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David C Hunt

This year Argentina hosted the second International Mathematical Olympiad (IMO) to be held in the Southern Hemisphere and the first ever in South America. The IMO was held in Mar del Plata, a resort city 400 kilometres south east of Buenos Aires, from 18th to 31st July 1997. It was attended by 460 competitors, all high school students, representing a record eighty two countries, seven more than in 1996. There were a number of new and returning countries from Central and South America, namely, Bolivia, Guatemala, Paraguay, Peru, Puerto Rico, Uruguay and Venezuela. In addition, there are now eleven teams representing constituent republics from the former Soviet Union. Hosting an IMO is clearly much more expensive than it was when we hosted the 29th IMO in Canberra in 1988.

As AMOC Director of Training and Australian Team Leader, the IMO is the highlight of my year and it represents the culmination of up to three years preparation for the six team members. The Australian Mathematical Olympiad Committee (AMOC) is a subtrust of the Australian Mathematics Trust of which the University of Canberra is trustee. The AMOC receives significant Commonwealth Government sponsorship but is totally dependent on academics, school teachers and ex-Olympians to carry out its programs. This year, for the first time, the Australian Mathematical Society made a welcome contribution to the team members' fares. A key part of the training is the ten day School of Excellence held in Melbourne in December and the Team Selection School held in Sydney in March or April each year. We invite twenty five students ranging from year 9 to year 12 to each school. The training does not end with team selection. Each week between Easter and early July the team members are sent a full length trial examination to attempt and discuss with their mentor, an ex-Olympian living in the same city. Three of this year's team were from Sydney, two from Melbourne and one from Adelaide.

There were several interesting and highly successful innovations at the 38th IMO. One of these was to invite experts from Brazil, Bulgaria, New Zealand, Poland and Russia to work with local mathematicians as the Problem Selection Committee. The committee worked long hours to sift through over 100 questions submitted by 42 countries to create a short list of 26 questions. These problems, with alternative formulations and solutions, were provided to the Jury, which comprises the eighty two team leaders. The shortlist contained 10 problems on geometry, 5 on number theory, 3 on polynomials, 3 on inequalities, 3 on recurrences and functional equations and 2 on combinatorics. The task of the Jury over the first three days was to reduce the shortlist to just six problems which would challenge the best in the world without completely demoralising the less well trained students. As most Jury members had hoped, the questions selected were significantly easier than those selected at Mumbai in 1996. The average score on the examination was 16.06 (out of the possible 42), well up on the 1996 average of 12.52. The exam paper is included and those interested in checking their own solutions can acquire a copy of the Australian Scene from the Trust. Question six is quite hard, but the rest are relatively straight forward.

While the Jury was hard at work Australia's Deputy Team Leader, Angelo di Pasquale of the University of Melbourne, was helping the team to relax in Buenos Aires in the hope of top performances from the team. On 21st July all the teams arrived and were housed at Hotel Treize de Julio, a very comfortable hostel which was well equipped with games rooms and computers. The examinations were held on successive mornings and, as usual, were a time of tension and hard work for all involved.

The marking and coordination phase of an IMO takes about three days. This process was made easier this year by another innovation - the organisers photocopied all the candidates scripts. This meant that before Angelo and I met the coordinators they had read our students' answers and marks were quickly agreed upon. As can be seen from the results outlined below Australia performed very well. The team came 9th overall and the individual students obtained two gold medals, three silver medals and a bronze medal. This is easily the best performance by any Australian team since we first entered the competition in 1981.

Congratulations are due to all participants, especially to the four students from Iran, Romania, United States of America and Vietnam who obtained perfect scores of 42. The organisation carried out by Professor Patricia Fauring and her Organising Committee ensured that the Olympiad was a resounding success. The team that Patricia had organised was huge and nothing that could have been done to make us welcome was left undone. As a Jury member I give special thanks to the co-chairmen Carlos Bosch and Luis Caffarelli and also to conference and travel coordinator Marta Joltac whose skills and patience made our stay in Mar del Plata so enjoyable.

