Excircles, tangencies, circumcircles

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Source: Own (2009 Korean Mathematical Olympiad Winter Training Program)

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#1 h Jan 30, 2009, 8:06 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Given an acute triangle $ABC$ with $AB \neq AC$ , let the excircle for $A$ meets lines $AB,AC,BC$ at $P,Q,R$ . Let $BC$ and $PQ$ meet at $X$ , and the circumcircles of $XCQ$ and $BCP$ meet at $Y \neq C$ . Prove that $\frac {RP}{PX} = \frac {RY}{YX}$ .

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#2 h Jan 31, 2009, 5:01 AM • 2 Y

Y by Adventure10 and 1 other user

$\angle XQY=\angle XCY =\pi -\angle RPY=\pi-(\angle RPY+\angle RPB)$
$=180-(\angle RPY +\angle RQP)\Leftrightarrow \angle RQY+\angle RPY =\pi \Leftrightarrow P,Q,R,Y$ is concyclic.
$\angle RXY =\pi -\angle CQY =\angle QRY$
and $\angle XRY=\angle RQY \Rightarrow \triangle XYR \sim \triangle RYQ$
$\angle XRP=\angle XQR\Rightarrow \triangle XPR \sim \triangle XRQ$

We have $\frac {RY}{YX}\frac{PX}{RP}=\frac{QY}{YR}\frac{RX}{QR}= \frac{QY}{QR} \frac{RX}{RY}=1$
$\Leftrightarrow \frac{RY}{YX}=\frac{RP}{PX}$ .

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