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Blaschke-Lebesgue Theorem

October 6, 2013 Leave a comment Go to comments

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.

References:

[1] H. G. Eggleston – Convexity

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

Categories: Geometry, shape optimization Tags: , , ,
  1. ciandivediliu
    August 7, 2018 at 7:47 pm
    Reply

    so

  2. yiannisvamvakas
    January 8, 2022 at 4:48 pm
    Reply

    Hello I would just like to ask where the root three in the last lines came from. I seem to not understand this part. Thank you for your time.

  3. yiannisvamvakas
    January 8, 2022 at 9:00 pm
    Reply

    I would like to ask why is 2 A(H) = root 3

  4. yiannisvamvakas
    January 9, 2022 at 9:07 am
    Reply

    why is A(H) root 3?

    • beni22sof
      January 9, 2022 at 8:13 pm
      Reply

      It is simply the area of a hexagon of width equal to one. Split it into 6 equilateral triangles and find the edge of the equilateral triangle. After a bit of work, the area can be quickly computed.

  5. yiannisvamvakas
    January 10, 2022 at 9:39 pm
    Reply

    Thank you for your response. However, isn’t the area of a hexagon 3root3/2 and if we multiply it by 2 3root3 and not root3 as A(H) seems to be?

    • beni22sof
      January 10, 2022 at 10:29 pm
      Reply

      Please do the computations! H is the hexagon of unit width. It can be split into 6 equilateral triangles with height 0.5. The total area of H is \sqrt{3}/2 after some computations.

      • yiannisvamvakas
        January 11, 2022 at 10:06 am

        Thank you very much I did not think that the height of 0.5.
        I am sorry for taking your time but I would like to ask one last thing.
        Why are they all equal to π? It is just that I am doing a math paper and i need to be sure.
        Thank you again.

      • beni22sof
        January 11, 2022 at 10:16 am

        Because the Minkowski sum of K and -K is a disk of radius equal to one whose area is Pi. You should consult the references indicated at the end of the post.

  6. beni22sof
    January 11, 2022 at 10:19 am
    Reply
  1. October 7, 2013 at 12:33 am
    Every constant width set has a circumscribed regular hexagon | Beni Bogoşel's blog
  2. November 26, 2020 at 11:25 pm
    Shapes with minimal area circumscribed around all bodies of constant width | Beni Bogoşel's blog

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