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You are on a rowboat in the middle of a large, perfectly circular lake. On the perimeter of the lake is a monster who wants to eat you, but fortunately, he can't swim. He can run (along the perimeter) exactly $4\times$ as fast as you can row, and he will always run towards the closest bit of shore to you. If you can touch shore even for a second without the monster already being upon you, you can escape. Suggest a strategy that will allow you to escape, and prove that it works.

Other notes:

  • If two paths take the monster to this location equally quickly, he will arbitrarily choose one.
  • The monster can reverse direction instantaneously, and you can turn your boat instantaneously.

Follow up: What is the minimum speed of the monster (relative to your boat) such that escape becomes impossible?

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Is this from the datagenetics blog? datagenetics.com/blog/october12013/index.html –  greg Aug 27 at 17:57
    
The basic problem is older than that. I don't know the origin, but two years before the datagenetics post it was already referred to as "a famous problem". –  Geobits Aug 27 at 18:15
    
I had heard it before too actually. I didn't make it up, I'll admit. –  Neil Aug 27 at 20:23
    
I heard about it. It was a duck in a pond and a cat running around. The duck needs to touch the ground to fly away. –  Florian F Sep 14 at 15:06
    
@FlorianF I've heard that variation too. I suppose I prefer this version a little more for the dramatic effect. –  Neil Sep 15 at 7:35

3 Answers 3

up vote 9 down vote accepted

First of all, row out to a radius $R/4$ (where the lake has radius $R$) keeping you, the centre of the lake and the monster in a straight line - with you on the far side to the monster. This is always possible; radius $R/4$ is the first point where the angular speed you can achieve just matches that of the monster as he runs round to get you.

You are now a distance $3R/4$ away from the shore, directly opposite the monster so he needs to run a distance $\pi R$ to get you. You will take time $3R/4V$ at speed $V$ if you now row directly towards the nearest shore, and he will take $\pi R/4V$, which is fractionally greater.


For the followup: If instead of $4\times$, the monster runs $N\times$ your speed... then you row out to radius $R/N$, you then take time $(N-1)R/NV$ to reach shore and he takes $\pi R/NV$ to reach the same point. You escape provided that $N < \pi + 1 \approx 4.1459$.

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For the followup - if instead of 4x, the monster runs Nx your speed... then you row out to radius R/N, you then take time (N-1)R/NV to reach shore and he takes pi R/NV to reach the same point. You escape provided N < pi + 1 = 4.1459 –  Julia Hayward Aug 27 at 10:34
    
You should add that comment as part of the answer. I think it's interesting and relevant enough, and my understanding is that it doesn't belong in the comments anyway. –  EnvisionAndDevelop Aug 27 at 14:42
    
Bear in mind that from your $3R/4$ distance the monster will not move until you cross the centre of the lake as up until that time the closest shore is still the one astern. This means you only need $2R/4$ compared to the monster's $\pi R/4$. –  Hugh Oct 25 at 6:45

It is possible to escape a monster that is a little more than $4.6$ times faster. Given a monster running at speed $X$ times rowing speed and a lake with radius $R$, you must first row to the circle that is $R/X$ from the centre of the lake by spiralling out while keeping the centre of the lake between you and the monster. Thereafter you should take a route that is tangential to this circle in the opposite direction of what the monster is currently running. This is longer than going the direct route to shore, but the monster also get a longer route to run.


Your path in red, monster path in green. The shape of the inner spiral is not exact.

Why is this exact solution best? Because of differential maths. That is a long explanation, personally I find that stuffing some equations into Wolfram Alpha works.

It turns out that the monster must run $R(\pi+\varphi)$ where $\varphi$ is given by:

$\tan(\varphi)=\pi+\varphi,\\0\leq\varphi\leq\pi/2$

From the circle you have to row a distance of $R\times\sin(\varphi)$, so the fastest monster that can be overcome is running $(\pi+\varphi)/\sin(\varphi)$ times faster than you row, which is approximately equal to $4.6033388487517$

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The monster can change direction if it is quicker to you in the opposite direction. In the last leg before you reach the shore, the monster will not necessarily continue in the same direction. –  Neil Aug 27 at 15:37
    
@Neil If the monster change direction, so do you, you can mirror your path at any point if necessary. But according to the described monster behaviour the monster will not change direction during the last leg as there is no point where you are closer or as close the other way round. –  eBusiness Aug 27 at 15:46
    
@EnvisionAndDevelop The inner circle is located exactly where you can only just keep up with the monster, therefore as soon as you are outside it there is no way to go so fast that the monster change direction. –  eBusiness Aug 27 at 16:35
    
@EnvisionAndDevelop In which case you must of course move down. –  eBusiness Aug 27 at 17:15
    
Actually, the spiral in the circle can be replace by a semi-circle. You can reach the circle in finite time! –  Florian F Sep 14 at 15:15

The monster wants to eat me, but he is also particuarly fond of my boat since "he will always run towards the closest bit of shore to your boat."

So I row the boat in one direction and then jump off and swim in the opposite direction. Even the fastest monsters would never catch me!

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3  
Maybe if the question were tagged "lateral thinking", but this pretty clearly wasn't the sort of solution the puzzle was looking for. –  EnvisionAndDevelop Aug 27 at 16:19
    
I can't fault you for finding a flaw in my wording, though no, this isn't the solution I had in mind. –  Neil Aug 27 at 20:25
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Yeah, but then a shark eats you. –  Ilmari Karonen Aug 27 at 23:20
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I s going to go with "it never says that you can swim", but i prefer the shark. –  Joel Rondeau Aug 29 at 0:27

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