|The IMO team at Komensky University, Bratislava, just after the
Ceremonial close of the proceedings:|
(from l. to r.) back: Hills, Edwards;
middle: Rodgers, Jackson, Bell;
front: Manton, Allwright, Vout.
Photo: P. Reynolds.
The amount of interest generated in the IMO in the host country rose to TV coverage, the flags of the nations flying outside the place of the competition, banners in the streets, and a specially created postmark in the Zilina, Czechoslovakia, post office. So seriously is the competition taken, particularly in the “Eastern” countries, that some school syllabuses are geared to it, and elaborate schemes of training are devised for the preparation of carefully selected teams. Moreover, the winning of a prize confers upon the student not so much a status symbol as an indelible mark of academic distinction which will be a passport for his future career,
Great Britain enters this professional arena as an amateur with only five years’ experience. It must be pointed out that the IMO is essentially a competition between individuals, limited to eight entrants from each participating country. Thus, from the point of view of the ethos of the IMO, the eight sixth-formers Great Britain sent to enter the 13th IMO did not constitute a team, any more than, say, the Australian entrants in the men’s singles at Wimbledon can be said to be one. Nevertheless, aggregates are computed, nations compared, league tables prepared, and post mortems held.
The following countries took part:
Each country sent eight competitors, with the exception of S (7) and C (4). This was Cuba’s first appearance in the Olympiad. As the cost of transport of teams from their country to the place of the competition is borne by the home country, Cuba’s expenditure for their half-size team and leader must have greatly exceeded that of all the others, with the possible exception of Outer Mongolia, whose journey takes a week each way. The generosity of some governments towards the furthering of opportunities of this sort for the young seems to know no bounds. Two other countries invited, Italy and Belgium, had not accepted.
The British team had been selected from the original 15,000 entrants to the 1971 National Mathematics Contest, the best 72 of whom had taken the BMO paper to decide the choice of team. They were: D. J. Allwright (Rugby S.), S. H. Bell (Dartford G.S.), D. M. Edwards (R.G.S. Newcastle upon Tyne), C. R. Hills (Dulwich College), D. J. Jackson (Perse S. Cambridge), N. Manton (Dulwich College), A. H. Rodgers (Royal Belfast Academical Institution), and C. Vout (Dulwich College).
The deputy leader, Mr. Peter Reynolds, Head of Mathematics at Doncaster College of Education, had charge of the team on its outward journey from London. They started at midday on 10 July, and survived a number of mishaps, including a lost air ticket from Prague to Bratislava, and absence of any reception upon arrival in Bratislava that evening. Having sorted out these problems and made contact with the other teams, they journeyed by chartered buses from Bratislava to the place of the competition in Zilina, a small, partly industrial, town on the fringe of the Low Tatras mountains in Slovakia. The teams of the 15 participating nations, comprising 115 students, including a handful of girls, resided in a students’ hostel in Zilina, and were thenceforward under separate supervision, as the deputy leader joined the “jury” which ran the competition.
The leaders of delegations had gathered in Bratislava on 7 July, and the first event was a dinner given by the Slovakian Ministry of Education. The speeches at this, as at other dinners and receptions later in the tour, were translated into English and Russian, which considerably protracted the proceedings. The salient message of the speakers seemed to be to convey the hope that the pursuit of mathematics would encourage international understanding and cooperation. Subsequently, we were to see this aspiration fulfilling itself, both between adults and also between members of the teams. True, discussion in the course of the jury sessions had been animated, but never acrimonious, and there developed a great amount of comradeship between colleagues. There is no doubt that conversations, comparisons of experiences and interchange of ideas are among the most valuable aspects of the mathematical get-together.
The heat wave was at its height on 8 July, when the party consisting of certain CS members of the IMO organisation, together with the leaders of delegations, set out by coach from Bratislava to what was mysteriously described as a “secret hide-out in the mountains”, there, in the Partizan Hotel, Talé, to prepare the competition. The jury consisted of heads of delegations, under the chairmanship of Prof. Stefan Schwarz, Academician, and head of mathematics in the University at Bratislava. Deliberations were conducted in four languages—English, French, German and Russian—the interpreters being all university mathematicians. Before departing for Zilina on 11 July, the jury had to select the questions for the two sessions each of four hours due to take place the following Tuesday 13th, and Wednesday 14th; to agree upon the formulation of the questions in the four master languages; and to decide upon the marks to be assigned to the questions which would correctly represent their relative weights.
