What is the maximum number of hypercube "faces" that a 2D plane can intersect at the same time?

Question

I have a $d$-dimensional hypercube in a $d$-dimensional space. I want to find out the number of "faces" of the hypercube that a 2D plane can intersect simultaneously.

The 2D plane can be written as the intersection of $d-2$ hyperplanes:

$$\langle \pmb{p} - \pmb{v_1}, \pmb{n_1} \rangle = 0, ..., \langle \pmb{p} - \pmb{v_{d-2}}, \pmb{n_{d-2}} \rangle = 0$$

So the intersection with a hyperplane, requires adding the hyperplane's equation to the system, and in the general case results in a 1D intersection set (in special cases it may be a 2D set, or a 0D set). The issue is that faces are bounded along each dimension (let's assume to $(0,1)$). So some of the resulting planar intersections need not be counted.

My conjecture is that at most 4 faces may be intersected (as in the 3d case), but I am unsure how to prove it/disprove it. If the intersection is of the plane coinciding with a face the intersection set is 2D, if the intersection is with a corner the set is 0D, in all other cases it is 2D.

My conjecture for a $k$-D plane is that the maximum number of intersected faces is $2^k$, but I am unsure how I can formalize that idea.

Answer 1

Let the cube be $0\le x,y,z\le1$. The plane $x+y+z=3/2$ cuts all six faces of the cube.

Answer 2

The main idea here is that of the Petrie polygon of the respective polytope.

It is defined to be a skeletal polygon contained within a regular polytope as pseudo face, which outlines in d-dimensional subelements a sequence of d+1 consecutive edges and then continues accordingly with the neighbouring d-dimensional subelement, where it already had outlined d consecutive edges. As induction start one clearly can take a flat regular polygon. Here the polygon itself and its Petrie polygon coincides. With respect to polyhedra e.g. the Petrie polygons thus are nothing but the equatorial antiprismatic zig-zags. In fact, these Petrie polygons quite generally are eqatorial non-flat zig-zags, allowing for some projection onto a 2-dimensional subspace. Thereby the polytope then will be contained completely within the projection of that zig-zag, which by itself then becomes a flat convex regular polygon.

Thus you would just have to use that 2D plane through the center of your nD hypercube and will readily derive the maximal possible section each.