Difference between revisions of "2013 USAJMO Problems/Problem 5"
Countingkg (talk | contribs) |
|||
Line 1: | Line 1: | ||
+ | ==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== |
Revision as of 09:26, 12 May 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
.