1. [4 points] On the table lies a piece of sugar, around which an ant and a beetle are crawling along two circles at the same speed. A rectangular coordinate system is introduced on the table plane, in which the sugar (the common center of the circles) is located at point $O(0 ; 0)$. The ant moves clockwise, while the beetle moves counterclockwise. At the initial moment, the ant and the beetle are at points $M_{0}(-1 ; \sqrt{3})$ and $N_{0}(2 \sqrt{3} ; 2)$, respectively. Determine the coordinates of all positions of the beetle where the distance between it and the ant will be the shortest.
Answer: $(2 ; 2 \sqrt{3}),(-4 ; 0),(2 ;-2 \sqrt{3})$.

Solution. Let's denote the points where the ant and the beetle are located as $M(\alpha)$ and $N(\beta)$ respectively, where $\alpha$ and $\beta$ are the angles that the radius vectors of points $M$ and $N$ form with the positive direction of the x-axis. Note that the angle between $\overrightarrow{A M_{0}}$ and $\overrightarrow{A N_{0}}$ is $\frac{\pi}{2}$, and at this point $\alpha_{0}=\frac{2 \pi}{3}, \beta_{0}=\frac{\pi}{6}$ - are the angles corresponding to the initial positions of the insects.

The distance between the ant and the beetle will be the smallest when the angle between the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ is zero. Since $\left|A M_{0}\right|=2$ and $\left|A N_{0}\right|=4$ are the radii of the circles, and $\left|A N_{0}\right|=2\left|A M_{0}\right|$, the angular velocity of the ant is twice the angular velocity of the beetle.

Let at the moment of the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ aligning, the beetle has moved by an angle $\omega$. Then $\alpha_{0}-2 \omega=\beta_{0}+\omega+2 \pi n$, where $n \in \mathbb{Z}$. Therefore, $\omega=\frac{\alpha_{0}-\beta_{0}}{3}-\frac{2 \pi n}{3}=\frac{\pi}{6}-\frac{2 \pi n}{3}$, where $n=0,-1,-2 \ldots$

There will be three different points (for $n=0,-1,-2$). For $n=0$ we get $\beta_{1}=\beta_{0}+\frac{\pi}{6}=\frac{\pi}{3}$. The coordinates of the beetle's position are found using the formulas $x_{N}=4 \cos \beta_{1}=2, y_{N}=4 \sin \beta_{1}=2 \sqrt{3}$.

The other points are obtained by rotating point $N_{1}$ around the origin by angles $\frac{2 \pi}{3}, \frac{4 \pi}{3}$ and have, respectively, coordinates $(-4 ; 0)$ and $(2 ;-2 \sqrt{3})$.