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Source: Macedonian JBMO TST 2013

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#1 h May 26, 2013, 8:07 AM • 2 Y

Y by Adventure10, Mango247

A triangle $ABC$ is given, and a segment $PQ=t$ on $BC$ such that $P$ is between $B$ and $Q$ and $Q$ is between $P$ and $C$ . Let $PP_1 || AB$ , $P_1$ is on $AC$ , and $PP_2 || AC$ , $P_2$ is on $AB$ . Points $Q_1$ and $Q_2$ аrе defined similar. Prove that the sum of the areas of $PQQ_1P_1$ and $PQQ_2P_2$ does not depend from the position of $PQ$ on $BC$ .

This post has been edited 1 time. Last edited by andrejilievski, Sep 7, 2013, 11:04 AM

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#2 h May 26, 2013, 10:50 AM • 2 Y

Y by Adventure10, Mango247

Solution. We denote $[ABC]$ be the area of triangle $ABC$ .
Since $P_2P \parallel AC, \; Q_2Q \parallel AC$ then $\triangle BPP_2 \sim \triangle BCA \; (AAA)$ and $\triangle BQQ_2 \sim \triangle BCA \; (AAA)$ .
Therefore $[BPP_2]= \dfrac{BP^2}{BC^2} \cdot [ABC]$ and $[BQQ_2]= \dfrac{(BP+t)^2}{BC^2} \cdot [ABC]$ .
We have $[PQQ_2P_2]= [BQQ_2]-[BPP_2]= [ABC] \cdot \dfrac{(BP+t)^2-BP^2}{BC^2}$ .
Similarly, we also have $[PQQ_1P_1]= [ABC] \cdot \dfrac{(t+QC)^2-QC^2}{BC^2}$ .
Thus, $[PQQ_2P_2]+[PQQ_1P_1]= [ABC] \cdot \dfrac{2t}{BC}$ .
So the sum of the areas of $PQQ_1P_1$ and $PQQ_2P_2$ does not depend from the position of $PQ$ on $BC$ .

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#3 h May 30, 2013, 5:00 AM • 2 Y

Y by Adventure10, Mango247

Let $P{ P }_{ 1 }$ and $Q{ Q }_{ 2 }$ meet at $O$ .
Now $\left[ PQQ_{ 1 }P_{ 1 } \right] +\left[ PQQ_{ 2 }P_{ 2 } \right] =2\left[ OPQ \right] +\left[ OQ{ Q }_{ 1 }{ P }_{ 1 } \right] +\left[ OP{ Q }_{ 2 }{ P }_{ 2 } \right]$
But $\triangle ABC\sim \triangle OPQ$
Therefore $\frac { BC }{ PQ } =\frac { AB }{ OP } =\frac { AC }{ OQ }$
$\Rightarrow \frac { a }{ t } =\frac { b }{ OP } =\frac { c }{ OQ }$
$\Rightarrow OP=\frac { tb }{ a } ;\quad OQ=\frac { tc }{ a }$
Also $\triangle OPQ\simeq \triangle Q{ Q }_{ 1 }O$
$\Rightarrow \left[ OPQ \right] =\left[ Q{ Q }_{ 1 }O \right] =\left[ P_{ 1 }O{ Q }_{ 1 } \right]$ ...... $(1)$

Similarly $\triangle OPQ\simeq \triangle PO{ P }_{ 2 }$
$\Rightarrow \left[ OPQ \right] =\left[ PO{ P }_{ 2 } \right] =\left[ { Q }_{ 2 }{ P }_{ 2 }O \right]$ ........... $(2)$

Therefore from $(1)$ and $(2)$ we have
$\quad \quad \left[ PQQ_{ 1 }P_{ 1 } \right] +\left[ PQQ_{ 2 }P_{ 2 } \right] =2\left[ OPQ \right] +\left[ OQ{ Q }_{ 1 }{ P }_{ 1 } \right] +\left[ OP{ Q }_{ 2 }{ P }_{ 2 } \right] \\ \Rightarrow \left[ PQQ_{ 1 }P_{ 1 } \right] +\left[ PQQ_{ 2 }P_{ 2 } \right] =6\left[ OPQ \right]$

But $\left[ OPQ \right] =\frac { 1 }{ 2 } (PQ)(OQ)sin(C)$
$\Rightarrow \left[ OPQ \right] =\frac { 1 }{ 2 } \frac { { t }^{ 2 }c }{ a } sin(C)$
Therefore $\left[ PQQ_{ 1 }P_{ 1 } \right] +\left[ PQQ_{ 2 }P_{ 2 } \right]=6\left[ OPQ \right] =3\frac { { t }^{ 2 }c }{ a } sin(C)$ which is independent of the position of $PQ$ on $BC$ ....

$Q.E.D$

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