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Do you know of any way to cut the corners of a cube by means of rotation assuming that the cube is centered in the origin of XoYZ?

For example if we have a square centered in the origin of XoY and we rotate this square 45 degrees the rotated square will cut all the corners of the original one with equal cuts. I do not know how to generelize this for higher dimensions.

Thanks, Bogdan.

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Well, I guess rotating the cube 45 degrees along the x axis, 45 degress along the z axis, and 45 degrees along the x=y, z=0 axis should do the job. –  roman Sep 27 '12 at 9:02
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2 Answers

up vote 1 down vote accepted

A square has $2\cdot2$ sides and $2^2$ vertices, but a three-dimensional cube $C$ has $2\cdot3$ faces and $2^3$ vertices; so you can never cut off all vertices at the same time, using a rotated copy of $C$.

You can, however, do the following: When $C=[{-1},1]^3$ then consider for a suitable $\epsilon>0$ the octahedron $$O:=\bigl\{(x,y,z)\ \bigm|\ |x|+|y|+|z|\leq 3-\epsilon\bigr\}\ .\qquad(*)$$ This will cut off the eight vertices of $C$ allright. In addition, this idea can be generalized to $n$ dimensions, $n\geq2$.

That $(*)$ describes an octahedron is seen as follows: The vertex $(1,1,1)$ in the first octant is cut off in a symmetric way with respect to the three axes by means of the condition $x+y+z\leq 3-\epsilon$. Writing $|x|+|y|+|z|\leq3-\epsilon$ instead of $x+y+z\leq 3-\epsilon$ makes the resulting solid symmetric with respect to all three reflections $x\mapsto -x$, $y\mapsto -y$, $z\mapsto -z$. In $n$ dimensions the corresponding condition would read as $|x_1|+|x_2|+\ldots+|x_n|\leq n-\epsilon$.

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The generalization can be:|x1|+|x2|+ ... |xn|≤n−ϵ ? –  Bogdan Sep 27 '12 at 10:35
    
@Bogdan: See my edit. –  Christian Blatter Sep 27 '12 at 11:11
    
Thank you for your answer. I am looking at the dual polyhedron from the linear programming theory point of view. The cube is a set of inequalities 0<=x<=1, 0<=y<=1, 0<=z<=1. This is something like A * [x, y, z]' <= b. I would like to know if the dual polyhedron can also be expressed in a similar form. –  Bogdan Sep 27 '12 at 11:19
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You can cut the corners of a cube by another concentric and congruent cube: taking it in random orientation, it will in most cases not contain any of the corners, thereby "cutting them all off". But the corners will never be cut off regularly in this case for a simple reason: the cutting is done by the faces of a cube of which there are $6$, but there are $8$ corners to cut, so necessarily some distinct corners will be cut off by the same face (which will cut off the whole edge between them as well).

You can cut off the corners of a cube regularly using an octahedron though. In general you need a dual polytope; the dual of a square happens to be a square. The same happens for any regular $n$-gon in the plane, and for the tetrahedron in space, but not for any other (regular) solids.

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A cube can be described by linear inequalities of the form: 0<=x<=1, 0<=y<=1, 0<=z<=1. Do you know of any transformation which takes this inequalities and gives the equations of the dual polyhedron? –  Bogdan Sep 27 '12 at 10:43
    
@Bogdan: You cannot obtain equations for the dual polyhedron directly from those of the polyhedron, as again their numbers do not match ($6$ for the cube, $8$ for the octahedron). The equations for the dual polyhedron come from the vertices of the original polyhedron, and vice versa. –  Marc van Leeuwen Sep 27 '12 at 11:19
    
Thanks! ... you are right! –  Bogdan Sep 27 '12 at 13:44
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