2013 USAJMO Problems/Problem 5

Revision as of 21:57, 31 August 2018 by Nivek (talk | contribs) (Solution 3)


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 \[\frac{\tan \alpha}{1-\tan (\alpha+\beta)\tan {\beta}}+\frac{\tan (\alpha+\beta)}{1-\tan (\alpha+\beta)\tan {\beta}}=\tan (2\alpha +\beta)=\frac{AY}{AX}.\]

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