Answer: 56

Small Hint:

Place $D$ at the origin and $E$ on the positive $x$-axis; then $R = (\cos 60^\circ, \sin 60^\circ)$ and the law of sines gives $DE = \frac{\sqrt{3}+1}{2}$

Big Hint:

Perpendicular slopes locate $C$ on the $x$-axis, and $CA = AR$ is a linear equation in the $x$-coordinate of $A$

Solution:

Since $\angle RDE = 180^\circ - 75^\circ - 45^\circ$ $= 60^\circ,$ place $D = (0,0)$ with $E$ on the positive $x$-axis, so $R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The law of sines gives $DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2},$ and $M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).$

The slope of $EM$ is $\frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}},$ so line $RC$ has slope $\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Descending from $R$ by $\frac{\sqrt{3}}{2}$ to the $x$-axis moves us left by $\frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22},$ so $C = (c, 0)$ with $c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.$

For $A = (t, 0),$ the condition $CA = AR$ reads $(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4},$ which is linear in $t:$ $t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}.$ Then $$\begin{aligned} AE &= t - \frac{\sqrt{3}+1}{2} \\ &= \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} \\ &= \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, \end{aligned}$$ so $a + b + c = 7 + 27 + 22 = 56.$