A regular $k$-gon is drawn, and on each of its sides, a regular $m$-gon is drawn outward. We know that the vertices of these $m$-gons, different from the vertices of the $k$-gon, form a regular $n$-gon. Determine the numbers $k, m$, and $n$.
We will show that $m \leq 4$. We know that every regular polygon can be circumscribed by a circle. Consider the $n$-gon and the circles circumscribed around the $m$-gons. The vertices of the $m$-gons, different from the vertices of the $k$-gon, lie on the circle circumscribed around the $n$-gon, and also on the circle circumscribed around some $m$-gon. These circles are clearly different, so they have at most two common points. Therefore, the $m$-gons have at most two vertices that are not vertices of the $k$-gon, which means $m \leq 4$.

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Figure 1

If $m=3$, then clearly $n=k$. In this case, a rotation by $\frac{360^{\circ}}{k}$ around the center of the $k$-gon maps the figure onto itself, so the $k$-gon is regular (Figure 1); thus, $k$ can be any integer greater than 2.

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Figure 2

If $m=4$, let $E, F$, and $G$ be consecutive vertices of the $k$-gon, and $A, B, C, D$ be the other vertices of the squares constructed on the sides $EF$ and $FG$ (Figure 2). The $n$-gon can only be regular if $AB = BC = CD$. But $AB = BF$ and $CD = CF$, so $\triangle BFC$ is equilateral. From this, we can calculate one of the angles of the $k$-gon: $\angle EFG = 360^{\circ} - (\angle EFB + \angle BFC + \angle CFG) = 360^{\circ} - (90^{\circ} + 60^{\circ} + 90^{\circ}) = 120^{\circ}$; thus, the $k$-gon can only be a hexagon. If we construct squares on the sides of a regular hexagon, the vertices of the squares different from the vertices of the hexagon form a dodecagon (12-sided polygon) where each side is equal and each angle is $150^{\circ}$ (Figure 3), so this dodecagon is indeed regular.

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Figure 3

Therefore, the possible values for $k, m, n$ are: $m=3$ and $n=k$, where $k$ is any integer greater than 2, or $m=4, k=6$, and $n=12$.