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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.

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  • $\begingroup$ Can you prove the first is a right triangle? $\endgroup$ – paparazzo Oct 19 '17 at 19:43
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    $\begingroup$ @Paparazzi: The sides have lengths sqrt(5), sqrt(45), sqrt(50). So the squares of the sides are 5, 45, and 50. $\endgroup$ – supercat Oct 19 '17 at 22:20
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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:

enter image description here

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  • $\begingroup$ What is the least amount of right triangles possible? $\endgroup$ – Carl Oct 21 '17 at 0:11

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