Bill Gosper <billgosper@gmail.com>
Keith F. Lynch wrote:
I am not a visual thinker. I'm not even sure what numbers of sides the polygon could have. It can obviously have 3, 4, or 6. I doubt it can have more than 6, but I'm not sure. I'm even less sure whether it can have 5.
gosper.org/pentagonal cubeslice.png
Thanks. I also see now that it can never have more than six, since each side is formed by the plane's intersection with a face of the cube, and cubes only have six faces, all of which are convex. I'm curious what the maximum numbers of sides are for the polygons formed by the intersection of a plane with the other regular polygons. I can obviously get all four with the tetrahedron, but I'm not at all sure what the number is for the other three regular solids. I'm guessing it's ten for the dodecahedron. I'm pretty sure it's not possible to get a polygon with fewer than four sides from a plane's intersection with an octahedron, or with fewer than five sides from a plane's intersection with an icosahedron. Since nobody else seems to have done so, I wrote a quick program to generate (pseudo-)random planes and count sides (for the cube only, not for other regular polygons). The program is not visual, so I'm largely taking it on faith that it's doing what I told it to do, but it did pass several sanity checks. For instance: * It only finds polygons with 3, 4, 5, or 6 sides. * If I set Max_D (the maximum distance from the center of the cube to the intersection with the plane) to 0, I only get 4 or 6 sides. * If I increase Max_D the proportions gradually change, and if I increase Max_D beyond the square root of 3 the proportions cease changing. The average number of sides is 4.0004. The proportion of each number of sides: 3: 27.97 % 4: 48.68 % 5: 18.70 % 6: 4.65 % The last digit should not be trusted, but the other digits should be good. Could the average be exactly 4? It's a couple sigmas out, but it's possible. The proportions for planes that pass through the center of the cube are 64.9% four sides and 35.1% six sides. Ideally, someone else will write their own program, and see if it gets similar numbers. Unfortunately, my program cannot easily be adapted to other regular polygons. For all three ways of choosing two of X, Y, and Z, I set those two to the four combinations of -1 and +1, then solve for the third. If the third is between -1 and +1, I've found an intersection between the plane and an edge of the cube. This would have to be completely rewritten for other regular solids. I could easily collect statistics on the side lengths and angles of the intersecting polygons, if anyone is interested.