The heptagon on the picture is a regular heptagon with side 1. What is the length of the dashed interval?
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…).
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 $d := |\overline{QP_3}|$.
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. $\square$