Here's the answer. The main claims (Claims 1-4) I am fairly sure I got right, but I could easily have missed a case (or counted an extra case) in the later enumeration. If anybody finds a mistake, please comment. Let me remark that I find Claims 1-4 much more interesting than the subsequent enumeration based on them.
The simplex centroid $e=(1,1,1,1,1,1,1,1)$ is included in all our hyperplanes, but I'm generally not counting it as a centroid in the discussion below (so centroid means centroid of a $k$-dimensional face, with $k < 7$).
We'll divide the question into cases.
The first case we deal with is when we don't have any unexpected centroids. We start
with three centroids $a$, $b$, and $c$. These will automatically generate $\bar{a}$, $\bar{b}$, $\bar{c}$. We will call any centroid other than these six an unexpected centroid. We represent our centroids as subsets of
{$1,2,\ldots,8$}. If we have the centroid
{$1,2,3$}$=\langle 1,1,1,0,0,0,0,0\rangle $,
then we automatically have the centroid {$4,5,6,7,8$}$=\langle 0,0,0,1,1,1,1,1\rangle$ corresponding to
the complement of the set. So, to summarize, the first case consists of hyperplanes
which pass through exactly six centroids: $a,b,c,\bar{a},\bar{b},\bar{c}$.
Now, let's represent this case by putting the numbers {$1,2,\ldots,8$} on
the vertices of a cube. The cube will have three faces
corresponding to $a,b,c$ and the three opposite faces will correspond to
$\bar{a},\bar{b},\bar{c}$. For example, if the sets were $a=${$1,2,3$},
$b=${ $1,5,6$} and $c=${$2,5,6,7$}, then the vertex $\bar{a}bc$ would contain
{$5,6$}, the vertex $\bar{a}\bar{b}\bar{c}$ would contain {$4,8$}, and the
vertex $\bar{a}b\bar{c}$ would be empty. It's not too hard to see that
whether there is an unexpected centroid only depends on the positions of the
empty vertices. It's also clear that rotations and reflections
of this cube give equivalent sections.
Claim 1: If two adjacent vertices are empty, there is an unexpected
centroid.
Proof: The two adjacent vertices form an edge. We might as well rotate
the cube so that the empty edge is the $ab$ edge. Then we have $a \cap b =
\emptyset$. This means that $a \cup b$ is an unexpected centroid
(here we have to use the fact $a \neq \bar{b}$). Example: if $a =${$1,2$}, $b =
${$3,4,5$}, then $a \cup b =${$1,2,3,4,5$} is in the linear span of $a$ and
$b$.
Claim 2: If two opposite vertices of the cube are empty, then there is an
unexpected centroid.
Proof: We can rotate the cube so the empty vertices correspond to
$\bar{a}\bar{b}\bar{c}$ and $abc$. Then, for example, if
$a=(1,1,1,0,0,0,0,0)$, $b=(0,0,1,1,1,1,0,0)$ and $c=(1,0,0,0,0,1,1,1)$,
we can take $a+b+c-e$ where $e$ is the all-ones vector, and get
$(1,0,1,0,0,1,0,0)$.
Claim 3: If the odd- or even-parity vertices of the cube are empty, then
there is an unexpected centroid.
Proof: Rotate the cube so that $\bar{a}\bar{b}\bar{c}$ is empty. Now, every
coordinate is in exactly 1 or 3 of $a,b,c$. Thus, $\frac{1}{2}(a+b+c-e)$
is an unexpected centroid. Example
$a=(1,1,1,0,0,0,1,1)$, $b=(0,0,0,1,1,0,1,1)$ and $c=(0,0,0,0,0,1,1,1)$, and
$\frac{1}{2}(a+b+c-e) = (0,0,0,0,0,0,1,1)$
Claim 4: If none of the situations in Claims 1,2,3 hold, then there is no
unexpected centroid.
