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Brazilian Mathematical Olympiad

Combinatorics Difficulty 5.0 AIME, harder Prove it Brazil

In a chess tournament each player plays every other player once. A player gets 1 point for a win, $\frac{1}{2}$ point for a draw and 0 for a loss. Both men and women played in the tournament and each player scored the same total of points against women as against men. Show that the total number of players must be a square.

Solution

Let $m$ and $w$ be the number of men and women in the tournament. Each game assigns a total of 1 point to the players, the total of points assigned to games between two men and two women are $\binom{m}{2}$ and $\binom{w}{2}$ . Since each player scored the same total of points against men as against women and the total of games played is $\binom{m+w}{2}$ ,
$2 \left( \binom{m}{2} + \binom{w}{2} \right) = \binom{m+w}{2} \iff m+w = (m-w)^2,$
that is, the total number of players, $m+w$ , is a perfect square.

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