Let $P$ be a closed polygon defined by the sequence  

$p\_0,\,\dots,\,p\_{n-1},p\_0$ of points.



> 
> **Question:**
> 
> 
> how can one construct, with straightedge and compass alone, another sequence of points $q\_0,\,\dots,q\_{n-1}$ such that:
> 
> 
> * $q\_i$ lies on the bisector of $p\_i$ and $p\_{i+1}$
> * $q\_i,p\_{i+1}$ and $q\_{i+1}$ are collinear
> 
> 
> 



Constructing a polygon from another with collinearity constraints
Let $L\_i$ be the bisector of $p\_i$ and $p\_{i+1}$, and let $f\_i \colon L\_i \to L\_{i+1}$ be the central projection through $p\_{i+1}$. This is a projective transformation with constructible coefficients. Your question is about the fixed points of the projective transformation $f\_n \circ \cdots \circ f\_1$. The coefficients of this projective transformation are also constructible. The fixed points are found by solving a quadratic equation, which can be done by compass and ruler.


It follows also that the number of solutions is among $0, 1, 2$ or $\infty$, although it is not clear whether there are examples with no solutions.


