Relation among the diagonals of a regular heptagon

Question

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 :-)

Answer 1

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 $$\begin{align} \color{red}{AC} + \color{blue}{AD} &= \color{green}{AD} \cdot \color{magenta}{AC},\\ \frac{1}{AD} + \frac{1}{AC} &= 1. \end{align}$$

Answer 2

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.

Answer 3

Here is a proof with just similarity between triangles. First consider the notations as in the figure down below

We define the angles $\alpha,\beta,\gamma$ as such $$\gamma=2\beta=4\alpha=\frac{4\pi}{7}$$ Indeed they appear as the angles of the arc supported by $4,2$ and $1$ side of the heptagon. Set $H$ the intersection of the two diagonals $AC$ and $BD$, as well as $C'$ the symetric image of $C$ about $(BD)$. Since $\widehat{ADB}=\widehat{BDC}$, the point $C'$ lies on $AD$. Thus $$DC=DC'.$$

The triangle $AHC'$ is isosceles in $A$, since the its angles are $\alpha,3\alpha,3\alpha$. So $$AH=AC'.$$

Thus $$AB+DH=DC+DH=DC'+AH=DC'+AC'=AD.$$

Dividing by $AB\cdot AD$ we get $$\frac{1}{AD}+\frac{DH}{AB\cdot AD}=\frac{1}{AB}.$$

Now the triangles $\Delta DHC'$ and $\Delta DAB$ are similar, so $$\frac{DH}{DA}=\frac{DC'}{DB}$$

in other words $$\frac{DH}{AB\cdot AD}=\frac{DH}{DA\cdot DC'}=\frac{1}{DB}=\frac{1}{AC}.$$

So we have shown that in a regular heptagon $ABCDEFG$ $$\frac{1}{AD}+\frac{1}{AC}=\frac{1}{AB}.$$