Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
In this solution, all angle measures are in degrees.
Let $M$ be the midpoint of $\overline{BC}$ so that $\overline{OM}\perp\overline{BC}$ and $A,G,M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$
Note that:


Since $\angle OGX = \angle OAX = 90,$ quadrilateral $OGAX$ is cyclic by the Converse of the Inscribed Angle Theorem.It follows that $\angle OAG = \angle OXG,$ as they share the same intercepted arc $\widehat{OG}.$
Since $\angle OGY = \angle OMY = 90,$ quadrilateral $OGYM$ is cyclic by the supplementary opposite angles.It follows that $\angle OMG = \angle OYG,$ as they share the same intercepted arc $\widehat{OG}.$

Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \angle XOY = 17k.$
Next, we express $\angle BAC$ in terms of $k.$ By angle addition, we have
\begin{align*} \angle AOM &= \angle AOB + \angle BOM \\ &= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\ &= 2\angle BCA + \angle BAC. &&\text{by Inscribed Angle Theorem} \end{align*}
Substituting back gives $17k=2(2k)+\angle BAC,$ from which $\angle BAC=13k.$
For the sum of the interior angles of $\triangle ABC,$ we get
\begin{align*} \angle ABC + \angle BCA + \angle BAC &= 180 \\ 13k+2k+13k&=180 \\ 28k&=180 \\ k&=\frac{45}{7}. \end{align*}
Finally, we obtain $\angle BAC=13k=\frac{585}{7},$ from which the answer is $585+7=\boxed{592}.$
~Constance-variance ~MRENTHUSIASM