Solution:

Let $O$ and $H$ be the circumcenter and orthocenter of $ABC$. Recall that
$$\begin{aligned} \angle BOC & = 2\angle A, \\ \angle BHC & = 180^\circ - \angle A, \\ \angle BIC & = 90^\circ + \frac{1}{2} \angle A. \end{aligned}$$
As $ABC$ is acute, $A$, $I$, $O$, $H$ all lie on the same side of $BC$.

- If $\angle A > 60^\circ$, then $\angle BOC > \angle BIC$, so $O$ lies inside $(BIC)$.
- If $\angle A < 60^\circ$, then $\angle BHC > \angle BIC$, so $H$ lies inside $(BIC)$.
- If $\angle A = 60^\circ$, then $O$ and $H$ are on $(BIC)$.

In all cases, two intersections are guaranteed.