Find 1975 points on the circumference of a unit circle such that the distance between each pair is rational, or prove it impossible.
Solution
Let x be the angle cos-14/5, so that cos x = 4/5, sin x = 3/5. Take points on the unit circle at angles 2nx for n integral. Then the distance between the points at angles 2nx and 2mx is 2 sin(n - m)x. The usual formula, giving sin(n - m)x in terms of sin x and cos x, shows that sin(n - m)x is rational. So it only remains to show that this process generates arbitarily many distinct points, in other words that x is not a rational multiple of π.
This is quite hard. There is an elegant argument in sections 5 and 8 of Hadwiger et al, Combinatorial geometry in the Plane. But we can avoid it by observing that there are only finitely many numbers with are nth roots of unity for n ≤ 2 x 1975, whereas there are infinitely many Pythagorean triples, so we simply pick a triple which is not such a root of unity.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
11 Oct 1998