In triangle $A B C$, the angle bisectors intersect at point $O$. Line $A O$ intersects the circumcircle of triangle $O B C$ at points $O$ and $M$. Find $O M$, if $B C=2$, and angle $A$ is $30^{\circ}$.
Prove that $O M-$ is the diameter of the specified circle.

## Solution

Let $M^{\prime}$ be the point of intersection of the bisectors of the external angles at vertices $B$ and $C$ of triangle $A B C$. Then $\angle O B M^{\prime} = \angle O C M^{\prime} = 90^{\circ}$. Therefore, a circle can be circumscribed around quadrilateral $O B M^{\prime} C$, and $O M^{\prime}$ is the diameter of this circle. The circle circumscribed around triangle $O B C$ coincides with the circumscribed circle of quadrilateral $O B M^{\prime} C$, since a unique circle can be circumscribed around a triangle. Therefore, point $M^{\prime}$ coincides with point $M$.

If $R$ is the radius of the circle, then $O M = 2 R$. From the Law of Sines, it follows that

$$
\begin{aligned}
O M = 2 R = & \frac{B C}{\sin \angle B O C} = \frac{B C}{\sin \left(90^{\circ} + \frac{1}{2} \angle B A C\right)} = \frac{2}{\sin 105^{\circ}} = \\
& = \frac{2}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{8}{\sqrt{6} + \sqrt{2}} = 2(\sqrt{6} - \sqrt{2})
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-58.jpg?height=504&width=260&top_left_y=2009&top_left_x=905)

## Answer

$2(\sqrt{6} - \sqrt{2})$.

## Problem