6. In 2022, it will be 65 years since the launch of the first artificial satellite of the Earth (ASZ). Currently, various types of satellites located at different orbits and at different altitudes are used to ensure the uninterrupted operation of cellular communication, television, and radio broadcasting systems.

The coverage area of a satellite is the part of the Earth's surface within which the signal level to and from the satellite is sufficient for reception with the required quality at a specific moment in time. As a rule, this part of the surface is limited by a circle passing along the visible horizon line. In the figure, the line passes through point $\Gamma$.

a) Determine the area of the Earth's surface (in km²) that is the coverage area of a satellite located at a height of $\mathrm{H}=500$ km relative to the Earth's surface, considering it a sphere with a radius of $\mathrm{R}=6400$ km and a center at point $\mathrm{O}$.

b) Find all values of $n>1$ for which on the Earth's surface, circles $\mathrm{C}_{1}, \ldots, \mathrm{C}_{\mathrm{n}}$ can be arranged, each of which externally touches the circle $\mathrm{C}_{0}$, with a center at point A and a radius $r<\mathrm{R}$, each being the boundary of the coverage area of an ASZ located at the same height $\mathrm{H}$ as the satellite with the coverage area $\mathrm{C}_{0}$. Each coverage area $\mathrm{C}_{\mathrm{i}}$ must externally touch circles $\mathrm{C}_{0}$ and $\mathrm{C}_{\mathrm{i}+1}, i=0,1, \ldots, n-1$, i.e., the first touches $\mathrm{C}_{0}$ and $\mathrm{C}_{2}$, the second touches $\mathrm{C}_{0}$ and $\mathrm{C}_{3}$, and so on. Circle $\mathrm{C}_{\mathbf{n}}$ must touch $\mathrm{C}_{0}$ and $\mathrm{C}_{1}$. (20 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-08.jpg?height=345&width=260&top_left_y=1381&top_left_x=1589)
Solution. a) The coverage area is the part of the sphere lying inside the cone.

$S=2 \pi R \cdot h$, where $h=$ AZ - the height of the segment. $h=R-R \cos \alpha$, here the angle $\alpha$ is the angle between the radius OG and the line OA, connecting the center of the sphere with the center of the circle, which is the line of intersection of the sphere and the cone.

Then the area is

$S=2 \pi R^{2}(1-\cos \alpha)=2 \pi R^{2}\left(1-\frac{R}{R+H}\right)=2 \pi R^{2} \cdot \frac{H}{R+H} \approx$

![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-08.jpg?height=385&width=323&top_left_y=1972&top_left_x=1552)

$\approx 6 \cdot 6400^{2} \cdot \frac{500}{6900}=6400^{2} \cdot \frac{10}{23} \approx \frac{4096}{23} \cdot 10^{5} \approx 178,09 \cdot 10^{5}=17809000$ km $^{2}$.

b) Let $\mathrm{O}$ be the center of the sphere, B be the point of tangency of the first and second circles, A and $\mathrm{A}_{1}$ be the centers of these circles, $3,3_{1}, 3_{2}$ be the points of intersection of the radii $R$ with the sphere. Denote $\alpha$ - the angle between OZ and OB. Then $\sin \alpha=\frac{r}{R}, 33_{1}=2 r$. In the regular pyramid O33 32, the planar angles at the vertex are equal to $2 \alpha$, the dihedral angle at the edge OZ is $360 / n$. Dropping perpendiculars from points $3_{1}$ and $3_{2}$ to the edge OZ at point $\mathrm{H}$, the triangles $\mathrm{O}_{1}{ }_{1}$ and $\mathrm{O}_{3} 3_{2}$ are equal (by three sides), since two sides are equal to R, and the third is $2 \mathrm{r}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_a494f1af39843ff219fbg-09.jpg?height=600&width=1670&top_left_y=241&top_left_x=182)

$$
H 3_{1}=H 3_{2}=2 r \cos \alpha=2 r \sqrt{1-\sin ^{2} \alpha}=2 r \sqrt{1-\left(\frac{r}{R}\right)^{2}}
$$

$\Rightarrow 2 r=33_{1}=3_{1} 3_{2}=2 \cdot 3_{1} H \cdot \sin (180 / n)=4 r \sqrt{1-\frac{r^{2}}{R^{2}}} \sin (180 / n)$. Therefore,

$2 \sqrt{1-\frac{r^{2}}{R^{2}}} \sin (180 / n)=1 \Rightarrow \sin (180 / n)=1 /\left(2 \sqrt{1-\frac{r^{2}}{R^{2}}}\right)>1 / 2 \Rightarrow n<6$.

