This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.
If you suspend a cube by one corner and slice it in half with a
horizontal plane through its centre of gravity, the section face is a
hexagon. Now suspend a tesseract (a four dimensional hypercube) by one
corner and slice it in half with a hyper-horizontal hyperplane through
its centre of hypergravity. What is the shape of the section
hyper-face?
geometry/dissections/tesseract.s
The 4-cube is the set of all points in [-1,1]^4 .
The hyperplane { (x,y,z,w) : x + y + z + w = 0 } cuts the 4-cube
in the desired manner.
Now, { (.5,.5,-.5,-.5), (.5,-.5,.5,-.5), (.5,-.5,-.5,.5) } is an
orthonormal basis for the hyperplane. Let (a,b,c) be a point on the
hyperplane with respect to this basis. (a,b,c) is in the 4-cube if and
only if |a| + |b| + |c| <= 2. The shape of the intersection is a
regular octahedron.
 
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