What is the area of the green triangle?

Let the interior point be $P$ .

Extend the sides $BC,DE,FA$ so that they meet and form an equilateral triangle. By Viviani's theorem , the sum of the perpendicular distances from $P$ to each of these sides is the same no matter where $P$ is inside the hexagon.

The same is true if we extend the sides $AB,CD,EF$ , and because the triangles have the same size, this constant is the same.

So the pink area is exactly half the area of the hexagon; the area of the hexagon is $32$ .

Now consider pairs of triangles whose bases are parallel (for example, $PEF$ and $PBC$ ). The sum of their heights is always the same (the distance between two parallel sides of the hexagon). So the sum of their areas is the same for each pair. Since there are three such pairs, and they fill the hexagon, each pair has a total area of $\frac{32}{3}$ .

So the green triangle has area $\frac{32}{3}-8=\boxed{\frac83}$ .

Let the side length of the regular hexagon be $a$ and its center be the origin $O(0,0)$ of the $xy$ -plane. Then $A \left(-\frac a2, \frac {\sqrt 3a}2 \right)$ , $B \left(\frac a2, \frac {\sqrt 3a}2 \right)$ , $C(a,0)$ , $D \left(\frac a2, -\frac {\sqrt 3a}2 \right)$ , $E \left(-\frac a2, - \frac {\sqrt 3a}2 \right)$ , and $F(-a,0)$ . Let the point inside the hexagon be $P(x,y)$ . We can find area using the coordinates. For $\triangle PBC$ :

$\begin{aligned} \frac 12 \left(\frac a2 - x \right) \left(\frac {\sqrt 3a}2 + y \right) + \frac 12 \left(a - \frac a2 \right) \left(0 - \frac {\sqrt 3a}2 \right) + \frac 12 \left(x - a \right) \left(y - 0 \right) & = 8 \\ \frac {\sqrt 3a^2}4 - \frac {\sqrt 3ax}2 + \frac {ay}2 - xy + \frac {\sqrt 3a^2}4 + xy - ay & = 16 \\ \implies \frac {\sqrt 3a^2}2 - \frac {\sqrt 3ax}2 - \frac {ay}2 & = 16 & ...(1) \end{aligned}$

For $\triangle PDE$ (just half base times height):

$\begin{aligned} \frac a2 \left(y+\frac {\sqrt 3a}2 \right) & = 5 \\ \implies \frac {\sqrt 3a^2}2 + ay & = 10 & ...(2)\end{aligned}$

For $\triangle PFA$ :

$\begin{aligned} \frac 12 \left(-a - x \right) \left(0 + y \right) + \frac 12 \left(- \frac a2 + a \right) \left(\frac {\sqrt 3a}2 + 0 \right) + \frac 12 \left(x + \frac a2 \right) \left(y + \frac {\sqrt 3 a}2 \right) & = 3 \\ - ay - xy + \frac {\sqrt 3a^2}4 + xy + \frac {ay}2 + \frac {\sqrt 3ax}2 + \frac {\sqrt 3 a^2}4 & = 6 \\ \implies \frac {\sqrt 3a^2}2 + \frac {\sqrt 3ax}2 - \frac {ay}2 & = 6 & ...(3) \end{aligned}$

From $\dfrac {(1)+(3)}2$ : $\implies \dfrac {\sqrt 3a^2}2 - \dfrac {ay} 2 = 11 \ \ ...(4)$ . From $(2)-(4)$ : $\implies \dfrac {3ay}2 = - 1 \implies \dfrac {ay}2 = - \dfrac 13 \ \ ...(5)$

For $\triangle PEF$ (let its area be $A$ ):

$\begin{aligned} A & = \frac 12 \left(-\frac a2 - x\right) \left(-\frac {\sqrt 3a}2 + y \right) + \frac 12 \left(-a + \frac a2\right)\left(0-\frac {\sqrt 3a}2 \right) + \frac 12 \left(x + a \right) \left(y+0\right) \\ 2A & = \frac {\sqrt 3a^2}4 + \frac {\sqrt 3 ax}2 - \frac {ay}2 - xy + \frac {\sqrt 3a^2}4 + xy + ay \\ & = \frac {\sqrt 3a^2}2 + \frac {\sqrt 3 ax}2 + \frac {ay}2 \\ & = \blue{\frac {\sqrt 3a^2}2 + \frac {\sqrt 3 ax}2 - \frac {ay}2} + 2 \times \red{\frac {ay}2} \quad \quad \small \implies \blue{(3)} + 2 \times \red {(5)} \\ & = \blue 6 \red{- \frac 23} = \frac {16}3 \\ \implies A & = \frac 83 \end{aligned}$

Therefore $a+b = 8+3 = \boxed{11}$ .