This puzzle was from a famous puzzle maker Dudeney. Don't give it away if you know the answer

 

In a rectangular room, 30' x 12' x 12', a spider is at the middle of an end wall (one with dimensions 12' x 12'), one foot from the ceiling.

 

The fly is at the middle of the opposite end wall, one foot above the floor. The fly is so scared it can't move.

 

What is the shortest distance the spider must crawl in the room to capture the fly? No web spinning. Hint: It is less than 42 '.

Impossible I say! What's the answer?

[hide]31ft it drops off the starting wall crawls 30ft to cross the floor and 1 foot up the wall to tuck into fly guts, never heard this one before[/hide]

That's a good answer. Now suppose that the only way the spider can move is if it crawls. Fragile little thing can't jump for fear of breaking a leg.

Are there any worm holes in this room? Can the spider undergoe quantum superposition?

Good question. I don't think so. Dudeney never knew about those.

Are we supposed to take the size of the spider or fly into consideration? Because if the spider crawled EXACTLY 42', he'd be right over top the fly. But if only needs to come close to the fly, he would need to crawl just under 42'.

 

That's my last desperate attempt, so if this is wrong, I give up and demand to know the answer. Do you know it?

spiders can drop from any height and not break a leg. their terminal velocity is very very low.

Hehe. This is supposed to be a more mathematical puzzle than a logic one. I didn't figure this one out either. It was in a book. I thought that the method to find the answer was really interesting, so I thought I'd post the puzzle here. I'd post the answer, but someone may want to figure it out first.

How do you do the spoiler thing and black out the answer?

[hide][ hide] [ /hide ][/hide]

[hide] Imagine the the box flattened so that the four 30' x 12' walls line up, connected by the 30' sides. One 12' x 12' wall will be connected to one of the four 30' x 12'. The last wall would be connected on the side two walls down and on the opposite side of the 30' x 12' wall. I'd draw a picture but I don't think that a spoiler would work on that. But imagine folding the box up again and placing the spider and fly where they're supposed to be. When you plug the numbers in, use the Pythagorean theorem, the answer is 40'. One leg of the triangle is 32 (1' + 1' + 30') and the other leg is 24 (12' + 6' + 6')[/hide]

thats not fair, you said distance not displacement.

I didn't write the puzzle! Don't blame me!

The answer works, and is entirely along the walls, actually - it's a weird answer...[hide]Didn't think that walking along 5 walls could afford a shorter path than 3...[/hide]

=uncool-