Belarusian Mathematical Olympiad

Answer: for odd $n$ first player wins, and for even $n$ the second player wins.

Let $n$ be even. Let's describe the winning strategy of the second player. First, similar to the solution of E1. of the problem of Category B, note that we can assume that the first player with his first move connected two adjacent vertices $P$ and $Q$. The second player with his first move connects one of them with adjacent vertex $R$. After that, the second player plays the game for $n-2$ vertices, assuming that the vertices $P$, $Q$ and $R$ "stick together" into one. It remains to check that the second player wins for $n=4$. Without loss of generality let the first player connect the vertices $A_0,A_1$. The second player with his first move will connect $A_2$ to the center of the circle. After that, there will be three components $A_0{A}_1,O{A}_2$ and $A_3$. The first player will connect some two of them with his move and the second player will win.

Let $n$ be odd. The solution is similar to the solution of the problem of Category B, since the first player connects the center at his first move.