Junior Macedonian Mathematical Olympiad

First we determine the total number of splitting triangles into which the hexagon is split. Let $A$ be one arbitrary point of the interior $m$ points. The sum of all angles at the point $A$ is $360^{\circ }$ (the sum of all angles at $A$ of all triangles having that point as a vertex). On the other hand, the sum of all angles at a vertex of the hexagon is $120^{\circ }$. Since the sum of angles in every triangle is $180^{\circ }$, the number of splitting triangles is: $\frac{m\cdot 360^{\circ }+6\cdot 120^{\circ }}{180^{\circ }}=2m+4$.

Let us assume the contrary to the statement, i.e. that the area of each of the splitting triangles is greater than $\frac{3\sqrt{3}}{4(m+2)}$. Then the sum of the areas of all the splitting triangles is greater than $(2m+4)\frac{3\sqrt{3}}{4(m+2)}=\frac{3\sqrt{3}}{2}$, which is impossible since the area of the given hexagon is $\frac{3\sqrt{3}}{2}$.