TITLE: Finding a solution to a simple geometric set of equalities
QUESTION [6 upvotes]: Let $p_1,\dots,p_n$ be a collection of points in the plane $\mathbb{R}^2$ and let $a$ be a positive number such that $a<1$.  Is there a good numerical algorithm to find points $x_1,\dots,x_n$ in the plane such that
$$\|x_i - x_{i+1}\| = a(\|x_i - p_i\|+\|p_i-x_{i+1}\|)$$ for all $i$, where we use $n+1\equiv 1$ to simplify notation?  A picture for $a=1/3$ is attached, in which the black dots indicate the $p_i$'s, the blue segments indicate the segments from $x_i$ to $x_{i+1}$, and the red segments indicate the $\|x_i - p_i\|+\|p_i-x_{i+1}\|$.

REPLY [7 votes]: It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi m/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $).
Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the assumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$,  define inductively  $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ corresponds then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I  won't spoil).
The same procedure produces, more  generally, for any $a$ and for any model triangle, with vertices $(0,1 ,u)$ such that  $|1-u|=a(1+|u|)$ and $u^n\neq1$,
 a unique solution such that the triangles with ordered vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with ordered vertices $(0,1,u)$.