A room is 30 ft long with square end walls 12 ft by 12 ft. A spider sits on one end wall, 1 ft below the ceiling and centred left to right. A fly sits frozen on the opposite end wall, 1 ft above the floor and centred. The spider may only walk on surfaces. What is the length of the shortest path?

A room is 30 ft long with square end walls 12 ft by 12 ft. A spider sits on one end wall, 1 ft below the ceiling and centred left to right. A fly sits frozen on the opposite end wall, 1 ft above the floor and centred. The spider may only walk on surfaces. What is the length of the shortest path?

Approach: Flatten the sequence of faces the path crosses into one plane, then the shortest route is a straight segment. Try the unfolding that uses five faces rather than three.

40. The shortest path is 40 ft. Unfold the five faces the path crosses, the near end wall, the ceiling, one side wall, the floor and the far end wall, into a single plane. In that flat picture the spider and the fly are separated by 1 + 30 + 1 = 32 along the length and by 6 + 12 + 6 = 24 across, so the straight line distance is sqrt(32^2 + 24^2) = 40, and 24, 32, 40 is a Pythagorean triple. The obvious route straight down the near wall, along the floor and up the far wall measures 11 + 30 + 1 = 42, which is longer. Every unfolding gives a valid upper bound, and checking the finitely many face sequences shows 40 is the smallest.

Follow-up: For a cube of side 1 with the spider and fly at opposite corners, what is the shortest surface path?

Key concepts: unfolding the surface, straight line distance, pythagorean triple.