jmc

In hexagon $ABCDEF$, let $AB=BC=CD=3$ and let $DE=EF=FA=5$. Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF=\angle BEF=60^{\circ }$. Similarly, $\angle CBE=\angle CFE=60^{\circ }$. Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$, $Q$ that of $\overline{BE}$ and $\overline{AD}$, and $R$ that of $\overline{CF}$ and $\overline{AD}$. Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral.

Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$. Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, $$\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.$$ It follows that $$\frac{\frac{AD-PQ}{2}}{PQ+5}=\frac{3}{5}\text{mbox}and\frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.$$ Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}$.