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

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Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>.  Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>.  Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>.  Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>.  Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>.  Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath>
 
Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>.  Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>.  Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>.  Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>.  Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>.  Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath>
  
==Solution==
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==Solution 1==
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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>.
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==Solution 2==
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First of all
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<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
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<cmath>\angle BXY =\angle BAY =\angle AXC</cmath> because <math>XABY</math> is cyclic and we have proved that
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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>
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<math>\blacksquare</math>
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==Solution 3==
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It is obvious that
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<cmath>\angle AXB=\angle CXY=\alpha</cmath>
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for some value <math>\alpha</math>. Also, note that <math>\angle BYA=\alpha</math>. Set
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<cmath>\angle BXC=\angle BYC=\beta.</cmath>
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We have
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<cmath>\frac{XC}{CY}=\tan {\angle CYZ}=\tan (90-\alpha)</cmath>
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and
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<cmath>\frac{CQ}{CY}=\tan {\angle CYQ}=\tan (\alpha+\beta).</cmath>
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This gives
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<cmath>\frac{CY}{XQ}=\frac{1}{\tan (90-\alpha)-\tan (\alpha+\beta)}.</cmath>
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Similarly, we can deduce that
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<cmath>\frac{BY}{XP}=\frac{1}{\tan (90-\alpha-\beta)-\tan {\alpha}}.</cmath>
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Adding gives
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<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>
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==Solution 4==
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First, since <math>XY</math> is the diameter and <math>A</math>, <math>B</math>, and <math>C</math> lie on the circle, <cmath>\angle {XAY} = \angle {XBY} = \angle{XCY} = 90</cmath>. Next, because <math>AZ</math> and <math>CY</math> are both perpendicular to <math>CX</math>, we have <math>AZ</math> to be parallel to <math>CY</math>.
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Now looking at quadrilateral <math>APZX</math>, we see that this is cyclic because <cmath>\angle {PAX} + \angle {PZX} = 90+90 = 180.</cmath> Set <math>\alpha = \angle{BXA} = \angle{BYA}</math>, and <math>\beta = \angle{BXC} = \angle{BYC}</math>.
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Now, <cmath>\angle{AYC} = \angle{YAZ}</cmath> since <math>AZ</math> and <math>CY</math> are parallel.
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Also, <cmath>\angle{PAZ} = \angle{PXZ} = \alpha + \beta.</cmath>
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That means <cmath>\angle{PXZ} = \angle{PXQ} + \angle{QXZ} = \beta + \angle{QXZ} = \alpha + \beta</cmath> so <cmath>\angle{QXZ} = \alpha.</cmath> This means <math>\angle{QXZ} = \angle{YBC} = \alpha</math>, so <math>BC</math> and <math>AY</math> are parallel.
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Finally, we can look at the equation.
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We know <cmath>XP\cos{\alpha} = AX,</cmath> so <math>XP = \frac{AX}{\cos{\alpha}}.</math>
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We also know <cmath>XQ\cos(\alpha+\beta) = AX,</cmath> so <math>XQ = \frac{AX}{\cos(\alpha+\beta)}.</math>
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Plugging this into the LHS of the equation, we get <cmath>\frac{BY\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}.</cmath>
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Now, let <math>H</math> be the point on <math>AY</math> such that <math>BH</math> is perpendicular to <math>AY</math>. Also, since <math>\angle{AYB} = \angle{CXY}</math>, their arcs have equal length, and <math>AB=CY</math>.
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Now, the LHS is simplified even more to <cmath>\frac{AB\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}</cmath> which is equal to <cmath>\frac{AH+YH}{AZ}</cmath> which is equal to <cmath>\frac{AY}{AX}.</cmath> This completes the proof.
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~jeteagle
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Latest revision as of 12:36, 24 July 2019

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 \[\frac{BY}{XP}=\frac{1}{\tan (90-\alpha-\beta)-\tan {\alpha}}.\] Adding gives \[\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.\]


Solution 4

First, since $XY$ is the diameter and $A$, $B$, and $C$ lie on the circle, \[\angle {XAY} = \angle {XBY} = \angle{XCY} = 90\]. Next, because $AZ$ and $CY$ are both perpendicular to $CX$, we have $AZ$ to be parallel to $CY$.

Now looking at quadrilateral $APZX$, we see that this is cyclic because \[\angle {PAX} + \angle {PZX} = 90+90 = 180.\] Set $\alpha = \angle{BXA} = \angle{BYA}$, and $\beta = \angle{BXC} = \angle{BYC}$. Now, \[\angle{AYC} = \angle{YAZ}\] since $AZ$ and $CY$ are parallel. Also, \[\angle{PAZ} = \angle{PXZ} = \alpha + \beta.\] That means \[\angle{PXZ} = \angle{PXQ} + \angle{QXZ} = \beta + \angle{QXZ} = \alpha + \beta\] so \[\angle{QXZ} = \alpha.\] This means $\angle{QXZ} = \angle{YBC} = \alpha$, so $BC$ and $AY$ are parallel. Finally, we can look at the equation. We know \[XP\cos{\alpha} = AX,\] so $XP = \frac{AX}{\cos{\alpha}}.$ We also know \[XQ\cos(\alpha+\beta) = AX,\] so $XQ = \frac{AX}{\cos(\alpha+\beta)}.$ Plugging this into the LHS of the equation, we get \[\frac{BY\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}.\] Now, let $H$ be the point on $AY$ such that $BH$ is perpendicular to $AY$. Also, since $\angle{AYB} = \angle{CXY}$, their arcs have equal length, and $AB=CY$. Now, the LHS is simplified even more to \[\frac{AB\cos\alpha}{AX}+\frac{CY\cos(\alpha+\beta)}{AZ}\] which is equal to \[\frac{AH+YH}{AZ}\] which is equal to \[\frac{AY}{AX}.\] This completes the proof.

~jeteagle


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