11.4. 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 $K N$.
Answer: $K N=R$.

First method (axial symmetry). Consider the point $K_{1}$, symmetric to the point $K$ with respect to the diameter $A B$ (see Fig. 11.4a). 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.

Triangle $K^{\prime} X K$ is isosceles with an angle of $120^{\circ}$ at the vertex. Therefore, the inscribed angle $K K^{\prime} N$ is $30^{\circ}$. Then the central angle $K O N$ is $60^{\circ}$. Thus, triangle $K O N$ is isosceles with an angle of $60^{\circ}$, meaning it is equilateral. Therefore, $K N=R$.

Second method (auxiliary circle). We will prove the following fact: if points $X$ and $O$ lie on the line $a$, rays $X K$ and $X N$ form equal angles with it, and point $O$ is equidistant from points $K$ and $N$, then points $K, N, X$, and $O$ lie on the same circle.

![](https://cdn.mathpix.com/cropped/2024_05_06_b57a61ad1a3a9eb6efc9g-2.jpg?height=436&width=562&top_left_y=249&top_left_x=359)

Fig. 11.4b

![](https://cdn.mathpix.com/cropped/2024_05_06_b57a61ad1a3a9eb6efc9g-2.jpg?height=441&width=487&top_left_y=247&top_left_x=1216)

Fig. 11.4c

Indeed, describe a circle around triangle $K X N$. Let it intersect the line $a$ again at point $O^{\prime}$ (see Fig. 11.4b). Then $\angle N K O^{\prime}=\angle N X O^{\prime}$ (inscribed angles subtending the same arc), and the external angle at vertex $X$ of the inscribed quadrilateral $N K X O^{\prime}$ is equal to the internal angle $K N O^{\prime}$. Therefore, in triangle $N K O^{\prime}$, the angles adjacent to side $N K$ are equal, so $O^{\prime} K=O^{\prime} N$, meaning point $O^{\prime}$ coincides with point $O$.

Applying the proven statement in our case, we get $\angle K O N=\angle K X N=60^{\circ}$. Thus, triangle $K O N$ is isosceles with an angle of $60^{\circ}$, meaning it is equilateral. Therefore, $K N=R$.

Third method (sine theorem). Let $\angle Х K O=\alpha ; \angle X N O=\beta$. By the sine theorem for triangles $X K O$ and $X N O$ (see Fig. 11.4c): $\frac{R}{\sin 120^{\circ}}=\frac{X O}{\sin \alpha}$ and $\frac{R}{\sin 60^{\circ}}=\frac{X O}{\sin \beta}$.

Since $\sin 120^{\circ}=\sin 60^{\circ}$, then $\sin \alpha=\sin \beta$. Since angles $\alpha$ and $\beta$ are acute, then $\alpha=\beta$. Therefore, points $X, K, O$, and $N$ lie on the same circle. The rest is described above.

Other methods of solution are also possible.

+ complete justified solution (by any method)

$\pm$ a generally correct solution with some minor gaps

- only the answer is provided