The Ant on a Cube

Shortest surface path = √5 ≈ 2.236. Found by unfolding two adjacent faces into a 1×2 rectangle and drawing a straight line from one corner to the diagonally opposite corner. The ant crosses exactly two faces, entering the second face not at a corner but at the midpoint of the shared edge.

Surface shortest paths on polyhedra are found by unfolding the faces into a flat plane and drawing a straight line. The challenge is finding which unfolding gives the minimum distance — different unfoldings correspond to different routes across the surface. Step 1: The ant starts at corner A and must reach corner G (opposite diagonal corner). Any path must cross at least 2 faces. Step 2: Option 1 — Travel along 3 edges: distance = 1+1+1 = 3. Step 3: Option 2 — Cross 2 faces via a fold. Unfold the bottom face and the front face into a flat 2×1 rectangle. Ant at bottom-left (0,0), target at top-right (1,2) after unfolding. Straight-line distance = √(1²+2²) = √5 ≈ 2.236. Step 4: Option 3 — Cross the side face and top face in a different orientation. Same calculation by symmetry: √5. Step 5: Option 4 — Cross 3 faces. Unfold 3 adjacent faces: creates a shape where straight-line distance = √(1.5² + 1.5²) = √4.5 ≈ 2.121. Unfolding front (1×1), top (1×1), back (1×1) into a strip 1×3: ant at (0,0), target at (1,2) gives √5. Or in a 2×1.5 arrangement: ant at (0,0), target at (2,1.5), distance = √(4+2.25) = √6.25 = 2.5. The 2-face unfolding with √5 ≈ 2.236 beats all others. Step 6: Verify √5 is optimal: no 1-face path exists (start and end not on same face). The 2-face unfolding giving √5 is the shortest. All 3-face unfoldings give ≥ √5. Key Insight: Unfolding a 3D surface into 2D converts a curved-path problem into a straight-line problem — one of the most elegant techniques in geometry. The key step is trying all topologically distinct unfoldings and computing straight-line distances. The winner here (√5) is not obvious until you physically unfold the cube and measure.