Semicircle in a Right Triangle!

We note that $\triangle CEF$ and $\triangle ABC$ are similar. Since $EF = r$ , then $FC = \frac 12 r$ . Let $\angle EFC = \theta$ , then

$\begin{aligned} A_p & = [CEF] - \frac \theta 2 r^2 & \small \blue{\text{where }[CEF] \text{ denotes the area of }\triangle CEF} \\ & = \frac 12 \cdot r \cdot \frac r2 - \frac \theta 2 r^2 \\ \frac {A_p}{r^2} & = \frac 14 - \frac \theta 2 = \frac 14 \left(1 - 2 \tan^{-1} \frac 12\right) \\ & = \frac 14\left(1 - 2 \cdot \frac \pi 2 + 2 \tan^{-1} 2 \right) \\ & = \frac 14 \left(2 \tan^{-1} 2 - (\pi - 1)\right) \end{aligned}$

Therefore $\alpha + \beta = 1+2 = \boxed 3$ .

Let $A$ be at $A(0, 0)$ and let the radius of the semi-circle be $r = 1$ .

Then $F$ is at $F(1, 0)$ and $E$ is at $E(1, 1)$ , and since the slope of the hypotenuse is $-2$ , it has an equation of $y = -2(x - 1) + 1$ , which intersects $C$ at $C(\cfrac{3}{2}, 0)$ .

Then the area of $\triangle EFC$ is $A_{\triangle EFC} = \cfrac{1}{2} \cdot \cfrac{1}{2} \cdot 1 = \cfrac{1}{4}$ , and the area of the sector of the $A_{EFC} = \cfrac{\frac{\pi}{2} - \arctan(2)}{2\pi} \cdot \pi \cdot 1^2 = \cfrac{\pi - 2\arctan(2)}{4}$ .

Therefore, $A_p = A_{\triangle EFC} - A_{EFC} = \cfrac{1}{4} - (\cfrac{\pi - 2\arctan(2)}{4}) = \cfrac{1}{2^2}(2 \arctan(2) - (\pi - 1))$ , so that $\alpha = 1$ , $\beta = 2$ and $\alpha + \beta = \boxed{3}$ .

$\triangle{BDE} \sim \triangle{EFC} \implies \dfrac{x - r}{r} = \dfrac{r}{2x - r} \implies 2x^2 - 3xr + r^2 = r^2 \implies$

$x(2x - 3r) = 0$ and $x \neq 0 \implies x = \dfrac{3r}{2} \implies x - r = \dfrac{r}{2} \implies$ $A_{\triangle{FEC}} =\dfrac{r^2}{4}$

and $\tan(\theta) = 2 \implies A_{sector(EFG)} = \dfrac{1}{2}(\dfrac{\pi}{2} - \arctan(2))r^2 \implies$

$A_{p} = A_{\triangle{FEC}} - A_{sector(EFG)} = \dfrac{1}{4}(2\arctan(2) - (\pi - 1))r^2 \implies$

$\dfrac{A_{p}}{r^2} = \dfrac{1}{4}(2\arctan(2) - (\pi - 1)) = \dfrac{\alpha}{\beta^{\beta}}(\beta\arctan(\beta) - (\pi - \alpha))$

$\implies \alpha + \beta = \boxed{3}$ .