6. Triangle $L O M$ with angle $\angle L O M=\alpha^{\circ}$ is rotated by some acute angle around point $O$. As a result, point $L$ moves to point $N$, which lies on side $L M$, and point $M$ moves to point $K$ such that $O M \perp N K$. Find the angle of rotation (in degrees).

Note: In different versions, different values were given for $\alpha: 21,24,27,33$, $39,42,48,51,54,57,63,66,69,72,78$.
Answer: $\frac{2}{3} \alpha \cdot$ Solution. Let the angle of rotation be denoted by $\varphi$. Then $\angle L O N=\angle M O K=\varphi$. Since $O N=O L$ (property of rotation), we have

$\angle O L N=\angle O N L=90^{\circ}-\frac{\varphi}{2}$.

Moreover, $\angle O N K=\angle O L N=90^{\circ}-\frac{\varphi}{2}$ (a consequence of the equality of $\triangle L O M$ and $\triangle N O K$).

![](https://cdn.mathpix.com/cropped/2024_05_06_adf4a392ca1a490a8c79g-07.jpg?height=451&width=549&top_left_y=360&top_left_x=1416)

Therefore, $\angle N O P=\frac{\varphi}{2}$. This means that $\alpha=\frac{3 \varphi}{2}$, or $\varphi=\frac{2 \alpha}{3}$.