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Consider the following figure with equilateral triangle $ABC$ and a cevian $AQ$ extended to $P$ on its circumcircle.

enter image description here

We are required to prove that:

$\frac{1}{PB} + \frac{1}{PC} = \frac{1}{PQ}$

Let $\angle PAC = \alpha$ and let length of $AB = s$

By the Law of Sines,

$\frac{PC}{\sin\alpha} = \frac{s}{\sin\angle CPA} = \frac{s}{\sin60^{\circ}} \implies PC = \frac{s\sin\alpha}{\sin60^{\circ}}$

Similarly, $PB = \frac{s\sin(60^{\circ} - \alpha)}{\sin60^{\circ}}$

$\frac{1}{PB} + \frac{1}{PC} = \frac{\sin60}{s}\left(\frac{\sin(60 - \alpha) + \sin\alpha}{\sin\alpha\sin(60 - \alpha)}\right)$

It remains to be proven that:

$PQ = \frac{\sin60\sin(60 - \alpha)\sin\alpha}{\sin60\sin(60 - \alpha) + \sin60\sin\alpha}$

I'm utterly lost from here.

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I have read that this problem appeared in South African mathematical competition in years 1994 and 1997. The proof I know is based on the areas of triangles.

As $\angle APB=\angle APC=60^\circ$ and $\angle BPC=120^\circ$, the areas of triangles $BPQ,$ $QPC$ and $BPC$ are $\frac{1}{2}BP\cdot PQ\cdot \sin 60^\circ,$ $\frac{1}{2}PQ\cdot PC\cdot \sin 60^\circ$ and $\frac{1}{2}BP\cdot PC\cdot \sin 120^\circ$ respectively. As $\sin 60^\circ=\sin 120^\circ$ and the sum of areas of triangles $BPQ$ and $QPC$ is the area of $BPC,$ we have that $BP\cdot PQ+PQ\cdot PC=BP\cdot PC.$ Now divide this equation by $BP\cdot PQ\cdot PC.$

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by Ptolemy , $AB\cdot PC+AC\cdot PB=BC\cdot AP$
$\therefore PA=PB+PC$
$\triangle PCQ\sim\triangle PAB$ $\Longrightarrow$ $PQ\cdot PA=PB\cdot PC$
$PQ\cdot PA=PQ\cdot (PB+PC)=PB\cdot PC$
$\therefore$ $\dfrac{1}{PB}+\dfrac{1}{PC}=\dfrac{1}{PQ}$

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