6. Astronomers have discovered a new celestial body beyond the planet Saturn, moving in a circular orbit, for the study of which a scientific research probe - an autonomous robot equipped with rocket engines, its own power plant, communication and navigation systems, scientific instruments, photo and video equipment, was directed. All of this is managed by onboard computers. To study the found object, it was decided to take photographs at two points of its orbit. After taking the first photograph, it was necessary to adjust the probe's speed to be able to take another photograph of the celestial body at another point of its orbit.

Let's consider a simplified model of the situation that arose. We will consider the object under study (the celestial body) and the research probe as material points, the celestial body is moving in a circular orbit with the center at point $O$ and radius $R=1.2 \cdot 10^{6}$ km with a constant angular velocity $\omega=0.25 \cdot 10^{-5} \text{rad} / \text{s}$. The projection of the probe onto the orbital plane will be called the sub-probe point. The speed of the sub-probe point is constant and equal to $V_{1}$, and its trajectory in the orbital plane is conditionally considered a straight line intersecting the circle at points $P$ and $K$. According to the embedded program, the photography of the celestial body by the probe is carried out at the moments of their closest approach, which correspond to the moments of the sub-probe point's trajectory intersecting the body's orbit (points $P$ and $K$). When the celestial body (point $T$) is exactly on the line between point $O$ and the sub-probe point (point $Z$), a timer is started ($t_{0}=0$). At point $P$, the celestial body and the sub-probe point are at the same time, and photography is carried out, after which the probe's speed is changed so that over point $K$ it again coincides with the body for its repeated photography. The speed of the sub-probe point on the segment $P K$ is constant.

Determine the distance between the sub-probe point and the studied body at the initial moment of time $t_{0}$, as well as the speed of the sub-probe point $V_{2}$ on the segment $P K$, if the central angle $P O K$ is equal to the angle $P Z O$ and is one and a half times smaller than the central angle $P O T$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-13.jpg?height=454&width=657&top_left_y=2166&top_left_x=1199)
In calculations, use the approximate value of the number $\pi$ - round it to the nearest integer value.
Solution. Let the angle POK be $\alpha$. Triangle PKO is isosceles, so the two angles at the base are equal to $\beta=90-\alpha / 2, \quad$ angle POT is equal to $\frac{3 \cdot \alpha}{2}$, then $\alpha+\left(\alpha+\frac{3 \alpha}{2}\right)+90-\frac{\alpha}{2}=180 \Rightarrow 3 \alpha=90 \Rightarrow \alpha=30$.

Using the Law of Sines:

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-14.jpg?height=257&width=1696&top_left_y=568&top_left_x=183)
$Z O^{2}-Z O \cdot R \sqrt{2}-R^{2}=0 \Rightarrow Z O=\frac{R \sqrt{2}+\sqrt{6} R}{2} \Rightarrow Z T=\frac{R \sqrt{2}(1+\sqrt{3})}{2}-R=\frac{R(\sqrt{2}+\sqrt{6}-2)}{2}$

Or using the Law of Sines

$\frac{Z P}{\sin 45}=\frac{P O}{\sin 30}=\frac{Z O}{\sin 105} \Rightarrow Z O=\frac{2 R}{1} \sin 105=2 R \cos 15=2 R \sqrt{(1+\cos 30) / 2}=R \sqrt{2+\sqrt{3}}$

$V_{2}=\frac{P K}{t_{2}}, \quad t_{2}=\frac{\hat{P K}}{V_{T}}=\frac{\hat{P K}}{\omega R}=\frac{\pi R}{180} \cdot \alpha \frac{1}{\omega R}=\frac{\pi \alpha}{180 \omega}=\frac{3 \cdot 30}{180 \cdot 0.25 \cdot 10^{-5}}=\frac{1}{2 \cdot 25 \cdot 10^{-7}}=\frac{10^{6}}{5}=2 \cdot 10^{5}$

