Pythagorean Theorem on Ice

By Mark Levi

In the spirit of winter fun, here is a short “skater’s proof” of the Pythagorean theorem. Starting in the corner \(O\) of a skating rink with the walls along the \(x\) and \(y\) axes, wearing perfectly slippery shoes on the perfectly slippery ice, I push away from the \(y\)-wall, acquiring speed \(a\) in the \(x\)-direction (see Figure 1). Next, I push away from the \(x\)-wall, gaining speed \(b\) in the \(y\)-direction. With the first push I acquired kinetic energy \(m a ^2 /2\), and with the second push I added \(m b ^2 /2\) to my kinetic energy. Indeed, the fact that I was sliding along the \(x\)-wall is irrelevant, because my gloves are perfectly slippery; it feels the same as if the wall were not sliding by at all, just like during the first push.¹ After the two pushes, my speed \(c\) is the hypotenuse of the velocity triangle, and thus my kinetic energy is \(m c ^2 /2\). But this energy is the accumulation of the two previous contributions:

\[\frac{mc^2}{2} = \frac{ma^2}{2}+ \frac{mb^2}{2}, \]

implying \(a ^2 + b ^2 = c ^2\). All this can be summarized by saying that the energies add as scalars, while the velocities add as vectors. Of course, the above is not meant as a rigorous proof and is rather an interpretation, or a physical incarnation, of the Pythagorean theorem.

¹ it is here that the orthogonality of the walls is used.

The figure in this article was provided by the author.

Mark Levi ([email protected]) is a professor of mathematics at the Pennsylvania State University.