6. The altitudes of an acute-angled, non-isosceles triangle \(ABC\) intersect at point \(H\). \(O\) is the center of the circumcircle of triangle \(BHC\). The center \(I\) of the inscribed circle of triangle \(ABC\) lies on the segment \(OA\). Find the angle \(BAC\).
Answer: $60^{\circ}$.

Solution. From the problem statement, it follows that point O lies at the intersection of the angle bisector of angle $A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, point O lies on this circle and is the midpoint of arc $BC$ (see Fig. 9.6a, b). Additionally, $\angle BHC = 180^{\circ} - \angle BAC$, since $H$ is the orthocenter of triangle $ABC$. We can reason in different ways from here.

![](https://cdn.mathpix.com/cropped/2024_05_06_e4dee7622667d7cedbf9g-3.jpg?height=468&width=429&top_left_y=177&top_left_x=1476)

First method. Let the angles of triangle $ABC$ be: $\angle A = \alpha, \angle B = \beta, \angle C = \gamma$. (see Fig. 9.6a) Then $\angle OCB = \angle OAB = \alpha / 2, \angle BHC = 90^{\circ} - \beta, \angle OCH = \angle OCB + \angle BHC = 90^{\circ} + \alpha / 2 - \beta$.

In triangle $OHC: OH = OC$ (radii of the same circle), so $\angle OHC = \angle OCH = 90^{\circ} + \alpha / 2 - \beta$. Similarly, $\angle OHB = \angle OBH = 90^{\circ} + \alpha / 2 - \gamma$. Therefore, $\angle BHC = \angle OHB + \angle OHC = 90^{\circ} + \alpha / 2 - \gamma + 90^{\circ} + \alpha / 2 - \beta = 180^{\circ} + \alpha - \gamma - \beta = 2\alpha$.

Since $\angle BHC = 180^{\circ} - \alpha$, we get the equation $2\alpha = 180^{\circ} - \alpha$, from which $\alpha = 60^{\circ}$.

Second method. We use the fact that the circumcircle of triangle $BHC$ is symmetric to the circumcircle of triangle $ABC$ with respect to line $BC$ (see Fig. 9.6b). Then the center $P$ of the circumcircle of triangle $ABC$ lies on arc $BHC$. Therefore, $\angle BHC = \angle BPC = 2 \angle BAC$. From $180^{\circ} - \angle BAC = 2 \angle BAC$, we get: $\angle BAC = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e4dee7622667d7cedbf9g-3.jpg?height=622&width=442&top_left_y=674&top_left_x=1475)

## Grading Criteria.

"+" - a complete and well-justified solution is provided

“$\pm$” - a generally correct solution with minor gaps in justification

"Ғ" - correct reasoning is provided, but there is an error in angle calculation leading to an incorrect answer

"Ғ" - it is proven that point O belongs to the circumcircle of triangle $ABC$, but there is no further progress

“-” - only the answer is provided

“-” - the problem is not solved or is solved incorrectly