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Difference between revisions of "2013 USAJMO Problems/Problem 5"
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==Problem==
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Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>.  Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>.  Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>.  Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>.  Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>.  Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}</cmath>.
 
Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>.  Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>.  Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>.  Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>.  Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>.  Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}</cmath>.
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==Solution==

Revision as of 08:26, 12 May 2013

Problem

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}\].

Solution

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