Geometry

Since $M$ is the midpoint of $PQ$, we have $PM = MQ = 18$. By the Angle Bisector Theorem, we have [\frac{PX}{QX} = \frac{PR}{QR} = \frac{22}{26} = \frac{11}{13}.] Therefore, $PX = \frac{11}{24}(QR) = \frac{11}{24}(26) = \frac{143}{12}$ and $QX = \frac{13}{24}(QR) = \frac{13}{24}(26) = \frac{169}{12}$. Since $MY$ is the perpendicular bisector of $PQ$, we have $PY = QY = \frac{1}{2}(PQ) = 18$. Then, since $AY$ is also the perpendicular bisector of $PQ$, we have $AP = AQ$, so [AX = AP + PX = AQ - QX + PX = (AP + PX) - QX = AX - QX.] Solving for $AX$, we find $AX = \frac{QX + PX}{2} = \frac{143}{24}$. Now, let $PR = a$, $PQ = b$, and $QR = c$. By Heron's formula, the area of triangle $PQR$ is [\sqrt{s(s-a)(s-b)(s-c)},] where $s = \frac{a+b+c}{2} = \frac{22+36+26}{2} = 42$ is the semiperimeter of the triangle. To find $c$, note that $c^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos \angle PQR$, so [c^2 = 36^2 + 22^2 - 2 \cdot 36 \cdot 22 \cdot \frac{6}{26} = \frac{4608}{13}.] Similarly, we have [b^2 = QR^2 + PQ^2 - 2 \cdot QR \cdot PQ \cdot \cos \angle QRP = 26^2 + 36^2 - 2 \cdot 26 \cdot 36 \cdot \frac{22}{26} = 1240] and [a^2 = PR^2 + QR^2 - 2 \cdot PR \cdot QR \cdot \cos \angle RPQ = 22^2 + 26^2 - 2 \cdot 22 \cdot 26 \cdot \frac{18}{26} = 400.] Thus, the area of triangle $PQR$ is \begin{align*} \sqrt{s(s-a)(s-b)(s-c)} &= \sqrt{42 \cdot 20 \cdot (42-22)(42-36)(42-26)} \ &= \sqrt{42 \cdot 20 \cdot 16 \cdot 8} \ &= 96\sqrt{35}. \end{align*} Finally, we need to find the length of $AY$. Since $AY$ is the perpendicular bisector of $PQ$, we have $PY = QY = 18$. Then, by the Pythagorean Theorem, we have [AY^2 = AP^2 + PY^2 = AQ^2 + QY^2 = 18^2 + \left(\frac{143}{24}\right)^2.] Therefore, [AY = \sqrt{18^2 + \left(\frac{143}{24}\right)^2} = \frac{169}{24}.] Finally, we use the fact that $MY = 8$ to find $AY - MY = AM$, so [AM = AY - MY = \frac{169}{24} - 8 = \frac{25}{24}.] Therefore, the area of triangle $PQR$ is $\boxed{96\sqrt{35}}$.

edit: the latex did not work lol