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Presumably a simple sphere would trivially qualify as being unable to pass through itself.
by biot 11mo ago
Presumably a simple sphere would trivially qualify as being unable to pass through itself.
- smokel 11mo agoThe puzzle applies only to convex polyhedra.
- biot 11mo agoThe article says: > The full menagerie of shapes is too diverse to get a handle on, so mathematicians tend to focus on convex polyhedra The phrase "tend to focus on" suggests it's not an exclusive thing. However, you're right -- it appears that the Rupert property only applies to convex polyhedra, so the article title and text is at the very least incomplete given that a sphere is a shape.
- LostMyLogin 11mo agoA sphere is not a convex polyhedron
- guelo 11mo agoAt the limit of faces they are.
- jibal 11mo agoA sphere has no faces so it's not a convex poloyhedron.
- burnt-resistor 11mo agoCorrection: a sphere has infinite faces so it's not an "convex poloyhedron [sic]." A convex polyhedron must have finite faces, so apeirotopes aren't allowed.
- jibal 11mo agoA sphere has no faces, not "infinite" faces.
- teraflop 11mo agoSure, and pi is the limit of a sequence of rational numbers, but lots of properties that hold for rational numbers don't hold for pi.
- guelo 11mo agoAs you approach sphere you lose Rupertness.
- akoboldfrying 11mo agoLimiting behaviour can be counterintuitive. As you add vertices to a polyhedron, some properties approach those of a sphere (volume, surface area), but others just get further and further away (number of surface discontinuities). It's not at all obvious which way "Rupertness" will go, or even whether it's monotone with respect to vertex addition.
- burnt-resistor 11mo agoConvex polyhedra are required to be finite polytopes.