9.4. The lines containing the altitudes of the non-isosceles triangle \(ABC\) intersect at point \(H\). \(I\) is the incenter of triangle \(ABC\), \(O\) is the circumcenter of triangle \(BHC\). It is known that point \(I\) lies on the segment \(OA\). Find the angle \(BAC\).
Answer: $60^{\circ}$

Solution. Note that point $O$ lies on the bisector (perpendicular through the midpoint) of segment $B C$. On the other hand, point $O$ lies on the bisector of angle $A$. However, these two lines intersect at a point lying on the circumcircle of triangle $B A C$, and dividing the arc $B C$ in half. Thus, point $O$ lies on the circumcircle of triangle $B A C$.

The angles at the vertices of triangle $B A C$ will be denoted by the same letter. Consider the isosceles triangle $O H C, O H=O C$. Then, from the observation about the location of point $O$ and the properties of inscribed angles, we immediately get that the angles at the base of this isosceles triangle are $90^{\circ}-B+A / 2$. Similarly, the angle at the base of the isosceles triangle $O H B(O H=$ $O B)$ is $90^{\circ}-C+A / 2$. But the sum of these two angles is equal to the angle $B H C$, which, in turn, is equal to the angle $180-A$ (by the theorem of vertical angles). Therefore, $180-B-C+A=180-A$, from which $3 A=180$.

Criteria.

(-.) The answer is incorrect and there is no complete solution, but there are significant advances in the solution.

(-/+) The correct answer is obtained, but the solution uses without proof that point $O$ belongs to the circumcircle of $\triangle A B C$.

(+/-) It is proven that point $O$ lies on the circumcircle of $\triangle A B C$.

(+.) The answer is incorrect due to an arithmetic error.

$(+)$ Correct answer + full justification.