Let the circle $\Omega$ with radius 1 and center $O$ touch the given lines at points $L$ and $M$, and the side $A B$ of triangle $A B C$ at point $K$ (see the left figure). By the problem's condition, the incircle of the triangle with center $I$ touches the side $A B$ at the same point $K$, and the base $A C$ at its midpoint $H$. Draw the line $O A$ until it intersects the height $B H$ of triangle $A B C$ at point $P$.

$A M = A K = A H$, so the right triangles $A M O$ and $A H P$ are equal. Therefore, $A$ is the midpoint of segment $O P$. $A I \perp A O$, since the angle bisectors of adjacent angles $M A B$ and $C A B$ are perpendicular, and the center of the circle lies on the angle bisector between the tangents drawn to it from the same point. Similarly, $O B \perp A O$. Thus, $A I \| O B$, meaning $A I$ is the midline of triangle $B P O$. Hence, $K$ is the centroid of triangle $B P O$, and

$I K = 1 / 2 O K = 1 / 2$.
![](https://cdn.mathpix.com/cropped/2024_05_06_11215e94233515718ac4g-59.jpg?height=582&width=1496&top_left_y=0&top_left_x=273)

## Answer

$1 / 2$.
$\underline{\text { Gordrin P.K. }}$.

Between two parallel lines, a circle of radius 1 is placed, touching both lines, and an isosceles triangle, the base of which lies on one of the lines, and the vertex on the other. It is known that the triangle and the circle have exactly one common point and that this point lies on the inscribed circle of the triangle. Find the radius of the inscribed circle of the triangle.