Areas !

Since $DE = \cfrac{1}{2}x$ and $DC = \cfrac{\sqrt{3}}{2}x$ , $\angle ECD = 30°$ .

$R_1$ is the difference between $\triangle CDE$ and sector $CDP$ , so $A_{R_1} = \cfrac{1}{2} \cdot DC \cdot DE - \cfrac{30°}{360°} \pi DC^2 = \cfrac{1}{2} \cdot \cfrac{\sqrt{3}}{2}x \cdot \cfrac{1}{2}x - \cfrac{30°}{360°} \pi \bigg(\cfrac{\sqrt{3}}{2}x\bigg)^2 = \cfrac{1}{16}(2\sqrt{3} - \pi)x^2$ .

Since $AF$ is the perpendicular bisector of $BD$ and $AB \perp AD$ , $AB = AD$ .

Since $\triangle ABC \sim \triangle DEC$ by AA similarity, $\cfrac{DE}{DC} = \cfrac{AB}{AC}$ . Letting $a = AB = AC$ , $\cfrac{\frac{1}{2}x}{\frac{\sqrt{3}}{2}x} = \cfrac{a}{a + \frac{\sqrt{3}}{2}x}$ , which solves to $a = \cfrac{1}{4}(3 + \sqrt{3})x$ .

That means $A_{R_2} = \cfrac{1}{2} \cdot DE \cdot AD = \cfrac{1}{2} \cdot \cfrac{1}{2}x \cdot \cfrac{1}{4}(3 + \sqrt{3})x = \cfrac{1}{16}(3 + \sqrt{3})x^2$ .

So $\cfrac{A_{R_1}}{A_{R_2}} = \cfrac{\frac{1}{16}(2\sqrt{3} - \pi)x^2}{\frac{1}{16}(3 + \sqrt{3})x^2} = \cfrac{2\sqrt{3} - \pi}{3 + \sqrt{3}} = \cfrac{(3 - \sqrt{3})^2(2\sqrt{3} - \pi)}{(3 - \sqrt{3})^2(3 + \sqrt{3})} = \cfrac{(2 - \sqrt{3})(2\sqrt{3} - \pi)}{3 - \sqrt{3}}$ .

Therefore, $\alpha = 2$ , $\beta = 3$ , and $\alpha + \beta = \boxed{5}$ .

$\angle{A}$ is a right angle and $\overline{AF}$ bisects $\overline{BD} \implies \triangle{ABD}$ is an isosceles right triangle

$\implies \overline{AD} = a \implies \overline{BD} = \sqrt{2}a$ and $\triangle{ABC}$ is a $(30,60,90)$ triangle $\implies$

$\overline{AC} = \dfrac{\sqrt{3}}{2}x = \dfrac{\sqrt{3}x + 2a}{2} = \sqrt{3}a \implies x =$ $\dfrac{2(\sqrt{3} - 1)}{\sqrt{3}}a = \dfrac{2(3 - \sqrt{3})}{3}a \implies$

$\overline{DE} = \dfrac{x}{2} = \dfrac{3 - \sqrt{3}}{3}a$ and $h^{*} = \overline{EG} = \dfrac{x}{2}\sin(45^{\circ}) = (\dfrac{3 - \sqrt{3}}{3})(\dfrac{1}{\sqrt{2}})a \implies$

$A_{\triangle{DBE}} = A_{R_{2}} = (\dfrac{1}{2})(\sqrt{2a})(\dfrac{3 - \sqrt{3}}{3})(\dfrac{1}{\sqrt{2}})a =$ $\boxed{\dfrac{3 - \sqrt{3}}{6}a^2}$

and

$A_{\triangle{DEC}} = \dfrac{\sqrt{3}}{8}x^2 = \dfrac{\sqrt{3}}{8}(\dfrac{4}{9})(12 - 6\sqrt{3})a^2 =$ $\dfrac{3\sqrt{3}}{9}(2 - \sqrt{3})a^2 = \dfrac{\sqrt{3}}{3}(2 - \sqrt{3})a^2$

and

$r = \dfrac{\sqrt{3}}{2}x = (\sqrt{3} - 1)a \implies$ the area of the sector $A_{1} = (\dfrac{1}{2})(\dfrac{\pi}{6})(\sqrt{3} - 1)^2a^2 =$

$\dfrac{\pi}{12}(4 - 2\sqrt{3})a^2 = \dfrac{\pi}{6}(2 - \sqrt{3})a^2$

$\implies A_{R_{1}} = A_{\triangle{DEC}} - A_{1} = (2 - \sqrt{3})(\dfrac{2\sqrt{3} - \pi}{6})a^2 \implies$

$\dfrac{A_{R_{1}}}{A_{R_{2}}} = \dfrac{(2 - \sqrt{3})(2\sqrt{3} - \pi)}{3 - \sqrt{3}} =$ $\dfrac{(\alpha - \sqrt{\beta})(\alpha\sqrt{\beta} - \pi)}{\beta - \sqrt{\beta}}$

$\implies \alpha + \beta = \boxed{5}$ .