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, 2016

Combinatorics Difficulty 5.6 AIME, harder Find the answer Slovenia

Three friends, Andrej, Blaž and Cene, were playing badminton. For each game, two of them were playing one against the other, and the third was free. After each game, the winner of the game played against the one that was free in the last game. Andrej played $17$ games and Blaž played $23$ games. At least how many games did Cene play?

Pick one

Solution

Let $n$ be the number of games played by Cene. Then the total number of games was $\frac{17+23+n}{2}$ , which implies that $n$ is even. The total number of games was at least $23$ since Blaž has played this many of them. Since Cene was free for at most one game in a row he had to have played at least $11$ games, but $n$ is even which means he played at least $12$ games. So the total number of games was at least $\frac{17+23+12}{2} = 26$ , which in turn implies that Cene has played at least $13$ games, or, due to parity, at least $14$ . In this case Andrej and Blaž have played $\frac{17+23-14}{2} = 13$ times, Andrej and Cene $\frac{17+14-23}{2} = 4$ times and Blaž and Cene $\frac{23+14-17}{2} = 10$ times. To show that this is indeed possible consider the following example: C-A, A-B, B-C, B-A, A-C, A-B, B-C, B-A, A-C, A-B, B-C, B-A, A-C, A-B, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C, B-A, B-C. The correct answer is (D).

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