Inscribed Hexagon

Please refer to the diagram in this posting . The only difference for this question is that $AF = FE = ED = 5$ and $AB = BC = CD = 3$ . I provided a solution Guiseppi's question already, so I will adapt that solution to the parameters given in this question.

Referring to the diagram, and letting the center be $O$ , let $\angle AOB = 2\alpha$ and $\angle AOF = 2\beta$ .

Now by symmetry $2\alpha + 2\beta = \frac{2\pi}{3}$ . By the Cosine Law we then have that

$(BF)^{2} = 3^{2} + 5^{2} - 2*3*5*\cos(\frac{2\pi}{3}) \Longrightarrow BF = 7$ .

Next, note that $\angle AFB = \frac{1}{2} \angle AOB = \alpha$ and $\angle ABF = \frac{1}{2} \angle AOF = \beta$ . So, using the Sine Law on $\Delta ABF$ we find that $\sin(\alpha) = \frac{3\sqrt{3}}{14}$ and $\sin(\beta) = \frac{5\sqrt{3}}{14}$ .

Now, we see that $\angle AFD = \frac{1}{2} \angle AOD = 3\alpha$ and $\angle ADF = \frac{1}{2} \angle AOF = \beta$ . So using the Sine Law on $\Delta ADF$ we have that

$\frac{sin(\beta)}{5} = \frac{\sin(3\alpha)}{AD} \Longrightarrow AD = \frac{5\sin(3\alpha)}{\sin(\beta)}$ .

Now $\sin(3\alpha) = 3\sin(\alpha) - 4\sin^{3}(\alpha) = \frac{180\sqrt{3}}{343}$ ,

and so $AD = \dfrac{\frac{5*180\sqrt{3}}{343}}{\frac{5\sqrt{3}}{14}} = \dfrac{360}{49}$ .

Thus $m + n = 360 + 49 = \boxed{409}$ .