PUZZLE #4: BUG IN A SHOEBOX VARIANT (SOLUTION)

Here is a variation on this puzzle that has an even more obvious wrong answer; I found it in a 1962 edition of "Mathematics for Pleasure," by Oswald Jacoby and William H. Benson: Instead of a shoebox, we have a room that is 12 ft. wide, 12 ft. high, and 30 ft. long. The bug, starting on one square wall 1 ft. from the ceiling and 6 ft. from either adjoining wall, wants to crawl to the opposite wall, 1 ft. from the floor and 6 ft. from either adjoining wall. How long is the shortest possible path?

The obvious answer is, of course, that the bug crawls straight up to the ceiling, straight across the ceiling, and down the opposite wall, for a total distance of 1+30+11 = 42 ft. Taking a clue from the previous problem, however, and by judicious cutting along edges and flattening, we arrive at a straight line distance that, when refolding takes place, gives rise to a broken-line path that touches five of the six sides of the room. While there are many ways to perform the cutting, the best is as follows: First cut along three of the edges of each square wall, in such a way that the bug's starting and ending points are closest to the uncut edge on each of the respective walls. Next cut along just one of the long edges and fold out flat. You'll find that the straight line between the two end points of the bug's trip does not cross any of the cuts you made, and it is the hypotenuse of a right triangle with side lengths 24 ft. and 32 ft. Hence the shortest path has length (24² +32²)^1/2 = 40 ft.