\begin{aligned} & \text { [The inscribed angle is half the central angle] } \\ & {[\quad \text { Symmetry helps solve the problem. }]}\end{aligned}

Point $X$ is located on the diameter $A B$ of a circle with radius $R$. Points $K$ and $N$ lie on the circle in the same half-plane relative to $A B$,

and $\angle K X A=\angle N X B=60^{\circ}$. Find the length of the segment $K N$.
Consider the point $K^{\prime}$, symmetric to point $K$ with respect to the diameter $A B$ (see the figure). It lies on the same circle and $\angle K X A=60^{\circ}$. Then the sum of the three angles with the vertex at point $X$ (marked on the drawing) is $180^{\circ}$. Therefore, points $K^{\prime}, X$, and $N$ lie on the same straight line.

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-22.jpg?height=529&width=554&top_left_y=2394&top_left_x=755)

Triangle $K'XK$ is isosceles with an angle of $120^{\circ}$ at the vertex. Therefore, the inscribed angle $KK'N$ is $30^{\circ}$, and the central angle $KON$ is $60^{\circ}$. Thus, $KN=R$.

## Answer

$KN=R$.