+0

# Geometry Question

0
24
1

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 $PYR.$

Apr 3, 2023

#1
0

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$, 18sin⁡�=36sin⁡(180∘−2�)=36sin⁡2�sinθ18​=sin(180∘−2θ)36​=sin2θ36​ Solving for$\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})

Apr 3, 2023