More of Circles and Squares.!

Let $a$ be a side of square $ABCD$ .

$\triangle{COD}$ is an isosceles triangle $\implies \overline{OQ}$ is the perpendicular bisector of base $\overline{CD} \implies$

$\overline{CQ} = \overline{QD} = \dfrac{a}{2}$

In right $\triangle{COQ}$ we have $r^2 = a^2 - 2ar + r^2 + \dfrac{a^2}{4} \implies a(5a - 8r) = 0$

and $a \neq 0 \implies r = \dfrac{5a}{8}$

$\overline{BM} = a - 2(a - r) = 2r - a = \dfrac{a}{4}$ and $\sin(\theta) = \dfrac{a}{2r} = \dfrac{4}{5} \implies$

Area of sector $MOP$ is $A_{s} = \dfrac{1}{2}\theta r^2 = \dfrac{1}{2}\arcsin(\dfrac{4}{5})(\dfrac{25}{64})a^2 =$ $\dfrac{25}{128}\arcsin(\dfrac{4}{5})a^2$

and

Area of trapezoid $PBMO$ is $A_{PBMO} = \dfrac{1}{2}(r + \overline{BM})(\dfrac{a}{2}) = \dfrac{1}{2}(\dfrac{5a}{8} + \dfrac{a}{4})(\dfrac{a}{2}) = \dfrac{7}{32}a^2$

$\implies A_{green} = 2(A_{PBMO} - A_{s}) = 2(\dfrac{7}{32} - \dfrac{25}{128}\arcsin(\dfrac{4}{5}))a^2 =$

$(\dfrac{28 - 25\arcsin(\dfrac{4}{5})}{64})a^2$

In $\triangle{MOC}$ we have $\sin(\theta^{*}) = \dfrac{a - r}{r} = \dfrac{3}{5} \implies \theta^{*} = \arcsin(\dfrac{3}{5}) \implies$

$2\theta^{*} = 2\arcsin(\dfrac{3}{5}) \implies$ area of sector $MOC$ is $A_{s^{*}} = \dfrac{1}{2}(2\theta)r^2 = \dfrac{25}{64}\arcsin(\dfrac{3}{4})a^2$

and $A_{\triangle{MOC}} = \dfrac{1}{2}(2(a - r))(\dfrac{a}{2}) = (\dfrac{3a}{8})(\dfrac{a}{2}) = \dfrac{3}{16}a^2$

$\implies A_{blue} = 2(A_{s^{*}} - A_{\triangle{MOC}}) = \dfrac{50\arcsin(\dfrac{3}{5}) - 24}{64}a^2$

From above $A_{s} = \dfrac{25}{128}\arcsin(\dfrac{4}{5})a^2$ and $A_{\triangle{COR}} = \dfrac{1}{2}(a - r)(\dfrac{a}{2}) = \dfrac{1}{2}(\dfrac{3a}{8})(\dfrac{a}{2}) =$

$\dfrac{3}{32}a^2 \implies A_{purple} = 2(A_{s} - A_{\triangle{COR}}) = \dfrac{25\arcsin(\dfrac{4}{5}) - 12}{64}a^2 \implies$

$\dfrac{A}{A_{ABCD}} = \dfrac{A_{green} + A_{blue} + A_{purple}}{A_{ABCD}} = \dfrac{25\arcsin(\dfrac{3}{5}) - 4}{32} =$ $\dfrac{\alpha\arcsin \left(\frac{\lambda}{\omega}\right) - \beta}{p} \implies$

$\alpha + \beta + \lambda + \omega + p = \boxed{69}$ .

$\textbf{Below is an alternative approach:}$

From above $A_{s} = \dfrac{25}{128}\arcsin(\dfrac{4}{5})a^2 \implies 2A_{s} = \dfrac{25}{64}\arcsin(\dfrac{4}{5})a^2$

and $A_{\triangle{MON}} = \dfrac{1}{2}(a)(a - r) = \dfrac{1}{2}(\dfrac{3}{8})a^2 = \dfrac{3}{16}a^2$

$\implies$ Area of segment is $MPN$ is $A_{s^{*}} = 2A_{s} - A_{\triangle{MON}} = \dfrac{25\arcsin(\dfrac{4}{5}) - 12}{64}a^2$

$A_{white} = A_{MCND} + A_{s^{*}} = 2(\dfrac{3}{8})a^2 + A_{s^{*}} =$ $(\dfrac{3}{4} + \dfrac{25\arcsin(\dfrac{4}{5}) - 12}{64})a^2 =$

$\dfrac{36 + 25\arcsin(\dfrac{4}{5})}{64}a^2 \implies A_{T} = \dfrac{25}{64}\pi a^2 - A_{white} =$ $\dfrac{25\pi - 25\arcsin(\dfrac{4}{5}) - 36}{64}a^2$

Form above $A_{green} = \dfrac{28 - 25\arcsin(\dfrac{4}{5})}{64}a^2 \implies$

$A = A_{T} + A_{green} = \dfrac{25\pi - 50\arcsin(\dfrac{4}{5}) - 8}{64}a^2$

Letting $\theta^{**} = \arcsin(\dfrac{3}{5}) \implies A = \dfrac{25\pi - 50(\dfrac{\pi}{2} - \theta^{**}) - 8}{64}a^2 =$

$\dfrac{25\arcsin(\dfrac{3}{5}) - 4}{32}a^2 \implies \dfrac{A}{A_{ABCD}} =$ $\dfrac{25\arcsin(\dfrac{3}{5}) - 4}{32} = \dfrac{\alpha\arcsin \left(\frac{\lambda}{\omega}\right) - \beta}{p}$

$\implies \alpha + \beta + \lambda + \omega + p = \boxed{69}$ .