An Inscribed Hexagon

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

$(BF)^{2} = 2^{2} + 4^{2} - 2*2*4*\cos(\frac{2\pi}{3}) \Longrightarrow BF = 2\sqrt{7}$ .

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 $\sin(\alpha) = \frac{\sqrt{21}}{14}$ and $\sin(\beta) = \frac{\sqrt{21}}{7}$ .

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

$\frac{sin(\beta)}{4} = \frac{\sin(3\alpha)}{AD} \Longrightarrow AD = \frac{4\sin(3\alpha)}{\sin(\beta)}$ .

Now $\sin(3\alpha) = 3\sin(\alpha) - 4\sin^{3}(\alpha) = \frac{9\sqrt{21}}{49}$ ,

and so $AD = \dfrac{\frac{36\sqrt{21}}{49}}{\frac{\sqrt{21}}{7}} = \dfrac{36}{7}$ .

Thus $m + n = 36 + 7 = \boxed{43}$ .

Let $FE \cap DC = M, ED \cap BC = N, AB \cap ED = L.$

Let $LN = y$ and $BL = z$

Let $EM = a, MD = b$

By the Power of a Point, we have

$a(a+4) = b(b+2)$ $\hspace{35pt}$ $(1)$

Notice that $\Delta EDM \sim \Delta NCD.$

So we can have

$DN = \dfrac{8}{a}, CN = \dfrac{2b}{a}$

By the Power of a Point, we have

$\dfrac{8}{a}(\dfrac{8}{a} + 4) = \dfrac{2b}{a}(\dfrac{2b}{a} + 2)$ $\hspace{35pt}$ $(2)$

Solving for $a$ and $b$ we get

$a = \dfrac{32}{5}$ and $b = \dfrac{28}{5}$

So $BN = \dfrac{15}{4}$ and $DN = \dfrac{5}{4}$

Notice that $\Delta ADL \sim \Delta BNL.$

So we have $\dfrac{z}{z+2} = \dfrac{y}{y+\dfrac{5}{4}}$ $\hspace{35pt}$ $(3)$

Also by the Power of a Point, we have

$(y+ \dfrac{5}{4})(y+ \dfrac{21}{4}) = z(z+2)$ $\hspace{35pt}$ $(4)$

Solving for $y$ and $z$ we get

$y = \dfrac{175}{52}$ and $x = \dfrac{70}{13}$

We then achieve $AD = \dfrac{36}{7}$

$\therefore 36 + 7 = \boxed{43}$ .

Length of the chord shared by two trapeziums each having three equal sides $a$ & $b$ & being inscribed in a circle is given by the following general formula

$\boxed{AD=\frac{3ab(a+b)}{a^2+b^2+ab}}$

As per question, setting $a=2$ & $b=4$ in above general formula we get

$AD=\frac{3\cdot 2\cdot 4(2+4)}{2^2+4^2+2\cdot 4}=\frac{36}{7}=\frac mn$

$\therefore m+n=36+7=43$

$Let ~2\alpha ~be ~the ~\angle ~subtended ~by ~the ~4 ~unit ~side~at ~the ~center\\ and ~2\beta ~by ~2 ~unit.\\ With ~R ~ circumradius ~~~~ RSin\alpha=4 ~~ and~ ~ RSin\beta=2,\\ we ~have ~~~6\alpha+6\beta=360 ~~ \implies ~\alpha+\beta=60.\\ \therefore ~ \dfrac{RSin\alpha}{RSin\beta} =\dfrac 4 2.\\ \implies ~\dfrac{Sin\alpha}{Sin(60-\alpha)}=2.\\ Solving ~Cot\alpha=\dfrac2{\sqrt3} ~~\implies ~\color{#3D99F6}{Cos2\alpha=\dfrac 1 7}. \\ \text{AD=Projection of AF on AD+ Projection of FE on AD+ Projection of ED on AD.}\\\ AD=4Sin\{180-2\alpha-90\}+4+ 4Sin\{180-2\alpha-90\}.\\ AD=4(1+2\color{#3D99F6}{Cos2\alpha}) = \dfrac{36} 7=\dfrac m n.\\ m+n= ~~~~\Huge \color{#D61F06}{43}$

Let O be the center of the circle and r the radius. Let 2t be the central angle subtended by the chord lengths of 4. Then sin(t) = 2/r. <AOF = <FOE = <EOD = 2t, so < AOD = 360 - 6t. Then <AOB = <BOC = <COD = 120 - 2t. In triangle AOB, by the Law of Cosines, 4 = 2r^2 - 2r^2 cos(120 - 2t).Cos(120 - 2t) = [sqrt(3)/2] sin(2t) - (1/2) cos(2t). Substituting r =2/sin(t), and simplifying, we derive cos^2(t) = 4/7. From triangle AOD, by the law of cosines, (AD)^2 = 2r^2(1 - cos(6t)). cos(t) = sqrt(4/7) = .755928946, t = 40.89339465, 6t = 245.3603679, cos(6t) = -.416909621. Substituting, (AD)^2 = (56/3) (1.416909621) = 26.44897959, AD = 5.14285714.... . We recognize the decimal part as 1/7. So AD = 36/7, ang m + n = 43. Ed Gray