If you suspend a cube by one corner and slice it in half with a horizontal plane through its centre of...

geometry/dissections/tesseract

Problem

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?

Show solution

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.