Answer: 11

Solution:

Center the hexagon at the origin with circumradius $1:$ $A = (1, 0),$ $B = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $C = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ and $F = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).$ Then $H = \left(0, \frac{\sqrt{3}}{2}\right)$ and $G = \left(\frac{3}{4}, \frac{\sqrt{3}}{4}\right).$ Rotation by $60^\circ$ permutes the six segments, so the smaller hexagon is regular and concentric, and the area ratio is the square of the ratio of distances from the center to a vertex.

One vertex is the intersection of $AH$ and $FG.$ Line $AH$ is $x + \frac{2}{\sqrt{3}}\,y = 1,$ and line $FG$ is $y = 3\sqrt{3}\,x - 2\sqrt{3}.$ Substituting gives $x + 6x - 4 = 1,$ so $x = \frac{5}{7}$ and $y = \frac{\sqrt{3}}{7}.$

That vertex has squared distance $\frac{25}{49} + \frac{3}{49} = \frac{4}{7}$ from the center, while $A$ is at distance $1.$ The ratio of areas is $\frac{4}{7},$ and $m + n = 4 + 7 = 11.$