Dynamic Geometry: P116

Let the dividing yellow chord be $AB$ , and the upper and lower moving points be $C$ and $C'$ respectively. From the compilation of calculations in Dynamic Geometry: P96 Series , we know that $\angle ACB = \pi - \tan^{-1} \frac {12}5$ and $\angle AC'B = \tan^{-1} \frac {12}5$ . Let the orange angle $\angle CAB = \angle C'AB = \theta$ . Then $\angle CBA = \tan^{-1} \frac {12}5 - \theta$ and $\angle C'BA = \pi - \tan^{-1} \frac {12}5 - \theta$ .

The slope of the Euler line of $\triangle ABC$ is given by:

$\begin{aligned} m_e & = - \frac {m_{AB}m_{BC}+m_{BC}m_{CA}+m_{CA}m_{AB}+3}{m_{AB}+m_{BC}+m_{CA}} & \small \blue{\text{where }m_{XY} \text{ is the slope of side }XY.} \\ & = - \frac {m_{BC}m_{CA}+3}{m_{BC}+m_{CA}} & \small \blue{\text{Since }m_{AB}=0} \\ & = - \frac {-\tan \left(\tan^{-1}\frac {12}5-\theta \right) \tan \theta + 3}{-\tan \left(\tan^{-1}\frac {12}5-\theta \right) + \tan \theta} & \small \blue{\text{Let }t = \tan \theta} \\ & = - \frac {\frac {t-\frac {12}5}{1+\frac {12t}5}t+3}{\frac {t-\frac {12}5}{1+\frac {12t}5}+t} \\ & = - \frac {5t^2+24t+15}{12t^2 + 10t -12} \end{aligned}$

Similarly, the slope of the Euler line of $\triangle ABC'$ is:

$\begin{aligned} m'_e & = - \frac {m_{BC'}m_{C'A}+3}{m_{BC'}+m_{C'A}} \\ & = - \frac {\tan \left(\pi - \tan^{-1}\frac {12}5 - \theta\right)(-\tan \theta) + 3}{\tan \left(\pi - \tan^{-1}\frac {12}5 - \theta\right)-\tan \theta} \\ & = - \frac {\frac {\frac {12}5+t}{1-\frac {12t}5}t+3}{-\frac {\frac {12}5+t}{1-\frac {12t}5}-t} \\ & = - \frac {5t^2-24t+15}{12t^2 - 10t -12} \end{aligned}$

When $m_e = m'_e$ :

$\begin{aligned} \frac {5t^2+24t+15}{12t^2 + 10t -12} & = \frac {5t^2-24t+15}{12t^2 - 10t -12} \\ 60 t^4 + 238 t^3 - 120 t^2- 438 t -180 & = 60 t^4 - 238 t^3 - 120 t^2 + 438 t -180 \\ 238 t^3 & = 438t & \small \blue{\text{Since }t \ne 0} \\ 119 t^2 & = 219 \\ \implies t & = \tan \theta = \sqrt{\frac {219}{119}} \end{aligned}$

Therefore $\sqrt{p-q} = \sqrt{219-119} = \boxed {10}$ .