jmc

We place the diagram in the coordinate plane so that the origin coincides with the midpoint of the bottom edge of the hexagon. Using the fact that the hexagon has side length 1, we can determine its vertices.



By symmetry, the equation of the trajectory is of the form $y=a{x}^2+c.$ Then $$\begin{array}{rl}\frac{a}{4}+c& =\sqrt{3},\\ a+c& =\frac{\sqrt{3}}{2}.\end{array}$$Solving, we find $a=-\frac{2}{\sqrt{3}}$ and $c=\frac{7}{2\sqrt{3}},$ so the equation of the trajectory is $$y=-\frac{2}{\sqrt{3}}{x}^2+\frac{7}{2\sqrt{3}}.$$Setting $y=0,$ we get $$-\frac{2}{\sqrt{3}}{x}^2+\frac{7}{2\sqrt{3}}=0.$$Then $x^2=\frac{7}{4},$ so $x=\pm \frac{\sqrt{7}}{2}.$ Thus, the distance $AB$ is $\frac{\sqrt{7}}{2}-\left(-\frac{\sqrt{7}}{2}\right)=\boxed{\sqrt{7}}.$