Let $\omega_0, \ldots, \omega_{n-1}$ be the $n$-th roots of unity, and $a_0, \ldots, a_{n-1}$ be real numbers in the $(0, 1]$ interval.
Define $z_i = a_i \omega_i$ as the vertices of a polygon on the complex plane, and $s_i$ as the area of a triangle whose vertices are $(0, z_i, z_{i+1})$, indices taken modulo $n$.
For the case illustrated below, with $n=4$, is it possible to determine the values of $(a_0, \ldots, a_3)$ given the values of $(s_0, \ldots, s_3)$?
What I suspect: for odd $n$, this problem can be reduced to a circulant linear system with a unique answer. For even $n$, the same approach is not adequate because it leads to an underdetermined system.
Best Answer
The area $s_k$ is $a_k a_{k+1}$ times the area of the triangle with the vertices $0, 1, \omega$, that is $$ \tag{*} s_k = \frac 12a_k a_{k+1} \sin\frac{2 \pi}{n} \, . $$ For odd $n$ is $$ \frac{s_0 s_2 \cdots s_{n-3}s_{n-1}}{s_1 s_3 \cdots s_{n-2}} = \frac 12a_0^2 \sin\frac{2 \pi}{n} $$ so that $a_0$, and by circular permutation, all $a_k$, are uniquely determined.
However, given positive numbers $s_0, \ldots, s_{n-1}$, the so computed lengths $a_0, \ldots, a_{n-1}$ are positive, but not necessarily $\le 1$.
If $n$ is even then both $s_0 s_2 \cdots s_{n-2}$ and $s_1 s_3 \cdots s_{n-1}$ are equal to $$ a_0 a_1 \cdots a_{n+1} \left( \frac 12 \sin\frac{2 \pi}{n}\right)^{n/2} \, , $$ so that $$ s_0 s_2 \cdots s_{n-2} = s_1 s_3 \cdots s_{n-1} $$ is a necessary condition for the existence of a solution. If this condition is satisfied then there is one degree of freedom. If one of the $a_k$ is given then all other $a_j$ are uniquely determined by equation $(*)$.
Again, the so computed lengths $a_0, \ldots, a_{n-1}$ are positive, but not necessarily $\le 1$.