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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.

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.

References:

[1] H. G. Eggleston – Convexity

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