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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.

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    "regular polygon" is the more standard terminology... :)2010-11-23
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    @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.2010-11-23
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    Ah, so we can't assume the polygon is convex? This should be very interesting...2010-11-23
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    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?2011-03-01
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    @user7530, I don't see the equivalence you claim.2011-05-16
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    I was trying to solve the following issue: Find the number of possible closed paths using one fifth of an arc (72 degrees), where at each time step we can move either clockwise or anti-clockwise. in that particular problem, it was assumed that the number of arcs should be 70. In my solution, I assumed that all closed paths have no loops. I then proved that the number of arcs should necessarilly be even in order to get a closed path. Then, I simplified the problem to finding the number of possible EQUILATERAL POLYGONS, with interior angles: 72, 2*72, 3*72, and 4*72. To do this I needed to find2011-05-16

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