Shapovalov A.V.

In a convex $n$-gon, several diagonals were drawn such that no three or more of them intersect at any point inside the polygon. As a result, the polygon was divided into triangles. What is the maximum possible number of triangles?
Let $n=2 k$ or $2 k+1$.

Example. We will cut off quadrilaterals from the $n$-gon by diagonals until it is no longer possible. In total, $k-1$ quadrilaterals will be obtained (and one more triangle for odd $n$). By drawing diagonals in each quadrilateral, we get $4 k-4=2 n-4$ triangles for even $n$ and $4 k-3=2 n-5$ triangles for odd $n$.

Estimate. No diagonal can intersect two others (consider the segment of the diagonal between adjacent intersection points; on the side where the sum of the angles is at least $180^{\circ}$, a triangle would adjoin it). Thus, intersecting diagonals are divided into pairs that intersect each other. Removing such pairs, we obtain a partition into quadrilaterals and triangles (let there be $l$ and $m$ of them, respectively). Calculating the sum of the angles in two ways, we get $2 l+m=n-2$. From this, the number of initial triangles is $4 l+m=2(2 l+m)-m=2 n-4-m \leq 2 n-4$.

For odd $n$, however, $m$ is also odd, so $m \geq 1$ and $2 n-4-m \leq 2 n-5$.

## Answer

$2 n-4$ for even $n$, $2 n-5$ for odd $n$.