jmc

Let $M$ be the intersection of $\overline{AH}$ and $\overline{BI}$ and $N$ be the intersection of $\overline{BI}$ and $\overline{CJ}$. Let $O$ be the center. Let $BC=2$ (without loss of generality). Note that $\angle BMH$ is the vertical angle to an angle of regular hexagon, and so has degree $120^{\circ }$. Because $△ABH$ and $△BCI$ are rotational images of one another, we get that $\angle MBH=\angle HAB$ and hence $△ABH\sim △BMH\sim △BCI$. Using a similar argument, $NI=MH$, and $$MN=BI-NI-BM=BI-(BM+MH).$$ Applying the Law of cosines on $△BCI$, $BI=\sqrt{2^2+1^2-2(2)(1)(\cos (120^{\circ }))}=\sqrt{7}$ $$\begin{array}{rl}\frac{BC+CI}{BI}& =\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH}\\ BM+MH& =\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}\\ MN& =BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}\\ \frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}& ={\left(\frac{MN}{BC}\right)}^2={\left(\frac{2}{\sqrt{7}}\right)}^2=\frac{4}{7}\end{array}$$ Thus, the answer is $4+7=\boxed{11}$.