Besicovitch constants and the Besicovitch-Kakeyo problem

This is an attempt to explain the Besicovitch constants and how their arbitrarily small size solves the Besicovitch-Kakeyo problem.

First, the Besicovitch-Kakeyo problem - this problem involves reversing a unit length line segment in the minimal amount of area.  One way to imagine it is to imagine the line segment oozing paint as it sweeps across the plane and attempting to make the painted area minimal.  Another way to think of it which I actually prefer is to think of the reversal taking place entirely within a region and attempting to make that region of minimal area.

It turns out that the minimal area cannot be zero, but it can be arbitrarily small.  This seems a bit counterintuitive, probably because if you count sweeps over the same area twice, the minimal area is Pi / 4.  So the magic takes place by sweeping over areas in just the right ways so that multiple sweeps keeps the area below whatever preassigned value you might choose.

Before we get started, it's important to note that we can translate the line segment along it's own infinite line sweeping out zero space.  We can also move the line to a parallel line at a specific distance using an arbitrarily small area by rotating by an arbitrarily small angle, sliding the segment along it's new line until it's endpoint makes contact with the parallel (infinite) line at which point we can rotate it back the same arbitrarily small angle until it lies on the new parallel line and slide it back into place.  This will become important in the discussion below.

In order to solve the Besicovitch-Kakeyo problem we need to introduce the Besicovitch constants which are defined as follows.  Suppose a triangle, ABC is divided on the BC leg into n equal parts.  Each of these parts is connected to C to form n smaller triangles as shown here...

We'll name the n'th triangle above s_n.  We now slide the triangles along the infinite line which contains the base BC as shown here...

The pointy area which represents the union of these translated triangles is obviously of smaller area than the original triangle since several of the s_n overlap.  The Besicovitch constant for n triangles, B_n , is the smallest area possible in this union if the area of the original triangle is 1.  We'll name the union of the triangles in this minimum configuration the Besicovitch figure for n triangles.

An important point to note is that we can move a unit length line segment in line AC of the original triangle to one in line AB - that is, we can rotate it by angle A and this can be done in an area arbitrarily close to or less than the B_n for any n. We can do this because after the translations, the right side of s_n is still parallel to the left side of triangle s_n+1. By the discussion above, then, we can move from the right side of the translated triangle s_n to the left side of the translated triangle s_n+1 using an arbitrarily small amount of sweep space. When we have moved the segment to the left side of triangle s_n+1 we sweep across it's area to the right side of the triangle and then repeat the process, moving to the parallel left side of the translated triangle s_n+2. In this process, we make at most n + 2 parallel moves (the extra two moves are to possibly move from the original position on the line defined by AC to the left side of s₁ and finally from the right side of s_n to the final position on the line defined by AB). These parallel moves can be done in an arbitrarily small amount of space (remember, we've set n, so we can arrange for the total parallel moves over all n to be arbitrarily small). Additionally, we sweep across every s_i in its translated position but these sweeps all take place within the confines of the Besicovitch figure for the n triangles so sweeps out an area less than or equal to B_n.

It's then easy to see that if angle A is a right angle and we place a similar triangle next to this one, we can reverse the line segment using this procedure twice - once for each triangle. The upshot is that if the B_n approach zero for arbitrarily large n, then we can reverse the line segment in an arbitrarily small area.

So guess what?  Perron showed that b_n does indeed approach zero as n approaches infinity. Well, not really.  Perron showed that B_2^n approaches 0 as n approaches infinity. Of course, this is enough to prove the Besicovitch-Kakeyo theorem since all we need is some B_n smaller than any preassigned value. In fact, as of 1982, it was only a conjecture that B_n > B_n+1 which seems pretty amazing.  It would seem more amazing yet if this turned out to be false. I'm unaware of whether it's been proven since then.  Anyway, Perron's proof isn't too hard to understand but is a bit longish and requires a lot of pictures so I'm not putting it up until I get a little more inspired, but the details can be found in "Mathematical Time Exposures", a really interesting book by Isaac Schoenberg with a lot of other interesting articles in it.  I think it's no longer published, but it's well worth the trouble to find it.  A very brief description of the process is also given in this Wikipedia article in the section on Besicovitch Sets.