Area Festival

$m_{OA} = \dfrac{1}{\sqrt{3}} \implies m_{\perp} = -\sqrt{3}$

$m_{OB} = -\dfrac{1}{\sqrt{3}} \implies m^{*}_{\perp} = \sqrt{3}$

Let $y = ax^2 + bx + c \implies \dfrac{dy}{dx} = 2ax + b$

$\implies \dfrac{dy}{dx}|_{(x = \dfrac{\sqrt{3}}{2})} = \sqrt{3}a + b = -\sqrt{3}$

and

$\dfrac{dy}{dx}|_{(x = \dfrac{-\sqrt{3}}{2})} = -\sqrt{3}a + b = \sqrt{3}$

Solving the above system we obtain $a = -1$ and $b = 0$ and using $A(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}) \implies$

$\dfrac{1}{2} = -1(\dfrac{\sqrt{3}}{2})^2 + c \implies c = \dfrac{5}{4} \implies$ $y = \dfrac{5}{4} - x^2$

and for the portion of the circle we have $y = \sqrt{1 - x^2}$

$\implies A = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} ((\dfrac{5}{4} - x^2) - \sqrt{1 - x^2}) dx$

$= 2(\dfrac{\sqrt{3}}{2} - \displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} \sqrt{1 - x^2} dx)$

Letting $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies$

$\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} \sqrt{1 - x^2} dx = \displaystyle\int_{0}^{\dfrac{\pi}{3}} (1 + \cos(2\theta)) d\theta =$ $\dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{0}^{\dfrac{\pi}{3}} =$

$\dfrac{1}{2}(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4})$

$\implies A = \sqrt{3} - (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}) = \dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3}$ $= \dfrac{a\sqrt{a}}{b} - \dfrac{\pi}{a} \implies a + b = \boxed{7}$ .