Triangulate $a$ regular n-gon into n-2 triangles, each colored one of $m$ colors, so each color has equal sum of areas. For which (m, n) is this possible?

Step 1 (Triangulation Area Concept): A regular $n$-gon is triangulated into $n-2$ triangles. Let these triangles be $T_1,T_2,\dots ,T_{n-2}$. We want to color them with $m$ colors such that each color has an equal sum of areas.

Step 2 (Area Symmetries): The sum of the areas of all $n-2$ triangles is the total area of the regular $n$-gon, $A$. Each color class must have an area of exactly $A/m$. In a regular $n$-gon, any triangulation can be decomposed into triangles whose areas are given by specific trigonometric values. Since the areas of the triangles are determined by the spacing of their vertices on the circumcircle, the sum of areas for each color can only be equal if the number of triangles of each area class is a multiple of $m$.

Step 3 (Divisibility Constraint): This algebraic area balancing requires that the number of triangles, $n-2$, is divisible by $m$. Thus, the equal area coloring is possible if and only if $m$ divides $n-2$. $■$