Dynamic Geometry: P107

Let the variable triangle be $ABC$ , where $AB=4$ and $\angle B = \tan^{-1} \frac 34$ , the purple triangle be $STU$ , and the locus of the orthocenter (the green line) intersects $BC$ at $N$ .

Since the orthocenter is the meeting point of altitudes from the three vertices, $AN$ must be perpendicular to $BC$ . Also since the circumcenter is the point where the perpendicular bisectors of the three sides meet, the locus of circumcenter (the red line) is the perpendicular bisector of the fixed side $AB$ .

Let the midpoint of $AB$ be $M(0,0)$ , the origin of the $xy$ -plane, and the variable vertex $C$ be $(x,y)$ . Then the line $BC$ is given by $\dfrac {y-0}{x-2} = - \dfrac 34 \implies 4y+3x = 6$ and the centroid $G(x_g,y_g)$ :

$\begin{cases} x_g = \dfrac {-2+x+2}3 = \dfrac x3 \\ y_g = \dfrac {0+y+0}3 = \dfrac y3 \end{cases} \implies 4y_g+3x_g = \frac {4y+3x}3 = \frac 63 = 2$

Therefore the equation of the locus of centroid (the orange line) is $4y + 3x = 2$ , which is parallel to $BC$ . Let the orange line intersect $AB$ and $CA$ at $D$ and $E$ respectively. When $y=0$ , $x = MD = \frac 23$ . Since $DU$ is parallel to $BC$ , $\triangle ADU$ and $\triangle ABN$ are similar and are $3$ - $4$ - $5$ right triangles. Similarly, $\triangle MDT$ and $\triangle STU$ are also $3$ - $4$ - $5$ right triangles. Then we have:

$UT = UD-TD = \frac 45 \cdot AD - \frac 54 \cdot MD = \frac 45 \left(2+\frac 23 \right) - \frac 54 \cdot \frac 23 = \frac {13}{10}$

Then $US = \dfrac 43 \cdot UT$ and the area of $\triangle STU$ , $[STU] = \dfrac {US \cdot UT}2 = \dfrac 12 \cdot \dfrac 43 \cdot \dfrac {13^2}{10^2} = \dfrac {169}{150}$ . And $p-q = 169-150 = \boxed{19}$ .