All-Soviet-Union Mathematical Olympiad

Solution:
Answer: 2, except for the equilateral triangle.
It is easy to find two. Take the two sides to be $AB$ and $AC$ with angle $BAC = 60^{\circ}$, and take the other vertices on the minor arc of the circle center $A$ radius $AB$ between $B$ and $C$.

Let the longest diagonal have length $k$. Suppose there are three sides with length $k$. Extend them (if necessary) so they meet at $A$, $B$, $C$. Suppose $\angle A > 60^{\circ}$. Take the vertices on side $AB$ to be $P$, $Q$ (where we may have $P = A$, or $Q = B$, or both). Take the vertices on side $AC$ to be $R$, $S$ (where we may have $R = A$, or $S = C$, or both). Then $AQ \geq k$, $AS \geq k$, so $QS > k$. Contradiction. Hence angle $A \leq 60^{\circ}$. The same is true for $\angle B$ and $\angle C$. Hence $\angle A = \angle B = \angle C = 60^{\circ}$. But now $QS > k$ unless $A = P = R$. Similarly, $B$ and $C$ must be vertices of the convex polygon, so that it is just an equilateral triangle.