On the heptagon

Given a heptagon of side $a$ and diagonals $b, c$ such that $b<c$ ,

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prove that:

$\frac{b^2}{a^2}+ \frac{c^2}{b^2} + \frac{a^2}{c^2} =5$

Solution

Let $a$ be the side of the heptagon and $b, c$ be its diagonals respectively. It holds that

(1) $\begin{equation*} b^2-a^2= ac \end{equation*}$

(2) $\begin{equation*} c^2-b^2 = ab \end{equation*}$

(3) $\begin{equation*} a^2-c^2 = -bc \end{equation*}$

(4) $\begin{equation*} \frac{c}{a} + \frac{a}{b} -\frac{b}{c} = 2 \end{equation*}$

Equation $(4)$ comes naturally from Vieta’s formulae since $\frac{c}{a}\; , \; \frac{a}{b}\;, \; -\frac{b}{c}$ are the roots of the equation $t^3-2t^2-t+1=0$ . Thus,

$\begin{aligned} \frac{b^2}{a^2} + \frac{c^2}{b^2} + \frac{a^2}{c^2} &=\frac{b^2-a^2+a^2}{a^2}+ \frac{c^2-b^2+b^2}{b^2}+ \frac{a^2-c^2+c^2}{c^2} \\ &=\left ( \frac{b^2-a^2}{a^2} + \frac{a^2}{a^2} \right ) + \left (\frac{c^2-b^2}{b^2}+\frac{b^2}{b^2} \right ) + \left ( \frac{a^2-c^2}{c^2}+ \frac{c^2}{c^2} \right ) \\ &= \left (\frac{ac}{a^2}+ 1 \right ) + \left ( \frac{ab}{b^2} + 1 \right ) + \left ( -\frac{bc}{c^2}+ 1 \right ) \\ &= \frac{c}{a} + \frac{a}{b} -\frac{b}{c}+ 3 \\ &=2 + 3 \\ &=5 \end{aligned}$