[ Examples and counterexamples. Constructions ]

Can 18 points be marked inside a convex pentagon so that the number of marked points inside each of the ten triangles formed by its vertices is the same?

#
Let's draw all the diagonals of a pentagon (see the figure). The ten triangles mentioned in the condition can be divided into two groups of five: 1) triangles formed by two adjacent sides and a diagonal of the pentagon; 2) triangles formed by a side and two diagonals of the pentagon.

![](https://cdn.mathpix.com/cropped/2024_05_06_dbba6af20ebd45cbbb54g-17.jpg?height=680&width=834&top_left_y=160&top_left_x=612)

Consider the pentagon formed by the points of intersection of the diagonals of the original pentagon and the five small triangles adjacent to the sides of the original. In each of these regions, mark three points. Any of the ten triangles mentioned above contains exactly two of the marked regions, so there are exactly six marked points inside each of them.

## Answer

It is possible.