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:
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 :)
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).