Isosceles Centers

Place the two base vertices of the isosceles triangle at $(-3, 0)$ and $(3, 0)$ , so that the third point is at $(0, h)$ .

By the Pythagorean Theorem, the two congruent sides have a length of $\sqrt{h^2 + 9}$ .

The radius of the incircle will be $r = \cfrac{A}{s} = \cfrac{6h}{6 + 2\sqrt{h^2 + 9}} = \cfrac{3\sqrt{h^2 + 9} - 9}{h}$ , so the coordinates of the incenter are $(0, \cfrac{3\sqrt{h^2 + 9} - 9}{h})$ .

The coordinates of the barycenter (or centroid) will be $(0, \cfrac{h}{3})$ .

The slope of the left side of the triangle is $\cfrac{h}{3}$ , so the altitude through that side through vertex $(3, 0)$ will have an equation of $y = -\cfrac{3}{h}(x - 3)$ , which intersects the vertical altitude of $x = 0$ at the orthocenter $(0, \cfrac{9}{h})$ .

Since the three points are collinear, if the orthocenter and barycenter are equidistant from its incenter, then the incenter is the midpoint of the orthocenter and barycenter, so $\cfrac{9}{h} + \cfrac{h}{3} = 2 \cdot \cfrac{3\sqrt{h^2 + 9} - 9}{h}$ , which solves to $h = 3\sqrt{3}$ or $h = 3\sqrt{15}$ for $h > 0$ .

If $h = 3\sqrt{3}$ then the incenter, barycenter, and orthocenter are all at $(0, \sqrt{3})$ (and the triangle is an equilateral triangle) which would make the distance between the points $0$ .

If $h = 3\sqrt{15}$ then the incenter is at $(0, \cfrac{3\sqrt{15}}{5})$ , the barycenter is at $(0, \sqrt{15})$ , and the orthocenter is at $(0, \cfrac{\sqrt{15}}{5})$ .

Therefore, the distance between the orthocenter and the barycenter is $\sqrt{15} - \cfrac{\sqrt{15}}{5} = \cfrac{4\sqrt{15}}{5} = \sqrt{\cfrac{48}{5}}$ , so $p = 48$ , $q = 5$ , and $p + q = \boxed{53}$ .