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Blaschke-Lebesgue Theorem

Blaschke-Lebesgue Theorem says that the Reuleaux triangle has the least area among plane figures of constant width.200px-ReuleauxTriangle.svg

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 {K} of the plane has constant width {\alpha} if one of the following happens:

  • Given any direction {\nu} the two lines having direction {\nu} which are tangent to {K} such that {K} lies in the strip delimited by them have distance between them equal to {\alpha}.
  • If {h} is the support function associated to {K} (i.e. {h(\nu)=\max_{x \in K} x \cdot \nu)} then {h(\nu)+h(-\nu)=\alpha} for every {\nu \in \Bbb{S}^1}.
  • As a direct consequence the Minkowski sum {K+(-K)} is equal to the ball {\alpha B_1}.

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 {A(K)} the area of {K} and {A(K_1,K_2)} the mixed area of the convex plane figures {K_1,K_2} in the sense that

\displaystyle A(K_1+K_2)=A(K_1)+2A(K_1,K_2)+A(K_2),

or equivalently considering the expansion

\displaystyle A(sK_1+tK_2)=s^2A(K_1)+2stA(K_1,K_2)+t^2A(K_2).

It is not hard to see that {A(K_1,K_2)} is monotonically increasing in each argument. Moreover, we have {A(K,K)=A(K)}.

Consider now a constant width figure {B} of width {1} and denote {H} the regular hexagon circumscribed about {B}. We can assume that {H} is symmetric by the origin so {H=-H}. Then we have

\displaystyle A(B,-B)\leq A(H,-H)=A(H,H)=A(H).

Then we obtain

\displaystyle \pi = A(B+(-B))=2A(B)+2A(B,-B) \leq

\displaystyle \leq 2A(B)+2A(H)=2A(B)+\sqrt{3}.

Therefore {A(B) \geq (\pi-\sqrt{3})/2} which is exactly the area of the Reuleaux triangle of width {1} (which you can calculate as the sum of the areas of three {\pi/3} sectors of the unit radius circle minus {2} times the area of an equilateral triangle of sidelength {1}).

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.


[1] H. G. Eggleston – Convexity

[2] G. D. Chakerian – Sets of Constant Width


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