jmc

Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel \overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel \overline{CD}$. Thus, $APME$ is a parallelogram and $AE=PM=\frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO=\frac{1}{2}$ and $EO=RO=\frac{\sqrt{3}}{2}$, so $D\left(\frac{1}{2},0\right)$, $E\left(-\frac{\sqrt{3}}{2},0\right)$, and $R\left(0,\frac{\sqrt{3}}{2}\right).$ $M=$ midpoint$(D,R)=\left(\frac{1}{4},\frac{\sqrt{3}}{4}\right)$ and the slope of $ME=\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4}+\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{1+2\sqrt{3}}$, so the slope of $RC=-\frac{1+2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO=x$ to the left, we go $\frac{x(1+2\sqrt{3})}{\sqrt{3}}=\frac{\sqrt{3}}{2}$ up. Thus, $x=\frac{\frac{3}{2}}{1+2\sqrt{3}}=\frac{3}{4\sqrt{3}+2}=\frac{3(4\sqrt{3}-2)}{44}=\frac{6\sqrt{3}-3}{22}.$ $DC=\frac{1}{2}-x=\frac{1}{2}-\frac{6\sqrt{3}-3}{22}=\frac{14-6\sqrt{3}}{22}$, and $AE=\frac{7-\sqrt{27}}{22}$, so the answer is $\boxed{56}$.