It's All Circles!

$\triangle{O'' OO'}$ is a right triangle $\implies (x - 1)^2 + x^2 = (x + 1)^2 \implies 2x^2 - 2x + 1 = x^2 + 2x + 1 \implies x(x - 4) = 0$ and $x \neq 0 \implies x = 4$

$\implies R_{1} + R_{2} = 5, R_{1} + R_{3} = 4, R_{2} + R_{3} = 3 \implies R_{2} = 5 - R_{1} = 3 - R_{3} \implies$

$R_{1} - R_{3} = 2$ and we have $R_{1} + R_{3} = 4 \implies R_{1} = 3 \implies R_{3} = 1 \implies R_{2} = 2$

and $\tan(\beta) = \dfrac{3}{4} \implies \beta = \arctan(\dfrac{3}{4})$

For the areas of the three sectors we have:

$A_{PO''P''} = \dfrac{1}{2}(\dfrac{\pi}{2}) = \dfrac{\pi}{4}$ .

$A_{POP'} = \dfrac{1}{2}(4)(\dfrac{\pi}{2} - \arctan(\dfrac{3}{4})) = \pi - 2\arctan(\dfrac{3}{4})$ .

$A_{P'O'P''} = \dfrac{1}{2}(9)\arctan(\dfrac{3}{4}) = \dfrac{9}{2}\arctan(\dfrac{3}{4})$

and $A_{\triangle{O''OO'}} = 6 \implies A_{R} = \dfrac{24 - 5\pi - 10\arctan(\dfrac{3}{4})}{4} =$

$\dfrac{a - b\pi - c\arctan(\dfrac{d}{e})}{e} \implies a + b + c + d + e = \boxed{46}$ .

By Pythagorean theorem ,

$\begin{aligned} OO'^2 - OO''^2 & = O'O''^2 \\ (x+1)^2-(x-1)^2 & = x^2 \\ 4x & = x^2 \\ \implies x & = 4 \end{aligned}$

Let the radii of the green, orange, and blue circles be $r_1$ , $r_2$ , and $r_3$ respectively. Then we have:

$\begin{cases} r_1+r_2 = 3 & ...(1) \\ r_2 + r_3 = 5 & ...(2) \\ r_3 + r_1 = 4 & ...(3) \end{cases} \implies (2)-(1)+(3): \ 2r_3 = 6 \implies r_3 = 3 \implies r_1 = 1, r_2 = 2$

Then the area of $R$

$\begin{aligned} A_R & = [OO'O''] - [O''PP''] - [O'P'P''] -[OPP'] \\ & = \frac {3\cdot 4}2 - \frac {\pi\cdot 1^2}4 - \frac {\beta}{2\pi} \cdot \pi \cdot 3^2 - \frac {\frac \pi 2 - \beta}{2 \pi} \cdot \pi \cdot 2^2 \\ & = 6 - \frac \pi 4 - \frac {9\beta}2 - \pi + 2\beta \\ & = \frac {24-5\pi - 10\beta}4 & \small \blue{\text{where }\beta = \tan^{-1} \frac 34} \end{aligned}$

Therefore, $a+b+c+d+e = 24+5+10+3+4 = \boxed{46}$ .