Find the Shaded Region

In $\triangle{EBD}$ we have:

$\lambda_{1} + 180 - \lambda_{3} + \lambda_{2} = 180 \implies \lambda_{3} = \lambda_{1} + \lambda_{2}$

and $\tan(\lambda_{3}) = \dfrac{h}{x}, \tan(\lambda_{2}) = \dfrac{h}{3x}$ and $\tan(\lambda_{1}) = \dfrac{h}{4x} \implies$

$\tan(\lambda_{1}) = \dfrac{1}{4}\tan(\lambda_{3}) = \dfrac{1}{4}\tan(\lambda_{1} + \lambda_{2})$

and

$\tan(\lambda_{2}) = \dfrac{1}{3}\tan(\lambda_{1} + \lambda_{2}) \implies 4\tan(\lambda_{1}) = 3\tan(\lambda_{2}) \implies$

$\tan(\lambda_{2}) = \dfrac{4}{3}\tan(\lambda_{1})$ and $4\tan(\lambda_{1}) = \tan(\lambda_{1} + \lambda_{2}) = \dfrac{7\tan(\lambda_{1})}{3 - 4\tan^2(\lambda_{1})} \implies$

$\tan(\lambda_{1})(16\tan^2(\lambda_{1}) - 5) = 0$ and $\tan(\lambda_{1}) \neq 0 \implies \tan(\lambda_{1}) = \dfrac{\sqrt{5}}{4} \implies$

$h = 4x\tan(\lambda_{1}) = \sqrt{5}x \implies 21x^2 = 36x^2 - 12 x + 1$ in right $\triangle{ABC}$

$\implies 15x^2 - 12x + 1 = 0 \implies x = \dfrac{6 \pm \sqrt{21}}{15}$ and $x = \dfrac{6 - \sqrt{21}}{15} \implies 6x - 1 < 0$

$\therefore$ we choose $x = \dfrac{6 + \sqrt{21}}{15}$ $\implies h = \sqrt{5}(\dfrac{6 + \sqrt{21}}{15}) \implies$

$A_{p} = xh =\sqrt{5}(\dfrac{6 + \sqrt{21}}{15})^2 = \dfrac{\sqrt{5}}{3 * 5^2}(19 + 2^2\sqrt{21}) =$

$\dfrac{\sqrt{\alpha}}{\beta * \alpha^{\gamma}}(\omega + \gamma^{\gamma}\sqrt{\lambda}) \implies \alpha + \beta + \gamma + \omega + \lambda = \boxed{50}$ .

AB² = BC² - AC²
= (6x - 1)² - (4x)²
= 20x² - 12x + 1

In the same way using Pythagoras, we get
BE² = 21x² - 12x + 1
(and
BD² = 29x² - 12x + 1
though not necessary)

Triangle BED is similar (not congruent, right?) to triangle CEB.
ED / BE = EB / CE
(ED)(CE) = BE²
6x² = 21x² - 12x + 1
15x² - 12x + 1 = 0
x = (2 ± √(7/3)) / 5
Since 6x - 1 > 0 ==> x > 1/6,
x = (2 + √(7/3)) / 5

Pink area
= x√(20x² - 12x + 1)
= √5 (19 + 2²√21) / ((3)(5²))
by substitution.

Answer
= 5 + 3 + 2 + 19 + 21
= 50

Let $\angle BCA = \angle EBD = \theta$ , $\angle ABE = \alpha$ , and $AB = h$ . By Pythagorean theorem ,

$\begin{aligned} AB^2 & = BC^2 - CA^2 \\ h^2 & = (6x-1)^2 + (4x)^2 = 20x^2 - 12x + 1 \end{aligned}$

Consider $\tan \angle ABD = \tan (\alpha + \theta)$ ,

$\begin{aligned} \tan (\alpha + \theta) & = \frac {\tan \alpha + \tan \theta}{1-\tan \alpha \tan \theta} & \small \blue{\text{Note that }\tan \alpha = \frac {AE}{AB} = \frac xh} \\ \frac {3x}h & = \frac {\frac xh + \frac h{4x}}{1-\frac xh \cdot \frac h{4x}} & \small \blue{\text{and }\tan \theta = \frac {AB}{CA} = \frac h{4x}} \\ \frac {3x}h & = \frac {4x^2 + h^2}{3xh} \\ 9x^2 & = 4x^2 + h^2 \\ 5x^2 & = \blue{h^2} & \small \blue{\text{Note that }h^2 = 20x^2 -12x + 1} \\ 15 x^2 - 12x + 1 & = 0 \\ \implies x & = \frac {6 + \sqrt{21}}{15} & \small \blue{\text{Since }6x-1>0} \\ \implies A_p & = \frac {2x \cdot h}2 = xh = \sqrt 5x^2 & \small \blue{\text{and }h^2 = 5x^2} \\ & = \sqrt 5 \left(\frac {6 + \sqrt{21}}{15}\right)^2 \\ & = \frac {\sqrt 5}{3\cdot 5^2}\left(19 \pm 2^2\sqrt{21} \right) \end{aligned}$

Therefore the required answer is $5+3+2+19+21 = \boxed{50}$