THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS

The required minimum is $5\sqrt{2}/6$: the circle of radius $5\sqrt{2}/6$, centered at $(5/6,5/6)$ passes through exactly three lattice points, namely, $(0,0)$, $(1,2)$ and $(2,1)$.

Next, consider a circle $\omega$ of radius at most $5\sqrt{2}/6$ passing through exactly three lattice points. We may and will assume that one of these lattice points is at the origin; let $(m,n)$ and $(p,q)$ be the other two. Since the diameter of $\omega$ does not exceed $5\sqrt{2}/3$, the integers $m^2+n^2$, $p^2+q^2$ and $(m-p{)}^2+(n-q{)}^2$ are all at most $5$.

If $m=0$, then $p\ne 0$ by non-collinearity, and $n\ne 0$ must be even and $q=n/2$, for otherwise $(p,n-q)$ would be a fourth lattice point on $\omega$. Hence $n=\pm 2$ and $q=\pm 1$ (corresponding signs), and $|p|$ is either $1$ or $2$. The case $|p|=1$ is ruled out by the fact that $(-p,q)$ would be a fourth lattice point on $\omega$, and the case $|p|=2$ is ruled out by the fact that the radius of $\omega$ would be $5/4>5\sqrt{2}/6$. Consequently, $m\ne 0$; similarly, $n\ne 0$, $p\ne 0$, $q\ne 0$, $m\ne p$, and $n\ne q$. It then follows that $(m,n)$ and $(p,q)$ are consecutive vertices of the hexagon with vertices at $(2,1)$, $(1,2)$, $(-1,1)$, $(-2,-1)$, $(-1,-2)$, $(1,-1)$ or of its reflection in one of the coordinate axes. The radius of $\omega$ is $5\sqrt{2}/6$ whatsoever the case.