Dynamic Geometry: P117

From the compiled calculations in Dynamic Geometry: P96 Series , the radius of the circle is $6.5$ . Let the center of the circle be $O$ , the yellow dividing chord be $AB$ , the upper moving and lower moving points be $C$ and $C'$ respectively.

Since the Euler line of a triangle passes through the circumcenter, the two Euler lines intersect at $O$ , the shared circumcenter of the two triangles. From the previous problem of the series, Dynamic Geometry: P101 , we concluded that the locus of the orthocenter of the inscribed triangle is a circle with the same radius of $6.5$ and its center $5$ above the inscribing circle such that the orthocenters of the upper and lower triangles are $H$ and $H'$ respectively. where $CH=C'H'=5$ .

Since the Euler line also passes through the orthocenter. $OH$ and $OH'$ are along the respective Euler lines. Let $M$ be the midpoint of $CC'$ . Since $\triangle COC'$ is isosceles, $OM$ is perpendicular to $CC'$ . Let $\angle COM=C'OM = \theta$ . When the Euler lines or $OH$ and $OH'$ are perpendicular to each other,

$\begin{aligned} \angle HOM & = \angle OH'M \\ \tan \angle HOM & = \tan \angle OH'M \\ \frac {HM}{OM} & = \frac {OM}{H'M} \\ \frac {6.5 \sin \theta + 5}{6.5\cos \theta} & = \frac {6.5 \cos \theta}{6.5 \sin \theta - 5} \\ \frac {169}4 \sin^2 \theta - 25 & = \frac {169}4 \cos^2 \theta \\ 338 \sin^2 \theta & = 269 \\ \implies \sin \theta & = \frac 1{13}\sqrt{\frac {269}2} \end{aligned}$

The distance between the two moving points $CC' = 2 \cdot 6.5 \sin \theta = \sqrt{\dfrac {269}2}$ . Therefore $p+q=269+2 = \boxed{271}$ .