50th Putnam 1989

Problem A5

Show that we can find α > 0 such that, given any point P inside a regular polygon with an odd number of sides which is inscribed in a circle radius 1, we can find two vertices of the polygon whose distance from P differs by less than 1/n - α/n3, where the polygon has 2n + 1 sides.



There are 2n+1 distances, all less than 2 (the diameter of the circle), so by the usual pigeon-hole argument there must be two distances differing by less than 2/2n. This is almost strong enough and suggests that we just need to refine this argument slightly.

The obvious refinement is that the diameter is slightly less than 2. In fact, it is a long diagonal of the polygon, length 2 cos(π/(4n+2)). So we need to show that cos(π/(4n+2)) < 1 - α/n2 for some positive α.

But it is easy to show that cos x < 1 - x2/8 for 0 < x < π/2 and so the result follows.



50th Putnam 1989

© John Scholes
1 Jan 2001