Sum of squares of ratios of diagonal of a regular heptagon

Question

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!

Answer 1

Note that $$1+\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0\quad\hbox{and}\quad \omega^7=1\ .\tag{$*$}$$ You can use this to get rid of denominators. Thus $$(1+\omega+\omega^2)(1+\omega^3)=1+\omega+\omega^2+\omega^3+\omega^4+\omega^5 =-\omega^6=-\omega^{-1}\ ,$$ so $$\frac1{1+\omega+\omega^2}=-\omega(1+\omega^3)\ .$$ Likewise $$\frac1{1+\omega}=-\omega(1+\omega^2+\omega^4)\ .$$ So both your expressions can be written as polynomials in $\omega$. Bit of work multiplying out the conjugates, but not too much. And though I must admit I haven't yet done the whole thing, I would be very surprised if it doesn't fall straight out using $(*)$.

Later: it works. But it's too much effort to type it all up, sorry :)

Answer 2

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 $$|1+\omega +\omega^2|=\left|\left(\color{red}{1}+\cos \frac{2\pi }{7}+\color{red}{\cos \frac{4\pi }{7}}\right)+i\left(\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}\right)\right|$$$$= \left|\left(\cos \frac{2\pi }{7}+2\cos^2\frac{2\pi}{7}\right)+i\left(\sin \frac{2\pi }{7}+2\sin \frac{2\pi }{7}\cos\frac{2\pi}{7}\right)\right|$$$$=1+2\cos\frac{2\pi}{7}.$$

Answer 3

Adding details to David's answer---too much algebra to fit in a comment. ;-)

Some milestones along the way:

• Taking the modulus-squared on both sides of David's expression for $\frac{1}{1+\omega}$ yields $$ \frac{1}{|1+\omega|^2} = |1+\omega^2 + \omega^4|^2 = (1+\omega^2+\omega^4) \overline{(1+\omega^2+\omega^4)} = (1+\omega^2+\omega^4)(1+\omega^5+\omega^3) $$
• Expanding the polynomial and reducing powers of $\omega$ modulo 7 yields $$ \frac{1}{|1+\omega|^2} = 1 + \omega^5 + \omega^3 + \omega^2 + \omega^7 + \omega^5 + \omega^4 + \omega^9 + \omega^7 = 3 + 2 \omega^2 + \omega^3 + \omega^4 + 2 \omega^5 $$
• Likewise, $$ |1+\omega+\omega^2| = (1+\omega+\omega^2)(1+\omega^6+\omega^5) = \cdots = 3 + 2 \omega + \omega^2 + \omega^5 + 2 \omega^6 $$
• Multiplying both sides of David's expression for $\frac{1}{1+\omega+\omega^2}$ by $1+\omega$ yields $$ \frac{1+\omega}{1+\omega+\omega^2} = - \omega(1+\omega)(1+\omega^3) $$ and taking modulus-squared on both sides, $$ \left|\frac{1+\omega}{1+\omega+\omega^2}\right|^2 = (1 + \omega)(1 + \omega^6)(1+\omega^3)(1+\omega^4) = (2 + \omega + \omega^6)(2 + \omega^3 + \omega^4) = \cdots = 4 + 2 \omega + \omega^2 + 3 \omega^3 + 3\omega^4 + \omega^5 + 2 \omega^6. $$ -- Adding the three equations together, $$ \frac{1}{|1+\omega|^2} + |1+\omega+\omega^2|^2 + \left|\frac{1+\omega}{1+\omega+\omega^2}\right|^2 = 10 + 4 \omega + 4 \omega^2 + 4 \omega^3 + 4 \omega^4 + 4 \omega^5 + 4 \omega^6. $$
• But that latter expression is $$ 6 + 4 (1 + \omega + \omega^2 + \cdots + \omega^6) = 6 + 4(0) = 6. $$