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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