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 either one empty cup or two cups that are symmetric with respect to the center of the $2 n$-gon, provided both are empty. The player who cannot make a move loses. For which $p$ will Petya win regardless of Vasya's actions?
Answer: for odd $n$.

Solution. Let $n$ be odd initially. Victory for Pete is ensured by the following strategy. His first move is to pour tea into two diametrically opposite cups. Then, on each step, he fills the cups that are symmetric relative to this diameter to the cups that Vasya filled on the previous move. Before Vasya's move, any two symmetric cups are either both free or both occupied. If Vasya fills two diametrically opposite cups, then the symmetric pair of diametrically opposite cups will remain empty, since with odd $n$ no diameter is symmetric to itself. Thus, Vasya's response move is always possible, and he will win.

Now let $n$ be even, that is, $n=2 m$. We divide the entire set of cups into $m$ quartets, forming the vertices of squares. To avoid losing, Vasya should follow this plan. If Pete fills two cups, then they are at the vertices of one of the squares, and Vasya fills the other two cups of the same square. If Pete pours tea into only one of the cups, then Vasya fills the cup in the adjacent vertex of the same square. In any case, Vasya will have a response move, so he will win.