Help with hard math problem!
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ImakeTargets wrote:
A more interesting question is how quickly the cannibal would have to run in order to catch his meal. We can give fairly crude solutions to the current problem.
Yes I made this very point earlier... the "how fast does he need to go" question is much harder to answer precisely.
I think depending on how quickly the missionery can get out of the water and start running he can get away. Please seee diagram below for position of missionery m1, m2, m3, m4, and m5 and corresponding position of cannibal c1, c2, c3, c4 and c5.
Elroch wrote:
The most interesting question is what is the critical speed ratio? i.e. above this ratio and the cannibal feasts, below the ratio and the missionary survives.
(sqrt2)
the most efficient way seems to be to always maintain a 90 degree angle from the cannibal and swim at full speed, which implies you're moving in a spiral. in other words, if the cannibal is in one corner, you start swimming towards either the first corner to the cannibal's right or left. at each step (or microstep 
) you adjust your direction a bit so that you're still at 90 degrees. that method guarantees you're always moving outwards and away from the cannibal. now, whether your speed is fast enough to get away is a different story. i haven't done the math.
EDIT: on second thought, it's probably smarter to start moving towards the corner opposite to the one the cannibal starts at. then based on which side the cannibal decides to run to chase you, you move 90 degrees away from him and keep it at that angle, still moving at full speed.
Assuming Missionary swims to the opposite corner of a square pool of sides "x"
Distance for Cannibal to travel = 2x
Distance for Missionary to travel = x / (sqrt2) (make 4 right angle triangles, calculate length using Pythagorean theorem)
Speed of Missionare = v
Speed of Cannibal = 3v
Time taken for Missionary = distance / velocity = (x / (sqrt2)) / v = x / v sqrt2
Time taken for Cannibal = distance / velocity = 2x / 3v
Compare the two, you can see the Cannibal is quicker
To make them equal, the Canibal Time Taken needs to be multiplied by 3 / (sqrt2)
Which means his velocity (the variable) needs to be multiplied by (sqrt2) / 3
Therefore the Cannibal will catch him if his speed is more than sqrt2 times faster than the missionary
* sqrt = Square Root of
Oct 15, 2009
0
#48
I posted the question as my Facebook status. Normally I get zero to 3 comments on my status updates. I've had 40+ today already 
There have been some excellent responses. I think a couple of my friends know waaay too much math.
Oct 15, 2009
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#49
The time to get out of the pool is a factor too, which is why the pool size matters.
If the missionary beats him to the pool's edge by only a second or 2, it still won't be enough time to run away.
[COMMENT DELETED]
Oct 16, 2009
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#51
Anyone who has been interested in this problem may like to solve the similar problem where the missionary starts at the centre of a circular pool, and the cannibal is 4.5 times as fast on land as he is in the water. (There is a solution, but it is pretty close).
Does the speed of the cannibal or shape of the pool even matter.
If the missionary swims directly towards the point that is equidistant from the cannibal's current position and keeps changing his direction to head for the new equidistant point as the cannibal moves then the missionary will always be getting closer to the edge and the cannibal never getting closer to the (always changing) exit point.
If the cannibal keeps running in the one direction the path of the swimmer will be a spiral. The relative speeds of the two will determine the shape of the spiral.
If the cannibal is very fast then the spiral will be tight and the swimmer will go round in slightly ever increasing circles but will eventually get out and the cannibal will always be on the opposite side. So the cannibal is never close to the missionary.
When the speeds are as in the original example he will get out quite quickly.
The swimmer would naturally have to swim backstroke to keep an eye on the cannibal.
Sorry that was wrong. There is an upper speed limit. I didn't take into account that as the swimmer changes direction his new exit point will on average increase the distance he has to swim thus meaning there will be a speed difference which will limit the ability to excape.
It's not actually a pure maths riddle. Keeping History in mind, by the time the missionary swims to the edge of the pool, he would have converted the cannibal into a good christian.
Oct 16, 2009
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#55
This one used to bug me when I was a kid:
3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS $30, SO EACH MAN PAID $10 AND WENT TO THE ROOM.
A WHILE LATER THE MAN BEHIND THE DESK REALIZED THE ROOM WAS ONLY $25, SO HE SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH $5.
ON THE WAY, THE BELLBOY COULDN'T FIGURE OUT HOW TO SPLIT $5 EVENLY BETWEEN 3, SO HE GAVE EACH MAN A $1 AND KEPT THE OTHER $2 FOR HIMSELF.
THIS MEANT THAT THE 3 MEN EACH PAID $9 FOR THE ROOM, WHICH IS A TOTAL OF $27, ADD THE $2 THAT THE BELLBOY KEPT = $29.
WHERE IS THE OTHER DOLLAR?
(Sorry about the 'shouting.' It's a cut-&-paste.)
