Problem: Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$

Solution: [asy] size(10cm); pair A, B, C, D, H, K, O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed);  label("$O$", O, ENE); label("$A$", A, NW); label("$B$", B, W); label("$C$", C, E); label("$H$", H, E); label("$H'$", K, NE); label("$X$", X, W); label("$Y$", Y, NE); label("$O'$", P, E); label("$M$", M, NE); label("$L$", L, NE); label("$D$", D, NNE);  label("$2$", X -- H, NW); label("$3$", H -- A, SW); label("$6$", H -- Y, NW); label("$R$", O -- Y, E);  dot(O); dot(P); dot(D); dot(H);  [/asy] Diagram not to scale.
We first observe that $H'$ , the image of the reflection of $H$ over line $BC$ , lies on circle $O$ .  This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$ .  This is a well known lemma.  The result of this observation is that circle $O'$ , the circumcircle of $\triangle BHC$ is the image of circle $O$ over line $BC$ , which in turn implies that $\overline{AH} = \overline{OO'}$ and thus $AHO'O$ is a parallelogram.  That $AHO'O$ is a parallelogram implies that $AO$ is perpendicular to $\overline{XY}$ , and thus divides segment $\overline{XY}$ in two equal pieces, $\overline{XD}$ and $\overline{DY}$ , of length $4$
Using Power of a Point, \[\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4\] This means that $\overline{HL} = \frac12 \cdot 4 = 2$ and $\overline{AL} = 2 + 3 = 5$ , where $L$ is the foot of the altitude from $A$ onto $BC$ .  All that remains to be found is the length of segment $\overline{BC}$
Looking at right triangle $\triangle AHD$ , we find that \[\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\] Looking at right triangle $\triangle ODY$ , we get the equation \[\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2\] Plugging in known values, and letting $R$ be the radius of the circle, we find that \[R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}\]
Recall that $AHO'O$ is a parallelogram, so $\overline{AH} = \overline{OO'} = 3$ .  So, $\overline{OM} = \frac32$ , where $M$ is the midpoint of $\overline{BC}$ .  This means that \[\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}\]
Thus, the area of triangle $\triangle ABC$ is \[\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}\] The answer is $3 + 55 = \boxed{058}$

Answer:  058