BlaschkeLebesgue Theorem
BlaschkeLebesgue Theorem says that the Reuleaux triangle has the least area among plane figures of constant width.
Before giving a proof of this theorem, I present a few interesting properties of bodies of constant width. Let’s begin with equivalent definitions:
A bounded subset of the plane has constant width if one of the following happens:
 Given any direction the two lines having direction which are tangent to such that lies in the strip delimited by them have distance between them equal to .
 If is the support function associated to (i.e. then for every .
 As a direct consequence the Minkowski sum is equal to the ball .
One interesting property which will be used below (which is very interesting itself) is the fact that any plane figure of constant width admits a circumscribed regular hexagon.
Let us now turn to the proof. I will denote the area of and the mixed area of the convex plane figures in the sense that
or equivalently considering the expansion
It is not hard to see that is monotonically increasing in each argument. Moreover, we have .
Consider now a constant width figure of width and denote the regular hexagon circumscribed about . We can assume that is symmetric by the origin so . Then we have
Then we obtain
Therefore which is exactly the area of the Reuleaux triangle of width (which you can calculate as the sum of the areas of three sectors of the unit radius circle minus times the area of an equilateral triangle of sidelength ).
Of course, all the above proof lays upon the fact that we can find a regular hexagon circumscribed about our constant width figure, and the proof of this is not trivial (at least in my opinion) and will be the subject of a future post.
References:
[1] H. G. Eggleston – Convexity
[2] G. D. Chakerian – Sets of Constant Width

October 7, 2013 at 12:33 amEvery constant width set has a circumscribed regular hexagon  Beni Bogoşel's blog