Ratio Problem!

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 $HE = \frac{\sqrt{3}}{3}(6 + 4\sqrt{3}) = 4 + 2\sqrt{3} = x = AD$ .

By the Pythagorean Theorem on $\triangle ADC$ , $DC = \sqrt{AC^2 - AD^2} = \sqrt{(8 + 4\sqrt{3})^2 - (4 + 2\sqrt{3})^2} = 6 + 4\sqrt{3} = EC$ .

By the Pythagorean Theorem on $\triangle HEC$ , $HC = \sqrt{HE^2 + EC^2} = \sqrt{(4 + 2\sqrt{3})^2 + (6 + 4\sqrt{3})^2} = 8 + 4\sqrt{3} = AC$ .

By the Pythagorean Theorem on $\triangle HDC$ , $HD = \sqrt{HC^2 - DC^2} = \sqrt{(8 + 4\sqrt{3})^2 - (6 + 4\sqrt{3})^2} = 4 + 2\sqrt{3} = AD = HE$ .

Since $HD = AD = HE$ , $AC = HC$ , and $DC = EC$ , triangles $\triangle ADC$ , $\triangle HDC$ and $\triangle HEC$ are all congruent to each other by SSS congruency.

By corresponding parts of congruent triangles, $\angle ACD = \angle HCD = \angle HCE$ .

Since $A_2$ spans exactly twice the triangles and angles of $A_1$ , $\cfrac{A_2}{A_1} = \boxed{2}$ .

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

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

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

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

$\implies \boxed{A_{2} = A_{DHEI} - A_{R_{2}} = 48 + 28\sqrt{3} - (14 + 8\sqrt{3})\pi = 2(24 + 14\sqrt{3} - (7 + 4\sqrt{3})\pi)}$

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

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

$\implies \dfrac{A_{2}}{A_{1}} = \boxed{2}$ .

$EH \perp BC$ and intersects $AB$ at $H$ :

$\begin{aligned} &CD = r = CE \\ \implies &RT\triangle CDH \cong RT\triangle CEH \\ \therefore &\angle ECH = \angle DCH = \dfrac 12 \angle BCD = 30\degree \\ \implies &RT\triangle CDH \cong RT\triangle CDA \cong RT\triangle CEH \\ \implies &A_2 = 2 A_1 \\ \therefore &\dfrac {A_2}{A_1} = \boxed{2} \end{aligned}$

Label the two parallel lines to $CA$ be $EF$ and $DG$ and $DH$ the perpendicular line to $CA$ .. By Pythagorean theorem ,

$\begin{aligned} \blue{CG}^2+DG^2 & = CD^2 & \small \blue{\text{Note that }CG = DH} \\ (AD \cdot \cos 30^\circ)^2 + (DB \cdot \sin 30^\circ)^2 & = (2x-2)^2 \\ \frac 34 x^2 + \frac 94 x^2 & = 4x^2 - 8x + 4 \\ x^2 - 8x + 4 & = 0 \\ (x-4)^2 & = 12 \\ \implies x & = 2\sqrt 3 + 4 \end{aligned}$

Then $CE=2x-2 = 4 \sqrt 3 + 6$ and $BC = AB \cdot \cos 30^\circ = 2 \sqrt 3 x = 8\sqrt 3 + 12 = 2 \cdot CE$ . Then the ratio of lengths $BE:BG:BC = 2:3:4$ . Let the area of $\triangle ADH$ , [ADH] = A). then $[BEF]=4A$ , $[ADGC] = 12A$ , $ADGC = 7A$ , $[DGCH] = 6A$ , and $[CDH] = 3A$ . Let $r = 2x-2$ . Then we have:

$\begin{cases} A_1 + A_2 = [AFEC] - \dfrac {\pi r^2}4 = 12a - \dfrac {\pi r^2}4 & ...(1) \\ A_2 = [ADGC]-[CDH] - \dfrac {\pi r^2}{12} = 4A - \dfrac {\pi r^2}{12} & ... (2) \end{cases}$

$\begin{aligned} (1) - (2): \quad A_1 & = 8A - \frac {\pi r^2}6 & \implies \frac {A_1}{A_2} = \boxed 2 \end{aligned}$