The Spider and the Fly

There is a room with the following dimensions: Length 30 ft, Width 12 ft, Height 12 ft.

On one of the 12x12 walls there is a spider, 1 ft from the ground and 6 ft from both long walls. On the opposing wall there is a fly, 1 ft from the ceiling and also 6 ft from both long walls. The spider can crawl over all walls, floor and ceiling.

What is the shortest path to get to the fly? The answer is not 42 (for a change) and there are no tricks.

Hint

To solve this puzzle, it helps to cut out pieces of graph paper representing the walls, ceiling, and floor and lay them flat to create a net of the room.

Here is a net of the room, with a white dot representing the spider and a black dot representing the fly. I excluded one of the long walls because the spider doesn't need to crawl on both of them.

Answer

Were the spider to crawl down to the floor, across the floor, and up the fly's wall, it would travel 42 feet. The same would be true if it were to start by climbing up to the ceiling.

But we can "unfold" the room differently to consider different paths. For example:

If you draw a straight line from the spider to the fly and measure it, you will see this route (spider's wall -> floor -> long wall -> fly's wall) is slightly shorter. You can also calculate the exact distance with the Pythagorean theorem. For a triangle with base 37 feet and height 17 feet, the hypotenuse is 40.7 feet. It's bit shorter, but let's try one more unfolding:

This route is even shorter. You can check with a ruler, or we can find the exact distance with Pythagorean theorem. This time, the base of the triangle is 32 feet and the height is 24. This means the spider's journey is only 40 feet. This is the shortest possible route!