Argentina is one of the most beautiful countries in the world although few who attended the IMO had the chance to travel widely. However, the friendliness of the people and the obvious safety in the cities as well as the natural attractions have encouraged me to try and arrange a later extended visit.

Some statistics:

Some placings:

1. China (223) 2. Hungary (219) 3. Iran (217) =4. USA and Russia (202)
6. Ukraine (195) =7. Bulgaria and Romania (191) 9. Australia (187)
10. Vietnam (183) 11. Japan (168) 12. Korea (164) 13. Germany (161)
14. Taiwan (148) 15. India (146) 16. UK (143) 17. Belarus (140)
18. Czech R. (139) 19. Sweden (127) 20. Poland (125) 30. Canada(107)
34. France (101) 39. South Africa (93) 41. Singapore (88) 48. New Zealand (71).

Australia's Individual Team Member's Results:

Name Q1 Q2 Q3 Q4 Q5 Q6 Total Award
Norman Do 4 7 7 7 7 0 32 Silver
Stephen Farrar 7 7 5 7 7 2 35 Gold
Justin Ghan 2 7 7 7 7 0 30 Silver
Jonathan Kusilek 4 0 4 7 7 1 23 Bronze
Thomas Lam 7 7 7 7 7 3 38 Gold
Daniel Mathews 0 7 7 7 7 1 29 Silver

The Questions

First day. Time allowed 4.5 hours. Each problem is worth 7 points.

  • 1. In the plane the points with integer coordinates are the vertices of unit squares. The squares are coloured alternately black and white (as on a chessboard). For any pair of positive integers m and n, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths m and n, lie along edges of the squares.
    Let S_1 be the total area of the black part of the triangle and S_2 be the total area of the white part. Let
    f(m,n) = |S_1 -S_2|.
    (a) Calculate f(m,n) for all positive integers m and n which are either both even or both odd.
    (b)Prove that f (m,n) \leq \frac{1}{2} \hbox{max}\{m,n\} for all m and n.
    (c) Show that there is no constant C such that f(m,n)<C for all m and n.

  • 2. Angle A is the smallest in the triangle ABC.
    The points B and C divide the circumcircle of the triangle into two arcs. Let U be an interior point of the arc between B and C which does not contain A.
    The perpendicular bisectors of AB and AC meet the line AU at V and W, respectively. The lines BV and CW meet at T.
    Show that
    AU = TB+ TC.

  • 3. Let x_1, x_2, x_3,... x_n be real numbers satisfying the conditions: |x_1 +x_2 + ... + x_n | = 1

    |x_i| \leq \frac{n+1}{2} \hbox{for} \quad i =1, 2, ..., n.
    Show that there exists a permutation y_1, y_2, ..., y_n of x_1, x_2,... x_n such that
    |y_1 + 2 y_2 + \ldots + ny_n| \leq \frac{n+1}{2}.

    Second day. \quad Time allowed 4.5 hours. Each problem is worth 7 points .

  • 4. A n\times n matrix (square array) whose entries come from the set S=\,2, ..., 2n-1\} is called a silver matrix if, for each i= 1, \ldots, n, the ith row and the ith column together contain all elements of S. Show that
    (a) there is no silver matrix for n= 1997;
    (b) silver matrices exist for infinitely many values of n.

  • 5. Find all pairs (a,b) of integers a\geq 1, b\geq 1 that satisfy the equation
    a^{b^2}= b^a.

  • 6. For each positive integer n, let f(n) denote the number of ways of representing n as a sum of powers of 2 with nonnegative integer exponents.
    Representations which differ only in the ordering of their summands are considered to be the same. For instance, f(4)=4 because the number 4 can be represented in the following four ways: 4; 2+2; 2+1+1; 1+1+1+1.
    Prove that, for any integer n\geq 3,
    2^{n^2/4} < f(2^n)< 2^{n^2/2}.

    IMO Team Leader
    AMOC Director of Training
    The University of New South Wales

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