Proof: We can rotate the cube so that the empty vertices are a subset of
$\bar{a}bc$, $a\bar{b}c$ and $ab\bar{c}$. For there to be an unexpected
centroid, you must be able to find $\alpha a + \beta b + \gamma c$ so that
the coordinates of this vector take on two values. One of these
coordinates is 0 (since $\bar{a}\bar{b}\bar{c}$ is not empty), meaning we
can assume wlog that $\alpha, \beta, \gamma$ are either 1 or 0. But
for $\alpha + \beta + \gamma$ to also be either 1 or 0, we need two
of $\alpha, \beta, \gamma$ to be 0, which means that we don't get an
unexpected centroid.
So now, we need to enumerate the number of ways of putting 8 elements
onto the vertices of a unit cube so that at least one element is on each of
the nonempty vertices. Since sections are equivalent under permutations of the coordinates, we should consider these to be 8 identical elements (so the only thing that matters is how many elements are on a vertex). There are four cases.
Case A: no empty vertices. There is just 1 way of doing this: putting
one element on each vertex.
Case B: one empty vertex. There are 3 ways of doing this. Exactly
one vertex will have two elements on it, and it can be either Hamming
distance one, two, or three from the empty vertex.
Case C: two empty vertices. In this case, these two vertices must have
Hamming distance 2. The two extra elements can either be on the same vertex
(3 ways) or two different vertices (7 ways).
Case D: three empty vertices. In this case, any pair of these three vertices
must have Hamming distance 2, so there's only one way of arranging them.
The three extra elements can either be all on the same vertex (3 ways),
divided two on one vertex and one on another (7 ways), or on three different
vertices (4 ways).
This gives 28 essentially different sections with no unexpected centroids.
We now must count the cases with unexpected centroids corresponding to Claims 1-3.
We'll deal with the situation in the Claims 1,2,3 separately.
Case of Claim 3
Let's start
with the situation in Claim 3. First, we can assume that there are no
empty vertices of the cube other than the two opposite ones (if this happens,
we are in the Claim 1 situation, and we take care of it there).
We now can choose another centroid $d$ in the linear span of $a,b,c,e$
so that $a \cup b \cup c \cup d$ covers every coordinate exactly twice.
By the criterion that there are six non-empty vertices, none of the
six intersections $a \cap b$, $a \cap c$, etc. can be empty. We need to
put the eight elements into these six intersections. This corresponds
to putting 8 elements on the edges of a tetrahedron so that every edge
corresponds to at least one element. There are 3 ways to do this (two
extra on one edge, one extra on each of two opposite edges, and one
extra on each of two adjacent edges). Claim 3 thus gives 3 more
non-equivalent sections. Notice that if we had analysed Claim 3 by just
looking at the symmetries of the cube (as we did for the cases without
unexpected centroids), we would have obtained four non-equivalent sections.
Case of Claim 2
What I'd like to claim here is that this is really the situation in Claim 1
disguised. Maybe the best way to do this is by example. If we have
$a=${$1,2,7,8$}, $b=${$3,4,7,8$}, $c=${$5,6,7,8$}, then the centroid {$7,8$}
is in our hyperplane, and the hyperplane is thus generated by $a'=${$1,2$}, $b'=${$3,4$},
$c'=${$5,6$}, which is covered by Claim 1.
Case of Claim 1
Here, there are three possibilities. In the first one, there are four
centroids $a$, $b$, $c$, $d$, with pairwise empty intersections so that
$a \cup b \cup c \cup d =${$1,2,\ldots,8$}. The number of ways of doing this
is the number of partitions of 8 into four non-empty parts, which is 5:
{$(5,1,1,1), (4,2,1,1), (3,3,1,1),(3,2,2,1),(2,2,2,2)$}.
In the second possibility, we have three pairwise disjoint
centroids $a$, $b$, $c$, with $a \cup b \cup c = e$,
and also another centroid $d$ so that
both $d\cap x$ and $\bar{d} \cap x$ are non-empty for $x=a,b,c$.
The cardinalities of $a,b,c$ could be {$4,2,2$} or {$3,3,2$}. In either
case, we get two non-equivalent sections, giving 4 total non-equivalent
sections.
