5. [5 points] A water strider and a water beetle slide counterclockwise around a float along 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 greatest.
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 denoted as $M(\alpha)$ 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 0, and at this point $0<\alpha_{0}<\frac{\pi}{2}$, where $\alpha_{0}=\beta_{0}$ is the angle corresponding to the initial positions of the insects.

The distance between the water strider and the beetle will be the greatest when the angle between the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ is $\pi$. 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 when the vectors $\overrightarrow{A M}$ and $\overrightarrow{A N}$ are oppositely directed, the beetle has moved by an angle $\omega$. Then $\alpha_{0}+\omega+\pi+2 \pi k=\alpha_{0}+5 \omega$, where $k=0,1,2, \ldots$ Therefore, $\omega=\frac{\pi+2 \pi k}{4}$, where $k=0,1,2, \ldots$

There will be four different points (for $k=0,1,2,3$). For $k=0$, 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 formulas for the cosine and sine of the sum of angles, we get

$$
\begin{aligned}
& \cos \beta_{1}=\cos \beta_{0} \cdot \cos \frac{\pi}{4}-\sin \beta_{0} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{2}}{4} \cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{14}}{4} \cdot \frac{\sqrt{2}}{2}=\frac{1-\sqrt{7}}{4} \\
& \sin \beta_{1}=\sin \beta_{0} \cdot \cos \frac{\pi}{4}+\cos \beta_{0} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{14}}{4} \cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{4} \cdot \frac{\sqrt{2}}{2}=\frac{1+\sqrt{7}}{4}
\end{aligned}
$$

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

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

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