Segments of a Circle

Let $\angle COD = \alpha$ and $\angle COB = \angle BOA = \beta$ . Then $\alpha + 2\beta = 180^\circ$ . Let the radius of the semicircle be $r$ . By cosine rule we have $2r^2(1-\cos \alpha) = 14^2$ and $2r^2 (1-\cos \beta) = 6^2$ . Therefore,

$\begin{aligned} \frac {1-\cos \alpha}{1-\cos \beta} & = \frac {14^2}{6^2} \\ 9 - 9 \cos \alpha & = 49 - 49 \cos \beta & \small \blue{\text{Putting }\alpha = 180^\circ - 2\beta} \\ - 9 \cos (180^\circ - 2\beta) + 49 \cos \beta - 40 & = 0 & \small \blue{\text{Note that }\cos (180^\circ -\theta)= - \cos \theta} \\ 9 \cos (2\beta) + 49 \cos \beta - 40 & = 0 & \small \blue{\text{and }\cos (2\theta)= 2\cos^2 \theta-1} \\ 18\cos^2 \beta + 49 \cos \beta - 49 & = 0 \\ \implies \cos \beta & = \frac 79 & \small \blue{\implies \sin \beta = \frac {4\sqrt 2}9} \\ \cos \alpha & = - \cos 2 \beta \\ & = \sin^2 \beta - \cos^2 \beta \\ & = - \frac {17}{81} & \small \blue{\implies \sin \alpha = \frac {56\sqrt 2}{81}} \end{aligned}$

From $2r^2(1-\cos \beta) = 6^2 \implies r^2 = 81$ . The area of the quadrilateral $A_{ABCD} = \dfrac 12r^2 \sin \alpha + r^2 \sin \beta = 64\sqrt 2$ and the area of the shaded regions $A = \dfrac 12 \pi r^2 - A_{ABCD} = \dfrac {81}2 \pi - 64\sqrt 2$ . Therefore $a+b+c = 81+2+64 = \boxed{147}$ .

Using the above diagram $2(180 - 2\theta) + \lambda = 180 \implies \lambda = 4\theta - 180 \implies$

$m = 180 - 2\theta$ .

For $\triangle{AOB}$ using the law of cosines we have:

$36 = 2r^2(1 - \cos(180 - 2\theta)) \implies 18 = r^2(1 + \cos(2\theta)) \implies r^2 = \dfrac{18}{1 + \cos(2\theta)}$

and

for $\triangle{COD}$ using the law of cosines we have:

$196 - 2r^2(1 + \cos(4\theta)) \implies 49 = r^2(\cos^2(2\theta)) \implies$

$49 + 49\cos(2\theta) = 18\cos^2(2\theta) \implies 18\cos^2(\theta) - 49\cos(2\theta) - 49 = 0 \implies$

$\cos(2\theta) = \dfrac{49 \pm 77}{2}$ and $|\cos(u)| \leq 1 \implies \cos(2\theta) = -\dfrac{7}{9}$

$\implies r^2 = \dfrac{18}{1 - \dfrac{7}{9}} = 81 \implies r = 9.$

Let $h_{1}$ be the height of $\triangle{AOB} \implies h_{1} = \sqrt{81 - 9} = \sqrt{72} = 6\sqrt{2} \implies$

$A_{1} = 2*A_{\triangle{AOB}} = 36\sqrt{2}$

Let $h_{2}$ be the height of $\triangle{COD} \implies h_{2} = \sqrt{81 -49} = \sqrt{32} = 4\sqrt{2} \implies$

$A_{2} = A_{\triangle{COD}} = 28\sqrt{2}$

$\implies$ Area of quadrilateral $ABCD$ is $A_{ABCD} = A_{1} + A_{2} = 64\sqrt{2} \implies$

The desired area is $A = \dfrac{81}{2}\pi - 64\sqrt{2} = \dfrac{a}{c}\pi - b\sqrt{c} \implies a + b + c = \boxed{147}$ .