The following problem seems a very hard one, is it known? It has a resemblance to the lonely runner conjecture. I am guessing.


In the plane let $v\_i$ be $n$ unit vectors no two of them are colinear. Take $P\_0$ any point in the plane and construct a successive set of points $P\_i$ such that for every $i$, $1\le i\le n$ the segments $P\_{i-1}P\_{i}=v\_j$ for some $j$. Every vector $v\_j$ is used once. Can we choose the vectors so that the obtained path from $P\_0\to P\_n$ is either a simple cycle or has no crossings (planar)?



Computational (conjecture) choices for a path
Let $S=\Sigma v\_i$. If $S=0$, sort the vectors according to their angle along the unit circle. Then the corresponding closed path traces the boundary of a convex polygon.


In fact, the vectors $v\_i$ can be of arbitrary length.


If $S\neq 0$, then add an auxiliary vector $v\_{n+1}=-S$ and proceed as in the first case. Finally remove the segment given by the vector $v\_{n+1}$.


