Sum of squares of ratios of diagonal of a regular heptagon A problem from the 1998 Lower Michigan Math Competition...
A regular heptagon has diagonals of two different lengths.  Let $a$ be the length of a side, $b$ the length of a shorter diagonal, and $c$ the length of a longer diagonal.  Prove that
$$
\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} = 6 \quad \text{ and } \quad \frac{b^2}{a^2} + \frac{a^2}{c^2} + \frac{c^2}{b^2} = 5.
$$
What I have so far:

*

*Introduce coordinates so that the heptagon consists of the seventh roots of unity in the complex plane.

*Let $\omega = e^{2 \pi i/7}$ so that the vertices are $1, \omega, \omega^2, \dots, \omega^6$.  Then $a = |1-\omega|$ and $b = |1-\omega^2| = |1-\omega||1+\omega|$ and $c = |1-\omega^3| = |1-\omega||1+\omega+\omega^2|$.

*The first requested expression then becomes
$$
\frac{1}{|1+\omega|^2} + \frac{|1+\omega|^2}{|1+\omega+\omega^2|^2} + |1+\omega+\omega^2|^2.
$$

*All those modulus-squared expressions can be written as (thingy) times (conjugate of thingy), and the conjugate of a power of $\omega$ is another power of $\omega$.  For example,
$$
|1+\omega|^2 = (1+ \omega)(1 + \overline{\omega}) = (1 + \omega)(1 + \omega^6).
$$
Using this, the first requested expression becomes
$$
\frac{1}{(1+\omega)(1+\omega^6)} + \frac{(1+\omega)(1+\omega^6)}{(1+\omega+\omega^2)(1+\omega^5+\omega^6)} + (1 + \omega + \omega^2)(1 + \omega^5 + \omega^6).
$$
Annnnd...that's about where I ran out of steam.  Not sure how much of this is progress, but I really want the complex-roots-of-unity approach to pan out!

Use $\displaystyle e^{\tfrac{2\pi i}{7}}=\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7} $ to get $$|1+\omega|=\left|1+\cos \frac{2\pi }{7}+i\sin \frac{2\pi }{7}\right|$$$$=\left|2\cos^2 \frac{\pi }{7}+2i\sin \frac{\pi }{7}\cos \frac{\pi }{7}\right|=2\cos \frac{\pi }{7}\left|\cos \frac{\pi }{7}+i \sin\frac{\pi }{7} \right|$$$$=2\cos \frac{\pi }{7}.$$
Also, $\omega^2= e^{\tfrac{4\pi i}{7}} =\cos \dfrac{4\pi }{7}+i\sin \dfrac{4\pi }{7} $ and  $$|1+\omega +\omega^2|=\left|\left(\color{red}{1}+\cos \frac{2\pi }{7}+\color{red}{\cos \frac{4\pi }{7}}\right)+i\left(\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}\right)\right|$$$$= \left|\left(\cos \frac{2\pi }{7}+2\cos^2\frac{2\pi}{7}\right)+i\left(\sin \frac{2\pi }{7}+2\sin \frac{2\pi }{7}\cos\frac{2\pi}{7}\right)\right|$$$$=1+2\cos\frac{2\pi}{7}.$$