Thus, $V_{2}=\frac{P K}{2 \cdot 10^{5}}$. PK is found using the Law of Cosines:

$P K^{2}=2 R^{2}-2 R^{2} \cos \alpha=2 R^{2}\left(1-\frac{\sqrt{3}}{2}\right)=R^{2}(2-\sqrt{3}) \Rightarrow P K=R \sqrt{2-\sqrt{3}}$

$V_{2}=\frac{R \sqrt{2-\sqrt{3}}}{2 \cdot 10^{5}}=\frac{1.2 \cdot 10^{6} \sqrt{2-\sqrt{3}}}{2 \cdot 10^{5}}=6 \sqrt{2-\sqrt{3}} \approx 3.11 \mathrm{kM} / \mathrm{c}$

Answer: $0.6 \cdot 10^{6}(\sqrt{2}+\sqrt{6}-2) \kappa \mathrm{m}, \quad 6 \sqrt{2-\sqrt{3}} \approx 3.11 \kappa \mathrm{m} / \mathrm{s}$

## Grading Criteria for the Olympiad Work

Subject: Mathematics

Class: 10, Variants 3 and 4

Task 1 (Maximum score 12 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed, and/or the correct answer is simply written. | 0 |
| The probabilities of "success" and "failure" in one procedure (one trial in a series) are correctly calculated. | 3 |
| The formula for calculating the probability of the required number of "successes" in the described series of trials is correctly written (with the probability of "success" in one procedure calculated correctly). | 6 |
| The task is solved with one computational error. | 9 |
| A fully justified solution is provided, and the correct answer is obtained. | 12 |

Task 2 (Maximum score 16 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed. | 0 |
| The number 2023 (Variant 3) or 2025 (Variant 4) is factored into mutually prime factors. Conditions are obtained for $t$ under which the given expression is divisible by these factors. | 4 |
| The connection between the conditions obtained in the previous point is established. | 8 |
| With correct reasoning, a computational error is made. | 12 |
| A fully justified solution is provided, and the correct answer is obtained. | 16 |

Task 3 (Maximum score 16 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed, and/or the correct answer is simply written. | 0 |
| It is proven that $BM$ is the bisector of triangle $LBK$. | 4 |
| Correct relationships of segment lengths necessary for solving the problem are obtained. | 8 |
| A computational error is made in the solution with correct reasoning. | 12 |
| A fully justified solution is provided, and the correct answer is obtained. | 16 |

Task 4 (Maximum score 16 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed, and/or the correct answer is simply written. | 0 |
| The equation is reduced to a system of equations and inequalities. | 4 |
| Using correct reasoning, all boundary points of the desired set of values for $a$ are obtained. | 8 |
| Using correct reasoning, the set of values for $a$ is obtained, differing from the desired set by a finite number of points and/or incorrect solutions are found. | 12 |
| A fully justified solution is provided, and the correct answers are obtained. | 16 |

Task 5 (Maximum score 20 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed, and/or the correct answer is simply written. | 0 |
| The section is correctly constructed with a full description of the construction. | 5 |
| The necessary relationships for solving the problem are found, in which the section plane divides the edges of the pyramid. The connection between the distance from the apex of the pyramid to the section plane and the angle between the section plane and the base plane of the pyramid is established. | 10 |
| With correct reasoning, one computational error is made. | 15 |
| A fully justified solution is provided, and the correct answers are obtained. | 20 |

Task 6 (Maximum score 20 points)

| Criterion (select one criterion) | Points |
| :--- | :---: |
| No points from the list below are completed, and/or the correct answer is simply written. | 0 |
| The angles between the lines connecting the objects are correctly found. | 5 |
| The distance between the celestial body and the sub-point is correctly found at the initial moment of time, and the distance PK is found. | 10 |
| With correct reasoning, one computational error is made. | 15 |
| A fully justified solution is provided, and the correct answers are obtained. | 20 |