Areas, Circles and Squares!

$R(\dfrac{1 - a}{2}, \dfrac{a}{2})$ is on the circle $(x - 1)^2 + y^2 = 1 \implies 2a^2 + 2a - 3 = 0 \implies$

$a = \dfrac{\sqrt{7} - 1}{2}$ dropping the negative root.

$\implies$ the area of the square $A_{s} = (\dfrac{\sqrt{7} - 1}{2})^2 = \dfrac{4 - \sqrt{7}}{2}$

Solving $x^2 + y^2 = 1$ and $x^2 - 2x + 1 + y^2 = 1 \implies 2x - 1 = 0 \implies x = \dfrac{1}{2} \implies y = \pm\dfrac{\sqrt{3}}{2} \implies$

The point of intersection of the two circles is $(\dfrac{1}{2},\pm\dfrac{\sqrt{3}}{2}) \implies$ the base, height and $m\angle{SOT}$ of $\triangle{SOT}$ are

$\overline{ST} = \sqrt{3}, height = \dfrac{1}{2}$ and $m\angle{SOT} = \dfrac{2\pi}{3} \implies A_{\triangle{SOT}} = \dfrac{\sqrt{3}}{4}$

and

$A_{sector{SOT}} = \dfrac{1}{2}(\dfrac{2\pi}{3}) = \dfrac{\pi}{3} \implies$ The area of the region between the two circles is

$I = 2(A_{sector{SOT}} - A_{\triangle{SOT}}) = \dfrac{4\pi - 3\sqrt{3}}{6} \implies$ the desired area is $A_{d} = I - A_{s} =$

$\dfrac{4\pi - 3\sqrt{3} - 12 + 3\sqrt{7}}{6} =$ $\dfrac{4(\pi - 3) + 3(\sqrt{7} - \sqrt{3})}{6} = \dfrac{a(\pi - b) + b(\sqrt{c} - \sqrt{b})}{d}$

$\implies a + b + c + d = \boxed{20}$ .