The Olympiad took place in Australia from 9th to 21st July 1988 and was designated an Australian Bicentennial Event, with the Prime Minister of Australia as Patron. The main venue was Canberra but teams and leaders spent two or three days in Sydney on arrival.
The team reached Sydney on 10th July accompanied by Paul Woodruff (Dulwich College), the Deputy Leader. I had arrived the previous day, in accordance with the programme for leaders, and I accompanied the team on the homeward journey, reaching London early on 25th July. Three days outside the official Olympiad dates were spent in Australia in order to take advantage of an excursion fare. Travel was by Qantas and all arrangements went smoothly.
The organisation of the Olympiad was extremely efficient, with a comprehensive programme of leisure activities to complement the academic work. Innovations included visits by teams to schools, students’ homes and to Embassies and High Commissions. There was a special programme for “accompanying persons”. The teams were accommodated, and the contest took place, at the Canberra College of Advanced Education. The first part of the Jury’s time was spent at University House and the Australian Academy of Sciences building on the campus of the Australian National University. The Jury moved to CCAE after the second examination. The team’s accommodation in Sydney on arrival, and for the extra days, was at the University of New South Wales.
The Jury met under the chairmanship of Professor Renfrey Potts of the University of Adelaide, who guided its proceedings with friendly efficiency. The selection, translation and typing of the problems were completed well within the time allocated. Of the five problems submitted by the United Kingdom, four were on the short list preselected by the Jury, and one of these was among that committee’s “starred” choices. This problem was selected as number 3 for the Competition. Judged by the number of contestants gaining the full mark of 7 for it, this proved to be the second hardest problem. The decisions on prizes were also made without undue difficulty. As Leader of a team submitting a contest problem, I was a member of the subcommittee to adjudicate on special prizes. Only one such prize was awarded, to a Bulgarian contestant for his solution to the extremely difficult, but very interesting, problem 6.
49 countries competed, several others being represented by observers. Our team, with a total mark of 121, gained three second prizes (Malcolm Law, Gareth McCaughan and Oliver Riordan) and two third prizes (Cohn Bell and Christopher Nash). We were 11th in “national order”, after USSR (217 out of 252), Romania (201), China (201), West Germany (174), Vietnam (166), USA (153), East Germany (145), Bulgaria (144), France (128) and Canada (124). This is much the same position as in recent years. In an attempt to improve it, we hope to improve our publicity to ensure that we attract the best candidates available into our selection procedure and to make some modifications to our tests.
Everybody greatly enjoyed the trip. Warmest thanks are due to the Olympiad organisers under Mr Peter J O’Halloran for their splendid arrangements and hospitality. I add especially my own thanks to Paul Woodruff for his mathematical work and careful attention to the well-being of the team. We are all most grateful to the sponsors who made our participation possible.
Department of Mathematics
University of Edinburgh
(King Edward’s S, Birmingham)
(Lincoln Christ’s Hospital S)
(King Edward’s S, Birmingham)
(St. Paul’s School)
(City of London School)
|out of a maximum|
total of 252
|Vietnam||166||Republic of Korea||79||Iceland (4)||37|
|East Germany (5)||145||Hong Kong||68||Spain||34|
Numbers () indicate teams of less than 6.
Consider two coplanar circles of radii R and r (R > r) with the same centre. Let P be a fixed point on the smaller circle and B a variable point on the larger circle. The line BP meets the larger circle again at C. The perpendicular to BP at P meets the smaller circle again at A (if is tangent to the circle at P then A = P).
Let n be a positive integer and let A1, A2, ...., A2n+1 be subsets of a set B. Suppose that
For which values of n can one assign to every element of B one of the numbers 0 and 1 in such a way that each Ai has 0 assigned to exactly n of its elements?
A function f is defined on the positive integers by
f(1) = 1, f(3) =
f(2n) = f(n)
f(4n+1) = 2f(2n+1) - f(n)
f(4n+3) = 3f(2n+1) - 2f(n)
for all positive integers n. Determine the number of positive integers n, less than or equal to 1988, for which f(n) = n.
Show that the set of real numbers x which satisfy the inequality
is a union of disjoint intervals, the sum of whose lengths is 1988.
ABC is a triangle right-angled at A, and D is the foot of the altitude from A. The straight line joining the incentres of the triangles ABD, ACD intersects the sides AB, AC at the points K, L respectively. S and T denote the areas of the triangles ABC and AKL respectively. Show that .
Let a and b be positive integers such that ab + 1 divides a2 + b2. Show that is the square of an integer.
Problems 1 – 3 first day; 4 – 6 second day. 4½ hours each day.
Each problem is worth 7 points.
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