Unit Cube Diagonal Path Explained

The question requires finding the shortest path length along the surface between the endpoints (A and B) of the longest diagonal of a unit cube.

Cube and Diagonal Definition

A unit cube has sides of length 1. The longest diagonal, or space diagonal, connects opposite vertices. Let vertex A be at coordinates $(0, 0, 0)$ and the opposite vertex B be at $(1, 1, 1)$.

Shortest Surface Path via Unfolding

To find the shortest path on the surface, we unfold the cube's faces into a 2D plane, forming a net. The path then becomes a straight line on this net.

Connecting A=(0,0,0) to B=(1,1,1) necessitates crossing at least two faces. We visualize unfolding two adjacent faces.

• Consider unfolding the bottom face (z=0) and the right face (x=1).
• Place A at the origin $(0,0)$ on the 2D plane representing the bottom face.
• Unfold the right face adjacent to the edge connecting $(1,0,0)$ and $(1,1,0)$.
• On this unfolded net, the endpoint B $(1,1,1)$ corresponds to a point effectively located at $(1+1, 1) = (2,1)$ relative to A.
• Alternatively, unfolding the bottom face (z=0) and the front face (y=1) places B effectively at $(1, 1+1) = (1,2)$ relative to A $(0,0)$.

This creates a rectangle on the net with dimensions either 2x1 or 1x2.

Shortest Distance Calculation

The shortest path length is the diagonal distance across this unfolded rectangle. We apply the Pythagorean theorem:

Path Length $= \sqrt{(\text{width})^2 + (\text{height})^2}$

Using the 2x1 dimensions: Length $= \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$

Using the 1x2 dimensions: Length $= \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$

Final Path Length

The shortest path length along the surface between A and B is $\sqrt{5}$.