Find The Angle

Note that $BCDE$ is a cyclic quadrilateral centered at $M,$ so $MB=MC=MD=ME.$ Let $H$ be the orthocenter of $\triangle ABC.$ Then, $AEHD$ is a cyclic quadrilateral centered at the midpoint of $AH.$ Now, $\angle MDH = \angle MDB = \angle DBM = 90^{\circ} - \angle BCA = \angle AHD,$ so $ME$ is tangent to the circumcircle of $\triangle AHD,$ which is also the circumcircle of $\triangle ADE.$ Similarly $MD$ is also tangent to the circumcircle of $\triangle AED$ (which shall be denoted as $(AED)$ throughout the rest of the solution).

We shall consider $M$ as a degenerate circle with radius zero. Any line passing through $M$ can be considered tangent to $M.$ Since $QE=EM$ and $PD=PM,$ $Q,P$ both lie on the radical axis of $M$ and $(AED),$ which implies $PQ$ is the radical axis of $M$ and $(AED).$

Now, note that $\angle XAC = \angle ACB = 180^{\circ} - \angle DEB = \angle AED,$ so $XA$ is tangent to $(AED).$ Since $X$ lies on the radical axis of $M$ and $(AED),$ $XA=XM,$ so $\angle XAM = \angle XMA.$

Then, $\angle XMA = \angle XAC + \angle MAC = \angle ACM + \angle MAC = 180^{\circ} - \angle AMC = \angle BMA,$ and $\angle CMX = 180^{\circ} - \angle BMX = 180^{\circ} - 2 \angle BMA.$

By Appolonius theorem on $\triangle ABC$ and median $AM,$ ,
$2(AM^2 + 5^2) = 12^2 + 8^2 \implies AM = \sqrt{79}.$ Then, by law of cosines in $\triangle ABM,$ $\cos \angle BMA = \dfrac{79 + 5^2 - 8^2 }{2 \cdot 5 \cdot \sqrt{79}} = \dfrac{4}{\sqrt {79}}.$ Finally, $\cos \angle CMX = - \cos \angle BMA = - (2 \cos ^2 \angle BMA - 1) = \dfrac{47}{79}.$ Hence, $x=47, y= 79,$ and $x+y-31= \boxed{95}.$