5. [5 points] A water strider and a water beetle slide counterclockwise around a float on two circles. A rectangular coordinate system is introduced on the water surface, in which the float (the common center of the circles) is at the point ( $0 ; 0$ ). The speed of the water strider is twice the speed of the beetle. At the initial moment, the water strider and the beetle are at points $M_{0}(-2 ;-2 \sqrt{7})$ and $N_{0}(5 ; 5 \sqrt{7})$ respectively. Determine the coordinates of all positions of the beetle at which the distance between the insects will be the shortest.
Answer: $\left(\frac{5}{\sqrt{2}}(1-\sqrt{7}) ; \frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right)$, $\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)$.

Solution. Let the points where the water strider and the beetle 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 $\pi$, and at the same time $-\pi<\alpha_{0}<-\frac{\pi}{2}, 0<\beta_{0}<\frac{\pi}{2}$, where $\alpha_{0}, \beta_{0}$ are the angles corresponding to the initial positions of the insects.

The distance between the water strider and the beetle will be the smallest when the angle between the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ is zero. Since $\left|A M_{0}\right|=4 \sqrt{2}$ and $\left|A N_{0}\right|=10 \sqrt{2}$ are the radii of the circles, and $\left|A N_{0}\right|=\frac{5}{2}\left|A M_{0}\right|$, the angular velocity of the water strider is 5 times 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 $\alpha+5 \omega=\beta+\omega+2 \pi n$, where $n=0,1, \ldots$ Therefore, $\omega=\frac{\beta-\alpha}{4}+\frac{\pi n}{2}=\frac{\pi}{4}+\frac{\pi n}{2}$, 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 beetle's position are found using the formulas $x_{N}=10 \sqrt{2} \cos \beta_{1}, y_{N}=10 \sqrt{2} \sin \beta_{1}$. Using the coordinates of point $N_{0}$, we find $\cos \beta_{0}=\frac{x_{N_{0}}}{10 \sqrt{2}}=\frac{1}{2 \sqrt{2}}$ and $\sin \beta_{0}=\frac{y_{N_{0}}}{10 \sqrt{2}}=\frac{\sqrt{7}}{2 \sqrt{2}}$. Using the cosine and sine sum formulas, we get

$$
\cos \beta_{1}=\cos \beta_{0} \cdot \cos \frac{\pi}{4}-\sin \beta_{0} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{2}}{4} \cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{14}}{4} \cdot \frac{\sqrt{2}}{2}=\frac{1-\sqrt{7}}{4}
$$

$$
\sin \beta_{1}=\sin \beta_{0} \cdot \cos \frac{\pi}{4}+\cos \beta_{0} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{14}}{4} \cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{4} \cdot \frac{\sqrt{2}}{2}=\frac{1+\sqrt{7}}{4}
$$

Thus, $x_{N_{1}}=10 \sqrt{2} \cdot \frac{1-\sqrt{7}}{4}=\frac{5}{\sqrt{2}}(1-\sqrt{7}), y_{N_{1}}=10 \sqrt{2} \cdot \frac{1+\sqrt{7}}{4}=\frac{5}{\sqrt{2}}(1+\sqrt{7})$.

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

$$
\left(-\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(1-\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(\sqrt{7}-1) ;-\frac{5}{\sqrt{2}}(1+\sqrt{7})\right),\left(\frac{5}{\sqrt{2}}(1+\sqrt{7}) ; \frac{5}{\sqrt{2}}(\sqrt{7}-1)\right)
$$