5. On a circle, $n$ points are marked ( $n \geqslant 5$ ). Petya and Vasya take turns (starting with Petya) to draw one chord connecting pairs of these points. Any two drawn chords must intersect (possibly at the endpoints). The player who cannot make a move loses. For which $n$ will Petya win regardless of Vasya's actions?
Answer: for odd $n$.

Solution. First, let's prove the following statement: if in the final set of chords a triangle can be identified, two sides of which are adjacent points, then for odd $n$ Petya will win, and for even $n$ Vasya will win. Indeed, let the triangle be denoted as $A B C$, where $A$ and $B$ are adjacent points on the circle. Then any of the other chords must intersect the sides of triangle $A B C$. Therefore, either it starts from point $A$ and intersects side $B C$, or it starts from $B$ and intersects $A C$. The number of such chords is equal to the number of points different from the vertices $A B C$, that is, $n-3$. Taking into account the sides of $A B C$, the total number of chords is $n$. Therefore, for odd $n$ the last move will be Petya's, and for even $n$ it will be Vasya's.

We will show that Petya can guarantee the presence of a triangle. His first move should be to connect two adjacent points (say, $A$ and $B$). Vasya's response move will be a chord starting from $A$ or $B$ (for example, $A C$). Then Petya draws the segment $B C$, and the triangle is constructed.

We will show that Vasya can always ensure the presence of the required triangle. Suppose the first move is the chord $A C$. In response, Vasya draws the segment $A B$, where $B$ is the point adjacent to $A$. If Petya draws the chord $B C$ - the triangle is ready. Otherwise, Petya's response will be the segment $B D$, intersecting $A C$. But then Vasya can himself draw the chord $B C$, as it intersects all three already constructed chords.