Skand wrote:
It's not actually a pure maths riddle. Keeping History in mind, by the time the missionary swims to the edge of the pool, he would have converted the cannibal into a good christian.
Rofl. Won't save the missionary though, the convert will just crucify him instead of eating him.
DeepGreene wrote:
This one used to bug me when I was a kid:
3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS $30, SO EACH MAN PAID $10 AND WENT TO THE ROOM.
A WHILE LATER THE MAN BEHIND THE DESK REALIZED THE ROOM WAS ONLY $25, SO HE SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH $5.
ON THE WAY, THE BELLBOY COULDN'T FIGURE OUT HOW TO SPLIT $5 EVENLY BETWEEN 3, SO HE GAVE EACH MAN A $1 AND KEPT THE OTHER $2 FOR HIMSELF.
THIS MEANT THAT THE 3 MEN EACH PAID $9 FOR THE ROOM, WHICH IS A TOTAL OF $27, ADD THE $2 THAT THE BELLBOY KEPT = $29.
WHERE IS THE OTHER DOLLAR?
(Sorry about the 'shouting.' It's a cut-&-paste.)
I think that the problem with that is at the end you are combining unlike terms. You are adding money the bellboy has to money the 3 men don'thave. It doesn't work because some of the money that the men paid went to the bellboy. It should say: "each man paid 9 dollars, which makes 27, and 2 of those dollars went to the bellboy and 25 to the man at the desk." or "Each man paid 10 dollars, which makes 30, then 25 went to the man at the desk, 2 to the bellboy, and 1 back to each of them." If you only add positives, everything makes sense: Each man has 1, the bellboy has 2, and the man at the desk has 25, so everything makes sense. Or if you think of it as each man paying 9 dollars, then that is the same as the man at the desk having 25, and the bellboy having 2 which makes 27.
edit: What marvellosity wrote works fine too. In fact his answer is simpler, which I think is better. Mine explains more, which somebody else may think is better. It's really just a matter of opinion. But both are correct.
DeepGreene wrote:
This one used to bug me when I was a kid:
3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS $30, SO EACH MAN PAID $10 AND WENT TO THE ROOM.
A WHILE LATER THE MAN BEHIND THE DESK REALIZED THE ROOM WAS ONLY $25, SO HE SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH $5.
ON THE WAY, THE BELLBOY COULDN'T FIGURE OUT HOW TO SPLIT $5 EVENLY BETWEEN 3, SO HE GAVE EACH MAN A $1 AND KEPT THE OTHER $2 FOR HIMSELF.
THIS MEANT THAT THE 3 MEN EACH PAID $9 FOR THE ROOM, WHICH IS A TOTAL OF $27, ADD THE $2 THAT THE BELLBOY KEPT = $29.
WHERE IS THE OTHER DOLLAR?
(Sorry about the 'shouting.' It's a cut-&-paste.)
The best way to answer this is to simply take the final line before the question and change it to
"THIS MEANT THAT THE 3 MEN EACH PAID $9 FOR THE ROOM, WHICH IS A TOTAL OF $27, MINUS THE $2 THAT THE BELLBOY KEPT = $25" - the cost of the room.
Oct 16, 2009
0
#59
oinquarki wrote:
DeepGreene wrote:
This one used to bug me when I was a kid:
3 MEN GO INTO A MOTEL. THE MAN BEHIND THE DESK SAID THE ROOM IS $30, SO EACH MAN PAID $10 AND WENT TO THE ROOM.
A WHILE LATER THE MAN BEHIND THE DESK REALIZED THE ROOM WAS ONLY $25, SO HE SENT THE BELLBOY TO THE 3 GUYS' ROOM WITH $5.
ON THE WAY, THE BELLBOY COULDN'T FIGURE OUT HOW TO SPLIT $5 EVENLY BETWEEN 3, SO HE GAVE EACH MAN A $1 AND KEPT THE OTHER $2 FOR HIMSELF.
THIS MEANT THAT THE 3 MEN EACH PAID $9 FOR THE ROOM, WHICH IS A TOTAL OF $27, ADD THE $2 THAT THE BELLBOY KEPT = $29.
WHERE IS THE OTHER DOLLAR?
(Sorry about the 'shouting.' It's a cut-&-paste.)
I think that the problem with that is at the end you are combining unlike terms. You are adding money the bellboy has to money the 3 men don't have. It doesn't work because some of the money that the men paid went to the bellboy. If you only add positives, everything makes sense: each man has 3, the bellboy has 2, and the man at the desk has 25, so everything makes sense. Or if you consider that each man did pay 9 dollars, then that is the same as the man at the desk having 25, and the bellboy having 2 which makes 27.
Well explained! :) Drove me nuts when I was a youngling. Cheers.
Does the pool have water in it?