HMMT February

Assume without loss of generality that the circumcenter is at the origin. By well known properties of the Euler line, the centroid $G$ is such that $O, G$, and $H$ are collinear, with $G$ in between $O$ and $H$, such that $G H=2 G O$. Thus, since $G=\frac{1}{3}(A+B+C)$, and we are assuming $O$ is the origin, we have $H=A+B+C$. This means that as long as $A, B$, and $C$ are integer points, $H$ will be as well. However, since $H$ needs to be distinct from $A, B$, and $C$, we must have $\triangle$ A B C not be a right triangle, since in right triangles, the orthocenter is the vertex where the right angle is. Now, if a circle centered at the origin has any integer points, it will have at least four integer points. (If it has a point of the form (a, 0)$, then it will also have$(-a, 0),(0, a)$, and$(0,-a)$. If it has a point of the form$(a, b)$, with$a, b $\neq$ 0, it will have each point of the form ( $\pm$ a, $\pm$ b)$.) But in any of these cases where there are only four points, any triangle which can be made from those points is a right triangle. Thus we need the circumcircle to contain at least eight lattice points. The smallest radius this occurs at is \sqrt{1^{2}+2^{2}}=\sqrt{5}$, which contains the eight points $( \pm 1, \pm 2)$ and $( \pm 2, \pm 1)$. We get at least one valid triangle with this circumradius: $$A=(-1,2), B=(1,2), C=(2,1)$$ The next valid circumradius is $\sqrt{1^{2}+3^{2}}=\sqrt{10}$ which has the valid triangle $A=(-1,3), B=(1,3), C=(3,1)$$