Subfield: Geometry


Question: A regular pentagon covers part of another regular polygon, as shown. This regular polygon has $n$ sides, five of which are completely or partially visible. In the diagram, the sum of the measures of the angles marked $a^{\circ}$ and $b^{\circ}$ is $88^{\circ}$. Determine the value of $n$.

(The side lengths of a regular polygon are all equal, as are the measures of its interior angles.)
![](https://cdn.mathpix.com/cropped/2023_12_21_17bd3319dc6e6567eee6g-1.jpg?height=620&width=1544&top_left_y=268&top_left_x=296)

Solution:
The angles in a polygon with $n$ sides have a sum of $(n-2) \cdot 180^{\circ}$.

This means that the angles in a pentagon have a sum of $3 \cdot 180^{\circ}$ or $540^{\circ}$, which means that each interior angle in a regular pentagon equals $\frac{1}{5} \cdot 540^{\circ}$ or $108^{\circ}$.

Also, each interior angle in a regular polygon with $n$ sides equals $\frac{n-2}{n} \cdot 180^{\circ}$. (This is the general version of the statement in the previous sentence.)

Consider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\circ}$ and $b^{\circ}$ emanate to form a hexagon.

![](https://cdn.mathpix.com/cropped/2023_12_21_0936dd99cfd990581a98g-1.jpg?height=399&width=309&top_left_y=1042&top_left_x=997)

This polygon has 6 sides, and so the sum of its 6 angles is $4 \cdot 180^{\circ}$.

Four of its angles are the original angles from the $n$-sided polygon, so each equals $\frac{n-2}{n} \cdot 180^{\circ}$.

The remaining two angles have measures $a^{\circ}+c^{\circ}$ and $b^{\circ}+d^{\circ}$.

We are told that $a^{\circ}+b^{\circ}=88^{\circ}$.

Also, the angles that measure $c^{\circ}$ and $d^{\circ}$ are two angles in a triangle whose third angle is $108^{\circ}$.

Thus, $c^{\circ}+d^{\circ}=180^{\circ}-108^{\circ}=72^{\circ}$.

Therefore,

$$
\begin{aligned}
4 \cdot \frac{n-2}{n} \cdot 180^{\circ}+88^{\circ}+72^{\circ} & =4 \cdot 180^{\circ} \\
160^{\circ} & =\left(4-\frac{4(n-2)}{n}\right) \cdot 180^{\circ} \\
160^{\circ} & =\frac{4 n-(4 n-8)}{n} \cdot 180^{\circ} \\
\frac{160^{\circ}}{180^{\circ}} & =\frac{8}{n} \\
\frac{8}{9} & =\frac{8}{n}
\end{aligned}
$$

and so the value of $n$ is 9 .
The angles in a polygon with $n$ sides have a sum of $(n-2) \cdot 180^{\circ}$.

This means that the angles in a pentagon have a sum of $3 \cdot 180^{\circ}$ or $540^{\circ}$, which means that each interior angle in a regular pentagon equals $\frac{1}{5} \cdot 540^{\circ}$ or $108^{\circ}$.

Also, each interior angle in a regular polygon with $n$ sides equals $\frac{n-2}{n} \cdot 180^{\circ}$. (This is the general version of the statement in the previous sentence.)

Consider the portion of the regular polygon with $n$ sides that lies outside the pentagon.

![](https://cdn.mathpix.com/cropped/2023_12_21_c36108b31727ccbc67b7g-1.jpg?height=404&width=309&top_left_y=638&top_left_x=997)

This polygon has 7 sides, and so the sum of its 7 angles is $5 \cdot 180^{\circ}$.

Four of its angles are the original angles from the $n$-sided polygon, so each equals $\frac{n-2}{n} \cdot 180^{\circ}$.

Two of its angles are the angles equal to $a^{\circ}$ and $b^{\circ}$, whose sum is $88^{\circ}$.

Its seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\circ}$, which equals $360^{\circ}-108^{\circ}$ or $252^{\circ}$.

Therefore,

$$
\begin{aligned}
4 \cdot \frac{n-2}{n} \cdot 180^{\circ}+88^{\circ}+252^{\circ} & =5 \cdot 180^{\circ} \\
340^{\circ} & =\left(5-\frac{4(n-2)}{n}\right) \cdot 180^{\circ} \\
340^{\circ} & =\frac{5 n-(4 n-8)}{n} \cdot 180^{\circ} \\
\frac{340^{\circ}}{180^{\circ}} & =\frac{n+8}{n} \\
\frac{17}{9} & =\frac{n+8}{n} \\
17 n & =9(n+8) \\
17 n & =9 n+72 \\
8 n & =72
\end{aligned}
$$

and so the value of $n$ is 9 .

Final answer: 9