Segment $K L$ is the diameter of a certain circle. Through its endpoints $K$ and $L$, two lines are drawn, intersecting the circle at points $P$ and $Q$, respectively, lying on the same side of the line $K L$. Find the radius of the circle if $\angle P K L=60^{\circ}$ and the point of intersection of the lines $K P$ and $Q L$ is 1 unit away from points $P$ and $Q$.
Triangle $K M L-$ is equilateral ( $M$ - the intersection point of lines $K P$ and $L Q$).

## Solution

Let $M$ be the intersection point of lines $K P$ and $L Q$. Point $M$ cannot lie on the circle. If $M$ is inside the circle, then $K M \cdot M P = L M \cdot M Q$. Therefore, $K M = M L$, which is impossible.

If point $M$ is outside the circle, then $M P \cdot M K = M Q \cdot M L$. Therefore, $K M = M L$. Then triangle $K M L-$ is equilateral. Its height $K Q$ is also a median. Therefore,

$$
K L = M L = 2 M Q = 2,
$$

and the desired radius is 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-11.jpg?height=415&width=420&top_left_y=1502&top_left_x=822)

## Answer

1.