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The question is about this problem (it is from a Math' Olympiad in Germany):

Prove that if a regular heptagon $ABCDEFG$ has side 1, then

$$\frac1{AC}+\frac1{AD}=1$$

I have found something: using the law of cosines I have derived a third degree equation that is satisfied if the statement is true, but this solution is long and ugly, and, in fact, it is not a solution since the equation has two more roots. I can search for my notes if anybody would need details.

In fact, I'm looking for a less algebraic kind of solution :-)

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2 Answers 2

up vote 2 down vote accepted

You can apply Ptolemy's theorem to quadrilateral $ACDE$: $$\color{red}{AC} \cdot DE + CD \cdot \color{blue}{AE} = \color{green}{AD} \cdot \color{magenta}{CE}.$$

By symmetry $DE = CD = 1$, $AE = AD$, $CE = AC$.
So $$\begin{align} \color{red}{AC} + \color{blue}{AD} &= \color{green}{AD} \cdot \color{magenta}{AC},\\ \frac{1}{AD} + \frac{1}{AC} &= 1. \end{align}$$

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My first thought is to place the vertices in the complex plane in the standard way: let $\zeta_7 = e^{2 \pi i/7}$ be a primitive $7^{\rm th}$ root of unity, and let $A = \zeta_7^0 = 1$, $B = \zeta_7$, $C = \zeta_7^2$, etc. Then the claim to be proven is equivalent to $$\frac{1}{|\zeta_7^2 - 1|} + \frac{1}{|\zeta_7^3 - 1|} = \frac{1}{|\zeta_7 - 1|}.$$ Then using the fact that $|z|^2 = z\bar z$ for any complex number $z$, $\bar \zeta_7 = \zeta_7^{-1}$, $\zeta_7^{7+k} = \zeta_7^k$, and $\sum_{k=0}^6 \zeta_7^k = 0$, you should be able to verify this identity.

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