Does there exist a space-filling rhombic disphenoid?
bump
Maybe you should ask your mom about how my thrombic fills the space between her isotittie.
>>16264456No.Source? Divine intuition
Yes, any object is space filling.What makes space filling curves special is that they're 1d, anything higher than 1d can obviously fill space by making it arbitrarily large.
>>16266308Wrong: https://mathworld.wolfram.com/Space-FillingPolyhedron.html
>>16264456Here's one in 4-dimensions.disphenoidal 288-cell
SAFE SPACEhttps://www.youtube.com/watch?v=TBvPIAWB_t0
>>16264466>Telephasic Workshophttps://www.youtube.com/watch?v=he8fMUmxHOU
It's huge
>>16264456Our math isn't advanced enough to answer this question
>>16264466Tesseracts have quadrilateral faces, moron
>>16264577>>16264981>>16265349>>16265851>>16266292>>16266708>>16267054>>16267484>>16267885>>16268325>>16268883>>16269438>>16269867>>16270317>>16270610>>16271455>>16271784>>16272967>>16273627>>16274496>>16275348>>16276871>>16279088Reporting this thread for spam. 35 replies and ~25 of them are from the same person.
>>16280310Maybe it's a 4D icosahedron?
>>16280327I bumped this thread twice because I find it interesting, and nobody has made any interesting posts yet.
>>16280595https://en.wikipedia.org/wiki/24-cell
>>16280327bump
>>16282677Ehm
I'll just analyze the obvious way to try to do it: join edges of the same length together. In order for this to work, at each edge the angle between the two faces that meet there must be [math]2\pi/n[/math] for some [math]n \geq 3[/math]. So let's try to calculate the angles between the faces.Line the x, y, and z axes up with the rotational axes of symmetry of the disphenoid, and let <a,b,c> be a normal vector to one of the faces. Then <a,-b,-c>, <-a,b,-c>, and <-a,-b,c> are normal vectors to the other faces. The possible angles between the normal vectors are [math]\cos^{-1}\left(\frac{a^2-b^2-c^2}{a^2+b^2+c^2}\right)[/math], [math]\cos^{-1}\left(\frac{-a^2+b^2-c^2}{a^2+b^2+c^2}\right)[/math], and [math]\cos^{-1}\left(\frac{-a^2-b^2+c^2}{a^2+b^2+c^2}\right)[/math], and the angles between the faces are the supplements of these angles. We obtain the result[math]\cos(\alpha) + \cos(\beta) + \cos(\gamma) = 1[/math]where [math]\alpha[/math], [math]\beta[/math], and [math]\gamma[/math] are the angles between faces, each of which occurs at two opposite edges.Now we just search for angles of the form [math]2\pi/n[/math] ([math]n \geq 3[/math]) that satisfy this equation. There is exactly one solution,[math]\cos(2\pi/4) + \cos(2\pi/6) + \cos(2\pi/6) = 1[/math].But this would give us a tetragonal disphenoid rather than a rhombic one.I don't know whether or not there might be a more clever way of doing it that bypasses this issue.
>>16283261So that's a no, then?
>>16283717No under the stated assumption. I would guess it's no in general, but I don't have a proof of that.
