<BOBOXdHHHHHHXHXXX$-Zl`xHH@R(,,,-^&'X`l /jXdXXjXm:Xl!(|DSETR XXXX((XXXBy far this has been the most intriguing chapter. Whether it was the lack of equations or the subject material , this chapter mad me think the most, and made the most sense. The projection of the 600 cell is awesome.
Anyway, my one main thought deals began with the duals of the regular polyhedra. If one takes the ratio of faces connected to a vertex to the total number of faces on a polyhedron, one comes up with a fraction over four. For the tetrahedron, three faces are connected to any vertex, and there are four faces, so the fraction is 3/4. For the octahedron and the cube, 1/2 of the faces are connected to any one vertex. And for the dodecahedron and icosahedron, 1/4 of the faces are connected to any one vertex.
I found this particularly interesting, for I conjectured that since an octahedron had more faces connected to each point then it would be connected to a greater percentage of the faces. However, logically, if every vertex is connected to more faces, then the number of faces will increase in direct proportion.
But if the progression goes 3/4, 2/4, 1/4, what happened to 0/4 and 4/4? 0/4 is obviously a point, since a single vertex is connected to zero of the faces, since there are no faces. But 4/4 is a different story. To imagine this, one must start with the angles of each of the polygons that makes up the polyhedrons. AS the polyhedrons increase in number of faces, so do these angles increase. If one were to imagine a cube balancing on a vertex upon a plane, and then an octahedron, and then an icosahedron, each time the faces would come a little bit close to touching the plane. Thus, the seventh (the point being the sixth) regular polyhedron would touch this plane. So what kind of space would this occupy? Half of infinity, I guess, the half being the side from which the other polyhedra descended from. The figures surface would be entirely composed of vertices, but entirely composed of faces. The only problem seems to be whether or not the edges of each face would meet. However, if one understands infinity as something without beginning or end, where two ends of a segment have no distance between them since their relative distance is infinitesimal, then all faces would fit together.
Very abstract, but I think it works out if you have faith. DSET|(H:XHl
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