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Define the diameter of a shape as the greatest distance between any two of its points. What diameter 1 shape has the greatest area?

Is it the circle?

I've been looking for the biggest little polyhedron for awhile, and looked at biggest little polygon again. A paper by Henrion and Messine extended results to the 16-gon. The trend shows that the best area is chaotic, with the optimal 14-gon having a better improvement over the regular 14-gon by a factor higher than shown in the 8, 10, and 12 gons.

optimal decagon

If the trend continues, there may be a unit thrackle that generates a 50-gon to 90-gon of diameter 1 with a total area greater than a circle of diameter 1.

It might be possible faster than that, by using Reuleaux polygon methods.

What diameter-1 shape has the greatest area? Diameter 1 polygons that won't fit in a diameter 1 circle are easy to find, but is there one with an outside area that exceeds the uncovered area?

triangle and circle

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up vote 9 down vote accepted

I think it is the circle by the following reasoning. I'm not totally sure about this though, so if there is an error hopefully someone can correct me.

The shape of maximal area must be bounded by a curve of constant width (with the diameter being the width), since any curve of varying width could be expanded in the directions where the width is deficient, increasing the area while keeping the diameter the same.

Now by Barbier's theorem, the perimeter of any curve of constant width $w$ is just $\pi w$, regardless of the shape. Thus all shapes that can be candidates for optimality must have the same perimeter.

But by the isoperimetric inequality, a circle is the curve enclosing maximal area among all shapes with a given perimeter. Thus the circle encloses the maximal area among all shapes with a given diameter.

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Candidate shape -> curve of constant width -> Barbier's theorem -> isoperimetric inequality. That's a very clever way of thinking about it. It wouldn't apply to the 3D case. – Ed Pegg Jan 8 at 23:46

The isodiametric inequality states $V(C) \le \left(\frac{\operatorname{diam}(C)}{2}\right)^n V(B_2^n)$ for any bounded set $C \subset \mathbb{R}^n$. This means that a ball of radius $\frac{1}{2}$ has the biggest volume of all shapes with diameter at most $1$.

A prove can be found in Gruber's book about convex geometry. Wlog you can assume $C$ to be convex and compact. It boils down to applying successive Steiner-symmetrizations with respect to the coordinate hyperplanes to $C$. The result is a convex body $D$ that has the same volume as $C$, the diameter of $D$ is at most the diameter of $C$ and $D$ is symmetric with respect to the origin. The last part implies that $D$ is a subset of $\frac{\operatorname{diam}(D)}{2} B_2^n$.

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Useful link. I like your answer for the 3D case. – Ed Pegg Jan 9 at 0:04

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