# 15 pawns on a chessboard

15 pawns are placed on the centers of distinct squares of a chessboard. Prove that there are three pawns which form a right triangle.

In the example board below, a couple of right triangles are illustrated. There are many others, but you only need to prove that one right triangle exists for any possible placement of pawns. • Can you prove the first is a right triangle? Oct 19, 2017 at 19:43
• @Paparazzi: The sides have lengths sqrt(5), sqrt(45), sqrt(50). So the squares of the sides are 5, 45, and 50. Oct 19, 2017 at 22:20

First note that:

If two pawns are on the same column, then placing a pawn in the same row as either of them forms a right triangle. I'll call any pawn in such a pair grouped.

Because there are 8 columns and 15 pawns:

The maximum number of pawns that are not grouped is 7. 8 is impossible, since there are only 8 columns and this would mean you would place at most one pawn in each, leaving nowhere to place the other 7.

This means that:

There are at least 8 grouped pawns. If there were two of those in the same row, they would form a right triangle, so they must be all in different rows. But then you can't place any of the remaining 7 pawns without forming a right triangle somewhere, since all 8 rows must have at least one grouped pawn.

This bound is actually tight, since we can place 14 pawns on the chessboard while not forming any right triangles: • What is the least amount of right triangles possible?
– Carl
Oct 21, 2017 at 0:11