In triangle ABC, bisectors AA1 and CC1 intersect at point O. AO = 12√3 cm, angle BAC = 120 degrees.

In triangle ABC, bisectors AA1 and CC1 intersect at point O. AO = 12√3 cm, angle BAC = 120 degrees. Find the radius of the circle written in the ABC triangle.

By the property of the bisectors of a triangle, the point of their intersection coincides with the center of the circle inscribed in the triangle.
From the center of the circle, draw the height OH to the AC side of the triangle.
In the formed right-angled triangle AOН, angle H is a straight line, angle A = BAC / 2 = 120/2 = 60, since AO is the bisector of angle A. The length of the hypotenuse OA = 12 * √3 cm, by condition.
Then SinOAH = OH / OA.
OH = OA * Sin60 = 12 * √3 * √3 / 2 = 18 cm.
The height to the side of the triangle, drawn from the center of the inscribed circle, is the radius of that circle.
R = OH = 18 cm.
Answer: The radius of the inscribed circle is 18 cm.