# Difference between revisions of "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

Using the Law of Sines and simplifying, we have $$\frac{BY}{XP}+\frac{CY}{XQ}=\frac{\sin \angle PXY \sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX}.$$

It is easy to see that $APZX$ is cyclic. Also, we are given $XQ\perp AZ$. Then we have \begin{align*} \frac{\sin \angle PXY\sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX} &= \frac{\sin \angle YAZ\sin \angle XZA+\cos \angle XZA\cos \angle YAZ}{\sin \angle AYX} \\ &= \frac{\cos(\angle XZA-\angle YAZ)}{\sin \angle AYX} \\ &= \frac{\cos \angle AYX}{\sin \angle AYX} \\ &= \cot AYX \\ &= \frac{AY}{AX}, \end{align*} and we are done.

## 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