Belarusian Mathematical Olympiad

Answers: $R = 0.5\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}$.

Let $A(l; l^2)$, $B(m; m^2)$, $C(n; n^2)$ be coordinates of the vertices of the triangle $ABC$. Then
$$AB = \sqrt{(m-l)^2 + (m^2-l^2)^2} = |m-l|\sqrt{1+(m+l)^2} = |m-l|\sqrt{1+4c^2},$$
since $c = 0.5(m + l)$.
Similarly, $BC = |m - n|\sqrt{1 + 4a^2}$ and $AC = |n - l|\sqrt{1 + 4b^2}$.

Further, it is easy to calculate that $S(ABC) = 0.5|(m - l)(l - n)(n - m)|$.

Now we find the circumradius using the formula
$$$$
R = $\text{AB BC}$ CA}{4S(ABC)} = $\frac{1}{2} \sqrt{(1}$ + 4a^2)(1 + 4b^2)(1 + 4c^2)}.