A Complicated Area

Let us first find the length of $CM$ . From cosine rule ,

$\begin{cases} \begin{aligned} 55 & = 7^2 + 8^2 - 2\cdot 7 \cdot 8 \cos B & ...(1) \\ \overline{CM}^2 & = 7^2 + 4^2 - 2 \cdot 7 \cdot 4 \cos B & ...(2) \end{aligned} \end{cases}$

For $2(2) - (1): \ \ 2 \overline{CM}^2 - 55 = 7^2 + 2(4^2) - 8^2 \implies \overline{CM} = 6$ . Then $\overline{CD} = \overline{CM} - \overline{DM} = 6 - 4 = 2$ . To find $\overline{CE}$ , we extend $CM$ to meet the circle at $F$ . Then by two secant theorem , we have:

$\begin{aligned} \overline{CA} \cdot \overline{CE} & = \overline{CF} \cdot \overline{CD} \\ \sqrt{55} \cdot \overline{CE} & = 10 \cdot 2 \\ \implies \overline{CE} & = \frac {20}{\sqrt{55}} \end{aligned}$

We note that the areas of $\triangle BCM$ and $\triangle ACM$ are the same and by Heron's formula , for $s = \dfrac {7+6+4}2 = 8.5$ , $[BCM] = \sqrt{8.5 \cdot 1.5 \cdot 2.5 \cdot 4.5} = \dfrac {3\sqrt{255}}4$ . Also that $\dfrac {\overline{CA}\cdot \overline{CM}\sin \angle MCA}2 = \dfrac {\sqrt{55}\cdot 6\sin \angle MCA}2 = \dfrac {3\sqrt{255}}4$ , $\implies \sin \angle MCA = \dfrac 14 \sqrt{\dfrac {51}{11}}$ .

Then the area of quadrilateral $AMDE$ is:

$\begin{aligned} [AMDE] & = [ACM] - [CDE] \\ & = \frac 12 \left(\overline{CM} \cdot \overline{CA} - \overline{CD} \cdot \overline{CE} \right) \sin \angle MCA \\ & = \frac 12 \left(6\sqrt{55} - 2 \cdot \frac {20}{\sqrt{55}} \right) \dfrac 14 \sqrt{\dfrac {51}{11}} \\ & = \frac {29\sqrt{255}}{44} \end{aligned}$

Therefore $a+b+c = 29+255+44 = \boxed{328}$ .

Same start as Chew-Seong Cheong, but I constructed segment $ME$ to form two isosceles triangles, both of which we can calculate the area by $\dfrac{1}{2}bh$ .

Finding $DE$ is an application of Stewart's theorem for $\triangle CEM$ , and finding $EA$ is trivial.

Calculations are all left as an exercise to the reader, but the answer is still the same. $\boxed{328}$