Every simple closed piecewise continuous curve will pass through the corners of some square. That is to say, every curve you can draw by hand, without picking up your pencil, that doesn’t intersect itself, has a square “in” it somewhere.
Now that’s cute! It would be nice to know the proof, but it’s probably too much to ask for anything elementary; the proof of the Jordan Curve Theorem (that any curve of the same type as the green one above partitions the plane into 3 regions - the curve itself and its ‘inside’ and ‘outside’) is very complicated, and on the face of it, this theorem is more impressive than the JCT. Although…perhaps you could get somewhere starting from the fact that any such curve has total curvature equal to 2π.
Can this illustration be extended to higher dimensions? (My favorite question numero uno when I encounter a nifty geometric property like this). My guess: yes! Assuming the version for Jordan curves works, it seems likely that any embedded, compact surface without boundary in R^3 will contain the corners of a cube. At least, I’m convinced this is true for easy examples like the sphere and ellipsoid.
The nice thing about this theorem is it connects topological properties (continuity, compactness, being embedded, having no boundary) with a geometric property. Here’s my
Conjecture: Let M, a topological manifold of dimension n - 1 (with n > 1), be compact, boundary-free, and embedded in R^n. Then M contains the corners of an n-dimensional hypercube.
A gold star to anybody who can prove (or disprove!) this…I have a feeling the generalized Gauss-Bonnet Theorem would be of great use, for starters.
