Are the followings true?

1. In every diagonal of every regular polygon, some diagonals of regular hexagon, whose lengths are twice as long as its edge length, are the only diagonals such that the length of a diagonal is an integer multiple of the edge length of the regular polygon.

2. In every diagonal of every regular polygon, some diagonals of regular hexagon are the only diagonals such that the length of a diagonal is a rational multiple of the edge length of the regular polygon.

Motivation : I reached these expectations by using computer. It seems true, but I can't prove them. Can anyone help? If you know something helpful, please let me know it.


2 Answers 2


So you are asking if for some $n \ge 4$ and $ 1 < k \le n/2$, $\frac{\sin(k\pi/n)}{\sin(\pi/n)}$ can ever be a rational, aside from $(k,n) = (3,6)$.

If this is a rational, so is its square. And its square is $\frac{1 - \cos(2k\pi/n)}{1 - \cos(2\pi/n)}$, which is a lot easier to work with.

This ratio is an algebraic number, and it is well known that the Galois group of $\cos(2\pi/n)$ is isomorphic to $(\Bbb Z/n\Bbb Z)^*/\{\pm 1\}$, by defining $\sigma_u(\cos(2j\pi/n)) = \cos(2ju\pi/n)$ for $u \in (\Bbb Z/n\Bbb Z)^*$ and any integer $j$.

The ratio is a rational $x$ if and only if it is invariant by this Galois group. Let us suppose this is the case. Then $\cos(2uk\pi/n) = (1-x) + x\cos(2u\pi/n)$ forall $u$ coprime with $n$.

In particular, the formula implies that $\Bbb Q(\cos(2\pi/n)) = \Bbb Q(\cos(2k\pi/n))$.

If $k$ is coprime with $n$, then letting $r$ be the order of $k$ modulo $n$, we get $\cos(2\pi/n) = \cos(2k^r\pi/n) = (1-x)(1+x+\ldots+x^{r-1}) + x^r\cos(2\pi/n)$. So we must have $x^r = 1$, hence $x = \pm 1$, and so the diagonal must have the same length as the edge of the regular polygon, so we are not at all looking at an interior diagonal.

So the interior diagonal must correspond to a diagonal (or an edge) of a strictly smaller regular polygon. Let $d$ be the gcd of $(k,n)$. Then let $m = n/d$. We get $\Bbb Q(\cos(2\pi/n)) = \Bbb Q(\cos(2k\pi/n)) = \Bbb Q(\cos(2\pi/m))$.

So the natural reduction map between the Galois groups $(\Bbb Z/n\Bbb Z)^*/\{\pm 1\} \to (\Bbb Z/m \Bbb Z)^*/\{\pm 1\}$ has to be an isomorphism.

The only way for this to happen, since $\phi(n) > \phi(m)$, is by having $\{\pm 1\} = \{1\} \pmod m$ and $\phi(n)/\phi(m) = 2$. So either $(n,m) = (4,2)$ or $(n,m) = (6,2)$.

In the $(4,2)$ case, we obtain that the diagonal of the square is $\sqrt 2$ times the side of the square, and the square of this ratio is rational.
In the $(6,3)$ case, we obtain the relationship you have found between the side and the diameter of the hexagon. Not only is the square of the ratio rational, but the ratio itself is rational then.

  • 1
    $\begingroup$ Thank you very much for nice answer! $\endgroup$
    – mathlove
    Oct 8, 2013 at 16:45

Here is a start. in a regular $n$-gon each side subtends an angle $\frac {2\pi}n$ at the centre. $k$ consecutive sides subtend an angle $\frac {2\pi k}n$ at the centre, which is also the angle subtended by the diagonal joining the first vertex in the chain with the last vertex in the chain. Let the distance from the centre to a vertex be $r$.

We then draw a perpendicular from the centre to the diagonal, and see that the length of the diagonal is $2r\sin \frac {k\pi}n$. When $k=1$ this gives the length of a side.

So your questions reduce to discovering when $\sin\frac {k\pi}n$ is an integer/rational multiple of $\sin\frac {\pi}n$ for positive integers $1\lt k \lt n$.

  • $\begingroup$ Thank you for your quick reply. The last sentence still seems difficult for me. Could you please explain more? $\endgroup$
    – mathlove
    Oct 3, 2013 at 15:13
  • $\begingroup$ @mathlove The length of a diagonal divided by the length of a side is $\cfrac {2r\sin\frac {k\pi}n}{2r\sin\frac {\pi}n}$. Cancel the $2r$ and you get an equation involving the two sines. Since you had not got to this point, I thought you might investigate this reformulated version of the problem for yourself. $\endgroup$ Oct 3, 2013 at 15:35

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