It's All Circles!

Descartes' circle theorem is the key here.

Let the radius of the orange and the blue semicircles are $r_1 = 2, r_2 = 1$ , and let $R = r_1 + r_2 = 3$ denote the radius of the green semicircle.

Denote the curvature of the pink circle as $k_3$ .

The curvatures of the blue, orange and green semicircles are $k_1 = 1, k_2 = \frac12, k_4 = -\frac13.$

Using Descartes' circle theorem, we get $\left(1 + \frac12 + k_3 - \frac13 \right)^2 = 2 \left( 1^2 + (\tfrac12)^2 + k_3^2 + (-\tfrac13)^2 \right)$

Solving the equation above gives $k_3 = \frac76$ , so the radius of the pink circle is $\frac67 .$

The area of the green shaded region is $\frac12 \pi \cdot 3^2 - \frac12 \pi \cdot 2^2 - \frac12 \pi \cdot 1^2 - \pi \cdot (\tfrac67 )^2 = \frac{62}{49} \pi$

The answer is $62 + 49 = \boxed{111} .$

Let the center of the large green semicircle be the origin $(0,0)$ , then the center of the red semicircle is $(-1,0)$ and the center of the blue semicircle is $(2,0)$ . Let the center of the purple circle be $(x,y)$ and its radius be $r$ . Then we have:

$\begin{cases} (x+1)^2 + y^2 = (2+r)^2 & ...(1) \\ (x-2)^2 + y^2 = (1+r)^2 & ...(2) \\ x^2 + y^2 = (3-r)^2 & ...(3) \end{cases}$

From $(1)-(2): \ 6x - 3 = 2r + 3 \implies 6x = 2r + 6 \quad ...(4)$ .

From $(1)-(3): \ 2x + 1 = 10r - 5 \implies 2x = 10r - 6 \quad ...(5)$ .

From $3\times (5)-(4): \ 0 = 28r - 24 \implies r = \dfrac 67$ .

Therefore the area of green region $A = \pi \left(\dfrac {3^2}2 - \dfrac {2^2}2 - \dfrac {1^2}2 - \left(\dfrac 67\right)^2 \right) = \dfrac {62}{49}\pi$ . $\implies a + b = 62+49 = \boxed{111}$ .

Let $(m,n)$ be center of the purple circle with radius $r$ .

$(1): \:\ (m - 2)^2 + n^2 = (2 + r)^2 \implies m^2 - 4m + 4 + n^2 = 4 + 4r + r^2$

$(2): \:\ (m - 5)^3 + n^2 = (1 + r)^2 \implies m^2 - 10m + 25 + n^2 = 1 + 2r + r^2$

$(3): \:\ (m - 3)^2 + n^2 = (3 - r)^2 \implies m^2 - 6m + 9 + n^2 = 9 - 6r + r^2$

Subtracting $2)$ from $(1)$ we obtain $3m - r = 12$

Subtracting $(3)$ from $(1)$ we obtain $m = 5r$

$\implies 14r = 12 \implies r = \dfrac{6}{7}$

$\implies A = (\dfrac{9}{2} - (2 + \dfrac{1}{2} + \dfrac{36}{49}))\pi = (2 - \dfrac{36}{49})\pi = \dfrac{62}{49}\pi = \dfrac{a}{b}\pi \implies a = b = \boxed{111}$ .

Label the diagram as follows, where C is the center of the large semi-circle:

The large semi-circle has a radius of $AC = CF = CJ = \frac{1}{2}AF = \frac{1}{2}(AB + BD + DE + EF) = \frac{1}{2}(2 + 2 + 1 + 1) = 3$ which means $CE = CF - EF = 3 - 1 = 2$ . Also, $BE = BD + DE = 2 + 1 = 3$ .

Let $r = IG = IH = IJ$ . Then $CI = CJ - IJ = 3 - r$ , $BI = BG + IG = 2 + r$ , and $EI = EH + IH = 1 + r$ .

By the law of cosines on $\triangle BEI$ and $\triangle CEI$ , $\cos \angle IEC = \cfrac{3^2 + (r + 1)^2 - (r + 2)^2}{2 \cdot 3 \cdot (r + 1)} = \cfrac{2^2 + (r + 1)^2 - (3 - r)^2}{2 \cdot 2 \cdot (r + 1)}$ , which solves to $r = \frac{6}{7}$ .

The area of the green region is then the area of the large semi-circle minus the sum of the areas of the two smaller semi-circles and the circle, which is $A = \frac{3^2}{2}\pi - (\frac{2^2}{2}\pi + \frac{1^2}{2}\pi + \frac{6^2}{7^2}\pi) = \frac{62}{49}\pi$ .

Therefore, $a = 62$ , $b = 49$ , and $a + b = \boxed{111}$ .