What is the smallest number of points that need to be marked inside a convex $n$-gon so that each triangle with vertices at the vertices of this $n$-gon contains at least one marked point?
Place points "near" the vertices of an $n$-sided polygon.

## Solution

Estimate. Since an $n$-sided polygon can be divided into $n-2$ triangles using diagonals, and each triangle must contain one point, at least $n-2$ points need to be marked.

Example. Let the vertices of the $n$-sided polygon be $A_{0}, A_{1}, \ldots, A_{n-1}$. Place the $k$-th point $B_{k} (k=1,2, \ldots, n-2)$ "near" vertex $A_{k}$ inside the angle $A_{0} A_{k} A_{n}$ (more precisely, in the intersection of this angle and the triangle $A_{k-1} A_{k} A_{k+1}$). Consider any triangle $A_{i} A_{k} A_{j} (i<k<j)$. This triangle contains the point $B_{k}$, since the angle $A_{i} A_{k} A_{j}$ contains the angle $A_{0} A_{k} A_{n}$, and the diagonal $A_{k-1} A_{k+1}$ intersects the segments $A_{i} A_{k}$ and $A_{k} A_{j}$.

## Answer

$n-2$ points.