Prove that there exists a convex 1990-gon such that all its angles are equal and the lengths of the sides are the numbers 12, 22, ... , 19902 in some order.
Solution
By Robin Chapman, Dept of Maths, Macquarie University, Australia
In the complex plane we can represent the sides as pn2wn, where pn is a permutation of (1, 2, ... , 1990) and w is a primitive 1990th root of unity.
The critical point is that 1990 is a product of more than 2 distinct primes: 1990 = 2·5·199. So we can write w = -1·a·b, where -1 is primitive 2nd root of unity, a is a primitive 5th root of unity, and b is a primitive 199th root of unity.
Now given one of the 1990th roots we may write it as (-1)iajbk, where 0 < i < 2, 0 < j < 5, 0 < k < 199 and hence associate it with the integer r(i,j,k) = 1 + 995i + 199j + k. This is a bijection onto (1, 2, ... , 1990). We have to show that the sum of r(i,j,k)2 (-1)iajbk is zero.
We sum first over i. This gives -9952 x sum of ajbk which is zero, and - 1990 x sum s(j,k) ajbk, where s(j,k) = 1 + 199j + k. So it is sufficient to show that the sum of s(j,k) ajbk is zero. We now sum over j. The 1 + k part of s(j,k) immediately gives zero. The 199j part gives a constant times bk, which gives zero when summed over k.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 1999