35. Given a regular 777-gon. Petya and Vasya take turns drawing its diagonals, Petya starts. The diagonals must not intersect inside the polygon. The player who, after their move, causes the diagonals to divide the 777-gon into parts, one of which is a quadrilateral, loses. Who wins, regardless of how the opponent plays?
$\square$ Answer: Vasya wins.

With his move, Petya splits the 777-gon into an $m$-gon and an $n$-gon, where $m < n$. In response, Vasya must cut off another $m$-gon from the $n$-gon. The remaining part will be a $(781 - 2m)$-gon, which is an odd-gon and therefore not a quadrilateral. Vasya then maintains the situation with one or several pairs of polygons with the same number of sides and a separate odd-gon. Specifically, when Petya makes a move in a polygon of a pair, Vasya responds with a similar move in the other polygon of the pair. When Petya makes a move in the odd-gon, Vasya responds by cutting off the second polygon for the pair. Since Vasya always has a move, he does not lose. And since the game ends (the number of diagonals in the 777-gon is finite!), Vasya wins.

(Rioplatense Olympiad 2013, Level 3)