Talk:Inscribed square problem

News

https://www.quantamagazine.org/new-geometric-perspective-cracks-old-problem-about-rectangles-20200625/ DTLT (talk) 01:53, 29 June 2020 (UTC)

You did see that this is already in the article? After two previous attempts at adding it that incorrectly implied that it solved the original problem? —David Eppstein (talk) 03:42, 29 June 2020 (UTC)

Now three attempts. What is it about the Quanta writeup that so easily leads editors to this mistake? —David Eppstein (talk) 18:09, 29 June 2020 (UTC)

Four and counting. —David Eppstein (talk) 22:25, 2 July 2020 (UTC)

@David Eppstein: Sorry for maybe being ignorant on the topic, but I saw that article just a few minutes ago. The Theorem in publication [1] states ″For every smooth Jordan curve γ and rectangle R in the Euclidean plane, there exists a rectangle similar to R whose vertices lie on γ.″ A square is a rectangle too, so isn't the problem now solved at least for smooth Jordan curves? --Maximilian Janisch (talk) 22:50, 4 July 2020 (UTC)

The problem for squares was solved for smooth Jordan curves by Snirelman and published at least as early as 1944 (the Quanta article says 1929 but I haven't been able to confirm this). As our article already states. I think we're now up to seven editors who came away from the Quanta article with completely the wrong impression. If you read it carefully it is accurate but something about how it is written is causing people to not read it carefully. —David Eppstein (talk) 22:58, 4 July 2020 (UTC)

Guggenheimer

@Benjamin Matschke: is it your own personal assessment that Guggenheimer is wrong, or is there a reliable source that has shown the result to be false? Or that we have mis-cited the journal and it never made this claim? -- Fyrael (talk) 21:37, 13 December 2021 (UTC)

My impression is that the result was quoted incorrectly here: it should be that smooth-enough surfaces include $2n$ of the vertices of an $n$-cube (what it says in the Guggenheimer reference), not that they include all $2^{n}$ of them (what the removed text here said). —David Eppstein (talk) 21:44, 13 December 2021 (UTC)

So it sounds like you're saying the statement should be fixed instead of removed. -- Fyrael (talk) 22:28, 13 December 2021 (UTC)

Actually, reading again, I still misquoted it. The correct statement is: smooth-enough ($C^{3}$) topological spheres contain all $2n$ vertices of a cross polytope. It's the dual of a hypercube, not a hypercube itself. I was confused by the paper calling it a $2^{n}$-cell; that is a name for the cross polytope, not for the hypercube. Yes, I think the corrected statement should be restored. —David Eppstein (talk) 23:19, 13 December 2021 (UTC)

I have no access to Guggenheimer's paper at the moment, but 1) the stated result about inscribed n-cubes (n-1) spheres embedded in R^n is just false for n>=3 simply for dimension reasons, and 2) his results about the analogous question on embedded regular cross polytopes are falsely proved (c.f. Section III.9 of https://www.math.u-bordeaux.fr/~bmatschke/DiplomaThesis.pdf). Up to today, as far as I know, there are only partial results about the cross polytope question, namely: a) (smooth case n=3) Vladimir V. Makeev. Universally inscribed and outscribed polytopes. PhD thesis, Saint-Petersburg State University, 2003. b) Roman N. Karasev. Inscribing a regular crosspolytope. arXiv:math/0905.2671v2, 2009. and c) Arseniy Akopyan and Roman N. Karasev. Inscribing a regular octahedron into polytopes. arXiv:math/1107.4428, 2011.) -- Benjamin Matschke (talk) 06:16, 14 December 2021 (UTC)

Ok, so I guess the answer to User:Fyrael's initial questions are yes, sort of, and yes: Guggenheimer is wrong, there is a reliable source pointing out a serious gap in his proof (but not really the falsity of what he was claiming), and we also mis-cited the journal and it never made the claim we cited it for. Given what the Matschke source says I guess we should just omit Guggenheimer. I'll remove it again since I was the last to add it back. —David Eppstein (talk) 06:48, 14 December 2021 (UTC)

Where is my mistake? I believe this is a simple curve without any squares inscribed.

