Four Turtles (or Bugs, or Dogs, etc.)

A well known pursuit problem has appeared in different guises: a certain number (say, $N)\;$ of animals of the same kind pursue each other in a sequence. If they start at the vertices of a regular $N\text{-gon},\;$ how long will it take for them to catch each other? What distance will each cover?

Four turtles

Most commonly $N = 4\;$ [Steinhaus, p. 136; Pappas, p. 228; Strogatz, p. 21; Chicken; p. 64, Wells, p. 201; Zeitz, p. 72; Rohrer, #49; Sharygin, p. 83, 141], although in [Gardner, p. 103] $N = 3.\;$ Any kind of animal capable of free motion and having an interest in its own kind will do for the purpose of the problem, although the slower kinds are more amenable to experimentation. Mathematically speaking, the requirement of them being of the same kind intends to justify an additional condition: all the animal actors move at the same constant speed. To make it more precise,

$N\;$ turtles are located at the corners of a regular $N\text{-gon}.\;$ On command they start moving with constant speed $V,\;$ each towards its counterclockwise neighbor. If the side of the polygon is $S,$

  1. How long will it take for the turtles to meet?
  2. What distance will each have covered?

The key observation here is that the turtles are all equal and none is more equal than the others. (This is of course an allusion to G. Orwell's classic Animal Farm.) For this reason, at any moment in time the turtles will be located at the vertices of a regular $N\text{-gon},\;$ exactly as they were at the start of the race. It may be difficult to point to an axiom that supports this conclusion, but the customary reasoning by symmetry applies: unless in time $t \gt 0\;$ they form a regular $N\text{-gon},\;$ either the turtles behaved differently or different things happen to them along the way. We tacitly assume that neither has happened.

The problem is not a difficult exercise in Calculus. However, both questions can be answered easily without Calculus. Graphically, if the polygons are drawn at regular intervals, the problem leads to the "whirl pattern" which, as D. Wells has observed, has a strong illusion of depth. Regretfully, the resulting prettiness is often the reason to peddle the problem in early grades. Paraphrasing Prof. Dumbledore, Head of Hogwarts School, the illusion of math education thus created will give us neither knowledge or truth.

Thus, while in motion, the turtles form regular $N\text{-gons}\;$ that share the same center. With time, the polygons rotate and shrink. Let's place an observer at the center of the polygons and make him rotate along with the turtles at their angular speed. What will such an observer see? He will see the turtles approaching him on straight lines, albeit moving sideways. The motion of the turtles along the radial lines is uniform. Indeed, their velocities that are directed along the side lines of the polygon have fixed projections on the radial lines joining the center with the vertices.

Let $OA\;$ and $OB\;$ be two such lines. $AB\;$ is the side length at time $t = 0,\;$ $S = AB;\;$ $OB\;$ is the distance to the center, $R = OB;\;$ $H\;$ is the midpoint of $AB.\;$ From $\Delta BOH,$

$\displaystyle\frac{S}{2} = R\cdot \sin(\alpha ),$

where $\displaystyle\alpha = \frac{\pi}{N}.\;$ The projection of velocity $\textbf{V}\;$ on $OB\;$ can be found from $\Delta BCD,\;$ where $CD\;$ is perpendicular to $OB.\;$ Taking $V = |\textbf{V}| = BC,\;$

$\displaystyle BD = V\cdot \cos(\frac{\pi}{2} - \alpha ) = V\cdot \sin(\alpha ).$

Thus from the view point of the rotating observer, the turtles move along straight lines, have to travel distance $R\;$ at the speed of $V\cdot \sin(\alpha ).\;$ The time need to accomplish this task is

$\displaystyle\begin{align} T&= \frac{R}{V\cdot \sin(\alpha )}\\ &= \frac{S}{2V\cdot \sin^2(\alpha )}\\ &= \frac{S}{V(1 - \cos(2\alpha ))}. \end{align}$

Thus the turtles will meet in time T as above. And, since they move with the constant speed $V,\;$ in time $T\;$ they will travel the distance of $D = VT:$

$\displaystyle D = \frac{S}{1 - \cos(2\alpha )}.$

For $N = 4,\;$ $\displaystyle 2\alpha = 2\frac{\pi}{4} = \frac{\pi}{2},\;$ so that $\cos(2\alpha ) = 0\;$ and $D = S,\;$ which explains the popularity of this case.

(The applet below allows you to experiment with the problem. You can choose either the stationary frame of reference or the one that rotates with the turtles. You'll have to press the "Make them move" button every time after changing your selection. Note that, to the rotating observer at the center of the polygons, the original polygon appears to rotate in the opposite direction.)


