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- Feb 14, 2012

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Let $ABCDEFG$ be a regular heptagon. Prove that $\dfrac{AD^3}{AB^3}-\dfrac{AB+2AC}{AD-AC}=1$.

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

Let $ABCDEFG$ be a regular heptagon. Prove that $\dfrac{AD^3}{AB^3}-\dfrac{AB+2AC}{AD-AC}=1$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Feb 14, 2012

- 3,894

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

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

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