dominicleejun wrote: consider a regular cube with side length 2. let A and B be 2 vertices that are furthest apart. construct a sequence of points on the surface of the cube $A_1,A_2,...,A_k$ so that $A_1=A$, $A_k=B$ and for any i = 1,..,k-1, the distance from $A_i$ to $A_{i+1}$ is 3. find the minimum value of k. Somebody please explain me this question because at each attempt to understand it in present wordings I'm getting more confused arriving at contradictions.

The sphere with centre $A$ and radius $3$ intersects the three edges at $B$ in three points $K, L, N$. By straightforward calculation using Pythagoras' Theorem, it is easy to show that they are the midpoints of the edges. Let $M$ be an interior point of any of the arcs $KL, LN, KN$ arising from the intersection with the sphere. The sphere with centre $M$ and radius 3 contains the point $A$. All the other points of the cube are in the interior of this sphere since the distance from $M$ to all the other vertices are $<3$. Thus $A_2$ must be one of $K, L, N$. From each of $K, L, N$ we can reach one of the vertices adjacent to $A$. Thus $A_3$ is a vertex connected to $A$ by a single edge. Now $A_4$ must be a midpoint of a side and $A_5$ is a vertex connected to $A$ by at most 2 edges. Since $A$ is connected to $B$ by a sequence of 3 edges, we see that $k\geq 7$. It is easy to construct a sequence of length 7 that works: \[(0,0,0)\rightarrow (2,2,1)\rightarrow (0,0,2)\rightarrow (2,1,0)\rightarrow (0,2,2)\rightarrow (1,0,0)\rightarrow (2,2,2).\]