M6. A symmetrical ring of $m$ identical regular $n$-sided polygons is formed according to the rules:
(i) each polygon in the ring meets exactly two others;
(ii) two adjacent polygons have only an edge in common; and
(iii) the perimeter of the inner region-enclosed by the ringconsists of exactly two edges of each polygon.
The example in the figure shows a ring with $m=6$ and $n=9$.
For how many different values of $n$ is such a ring possible?
Solution
Let the exterior angle in each polygon be $\alpha^{\circ}$, so that $\alpha=\frac{360}{n}$, and let $P, Q, R, S$ and $T$ be five consecutive vertices of one of the polygons, as shown, where $Q R S$ is part of the perimeter of the inner region.

From the symmetry, the point $O$ where $P Q$ and $T S$ meet is the centre of the inner region and hence
$$
\angle Q O S=\frac{360^{\circ}}{m} .
$$

Now the sum of the angles in quadrilateral $O S R Q$ is $360^{\circ}$, so we have
$$
360=\frac{360}{m}+\alpha+(180+\alpha)+\alpha
$$
and therefore
$$
1=\frac{2}{m}+\frac{6}{n}
$$

Multiplying each term by $m n$, we obtain
$$
m n=6 m+2 n,
$$
so that
$$
(m-2)(n-6)=12 \text {. }
$$

Now $m>2$ for the ring to exist. Hence $(m-2)(n-6)$ is a product of two positive integers, so that $n-6$ is a positive factor of 12 . The only possibilities for $n-6$ are therefore $1,2,3,4,6$ and 12, and thus the only possibilities for $n$ are 7, 8, 9, 10, 12 and 18. The following figures show the corresponding rings.

Thus a ring of polygons of the required form is possible for six different values of $n$.