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: http://en.wikipedia.org/wiki/Curve_of_constant_width.
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 parallel 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 (http://en.wikipedia.org/wiki/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.
