# A geometry problem from maths competition

I have a problem understanding the proof.

Given an acute angle $A$. Choose an arbitrary point $P$ from the bisector of $A$ and another point $B$ from the side of angle $A$. Draw a line $l$ going through points $P$ and $B$. Now $l$ intersects another side of the angle $A$ at point $C$. Prove that no matter how we choose $B$ and keep $P$ fixed, the expression $\frac{1}{\mid AB\mid}+\frac{1}{\mid AC\mid}$ is constant.

Proof.

Let $\angle A=2\theta$, $\angle ACB = \phi$, and $PD$ is perpendicular to $AC$. Because $AP$ bisects $\angle A$ we have ${AC\over AB}={PC\over PB}$. Thus:

${1\over AB}+{1\over AC}={1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}={\sin 2\theta\over {PC\cdot\sin\phi}}={\sin 2\theta\over PD}={\sin 2\theta\over {PA\cdot\sin \theta}}={2\cos\theta\over PA}={\rm constant,}$

because both $\theta$ and $PA$ are given.

Why do we have that ${1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}$ and is that proof correct?

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In virtue of the angle bisector theorem, $$\frac{PB}{PC}=\frac{AB}{AC},$$ then: $$\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AB}+\frac{PB}{AB\cdot PC}=\frac{1}{AB}\left(1+\frac{PB}{PC}\right)=\frac{1}{AB}\cdot\frac{BC}{PC}.$$

It is interesting to note that, if we take $B'$ as the symmetric of $B$ wrt to $AP$, $C'$ as the symmetric of $C$ wrt to $AP$, the diagonals of the isosceles trapezoid $BB'CC'$ meet in $P$. If we take the perpendicular to $AP$ through $P$, the length of the segment cut by the angle is equal to the harmonic mean of $BB'$ and $CC'$, so the result follows.

Another way is to consider the angle as a skew reference system and the point $P$ as $(1,1)$. In this setting, all the lines throug $P$ have equation $ax+by=a+b$, so they intersect the coordinate axis in $y_0=\frac{a+b}{b}$ and $x_0=\frac{a+b}{a}$: this gives $\frac{1}{x_0}+\frac{1}{y_0}=1$, QED.

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@student: if the answer is useful, you should consider upvoting. – robjohn Nov 16 '12 at 15:28
I haven't registered to the forum. – student Nov 16 '12 at 17:59

Area APB + Area APC = Area ABC $$\frac{1}{2}PB\sin(θ)+\frac{1}{2}PC\sin(θ) = \frac{1}{2}BC\sin(2θ)$$ $$\sin(2θ)= 2\sin(θ)cos(θ) \\ \Rightarrow PB+PC= 2BC\cos(θ)$$ (divide by $BCP$) $$\Rightarrow \frac{1}{C}+\frac{1}{B} =2\frac{\cos(θ)}{P}$$

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Please try to learn some Latex, it makes proofs much easier to follow. – Simon Hayward Nov 17 '12 at 13:47