Obviously, the circumcenter of $\triangle ABC$ is the orthocenter of $\triangle A_{1} B_{1} C_{1}$. Points $A_{2}, B_{2}, C_{2}$ are the reflections of point $O$ over $B_{1} C_{1}$, $C_{1} A_{1}$, and $A_{1} B_{1}$, respectively.

(1) $\triangle ABC$ is an acute triangle.
As shown in Figure 6, draw lines through $A_{2}, B_{2}, C_{2}$ parallel to $B_{1} C_{1}$, $C_{1} A_{1}$, and $A_{1} B_{1}$, respectively, and let the triangle formed by their intersections be $\triangle A_{3} B_{3} C_{3}$. Then $\triangle A_{3} B_{3} C_{3}$ is similar to $\triangle A_{1} B_{1} C_{1}$ with respect to point $O$, with a similarity ratio of 2.
Thus, $O$ is the orthocenter of $\triangle A_{1} B_{1} C_{1}$
$\Leftrightarrow O$ is the orthocenter of $\triangle A_{3} B_{3} C_{3}$
$\Leftrightarrow O$ is the incenter of $\triangle A_{2} B_{2} C_{2}$.

(2) $\angle A$ is an obtuse angle.
As shown in Figure 7, let the projections of point $O$ onto $B_{1} C_{1}$, $C_{1} A_{1}$, and $A_{1} B_{1}$ be $A_{3}$, $B_{3}$, and $C_{3}$, respectively. Then $\triangle A_{3} B_{3} C_{3}$ is similar to $\triangle A_{2} B_{2} C_{2}$ with respect to point $O$, with a similarity ratio of $\frac{1}{2}$.
Thus, $O$ is the excenter of $\triangle A_{2} B_{2} C_{2}$ at $\angle A_{2}$
$\Leftrightarrow O$ is the excenter of $\triangle A_{3} B_{3} C_{3}$ at $\angle A_{3}$
$\Leftrightarrow A_{1}$ is the incenter of $\triangle A_{3} B_{3} C_{3}$
$\Leftrightarrow A_{1}$ is the orthocenter of $\triangle O B_{1} C_{1}$
$\Leftrightarrow O$ is the orthocenter of $\triangle A_{1} B_{1} C_{1}$.