A symmetric ring with $m$ regular polygons, each with $n$ sides, is formed according to the rules:

i) each polygon in the ring meets two others;

ii) two adjacent polygons share only one side;

iii) the perimeter of the region inside the ring consists of exactly two sides of each polygon.

The example in the following figure shows a ring with $m=6$ and $n=9$. For how many different values of $n$ is it possible to construct this ring?

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-06.jpg?height=836&width=859&top_left_y=478&top_left_x=480)
Solution

Let $\alpha=\frac{360^{\circ}}{n}$ be the measure of each exterior angle of a regular polygon $A B C D E \ldots$. Suppose that the path $B C D$ is part of the perimeter of some internal region of a ring. If $O$ is the intersection of the extensions of $A B$ and $D E$, by symmetry, it is the center of the internal region. Since there are $m$ regular polygons, we conclude that $\angle B O D=\frac{360^{\circ}}{m}$. Now, the sum of the interior angles of $B C D O$ is

$$
\begin{aligned}
360^{\circ} & =\frac{360^{\circ}}{m}+\alpha+(180+\alpha)+\alpha \\
360^{\circ} & =\frac{360^{\circ}}{m}+\frac{3 \cdot 360^{\circ}}{n}+180^{\circ} \\
1 & =\frac{2}{m}+\frac{6}{n}
\end{aligned}
$$

Thus, multiplying the above equation by $m n$ and factoring the resulting expression, we have

$$
\begin{aligned}
m n & =6 m+2 n \\
(m-2)(n-6) & =12
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-07.jpg?height=756&width=685&top_left_y=523&top_left_x=757)

For the ring to exist, we need $m>2$. Additionally, $(m-2)$ and $(n-6)$ must divide 12 and $n>0$, so $(n-6) \in\{1,2,3,4,6,12\}$, which means $n \in\{7,8,9,10,12,18\}$, giving 6 different possible values for $n$. To verify that all of them are solutions, it suffices to consider the following examples of rings:
![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-07.jpg?height=850&width=1232&top_left_y=1598&top_left_x=484)