, 2010

The required numbers are of the form $n = 2^r(2^s+1)$, where $r$ and $s$ are nonnegative integer numbers which do not vanish simultaneously. Clearly, any such $n$ works.

To establish the converse, let $K$ be a regular $n$-gon, $n \ge 4$, which can be decomposed into isosceles triangles by noncrossing diagonals. Begin by noticing that each edge $e$ of $K$ must be an edge of a unique isosceles triangle $T_e$ in the decomposition. Two cases are possible: either $e$ is opposite the apex of $T_e$ or $e$ and one of the adjacent edges of $K$ are the edges of $T_e$ issuing from the apex. (Since $n \ge 4$, $T_e$ cannot be equilateral, so the apex is well defined.)

If $n$ is even, no vertex of $K$ lies on the perpendicular bisector of an edge of $K$, so the first case is ruled out. Consequently, the decomposition must contain exactly one of the two bracelets of $n/2$ isosceles triangles clipped off by short diagonals joining consecutive vertices of $K$ of likewise parity. These short diagonals are the edges of a regular $n/2$-gon which is also decomposed into isosceles triangles by noncrossing diagonals and the conclusion follows by induction.

If $n$ is odd, then $K$ has a unique edge $e$ opposite the apex of $T_e$: Since $n$ is odd and each short diagonal clips off two edges of $K$, at least one such $e$ exists. The apex of $T_e$ lies on the perpendicular bisector of $e$, so it must be the vertex of $K$ opposite $e$. Uniqueness of $e$ should now be clear: were there another such $e'$, the interiors of $T_e$ and $T_{e'}$ would overlap. Consequently, $e$ is unique and $K$ splits into $T_e$ and two polygons $L$ and $L'$ which are reflections of one another in the perpendicular bisector of $e$.

To complete the proof, it is sufficient to show that the number of vertices of $L$ is one plus a power of 2. Begin by noticing that $L$ inherits by restriction a decomposition into isosceles triangles by noncrossing diagonals. Let $x_0, \dots, x_m$ be a circular labelling of the vertices of $L$ around the boundary, where $x_0$ is the vertex of $K$ opposite $e$ and $x_m$ is a vertex of $e$. Since $\text{dist}(x_i, x_j) < \text{dist}(x_0, x_m)$ if $\{i, j\} \neq \{0, m\}$, it follows that $x_0x_m$ is an edge of an isosceles triangle with apex at some $x_k$, $0 < k < m$. Notice that $\text{dist}(x_0, x_i) < \text{dist}(x_i, x_m)$ if $0 < i < m/2$ and $\text{dist}(x_0, x_i) > \text{dist}(x_i, x_m)$ if $m/2 < i < m$, to deduce that $m$ must be even, $k = m/2$, and $L$ splits into an isosceles triangle, $x_0x_{m/2}x_m$, and two polygons, $x_0 \cdots x_{m/2}$ and $x_{m/2} \cdots x_m$, which are reflections of one another in the perpendicular bisector of the segment $x_0x_m$. Now we are essentially back in the situation that arose above. Repeat the same argument verbatim to infer that $m/2$ must be even and so on all the way down to conclude that $m$ must be a power of 2.