\section*{Problem 3}

What is the greatest number of sides of a convex polygon that can equal its longest diagonal?


\section*{Solution}

Answer: 2, except for the equilateral triangle.

It is easy to find two. Take the two sides to be \(\mathrm{AB}\) and \(\mathrm{AC}\) with angle \(\mathrm{BAC}=60 \mathrm{deg}\), and take the other vertices on the minor arc of the circle center \(\mathrm{A}\) radius \(\mathrm{AB}\) between \(\mathrm{B}\) and \(\mathrm{C}\).

Let the longest diagonal have length k. Suppose there are three sides with length k. Extend them (if necessary) so they meet at A, B, C. Suppose \(\square A>60^{\circ}\). Take the vertices on side AB to be \(\mathrm{P}, \mathrm{Q}\) (where we may have \(\mathrm{P}=\mathrm{A}\), or \(\mathrm{Q}=\mathrm{B}\), or both). Take the vertices on side \(\mathrm{AC}\) to be R, \(S\) (where we may have \(R=A\), or \(S=C\), or both). Then \(A Q \geq k, A S \geq k\), so \(Q S>k\). Contradiction. Hence angle A \(\leq 60^{\circ}\). The same is true for \(\square \mathrm{B}\) and \(\square \mathrm{C}\). Hence \(\square \mathrm{A}=\square \mathrm{B}=\square \mathrm{C}\) \(=60^{\circ}\). But now \(\mathrm{QS}>\mathrm{k}\) unless \(\mathrm{A}=\mathrm{P}=\mathrm{R}\). Similarly, \(\mathrm{B}\) and \(\mathrm{C}\) must be vertices of the convex polygon, so that it is just an equilateral triangle.

