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Did you know it's possible to cut a hole in a cube such that an identical cube can fit inside it? Really! It's called "Rupert's Property." *All Platonic solids are Rupert!* Except one new shape, which *cannot* fit inside itself. This eldritch polygon is called a Noperthedron!

arxiv.org/pdf/2508.18475

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in reply to Cory Doctorow

naturally, this only works in that same ideal/purely theoretical sense (ask any carpenter or mechanic) in which all cows are orbs for Physicists. =)

It's not that Mathematicians don't have a sense of humor, it's just that most of it is not perceivable by humans, such as - say - that of Astrophysicists.

in reply to Cory Doctorow

I think the term they coined is "Noperthedron". Like "Ru_pert", but not ;)

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in reply to Cory Doctorow

A small correction to this description— this new polyhedron is convex, not platonic. All platonics ARE Rupert; this new solid disproves the conjecture that all CONVEX 3d polyhedra are Rupert.

And I agree, Noperthedron is a great name for it. I wouldn’t have noticed the name if you hadn’t pointed it out.

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in reply to Matt Diamond

A second correction (sorry):

Rupert’s property is that an identical cube can •pass through• the hole, not •fit inside• it.

(The latter would is trivially true if we read it as “at least some of it can fit inside the hole” — just remove any point on the surface — and would clearly be impossible if we read it as “fitting fully inside the hole” unless we allow the “hole” to consist of the entire polyhedron’s volume.)

The more formal definition of Rupert’s property is that there there are two different isometric 2D projections of a cube such that one of the resulting 2D shapes is a subset of the other.

This entry was edited (4 days ago)

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in reply to John Carlos Baez

@johncarlosbaez

I’m not sure what “interior” means in a mathematical sense here. Can you clarify?

in reply to Paul Cantrell

- a point 𝑥 is in the 'interior' of a subset S of the plane if for some number ε>0, every point that has distance < ε to x is in S.

If we drop my legalistic caveat, a the cube would obey Rupert's property simply because it's easy to find two different 2d projections of the cube that are both equal to each other: both a square, for example.

This caveat forces us to find one 2d projection that fits 'truly inside' another projection.

This entry was edited (3 days ago)
in reply to John Carlos Baez

@johncarlosbaez
Ah, OK! The paper in question just defines it using a proper subset (definition 1) to rule out two projections with identical results:
arxiv.org/pdf/2508.18475

I wonder if there are shapes where your topological constraint rules something out, e.g. a shape where the only projection that satisfies the Rupert property touches the boundaries.

in reply to Paul Cantrell

@inthehands - oh, my bad!

Then a round ball is Rupert by that paper's definition, but it's ruled out by my extra constraint.

(The hole you have to cut out of the ball, to fit an identical ball through, is the entire ball!)

But they were only talking about convex polyhedra, not a ball. I don't know if there are convex polyhedra ruled out by my extra constraint, that their paper would allow!

in reply to John Carlos Baez

@johncarlosbaez

Re ball: I don’t think so? Entire ball is not a proper subset, right? (They say ⊂ and not ⊆.) Am I missing something?

in reply to Paul Cantrell

@inthehands @johncarlosbaez the paper is using the topological interior, I think. That is what the little superscript circle in equation (1) means.
in reply to Ballooniper 🎈🏳️‍⚧️

This entry was edited (3 days ago)
in reply to John Carlos Baez

@johncarlosbaez @eruonna

It’s possible they might sharpen up these details in peer review. In any case, the alternative definitions do pose some interesting questions!

in reply to Paul Cantrell

- I think they did everything exactly fine - except that, alas, a substantial percentage of mathematicians still use ⊂ to mean ⊆, so we can't tell what they mean here. If I were the referee I'd demand that this be clarified, but I bet the actual referee won't. Luckily, as I mentioned, it doesn't make a bit of difference in this particular definition.
This entry was edited (3 days ago)
in reply to John Carlos Baez

@johncarlosbaez @eruonna
Not for polyhedra, anyway!

(…through I’m pretty sure that proving even that rigorously requires more topology than I’ve got)

in reply to Cory Doctorow

@Cory Doctorow
Oh good heavens, now we know: the Gallifreyans got their physics from Rupert. Who knew. Who knew?

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