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Pythagorean Theorem on Ice

By Mark Levi

Figure 1. A kinetic energy “proof” of the Pythagorean theorem.
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 (levi@math.psu.edu) is a professor of mathematics at the Pennsylvania State University.

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