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# Pythagorean Theorem on Ice 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 ([email protected]) is a professor of mathematics at the Pennsylvania State University.