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.

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?

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.

I would guess the margin goes toward 0.

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

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.

Wouldn't you need a little material "left over" to claim that it can pass through itself? Two spheres of equal size wouldn't work because they would occupy exactly the same space.

Yes!

The "pass through itself" criteria is the same as "has one shadow that fits entirely inside another shadow". If you allow "one shadow equals another shadow" then it's trivially true for every shape because a shadow equals itself.

Note that this "shadow" language assumes a point light source at infinity, i.e. all the rays are parallel.

That's trivially true for every shape, so it's probably not interesting in the context of this puzzle.

I think Sphere is a outlier for this context.

Yeah I’m confused

The shadow has to be bigger for the other shape to pass through. There's no way to orient a sphere so its shadow becomes bigger. For a cube there is.

Make a 2” inner diameter cylindrical hole in a 2” diameter sphere.

That would depend on the light source and its size and distance.