A related question is, what is the max number of pieces you can cut a bagel into with 3 cuts? Answer: 13 For n cuts, see https://oeis.org/A003600 But what about a solid figure-8, or a 4-D bagel (or torus)? Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Sun, Sep 16, 2018 at 8:07 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 3 there are 12 pieces: 3 regular octahedra centred in faces of the cube; 1 regular octahedron centred at vertices of the cube; 8 regular tetrahedra meeting octahedra in their faces. Note piece edge-length = sqrt(2) if cube edge-length = 2 .
See the corresponding solid lattice of planes at https://en.wikipedia.org/wiki/Tetrahedral-octahedral_honeycomb
WFL
On 9/16/18, Dan Asimov <dasimov@earthlink.net> wrote:
Let the n-dimensional torus T^n be defined as the n-cube
Q_n = [-1, 1]^n
with its opposite faces identified. (That is, any point x of Q_n with coordinate x_k = ±1 for some k is identified with the point having the same coordinates except with x_k changed to -x_k.)
The vertices of Q_n are the 2^n points {-1, 1}^n.
Now we are going to cut T^n along each of the 2^(n-1) hyperplanes defined as the perpendicular bisectors of each pair of antipodal vertices of Q_n. This requires extending each hyperplane throughout the n-torus.
Puzzle: How many pieces do these planes cut the torus T^n into?
E.g., for n = 2 there are two pieces.
—Dan
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