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

m (Solution 1)
(Solution 2)
Line 7: Line 7:
  
 
==Solution 2==
 
==Solution 2==
First <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.
+
First of all
 +
 
 +
<cmath>\angle BXY = \angle PAZ =\angle AXQ =\angle AXC</cmath> 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  
 +
 
 +
<cmath>\angle BXY =\angle BAY =\angle AXC</cmath> because <math>XABY</math> is cyclic and we have proved that  
 +
 
 +
<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>
  
  
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:49, 25 April 2015

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 \[\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}\] $\blacksquare$


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

Invalid username
Login to AoPS