## Task 1 - 090931

Let $A B C D E F G H$ be a regular octagon. Consider all triangles formed by any three of the points $A, B, C, D, E, F, G, H$.

Someone wants to list some of these triangles, with the condition that no two of the listed triangles are congruent to each other.

Determine the maximum number of triangles he can list under this condition!
There are only five different types of triangles, i.e., every possible triangle belongs to exactly one of these types.

Justification: For a triangle with vertices $X, Y, Z \in\{A, \ldots, H\}$, let $x, y$, and $z$ be the distances between each pair of vertices. (For the triangle with vertices $B, C, H$, for example, $x=1, y=5, z=2$.)

Two triangles are congruent if and only if the triples of their corresponding distances are equal. However, the order of the distances does not matter. (For the triangle $E, G, D$, for example, $x=2, y=5, z=1$, and the triangles $E, G, D$ and $B, C, H$ are congruent.)

The sum $x+y+z$ is always equal to 8. Therefore, the question is equivalent to how many ways 8 can be expressed as the sum of three natural numbers, where the order of the summands does not matter.

There are only the five possibilities $1+1+6=1+2+5=1+3+4=2+2+4=2+3+3$