Berkeley Math Circle: Monthly Contest 5

Solution:

Draw a diagonal of the small $n$-gon to form the red triangle $\triangle ABC$ as shown in the diagram, where $A$ is a vertex of the large $n$-gon and $B$ and $C$ are the vertices of the small $n$-gon closest to $A$ on the two diagonals from $A$. Also draw the diagonal $AD$ of the large $n$-gon such that $\angle DAB = \pi / n$.

Let $E$ be the vertex of the large $n$-gon adjacent to $A$ on the same side of $A$ as $D$. Then $\angle EAB = (k-1)\pi / n$, and since the angles of the $n$-gon are $(n-2)\pi / n$,
$$\angle BAC = \frac{(n-2)\pi}{n} - 2 \cdot \frac{(k-1)\pi}{n} = \frac{(n-2k)\pi}{n},$$
so since $\triangle ABC$ is isosceles, the base angle is
$$\angle ABC = \angle ACB = \frac{k\pi}{n}.$$
Thus we have
$$\frac{BC}{AB} = 2 \cos (k\pi / n).$$
Since $\angle BAD = \pi / n$,
$$\frac{AB}{AD} = \frac{1}{2 \cos (\pi / n)} \Longrightarrow \frac{BC}{AD} = \frac{\cos (k\pi / n)}{\cos (\pi / n)}.$$
But the green shapes are similar, so the ratio of the side length of the small $n$-gon to the side length of the large $n$-gon is equal to the ratio $BC / AD$. This proves our claim.