Final Round of National Olympiad

Let $\triangle$ be the triangle containing the center $O$ of the $n$-gon (colored in Fig. 23; if the center lies on a diagonal then choose either of the triangles having this diagonal as a side). Let $d$ be any side of $\triangle$. If $d$ is a diagonal of the $n$-gon then $d$ separates $\triangle$ from a neighboring triangle whose other two sides are shorter than $d$. By assumption, $d$ has to be a side of minimal length in $\triangle$. But if $d$ is a side of the $n$-gon then $d$ is also a side of minimal length in $\triangle$. Thus all
sides of $\triangle$ are equally minimal, meaning that $\triangle$ is equilateral. Consequently, there is a constant number of sides of $n$-gon between the endpoints of every side of $\triangle$. Hence $n = 3s_0$ for a positive integer $s_0$.

Fig. 23

Consider now an arbitrary triangle $\Delta'$ neighboring $\Delta$. Let $d'$ be any of its two sides not common with $\Delta$. If $d'$ is a diagonal of the $n$-gon then $d'$ separates $\Delta'$ from a third triangle whose other sides are shorter than $d'$. Thus $d'$ must be a side of minimal length in $\Delta'$. But if $d'$ is a side of the $n$-gon then $d'$ is also a side of minimal length in $\Delta'$. Hence the sides of $\Delta'$ not common with $\Delta$ have equal length and there must be the same number of sides of the $n$-gon between the endpoints of these sides of $\Delta'$. Consequently $s_0 = 2s_1$ where $s_1$ is a positive integer.
If the sides of equal length of $\Delta'$ are sides of the $n$-gon then $s_1 = 1$ and $n = 3 \cdot 2$. Otherwise we can continue similarly to get $s_1 = 2s_2$ where either $s_2 = 1$ or $s_2 = 2s_3$, etc. Thus $n = 3 \cdot 2^k$ for a natural number $k$. A construction for every $n$ of the form $3 \cdot 2^k$ follows from the argumentation.