USAMO 2013/1

[tanmoy]

In triangle $ABC$,points $P,Q,R$ lie on sides $BC,CA,AB$ respectively.Let $w_{A},w_{B},w_{C}$ denote the circumcircles of triangles $AQR,BRP,CPQ$ respectively.Given the fact that segment $AP$ intersects $w_{A},w_{B},w_{C}$ again at $X,Y,Z$ respectively,prove that $\frac{YX} {XZ}=\frac{BP} {PC}$.

[Tahmid]

let, $M=w_{A}\cap w_{B}\cap w_{C}$

$\frac{XY}{XZ}=\frac{sin\angle YMX \cdot YM}{sin\angle ZMX\cdot ZM}$

or, $\frac{XY}{XZ}=\frac{sin\angle YMX \cdot sin\angle YZM}{sin\angle ZMX\cdot sin\angle ZYM}$

now , $\angle YMX=180-\angle MYX-\angle MXY=\angle MRB-\angle MRP=\angle PRB=\angle PMB$
and let $XM\cap w_{C}=D$
so , $\angle ZMX=180-\angle MZX-\angle MXZ=\angle XDP+\angle XPD-XDP=\angle XPD$
as $AP\parallel DC$ so, $\angle XPD=\angle CDP=\angle CMP$

$\therefore \frac{sin\angle YMX}{sin\angle ZYM}=\frac{sin\angle BMP}{sin\angle MBP}=\frac{BP}{PM}$

and , $\frac{sin\angle YZM}{sin\angle ZMX}=\frac{sin\angle MCP}{sin\angle CMP}=\frac{MP}{PC}$

multiplying previous two results we have ,
$\frac{sin\angle YMX \cdot sin\angle YZM}{sin\angle ZMX\cdot sin\angle ZYM}=\frac{BP}{PC}$

or, $\frac{XY}{XZ}=\frac{BP}{PC}$

[tanmoy]

$\frac{XY}{XZ}=\frac{sin\angle YMX \cdot YM}{sin\angle ZMX\cdot ZM}$
or, $\frac{XY}{XZ}=\frac{sin\angle YMX \cdot sin\angle YZM}{sin\angle ZMX\cdot sin\angle ZYM}$

now , $\angle YMX=180-\angle MYX-\angle MXY=\angle MRB-\angle MRP=\angle PRB=\angle PMB$
and let $XM\cap w_{C}=D$
so , $\angle ZMX=180-\angle MZX-\angle MXZ=\angle XDP+\angle XPD-XDP=\angle XPD$
as $AP\parallel DC$ so, $\angle XPD=\angle CDP=\angle CMP$

$\therefore \frac{sin\angle YMX}{sin\angle ZYM}=\frac{sin\angle BMP}{sin\angle MBP}=\frac{BP}{PM}$
and , $\frac{sin\angle YZM}{sin\angle ZMX}=\frac{sin\angle MCP}{sin\angle CMP}=\frac{MP}{PC}$

multiplying previous two results we have ,
$\frac{sin\angle YMX \cdot sin\angle YZM}{sin\angle ZMX\cdot sin\angle ZYM}=\frac{BP}{PC}$
or, $\frac{XY}{XZ}=\frac{BP}{PC}$

You did not state what is the point $M$ is.BTW,here is my proof:
The circumcircles of triangles $AQR,BRP,CPQ$ must meet at a point($\text{Miquel's Theorem}$),say $M$.
Let $XM$ meet $\omega_B, \omega_C$ again at $S$ and $T$, respectively. Then by Power of a Point, we have $XM \cdot XT = XZ \cdot XP \quad\text{and}\quad SX \cdot XM = YX \cdot XP$.$\therefore \frac{YX}{XZ} = \frac{SX}{XT}$.Now,$\angle XSP = \angle MSP = \angle MBP$ and $\angle SXP = \angle MXY = \angle MXA = \angle MRA =\angle MPB$.$\therefore \triangle XSP \sim \triangle PBM$.$\therefore \frac{SX}{XP} = \frac{PB}{PM}$.In the same way we find that $\frac{XT}{XP} = \frac{PC}{PM}$.$\therefore \frac{SX} {XT}=\frac{BP} {PC}$.Or,$\frac{YX}{XZ}=\frac{BP}{PC}$.

[Tahmid]

You did not state what is the point M is

oppsss sorry . didn't notice . now it is edited