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

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− | + | First of all <math> \angle BXY = \angle PAZ =\angle AXQ =\angle AXC</math>, since the quadrilateral <math>APZX</math> is cyclic, and triangle <math>AXQ</math> is rectangle, and <math>CX</math> is orthogonal to <math>AZ</math>. Now <math>\angle BXY =\angle BAY =\angle AXC</math> because <math>XABY</math> is cyclic and we have proved that <math>\angle AXC = \angle BXY</math>, so <math>BC</math> is parallel to <math>AY</math>, and <math>AC=BY</math>, <math>CY=AB</math>. Now by Ptolomey's theorem on <math>APZX</math>, we have <math>(AX)(PZ)+(AP)(XZ)=(AZ)(PX)</math>, we see that triangles <math>PXZ</math> and <math>QXA</math> are similar since <math>\angle QAX= \angle PZX= 90</math> and <math>\angle AXC = \angle BXY</math>, already proven, so <math>(AX)(PZ)=(AQ)(XZ)</math>, substituting we get <math>(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)</math>, dividing by <math>(PX)(XZ)</math>, we get <math>\frac {AQ+AP}{XP} = \frac {AZ}{XZ}</math>. Now triangles <math>AYZ</math>, and <math>XYP</math> are similar so <math>\frac {AY}{AZ}= \frac {XY}{XP}</math>, but also triangles <math>XPY</math> and <math>XZB</math> are similar and we get <math>\frac {XY}{XP}= \frac {XB}{XZ}</math>, comparing we have, <math>\frac {AY}{XB}= \frac {AZ}{XZ}</math> substituting, <math>\frac {AQ+AP}{XP}= \frac {AY}{XB}</math>. Dividing the new relation by <math>AX</math> and multiplying by <math>XB</math> we get <math>\frac{XB(AQ+AP)}{(XP)(AX)} = \frac {AY}{AX}</math>, but <math>\frac {XB}{AX}= \frac {XY}{XQ}</math>, since triangles <math>AXB</math> and <math>QXY</math> are similar, because <math>\angle AYX= \angle ABX</math> and <math>\angle AXB= \angle CXY</math> since <math>CY=AB</math>. Substituting again we get <math>\frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =\frac {AY}{AX}</math>. Now since triangles <math>ACQ</math> and <math>XYQ</math> are similar we have <math>XY(AQ)=AC(XQ)</math> and by the similarity of <math>APB</math> and <math>XPY</math>, we get <math>AB(CP)=XY(AP)</math> so substituting, and separating terms we get <math>\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}</math>, but in the beginning we prove that <math>AC=BY</math> and <math>AB=CY</math> so <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>, and we are done. | |

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## Revision as of 16:21, 2 April 2015

## 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 1

Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants and set A and B . Now, let's use our coordinate tools. It is easily derived that the equation of is and the equation of is , where and are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, , is . Also, . It shall be left to the reader to find the slope of , the coordinates of Q and C, and use the distance formula to verify that .

## Solution 2

First of all , since the quadrilateral is cyclic, and triangle is rectangle, and is orthogonal to . Now because is cyclic and we have proved that , so is parallel to , and , . Now by Ptolomey's theorem on , we have , we see that triangles and are similar since and , already proven, so , substituting we get , dividing by , we get . Now triangles , and are similar so , but also triangles and are similar and we get , comparing we have, substituting, . Dividing the new relation by and multiplying by we get , but , since triangles and are similar, because and since . Substituting again we get . Now since triangles and are similar we have and by the similarity of and , we get so substituting, and separating terms we get , but in the beginning we prove that and so , and we are done.

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