In the interior of a 7-sided convex polygon, we select 10 points such that in the 17-element set $H$ consisting of the 7 vertices of the polygon and the selected points, no 3 points lie on the same line. We then partition the heptagon into triangles such that the set of vertices of the triangles is exactly $H$. What can we say about the number of triangles formed?
In the prescribed tiling, the triangles uniformly and without gaps cover the heptagon. Instead of counting the triangles, we calculate the sum of the angles of the triangles. The triangles, on one hand, contain the interior angles of the heptagon, the sum of which, since the heptagon is convex, is \((n-2) 180^{\circ}\), which for \(n=7\) is \(5 \cdot 180^{\circ}\). On the other hand, every interior point is a vertex of a triangle, and since no three points lie on a straight line, the full \(360^{\circ}\) around each of the 10 interior points contributes to the sum. Thus, the total sum is \(5 \cdot 180^{\circ} + 10 \cdot 360^{\circ}\). If we divide this by the sum of the angles of one triangle \((180^{\circ})\), we get the number of triangles formed.

This number is always 25, regardless of the tiling.

Miklós Bóna (Székesfehérvár, Petőfi S. Secondary School, 8th grade)

Remarks. 1. Several people tried to connect all points of \(H\) in every possible way and thus count the triangles. This is overly complicated and leaves no room for oversight. Moreover, even if it was possible to count the triangles formed by a specific arrangement of points, it is not certain that the number would be the same for a different arrangement. This would need to be proven separately.

2. The problem can be easily generalized. If we mark \(m\) points inside an \(n\)-sided convex polygon and decompose it into triangles according to the conditions of the problem, the number of triangles formed is \(n-2+2m\).