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There is one rocket in each corner of a square. At some point they start moving towards the rocket in the neighbouring corner in (say) clockwise direction. Their subsequent motion is such that they move exactly in the direction of the rocket they "follow" at any moment. The magnitude of their speed is constant (and is the same for any rocket). They all collide at the center of the square (as can be seen intuitively). What is the length of their path taken before the collision? I heard that it has a very easy and a complicated solution too (with integrals). Could you give both and tell if it can be generalized for regular n-gons?

The answer is the length of the sides of the square but I don't know how we arrive at it.

Matthew Panks
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  • Why can it be seen intuitively that they meet in the center of the square? I could understand that they can't meet at any other place by some (possibly easy) argument from symmetry, but why is it obvious that they meet at all? – DRF Jun 26 '15 at 13:50
  • I'm not even convinced the rockets crash. Sure, I think they probably do, but until I either calculate it or see a convincing strict calculaton showing that's the case, I wouldn't go around explaining how that's "intuitively" true. – 5xum Jun 26 '15 at 13:51
  • http://mathworld.wolfram.com/MiceProblem.html – Narasimham Jun 26 '15 at 13:52
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    They will collide beacause the only case they are not getting nearer to each other would be when they are moving in the same direction which of course can't happen. – Matthew Panks Jun 26 '15 at 13:56
  • Narasimham thanks but how do we arrive at the formula for their path? – Matthew Panks Jun 26 '15 at 13:58
  • @MatthewPanks But there is no "intuitive" reasoning that tells you that the rockets will collide and not simply get nearer and nearer asymptotically – 5xum Jun 26 '15 at 14:34

2 Answers2

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By putting the origin in the center of the square, we have that the position of a rocket/mice and its velocity always make an angle equal to $45^\circ$, so it is not difficult to guess that the trajectory is given by a logarithmic spiral

$$ r(\theta) = r_0 \exp\left(-\frac{\theta}{\tan 45^\circ}\right)=r_0 \exp(-\theta). $$

Assuming that the square has unit side length, we have $r_0=\frac{1}{\sqrt{2}}$ and the length of the logarithmic spiral is given by:

$$ L = \int_{0}^{+\infty}\sqrt{2 r_0^2 e^{-2\theta}}\,d\theta = 1. $$ The polygonal case is similar.

Jack D'Aurizio
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  • Thanks, but wikipedia says about the square case: " In this case they meet after a time of one unit, because the distance between two neighboring mice always decreases at a speed of one unit." and I don't get this, could you explain it too? – Matthew Panks Jun 26 '15 at 14:42
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    @MatthewPanks: that is not related to my solution, so I do not see a reason to explain it. Anyway, you can check it by computing the distance between a rocket and its neighbour as a function of the time. – Jack D'Aurizio Jun 26 '15 at 14:56
  • OK, my wording wasn't good (I shouldn't have written "but"). Still, I would be interested in how the square case has also a simpler solution. – Matthew Panks Jun 27 '15 at 12:29
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Draw a diagram to show neighbor positions and slopes to obtain a nice differential equation:

$$ \dfrac{dy}{dx} = \dfrac{x-y}{x+y} $$

Narasimham
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