The heptagon on the picture is a regular heptagon with side 1. What is the length of the dashed interval?

the problem

This is a (kind of) 'geometric version of a quadratic Gauss sum for p=7' (this observation — and the problem — is due to Dušan Djukić, I believe): once you know that $\sin\frac{2\pi}7+\sin\frac{4\pi}7+\sin\frac{8\pi}7=\frac{\sqrt7}2$ finding the answer is easy. But I'm interested in going in the opposite direction — so I'm asking for a geometric proof (not using trigonometry / complex numbers).

It would also be nice to have a generalisation to 11-gon (and beyond). One possible geometric interpretation of quadratic Gauss sum for p=11 is on the picture below (but maybe there is a better choice…).


  • 1
    $\begingroup$ For the curious (and/or those not wanting to spend time on the "easy" part): The answer for the heptagon is $\sqrt{2}$. $\endgroup$ – Blue Jan 20 '18 at 20:47
  • $\begingroup$ For $x = 2 \cos (2 \pi/11),$ we get $x^5 + x^4 - 4 x^3 - 3 x^2 + 3 x + 1 = 0.$ Given a primitive 11th root of unity $\alpha,$ we find that $\eta = \alpha + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^9$ solves $\eta^2 + \eta + 3 = 0,$ meaning not actually real in the first place. See pages 9 and 10 in Reuschle books.google.com/… $\endgroup$ – Will Jagy Jan 20 '18 at 21:15
  • 1
    $\begingroup$ @WillJagy Of course, the value of Gauss sum for $p=4k+3$ is not real. But it gives an identity $$\sin(2\pi/11)+\sin(6\pi/11)+\sin(8\pi/11)+\sin(10\pi/11)+\sin(18\pi/11)=\sqrt{11}/2$$ which one can try to prove geometrically. $\endgroup$ – Grigory M Jan 20 '18 at 21:27

Let $P_0 P_1 P_2 P_3 P_4 P_5 P_6$ be a regular heptagon with side-length $p$ and $\theta := \angle P_5 P_1 P_2 = \angle P_6 P_2 P_3$. Let $\overline{P_1 P_5}$ meet $\overline{P_2 P_6}$ at $Q$. Define $q :=|\overline{QP_2}| = |\overline{QP_5}|$ and $d := |\overline{QP_3}|$.

enter image description here

Claim. $\quad d = p\sqrt{2}$

Because $\overline{P_1 P_5}\parallel\overline{P_0P_6}$ and $\overline{P_2 P_6}\parallel\overline{P_1P_0}$, we have that $\square P_0 P_1 Q P_6$ is a parallelogram; in particular, it is a rhombus, so that $|\overline{QP_1}| = p$.

Using a pinch of trig in the form of the Law of Cosines, we observe

$$\begin{align} \triangle QP_2 P_3: \quad d^2 &= p^2 + q^2 - 2 pq \cos\theta \tag{1a}\\ \triangle QP_1 P_2: \quad q^2 &= 2p^2-2p^2\cos\theta \tag{1b} \end{align}$$

Thus, replacing $q^2$ in $(1a)$ using $(1b)$,

$$d^2 = 3 p^2 - 2p(p+q)\cos\theta \tag{2}$$

However, writing $M$ for the midpoint of $\overline{P_1 P_2}$, we see in $\triangle P_5 P_1 M$ that $$(p+q)\cos\theta = \frac{1}{2}p \tag{3}$$ which implies $$d^2 = 2 p^2 \tag{4}$$ and proves the Claim. $\square$


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