# 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,

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

!

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

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

Answer: 1.