Is there a regular polygon in which the length of the shortest diagonal is equal to the radius of the circumscribed circle of the polygon?
Solution. Let the vertices of the regular $n$-gon be $A_{1}, A_{2}, A_{3}, \ldots, A_{n}$. Starting from vertex $A_{1}$, draw the shortest diagonal from $A_{1}$, which is clearly the segment $A_{1} A_{3}$. The radius of the circumscribed circle of the polygon is $r$, and its center is $O$. According to the condition, $A_{1} A_{3}=r$. Then the triangle $O A_{1} A_{3}$ is equilateral, and thus $\angle A_{1} O A_{3}=60^{\circ}$. The line $O A_{2}$ is perpendicular to $A_{1} A_{3}$ and bisects the angle $\angle A_{1} O A_{3}$. Therefore, $\angle A_{1} O A_{2}=30^{\circ}$, and the sought polygon is the regular 12-gon.

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