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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>
 +
This simplifies into
 +
<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 21:53, 31 August 2018

Problem

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

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

Solution 2

First of all

\[\angle BXY = \angle PAZ =\angle AXQ =\angle AXC\] since the quadrilateral $APZX$ is cyclic, and triangle $AXQ$ is rectangle, and $CX$ is orthogonal to $AZ$. Now

\[\angle BXY =\angle BAY =\angle AXC\] because $XABY$ is cyclic and we have proved that

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

Solution 3

It is obvious that \[\angle AXB=\angle CXY=\alpha\] for some value $\alpha$. Also, note that $\angle BYA=\alpha$. Set \[\angle BXC=\angle BYC=\beta.\] We have \[\frac{XC}{CY}=\tan {\angle CYZ}=\tan {90-\alpha}\] and \[\frac{CQ}{CY}=\tan {\angle CYQ}=\tan {\alpha+\beta}.\] This gives \[\frac{CY}{XQ}=\frac{1}{\tan (90-\alpha)-\tan (\alpha+\beta)}.\] Similarly, we can deduce that \[\frac{BY}{XP}=\frac{1}{\tan (90-\alpha-\beta)-\tan {\alpha}}.\] This simplifies into \[\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.\]


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

Solution 3

It is obvious that \[\angle AXB=\angle CXY=\alpha\] for some value $\alpha$. Also, note that $\angle BYA=\alpha$. Set \[\angle BXC=\angle BYC=\beta.\] We have \[\frac{XC}{CY}=\tan {\angle CYZ}=\tan {90-\alpha}\] and \[\frac{CQ}{CY}=\tan {\angle CYQ}=\tan {\alpha+\beta}.\] This gives 

\[\frac{CY/XQ}=\frac{1}{\tan {90-\alpha}-\tan {\alpha+\beta}.\] (Error compiling LaTeX. Unknown error_msg)

Similarly, we can deduce that 

\[\frac{BY/XP}=\frac{1}{\tan {90-\alpha-\beta}-\tan {\alpha}.\] (Error compiling LaTeX. Unknown error_msg)

This simplifies into \[\frac{\tan \alpha}{1-\tan {\alpha+\beta}\tan {\beta}}+\frac{\tan {\alpha+\beta}}{1-\tan {\alpha+\beta}\tan {\beta}}=\tan {2\alpha +\beta}=\frac{AY}{AX}.\]


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

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