THEOREM C: Every simple closed curve has at least one inscribed rectangle.

Proof: This proof is a bit more sophisticated than the others, but it is one of the most beautiful proofs I have ever seen.

Begin by noting that since J is topologically the same as the circle C, the Cartesian product J^_xJ is topologically equivalent to C^_xC -- a torus! (If you've had a course in topology this will make sense -- if not, picture C^_xC as a union of circles all placed around another circle to give the surface of a donut.)

Now the most common way for a mathematician to think about a torus is to draw it as the square [0,1]^_x[0,1], with the understanding that the pairs of opposite edges are to be sewn together (so that (0,2/3) and (1,2/3) actually represent the same point of the torus) -- see the diagram.

Now let X represent the set of all (unordered) pairs of points on the curve J. In other words, a "point" in X will be a pair {P,Q} of points on J. Now the points on our torus represent ordered pairs of points on J (since the torus is J^_xJ). But both (P,Q) and (Q,P) in J^_xJ represent the "point" {P,Q} in X, so topologically we can think of X as what we get when we "fold" our torus over onto itself, matching up (P,Q) and (Q,P). Using our representation of the torus as the square [0,1]^_x[0,1], this gives us a picture of X as what we get when we fold this square across its diagonal so as to identify the points (s,t) and (t,s) into a single "point" {s,t}. Notice that the diagonal of the square would correspond to ordered pairs (P,P), and thus to one-element sets {P} of points on J (instead of two-element sets like the ones in X).

Now a little bit of cutting and pasting back together shows us that X is actually a Mobius strip (more specifically an open Mobius strip since its bounding curve is the missing diagonal of "singleton sets") -- see the figure below. (Note that it is a Mobius strip because in the last step the sewing is done only after aligning the blue arrows on the opposite edges, which requires a twist.)

Imagine R² as the plane z=0 sitting inside x,y,z-space R³, and imagine J lying on that plane. Define a function f:X ---> R³ by the following rule: f({P,Q}) is the point in R³ lying directly over the midpoint of the segmentPQ but with z-coordinate equal to the distance from P to Q. This function clearly takes our Mobius strip X into the region above the x,y-plane, and sews its boundary onto the curve J (since as the "points" {P,Q} of X get close to the boundary of X, the distance from P to Q becomes very small and so the z-coordinate of f({P,Q}) becomes very close to zero).

It doesn't take much convincing for most people to accept that you can't do this in a one-to-one way. That is, the image set f(X) must have some collisions where more than one point of X is carried to the same point of R³. (Try building yourself a Mobius strip and sewing its boundary to a circle on a tabletop. You'll be convinced!) A topologist might require more convincing, but would believe it as soon as you remind them that you can't embed the space you get by sewing a disk to a Mobius strip (called the "projective plane") into three dimensions.

So, what does this show? Well, we now know that there must be two pairs of points {P,Q} and {P',Q'} so that f({P,Q}) = f({P',Q'}). This means that the segments PQ and P'Q' meet at their midpoints and have the same length. So, PP'QQ' is a rectangle!