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In the YouTube video "5-Sided Square" from Numberphile, Cliff Stoll states that, if a "square" is a shape with sides of equal length whose angles are all $\pi/2$, then we can find:

  • a three sided "square" on the sphere.
  • a five sided "square" on the pseudosphere.

The Question:

Do there exist "squares" with six or more sides?

Context:

I don't know much about geometry. Therefore, please pitch your answers at or below undergraduate level. If things are beyond that, then a summary would suffice.

This question arose naturally out of wondering whether five is the maximum.

Thoughts:

My guess is that, yes, they exist. A small section of one might look like this:

enter image description here

I imagine them to live on the surface of some warped shape, perhaps in a high number of dimensions.

Further Context:

Looking for stuff on this online seems difficult, particularly because of the high volume of sites out there aimed at GCSE students, covering basic properties of squares and rectangles. My searches were typically like these:

  • squares with more than four corners.
  • regular n-gons with right angled corners for n greater than 4.
  • regular n-gons on non-Euclidean surfaces.

Please help :)

Blue
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Shaun
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  • Would someone please explain the downvote? – Shaun Jul 07 '23 at 13:14
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    For any $n>4$ we can get orthangles (= polygons with all right angles) with $n$ sides in the hyperbolic plane: take a very small regular $n$-gon and gradually expand it until the angle defect gets bad/good enough. – Noah Schweber Jul 07 '23 at 13:20
  • Awesome! Thank you, @NoahSchweber. Where can I find out more? Because, again, Google isn't very helpful here, due to all the elementary stuff out there. Also, I invite you to write an answer. – Shaun Jul 07 '23 at 13:24
  • https://www.d.umn.edu/~ddunham/isis4/figure2.gif – Eric Jul 07 '23 at 13:29
  • I really do not get the downvotes. Please give some feedback. – Shaun Jul 07 '23 at 13:31
  • You can use this chatroom to give me feedback. – Shaun Jul 07 '23 at 14:17
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    @Shaun: "All you have to do" is convince yourself that you can make an isosceles triangle with base angles of $45^\circ$ and a vertex angle of $360^\circ/n$. (Then, $n$ of this would fit around the vertex point to make the desired polygon.) But, the angle-sum of such a triangle is $90^\circ (n+4)/n$; for $n>4$, this sum is less than $180^\circ$, so that the angles correspond to a valid hyperbolic triangle. (You could, for instance, use the Hyperbolic Law of Cosines to calculate the corresponding side-lengths.) – Blue Jul 07 '23 at 14:44
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    Related: https://math.stackexchange.com/questions/275079/books-for-hyperbolic-geometry and https://math.stackexchange.com/questions/412211/hyperbolic-geometry-reference-request?noredirect=1&lq=1 and https://math.stackexchange.com/questions/2678194/requesting-book-for-hyperbolic-manifold?noredirect=1&lq=1 – Lee Mosher Jul 07 '23 at 14:48

1 Answers1

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For any $n>4$ we can get orthangles (= polygons with all right angles) with $n$ sides in the hyperbolic plane: take a very small regular $n$-gon and gradually expand it until the angle defect gets bad/good enough. (I don't know a source about this specifically, but it's a consequence of the general angle defect behavior of the hyperbolic plane.)

More precisely, for $r>0$ let $C_r$ be a circle in the hyperbolic plane with radius $r$ and let $P_r$ be an equally-spaced set of $n$ points on $C_r$. Let $\alpha_r$ be the interior angle at any vertex in (the polygon formed by) $P_r$. By the angle defect formula for the hyperbolic plane, we have $\lim_{r\rightarrow\infty}\alpha_r=0$ and $\lim_{r\rightarrow 0^+}\alpha_r={(n-2)\pi\over n}$, and moreover $\alpha_r$ is continuous as a function of $r$. By the intermediate value theorem, some $r_0$ has $\alpha_{r_0}={\pi\over 2}$ (note that we need ${(n-2)\pi\over n}>{\pi\over 2}$ here, which is where the assumption $n>4$ comes in).

Noah Schweber
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