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Concise question: In two dimensions, do all shapes of constant width admit a measure over their interior such that for any two parallel lines intersecting the shape, the area between them under the measure is proportional to their distance?

Clarification: If you're not sure what I mean by "shapes of constant diameter", what I mean is the interior of one of these curves:

I've probably slightly misstated my question, so allow me to give some motivating context.

First, a puzzle someone asked me: Given a disc of diameter n, what's the minimum number of rectangles of length n and width 1 that can cover the disc?

Answer: Obviously, n such rectangles suffice. To prove that n are necessary, create a measure of the disc proportional to the area of a sphere with the same center and width, and observe that any such rectangle can contain at most 1/n of the total area of that sphere. (The area of a sphere in between two parallel planes intersecting the sphere is proportional to the distance between the planes.)

How this motivates the question:

This caused me to wonder for what set of shapes in the plane is their such a measure, i.e. a measure such that the part of the shape between any two lines intersecting the shape always has measure proportional to the distance between the lines. One can immediately rule out a square, for example, because it takes more strips of a given width to cover a rectangle diagonally than in an axis-aligned direction. In fact, this argument rules out such a measure for most shapes, but leaves open the question of whether shapes of constant diameter admit such measures.

My feeble attempts at an answer: Consider the special case of the Reuleaux triangle ( In this case, clearly the measure will have to approach infinity much faster near the corners than near the smooth edges. I originally hoped to construct a measure by some combination of three of the measures that work for the disc, one centered at each corner, but I haven't yet been able to make this work. I believe that the measure for the Reuleaux triangle, if it exists, must be very similar to the one for the disc near the smooth edges.

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The same question appeared in the comments here: no answer was given. – Anton Petrunin Aug 20 '13 at 17:44
Thanks. I will read the wikipedia page on Radon transforms and perhaps that will make it obvious. – user38858 Aug 22 '13 at 23:58

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