Problem: For which integers $n \ge 3$ can one find a triangulation
of regular $n$-gon consisting only of isosceles triangles?

Solution: The answer is $n$ of the form $2^a(2^b+1)$
where $a$ and $b$ are nonnegative integers not both zero.

Call the polygon $A_1 \dots A_n$
with indices taken modulo $n$.
We refer to segments $A_1 A_2$, $A_2 A_3$, \dots, $A_n A_1$
as \emph{short sides}.
Each of these must be in the triangulation.
Note that
\begin{itemize}
  \ii 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 \emph{small}.
  \ii when $n$ is odd, in addition to the small triangles,
  we have $\triangle A_{\half(n+3)} A_1 A_2$,
  which we call \emph{big}.
\end{itemize}

This leads to the following two claims.
\begin{claim*}
  If $n > 4$ is even, then $n$ works iff $n/2$ does.
\end{claim*}
\begin{proof}
  All short sides must be part of a small triangle;
  after drawing these in, we obtain an $n/2$-gon.
  \begin{center}
  \begin{asy}
    size(3cm);
    draw(unitcircle, dotted);
    picture p;
    dot(p, dir(0));
    dot(p, dir(30));
    dot(p, dir(60));
    fill(p, dir(0)--dir(30)--dir(60)--cycle, opacity(0.1)+lightred);
    draw(p, dir(0)--dir(30)--dir(60), blue);
    draw(p, dir(0)--dir(60), red);
    for (int i=0; i<6; ++i) {
      add(rotate(60*i)*p);
    }
  \end{asy}
  \end{center}
  Thus the sides of $\mathcal P$ must pair off,
  and when we finish drawing we have an $n/2$-gon.
\end{proof}
Since $n = 4$ works, this implies all powers of $2$ work
and it remains to study the case when $n$ is odd.

\begin{claim*}
  If $n > 1$ is odd, then $n$ works if and only if $n = 2^b+1$
  for some positive integer $b$.
\end{claim*}
\begin{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 $\half(n-1)$ 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.
\end{proof}