Answer: an equilateral triangle with side $a$. Consider a convex $n$-gon $A_{1} \ldots A_{n}$ with side $A_{1} A_{\mathrm{n}}$ $=a$, possessing the given property. If $n \geq 4$, then this $n$-gon can be replaced by an $(n-1)$-gon

$A_{1} \ldots A_{n-3} A_{n-2} 2^{\prime} A_{n-1}$ ', where $A_{n-1}$ ' $=A_{n}$ and $A_{n-2}$ - is the point of intersection of the rays $A_{n-3} A_{n-2}$ and $A_{\mathrm{n}} A_{n-1}$ (these rays intersect because the sum of the exterior angles at vertices $A_{n-2}$ and $A_{n-1}$ is less than $180^{\circ}$). The new polygon has a strictly larger area. Therefore, it is sufficient to consider the case of a triangle. For the considered triangles, the side $B C=a$ is fixed and the opposite angle $\angle A$ (which is $60^{\circ}$). Point $A$ is located on the arc of a circle from which the segment $B C$ is seen at an angle of $60^{\circ}$. Therefore, the height dropped from point $A$ is maximal in the case of an isosceles triangle.