THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Question

On each side of a triangle we consider 9 distinct points, different from the triangle's vertices. Determine the number of triangles having the vertices in three of these 3 9 points. Lucian Dragomir

Accepted Answer

The triangles can be of two types: with all three vertices on different sides or with two vertices on one side and the third vertex on another side.

In case all vertices are on different sides, each vertex may be chosen in one of

9

ways, accounting for a total of

9^{3} = 729

such triangles.

In the other case, there are

\frac{9 \cdot 8}{2} = 36

ways of choosing two vertices from a given side. The third vertex can be chosen from the other two sides in

2 \cdot 9 = 18

ways, hence there will be a total of

3 \cdot 2 \cdot 9 \cdot 36 = 3 \cdot 648 = 1944

such triangles.

So there are

729 + 1944 = 2673

triangles with vertices among the given

27

points.