Dynamic Geometry: P137

The Euler line of a triangle $\triangle ABC$ passes through its orthocenter $(H)$ , centroid $(G)$ and circumcenter $(O)$ .

Also, point $G$ , divides line segment $OH$ in the ratio $1:2$ $\implies \dfrac{{OG}}{{GH}}=\dfrac{1}{2}$

Let $M$ be the midpoint of $BC$ and $D$ be the foot of the perpendicular from $A$ to $BC$ . So, $MB=MC=2\;$ and $\;OMDH$ is a rectangle.

In $\triangle OMC$ , $\begin{aligned} OM^2+MC^2&=OC^2 \\ OM^2+4&=R^2 \\ \therefore \; OM=HD&=\sqrt{R^2-4} \end{aligned}$ Since $\triangle OGM\sim \triangle HGA$ , $\begin{aligned} \frac{OG}{GH}=\frac{OM}{AH} \implies \frac{1}{2}=\frac{\sqrt{R^2-4}}{AH}\implies AH&=2\sqrt{R^2-4} \end{aligned}$ Since the length of the altitude $AD$ is $3$ , $\;AD=AH+HD=3\sqrt{R^2-4}=3\implies \boxed{R=\sqrt{5}}$

Let the triangle be $ABC$ , where $AB=4$ is the horizontal base. Let the midpoint of $AB$ , $M$ be the origin $(0,0)$ of the $xy$ -plane. Then $A=(-2,0)$ , $B=(2,0)$ , and $C=(u,3)$ . Then the centroid of $\triangle ABC$ ,

$G = \left(\frac {-2+u+2}3. \frac 33 \right) = \left(\frac u3, 1 \right)$

Therefore $G$ is always on $y=1$ . Since the circumcenter $O$ is the meeting point of perpendicular side bisectors, it is always on the $x=0$ , the $y$ -axis. As $G$ and $O$ are on the Euler line, the Euler line is parallel to $AB$ , when $O$ is on $y=1$ that is $O=(0,1)$ . Then the circumradius of $\triangle ABC$ .

$R = OB = \sqrt{(2-0)^2+(0-1)^2} = \sqrt 5$

The required answer $q = \boxed 5$ .