Centroid

$\dfrac{AP}{AB}=\dfrac{1}{1+2}=\dfrac 1 3~~\implies~AP=\dfrac 1 3 *AB .\\\dfrac{AG}{AP}=\dfrac 2 3 * \dfrac{\sqrt3} 2 =\dfrac 1 {\sqrt 3}.~~\implies AG=\dfrac 1 {\sqrt 3}*AP= \dfrac 1 {\sqrt 3}*AB*\dfrac 1 3. \\AR=\dfrac{\sqrt3} 2 AB.\\GR=AR-AG=(\dfrac {\sqrt3} 2 -\dfrac 1 {3*\sqrt3})*AB=\dfrac{7*\sqrt3}{18}*AB.\\\therefore~\dfrac{AG}{GR}=\dfrac {18}{7*\sqrt3}* \dfrac 1 {3*\sqrt3} = \dfrac 2 7 =\dfrac x y .\\x+y =~~~\large\color{#D61F06}{9}$

Image given above is not as per the question asked

This is clearly not required for this problem, but since I was learning barycentric coordinates, I decided to test it out on this problem.

Let $A=(1,0,0), B=(0,1,0), C=(0,0,1)$ . We then have $P=(\frac{2}{3},\frac{1}{2},0)$ and $Q=(\frac{2}{3},0,\frac{1}{3})$ .

$G$ is the centroid of $APQ$ , so it is the average of the coordinates of $A$ , $P$ , and $Q$ . Thus, $G=(\frac{7}{9},\frac{1}{9},\frac{1}{9})$ . R is the mid-point of BC, so $R=(0,\frac{1}{2},\frac{1}{2})$ .

$\overrightarrow{AG}=(-\frac{2}{9},\frac{1}{9},\frac{1}{9})$ and $\overrightarrow{GR}=(-\frac{7}{9},\frac{7}{18},\frac{7}{18})$

It is clear that $\frac{\overrightarrow{AG}}{\overrightarrow{GR}} = \frac{2}{7}$ .

And so our desired answer is $2+7=\boxed{9}$