Equilateral polygon in a plane Let $n$ be a positive integer. Suppose we have an equilateral polygon in the Euclidean plane with the property that all angles except possibly two consecutive ones are an integral multiple of $\frac{\pi}{n}$, then all angles are an integral multiple of $\frac{\pi}{n}$.
This problem is #28 on page 61 in these notes restated here for convenience:
http://websites.math.leidenuniv.nl/algebra/ant.pdf
I have seen a number-theoretic proof of this. I was wondering if there are any geometric (or at least non number-theoretic) proofs of this result.
 A: Maybe this is a start: Given: an n-step walk in the plane with each step of length 1 that begins with a step to (1,0) and ends at the origin, and all angles between steps being a multiple of $\pi/n$ except perhaps those between three adjacent steps (i.e. two corners). Show that all the angles are multiples of $\pi/n$.
Label the angles by $\{\alpha_k|1\le k \le n\}$ with the angles that are not multiples of $\pi/n$ being $\alpha_{n-1}$ and $\alpha_n$. Define each consecutive step in vector form, that is, relative to this coordinate system. We have steps of form:
$$(\cos(\beta_k),\sin(\beta_k))$$
where $\beta_k = \Sigma_{j\le k}\;\alpha_j$ and $\beta_1=0$. So $\beta_k$ is a multiple of $\pi/n$ except for $k=n-1$ or $n$.
That the path returns to the origin requires:
$$\Sigma_k(\cos(\beta_k),\sin(\beta_k)) = (0,0).$$
Now note that $\cos(k\pi/n)$ can be written as a polynomial over $Z$ in $\cos(\pi/n)$ and that $\sin(k\pi/n)$ can be written as $\sin(\pi/n)$ times a polynomial over $Z$ in $\cos(\pi/n)$. These apply to all the $\beta_k$ except the last two.
So it looks like a problem in $Z[\cos(\pi/n)]$. And maybe you should divide the "y" restriction by $\sin(\pi/n)$.
