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Comment by nyrikki | Hacker News Reader

A polyhedron has the Rupert property if a polyhedron of the same or larger size and the same shape as can pass through a hole in the original polyhedron.

A sphere is a surface of constant width, which the polyhedron approximation is not.

> The projected shadow has the same size as its diameter

Thus this is exactly why the sphere doesn't have the Rupert property.

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Ok, so by that definition a geodesic sphere has the Rupert property, as the sphere is an approximation made up of equilateral triangles. What if we perform isotropic subdivision on the equilateral triangles, such that each inserted point lies on the sphere, centred on each base triangle. We then subdivide each base triangle by constructing 3 new triangles around the inserted point. Thus at each iteration, geodesic sphere of N triangles is subdivided into 3*N triangles. If we continue with the subdivision, each iteration is a refinement of the geodesic sphere, and the geometric approximation gets closer to the shape of a true sphere. As N approaches infinity, the Rupert property holds true (according to the definition). What happens at infinity?

I would guess the margin goes toward 0.

Why do you say that the Rupert property applies for all finite N?

At infinity, the shape becomes a sphere and all orientations of it are identical. It is no longer a convex polyhedron and, thus, not subject to consideration.

A sphere is not an infinity-sided polyhedron. It's a sphere. It's also the limit of a sequence of polyhedra, each of which does not have infinity sides. Just like aleph-null is the limit of the sequence of natural numbers, but is not a natural number.

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