Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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 :-)

share|cite|improve this question
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 $$\begin{align} \color{red}{AC} + \color{blue}{AD} &= \color{green}{AD} \cdot \color{magenta}{AC},\\ \frac{1}{AD} + \frac{1}{AC} &= 1. \end{align}$$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.