Squares and Circles!

Let the side length of square $ABCD$ be $1$ Then the area of square $ABCD$ , $[ABCD] = 1$ . Let $BE=x$ . By tangent-secant theorem , we have $BE\cdot BC = BP^2 \implies x \cdot 1 = \left(\dfrac 12\right)^2 \implies x = \dfrac 14$

Due to symmetry, we note that $BE=PG=HQ = \dfrac 14$ . Since the diameter of the circle $PQ = PH + HQ = 1 + \dfrac 14 = \dfrac 54$ , the radius of the circle $r = \dfrac 58$ .

Then the area of the red regions

$\begin{aligned} A_R & = \overbrace{[ABEF]}^{\rm rectangle} - \overbrace{[EPFG]}^{\rm segment} \\ & = \overbrace{[ABEF]}^{\rm rectangle} - \bigg(\overbrace{[EPFO]}^{\rm sector} - \overbrace{[EFO]}^{\rm triangle} \bigg) \\ & = AF \cdot AB - r^2 \sin^{-1} \left(\frac {EG}{OE} \right) + \frac 12 \cdot EF \cdot OG \\ & = 1 \cdot \frac 14 - \frac {5^2}{8^2} \sin^{-1} \left( \frac 12 \cdot \frac 85 \right) + \frac 12 \cdot 1 \cdot \left(\frac 58 - \frac 14 \right) \\ & = \frac 14 - \frac {25}{64} \sin^{-1} \left(\frac 45 \right) + \frac 3{16} \\ & = \frac {28 - 25\sin^{-1} \left(\frac 45 \right)}{64} \end{aligned}$

Since $[ABCD]=1$ , $\dfrac {A_R}{[ABCD]} = A_R$ and the required answer is $28+25+4+5+64 = \boxed{126}$ .

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

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

$(\dfrac{28 - 25\arcsin(\dfrac{4}{5})}{64})a^2 \implies \dfrac{A_{R}}{A_{ABCD}} =$ $\dfrac{28 - 25\arcsin(\dfrac{4}{5})}{64} = \dfrac{\alpha - \beta\arcsin(\dfrac{\lambda}{\omega})}{p}$

$\implies \alpha + \beta + \lambda + \omega + p = \boxed{126}$ .