4. A spaceship is located in a half-space at a distance $a$ from its boundary. The crew knows this but has no idea in which direction to move to reach the boundary plane. The spaceship can fly along any trajectory in space, measuring the length of the path traveled, and has a sensor that signals when the boundary is reached. Can the spaceship guarantee reaching the boundary, traveling a distance of no more than $14a$?
(M. A. Evdokimov) Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_c9eb608fba8b4c875be2g-4.jpg?height=650&width=609&top_left_y=450&top_left_x=758)

Let the ship be at some point $O$. Consider a regular octahedron $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ circumscribed around a sphere of radius $a$ with center at point $O$. We will prove that the path $O \rightarrow A_{1} \rightarrow A_{2} \rightarrow A_{3} \rightarrow A_{4} \rightarrow A_{5} \rightarrow A_{6}$ will definitely allow reaching the boundary plane. Assume the opposite. Then the vertices of the octahedron, and thus the octahedron itself (the convex hull of its vertices), lie strictly inside the half-space. Therefore, the inscribed sphere of the octahedron, whose radius is $a$, also lies strictly inside the half-space. This leads to a contradiction, since the distance to the boundary plane of the half-space is $a$ by the problem's condition. Now we will show that the length of the path $O \rightarrow A_{1} \rightarrow A_{2} \rightarrow A_{3} \rightarrow A_{4} \rightarrow A_{5} \rightarrow A_{6}$ is less than $14a$. Let $O A_{1}=O A_{2}=O A_{3}=x$. We can write the volume of the pyramid $O A_{1} A_{2} A_{3}$ in two ways: $V=\frac{1}{6} x^{3}=\frac{1}{3} \cdot O H \cdot S_{A_{1} A_{2} A_{3}}=\frac{1}{3} \cdot a \cdot \sqrt{3} \cdot \frac{(x \sqrt{2})^{2}}{4}$. From this, $x=a \sqrt{3}$, and the length of the edge of the octahedron is $a \sqrt{6}$. Therefore, the length of the path is $(\sqrt{3}+5 \sqrt{6}) a < 14 a$, since $\sqrt{2} < 43 / 30$.

Answer. Yes, it can.

Comment. There are many other ways. Also, an example can be constructed where the length of the path is less than $13a$. For this, in the original solution, replace the path segment $A_{2} \rightarrow A_{3} \rightarrow A_{4} \rightarrow A_{5}$ with the one shown in the figure below (half the side of a square $\rightarrow$ semicircle $\rightarrow$ half the side of a square). In this case, the total length of the path will decrease by $\left(2-\frac{\pi}{2}\right) \cdot a \sqrt{6}$ and will be slightly less than $13a$. This example works because the convex hull of the points of the path still contains the inscribed sphere of the octahedron with radius $a$, since this sphere touches the generating line of the cone with vertex $A_{1}$ and base, which is the inscribed circle $\omega$ of the square $A_{2} A_{3} A_{4} A_{5}$.

The estimate can be improved to approximately $12.75a$ if, in the plane of the square $A_{2} A_{3} A_{4} A_{5}$, we consider a regular hexagon $A B C D E F$ circumscribed around the same circle $\omega$. The path will be as follows: along the segment $O A_{1}$ in a straight line to the vertex of the hexagon $A$, then along the tangent to the point of tangency with $\omega$, then along the arc of the circle $\omega$, then along the tangent to the point $F$, and finally along the segment $F A_{6}$. Computer modeling allows for a slight improvement in this construction if the distances from the points $A_{1}$ and $A_{6}$ to the plane of the circle $\omega$ and the path in this plane are adjusted so that the union of the two corresponding cones contains the sphere with center at point $O$ and radius $a$.

However, the improvement is minor. If any participant in the olympiad manages to significantly improve the upper bound or obtain a good lower bound, please write to the author of the problem at: mmo2022kosmos@mail.ru
![](https://cdn.mathpix.com/cropped/2024_05_06_c9eb608fba8b4c875be2g-5.jpg?height=858&width=1112&top_left_y=232&top_left_x=539)