, 2022

Solution:

If either $a = 0$ the problem statement is clearly true. Thus, assume that $a \neq 0$. Let the roots be $0, z_{1}, z_{2}, z_{3}$, and let the circle through these points be $C$. Note that we have
$$\begin{aligned} & \frac{3}{z_{1} + z_{2} + z_{3}} = -\frac{3}{a} \\ & \frac{\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}}{3} = -\frac{b}{3} \end{aligned}$$
Note that the map $z \rightarrow \frac{1}{z}$ maps $C$ to some line $L$. Thus, the second equation represents the average of three points on $L$, which must be a point on $L$, while the second equation represents the reciprocal of the centroid of $z_{1}, z_{2}, z_{3}$. Since this centroid doesn't lie on $C$, we must have its reciprocal doesn't lie on $L$. Thus, we have
$$-\frac{3}{a} \neq -\frac{b}{3} \Longrightarrow a b \neq 9$$