## Task 2.

Let $n \geqslant 3$ be a natural number. Determine the minimum number of points that need to be marked inside any convex $n$-gon so that every triangle whose vertices are also vertices of that $n$-gon contains at least one marked point in its interior.
## Solution.

Since the diagonals from one vertex divide the $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points need to be marked.

We claim that it is possible to choose $n-2$ points such that the condition of the problem is satisfied. Let the vertices of the given $n$-gon be denoted by $A_{1}, A_{2}, \ldots, A_{n}$. Draw all the diagonals of the given polygon and color the regions determined by the diagonals $A_{1} A_{k}, A_{k} A_{n}$, and $A_{k-1} A_{k+1}$ for each $k \in\{2,3, \ldots, n-1\}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_59f140c89de2d58a74e9g-29.jpg?height=711&width=859&top_left_y=1712&top_left_x=587)

If we mark one point in each colored region, every triangle whose vertices are also vertices of the $n$-gon will contain at least one marked point in its interior. Indeed, the triangle $A_{l} A_{k} A_{m}$ for any $1 \leqslant l<k<m \leqslant n$ contains the entire colored region determined by the diagonals $A_{1} A_{k}, A_{k} A_{n}$, and $A_{k-1} A_{k+1}$, and thus the corresponding marked point.