When $n$ points on the unit circle are distributed so nice and symmetrically that they form the corners of a regular $n$-gon, their sum (as vectors in the plane, or as complex numbers) equals 0.
Geometrically this is intuitively clear, but to prove it I think it is easiest algebraically, by viewing the points as complex numbers. I put the proof at the end of the post for readers who haven't seen it earlier in their life.
It follows that any set of points on the unit circle that is a union of vertices of regular $k$-gons for various $k$ also sums to zero. This still remains true if, for the sake of this question, we define the vertices of a regular $2$-gon to be a pair of the form $\{x, -x\}$.
My question is: does the converse of this also hold?
In other words:
If a set of set of $n$ points on the unit circle sums to zero, is it necessarily the union of the vertices of regular $k$-gons for some multiset of $k$'s (that may include 2 with the above convention) or are there other configurations that sum to 0
In yet other words:
Is it true that a set of $n$ points on the unit circle that sum to zero is either the set of vertices of a regular $n$-gon or contains a a proper, non-empty subset of points that also sums to zero (or both)?
I checked that this last statement is true for $n = 2, 3, 4$ by looking at on which lines in the plane the sums of two of the points must lie, but it is not so clear how this type of reasoning extends to $n = 5$ or more.
Any help is appreciated.
Proof that vertices of a regular $n$-gon whose circumcircle is the unit circle in $\mathbb{C}$ sum to 0:
In order to form a perfect $n$-gon, the $n$ numbers must be of the form $\xi\zeta, \xi\zeta^2, \ldots, \xi\zeta^n = \xi$ where $\zeta = e^{2\pi i/n}$ is a primitive $n$'th root and $\xi$ is just one of the points. Now their sum equals $\xi(\zeta^{n-1} + \ldots + \zeta + 1)$. Multiplying this by the non-zero number $\zeta – 1$ we obtain $\xi(\zeta^n – 1)$ and by definition of $\zeta$ this equals $\xi0 = 0$
Best Answer
An equivalent question is: are there paths with unit-length steps that start and end at the origin but whose steps cannot be reordered so that the path returns more than once on the origin? This is an easy counterexample. Notice that there clearly are no proper nonempty subsets of vectors that sum to zero, otherwise you'd have two parallel vectors in the picture, which you have not.
Converting from path to points on a circle gives the following picture.