What's the Pink Area?

Referring to the figure, let . Then, on right $\triangle DJG$ , $\sin \theta =\dfrac{r}{a-r}$ Applying cosine rule on $\triangle AGD$ ,

$\begin{aligned} & A{{G}^{2}}=D{{A}^{2}}+D{{G}^{2}}-2DA\cdot DG\cdot \cos \left( \angle ADG \right) \\ & \Rightarrow {{\left( a+r \right)}^{2}}={{a}^{2}}+{{\left( a-r \right)}^{2}}-2a\left( a-r \right)\cos \left( 90{}^\circ -\theta \right) \\ & \Rightarrow \cancel{{{a}^{2}}}+2ar+\bcancel{{{r}^{2}}}=\cancel{{{a}^{2}}}+{{a}^{2}}-2ar+\bcancel{{{r}^{2}}}-2a\left( a-r \right)\sin \left( \theta \right) \\ & \Rightarrow 4ar={{a}^{2}}-2a\cancel{\left( a-r \right)}\dfrac{r}{\cancel{a-r}} \\ & \Rightarrow 6ar={{a}^{2}} \\ & \Rightarrow r=\dfrac{a}{6} \\ \end{aligned}$ Now, for the area ${{A}_{p}}$ of the pink region,

$\begin{aligned} {{A}_{pink}} & ={{A}_{\left( D,\overset\frown{EC} \right)}}-\left( {{A}_{\left( A,\overset\frown{ED} \right)}}-\left[ AED \right] \right)-{{A}_{red\text{ }circle}} \\ & =\dfrac{1}{2}\cdot \dfrac{\pi }{6}\cdot {{a}^{2}}-\left( \dfrac{1}{2}\cdot \dfrac{\pi }{3}\cdot {{a}^{2}}-\dfrac{{{a}^{2}}\sqrt{3}}{4} \right)-\pi {{\left( \dfrac{a}{6} \right)}^{2}} \\ & =\ldots \\ & =\left( \dfrac{\sqrt{3}}{4}-\dfrac{\pi }{9} \right){{a}^{2}} \\ \end{aligned}$

Hence, $\dfrac{{{A}_{p}}}{{{a}^{2}}}=\dfrac{\sqrt{3}}{4}-\dfrac{\pi }{9}$ For the answer, $\alpha =3$ , $\beta =4$ , $\lambda =9$ , thus, $\alpha + \beta + \lambda = \boxed{16}$ .

Using the above diagram $h^2 = (a + r)^2 - (a - r)^2 = (a - r)^2 - r^2 \implies$

$4ar = a^2 - 2ar \implies 6ar = a^2 \implies r = \dfrac{a}{6} \implies$ the area of the red circle is $A_{c} = \dfrac{a^2}{36}\pi$

The equations of the blue and green quarter circles are:

$x^2 + y^2 = a^2$

$(x - a)^2 + y^2 = a^2$

Solving the two equations above we obtain $x = \dfrac{a}{2} \implies$

$I = \displaystyle\int_{\frac{a}{2}}^{a} \sqrt{a^2 - (x - a)^2} - \sqrt{a^2 - x^2} \:\ dx$

For $I_{1} = \displaystyle\int_{\frac{a}{2}}^{a} \sqrt{a^2 - (x - a)^2} \:\ dx$

Let $x - a = a\sin(\theta) \implies dx = a\cos(\theta) \:\ d\theta \implies$

$I_{1} = \dfrac{a^2}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{6}}^{0} =$ $\dfrac{a^2}{2}(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{4})$

For $I_{2} = \displaystyle\int_{\frac{a}{2}}^{a} \sqrt{a^2 - x^2} \:\ dx$

Let $x = a\sin(\theta) \implies dx = a\cos(\theta) \:\ d\theta \implies$

$I_{2} = \dfrac{a^2}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{-\frac{\pi}{6}}^{\frac{\pi}{2}} =$ $\dfrac{a^2}{2}(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4})$

$\implies I = \dfrac{a^2}{2}(\dfrac{\sqrt{3}}{2} - \dfrac{\pi}{6}) \implies A_{p} = I - A_{c} = (\dfrac{\sqrt{3}}{4} - \dfrac{\pi}{9})a^2 \implies$

$\dfrac{A_{p}}{a^2} = \dfrac{\sqrt{3}}{4} - \dfrac{\pi}{9} = \dfrac{\sqrt{\alpha}}{\beta} - \dfrac{\pi}{\lambda} \implies \alpha + \beta + \lambda = \boxed{16}$ .