## Saturday, May 22, 2010

### Polygons and vertex orbits

Last Christmas, I went with my family to a road trip. I drove from Waco to New Orleans, a 8 hrs drive more or less, and when we reached there, I was a little tired. I laid down in the bed and stared at the roof of our room, and then, I started imaging a rubber ball bouncing all over a vertical cross section of our room, like a billiards ball, and then I wondered which kind of orbits could be ball have, depending of the incident angle of the first bounce.

After thinking a little, one can realize that for a rectangular cross section (my room's) this problem is not really hard, and the possible answers are quite few, depending on the ratio of the lengths of the rectangle. Then I though that a more interesting question would be when the cross section its a circle.

Basically two thing can happen, one is that the incident angle is such that the ball bounces only in finitely many places, and hence, the bouncing points make a periodic sequence on the circumference, and the other is that the bouncing points are dense in the circumference.

For instance, if the initial angle happens to be $\pi/2$, we are going to have only 2 bouncing points. Suppose that we have an initial angle $a$ and we are working with a unit circle, so that the arc length is the same as the angle value. If we look look up for periodic points, we require that $na=m\pi$, for some $n,m\in\mathbb{Z}^+$. That means that $a= \frac{m}{n}\pi$, and with out loss of generality, we can require $m/n \leq 1$. Then, one can draw the path of the ball as follows:
• Step 1: Start drawing the path at vertex $k=1$
• Step 2: Draw a line from the vertex $k$ to the vertex $k+m$ (sums are taken modulo $n$)
• Step 3: Go to step 2 until you hit vertex 1
For instance, in the case $n=5$, we will have two different patterns

The orange path is obtained for $m=1$ and $m=4$, and the black one, for $m=2$ and $m=3$. In general, $m$ and $n-m$ will lead to the same orbit, but made in different directions (clockwise and counterclockwise).

So in the remaining case, when $a$ is not a rational multiple of $\pi$, we have that the bouncing points are dense in the circumference.

This result is not surprising, as this picture is related with the problem of wrapping a line around a torus, and it is well know that this wrapping is dense if it has an irrational slope, and periodic otherwise.

When glazed in this perspective, it turns out that the original problem, with the rectangle, and the one with the circle are exactly the same, since a torus is just the complex plane modulo a rectangle ( a lattice ), so at the end, my more interesting problem turned out to be as simple as the original one. It is quite interesting how apparently different problems became just two perspectives of the same phenomenon, one could seem totally boring and the other one, completely attractive and challenging. Some would say, beauty is in the eye of the beholder, but sometimes what happens is that differences are in the eye of the beholder.

#### 1 comment:

1. Sos mono extra va... jajaja... me llegas por vivoras! :)