Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\pi/n$, then all angles are an integral multiple of $\pi/n$.

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.

share|improve this question
    
"regular polygon" is the more standard terminology... :) –  J. M. Nov 23 '10 at 3:41
1  
@JM: Not quite. It is only implied that the polygon has equal sides, not equal angles. Also the problem is almost verbatim from the notes, so not my words. –  Timothy Wagner Nov 23 '10 at 3:42
    
Ah, so we can't assume the polygon is convex? This should be very interesting... –  J. M. Nov 23 '10 at 3:57
    
Hmm, I'm missing something... this problem is equivalent to proving that the only elements of $\mathbb{Z}[e^{2\pi i/n}]$ with unit norm are the nth roots of unity, right? But isn't that known to be false for n > 6? –  user7530 Mar 1 '11 at 5:30
    
@user7530, I don't see the equivalence you claim. –  Gerry Myerson May 16 '11 at 12:49

1 Answer 1

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 $\sin(\pi/n)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.