## Task A-4.5.

Let $ABC$ be a triangle and $O$ the center of its circumscribed circle. The line $p$ is perpendicular to the bisector of angle $\varangle BAC$, passes through the midpoint of side $\overline{BC}$, and the midpoint of segment $\overline{AO}$. Determine the measure of angle $\varangle BAC$.
## Solution.

Let $A^{\prime}$ be the second intersection of the angle bisector of $\angle B A C$ and the circumcircle of triangle $A B C, P$ the midpoint of $\overline{A O}$, and $M$ the midpoint of $\overline{B C}$. Denote by $M^{\prime}$ and $O^{\prime}$ the feet of the perpendiculars from $M$ and $O$ to $A A^{\prime}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_30_5917997a941cee38e810g-26.jpg?height=745&width=720&top_left_y=1501&top_left_x=605)

The lines $p$ and $O O^{\prime}$ are perpendicular to the angle bisector of $\angle B A C$, so they are parallel. By applying Thales' theorem to these lines and the angle $\angle O A A^{\prime}$, we get

$$
\frac{\left|A M^{\prime}\right|}{\left|A O^{\prime}\right|}=\frac{|A P|}{|A O|}=\frac{1}{2}
$$

From this, it follows that $\left|M^{\prime} O^{\prime}\right|=\frac{1}{2}|A O|^{\prime}$.

1 point

Since triangle $A A^{\prime} O$ is isosceles and $O^{\prime}$ is the orthogonal projection of point $O$ onto $A A^{\prime}$, we have $\left|A O^{\prime}\right|=\left|A^{\prime} O^{\prime}\right|$.

It is known that $A^{\prime}$, as the intersection of the angle bisector of $\angle B A C$ and the circumcircle of triangle $B A C$, lies on the perpendicular bisector of segment $\overline{B C}$, so points $A^{\prime}, O$, and $M$ are collinear.

By applying Thales' theorem to the lines $p$ and $O O^{\prime}$ and the angle $\angle O A^{\prime} A$, we get

$$
\frac{\left|O^{\prime} M^{\prime}\right|}{\left|A^{\prime} O^{\prime}\right|}=\frac{|O M|}{\left|A^{\prime} O\right|}
$$

From all the obtained results, we conclude that $|O M|=\frac{1}{2}\left|A^{\prime} O\right|=\frac{R}{2}$, where $R$ is the radius of the circumcircle of triangle $A B C$.

Triangle $O M C$ is a right triangle and $|O M|=\frac{1}{2} R=\frac{1}{2}|O C|$, so this triangle is half of an equilateral triangle, i.e., $\angle M C O=30^{\circ}$.

Triangle $B O C$ is isosceles, so $\angle B O C=180^{\circ}-2 \angle M C O=120^{\circ}$.

Finally, by the theorem of the inscribed and central angles, we have

$$
\angle B A C=\frac{1}{2}\left(360^{\circ}-\angle B O C\right)=120^{\circ} .
$$