# Problem Of The Week #445 December 8th, 2020

Status
Not open for further replies.

#### anemone

##### MHB POTW Director
Staff member
• Greg and topsquark

#### anemone

##### MHB POTW Director
Staff member
No one answered last week's POTW. However, you can check out the solution of other as follows:
Let $ABCDEDG$ be a regular heptagon having sides of length $a$, short diagonal (e.g. $AC$) of length $b$ and long diagonals (e.g. $AD$) of length $c$. Let $\theta=\dfrac{\pi}{7}$ so that $a=2R\sin \theta,\,b=2R\sin 2\theta$ and $c=2R\sin 3 \theta$, where $R$ is the circumradius.

Applying the Ptolemy's theorem to the respective cyclic quadrilaterals $ABCD,\,ACEG,\,ADEG,\,ADFG$, we have

$a^2+ac=b^2\\b^2+ab=c^2\\a^2+bc=c^2--(1)\\ac+ab=bc--(2)$

We have to show that

$\dfrac{c^3}{a^3}-\dfrac{a+2b}{c-b}=1$, or $\dfrac{c^3}{a^3}=\dfrac{a+b+c}{c-b}$

From (1) and (2), we obtain that

$\dfrac{c^3}{a^3}=\dfrac{c}{a^2}\left(a+\dfrac{bc}{a}\right)=\dfrac{1}{c-b}(a+b+c)\,\,\, \text{Q.E.D.}$

• topsquark
Status
Not open for further replies.