Dynamic Geometry: P95

Let the center of the unit semicircle be $O(0,0)$ , the origin of the $xy$ -plane, and its diameter $AB$ , the center of the purple circle be $P$ , $\angle POB = \theta$ , and $t = \tan \frac \theta 2$ . Also let an arbitrary point on the locus or the tangent point of the cyan semicircle and orange circle be $T_1(x,y)$ . From the calculations in Dynamic Geometry: P80 Series , we have:

$\begin{cases} x = \dfrac {13-t^2}{t^2+4t+13} \\ y = \dfrac {6t}{t^2+4t+13} \end{cases}$

Then the area under the locus is given by

$\begin{aligned} A & = 2 \int_0^{t=\sqrt{13}} x \ \ce dy \\ & = 2 \int_0^{t=\sqrt{13}} \frac {13-t^2}{t^2+4t+13} \ce d \left(\frac {6t}{t^2+4t+13} \right) \\ & = 2 \int_0^{\sqrt{13}} \frac {13-t^2}{t^2+4t+13} \cdot \frac {6(13-t^2)}{(t^2+4t+13)^2} \ce dt \\ & = 12 \int_0^{\sqrt{13}} \frac {(13-t^2)^2}{(t^2+4t+13)^3} \ce dt \\ & = \frac {13}9 \tan^{-1} \frac 32 - \frac 23 \end{aligned}$

Therefore $\sqrt[3]{p+q+m+n} = \sqrt[3]{13+9+3+2} = \boxed 3$ .