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The curves traced by the turtles are logarithmic spirals which probably can't be established without at least some Calculus. Unlike the radial motion, the angular component of the motion is not uniform. As the spirals approach their common pole - the center of the polygons - the rotation speeds up.

In passing, were the angular motion uniform, the turtles would trace a different curve: Archimedes' spiral with the equation $r = ct\;$ in polar coordinates $(r, t),\;$ where $c\;$ is a constant. The equation of the logarithmic spiral in polar coordinates is $r = ce^{kt},\;$ where $c\;$ and $k\;$ are constant. As we already observed, the smaller $r,\;$ the faster angle $t\;$ changes. On the other hand, the angle formed by the tangent at a point on the logarithmic spiral and its radius vector is constant. [Zwikker, p. 210] mentions that insects approach a burning candle along a logarithmic spiral because they try to see the light at a constant angle with the direction they fly in just as they do when they fly in the sunshine along a straight line.

Remark

Changing a frame of reference is a useful problem solving device. As an additional example, finding a suitable frame of reference renders the Four Travelers Problem almost trivial.

The foregoing derivation is based on [Zeitz, p. 72]. Steven Strogtaz also promotes changing a reference frame. But his is associated with a particular dog:

Put a camera on the chaser's head. It swivels in such a way as to always keep the chasee in the center of the picture. (Actually, that's wrong - it won't have to swivel at all, will it?)

The chasee will appear in the center of the picture and won't seem to be running away. You're just moving closer. Could you distinguish this movie from the following one (assuming no background!)? The chasee stays still in its corner, the chaser runs the length of the side of the square $S\;$ and catches the chasee? You couldn't distinguish, so in both cases, $S\;$ is the answer.

It is edifying to consider other arguments, some of which strongly depend on intensive arm waving.

Mathematical Mosaic (I. F. Sharygin)

At every moment the turtles are located at the vertices of a square. Additionally, since the direction of the motion of turtle $B\;$ is perpendicular to the directions of the motion of turle $A,\;$ the rate at which the two approach equals the speed of the turtle $A.\;$ The same is true of the pairs $B\;$ and $C,\;$ $C\;$ and $D,\;$ $D\;$ and $A.\;$ It then follows that the turtles will crawl $3m\;$ each and meet in $5\;$ min. (AB: the side of the square is $3 m\;$ and the speed of the turtles is $1 cm/sec.)$

More Thought Provokers (D. Rohrer)

... since each creature is running directly to its target, the paths of a chaser and its chasee are always at right angles. That is, the distance between the chaser and its chasee depends only on the movement of the chaser, not its chasee. Therefore, the distance that the chaser runs is the same as if the chasee did not run at all.

The Joy of Mathematics (T. Pappas)

Notice as the spiders move the size of the square they form shrinks, but it always remain a square. Each spider's path is perpendicular to the spider's path on its right. A spider will reach the spider on its right in the same time it would take if the spider on the right had not moved.

The Chicken from Minsk (Y. Chernyak)

Symmetry is very important here. Due to the symmetry of the arrangement, the turtles will always be arranged at the corners of a square.

The square will rotate and shrink, and the turtles will spiral into the center of the original square. As you can see directly from the figure, the rate of the shrinkage of the side will be the turtles' constant speed $V,\;$ so we can write $S = S_{0} - Vt\;$ for the length of the side $S\;$ as a function of time with the initial side length $S_{0}\;$ The shape does not change, but but the side of the square shrinks in a linear fashion with time, and they meet at the center at $\displaystyle t = \frac{S_{0}}{V}.\;$ If you wish a more formal solution with more quantitative details, look at the second figure.

References

  1. Yuri Chernyak, Robert M. Rose, The Chicken from Minsk: And 99 Other Infuriatingly Challenging Brain Teasers from the Great Russian Tradition of Math and Science, Basic Books, 1995
  2. M. Gardner, The Unexpected Hanging and Other Mathematical Diversions, The University of Chicago Press, 1991
  3. T. Pappas, The Joy of Mathematics, Wide World Publishing, 1989
  4. D. Rohrer, More Thought Provokers, Key Curriculum Press, 1994, #19
  5. I. F. Sharygin, Mathematical Mosaic, Mir, 2002 (in Russian)
  6. H. Steinhaus, Mathematical Snapshots, umpteen edition, Dover, 1999
  7. S. Strogatz, The Calculus of Friendship, Princeton University Press, 2009
  8. E. W. Weisstein, Mice Problem, MathWorld--A Wolfram Web Resource
  9. D. Wells, The Penguin Dictionary of Curious and Intersting Geometry, Penguin Books, 1991
  10. P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999
  11. C. Zwikker, The Advanced Geometry of Plane Curves and Their Applications, Dover, 2005

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