In his remarkable book [On the Sphere and Cylinder][1], 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. This fact, which has been dubbed [Archimedes 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 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 $d$ and intersecting $S$ is constant. Does it follow then that $S$ is a sphere? Although this is known to be an open problem, 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