From:           zare@cco.caltech.edu (Douglas J. Zare)
Newsgroups:     sci.math
Subject:        Re: On simplex/hyperplane intersection
Date:           3 Jan 1996 00:33:08 GMT
Organization:   California Institute of Technology, Pasadena

James C. Evans <jce@seaice.geol.scarolina.edu> wrote:
>Suppose that a "solid" simplex of dimension n that has of n+1
>vertices, each of which is a point having n coordinates, is
>intersected by a hyperplane of integer dimension m, 0 < m < n,
>namely
>	sum_{i=0}^m a_i * x_i = b .
>By solid, I mean that the simplex should be regarded as including
>its interior.
>
>Is it true that the intersection is either null, or is another
>solid simplex of dimension p, where 0 < p < m and where a point
>and a line segment here are defined as degenerate simplexes of
>dimension 0 and 1, respectively?  
>[...]

No, you can get a square by cutting a tetrahedron in half such that two 
vertices are "above" the plane and two are below.

In slightly more generality, if there are n points above the plane and m 
below, then the intersection is the Cartesian product of an n-point 
simplex and an m-point simplex. For example, (n,m)=(2,3) yields a 
triangular prism. Note that there are nm vertices of this figure, yet in 
the most symmetric case there are only 2 distances between the vertices. 

If there are p points in the plane, n above, and m below, then the result 
is the p-times iterated cone over the product of n-point and m-point 
simplices. For example, when (p,n,m) = (1,2,2), the result is a 
square-based pyramid; when (p,n,m) = (2,2,2), the result is a hyper 
pyramid whose base is a square-based pyramid. This even makes sense when 
one or more of p, n, and m equal 0.

The proofs are straightforward, so I hope I did them correctly: First, 
note that the combinatorial type is determined by p, n, and m. Second, 
use the coordinatization of an k-point simplex as nonnegative real 
ordered k-tuples which sum to 1. For the p=0 case, it suffices to 
consider the plane defined by setting the sum of the first n coordinates 
to 1/2; this is also implies that the sum of the last m coordinates is 
1/2, hence the intersection is the Cartesian product of 1/2-size n-point 
and m-point simplices. The case of p>0 is similar.

This is a really neat problem. I'll have to think about the higher
co-dimension cases and cubes, cross-polytopes, and hemi-cubes instead of
simplices. I'll also post to the Geometry Forum's geometry.puzzles,
geometry.college, and/or goemetry.research
(http://forum.swarthmore.edu/~sarah/HTMLthreads/index.html), but I don't
see how to link those to posts to sci.math. 

Douglas Zare
http://www.cco.caltech.edu/~zare