Some 80 questions had been submitted by the participating countries, from which the CS committee had selected 17. There were two hard, full days of discussion before the final programme of six questions was agreed upon.
The discussion centred around the following points:
In fact, it is extremely difficult to find suitable questions to satisfy the requirements implicit in these principles, and to be acceptable to all participating countries. So far as (1) is concerned, great care was taken not to accept questions on work (such as coordinate geometry) which would be unfamiliar to the “Eastern” countries, whereas questions unrelated to British syllabuses (such as no. 3 on number theory) were included. While it would be an exaggeration to say that SU ran the show, there was no doubt that Soviet views were listened to with great respect, and deference paid to their proposals and opinions. At one point in the discussion about keeping to the “syllabus”, one delegate was heard to say “Let’s give them a surprise this year!”.
Nothing of a very surprising nature was included; the range of work covered was rather narrow; and the absence of certain topics caused some consternation. Several leaders were not happy about the “syllabus”, which excludes entirely coordinate geometry, calculus, probability, complex numbers, and mechanics. There was some mention of the possibility of a choice of questions to enable all candidates to work on familiar ground.
If this ever came into being—and the mood at the present moment does not inspire optimism—the British team would be in a stronger position; provided a great deal of care was taken to select the best team, and a certain amount of preliminary work was done, there is no reason at all why we could not offer a strong challenge to H and SU. As things stand, there is no doubt that the type of questions selected operate to GB’s disadvantage and favour the leading countries. It is rather like British spin bowlers having to play on matting wickets.
So far as difficulty was concerned, there is no doubt that, though the six questions finally selected were good ones, the papers were very difficult. Performances confirm this, with every country, except H and SU, doing very much worse than in previous years. A sad result of this increased difficulty was that no fewer than 39 of the 115 candidates obtained five or less marks out of the possible 42. The maximum marks for each question, and the country of origin were not known by the candidates.
Another issue was the coarseness of the marking, arising from the allocation of such low points for each question. A slight concession was made by the limited use of half marks.
The journey from Talé to Zilina provided opportunity for some spectacular sight-seeing, and there were other coach excursions arranged for the visitors, as well as for the boys. These trips had the dual purpose of giving some idea of the great natural beauties of Slovakia, and ensuring that teams and their leaders were kept apart in the interest of security. The climax of the sequence of excursions was the visit to the High Tatras mountains, near the Polish border, during the whole of the final Saturday.
From Monday, 12 July, the work of the augmented jury consisted first in preparing the actual papers for the examinations, and then, starting on Wednesday morning, marking and coordination occupied two days solidly. Each pair of delegates marked the papers of his own team according to his private mark scheme. These were then scrutinised by a team of CS coordinators, two being allocated to each of the six questions. Their difficult job was to ensure uniformity between the marking of the 15 teams.
The difficulty was increased by the requirement that they had to examine the solutions of each team in its own language. The coordinator of one question added to his difficulties by sitting on the opposite side of the table and reading the British answers upside down! This self-imposed handicap did not prevent his making shrewd comments, and detecting an error which had escaped the present writer during his original marking in his own language—the right way up!
The result of the coordinators’ adjudication was generally to cause a lowering of marks, especially in the case of Qs. 3 and 6, where they tended to require complete solutions and were not disposed to give any credit for a good try. The leaning towards an all-or-nothing marking scheme is really the only criticism one can make of them. Otherwise, their work was admirable and they are greatly to be praised for their tact, firmness, and integrity. Many of them also acted as interpreters, and tribute must be paid to their part in the proceedings, which called for stamina, concentration and courtesy; upon them hung the success of the whole operation.
One country held out stubbornly against some of the coordinators’ decisions, and this prolonged the final sessions of the jury, which were not completed till the early hours following the meetings on the evenings of Thursday and Friday. The final task of the jury had been to decide upon the award of prizes.
There were seven 1st prizes; 12 2nd prizes (including Jackson); and 29 3rd prizes (including Allwright, Hills, Rodgers, and Vout).
|1st prize||42–35||( 7 candidates)|
|2nd prize||34–23||(12 candidates)|
|3rd prize||22–11||(29 candidates)|
In addition there were four prizes for solutions of special merit to individual questions; here the jury rubber-stamped the selections made by coordinators.
On Sunday 18th, all involved in the IMO moved by train from Zilina to Bratislava. So tight was the programme that it was only while in transit that time could be found to mix with the teams; during the journey, students were informed by their respective leaders of their results, and their papers returned to them. Indeed, much mathematics was being done on the train by many of the students, and a number of unsolved teasers were released among this concentration of mathematical talent, the result of which was the solution by SU of a problem which has been troubling Oxford for years!
