# Problem 7.

Inside a convex 13-gon, 200 points are placed such that no three of these 213 points (including the vertices of the polygon) are collinear. The polygon is divided into triangles, with the vertices of each triangle being any three of the 213 points. What is the maximum number of triangles that could result?
Answer: 411

Solution. If each of the points is a vertex of some triangle, then the total sum of the angles will be equal to $180^{\circ} \cdot(13-2)+360^{\circ} \cdot 200=180^{\circ} \cdot(11+400)=180^{\circ} \cdot 411$. This means 411 triangles are formed. If some points do not participate in the cutting, then the number of triangles will be less.

## B-2

Inside a convex 100-gon, 17 points are placed such that no three of these 117 points (including the vertices of the polygon) lie on the same straight line. The polygon is divided into triangles, with the vertices of each being 3 of the given 117 points. What is the maximum number of triangles that can result?

Answer: 132

## B-3

Inside a convex 200-gon, 13 points are placed such that no three of these 213 points (including the vertices of the polygon) lie on the same straight line. The polygon is divided into triangles, with the vertices of each being 3 of the given 213 points. What is the maximum number of triangles that can result?

Answer: 224

## B-4

Inside a convex 17-gon, 100 points are placed such that no three of these 117 points (including the vertices of the polygon) lie on the same straight line. The polygon is divided into triangles, with the vertices of each being 3 of the given 117 points. What is the maximum number of triangles that can result?

Answer: 215

## Lomonosov Olympiad for Schoolchildren in Mathematics

Preliminary stage $2020 / 21$ academic year for $7-8$ grades

## B-1

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