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In his remarkable book On the Sphere and Cylinder, where he came tantalizingly close to discovering calculus, Archimedes showed that the area of the portion of the sphere contained between a pair of parallel planes cutting the surface depends only on the distance between the planes (see p. 625 of this paper of King). This fact, which has been dubbed Archimedes hatbox theorem, is now a standard exercise in many calculus texts, and is even commemorated on the back of the Fields medal (if you look closely there, you can see the figure below in the background).

                                                     

Conversely, Blaschke showed that the only convex surface with this slab area property is the sphere. Indeed it is a simple exercise in classical differential geometry to check that any smooth convex surface with the slab area property must have constant curvature (Let $A(h)$ be the area trapped between a tangent plane and a prallel plane at distance $h$, and compute the limit of $A(h)/h$ as $h\to 0$). But can one still characterize the sphere if we fix the distance between the planes:

Question: Let $S\subset R^3$ be a convex surface with diameter $d$. Suppose that for some constant $0<h<d$, the area of the portion of $S$ trapped between every pair of parallel planes separated by the distance $h$ is constant, whenever both planes intersect $S$. Does it follow then that $S$ is a sphere?

A convex surface is the boundary of a compact convex set with interior points, and its diameter is the distance between its farthest points in the ambient space. Although this is known to be an open problem, I am not aware if it has been explicitly mentioned anywhere.

Addendum: See this question for a closely related problem on plank invariant measures for convex planar's domains, and more references in this area.

  • For reasons of non-triviality, do we impose that the solid cannot be bounded between two parallel planes separated by distance $h$? – Adam P. Goucher Oct 10 '17 at 11:27
  • ...and by 'smaller than the width' you mean 'smaller than every width' (since width depends on the direction of projection)? – Adam P. Goucher Oct 10 '17 at 13:33
  • Yes, by the "width", I mean minimal width. – Mohammad Ghomi Oct 10 '17 at 13:43
  • Re "so the current statement is not vacuous": you are right; this is a nice example of a substantial use of the word 'only' in "depends only on" in the OP; without the 'only', it'd be vacuous; never looked at this theorem of Archimede like this; learned something new. – Peter Heinig Oct 10 '17 at 18:23
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    I think that, although you have commented to reflect your intent, the original statement still doesn't reflect @AdamP.Goucher's comment. Indeed, if the surface is too small, then there is no pair of parallel planes at distance $h$ that (both, I assume) cut it, so the area bounded by any such pair of parallel planes is (vacuously) constant. – LSpice Oct 10 '17 at 18:58
  • OK, reedited to further clarify this point. – Mohammad Ghomi Oct 10 '17 at 20:43
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    I believe that there is a typo in the actual Question (the part in the grey box): shouldn't "trapped between every pair of parallel planes separated by the distance $d$" be "trapped between every pair of parallel planes separated by the distance h"?? – Vincent Oct 13 '17 at 09:18
  • @Vincent: the opening poster has in the meantime corrected this error which you correctly pointed out. – Peter Heinig Oct 13 '17 at 14:24
  • Please bear in mind that each edit bumps the question to the top of the front page, and thus displaces other questions which are vying for attention. 35 edits or more in a short period of time is rather excessive. – Todd Trimble Oct 13 '17 at 18:09
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    @MohammadGhomi: That has been proposed many times and always declined by the site developers. See for instance https://meta.stackexchange.com/questions/26483/could-we-have-the-ability-to-mark-a-change-as-minor-in-questions-or-answers – Nate Eldredge Oct 13 '17 at 20:18
  • "he came tantalizingly close to discovering calculus" What exactly would constitute "discovering calculus"? The biggest difference between Archimedes's work and "calculus" seems to be that he did not have the "fundamental theorem". Did he come close to discovering that? When he showed that the center of gravity of a hemisphere is $5/8$ of the way from the pole to the sphere's center, was that "calculus"? – Michael Hardy Feb 25 '22 at 01:53

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