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

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

<math>\blacksquare</math> | <math>\blacksquare</math> | ||

+ | |||

+ | ==Solution 3== | ||

+ | It is obvious that | ||

+ | <cmath>\angle AXB=\angle CXY=\alpha</cmath> | ||

+ | for some value <math>\alpha</math>. Also, note that <math>\angle BYA=\alpha</math>. Set | ||

+ | <cmath>\angle BXC=\angle BYC=\beta.</cmath> | ||

+ | We have | ||

+ | <cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath> | ||

+ | and | ||

+ | <cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath> | ||

+ | This gives | ||

+ | <cmath>\frac{CY}{XQ}=\frac{1}{\tan (90-\alpha)-\tan (\alpha+\beta)}.</cmath> | ||

+ | Similarly, we can deduce that | ||

+ | <cmath>\frac{BY}{XP}=\frac{1}{\tan (90-\alpha-\beta)-\tan {\alpha}}.</cmath> | ||

+ | Adding gives | ||

+ | <cmath>\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.</cmath> | ||

+ | |||

+ | |||

+ | |||

+ | {{MAA Notice}} | ||

==Solution 3== | ==Solution 3== |

## Revision as of 21:55, 31 August 2018

## 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 is already proven, so Substituting yields 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 In the beginning we prove that and so

## Solution 3

It is obvious that for some value . Also, note that . Set We have and This gives Similarly, we can deduce that Adding gives

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

## Solution 3

It is obvious that for some value . Also, note that . Set We have and This gives Similarly, we can deduce that This simplifies into

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

## Solution 3

It is obvious that for some value . Also, note that . Set We have and This gives Similarly, we can deduce that This simplifies into

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

## Solution 3

It is obvious that for some value . Also, note that . Set We have and This gives

\[\frac{CY/XQ}=\frac{1}{\tan {90-\alpha}-\tan {\alpha+\beta}.\] (Error compiling LaTeX. ! File ended while scanning use of \frac .)

Similarly, we can deduce that

\[\frac{BY/XP}=\frac{1}{\tan {90-\alpha-\beta}-\tan {\alpha}.\] (Error compiling LaTeX. ! File ended while scanning use of \frac .)

This simplifies into