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Four Dogs Running


Date: 08/08/2001 at 04:28:09
From: Devon King
Subject: Geometry/algebra

Four dogs are at four corners of a field. Each dog chases the dog to 
its right; all four run at the same speed and no acceleration is 
assumed. The questions are:

1. Where will they meet? 
2. How far will they have run when they meet?
3. How long will it take them to meet?

I can only assume the they will run a perfect arc before meeting in 
the middle. If we assume that the length of the field is x, the 
distance run will be 0.25 pi x. If we assume that their speed is s
then the time taken will be 0.25 pi x/s. Is this correct, or is my 
first assumption wrong?


Date: 08/08/2001 at 12:44:24
From: Doctor Peterson
Subject: Re: Geometry/algebra

Hi, Devon.

Your assumption goes too far; we can't assume something just because 
it makes the problem easy to solve. You can hope it will work out that 
way, and look for ways to prove it, but in this case that won't work. 
The "pursuit curve" is not just an arc of a circle. But in fact, you 
don't need to know the shape of the curve.

Properly, this kind of "pursuit problem" belongs in the study of 
calculus. Even the trick answer I'm familiar with depends on the sort 
of thinking you do in calculus. But here's the basic idea; see if you 
can follow the reasoning.

Picture a short interval of time after the dogs start running. Each 
dog will have run a short distance almost along the edge of the 
square, since in that short time the dog it is chasing will not have 
moved much. If you draw this (exaggerating how far they have moved, so 
you can see it), you will find the dogs are still at the corners of a 
square, but the square is tilted slightly. The distance each dog has 
run, plus its distance to the dog it is chasing, is (approximately, if 
the time is short enough) the length of a side of the original square.

If you continue this for repeated small intervals, you will see a 
sequence of more and more tilted, smaller and smaller squares. But the 
sum of the distance a dog has gone, and the distance it has left to 
go, will not change, because the dog in front is always moving 
perpendicular to the direction your dog is going. See if this gives 
you an idea as to how far the dog ran.

It's hard to be fully convincing about this without some pretty deep 
thinking; but you can find a fuller discussion of both the calculus 
methods (which you can skip) and the trick, in this answer from our 
archives to a harder question:

   Catching a Pig
   http://mathforum.org/dr.math/problems/udo5.9.97.html   

(Look for the discussion of the four turtles in Martin Gardner's book 
_Aha! Gotcha: Paradoxes to Puzzle and Delight_ in the second answer.)

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
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