On 2016-09-25 16:18, James Propp wrote:
Kate Fractal should be invited to join math-fun!
Jim Propp
On Sunday, September 25, 2016, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Sep 25, 2016 at 12:01 PM, Hans Havermann <gladhobo@bell.net <javascript:;>> wrote:
"The volume of a regular tetrahedron (triangular pyramid with unit edges) is exactly half the volume of a square pyramid with unit edges."
This is (I think) equivalent to stating that the volume of an octahedron is four times the volume of a tetrahedron (unit edges assumed).
Yes because you can slice an octahedron into two square pyramids.
The geometrical "proof" of this is that you can assemble 4 unit tetrahedrons and a unit octahedron to produce a tetrahedron with edge length two, and therefore area equal to 8 tetrahedrons.
I don't have the ability to draw a picture of this here, but start with an equilateral triangle of side 2, and divide it into 4 equilateral triangles of side 1. Place tetrahedrons on the 3 of these that are in the same orientation, and then add as a second layer a fourth tetrahedron whose bottom three vertices are the points of the first 3 tetrahedrons. These four tetrahedrons outline the shape of a side-2 tetrahedron, and the space in the large tetrahedron that isn't in any of the small ones is exactly a unit octahedron.
Thanks to Kate Fractal for the idea of this construction.
Andy
https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb#/media/File:H... (Stick a tetrahedron onto the foremost face to complete a double size tetrahedron.) But don't we need to prove that it _is_ tetrahedral, i.e. the tetrahedron and octahedron dihedrals are supplementary? ("Behold!") Hans Moravec designed a version of hashlife based on this recursive partitioning of space. Time ran vertically, and the "initial" conditions were the lower faces, which completely determined the interiors and upper faces. I think he even got some unfortunate grad student to implement it. It was awkward because you normally start with a horizontal square grid initial condition. I think their scheme was to treat this as a "waffle iron" of minimal octahedra and then fill in the next layer of tetrahedra to make an "endo waffle iron", which then determined the next layer of octahedra. But this needs to be done cleverly, so as to halve the spatial frequency at each step, to get the crucial geometric growth of cell sizes, and hopefully, running speed. The hash table records upper faces as results of lower ones. I have no idea by how much this beats my inelegant scheme of overlapping frusta: https://www.lri.fr/~filliatr/m1/gol/gosper-84.pdf --rwg