There is a problem I need you to solve.

"In the centre of a square pool, there is a missionary. On the corner of the pool there is a cannibal waiting to eat the missionary. The cannibal can run thrice as fast as the missionary can swim, but the missionary can run faster than the cannibal. The cannibal cannot swim. Assuming that the characters are infinitely manuverable, can the missionary escape in finite time?"

Oh thats a doozy.  Thatll keep me up at night...  Is it yes?

I dunno! That'a why I'm asking you.

If the cannibal can't swim, and the characters are infinitely maneuverable, the missionary should just stay in the water.

True. I think the word escape requires further definition. If it means never being eaten, then as kim stated the missionary should just never leave the water. 

I think that if the missionary swims halfway to the corner opposite the cannibal and get halfway there, by then the cannibal will be slightly past the next corner and be about 94% of the way to the corner the missionary was heading towards. But for the missionary to go there will be a distance of about 35% of the square length (sqrt2/4). So then the missionary turns and goes to the side of the pool (the one further from the cannibal of course) and that is a distance of 25% the square length (it's pretty obvious why if you draw the square). But the cannibal needs to go an extra 25% in addition to the 94% he needs to reach the corner because he is going past the corner. So the cannibal is not fast enough and the missionary gets out alive.

Hey, he must try to escape! Both of them must keep moving.

I did a rough sketch, and assuming the missionary can't just stay in the pool as others have said, it appears that the missionary can escape by swimming for the corner of the pool opposite the corner the cannibal is on.

Edit: Actually I think oinquarki is right...

No, he cannot go directly to the corner. 3(sqrt2 / 2) > 2

Unless there is some trickery, no.  The missionary and the cannibal will either meet at the endge of the pool at the same time - or the cannibal will beat him.  The shortest path for missionary is to swim directly for one of the walls.  Assume a 10 m x 10 m square.  The missionary would swim 5m - the furthest distance the canniblal would need to run is 15m - so if he's three times faster they meet at the edge.  Swimming to the far corner would  mean the cannibal has to run 20 m - but the distance for the missionary would be over 7m - so the cannibal would beat him there.

So the missionary should play dead trick the cannibal into the water to get him - then swim away and run to safety :-)  Either that or convert him.

Now if the missionary can fake the canibal into one direction as has been stated above, then he should be able to escape - but that requires some fairly difficult math to do in your head while being chased to death.  And come on won't nearly everyone try to go in a straight line initially?

This is precisely why I never understood algebra.

 Me? Right about something? Yeah and a flying pig just painted Mars purple...

Wasabi_Kid: In the time that the missionary swims to the opposite corner (distance = sqrt(2)/2), the cannibal will have enough time to run the distance (3/2)*sqrt(2) ~= 2.123, which is enough to intercept the missionary.

The cannibal is not fastr enough, 3x is just too slow, the ratio needs to be larger than pi, so 3.2 times as fat would do it. I think... Don't try this at home.

Yes the missionary can escape. Give me a little time to present the answer clearly.

My bad...

I'm still young, haven't really learned algebra yet. Also this was just a rough sketch that I did.

oinquarki is right. the missionary escapes easily by going halfway to a corner and then turning away from the side the cannibal chose to run along. He can even take a longer path (to the midpoint of the side).

 I think that even if the cannibal could run 3.2 times as fast as the missionary swims the missionary can still escape using the method I proposed. The missionary swims about 60% of the side length but the cannibal needs to run 225%.

Times as fast?

And yes, I still believe oinquarki is right, even if a flying pig did just paint Mars purple.

Yes. I see son_of_pushwood's solution is a bit more efficient. But mine is easier to calculate and still works

edit; I just realized that it doesn't work because the cannibal can go around the other way.