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.