\section*{Problem 6}

The difference between the longest and shortest diagonals of the regular n-gon equals its side. Find all possible \(n\).


\section*{Solution}

Answer: \(\mathrm{n}=9\).

For \(\mathrm{n}7\) the side has length \(2 R \sin \pi / n\), the shortest diagonal has length \(2 R \sin 2 \pi / n\), and the longest diagonal has length \(2 R\) for \(n\) even and \(2 R \cos \pi / 2 n\) for \(n\) odd (where \(R\) is the radius of the circumcircle). Thus we require:

\(\sin 2 \pi / n+\sin \pi / n=1\) and \(n\) even, or
\(\sin 2 \pi / n+\sin \pi / n=\cos \pi / 2 n\) and \(n\) odd.

Evidently the lhs is a strictly decreasing function of \(\mathrm{n}\) and the rhs is an increasing function of \(\mathrm{n}\), so there can be at most one solution of each equation. The second equation is satisfied by \(\mathrm{n}\) \(=9\), although it is easier to see that there is a quadrilateral with the longest diagonal and shortest diagonals as one pair of opposite sides, and 9 -gon sides as the other pair of opposite sides. The angle between the longest side and an adjacent side is 60 , so that its length is the length of the shortest diagonal plus \(2 \times 9\)-gon side \(\mathrm{x} \cos 60\). Hence that is the only solution for n odd.

For \(\mathrm{n}=8\) we have the same quadrilateral as for the 9 -gon except that the angle is 67.5 and hence the difference is less than 1 . For \(n=10\), \(\sin 2 \pi / 10+\sin \pi / 10=\sin \pi / 10(2 \cos \pi / 10+1)\) \(<3 \sin \pi / 10<3 \pi / 10<1\). So there are no solutions for \(n\) even \(\geq 10\), and hence no solutions for n even.

