1. [4 points] On the floor, there is a saucer of milk, around which a kitten and a puppy walk along two circles. The kitten's speed is half the puppy's speed. A rectangular coordinate system is introduced on the floor plane, with the saucer (the common center of the circles) located at point $O(0 ; 0)$. The kitten moves clockwise, while the puppy moves counterclockwise. At the initial moment, the kitten and the puppy are at points $M_{0}(-6 ; 0)$ and $N_{0}(2 ; 2 \sqrt{3})$ respectively. Determine the coordinates of all positions of the kitten where the distance between the animals will be the shortest.
Answer: $(-3 \sqrt{3} ; 3),(3 ; 3 \sqrt{3}),(3 \sqrt{3} ;-3),(-3 ;-3 \sqrt{3})$.

Solution. Let the points where the kitten and the puppy are located be $M(\alpha)$ 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 $\frac{2 \pi}{3}$, and at this point $\alpha_{0}=\pi$, $\beta_{0}=\frac{\pi}{3}-$ are the angles corresponding to the initial positions of the animals.

The distance between the puppy and the kitten will be the smallest when the angle between the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ is zero. Since $\left|A M_{0}\right|=6$ and $\left|A N_{0}\right|=4$ are the radii of the circles, and $\left|A N_{0}\right|=\frac{1}{2}\left|A M_{0}\right|$, the angular velocity of the puppy is three times the angular velocity of the kitten.

Let at the moment of the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ aligning, the kitten has moved by an angle $\omega$. Then $\alpha_{0}-\omega+2 \pi k=\beta_{0}+3 \omega$, where $k \in \mathbb{Z}$. Therefore, $\omega=\frac{\alpha_{0}-\beta_{0}}{4}+\frac{\pi k}{2}=\frac{\pi}{6}+\frac{\pi k}{2}$, where $k=0,1,2 \ldots$ There will be four different points (for $k=0,1,2,3$). For $k=0$, we get $\alpha_{1}=\alpha_{0}-\frac{\pi}{6}=\frac{5 \pi}{6}$. The coordinates of the kitten's position are found using the formulas $x_{N}=6 \cos \alpha_{1}=-3 \sqrt{3}, y_{N}=6 \sin \alpha_{1}=3$.

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