Intersection between hyperplane within a simplex

[Debashis]

Hi,

If we contract a (n-1)d hyperplane with a n-simplex, then what is maximum number of intersection points with the egdes of the simplex and the hyperplane ?

For, if we draw a line within a 2-simplex (there are 3 edges), it will have a intersection of maximum two edges. For 3-simplex, any plane within the tetrahedron (where there are 6 egdes) can have maximum intersection with 4 edges. What is the value for n-simplex ?

 

[homeomorphic]

The question is a bit unclear. A line can intersect the edges of a 2-simplex in infinitely many points if you just make the edge a subset of the line. Do you mean to ask what is the maximum number of edges that can be intersected, rather than the maximum number of points?

 

[Ben Niehoff]

Think of this as a graph theory problem instead. An n-simplex is just a complete graph with n vertices; i.e., each vertex is connected to every other vertex. Then the question is, how should we partition a complete graph on n vertices into two subgraphs, such that the number of edges crossing between the two subgraphs is maximized.

A complete graph has $n!$ edges. If we take $k$ vertices and call them subgraph A, while the remaining $n-k$ vertices we call subgraph B. Now we want to know the number of edges that go between A and B. Since each vertex has an edge connecting to every other vertex, then the number of edges going between A and B must be given by

$$k (n-k)$$
To maximize this for a given $n$, we must clearly choose the $k$ that is closest to $n/2$. Therefore the maximum number of (isolated) points in which a hyperplane can intersect a simplex is given by

$$f(n) = \begin{cases} \frac{n^2}{4} & n \; \text{even} \\ \frac{n^2 - 1}{4} & n \; \text{odd} \end{cases}$$
which gives the expected result for $n=3$ (triangle) and $n=4$ (tetrahedron).

 

[Debashis]

thank you very much for the answer. How can we represent that hyperplane in the respective graph of n-vertices ?