IMO 1971

Let P₁ be a convex polyhedron with vertices A₁, A₂, ... , A₉. Let P_i be the polyhedron obtained from P₁ by a translation that moves A₁ to A_i. Prove that at least two of the polyhedra P₁, P₂, ... , P₉ have an interior point in common.

Solution

The result is false for 8 vertices - for example, the cube. We get 8 cubes, with only faces in common, forming a cube 8 times as large.

This suggests a trick. Each P_i is contained in D, the polyhedron formed from P₁ by doubling the scale. Take A₁ as the origin and take the vertex B_i to have twice the coordinates of A_i. Given a point X inside P₁, the midpoint of P_iX must lie in P₁ by convexity. Hence the point with doubled coordinates, which is obtained by adding the coordinates of A_i to the coordinates of X, lies in D. In other words every point of P_i lies in D. But the volume of D is 8 times the volume of P₁, which is less than the sum of the volumes of P₁, ... , P₉.

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.

13th IMO 1971

© John Scholes
jscholes@kalva.demon.co.uk
8 Oct 1998
Last updated/corrected 14 Mar 03