Answer: a) 17809000,

b) $2,3,4,5$.

## School Olympiad "Step into the Future"

Profile: Mathematics

Variant: 4

Class: 11

Problem 1 (12 points). A numerical sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is defined such that $a_{1}=$ $\log _{2}\left(\log _{2} f(2)\right), a_{2}=\log _{2}\left(\log _{2} f(f(2))\right), \ldots, a_{n}=\log _{2}(\log _{2} \underbrace{f(\ldots f}_{n}(2)))), \ldots$, where $f(x)=x^{x}$. Determine the index $n$ for which $a_{n}=2059+2^{2059}$.

Problem 2 (16 points). Clowns Pliukh and Shmyak have six pairs of felt boots between them. Each pair of boots is painted in a unique color, and the boots in each pair are identical (not distinguished as left or right). In how many ways can both clowns be simultaneously wearing mismatched boots?

Problem 3 (16 points). Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, such that $A M=2, M C=6$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Find the length of segment $D C$, if the tangent of angle $A C H$ is $1 / 7$.

Problem 4 (16 points). Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{c}
\left(a\left|\log _{2} y\right|+a\left|\log _{2} x\right|-2\right)\left(\left(\log _{2} x\right)^{2}+\left(\log _{2} y\right)^{2}-48 a^{2}\right)=0 \\
\left(\log _{a} x\right)^{2}\left(\log _{a} y\right)^{2}\left(\log _{2} a^{2}\right)^{4}=256 a^{2}
\end{array}\right.
$$

has exactly eight distinct solutions.

Problem 5 (20 points). Inside a regular quadrilateral pyramid $S A B C D$ with base $A B C D$, a regular quadrilateral prism $K L M N K_{1} L_{1} M_{1} N_{1}$ is located, the base $K L M N$ of which lies in the plane $A B C$. The center of the base $K L M N$ of the prism is located on the segment $A C, K L\|A C, K N\|$ $B D$ (points $K$ and $B$ lie on the same side of $A C$), the side of the base of the prism is 2, and the lateral edge $K K_{1}$ of the prism is 1. The vertices $L_{1}$ and $M_{1}$ of the upper base of the prism $K L M N K_{1} L_{1} M_{1} N_{1}$ lie on the lateral faces $S B C$ and $S C D$ of the pyramid $S A B C D$ respectively. The plane $\gamma$ passes through points $B, K_{1}$, and $N_{1}$. Find the volumes of the parts into which the plane $\gamma$ divides the pyramid $S A B C D$, if the side of the base of the pyramid is $8 \sqrt{2}$, and its height is 4.

Problem 6 (20 points). More and more countries are exploring space. There are already 12 states that launch their satellites using their own launch vehicles. There are also countries that use the services of the main space powers to launch their satellites for civilian purposes. Due to the increasing number of participants in space activities and the intensification of these activities, issues of ensuring the safety of space operations are arising. For example, NASA has asked the corporation "Energia" to reduce the altitude of the ISS orbit.

a) Assuming the Earth is a sphere with radius $R$, determine the maximum number of satellites that can simultaneously be in orbits around the Earth at the same height $H$ from its surface such that the distance between the satellites is greater than $\sqrt{2}(R+H)$.

b) For the maximum number of satellites found, specify the coordinates of their possible positions in a coordinate system with the origin at the center of the Earth and the x-axis directed along the vector connecting the center of the Earth with one of the satellites.

## Solution to Variant 4 (Mathematics - 11th Grade)