21st Putnam 1960

Problem A2

Let S be the set consisting of a square with side 1 and its interior. Show that given any three points of S, we can find two whose distance apart is at most √6 - √2.

Solution

Suppose we can find three points of S so that their three distances exceed k = √6 - √2. If one of the points is not on the top side of S and another on the bottom side, we can expand S by a combination of translation and expansion in the direction perpendicular to those sides to get three points with larger minimum distance. Similarly, if one of the points is not on each of the other two sides of S, we can expand S by a combination of expansion and translation to get three points with larger minimum distance. So we end up with a point on each side. But there are only three points, so one point must be at a vertex of the square. The other points on the two sides meeting at the vertex are all a distance <= 1 < k from the vertex, so the other two points must be on the other two sides. Let the vertices be A, B, C, D (in order). Assume the vertex point is at A. Let X be the point on BC a distance 2 - √3 from B. Then AX = k, so the point on BC must lie on the segment XC. Similarly, let Y be the point on DC a distance 2 - √3 from D. Then AY = k, so the point on DC must lie on the segment YC. But XY = k, so the two points will be closer than k unless one is at X and the other at Y but in that case the distances AX, AY do not exceed k. So it is not possible to arrange for the three points to have all distances > k.

© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002