Let A be a convex compact set in the plane (with a piecewise smooth boundary, say). We want to `inflate' it in such a way that the diameter does not increase.

More accurately, we are looking for all sets C such that

a) A is a subset of C; b) diam(A)=diam(C)

Let now B is the largest possible set C which satisfies these two properties.

By `largest' I mean either that it m(B) = max m(C), where m is the Lebesgue measure; or that B actually contains any C with these properties. Let us call B the isodiametric hull of A.

The simplest example of A is of course the square: here B is the superscribed disc, and it is the isodiametric hull of A in the strong sense.

Another example is the equilateral triangle, for which B is the Reuleaux triangle. Similarly, for any regular 2n-gon we have the disc, and for any regular (2n+1)-gon its isodiametric hull is a Reuleaux polygon.

The first non-trivial example that comes to mind is an isosceles triangle that isn't equilateral. It is clear that the hull is always a set of constant diameter but how does one actually obtain it? It seems that its boundary - a curve of constant width - is not a finite union of circular arcs.

I wonder if all this is well known (being such a natural question!). In particular, does the isodiametric hull of a set always exists in the strong sense?

Added: of course, if there is no IDH in the strong sense, B may not be unique. Its area is unique, though. How does one find it?


2 Answers 2


For any shape that is not constant width, there are many different maximal elements of the same diameter containing it.

Constant width shapes are maximal. They are regions swept out by line segments of length $D$ swept by their midpoints moving perpendicularly along curves perpendicular to along smooth curves (with cusps) whose tangent line turns 180 degrees

Any shape that is not constant width has some projections that do not have maximal diameter. For each projection that has maximal diameter, there is a diameter as above perpendicular to the line of projection.

Any method of interpolating a positively turning curve between the existing diameters will create a maximal element of the set of shapes of the given diameter. There are always infinitely many ways to do this unless the set is already maximal --- any one solution for an interpolating curve can be perturbed anywhere it is not constrained.

This observation can be applied, for exmaple, to the case of squares mentioned in the question. For example, you can begin enlargement of a square by adding an arc of a circle centered at one corner and passing through an opposite corner, then take the convex hull. More systematically: the square only defines two diameters. There are many Legendrian sections of tangent line bundle of plane $\to \mathbb {RP|^2}$ (cf. Is the sphere the only surface all of whose projections are circles? Or: Can we deduce a spherical Earth by observing that its shadows on the Moon are always circular?) interpolating these two elements.

The total area is locally given by an easy formula, but globally it seems like a highly irregular function, since it depends on how the sweeping diagonal overlaps itself. I doubt if there is a good theory of a maximal area shape of the same diameter containing the given one.


This isodiametric hull is not well defined. The first observation is that any curve of constant width is its own isodiametric hull. Now let's take an isosceles triangle ABC with $\angle A=20^{\circ}$, $|AB|=|AC|=r$ for example. We can place its vertices on a Ruleaux triangle of width $r$ so that $A$ is in one of the the corners, however the triangle can slide into different positions each of which gives a different maximal isodiametric hull.

  • $\begingroup$ Sorry, how do you know that such a Reuleaux triangle yields the maximal area for a hull? I think I can do better with a pentagon, for example. $\endgroup$ Sep 18, 2010 at 0:10
  • $\begingroup$ In your question you ask about isodiametric hulls in the strong sense. The pentagon will not contain the Ruleaux triangles described above. $\endgroup$ Sep 18, 2010 at 0:41
  • $\begingroup$ In fact any figure which contains them will have larger diameter. $\endgroup$ Sep 18, 2010 at 0:43

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