A Converging Rabbit

I’ve decided to make a page of some fun problems. I’m not sure how successful this will be, but in theory, there will eventually be a nice long list of my favorite problems. Here’s the first one:

A rabbit climbs out at its hole, and walks 1 mile in a straight line. Then, the rabbit repeatedly turns \pi/3 radians and walks half of the distance it just walked,
as pictured below.

How far away from the rabbit’s hole is the point at which the rabbit converges?

This is an ARML-style problem: There are lots of ways of solving this, but some methods are cleaner and faster than others. Click to see some solutions.

One way to do this is to break the rabbit’s path into x and y components and compute each separately as a geometric series that repeats every three turns. A simpler way would be to consider the rabbit’s path as a geometric series of complex numbers in the complex plane; the sum of the geometric series yields the rabbit’s endpoint.

The easiest way doesn’t use infinite series at all. Instead, let the initial point be A and let the location of the first turn be B, and let the converging point be X. Then the path from B to X is a rotation and dilation of the path of A to X, so that BX = \frac12 AX and \angle AXB = \frac\pi3. The Law of Cosines on \triangle AXB yields the desired result.


5 responses to “A Converging Rabbit

  1. Américo Tavares June 13, 2011 at 2:09 pm

    Fun problem indeed. Could you please explain why \angle AXB=\frac{\pi}{3}?

    • Moor Xu June 13, 2011 at 2:55 pm

      The path from B to X can be obtained by taking the path from A to X and applying a dilation by 1/2 and a rotation by \frac\pi3. This same dilation and rotation also applies to transform \overline{AX} into \overline{BX}. This means that \overline{BX} is simply \overline{AX} rotated by \frac\pi3 about X and shrunk by a factor of 2; that’s why we have BX = \frac12 AX and \angle AXB = \frac\pi3.

  2. Pingback: Visualização da convergência de uma série geométrica complexa | problemas | teoremas

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