Track / Stage 6 / 39 of 400 #1039 of 1964

Problem 1039

National olympiad, first round

Combinatorics Difficulty 6.0 Prove it

In a school chess tournament, everyone played against everyone else exactly once. Each player scored the same number of points against the girls as against the boys. Prove that the number of participants is a perfect square. (1 point for a win; 0.5 points for a draw; 0 points for a loss.)

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This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution. Let the number of girls be $L$ , and the number of boys be $F$ . The girls played a total of $\binom{L}{2}$ matches among themselves. Since exactly 1 point is awarded in each match, the girls scored a total of $\binom{L}{2}$ points against each other. According to the problem, the girls also scored a total of $\binom{L}{2}$ points in their matches against the boys. Similarly, the boys scored a total of $\binom{F}{2}$ points against the girls. Since there were $L F$ girl-boy matches, we have

$L F=\binom{L}{2}+\binom{F}{2}$

which simplifies to $2 L F=L(L-1)+F(F-1)$ . From this equality, the total number of boys and girls participating in the competition is $L+F=L^{2}+F^{2}-2 L F=(L-F)^{2}$ , which is indeed a perfect square.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.