Here we have a rectangular room, measuring 30 feet by 12 feet, and 12 feet high.

There is a spider in the middle of one of the end walls, 1 foot from the ceiling (A).

There is a fly in the middle of the opposite wall, 1 foot from the floor (B).

What is the shortest distance that the spider must **crawl** in order to reach the fly?

The Spider and the Fly – The Canterbury Puzzles, Henry Ernest Dudeney.

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Hint

Going down and across the floor isn't the only route.

Answer

40 feet.

Explanation Diagram

If you imagine the room to be a cardboard box, you can 'unfold' the room in various ways, and each route gives a different answer.

We can use Pythagoras' theorem (a^{2} + b^{2} = c^{2}) to calculate the distances:

distance^{2} = horizontal^{2} + vertical^{2}

distance = √(horizontal^{2} + vertical^{2})

Route #1

distance = 1 + 30 + 11 = 42 feet.

Route #2

horizontal = 6 + 30 + 6 = 42 feet.

vertical = 10 feet.

distance = √(42^{2} + 10^{2}) ≈ 43.174 feet.

Route #3

horizontal = 1 + 30 + 6 = 37 feet.

vertical = 6 + 11 = 17 feet.

distance = √(37^{2} + 17^{2}) ≈ 43.178 feet.

Route #4

horizontal = 1 + 30 + 1 = 32 feet.

vertical = 6 + 12 + 6 = 24 feet.

distance = √(32^{2} + 24^{2}) = 40 feet.