Task 4. (20 points) To protect an oil platform located at sea, it is necessary to distribute 5 radars around it, each with a coverage radius of $r=13$ km. Determine the maximum distance from the center of the platform at which they should be placed to ensure that the radars cover a ring around the platform with a width of 10 km. Calculate the area of this ring of coverage.
Solution.

To ensure radar coverage of a ring around the platform, the radars should be placed at the vertices of a regular polygon, the center of which coincides with the center of the platform (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-03.jpg?height=656&width=694&top_left_y=1902&top_left_x=310)

Fig. 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-03.jpg?height=551&width=554&top_left_y=2009&top_left_x=1088)

Fig. 3.

Point $O$ is the center of the oil platform, and points $A$ and $B$ are the locations of the radars. The circles represent the radar coverage. Consider the fragment of Fig. 2, $\triangle A O B$ (Fig. 3). In the right triangle $\triangle B C D$, by the Pythagorean theorem, we find $B D$:

$$
B D=\sqrt{13^{2}-5^{2}}=12
$$

Then $A B=24$, so the distance from the center of the platform to the radars is equal to the radius of the circumscribed circle around the regular pentagon:

$$
O B=\frac{A B}{2 \sin (180 / 5)}=\frac{24}{2 \sin 36^{\circ}}=\frac{12}{\sin 36^{\circ}}
$$

To find the area of the ring coverage, we need to subtract the area of the circle with radius $O C$ from the area of the circle with radius $O E$.

$$
\begin{aligned}
& S_{\text {ring }}=\pi\left(O E^{2}-O C^{2}\right) \\
& O D=\frac{D B}{\operatorname{tg} 36^{\circ}}=\frac{12}{\operatorname{tg} 36^{\circ}} \\
& O C=O D-5=\frac{12}{\operatorname{tg} 36^{\circ}}-5 \\
& O E=O D+5=\frac{12}{\operatorname{tg} 36^{\circ}}+5 \\
& S_{\text {ring }}=\pi\left(\left(\frac{12}{\operatorname{tg} 36^{\circ}}+5\right)^{2}-\left(\frac{12}{\operatorname{tg} 36^{\circ}}-5\right)^{2}\right)=\frac{240 \pi}{\operatorname{tg} 36^{\circ}}
\end{aligned}
$$

Answer. $\frac{12}{\sin 36^{\circ}} ; \frac{240 \pi}{\operatorname{tg} 36^{\circ}}$.