Dynamic Geometry: P126

From the previous similar problem Dynamic Geometry: P107 , we find that the locus of the circumcenter $O$ of the variable $\triangle ABC$ is the red line which is the perpendicular bisector of $AB$ at $M$ , the locus of the orthocenter $H$ is the green line, and the locus of the centroid $G$ is the orange line which is parallel to the side $BC$ . If the centroid line intersect $AB$ and $CA$ at $D$ and $E$ , then $MD = \frac 23$ .

Since the Euler line passes through $O$ and $G$ , it is parallel to $BC$ only when $O$ is on the centroid line as shown in the figure (note that $H$ is also on that line as shown by the perfect animation of @Valentin Duringer ). That is the centroid line becomes the Euler line. When that happens, let the perpendicular bisector of $BC$ , which passes through $O$ , intersects $AB$ at $K$ and $BC$ at $F$ . Note that $\triangle BFK$ , $\triangle DKM$ , and $\triangle DOM$ are similar and are all $3$ - $4$ - $5$ right triangle. Since $MD = \frac 23$ , $OM = \frac 34 \cdot MD = \frac 12$ , $KM = \frac 34 \cdot OM = \frac 38$ , $KB = KM + MB = \frac {19}8$ , $FB = \frac 45 \cdot KB = \frac {19}{10}$ , and $BC = 2 \cdot FB = \frac {19}5$ .

Now note that $\triangle AED$ and $\triangle ABC$ are similar. Then

$\frac {DE}{BC} = \frac {AD}{AB} \implies DE = \frac {AD}{AB} \cdot BC = \frac {2+\frac 23}4 \cdot \frac {19}5 = \frac {38}{15}$

Therefore $p+q = 38+15 = \boxed{53}$ .