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's call the perpendicular bisector of PQ as line l. Since MY is 8, we know that PY = PM = 36 / 2 = 18.
Since PX bisects angle QPR, we have QPR = PXR / 2. By the Law of Cosines on triangle PQR, we have:
cos(QPR) = (PQ^2 + QR^2 - PR^2) / (2 * PQ * QR)
Substituting the values for PQ and PR, we find that cos(QPR) = 0.95. Hence, QPR = cos^-1(0.95) = 11.75 degrees. So PXR = 2 * 11.75 = 23.5 degrees.
Let PX = a. Since PX bisects angle QPR, we have QRX = QPR / 2 = 11.75 degrees. By the Law of Sines on triangle PXR, we have: sin(QRX) / PX = sin(QPR) / PR
Hence, a / 18 = sin(11.75) / 22
Solving for a, we find that a = 19.5.
Finally, by the Pythagorean Theorem on right triangle PYR, we have:
RY^2 = PX^2 - MY^2 = (19.5)^2 - 8^2 = 289
Hence, RY = sqrt(289) = 17. So the area of triangle PYR = (1/2) * PY * RY = (1/2) * 18 * 17 = 153.
