In an isosceles triangle $A B C(A C=B C)$, a circle with radius 3 is inscribed. A line $l$ is tangent to this circle and parallel to the line $A C$. The distance from point $B$ to line $l$ is 3. Find the distance between the points where the given circle touches the sides $A C$ and $B C$.
Consider the right triangle $O C Q$, where $O$ is the center of the circle, and $Q$ is the point of tangency of the circle with side $B C$.

## Solution

Let $P$ and $Q$ be the points of tangency of the circle with sides $A C$ and $B C$ of triangle $A B C$, $O$ be the center of the circle, $E$ be the point of tangency of the circle with line $l$, and $M$ and $N$ be the points of intersection of line $l$ with sides $A C$ and $B C$. From symmetry, it is clear that point $E$ lies on segment $C O$ and $C O = C E + O E = 3 + 3 = 6$.

From the right triangle $C Q O$, we find that $\angle O C Q = 30^{\circ}$. Therefore, $P Q = 2 O Q \sin 60^{\circ} = 3 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-11.jpg?height=400&width=558&top_left_y=1967&top_left_x=750)

## Answer

$3 \sqrt{3}$.