It's All Areas!

Let the point to the left of $E$ be $H$ , and let $BE = y$ .

Since $\angle ABC = 30°$ and $AB = 4x$ , from $\triangle ABC$ we have $AC = 2x$ and $BC = 2\sqrt{3}x = 2x - 2 + y$ .

By the tangent-secant theorem , $(3x)^2 = y(4x - 4 + y)$ .

These two equations solve to $x = 4 + 2\sqrt{3}$ and $y = 6 + 4\sqrt{3}$ .

That means $AC = 2x = 2(4 + 2\sqrt{3}) = 8 + 4\sqrt{3}$ and $EC = 2x - 2 = 2(4 + 2\sqrt{3}) - 2 = 6 + 4\sqrt{3}$ .

Since $\angle HBE = 30°$ and $BE = y = 6 + 4\sqrt{3}$ , from $\triangle HBE$ we have $HE = \frac{\sqrt{3}}{3}(6 + 4\sqrt{3}) = 4 + 2\sqrt{3}$ .

The sum of the areas $A_1 + A_2$ is the difference between the areas of trapezoid $AHEC$ and the quarter circle, so:

$A_1 + A_2 = \frac{1}{2}(AC + HE) \cdot EC - \frac{1}{4}\pi\cdot EC^2$

$A_1 + A_2 = \frac{1}{2}(8 + 4\sqrt{3} + 4 + 2\sqrt{3}) \cdot (6 + 4\sqrt{3}) - \frac{1}{4}\pi\cdot (6 + 4\sqrt{3})^2$

$A_1 + A_2 = 3(24 + 14\sqrt{3} - (7 + 4\sqrt{3})\pi)$ .

Therefore, $a = 3$ , $b = 24$ , $c = 14$ , $d = 7$ , $e = 4$ , and $a + b + c + d + e = \boxed{52}$ .

$AF = 2x - (2x - 2) = 2, \overline{GD} = \dfrac{\sqrt{3}}{2}x, \overline{AG} = \dfrac{x}{2} \implies$

$(\overline{CD})^2 = 4(x - 1)^2 = 16x^2 - 32x + 16 = (2x - \dfrac{x}{2})^2 + \dfrac{3}{4}x^2 = 12x^2 \implies x^2 - 8x + 4 = 0 \implies x = 4 \pm 2\sqrt{3}$

$x = 4 - 2\sqrt{3} \implies 2(x - 1) = 2(3 - 2\sqrt{3}) < 0 \therefore$ choose $x = 4 + 2\sqrt{3}$

$\implies$ radius $r = 2x - 2 = 2(3 + 2\sqrt{3})$

For $A_{2}$ :

Let $\theta = m\angle{DCE}$

$\implies \tan(\theta)= \sqrt{3} \implies \theta = \dfrac{\pi}{3} \implies A_{sector{CDE}} = \dfrac{1}{2}\theta r^2 = \dfrac{1}{2}(\dfrac{\pi}{3})(4)(3 + 2\sqrt{3})^2 =$

$2\pi(7 + 4\sqrt{3})$

and $A_{\triangle{CDI}} = \dfrac{1}{2}(\dfrac{3x}{2})(\dfrac{\sqrt{3}}{2}x) = \dfrac{3\sqrt{3}}{8}x^2 =$ $\dfrac{3\sqrt{3}}{8}(4 + 2\sqrt{3})^2$ $= \dfrac{3\sqrt{3}}{2}(7 + 4\sqrt{3})$

Let $R_{2}$ be region $DEI \implies A_{R_{2}} = A_{sector{CDE}} - A_{\triangle{CDI}} = \dfrac{(7 + 4\sqrt{3})(4\pi - 3\sqrt{3})}{2}$

$\overline{BC} = 2\sqrt{3}x \implies \overline{BE} = 2\sqrt{3}x - (2x - 2) = 2((\sqrt{3} - 1)x + 1) =$

$2(2\sqrt{3} + 3)$ and $\triangle{ADG} \sim \triangle{HBE} \implies \dfrac{1}{\sqrt{3}} = \dfrac{\overline{HE}}{2(2\sqrt{3} + 3)} \implies \overline{HE} = \dfrac{4\sqrt{3} + 6}{\sqrt{3}} =$ $4 + 2\sqrt{3} = x$ and $DI = \dfrac{3x}{2} = 3(2 + \sqrt{3})$ and $IE = \overline{CE} - \overline{CI} =$

$(\dfrac{4 - \sqrt{3}}{2})x - 2 = 3 + 2\sqrt{3} \implies$ area of trapezoid

$A_{DHEI} = \dfrac{1}{2}(5)(2 + \sqrt{3})(3 + 2\sqrt{3}) = \dfrac{5}{2}(12 + 7\sqrt{3})$

$\implies \boxed{A_{2} = A_{DHEI} - A_{R_{2}} = 48 + 28\sqrt{3} - (14 + 8\sqrt{3})\pi}$

For $A_{1}:$

$m\angle{FCD} = \dfrac{\pi}{6} \implies A_{sector{CDF}} = \dfrac{1}{2}(\dfrac{\pi}{6})(4)(3 + 2\sqrt{3})^2 = \pi(7 + 4\sqrt{3})$

and $A_{\triangle{DGC}} = \dfrac{3\sqrt{3}}{2}(7 + 4\sqrt{3})$

Let $R_{1}$ be region $FDG \implies A_{R_{1}} = A_{sector{CDF}} -A_{\triangle{DGC}} =$ $\dfrac{(7 + 4\sqrt{3})(2\pi - 3\sqrt{3})}{2}$

and $A_{\triangle{ADG}} = \dfrac{\sqrt{3}}{8}x^2 = \dfrac{\sqrt{3}}{2}(7 + 4\sqrt{3})$

$\implies \boxed{A_{1} = A_{\triangle{ADG}} - A_{R_{1}} = (7 + 4\sqrt{3})(2\sqrt{3} - \pi) = 24 + 14\sqrt{3} - (7 + 4\sqrt{3})\pi}$

$\implies A_{1} + A_{2} = 72 + 42\sqrt{3} - (21 + 12\sqrt{3})\pi = 3(24 + 14\sqrt{3} - (7 + 4\sqrt{3})\pi) =$

$a(b + c\sqrt{a} - (d + e\sqrt{a})\pi)$ $\implies a + b + c + d + e = \boxed{52}$ .