TECH-MUSINGS

Thoughts On Algorithms, Geometry etc...

Wednesday, October 03, 2012

Maximizing and Minimizing the Diameter

This a digression from the isoperimetric/isodiametric problems (I don't know of any work on most of these questions and am making some enquiries).

Given a planar convex shape, the diameter is defined to be the largest distance that can be formed between two opposite parallel lines tangent to its boundary, and the width is defined to be the smallest such distance.

Question: In 2D, if both area and perimeter are specified, which convex shape has maximum/minimum diameter?

Obviously, for given area A, perimeter has to be more than a criticial value, say Pc, for a convex shape to be at all possible. And at that criticial perimeter, the only shape possible is a circle (and it gives both max and min diameter).

Minimum diameter: With specified perimeter above Pc, the convex shape with least diameter for specified area appears to be an ellipse.

Maximum diameter: Guess: When the specified perimeter P is higher than Pc, the convex shape with highest diameter appears necessarily polygonal. For P above a sufficiently large value, say Pm, the convex shape will be a triangle. As P is reduced from Pm, the shape will be a quadrilateral, pentagon, and so on until when P = Pc, it becomes a smooth circle.

Question: The same question in 3d. Volume fixed. Surface area specified, diameter to be maximized.

Guess: For minimum diameter, the convex shape is probably an oblate spheroid (with semi-axes, a =b >c and hence, there are an infinity of diameter segments). In an extreme case, it will be a sphere. For maximum diameter, the required convex region is polyhedral and for sufficiently large specified surface area, it will be a tetrahedron.

Generalizations of these questions and guesses to higher dimensions appear obvious.

Update (14th October 2012)

In 2D, there are six problems one could think of (they may not all be independent or even non-trivial but here goes): Specify 2 of the three quantities {diameter, area, perimeter} and then maximize or minimize the 3rd quantity.

Likewise, in any dimension, there are at least 6 new questions. At the time of writing this, I dont know if any has been conclusively settled. I have approached a few experts with them and now it is wait and watch. Most, if not all the above *guesses* appear very shaky but that is fine by me!

And one could also ask if the convex figure maximizing the diameter also has the least possible *width* and vice versa. Looks plausible! -------------

Background: Here is a slightly trimmed quote from the paper 'Isodiametric Problems for Polygons' by Michael Mossinghof (available online):

The isoperimetric problem in the plane asks for the maximal area of a closed curve with fixed perimeter. If one fixes the diameter instead of the perimeter, one obtains two similar isodiametric problems: first, maximize the area, and second, for convex curves, maximize the perimeter. The circle is optimal in all three of these problems. Its optimality in the isodiametric area problem was established by Bieberbach, and in the perimeter problem by Rosenthal and Sz´asz.

We can ask the same three questions for polygons. In the isoperimetric problem, it is again well known that the regular n-gon alone has maximal area among all n-gons with the same perimeter.

The isodiametric problems for polygons however are more complicated.In the area problem, Reinhardt proved that the regular n-gon alone is optimal when n is odd. When n = 4, the square achieves the largest possible area, but in this case there are an infinite number of quadrilaterals with the same diameter and area.

Lenz posed the problem of determining the maximal area when n is even and n > 6. Reinhardt in fact proved that the regular n-gon is never optimal for such n, and Sch¨affer published another proof of this fact. Neither proof however establishes a quantitative improvement in the area problem when n is even.


Needless to say, surprises begin with: (in isodiametric) regular n-gon is never optimal for even n >6!

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