This solution, by Titu Zvonaru, is given in the February 2008 issue of Crux Mathematicorum:
Suppose that the most unfortunate player beat 3 others. Label the players so that this player is P1 and that the players he beat follow him immediately in the lineup, here as P2, P3, and P4. Since the minimum number of wins is 3, there must be some player Pk such that player P2 beat player Pk and k is greater than 4 (otherwise player P2 could have beaten only players P3 and P4, which would give him the minimum number of wins, contrary to our supposition). This means that player Pk must have beaten player P1, and we can declare that A = P1, B = P2, and C = Pk.
The same argument holds for any minimum number of wins. (It wouldn’t work if the minimum were zero, but we’ve been told that’s not the case.)
09/02/2024 Reader Robert Filman offers an inductive proof:
The theorem is obviously true for n = 1, 2 and 3. (n=1 is not a possible situation, because everyone won a game, and with n=1, there are no games; with n=2, there’s only one game, so the loser of that game had no one to beat. With n=3, the conditions require A>B, B>C, and C>A.)
Assume hypothesis is true for 1, 2, 3, 4, … n – 1 players, and prove it true for n players.
Pick any player, A. Divide the players into the ones that beat A (set W) and the ones that lost to A (set L). L is not empty, because A beat someone.
If any player x in L beat any player y in W, we have a triangle <a, x, y>
If not, consider the set L. No one in L beat anyone outside of L. Every player in L’s win came against another player in L. But the size of L is at most n – 1. So by induction, the theorem is true for L.
Hence, the theorem is always true.
(Thanks, Robert.)