In Bratislava, after final tours for shopping and photography, all attended the final ceremony prize-giving in a hall of the university. This was presided over by Prof. Schwarz, while his fellow Academician, Prof. J. Novak, distributed the certificates. The actual prizes—which varied from opulent at the top end to generous at the lower end of the prize list, were given out by three young women in colourful Slovakian national costume. Apart from this charming touch, and also the intrusion of TV, the ceremony somewhat resembled an ordinary school prizegiving. It was concluded by the Polish leader inviting the 14th IMO to be held in Warsaw next year.
In the evening, there was a final ceremonial dinner, and much interchange of gifts between representatives of different countries. Thanks largely to the generosity of Messrs. Longmans Green and Co., the Mathematical Association, and the Science Teacher, the British delegates were able to make substantial reciprocation of gifts.
The GB team left early the next morning by planes from Bratislava to Prague. Looking back on the tour, the only criticism of the organisation, which was otherwise very good, was in regard to arrangements for transport to and from Prague. Once again we experienced difficulty in the bookings for these planes, but in the end we made our connection with the flight from Prague and arrived in London soon after midday.
Some of the more important statistics are given below. The performance of the British boys must be accounted a success on the whole, in view of the difficulty of the papers and the handicaps under which they worked as pointed out above.
Thanks must be expressed to the Guinness Awards, which made available once again the organisational and administrative services of Maurice Goldsmith and his staff.
The country of origin of the Q. and the number of marks awarded to it appear at the end of each Q.
Prove that the following assertion is true for n = 3 and for n = 5, and is false for every other integer n greater than 2:
“Whenever a1, a2, ...,
an are real numbers, the inequality below is
A convex polyhedron P1 with exactly nine vertices A1, A2, ..., A9 is given. Denote by P2, P3, ..., P9 the polyhedra which arise from P1 by the translations defined by the vectors , , ..., respectively. Prove that at least two of the polyhedra P1, P2, ..., P9 must have some of their interior points in common.
Prove that the set of integers of the form 2n - 3, n = 2, 3, 4, ... contains at least one infinite subset every two members of which are relatively prime.
The tetrahedron ABCD has all its faces acute-angled triangles. We consider all closed polygonal lines XYZTX defined as follows: X is any point on the edge AB, coinciding with neither A nor B. Similarly Y, Z, T are interior points of the edges BC, CD, DA respectively. Prove that:
Prove that, for any positive integer m, there exists a non-empty finite set S of points in a plane with the following property: Every point of S is at unit distance from exactly m other points of S.
Consider an n × n array of non-negative integers:
with the following property: Whenever an element aij is zero, then the combined sum of all the elements of the ith row and the jth column is greater or equal to n. Prove that the sum of all the elements of the array is greater or equal to ½n2.
Solutions to appear in the December 1971 issue of ‘Science Teacher’.
Report reproduced from Science Teacher volume 15 number 1 (October 1971) pages 1, 3, 4 and 6.
n = 3. .
n = 5. Without loss of generality, since expression is symmetrical, assume . Then first and second terms each contain the factor a - b, and together equal (a-b) [(a-c) (a-d) (a-e) - (b-c) (b-d) (b-e)] which is since , etc. Similarly the last two terms are together . The middle term , since it is the product of two non-negative and two non-positive terms.
n even. Since each term of odd degree, when all the a’s change sign, the expn changes sign. But we can certainly find values of the a’s to make it positive, and therefore we can also make it negative. OR, consider the values 0, 0, 0, ...., 0, -1, for which the expn = (-1)n-1 < 0 when n is even.
n odd > 6. Choose the values 0, 0, 0, ..., 0, 1, 1, 1, a. Then expn = (a - 1)3 an-3 and this is negative when 0 < a < 1. Note this counter-example does not work when n = 3 and when n = 5. Other c-examples possible.
Denote by P' the polyhedron derived from P by an enlargement 2 : 1, centre A1. Then all of P1, P2, P3, ..., P9 are contained within P'. For if the translation takes the point X (on or within P1) into Y (on or within Pr), then , so A1Y and ArX have the same mid-point Z. But Z lies on XAr, and is interior to P1 (being convex), and so Y is interior to P'. Since volume of P' = 8 × volume of P1, while combined volume of P1, ..., P9 is 9 × volume of P1, it follows that at least two of the latter have interior points in common.
