2013 USAJMO Problems/Problem 5

Revision as of 23:11, 23 April 2015 by CaptainFlint (talk | contribs) (Solution 2)


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 $(\frac{u-v}{u+v}, \frac{2uv}{u+v})$. Also, $Z(\frac{u-v}{u+v}, 0)$. 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 $\angle BXY = \angle PAZ =\angle AXQ =\angle AXC$, 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 $\angle AXC = \angle BXY$, already proven, so $(AX)(PZ)=(AQ)(XZ)$, substituting we get $(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)$, 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 $\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}$, but in the beginning we prove that $AC=BY$ and $AB=CY$ so $\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}$, and we are done.

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