[Geometry]
Problem: Polygon $A_{1} A_{2} \ldots A_{n}$ is a regular $n$-gon. For some integer $k<n$, quadrilateral $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle of area 6 . If the area of $A_{1} A_{2} \ldots A_{n}$ is 60 , compute $n$.

Solution: Because $A_{1} A_{2} A_{k} A_{k+1}$ is a rectangle, $n$ must be even, and moreover, $k=\frac{n}{2}$. Also, the rectangle's diagonals meet at the center $O$ of the circumscribing circle. $O$ is also the center of the $n$-gon. The diagram below shows the case $n=16$.



<img_3867>

Then $\left[A_{1} A_{2} O\right]=\frac{1}{4}\left[A_{1} A_{2} A_{k} A_{k+1}\right]=\frac{1}{n}\left[A_{1} A_{2} \ldots A_{n}\right]=60$. So $\frac{1}{4}(6)=\frac{1}{n}(60)$, and $n=40$.

Final answer: 40