# CTK Insights

• ## Pages

01 May

### A wrapping surprise

Jim Henle's book "The Proof and the Pudding" that I have recently reviewed contains a good deal of surprises. Across several chapters of the book the author looks into the billiard problem (that is very much like the Two Pails puzzle) and its modifications. One of these involves wrapping a ribbon around a 3D box. Start at a corner and move within one of the faces along the diagonal of that corner. When you hit an edge, just bend the ribbon around the edge to the adjacent face, keeping the same angle $(45^{\circ}).$ Here's an example with an incredible box of dimensions $\sqrt{2}\times e\times\pi:$

Can you figure out how that path will proceed? In a 2D billiard with the crazy dimensions like $\sqrt{2}$ or $\pi$, the path will circle without end. Most likely, this is what your intuition may suggest happens in the 3D case at hand. My intuition did. So here comes the first surprise:

The path that consists of five legs ends up right at the corner where it started. There is an easy explanation why this is so. As in the study of the regular billiard, we may - instead of bending around an edge - flatten two adjacent faces of the cube and let the path pass on a straight line:

The sequence of five successive face traversals snugly fit into a square of size $\sqrt{2}+e+\pi,$ with the (image of the) path going from one corner of the square to its opposite along the diagonal. Note one face outside the square not touched by the path.

Now, that you may think that you understand and can explain why the path returns to the starting corner, consider wrapping exactly same box but starting in a different face. Before, our first leg lay in a $\sqrt{2}\times\pi$ face. Now, let's start within a $\sqrt{2}\times e$ face. Can you predict what will happen to the path?

As you may surmise, the path will behave - if I may say so - in a more rational way. Given the incommensurate dimensions of the box it was rational to expect an endless path. This is what you get on the second attempt. But there remains a question to ponder: Why was the first path so short? Jim Henle leaves it to his readers to find the answer. So do I, though I'd like to suggest also starting the wrapping in the $e\times\pi$ face.