# Task 4. (12 points)

The angle bisectors of angles $A, B$, and $C$ of triangle $A B C$ intersect the circumcircle of this triangle at points $A_{1}, B_{1}$, and $C_{1}$, respectively. Find the distances between point $A_{1}$ and the center of the inscribed circle of triangle $A B C$, given that $\angle A_{1} B_{1} C_{1}=50^{\circ}, \angle A_{1} C_{1} B_{1}=70^{\circ}, B_{1} C_{1}=\sqrt{3}$.
# Solution.

In the figure, identical numbers mark equal angles (this follows from the fact that $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of triangle $ABC$, the angle marked "1+2" near point $O$ (which is the center of the inscribed circle) is equal to $\angle O A B + \angle O B A = \angle 1 + \angle 2$ by the exterior angle theorem of a triangle. Therefore, triangle $O B A_{1}$ is isosceles and $A_{1} B = A_{1} O$ is the desired segment.

$\angle A B_{1} C_{1} = 50^{\circ}, \angle A C_{1} B_{1} = 70^{\circ}$, hence $\angle B_{1} A_{1} C_{1} = 60^{\circ}$. By the Law of Sines,

$\frac{B_{1} C_{1}}{\sin \angle B_{1} A_{1} C_{1}} = 2 R, \frac{\sqrt{3}}{\sin 60^{\circ}} = 2 R, R = 1$. Next,

$\angle A = \angle A_{1} B_{1} C_{1} + \angle A_{1} C_{1} B_{1} - \angle B_{1} A_{1} C_{1} = 60^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_d97c98d1a88f3e241cd1g-4.jpg?height=751&width=697&top_left_y=1024&top_left_x=1182)

from which $\angle 1 = 30^{\circ}$. Then

$A_{1} O = A_{1} B = 2 \cdot R \cdot \sin \angle 1 = 1$.

Answer: 1.

| Criterion for Evaluation  Full solution. | Rating  + | Points |
| :---: | :---: | :---: |
| The main logical steps of the solution are presented. The solution  lacks some justifications or has a computational error or typo. | $\pm$ | 9 |
| The idea of the solution is found, but it is not completed. However,  a significant part of the task is performed. | $+/ 2$ | 6 |
| The solution is generally incorrect or incomplete, but contains  some progress in the right direction. | $\mp$ | 2 |
| The solution does not meet any of the criteria described |  |  |
| above. |  |  |

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