# Polygon neighbors

This question is from the German mathematics competition Känguru der Mathematik. In this competition students have to solve 30 mathematical tasks like this in 90 minutes without calculator. Actually they are given 5 possible answers, but for this community a bit of additional challenge does not hurt.

A regular $15$-gon $A_1 \, A_2 \, ... \, A_{15}$ and a regular $n$-gon $B_1 \, B_2 \, ... \, B_n$ with side lengths of $1$ have a common side: $B_1 B_2 = A_2 A_1$. What value of $n$ makes the distance between $B_3$ and $A_{15}$ be $1$?

First, note that

The angles around $A_1$ should sum to 360º.

For the angle inside A,

Since A is a regular 15-gon, we know each of the interior angles of A measures $\frac{15-2}{15} \times 180^{\circ} = 156^{\circ}$.

For the angle between A and B,

If $B_3A_{15} = 1$, as we have $B_2B_3 = A_1A_{15} = 1$, then $B_3A_{15}A_2$ is an equilateral triangle, so each of its angles measures 60º.

Therefore, for the remaining angle in B

The remaining angle from B must measure 360º - 156º - 60º = 144º. Since it is a regular n-gon, all of its interior angles measure $\frac{n-2}{n} \times 180^{\circ}$.

$\frac{n-2}{n} \times 180^{\circ} = 144^{\circ} \Rightarrow \fbox{n=10}$

By easy angle chasing, the answer is 10.