In triangle $PQR,$ $M$ is the midpoint of $\overline{PQ}.$ Let $X$ be the point on $\overline{QR}$ such that $\overline{PX}$ bisects $\angle QPR,$ and let the perpendicular bisector of $\overline{PQ}$ intersect $\overline{PX}$ at $Y.$ If $PQ = 36,$ $PR = 22,$ and $MY = 8,$ then find the area of triangle $PYR.$
Let $PX$ meet $RY$ at $Z$. Since $RY\parallel PQ$ and $MY$ is the perpendicular bisector of $PQ$, we have $MY=ZY$. Also, $PY=QY$ since $PQ$ is perpendicular to $RY$ and $PYQ$ is isosceles. Thus, $PY=QY=18$.
Let $\angle PQY=\theta$. Since $PX$ bisects $\angle QPR$, we have $\angle PQX=\angle PRX=\frac{1}{2}(180^\circ-\theta)$. Also, $\angle QPY=\angle QMY=\frac{1}{2}\theta$, so $\angle QPX=\frac{1}{2}(180^\circ-\theta)-\frac{1}{2}\theta=90^\circ-\frac{1}{2}\theta$. Therefore, $\angle PZY=180^\circ-\angle QPX=90^\circ+\frac{1}{2}\theta$.
Let $PZ=x$. Then, $QZ=36-x$. By the Law of Cosines in $\triangle PZY$,
182=�2+64−16�cos�182=(36−�)2+64−16(36−�)cos�182182=x2+64−16xcosθ=(36−x)2+64−16(36−x)cosθ
Simplifying the second equation gives $\cos\theta=\frac{5}{8}-\frac{x}{32}$.
Substituting this into the first equation and simplifying gives $8x^2-32x-23=0$, which has solutions $x=\frac{1}{4}(8+\sqrt{39})$ and $x=\frac{1}{4}(8-\sqrt{39})$. Since $x
Now, we have $PZ=\frac{1}{4}(8+\sqrt{39})$, $QZ=\frac{1}{4}(24-\sqrt{39})$, $PY=QY=18$, and $\angle PQY=\theta$. By the Law of Sines in $\triangle PQY$,
18sin�=36sin(180∘−2�)=36sin2�sinθ18=sin(180∘−2θ)36=sin2θ36
Solving for $\sin\theta$ gives $\sin\theta=\frac{1}{2}\sin 2\theta=\frac{1}{2}\cdot\frac{2PY\cdot QY}{PQ\cdot QZ}=\frac{9}{(24-\sqrt{39})\sqrt{39}}$.
Finally, the area of $\triangle PYR$ is \frac{1}{2}\cdot PY\cdot PR\cdot\sin\theta=18\cdot 22\cdot\frac{9}{(24-\sqrt{39})\sqrt{39}}=\frac{594}{8+\sqrt{39}}}
=\frac{594}{25}*(8 - \sqrt{39})
