XIV OBM

The answer is the convex polygon inscribed in a semicircle of diameter $A_0{A}_n$, that is, $\angle A_0{A}_i{A}_n=90^{\circ }$ for all $i=1,2,\dots ,n-1$.

First of all, the polygon must be convex because if it isn't the case one can obtain a polygon of greater area by reflecting the sides of the polygon inside of its convex hull.

Consider an arbitrary vertex $A_i$ and fix the polygons $A_0{A}_1\dots A_i$, $A_i{A}_{i+1}\dots A_n$; in particular, $A_0{A}_i$ and $A_i{A}_n$ are fixed. Now the area of $A_0{A}_1{A}_2\dots A_n$ is the sum of the areas of the fixed polygons $A_0{A}_1\dots A_i$, $A_i{A}_{i+1}\dots A_n$ and the triangle $A_0{A}_i{A}_n$, which attains its maximum if and only if $\angle A_0{A}_i{A}_n=90^{\circ }$, since the area of the triangle is $\frac{1}{2}{A}_i{A}_0\cdot A_i{A}_n\sin \angle A_0{A}_i{A}_n$ and $A_i{A}_0$ and $A_i{A}_n$ are both fixed.

It remains to show that such polygon is unique. Indeed, if $A_0{A}_n=2r$ and $O$ is the midpoint of $A_0{A}_n$ and center of the semicircle, then $\angle A_{i}O{A}_{i+1}=2\arcsin \frac{x_i+x_{i+1}}{2r}$. But ${\sum }_{i=1}^{n}2\arcsin \frac{x_i+x_{i+1}}{2r}=\pi$ and $\arcsin$ is increasing in $[0,1]$, so $r$ is unique and hence the polygon is unique.