Show that three lines don't intersect in one point

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Show that three lines don't intersect in one point

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Source: 40th Austrian Mathematical Olympiad 2009, round 1, problem 3

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FelixD

#1 h Aug 5, 2009, 2:14 PM • 2 Y

Y by Adventure10, Mango247

Let $D$ , $E$ , $F$ be the feet of the altitudes wrt sides $BC$ , $CA$ , $AB$ of acute-angled triangle $\triangle ABC$ , respectively. In triangle $\triangle CFB$ , let $P$ be the foot of the altitude wrt side $BC$ . Define $Q$ and $R$ wrt triangles $\triangle ADC$ and $\triangle BEA$ analogously. Prove that lines $AP$ , $BQ$ , $CR$ don't intersect in one common point.

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#2 h Aug 5, 2009, 2:56 PM • 2 Y

Y by Adventure10, Mango247

Something is not right. See figure!

Attachments:

Figure.pdf (19kb)

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FelixD

#3 h Aug 5, 2009, 3:03 PM • 2 Y

Y by Adventure10, Mango247

@mihai miculita: That is a problem of the Austrian Mathematical Olympiad, thus, it is easy

(btw., at AopsWiki the first round is rated with 2). So don't wonder

But during the contest I wondered too. I thought, it can't be that easy...^^

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#4 h Aug 5, 2009, 4:58 PM • 1 Y

Y by Adventure10

1). $AD\cap BE\cap CF=\{H\}\Rightarrow \frac{|DB|}{|DC|}.\frac{|EC|}{|EA|}.\frac{|FA|}{|FB|}=1.$
2). $\frac{|PB|}{|PC|}<\frac{|DB|}{|DC|};\frac{|QC|}{|QA|}<\frac{|EC|}{|EA|};\frac{|RA|}{|RB|}<\frac{|FA|}{|FB|}\Rightarrow$
$\Rightarrow \frac{|PB|}{|PC|}.\frac{|QC|}{|QA|}\frac{|RA|}{|RB|}<\frac{|DB|}{|DC|}.\frac{|EC|}{|EA|}.\frac{|FA|}{|FB|}=1\Rightarrow\\ \Rightarrow \frac{|PB|}{|PC|}.\frac{|QC|}{|QA|}\frac{|RA|}{|RB|}<1\Rightarrow \mbox AP\cap BQ\cap CR =\emptyset$

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