# Beyond 3-D, Chapter 8

## David Akers

### Slicing the cube

If you have a JAVA capable browser, check out my new applet. You can
now slice a cube from a vertex...
### Coordinates of hypercubes . . .

The combinatorial explanation for the number of edges in a hypercube makes
perfect sense to me now. Obviously the number of points in an n-cube will
be 2^n, since we have exactly two choices for each coordinate (1 or 0).
For some reason I hadn't thought of it in this way before.
The discussion of hypercube slicing was a bit more mystifying,
unfortunately. I guess I just can't seem to visualize it very well
without that movie in front of me. I sometimes wished that the book had a
few more of the mathematical details. I found myself forced to make a lot
of assumptions and (in some cases) guesses about the math. Here's some of
what I came up with (please correct me if I am wrong on any of this).

**Regarding slices perpendicular to the longest diagonal:**
In three space, the slice which goes through (1, 0, 0), (0, 1, 0), and (0,
0, 1) gives us the coordinates of a two-simplex which lies on the barrier
between slices which are triangles and slices which are truncated
triangles (6 sided). Similarly, in four space, the hyperslice which
contains (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) should
be on the boundary between hyperslices which are tetrahedrons and
hyperslices which are truncated tetrahedrons (8 sided). If, then, we could
find the point which lies on the longest diagonal of the hypercube, and is
contained by this hyperslice, we would then know the distance we had
travelled from the vertex, right? I am going on the assumption that there
is only one hyperplane which slices the hypercube perpendicular to its
longest diagonal and goes through a particular point on this diagonal.
This would seem logical, but I'm not quite sure. Does this make sense?

In a cube, the division between tetrahedra and truncated tetrahedra
occurs exactly 1/3 of the way down the object. Is this also true for the
hypercube? Maybe calculating this result comes down to finding the center
of the tetrahedron formed by the four points, then applying the extended
version of the pythagorean theorem to find the distance from this center
to the vertex.

### Links

Dan
Margalit's week 11 paper.

Michael
Matthews' week 11 paper.

Jeremy
Kahn's week 11 paper.

Prof. Banchoff's Response.

David Akers