# Problem 8.

Inside a convex $n$-gon, 100 points are placed such that no three of these $n+100$ points lie on the same line. The polygon is divided into triangles, each of whose vertices are 3 of the given $n+100$ points. For what maximum value of $n$ can there not be more than 300 triangles?
Answer: 102

Solution. If each point is a vertex of some triangle, then the sum of the angles of all the obtained triangles is $180^{\circ} \cdot(n-2)+360^{\circ} \cdot 100=180^{\circ} \cdot(n-2+200)=180^{\circ} \cdot(n+198)$. Therefore, $\frac{180^{\circ} \cdot(n+198)}{180^{\circ}}=n+198$ triangles are obtained. If one of the points does not participate in the cutting, there will be fewer triangles. Therefore, $n+198 \leqslant 300, n \leqslant 102$.

## B-2

Inside a convex 200-gon, $n$ points are placed such that no three of these $n+200$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+200$ points. For what maximum value of $n$ can there not be more than 400 triangles?

Answer: 101

## B-3

Inside a convex $n$-gon, 200 points are placed such that no three of these $n+200$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+200$ points. For what values of $n$ can there not be more than 600 triangles?

Answer: 202

## B-4

Inside a convex 100-gon, $n$ points are placed such that no three of these $n+100$ points lie on the same line. The polygon is cut into triangles, the vertices of each of which are 3 of the given $n+100$ points. For what values of $n$ can there not be more than 500 triangles?

Answer: 201

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary Stage 2020/21 academic year for $10-11$ grades

## B-1