Trapezoids and Hexagons !

$h^2 = (a + 2)^2 - x^2 = a^2 - (8 - x)^2 \implies$

$a^2 + 4a + 4 - x^2 = a^2 - 64 + 16x - x^2 \implies$ $a + 1 = 4x - 16 \implies$

$x = \dfrac{a + 17}{4} = DE \implies FE = \dfrac{15 - a}{4}$ and $h = \dfrac{\sqrt{15(a + 5)(a - 3)}}{4}$

$\implies A_{Trapezoid} = \dfrac{1}{2}(2(a + 10))\dfrac{\sqrt{15(a + 5)(a - 3)}}{4} = \boxed{(a + 10)\dfrac{\sqrt{15(a + 5)(a - 3)}}{4}}$

We want: $A_{ABQCDP} = A_{Trapezoid} - (A_{\triangle{BQC}} + A_{\triangle{APD}})$

$AB \parallel DC \implies m\angle{B} + m\angle{C} = 180^{\circ}$ and $m\angle{A} + m\angle{D} = 180^{\circ} \implies$

$\dfrac{1}{2}m\angle{B} + \dfrac{1}{2}m\angle{C} = 90^{\circ}$ and $\dfrac{1}{2}m\angle{A} + \dfrac{1}{2}m\angle{D} = 90^{\circ}$

$\implies \triangle{APD}$ and $\triangle{BQC}$ are right triangles $\implies$ $\angle{APD} = 90^{\circ}$ and $\angle{BQC} = 90^{\circ}$

For $\triangle{BQC}:$

$\cos(2\lambda) = \dfrac{15 -a}{4a} \implies \cos(\lambda) = \sqrt{\dfrac{1 + \dfrac{15 - a}{4a}}{2}} =$ $\sqrt{\dfrac{3(a + 5)}{8a}} = \dfrac{1}{2}\sqrt{\dfrac{3(a + 5)}{2a}}$

$\implies \sin(\lambda) = \dfrac{1}{2}\sqrt{\dfrac{5(a - 3)}{2a}} \implies BQ = \dfrac{a}{2}\sqrt{\dfrac{5(a - 3)}{2a}}$ and $QC = \dfrac{a}{2}\sqrt{\dfrac{3(a + 5)}{2a}}$

$\implies \boxed{A_{\triangle{BQC}} = \dfrac{\sqrt{15(a - 3)(a + 5)}}{16}a}$

For $\triangle{APD}:$

$\cos(2\theta) = \dfrac{a + 17}{4(a + 2)} \implies \cos(\theta) = \sqrt{\dfrac{1 + \dfrac{a + 17}{4(a + 2)}}{2}} =$ $\sqrt{\dfrac{5(a + 5)}{8(a + 2)}} =$

$\dfrac{1}{2}\sqrt{\dfrac{5(a + 5)}{2(a + 2)}}$

$\implies \sin(\theta) = \dfrac{1}{2}\sqrt{\dfrac{3(a - 3)}{2(a + 2}} \implies$ $AP = \dfrac{a + 2}{2}\sqrt{\dfrac{3(a - 3)}{2(a + 2}}$

and $DP = \dfrac{a + 2}{2}\sqrt{\dfrac{5(a + 5)}{2(a + 2)}} \implies$ $\boxed{A_{\triangle{APD}} = \dfrac{\sqrt{15(a - 3)(a + 5)}}{16}(a + 2)}$

$\implies A_{ABQCDP} = A_{Trapezoid} - (A_{\triangle{BQC}} + A_{\triangle{APD}}) =$ $\dfrac{a + 19}{8}\sqrt{15(a - 3)(a + 5)}$

$= \dfrac{a + 19}{8} \implies$ $15(a^2 + 2a - 15) = 1 \implies 15a^2 + 30a - 226 = 0$

dropping the negative root $\implies a = \dfrac{-15 + \sqrt{3615}}{15} =$ $\dfrac{-15 + \sqrt{15}\sqrt{241}}{15} =$

$\dfrac{\sqrt{15}(-\sqrt{15} + \sqrt{241})}{15} =$ $\dfrac{-\sqrt{15} + \sqrt{241}}{\sqrt{15}} =$

$\dfrac{-\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha}} \implies \alpha + \beta = \boxed{256}$ .

Label the diagram as follows:

Since $CQ$ is an angle bisector, $\angle BCF = 2\lambda$ , and from right $\triangle BCF$ , $h = a \sin 2\lambda$ and $FC = \sqrt{a^2 - h^2}$ . Similarly, on right $\triangle ADE$ , $h = (a + 2) \sin 2\theta$ and $DE = \sqrt{(a + 2)^2 - h^2}$ . $DE + FC = DC - AB = (a + 14) - (a + 6) = 8$ , so:

$\sqrt{(a + 2)^2 - h^2} + \sqrt{a^2 - h^2} = 8$

The area of trapezoid $ABCD$ is $A_{\text{trap}} = \frac{1}{2}(a + 6 + a + 14)h$ or $A_{\text{trap}} = (a + 10)h$ .

Since $AB || DC$ , $\angle ABC = 180° - 2\lambda$ , and since BQ is an angle bisector, $\angle CBQ = 90° - \lambda$ . Then by the angle sum of $\triangle BQC$ , $\angle BQC = 90°$ . Therefore, $BQ = a \sin \lambda$ and $CQ = a \cos \lambda$ , so that the area of $\triangle BQC$ is $T_1 = \frac{1}{2} \cdot BQ \cdot CQ = \frac{1}{2} \cdot a \sin \lambda \cdot a \cos \lambda = \frac{1}{4} a^2 (2 \sin \lambda \cos \lambda) = \frac{1}{4} a^2 \sin 2 \lambda$ which after substituting $h = a \sin 2\lambda$ from above, $T_1 = \frac{1}{4} ah$ .

By a similar argument, the area of $\triangle ADE$ is $T_2 = \frac{1}{4} (a + 2)h$ .

The area of hexagon $ABQCDP$ is the area of the trapezoid minus the two right triangles, or $A_{ABQCDP} = A_{\text{trap}} - T_1 - T_2 = \frac{1}{8}(a + 19)$ , so that:

$(a + 10)h - \frac{1}{4} ah - \frac{1}{4} (a + 2)h = \frac{1}{8}(a + 19)$

The two equations solve to $h = \cfrac{1}{4}$ and $a = \cfrac{-\sqrt{15} + \sqrt{241}}{\sqrt{15}}$ for $a > 0$ .

Therefore, $\alpha = 15$ , $\beta = 241$ , and $\alpha + \beta = \boxed{256}$ .