# Difference between revisions of "2013 USAJMO Problems/Problem 5"

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==Problem== | ==Problem== | ||

− | 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> |

==Solution== | ==Solution== | ||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 18:05, 3 July 2013

## Problem

Quadrilateral is inscribed in the semicircle with diameter . Segments and meet at . Point is the foot of the perpendicular from to line . Point lies on such that line is perpendicular to line . Let be the intersection of segments and . Prove that

## Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.