5. [5 points] Around a hook with a worm, in the same plane as it, a carp and a minnow are swimming along two circles. In the specified plane, a rectangular coordinate system is introduced, in which the hook (the common center of the circles) is located at the point $(0 ; 0)$. At the initial moment of time, the carp and the minnow are at points $M_{0}(-1 ; 2 \sqrt{2})$ and $N_{0}(2 ;-4 \sqrt{2})$ respectively. The speed of the carp is two and a half times the speed of the minnow, and both are moving clockwise. Determine the coordinates of all positions of the minnow at which the distance between the fish will be the shortest.
Answer: $(\sqrt{2}-4 ;-4-\sqrt{2}),(-4-\sqrt{2} ; 4-\sqrt{2}),(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4)$.

Solution. Let the points where the carp and the minnow are located be $M(\alpha)$ and $N(\beta)$, respectively, where $\alpha$ and $\beta$ are the angles formed by the radius vectors of points $M$ and $N$ with the positive direction of the x-axis. Note that the angle between $\overrightarrow{A M_{0}}$ and $\overrightarrow{A N_{0}}$ is $\pi$, and at this point $\frac{\pi}{2}<\alpha_{0}<\pi$, $-\frac{\pi}{2}<\beta_{0}<0$, where $\alpha_{0}, \beta_{0}$ are the angles corresponding to the initial positions of the fish.

The distance between the carp and the minnow will be the smallest when the angle between the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ is zero. Since $\left|A M_{0}\right|=3$ and $\left|A N_{0}\right|=6$ are the radii of the circles, and $\left|A N_{0}\right|=2\left|A M_{0}\right|$, the angular velocity of the carp is 5 times the angular velocity of the minnow.

Let at the moment of the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ aligning, the minnow has moved by an angle $\omega$ clockwise. Then $\alpha-5 \omega=\beta-\omega-2 \pi n$, where $n=0,1, \ldots$ Therefore, $\omega=\frac{\alpha-\beta}{4}+\frac{\pi n}{2}=\frac{\pi}{4}+\frac{\pi n}{2}$, where $n=0,1, \ldots$

There will be four different points (for $n=0,1,2,3$). For $n=0$, we get $\beta_{1}=\beta_{0}-\frac{\pi}{4}$. The coordinates of the minnow's position are found using the formulas $x_{N}=6 \cos \beta_{1}, y_{N}=6 \sin \beta_{1}$. Using the coordinates of point $N_{0}$, we find $\cos \beta_{0}=\frac{x_{N_{0}}}{6}=\frac{1}{3}$ and $\sin \beta_{0}=\frac{y_{N_{0}}}{6}=-\frac{2 \sqrt{2}}{3}$. Using the cosine and sine sum formulas, we get

$$
\begin{gathered}
\cos \beta_{1}=\cos \beta_{0} \cdot \cos \frac{\pi}{4}+\sin \beta_{0} \cdot \sin \frac{\pi}{4}=\frac{1}{3} \cdot \frac{1}{\sqrt{2}}-\frac{2 \sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}}=\frac{1-2 \sqrt{2}}{3 \sqrt{2}} \\
\sin \beta_{1}=\sin \beta_{0} \cdot \cos \frac{\pi}{4}-\cos \beta_{0} \cdot \sin \frac{\pi}{4}=-\frac{2 \sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}}-\frac{1}{3} \cdot \frac{1}{\sqrt{2}}=\frac{-1-2 \sqrt{2}}{3 \sqrt{2}}
\end{gathered}
$$

From this, $x_{N_{1}}=6 \cdot \frac{1-2 \sqrt{2}}{3 \sqrt{2}}=\sqrt{2}-4, y_{N_{1}}=6 \cdot \frac{-1-2 \sqrt{2}}{3 \sqrt{2}}=-\sqrt{2}-4$.

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

$(-4-\sqrt{2} ; 4-\sqrt{2}),(4-\sqrt{2} ; 4+\sqrt{2}),(4+\sqrt{2} ; \sqrt{2}-4)$.