 125 candidates found the solution using the minimum number of points (10). This is a cube, two opposite faces of which are surmounted by congruent right pyramids; and any affine transformations of the above will do equally well.

3. (Based on solution by M. Beasley)

   x⁴ + ax³ + bx² + ax + 1 = 0. (i) Replace x, a, b by t, x, y giving t⁴ + xt³ + yt² + xt + 1 = 0. (ii) which may be re-written x(t³ + t) + yt² + (t⁴ + 1) = 0, and this represents a family of straight lines with real parameter t. It is required to minimise x² + y² = r², i.e. the square of the distance from the origin to any of the lines. Now the minimum distance from 0 to lx + my + n = 0 is given by

   

   so in our case, we have

   

   Beasley proceeded by differentiation, but the work can be greatly reduced by using k = t² + t^-2, where for real t, and

   

   where K = k + 3, and .

   So p² = K + 9/K - 6, and since K + 9/K is minimum when K = 3 and increases for K > 3, the minimum p occurs when K = 5, when p takes the value 4/5.

4. This proved to be the easiest Q. overall, but many points were lost because solvers having found the correct minimal route, failed to check for complete coverage.

   Circular arcs are shown having radii equal to the range of the detector (= ½h). Having reached P on the boundary of circle centre C, he must then proceed directly towards B.

   Total path AP + PB must be minimised. This is achieved by having P on the altitude from C. (Proof by reflection, confocal ellipses, etc.) Checking for complete coverage is comparatively simple but must be carried out exhaustively.

   

5. (Solution by D. Pritchard)

   Consider any two arbitrary functions from G: f(x) = ax + b and g(x) = cx + d, where , .

   Now f(x_f) = ax_f + b = x_f (by defn.) x_f = b/(1 - a), . Similarly, x_g = d/(1 - c), . Let F(x) = f^-1g^-1fg(x) = x + (ad - bc + b - d)/ac (by computation), (, ). But by condition (i), , and so by condition (iii), there exists x_F such that x_F = F(x_F), i.e. x_F = x_F + (ad - bc + b - d)/ac ad - bc + b - d = 0 b/1-a = d/1-c (, ), i.e. x_f = x_g.

   But f and g were arbitrarily chosen, so x_f is constant for all , i.e. there exists a real k so that f(k) = k for all .

6. Let

   bk	=	, the summation being taken over all admissible i, with .
   	=	a1qk-1 + a2qk-2 + ... + ak-1q + ak + ak+1q + ak+2k2 + ... + anqn-k

   Then

   1. a_k < b_k is obvious (k = 1, 2, ..., n)

   2. qbk - bk-1	=	q(a1qk-1 + a2qk-2 + ... +ak-1q + ak + ak+1q + ... + anqn-k)
      		- (a1qk + a2qk-1 + ... + ak-1q2 + akq + ak+1 + ... + anqn-k-1)
      	=	(ak+1 + ak+2q + ... + anqn-k-1) (q2 - 1) < 0

      and

      qbk+1 - bk	=	q(a1qk + a2qk-1 + ... + akq + ak+1 + ak+2q + ... + anqn-k-1)
      		- (a1qk-1 + a2qk-2 + ... + ak + ak+1q + ak+2q2 + ... + anqn-k)
      	=	(q2 - 1) (a1qk-1 + a2qk-2 + ... + ak) < 0. So (ii) is proved.

   3. b1	+ b2 + b3 + ... + bn	=	a1 + a2q + a3q2 + ...	... + anqn-1
      			+ a1q + a2 + a3q + ...	... + anqn-2
      			+ a1q2 + a2q + a3 + ...	... + anqn-3
      			...	...
      			...	...
      			a1qn-1 + a2qn-2 + a3qn-3 + ...	... + an
      =	a1(1 + q + q2 + q3 + ... + qn-1) + a2(q + 1 + q + q2) +
      	a3(q2 + q + 1 + q + q2 + ... + qn-3) +
      	a4(q3 + q2 + q + 1 + q + ... + qn-4) + ... + an(qn-1 + qn-2 + ... + 1)
      <	(a1 + a2 + a3 + ... + an) (1 + 2q + 2q2 + 2q3 + ... + 2qn-1) =

   Other methods based on b_k = Max(a_k+i q^|i|), or Max(a_i q^|k-i|), a small being to ensure the strict inequality.

F.J.B.

Solutions reproduced from Science Teacher volume 17 number 3 (February 1974) pages 5 and 7.

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