For which integers $n \geq 3$ can one find a triangulation of regular $n$-gon consisting only of isosceles triangles?
The answer is $n$ of the form $2^{a}\left(2^{b}+1\right)$ where $a$ and $b$ are nonnegative integers not both zero.

Call the polygon $A_{1} \ldots A_{n}$ with indices taken modulo $n$. We refer to segments $A_{1} A_{2}$, $A_{2} A_{3}, \ldots, A_{n} A_{1}$ as short sides. Each of these must be in the triangulation. Note that
- when $n$ is even, the isosceles triangles triangle using a short side $A_{1} A_{2}$ are $\triangle A_{n} A_{1} A_{2}$ and $\triangle A_{1} A_{2} A_{3}$ only, which we call small.
- when $n$ is odd, in addition to the small triangles, we have $\triangle A_{\frac{1}{2}(n+3)} A_{1} A_{2}$, which we call big.
This leads to the following two claims.
Claim - If $n>4$ is even, then $n$ works iff $n / 2$ does.

Proof. All short sides must be part of a small triangle; after drawing these in, we obtain an $n / 2$-gon.
Thus the sides of $\mathcal{P}$ must pair off, and when we finish drawing we have an $n / 2$-gon.
Since $n=4$ works, this implies all powers of 2 work and it remains to study the case when $n$ is odd.
Claim - If $n>1$ is odd, then $n$ works if and only if $n=2^{b}+1$ for some positive integer $b$.

Proof. We cannot have all short sides part of small triangles for parity reasons, so some side, must be part of a big triangle. Since big triangles contain the center $O$, there can be at most one big triangle too.

Then we get $\frac{1}{2}(n-1)$ small triangles, pairing up the remaining sides. Now repeating the argument with the triangles on each half shows that the number $n-1$ must be a power of 2 , as needed.