3. All vertices of a broken line lie on the faces of a cube with an edge length of 2, and each segment of the broken line is 3 units long. This broken line connects two farthest vertices of the cube. How many segments does such a broken line have at least?

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Solve: Consider the cube circumscribing a sphere with center $A$ and radius 3, intersecting the cube's faces at three arcs: $K \hat{L}, \widehat{L N}, \widehat{N K}$ (Figure 3). The points $K, L, N$ on the edges bisect these three edges. In fact, $A D_{1}=\sqrt{8}$, so $L D_{1}=\sqrt{A} L^{\overline{2}-\overline{D_{1}^{2}}}=1$, and similarly for points $K$ and $N$.

Let point $M$ be any point on the aforementioned arcs, for example, on the arc $K L$, and $M$ is different from any endpoint of the arc. We consider a sphere with center $M$ and radius 3, and point $A$ lies on this sphere. It is easy to prove that all other points of the cube are inside this sphere.

From this, it follows that point $M$ can only be connected to point $A$, and the length of the segment is 3. Therefore, the first node of the broken line can only be one of the points $L, K$, and $N$. From point $K$, only the distance to point $D$ is 3 (except for point $A$). Similarly, the points that can be reached from points $L$ and $N$ can only be points $B$ and $A_{1}$, so all nodes are vertices of the cube or adjacent to point $A$. Similarly, the next node of the broken line must be the midpoint of one of the cube's edges, and the next node will be one of the points $C, B_{1}$, $D_{1}$, and the next node will again be the midpoint of an edge, and the final endpoint of the sixth segment can fall on point $C_{1}$, the opposite vertex of point $A$ (Figure 4 is an example). Thus, the minimum number of segments in a broken line that satisfies the conditions of the problem is 6.