Dynamic Geometry: P106

Let the center of the circle be $O$ , the triangle $ABC$ , where $AB$ is the horizontal chord, the center and radius of the incircle at any moment be $P$ and $r$ , and $PN$ and $OM$ be perpendicular to $AB$ . Since the heights of two circular segments are $4$ and $9$ , the diameter of the circle is $13$ , hence its radius $6.5$ . Then $AM = BM = 6$ and $OM = 2.5$ . The smaller $\angle AOB = 2 \tan^{-1} \frac {12}5$ and the larger $\angle AOB = 2\pi - 2 \tan^{-1} \frac {12}5$ . When $C$ is on top, $\angle ACB = \frac 12 \left(2\pi - 2 \tan^{-1} \frac {12}5 \right) = \pi - \tan^{-1} \frac {12}5$ . Let $\angle CAB = \theta$ , then $\angle CBA = \tan^{-1} \frac {12}5 - \theta$ , and

$\begin{aligned} AN + NB & = AB \\ r \cot \frac \theta 2 + r \cot \left(\frac 12 \tan^{-1} \frac {12}5 - \frac \theta 2 \right) & = 12 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + \frac {1+\frac 23t}{\frac 23 -t} \cdot r & = 12 \\ \implies r & = \frac {12}{\frac 1t + \frac {3+2t}{2-3t}} = \frac {6t(2-3t)}{1+t^2} \end{aligned}$

Let the midpoint of $AB$ be the origin of the $xy$ -plane and an arbitrary point on the upper locus be $P(x,y)$ . Then

$\begin{cases} y = r = \dfrac {6t(2-3t)}{1+t^2} \\ x = \dfrac rt - 6 = \dfrac {6(2-3t)}{1+t^2} - 6 \end{cases}$

Then the area under the locus is

$\begin{aligned} A_1 & = \int_{-6}^6 y \ \ce dx \\ & = \int_{-6}^6 \frac {6t(2-3t)}{1+t^2} \ce d \left(\dfrac {6(2-3t)}{1+t^2} - 6 \right) \\ & = \int_0^\frac 23 \frac {36t(2-3t)(3+4t-3t^2)}{(1+t^2)^3} \ce dt \\ & = 117 \tan^{-1} \frac 23 - 54 \end{aligned}$

Similarly, when $C$ is at the bottom, $\angle ACB$ is half the smaller $\angle AOB$ , $\implies \angle ACB = \tan^{-1} \dfrac {12}5$ , $r = y = \dfrac {4t(3-2t)}{1+t^2}$ , $x = \dfrac {4(3-2t)}{1+t^2} - 6$ , and the area under the locus is

$\begin{aligned} A_2 & = \int_{-6}^6 y \ \ce dx \\ & = \int_{-6}^6 \frac {4t(3-2t)}{1+t^2} \ce d \left(\dfrac {4(3-2t)}{1+t^2} - 6 \right) \\ & = \int_0^\frac 32 \frac {36t(2-3t)(3+4t-3t^2)}{(1+t^2)^3} \ce dt \\ & = 52 \tan^{-1} \frac 32 - 24 \end{aligned}$

Therefore $A_1 + A_2 = 117 \tan^{-1} \frac 23 + 52 \tan^{-1} \frac 32 - 78$ and $p+q+m+n+1 = 117+52+2+3+78 = \boxed{252}$