5. At the vertices of a regular $2 n$-gon, empty cups are placed. Petya and Vasya take turns (starting with Petya) pouring tea into the cups. In one move, a player can pour tea into two empty cups that are either adjacent or symmetric relative to the center of the $2 n$-gon. The player who cannot make a move loses. For which $n$ will Petya win regardless of Vasya's actions?
Answer: for odd $n$.

Solution. Let's call a pair two cups that are symmetric relative to the center of the $2n$-gon. Suppose $n$ is even. This means the number of pairs is even. Vasya's winning strategy is as follows. If Petya pours tea into cups from one pair, then Vasya fills the cups from some other pair. Suppose Petya fills two adjacent cups. Then the cups symmetric to them are also adjacent, and Vasya pours tea into them. After an even number of moves, the cups of any pair will be either simultaneously full or simultaneously empty. Therefore, the process will continue until all cups are filled. Due to the evenness of $n$, there will be no empty cups left after Vasya's move, and he will win.

Now suppose $n$ is odd. Petya should first fill the cups from some pair. After this, the number of free pairs will become even, and the problem will reduce to the already discussed case with the roles of Petya and Vasya reversed. Using the previously described strategy of Vasya, Petya will achieve victory.