The inverse of the "Given vertices find area of polygon in complex plane" problem

Question

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.

Find vertices $z_i$ given areas $s_i$

Answer 1

Equations are ($n \ge 3$):
$\forall i=0 \dots n-1, a_i a_{i+1} = 2s_i / \sin(\frac {2\pi} n)$, with $a_n = a_0$.
Taking the $\log$ transforms these into linear equations: if $l_i = \log a_i$, the system is equivalent to
$\forall i=0 \dots n-1, l_i + l_{i+1} = \log(2s_i / \sin(\frac {2\pi} n))$.

The matrix has its diagonal made of $1$, plus a second diagonal above of $1$, and a $1$ at the lower left corner, and zeroes everywhere else.

Exemple for $n=4$: $\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{pmatrix}$

When developing the determinant by the first line:

• If we take the $1$ on $(0,0)$, then on column $1$ we cannot take the $1$ on $(0,1)$ so we take the one on $(1,1)$, and then we are forced to take the $1$ on $(2,2)$, etc.: this is the principal diagonal.
• If we take the $1$ on $(0,1)$, then on line $1$ we cannot take the $1$ on $(1,1)$ so we take the $1$ on $(1,2)$, and then, etc.: we take the second diagonal of $1$s, including the $1$ on the lower left corner as last element.

The first diagonal has the identity permutation, so it has even parity, which makes a $+1$.
The second diagonal above has a cycle permutation, so its parity is even if $n$ is odd, and odd if $n$ is even - this is because an $n$ cycle decomposes into $n-1$ transpositions. Which makes $(-1)^{n-1}$

So the determinant is $1 + (-1)^{n-1}$, i.e. $0$ when $n$ is even and $2$ when $n$ is odd: the system has a unique solution for $n$ odd, and either no solution or an infinity, for $n$ even.

For $n$ even, another way to see it is to use an alternate sum of equations $i=1 \dots n-1$. This gives the left handside of equation $i=0$, so the system is underdetermined: either no solution or an infinity.

The case $n=4$ allows still another proof: one can apply a scaling by $\alpha > 0$ on the $x$ axis and a scaling by $1/ \alpha$ on the $y$ axis, this changes the $(a_i)$ but not the $(s_i)$.

Answer 2

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 $\le 1$.