What's the Area of the Region?

$\triangle{ABE} \sim \triangle{FME} \implies \dfrac{2}{1} = \dfrac{a}{\overline{FM}} \implies \overline{FM} = \dfrac{a}{2} \implies A_{\triangle{FME}} = \dfrac{a^2}{4}$

and $\tan(\theta) = \dfrac{1}{2} \implies \theta = \arctan(\dfrac{1}{2}) \implies A_{sector(FPE)} = \dfrac{1}{2}\arctan(\dfrac{1}{2})a^2 \implies$

$A_{B} = A_{\triangle{FME}} - A_{sector(FPE)} = (\dfrac{1}{4} - \dfrac{1}{2}\arctan(\dfrac{1}{2}))a^2 \implies$

$A_{R} = a^2 - \dfrac{\pi}{4}a^2 - A_{B} = (\dfrac{3}{4} + \dfrac{1}{2}\arctan(\dfrac{1}{2}) - \dfrac{\pi}{4})a^2$

To Find $a^2$ using $A_{P} = \dfrac{1}{2}$ :

$\overline{BE} = \sqrt{5}$ and $\tan(\lambda) = \tan(\dfrac{\pi}{2} - \theta) = \dfrac{1}{\tan(\theta)} = 2 \implies$

$A_{sector(PDE)} = \dfrac{1}{2}\arctan(2)a^2$ and $h = a\sin(\lambda) = a\cos(\theta) = \dfrac{2}{\sqrt{5}}a \implies$

$A_{\triangle{PDE}} = \dfrac{1}{2}(a)(\dfrac{2a}{\sqrt{5}}) = \dfrac{a^2}{\sqrt{5}} \implies$ $A_{P} = \dfrac{1}{2} = A_{sector(PDE)} - A_{\triangle{PDE}} =$

$(\dfrac{\sqrt{5}\arctan(2) - 2}{2\sqrt{5}})a^2 \implies a^2 = \dfrac{\sqrt{5}}{\sqrt{5}\arctan(2) - 2}$

Using the fact that $\theta = \arctan(\dfrac{1}{2})$ and $\lambda = \dfrac{\pi}{2} - \theta = \arctan(2) \implies$

$\theta = \dfrac{\pi}{2} - \arctan(2) \implies$ $A_{R} = \dfrac{\sqrt{5}}{4}(\dfrac{3 - 2\arctan(2)}{\sqrt{5}\arctan(2) - 2}) =$

$\dfrac{\sqrt{\alpha}}{\beta^{\beta}} (\dfrac{\omega - \beta\arctan(\beta)}{\sqrt{\alpha}\arctan(\beta) - \beta}) \implies$ $\alpha + \beta + \omega = \boxed{10}$ .

By the Pythagorean Theorem on $\triangle BDE$ , $BE = \sqrt{BD^2 + DE^2} = \sqrt{(2a)^2 + a^2} = \sqrt{5}a$ , so $\tan \angle BED = \cfrac{2a}{a} = 2$ and $\sin \angle BED = \cfrac{2a}{\sqrt{5}a} = \cfrac{2}{\sqrt{5}}$ .

The area of the sector formed by $EBD$ is $A_{\text{sec} EBD} = \cfrac{m\angle BED}{2\pi} \pi a^2 = \cfrac{1}{2}a^2\arctan(2)$ .

The area of $\triangle EBD$ is $A_{\triangle EBD} = \cfrac{1}{2}a^2 \sin \angle BED = \cfrac{1}{\sqrt{5}}a^2$ .

Then $A_p = A_{\text{sec} EBD} - A_{\triangle EBD} = \cfrac{1}{2}a^2\arctan(2) - \cfrac{1}{\sqrt{5}}a^2 = \cfrac{1}{2}$ , which solves to $a^2 = \cfrac{\sqrt{5}}{\sqrt{5} \arctan (2) - 2}$ .

And $A_R = A_{FCDE} - A_{\triangle FBE} - A_{\text{sec} EBD} = a^2 - \cfrac{1}{4}a^2 - \cfrac{1}{2}a^2\arctan(2) = \cfrac{3 - 2 \arctan(2)}{4} \cdot \cfrac{\sqrt{5}}{\sqrt{5} \arctan (2) - 2} = \cfrac{\sqrt{5}}{2^2}\bigg(\cfrac{3 - 2 \arctan (2)}{\sqrt{5} \arctan (2) - 2}\bigg)$ .

Therefore, $\alpha = 5$ , $\beta = 2$ , $\omega = 3$ , and $\alpha + \beta + \omega = \boxed{10}$ .