Let $M$ be the midpoint of $\overline{BC},$ so $A,$ $G,$ $M$ are collinear along the median, while $X,$ $G,$ $Y$ are collinear by definition. Since $OA \perp AX$ (tangent and radius) and $OG \perp GX,$ quadrilateral $OAXG$ is cyclic with diameter $\overline{OX}.$ Since $OG \perp GY$ and $OM \perp MY$ (the segment from the center to the midpoint of a chord is perpendicular to it), quadrilateral $OGYM$ is cyclic with diameter $\overline{OY}.$

In each circle the chord $\overline{OG}$ subtends equal angles, so $\angle OXY = \angle OXG$ $= \angle OAG = \angle OAM$ and $\angle OYX = \angle OYG$ $= \angle OMG = \angle OMA.$ Triangles $OXY$ and $OAM$ therefore have the same angle sums at their bases, giving $$\begin{aligned} \angle XOY &= 180^\circ - \angle OXY \\ &\quad {}- \angle OYX \\ &= 180^\circ - \angle OAM \\ &\quad {}- \angle OMA \\ &= \angle AOM. \end{aligned}$$

Write $\angle ABC = 13k$ and $\angle BCA = 2k,$ so $\angle BAC = 180^\circ - 15k.$ Central angles give $\angle AOB = 2\angle BCA = 4k,$ and $\overline{OM}$ bisects $\angle BOC = 2\angle BAC,$ so on the side of $B$ (nearer to $A$'s arc since $\angle ABC \gt \angle BCA$), $$\begin{aligned} \angle AOM &= \angle AOB + \angle BOM \\ &= 4k + (180^\circ - 15k) \\ &= 180^\circ - 11k. \end{aligned}$$ Setting $180^\circ - 11k = \angle XOY = 17k$ gives $k = \frac{45}{7},$ so $\angle BAC = 180^\circ - 15 \cdot \frac{45}{7} = \frac{585}{7}$ degrees, and all three angles are acute as required. Then $m + n = 585 + 7 = 592.$