2013 USAJMO Problems/Problem 5


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 \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]


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.


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