Suppose that a regular $n$-gon has integer side length $m$. Is the lengths of its diagonals always algebraic numbers? If yes and if $n,m$ are given, is there an easy way to compute the diagonal lengths as a root of some polynomial?
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It appears to me that WLOG $n=1$ suffices, since the diagonals are linearly dependent on the side length (by intercept theorem) – AlexR Feb 23 '14 at 16:59
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See also the paper Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations. – lhf Feb 02 '18 at 00:37
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All diagonals are either diameters, or sides of a triangle whose other two legs are segments uniting the center of the polygon to the diagonal's two extremities. As such, their lengths can be computed using the generalized Pythagorean theorem, also known as the law of cosines. The angle between these two sides is a rational multiple of π, hence its sine and cosine are both algebraic, being the real and imaginary part of the n complex roots of unity, per de Moivre's and Euler's formula. Now all that's left to show is that these radii are themselves algebraic: Indeed, $r^2+r^2$$-2rr\cos\dfrac{2\pi}n=m^2$ implying $r=\dfrac m{\sqrt{2\Big(1-\cos\frac{2\pi}n\Big)}}\in\mathbb{A}$. Then $d_j=r\sqrt{2\bigg(1-\cos\dfrac{2k\pi}n\bigg)}\in\mathbb{A}$. QED.
Lucian
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and in fact, if $r$ is an algebraic integer (such as 1) then so are the diagonals for by the same principle! – Thiago Dec 27 '19 at 21:28
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To find the shortest diagonal of a regular $n$-gon, it's
$$
s\sqrt{\frac{n}{2}}
$$
For a triangle you'll just get the side length.
scratchminer
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1This is not correct; it fails for all $n$ other than $n=4$ and $n=6$. For instance, when $n=5$ all diagonals of a unit regular pentagon have length $\frac{1+\sqrt{5}}2\neq \sqrt{\frac52}$. – RavenclawPrefect Feb 23 '21 at 23:47