The Inscribed Squares Problem

THEOREM A: Every simple closed curve that is symmetric about the origin has an inscribed square.

Proof: Let J be a simple closed curve symmetric about the origin (so that for each point P on J the point -P -- with coordinates that are negatives of the coordinates of P -- is also on J). Also let f: R2 ---> R2 be the function that rotates each point in the plane 90 degrees about the origin.

Notice in the figure above that it appears that J and f(J) cross each other. We'll see shorly that this must actually happen -- these two curves have to share a point in common. So, assuming for the moment that there is such a point, choose a point f(P) on f(J) that is also on J (as in the figure above).

Then it isn't hard to see that the four points P, f(P), -P, and -f(P) (all points on J) form the vertices of a square.

So, all we need to do now is convince ourselves that J and f(J) really do have points in common. But this isn't too hard: let Pnear and Pfar be points of J that are at minimum and maximum distance from the origin O. Then:

Note: A similar theorem (from the same reference) states that any simple closed curve with symmetry across a line of reflection will have an inscribed square. See Exercise 4.