The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » geometry.puzzles

Topic: Tesseract Projected Along a Main Diagonal
Replies: 2   Last Post: Sep 16, 1994 1:44 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Daniel A. Asimov

Posts: 35
Registered: 12/6/04
Re: Tesseract Projected along a Main Diagonal
Posted: Sep 16, 1994 1:44 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <35ar9f$> (HopD) writes:
>In article <348a6r$>,
>(Daniel A. Asimov) writes:
>Re: Tesseract Projected along a main diagonal.
>Take the points of the tesseracts vertices to be 16 ordered quadruples
>whose elements were either one or zero (0,0,0,0), (1,0,0,0), (0,1,0,0),
>etc. Describe these quadruples in terms of 4 unit vectors, the first
>pointing in the direction of the diagonal and the other 3 perpendicular to
>the first and each other. Take the vertices described with this basis and
>eliminate the components pointing in the direction of the diagonal. I
>believe you are left with ordered triples in a space perpendicular to the
>diagonal. Sketching it these triples seem to form a rhombic dodecahedron.

From Hop David Ajo, Arizona

In case anyone forgot my original question, I'll paraphrase it as follows:

If an ordinary cube is projected orthogonally onto the plane perpendicular
to its main diagonal, the result is a regular hexagon (as is well-known).

What is the result of projecting a 4-dimensional cube (also called a tesseract)
onto a hyperplane (which is a copy of 3-space) that is perpendicular to one of
its main diagonals?

* * *

The answer given above by Hop David is correct. This was the only response
offered by anyone, so this was apparently a fairly difficult problem.

My solution (partial):

Assume that the hypercube's 16 vertices lie at the points (x,y,z,w) where
each of x,y,z,w take the values 0 or 1 in all combinations.

I.e., the vertices are (0,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1),
and the 11 other vectors obtained from these by permuting their coordinates.

A vector in the direction of a main diagonal is (1,1,1,1). We can project
any point onto the 3-plane P perpendicular to this vector if we can find an
orthonormal basis for P.

Fortuitously, this is easier than it may sound. In fact, the four vectors

( 1, 1, 1, 1),
( 1, 1, -1, -1),
( 1, -1, 1, -1),
( 1, -1, -1, 1)

are mutually orthogonal, since the dot product of any two of them is 0.

These four vectors each have length = 2, so { a, b, c } form an orthonormal
basis for the hyperplane P, where

a = ( 1/2, 1/2, -1/2, -1/2),
b = ( 1/2, -1/2, 1/2, -1/2),
c = ( 1/2, -1/2, -1/2, 1/2).

We can choose these vectors a, b, c as the directions for a set of three
orthogonal axes for the plane P.

So, using this coordinate system, the projection of any vertex v of the
hypercube onto the hyperplane P is ( v*a, v*b, v*c ).

Now we can calculate the projections of all 16 vertices of the hypercube in
this way. (For example, v = (1,0,1,1) is projected onto (v*a, v*b, v*c) =
(-1/2, 1/2, 1/2).)

As it turns out, (0,0,0,0) and (1,1,1,1) project to (0,0,0); the other fourteen
vertices of the hypercube project to the vertices of a rhombic dodecahedron in
3-space. A little more thought will convince you that this (solid) rhombic
dodecahedron is in fact the projection of the hypercube onto P.

Dan Asimov
Mail Stop T27A-1
NASA Ames Research Center
Moffett Field, CA 94035-1000
(415) 604-4799 w
(415) 604-3957 fax

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.