Find the ratio of the areas of these two regions

Since $\triangle ABC$ is equilateral, the center of its circumcircle is also the centroid of $\triangle ABC$ . Due to symmetry, $BR$ is the diameter of the circumcircle. Let the radius of the circumcircle be $1$ . Then diameter $BR=2$ . Also let the center and radius of the green circle be $O$ and $r$ respectively. We note that $\triangle BPO$ is a right triangle with $\angle BPO = 90^\circ$ and $\angle PBO = 30^\circ$ . Then we have:

$\begin{aligned} BO+OR & = BR \\ \frac r{\sin 30^\circ} + r & = 2 \\ 2r+r & = 2 \\ \implies r & = \frac 23 \end{aligned}$

Then the ratio of the areas of the orange crescent and the green circle is $\dfrac {\pi (1)^2 - \pi r^2}{\pi r^2} = \dfrac 1{r^2} - 1 = \dfrac 94-1 = \dfrac 54$ . Therefore, $a+b=5+4 = \boxed 9$ .