In his remarkable book [On the Sphere and Cylinder][1], where he came tantalizingly close to discovering calculus, Archemedes showed that the area of the portion of the sphere contained between a pair of parallel planes cutting the sphere depends only on the distance between the planes. This fact, which has been dubbed [Archemedes hatbox theorem][2], is now a standard exercise in many calculus texts, and is even commemorated on [the back of the Fields medal][3] (if you look closely, you will see a sphere inside a cylinder in the background). <img src="https://i.sstatic.net/ZKPoU.png" width="160"> Conversely, [Blaschke][5] showed that the only convex surface with this slab area property is the sphere. Indeed it is a simple exercise in differential geometry to check that any smooth convex surface with the slab area property must have constant curvature. But can one still characterize the sphere if we fix the distance between the planes: >**Question:** Suppose that, for some fixed $h$, the area trapped between every pair of parallel planes, separated by the distance $h$, cutting a convex surface is constant. Does it follow then that the surface is a sphere? By "cutting" here we mean that both planes intersect the surface, and at least one of the planes contains an interior point of the convex body bounded by the surface. In other words, $h$ is small enough so that the surface is never contained entirely in between the planes. Although this is known to be an open problem, and one would assume was in the back of Blaschke's mind, I am not aware if it has been explicitly mentioned anywhere. [1]: https://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder [2]: http://mathworld.wolfram.com/ArchimedesHat-BoxTheorem.html [3]: http://www.mathunion.org/general/prizes/fields/details/ [4]: https://i.sstatic.net/ZKPoU.png [5]: https://link.springer.com/book/10.1007%2F978-3-642-47392-0