# 2013 USAJMO Problems/Problem 5

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