Sukhov K.

Two players take turns drawing diagonals in a regular $(2 n+1)$-gon $(n>1)$. A diagonal can be drawn if it intersects (at interior points) with an even number of previously drawn diagonals (and has not been drawn before). The player who cannot make a move loses. Who will win with correct play?
Note that on one side of each diagonal there is an even number of vertices, and on the other side - an odd number. Therefore, each diagonal intersects an even number of other diagonals of the $(2 n+1)$-gon. Suppose at some point in the game it is impossible to make a move, then each unmarked diagonal intersects an odd number of already marked ones, and consequently, an odd number of unmarked diagonals. Such a situation is possible only when the number of unmarked diagonals is even (see, for example, problem 30425).

Thus, if the total number of diagonals in the polygon is odd, the first player wins, and if it is even - the second. In a $(2 n+1)$-gon, the number of diagonals is $(2 n+1)(n-1)$ (see problem $\underline{60391})$, which is odd when $n$ is even and even when $n$ is odd.

## Answer

When $n$ is odd, the second player wins; when $n$ is even, the first player wins.

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