If this really has no squares inscribed, isn't that proof that the answer must be "No, not every plane simple closed curve contains all four vertices of some square."? Please tell me where I am or what I got wrong, where the square in my curve is or where I can hand my proposal for proofing. Thank you very much.

Here's my curve:

Take a less than 90° section of a Logarithmic_spiral, copy it, rotate the copy by 180° and connect it with the original section to form a closed curve. If the sharp < 90° corners invalidate the curve for the problem, just put some fitting > 90° sections of a circle (radius of the circle much smaller than half the distance of the corners) in it to close it off in a smooth manner. Maybe the rounded off by circles edges make it possible to spawn a rectangle on the curve (pretty sure they do), but as the circle radius is much smaller than any of the radii of the spiral sections, this may not be a square. Lets assume that the corners are on a horizontal line. Because this is a slim form (much wider than high), it is not high enough for a square to form with 2 points on a square section and 2 points on a spiral section. The diagonals could not be of the same length in this case. So the only chance would be that each spiral section contains two of the corners of the quare. Each two points on the same section of the spiral would have to form a side of the square. But because of the nature of the spiral, those two opposite sides of the square could never line up parallel AND in line with each other, in a way that their centered verticals would be identical. Due to the reflectional symmetry around the center point of the structure, a parallelogram would form, but not one with rectangular corners, a square.

I hope that this description is depictive enough to understand. I made a picture of it from a bitmap spiral, but without an account, I believe I cannot upload it to wikimedia to embed it here, unfortunately. As I do not have an account, just to leave some information on who I am and how to contact me: I am the admin of videobasics.de (no link on purpose so it is not deleted because of avertisement), which also has a contact form on it. But I will check back here in the next days at least for some answers. 176.3.83.131 (talk) 17:55, 11 April 2024 (UTC)

You know that the inscribed square problem allows tilted squares, not just axis-parallel ones, right? Also, this discussion page is for discussing improvements to the article based on published sources, not for asking for help with a mathematics argument. In future, Wikipedia:Help desk would be a more appropriate place to ask this sort of question. —David Eppstein (talk) 18:14, 11 April 2024 (UTC)

Hi David, thanks for your answer. Wikipedia Help Desk says it is for "how to use or edit Wikipedia". That does not sound more fitting for this question than here, next to the article itself. I would be pleased to know, where this question really fits to and ask it there (maybe there is a mathematics board somewhere?).

If you mean tilting like in rotating: yes, I am aware, and it won't fit, because the whole structure has a skew in it from the spiral. — Ralf from videobasics.de 176.3.64.145 (talk) 05:49, 12 April 2024 (UTC)

Your curve C is a simple closed curve, with 180° rotational symmetry, and with exactly one point of C on each ray from its center of symmetry. Finding an inscribed square on a curve with these properties is easy. (Note that "inscribed" means only that the square's four corners all lie on the curve; it is allowed to cross C.) Follow these steps:

1. place C so that the origin of the Cartesian coordinate system is at the center of symmetry, inside C.
2. the x-axis must cross C at at two symmetric points (±a,0) and the y-axis must cross C at two symmetric points (0,±b). If a = b you are done; these four points form a square.
3. If you don't already have a = b, consider continuously rotating C through 90° so that as you rotate it a and b vary continuously. After 90° of rotation, a and b will have swapped their values with each other. If you started with a > b, you will end the rotation with a < b, and vice versa. By the intermediate value theorem, somewhere in the middle of the rotation you will see a = b.

—David Eppstein (talk) 07:32, 12 April 2024 (UTC)

Thank you very much. Didn't think of that. It's so easy. So my understanding deceived me... :/ Well, it was expected. ;-) I will delete this question in a few days, so everyone has a chance to read this last answer. — Ralf 176.3.91.173 (talk) 18:42, 12 April 2024 (UTC)

Seems like I cannot delete this question whatever I try... please anyone feel free to get rid of it as it is out of place here...

Ralf 176.3.76.4 (talk) 18:09, 20 April 2024 (UTC)