Answer: 59

Solution:

The four sticks are chords of the unit circle subtending fixed arcs $\alpha,$ $\beta,$ $\gamma,$ $\delta$ with $\alpha + \beta + \gamma + \delta = 360^\circ.$ The three quadrilaterals are the three distinct cyclic orders of the sides: say $A$ has arcs in order $\alpha, \beta, \gamma, \delta;$ then $B$ (order $\alpha, \gamma, \beta, \delta$) and $C$ (order $\alpha, \beta, \delta, \gamma$) are the other two. The angle between the diagonals of a cyclic quadrilateral is half the sum of the arcs subtended by either pair of opposite sides, so $\sin\varphi_B = \sin\frac{\alpha + \beta}{2}$ and $\sin\varphi_C = \sin\frac{\alpha + \delta}{2} = \sin\frac{\beta + \gamma}{2}.$

In a circle of radius $1,$ a chord spanning an arc $\theta$ has length $2\sin\frac{\theta}{2}.$ The diagonals of $A$ span the arcs $\alpha + \beta$ and $\beta + \gamma,$ so their lengths are $2\sin\varphi_B$ and $2\sin\varphi_C.$ Hence $$\begin{aligned} &K = \frac{1}{2}\,d_1 d_2 \sin\varphi_A \\ &= 2\sin\varphi_A \sin\varphi_B \sin\varphi_C, \end{aligned}$$ a formula symmetric in the three quadrilaterals, which is why all three areas are equal.

Therefore $K = 2 \cdot \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{6}{7} = \frac{24}{35},$ and $m + n = 24 + 35 = 59.$