In a regular $n$-gon, $A_{1} A_{2} A_{3} \cdots A_{n}$, where $n>6$, sides $A_{1} A_{2}$ and $A_{5} A_{4}$ are extended to meet at point $P$. If $\angle A_{2} P A_{4}=120^{\circ}$, determine the value of $n$.

![](https://cdn.mathpix.com/cropped/2024_04_30_fa78072b1e75930324fdg-2.jpg?height=239&width=675&top_left_y=1824&top_left_x=779)
Since $A_{1} A_{2} A_{3} \cdots A_{n}$ is a regular $n$-gon, then $\angle A_{1} A_{2} A_{3}=\angle A_{2} A_{3} A_{4}=\angle A_{3} A_{4} A_{5}=180^{\circ}-\frac{360^{\circ}}{n}$. (This comes from the fact that the sum of the interior angles in a regular $n$-gon is $180^{\circ}(n-2)$.)

Therefore, $\angle P A_{2} A_{3}=\angle P A_{4} A_{3}=180^{\circ}-\left(180^{\circ}-\frac{360^{\circ}}{n}\right)=\frac{360^{\circ}}{n}$.

Also, the reflex angle at $A_{3}$ equals $360^{\circ}-\left(180^{\circ}-\frac{360^{\circ}}{n}\right)=180^{\circ}+\frac{360^{\circ}}{n}$.

Since the sum of the angles in quadrilateral $P A_{2} A_{3} A_{4}$ equals $360^{\circ}$, then

$$
120^{\circ}+2\left(\frac{360^{\circ}}{n}\right)+180^{\circ}+\frac{360^{\circ}}{n}=360^{\circ}
$$

or $\frac{1080^{\circ}}{n}=60^{\circ}$ and so $n=18$.

ANSWER: 18