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

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==Solution 1== | ==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 <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>(\frac{u-v}{u+v}, \frac{2uv}{u+v})</math>. Also, <math>Z(\frac{u-v}{u+v}, 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>. | + | 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 <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>\left(\frac{u-v}{u+v}\right), \frac{2uv}{u+v})</math>. Also, <math>Z\left(\frac{u-v}{u+v}\right), 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>. |

==Solution 2== | ==Solution 2== |

## Revision as of 20:48, 25 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 , 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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