Area Problem.

Let the height $AD=CE = h$ . Then the base of the triangle and rectangle $AC=DE = 2\sqrt{1-\left(\dfrac h2\right)^2} = \sqrt{4-h^2}$ . Since the area of $\triangle ABC$ and the area of rectangle $ACDE$ are equal, then

$\begin{aligned} [ABC] & = [ACDE] \\ \frac 12 \left(1-\frac h2\right)\sqrt{4-h^2} & = h\sqrt{4-h^2} \\ \frac 12 \left(1-\frac h2\right) & = h \\ 2 - h & = 4h \\ \implies h & = \frac 25 \end{aligned}$

Let the center of the circle be $O$ . Then the area of the pink regions is

$\begin{aligned} A & = \text{area of sector OABC} - \text{area of }\triangle AOC - \text{area of }\triangle ABC \\ & = \frac {\angle AOC}2 \cdot \blue 1 - \frac 12 \cdot \frac h2 \cdot AC - \frac 12 \left(1-\frac h2\right) \cdot AC & \small \blue{\text{The radius of the circle }= 1} \\ & = \sin^{-1} \left(\frac {\frac 12 AC}{OC} \right) - \frac 12 \cdot AC \\ & = \sin^{-1} \left(\frac {\sqrt{4-h^2}}2 \right) - \frac {\sqrt{4-h^2}}2 \\ & = \sin^{-1} \left(\frac {2\sqrt 6}5 \right) - \frac {2\sqrt 6}5 \end{aligned}$

Therefore $a+b+c = 2+6+5 = \boxed{13}$ .

$A_{\triangle{ABC}} = \dfrac{1}{2}(1 - \dfrac{x}{2})y = A_{ACDE} = xy \implies 2 - x = 4x \implies x = \dfrac{2}{5}$

$\dfrac{1}{5} = \cos(m)$ and $\dfrac{y}{2} = \cos(\dfrac{\pi}{2} - m) = \sin(m) \implies y = 2\sin(m) = 2\sqrt{\dfrac{24}{25}} = \dfrac{4}{5}\sqrt{6}$

$\implies A_{ACDE} = xy = \dfrac{8}{25}\sqrt{6} = A_{\triangle{ABC}}$

The area $I$ of the bounded circular region above the line $y = \dfrac{1}{5}$ and the circle

$x^2 + y^2 = 1$ is:

$I = \int_{\frac{-2\sqrt{6}}{5}}^{\frac{2\sqrt{6}}{5}} \sqrt{1 - x^2} - \dfrac{1}{5} dx$ ,

Letting $x = \sin(\theta) \implies dx = \cos(\theta) \implies$ $I = \arcsin(\dfrac{2\sqrt{6}}{5}) - \dfrac{2\sqrt{6}}{25} \implies$

The desired area $A = I - \dfrac{8\sqrt{6}}{25} =\arcsin(\dfrac{2\sqrt{6}}{5}) - \dfrac{2\sqrt{6}}{5}$ $= \arcsin(\dfrac{a\sqrt{b}}{c}) - \dfrac{a\sqrt{b}}{c} \implies$

$a + b + c = \boxed{13}$ .

Note: $A_{\triangle{ABC}} = \dfrac{1}{2}y(1 - \dfrac{x}{2}) = \dfrac{1}{2}(\dfrac{4}{5}\sqrt{6})(\dfrac{4}{5}) = \dfrac{8}{25}\sqrt{6} = xy$