Task 4. (20 points) To protect an oil platform located at sea, it is necessary to distribute 8 radars around it, each with a coverage radius of $r=17$ 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 16 km. Calculate the area of this ring of coverage.
# Solution.

To ensure radar coverage of a ring around the platform, they need to 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-27.jpg?height=714&width=737&top_left_y=1773&top_left_x=451)

Fig. 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-27.jpg?height=603&width=511&top_left_y=1886&top_left_x=1258)

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 $B D$:
$B D=\sqrt{17^{2}-8^{2}}=15$.

Then $A B=30$, so the distance from the center of the platform to the radars is equal to the radius of the circumscribed circle around the regular octagon:

$$
O B=\frac{A B}{2 \sin (180 / 8)}=\frac{30}{2 \sin 22.5^{\circ}}=\frac{15}{\sin 22.5^{\circ}}
$$

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 $O E$.

$$
\begin{aligned}
& S_{\text {ring }}=\pi\left(O E^{2}-O C^{2}\right) \\
& O D=\frac{D B}{\tan 22.5^{\circ}}=\frac{15}{\tan 22.5^{\circ}} \\
& O C=O D-8=\frac{15}{\tan 22.5^{\circ}}-8 \\
& O E=O D+8=\frac{15}{\tan 22.5^{\circ}}+8 \\
& S_{\text {ring }}=\pi\left(\left(\frac{15}{\tan 22.5^{\circ}}+8\right)^{2}-\left(\frac{15}{\tan 22.5^{\circ}}-8\right)^{2}\right)=\frac{480 \pi}{\tan 22.5^{\circ}}
\end{aligned}
$$

$$
\text { Answer. } \frac{15}{\sin 22.5^{\circ}} ; \frac{480 \pi}{\tan 22.5^{\circ}} \text {. }
$$