Notes. A similar argument in n-dimensions shows that a polytope with 2n + 1 vertices suffering these translations must result in polytopes with some common volume.
In 2D, argument shows pentagons will not work. Easy to show that the only quadrilaterals which work are parallelograms. Hence faces of solid must be triangles and parallelograms.
Conjecture: only polyhedron with 8 vertices which works is parallelopiped. Finally, for polyhedron with >9 vertices, obviously true provided convex.
Note: . It is not known for any m whether the sequence contains an infinite number of primes; e.g. for m = -1, we have 2n - 1, and the Mersenne primes may be finite in number.
22n is not a solution, for the numbers 2222, etc., are all = 3 (mod. 13).
Solution. Suppose 2n1 - 3, 2n2 - 3, ...., 2nk - 3 are all coprime pairs, with prime factors p1, p2, ...., pn. Let nk+1 = 1 + (p1 - 1)(p2 - 1) .... (pn - 1). By Fermat, 2pi = 1 (mod. pi), i = 1, 2, ..., n.
|Hence||2nk+1 - 3|
|=||2.2(p1-1)(p2-1).....(pn-1) - 3|
|=||2.[2(p1-1)](p2-1)(p3-1)....(pn-1) - 3|
|=||2. 1(p2-1)(p3-1)...(pn-1) - 3 (mod. p1)|
|=||2.1 - 3 = -1 (mod. p1).|
Similarly, it also = -1 (mod. pi, i = 2, 3, ..., n). Hence 2nk+1 - 3 contains none of the prime factors of 2ni - 3 (= 1, 2, ..., k), and the result follows.
Using the ‘net’ of the tetrahedron (see Fig. 1). If Z, Z' are positions of the point on CD, C'D', so that CZ = C'Z', then for minimum perimeter we must have X, Y, T lying on str. line ZZ'. Clearly . In this case, ZZ' is constant for all positions of Z, Z', and these infinitely many are all of length (from ) = , where , etc. Note that .
When CD not parallel to C'D' (Fig. 2, showing the case CC' > DD'), then ZZ' diminishes as Z moves from C to D, for Z'Z''Z < Z'Z''D'' = Z''Z'D' < Z''Z'Z ZZ' < ZZ'' = CC', and similarly ZZ' > DD'. Here the perimeter has no minimum value, since the set of values is open, the end points being excluded.
Necessity of acute-angled triangles: ZZ' must cut the edges CB, BA, AD', and this may fail if both A and B lie outside parallelogram CDD'C' and on the same side of CC' or of DD'. This is impossible in the case when the triangles are acute-angled, since it would require either (when ), or (when ).
The ‘official’ solution gave 2m points, but working inductively, Hills gave the following solution with only 3.2m-2 points (conjecture: is this the minimum possible?). Call the required figure Mm (see Fig. 3). Then M2 in simplest form is an equilateral triangle. M3 may be obtained from M2 by giving it a translation through unit distance in a suitable direction. Similarly M4 may be obtained by translating M3 through unit distance, the number of points being doubled at each stage, leading to Mm with 3.2m-2 points. It remains to specify what we mean by a suitable direction. This is such as to avoid any unwanted equilateral triangles. Hills did this by taking (see Fig. 3) to be rational multiples of , successive values of being halved, and this would appear to guarantee no possibility of unit distances other than the wanted ones, by the cosine formula. A simpler way is to say that unwanted equilateral triangles can be made in only a finite number of ways, using a finite number of possible values of , so take (at each stage) to have a value other than one of these ..... (F.J.B. and P.R. considered this solution to be worthy of consideration for a special prize, having the merit of doing the job with only three-quarters of the points of the ‘official’ solution, and also being quite distinct from it.)
Let r = smallest of all row sums and column sums. If , then S, the sum of all the elts. of the array, is , and the assertion is true. So let r < ½n. Consider the row (or column) whose sum is r, and suppose it contains z zeroes. Let the sum of the z columns containing those zeroes be S0 and let the sum of the n-z columns containing the non-zeroes be S+. Then, adding the given criteria for the z zero elts, . Also, since each remaining column has sum , we have .
|By addition, S||=|
|=||½n2 + nz - 2rz + nr - ½n2|
|=||½n2+ ½(n-2r) (2z-n).|
But n > 2r, by assumption, so n-2r > 0, and, since the row contains n-z non-zero terms, each , we have also r > n-z, so that n > 2(n-z) 2z-n > 0, and the result follows.
Solutions reproduced from Science Teacher volume 15 number 2 (December 1971) pages 40 and 41.
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