IMO 1999

Find all finite sets S of at least three points in the plane such that for all distinct points A, B in S, the perpendicular bisector of AB is an axis of symmetry for S.

Solution

by Gerhard Woeginger

The possible sets are just the regular n-gons (n > 2).

Let A₁, A₂, ... , A_k denote the vertices of the convex hull of S (and take indices mod k as necessary). We show first that these form a regular k-gon. A_i+1 must lie on the perpendicular bisector of A_i and A_i+2 (otherwise its reflection would lie outside the hull). Hence the sides are all equal. Similarly, A_i+1 and A_i+2 must be reflections in the perpendicular bisector of A_i and A_i+3 (otherwise one of the reflections would lie outside the hull). Hence all the angles are equal.

Any axis of symmetry for S must also be an axis of symmetry for the A_i, and hence must pass through the center C of the regular k-gon. Suppose X is a point of S in the interior of k-gon. Then it must lie inside or on some triangle A_iA_i+1C. C must be the circumcenter of A_iA_i+1X (since it lies on the three perpendicular bisectors, which must all be axes of symmetry of S), so X must lie on the circle center C, through A_i and A_i+1. But all points of the triangle A_iA_i+1X lie strictly inside this circle, except A_iand A_i+1, so X cannot be in the interior of the k-gon.

40th IMO 1999

(C) John Scholes
jscholes@kalva.demon.co.uk
22 Aug 1999
Last corrected/updated 19 Aug 03