2. The broken line, where each segment has a length of 3 and whose vertices belong to the vertices of a cube with an edge of 2, connects two of the most distant vertices of that cube. What is the smallest number of segments that such a broken line can have?
Solution. Let the broken line start at vertex $A$ and end at vertex $C^{\prime}$. The sphere with radius 3 and center at $A$ intersects the surface of the cube in three arcs: $K L, L N, N K$, where $K, L, N$ are the midpoints of the edges $B^{\prime} C^{\prime}, C^{\prime} D^{\prime}, C^{\prime} C$, respectively, as shown in the right diagram. This means that the next point $M$ on the broken line lies on one of these arcs. If point $M$ does not coincide with any of the points $K, L, N$, then the sphere with radius 3 and center $M$ passes through point $A$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_3665c50b48bf37a35b20g-3.jpg?height=448&width=463&top_left_y=1218&top_left_x=999)
contains the remaining points of the cube in its interior. Therefore, the broken line passes through one of the points $K, L, N$. Let this be point $K$. Now, the only segment of length 3 is obtained if the broken line passes through vertex $D$ (Why?). Continuing this process, we find that the broken line from $A$ to $C^{\prime}$ must contain at least six segments. One such broken line is shown in the left diagram.