convex polyhedron

(but your point about the title is valid)

A sphere can be approximated by a polyhedron. Somewhat obviously, all such polyhedra would seem to have the Rupert property. This new Nopert seems to differ in one key detail: some of the vertices near the flat top/bottom are at a shallower angle to the vertical axis than the vertices below/above them.

Can you pass the T-shaped tetromino through itself?

The T-shaped tetromino is not convex, so not part of the conjecture. There are many nonconvex shapes that don't have the Rupert property.

Nevertheless, the t-shaped tetronimo (assuming four glued cubes) has a shadow shaped like a bar of length two. I believe that such a shadow will pass through a bar of length three, with a tilt similar to the cube's.

If the remaining edge has exactly zero thickness , it means it doesn’t fit. I think that would be the case in that example?

It fits if it is tilted. If the 3-cube bar of the T is tilted 45° there will be a 3×√2 rectangle part of the shadow, which the 1×2 shadow fits through.

Somewhat unrelated question, a lot of the folks replying to the parent comment read to me like they're really good at visualizing things in their minds eye, when you talk like this is it because you can think about math really well? Can you visualize what you're saying? Sorry if this question doesn't make sense!

I do not have a mind's eye*, outside of the dimmest, briefest flash of people's faces if I've known them for several years. I do have a peculiarly strong sense of imaginary touch, which gets used when contemplating problems like these. Also, a significant component of my job is arranging things in 3-space. I'd be one of those people who say "I'm good at tetris" while packing a moving truck, but I am not actually good at tetris per se.

* https://en.wikipedia.org/wiki/Aphantasia

Thank you for your reply. Extremely interesting to me. My thinking and memory is pure motion picture and sound, it makes me think that I could be good at geometry as I can do spatial thinking well, however the downside of my thinking style is I've never been able to find a framework that allows me to hold fine detail symbols in space and work with them usefully, I suppose hence dyslexia + dyscalculia diagnosis. Maybe people like me who are in math use whiteboards and notes a lot or something? Maybe people like me don't go into maths so much. Imagination of touch is also very interesting, I've read about kinetic leaners before, I like to touch things when I'm learning as it helps with the recall later, but absolutely zero sense of touch is present in the recall. Sorry for the massive tangent, I'm just very curious about this stuff!

I'm a mathematician, so I can think about the maths well, and I figured out that particular maths by visualising the shapes and seeing the dimensions and the way the shadows fit together.

Cool. I appreciate your reply. If by chance you happened to do any art of any type by chance, I would be extremely interested in viewing it. :)

I don't think that's the case, how can you turn a bar of length two but still have it fit within the width of the bar of length three?

The long side is three cubes long, the short side two. You can easily move the short side through the long side's shadow if you tilt the latter so it becomes wider than a cube's side.

You are of course right, thanks.

That’s true. I digress, but you could give tetrominos convex hulls, and the result would still be somewhat Tetris-compatible.

For laymen's sake I think the title should say "First shape (without curves) found that [...]"

And not "pass through itself" but "pass through its copy"

The article isn't really for the layperson. It's confusing why several people are nitpicking at the title.

Because not-laypoeople ale precisely the kind of people who would nitpick the technically incorrect title.

I'm a lay person w.r.t. this topic, and I assumed the great exposition was meant exactly for persons like me.

This is quanta magazine. It is for lay people. The reason people are "nitpicking" the title is that "shape" is not a technical term. The technical term for what was found is "convex polyhedron". I read so much of the article before I was sure that it was talking about convex polyhedra specifically because the title is so ambiguous.

Quanta Magazine is very much designed for non-technical lay people.

From their About page: Quanta Magazine is an editorially independent online publication launched by the Simons Foundation in 2012 to enhance public understanding of science.

Oh come on. Quanta Magazine basically writes for HN. They have very little online footprint elsewhere, but they feature here multiple times a week and I'm sure they know it. The titles are almost always in this mold, implying some profound yet vague discovery. Some real, recent examples:

  - "Researchers Discover the Optimal Way to Optimize"
  - "Origami Patterns Solve a Major Physics Riddle"
  - "A simple way to measure knots has come unraveled"
  - "The Hidden Math of Ocean Waves Crashes Into View"

I don't necessarily mind it, even if I don't find the articles very informative. But it's certainly fair game to nitpick this borderline-clickbait style.

Why wouldn't a sphere pass through itself? The projected shadow has the same size as its diameter

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.

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

I would guess the margin goes toward 0.

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.

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

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

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.

Does a sphere not pass through itself (with zero margin?)

There needs to be nonzero margin, else the question is pretty trivial.

Ah yes correct you just use same orientation and you can get anything through itself.

You can do it regardless of orientation given you can apply enough pressure

How thick would the remaining walls be after you’ve made the hole required?

Infinitesimal I guess! You can fit an arbitrarily slightly smaller sphere for vanishing values of smaller.