[Geometry]
Problem: A regular hexagon has side length 1. Compute the average of the areas of the 20 triangles whose vertices are vertices of the hexagon.

Solution: There are 6 triangles of side lengths $1,1, \sqrt{3} ; 2$ equilateral triangles of side length $\sqrt{3}$; and 12 triangles of side lengths $1, \sqrt{3}, 2$. One triangle of each type is shown in the diagram below.
<img_3233>

Each triangle in the first set has area $\sqrt{3} / 4$; each triangle in the second set has area $3 \sqrt{3} / 4$; and each triangle in the third set has area $\sqrt{3} / 2$. The average is

$$
\frac{6\left(\frac{\sqrt{3}}{4}\right)+2\left(\frac{3 \sqrt{3}}{4}\right)+12\left(\frac{\sqrt{3}}{2}\right)}{20}=\frac{\frac{6 \sqrt{3}}{4}+\frac{6 \sqrt{3}}{4}+\frac{24 \sqrt{3}}{4}}{20}=\frac{\mathbf{9} \sqrt{\mathbf{3}}}{\mathbf{2 0}} .
$$

Final answer: $\frac{9 \sqrt{3}}{20}$