Argentina_2018

Call $n$ good if the regular $n$-gon can be triangulated with isosceles triangles. By segments we mean the sides and the diagonals of the $n$-gon; the sides are the shortest among all segments.

Let $n$ be good and $T$ an isosceles triangulation of the regular $n$-gon $P$. Suppose that the base of a triangle $\Delta \in T$ is a side $a$ of $P$. Then the vertex of $\Delta$ opposing $a$ is on the perpendicular bisector of $a$, which passes through the center of $P$, and also the circumcircle of $P$. Hence $n$ is odd and the center of $P$ is interior to $\Delta$, implying that such a triangle $\Delta$ is unique.

Let $n$ be even. Then sides are the shortest segments and none of them is a base of a triangle of $T$. So all of them are divided into pairs of consecutive ones, and each pair contains the equal sides of a triangle of $T$. Deleting these $\frac{n}{2}$ isosceles triangles leaves a regular $\frac{n}{2}$-gon which therefore also admits of an isosceles triangulation. It follows that an even $n\ge 6$ is good if and only if so is $\frac{n}{2}$.

Let $n$ be odd. Then the sides cannot be paired up like in the even case, one of them must be a base of a triangle $\Delta$ from $T$ as explained earlier. The equal sides of $\Delta$ are diagonals of $P$ (longest ones). Removing $\Delta$ leaves two congruent polygons which must have isosceles triangulations. Let $P_1$ be one of them. It has $n_1=\frac{1}{2}(n(n+1))$ sides; thus $n_1$ is good. One of the sides is a diagonal $d_1$, the rest are sides of $P$, hence shorter. So $d_1$ is a base of a triangle ${\Delta }_1$ of $T$, and its opposite vertex divides the remaining $n_1-3$ vertices into two equal halves. It follows that $n_1$ is odd. Remove ${\Delta }_1$ from $P_1$ and denote by $P_2$ one of the two obtained congruent polygons with $n_2=\frac{1}{2}(n_4+1)$ sides; $n_2$ is good. The same argument applies to $P_2$ because one of its sides is a diagonal $d_2$, and the rest are sides of $P$. We conclude that $n_2$ is odd then define $n_3=\frac{1}{2}(n_2+1)$, and so on. Thus each of the numbers $n>n_1>n_2>\dots$ is odd and good, as long as it is $\ge 3$. Let $k$ be such that $n_k\ge 3>n_{k+1}$. If $n_{k+1}=1$ then $n_k=1$ which is false. So $n_{k+1}=2$ and so $n_k=3$. Write $n_k=3=2^1+1$ and backwards to obtain $n_{k-1}=2^2+1$ and likewise $n_{k-2}=2^3+1$, ..., $n_1=2^k+1$, $n=2^{k+1}+1$. Therefore $n-1$ is a power of 2. In addition the steps of the argument imply a construction showing that the converse is also true.

To sum up, consider two cases for a general $n$. If $n\ge 4$ is a power of 2, $n=2^m$ with $m\ge 2$, then it is good if and only if so are $2^{m-1},2^{m-2},\dots ,2^2=4$. Since the square has an isosceles triangulation, the powers of 2 are good. If $n\ge 3$ is not a power of 2 then $n=2^m$ with $k\ge 3$ odd and $m\ge 0$. By the above, $n$ is good if and only if so is $k$, and the latter holds if and only if $k$ is of the form $k=2^l+1$ with $l\ge 1$. Hence $n=2^n+2^l$ with $u>v\ge 0$. In conclusion the good numbers are $2^m$ with $m\ge 2$ and $2^n+2^l$ with $u>v\ge 0$.