Franklin 5. 

A convex $n$-gon $P$, where $n>3$, is cut into equal triangles by diagonals that do not intersect inside it.

What are the possible values of $n$, if the $n$-gon is cyclic?
Lemma. Let a convex $n$-gon be cut into equal triangles by diagonals that do not intersect inside it. Then, for each of the triangles in the partition, at least one side is a side (not a diagonal) of the $n$-gon.

Proof. Let a triangle in the partition have angles $\alpha \leq \beta \leq \gamma$ with vertices $A, B, C$ respectively, and $A C$ and $B C$ are diagonals of the $n$-gon. At vertex $C$, there are at least two more angles of the triangles in the partition. If at least one of these angles is greater than $\alpha$, then the sum of the angles at vertex $C$ is not less than $\gamma+\beta+\alpha=\pi>\angle C$. Contradiction. Therefore, all angles at vertex $C$, except $\angle A C B$, are equal to $\alpha$, and $\alpha<\beta$.

Consider the second triangle in the partition adjacent to $B C$. Since it is equal to triangle $A B C$, the angle opposite side $B C$ in it is equal to $\alpha$. But the angle at vertex $C$ in this triangle is also equal to $\alpha$. Contradiction.

Since the sum of the angles of the polygon $P$ is $\pi(n-2)$ and they are composed of all the angles of the triangles in the partition, the number of these triangles is $n-2$. By the lemma, in each of these triangles, at least one side is a side of $P$. From this, it follows that two of the triangles in the partition have two sides that are sides of $P$.

Let $K L M$ be one of these triangles, with $K L$ and $L M$ being sides of $P$. A side $K M$ is adjacent to another triangle in the partition $K M N$. One of its sides (for definiteness, $K N$) is a side of $P$. Since the triangles in the partition are equal, the angle $NKM$ is equal to either the angle $LKM$ or the angle $KML$. In the first case, $KM$ is the bisector of the angle of the described polygon $P$ and therefore contains the center $I$ of the inscribed circle. In the second case, $K N \parallel L M$. Then $I$ lies on the common perpendicular to these segments and therefore is contained (by convexity) in the parallelogram $K L M N$, and thus in at least one of the triangles $K L M, K M N$.

Let $K' L' M'$ be another triangle in the partition, two sides of which are sides of $P$. Similarly, $I$ is contained either in this triangle or in an adjacent triangle in the partition. If $I$ is contained in at least one of the triangles $K L M, K' L' M'$, then they have a common side, and then $n=4$. Otherwise, the triangle $K M N$ is adjacent to both of these triangles and contains $I$. In this case, the side $M N$ is common with the triangle $K' L' M'$; we can set $M=M'$, $N=K'$. Reasoning as above, we get that $L M, K N, L' M$ are parallel to each other. But then the adjacent sides $L M$ and $M L'$ lie on the same line. Contradiction.

## Answer

$n=4$.

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