For the third possibility, we have three pairwise disjoint
centroids $a$, $b$, $c$, with $a \cup b \cup c = e$, and
we have two more pairwise disjoint centroids $f$ and $g$ so that $f \cup g = a \cup
b$. In this case, the cardinality of $c$ can range from 1 to 4.
I'll just list representative vectors for these possibilities. The coordinates
considered are those not in $c$.
$c=4$
$(1,1,0,0),(0,1,1,0),(0,0,1,1),(1,0,0,1)$
$c=3$
$(1,1,1,0,0),(0,0,1,1,0),(0,0,0,1,1),(1,1,0,0,1)$
$c=2$
$(1,1,1,0,0,0),(0,0,1,1,1,0),(0,0,0,1,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,0,1,1,1,0),(0,0,0,0,1,1),(1,1,0,0,0,1)$
$(1,1,1,1,0,0),(0,1,1,1,1,0),(0,0,0,0,1,1),(1,0,0,0,0,1)$
$c=1$
$(1,1,1,1,1,0,0),(0,1,1,1,1,1,0),(0,0,0,0,0,1,1),(1,0,0,0,0,0,1)$
$(1,1,1,1,1,0,0),(0,0,1,1,1,1,0),(0,0,0,0,0,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,0,1,1,1,1,0),(0,0,0,0,1,1,1),(1,1,0,0,0,0,1)$
$(1,1,1,1,0,0,0),(0,1,1,1,1,0,0),(0,0,0,0,1,1,1),(1,0,0,0,0,1,1)$
This gives 9 more non-equivalent sections, making 49 altogether.
Best Answer
For every simplex $\Delta\subset\Bbb R^n$ with $n+1$ vertices and every number $m\le n$ there is an $m$-dimensional subspace $W\subseteq\Bbb R^n$ so that the orthogonal projection of $\Delta$ onto $W$ has $n+1$ vertices.
Proof.
Fix a polytope $P\subset\Bbb R^m$ with $n+1$ vertices. First, we show that $P$ is the orthogonal projection of some simplex $\Delta_P \subset\Bbb R^n$: if $P$ has vertices $v_0,...,v_n\in\Bbb R^m$, then for $i\in\{0,...,n\}$ the
$$w_i:=\underset{\!\!\!\!\!\!\!\!\begin{array} .\uparrow\\ \!\!\!\!\!\!\!\!\text{at index $m$}\end{array}}{\begin{cases} (v_i,0,...,0) & \text{for $i\le m$} \\ (v_i,0,...,1,...,0) & \text{for $i > m$} \end{cases}}\quad\in\Bbb R^n$$
are $n+1$ points in $\Bbb R^n$ in convex position. They necessarily form the vertices of some simplex $\Delta_P$. Also, the projection onto the first $m$ coordinates (which is orthogonal) yields the initial polytope $P$. Alternatively, we can say that the intersection of the subspace $\Bbb R^m\subset\Bbb R^n$ with the cylinder $C:=P+\Bbb R e_{m+1}+...+\Bbb R e_n$ is $P$.
Now, any two simplices in $\Bbb R^n$ with $n+1$ vertices are affinely equivalent, and so we can choose an affine transformation $X$ with $X\Delta_P= \Delta$. Clearly, the intersection of the transformed cylinder $X C$ with the transformed subspace $W':=X\Bbb R^m$ is still a polytope with $n+1$ vertices. The same is true if you replace $W'$ by a subspace $W\subseteq\Bbb R^n$ orthogonal to $Xe_i,i\in\{m+1,...,n\}$, as all cross-sections of the cylinder are affinely equivalent. But this intersection can be interpreted as the orthogonal projection of $\Delta=X\Delta_P$ onto $W$, which then has $n+1$ vertices. $\quad$ $\square$
As a side note, the intersection $XC\cap W$ is an affine transformation of $P$. Thus, given any simplex $\Delta\subset\Bbb R^n$ and any polytope $P\subset\Bbb R^m$ with $n+1$ vertices, $P$ is affinely equivalent to an orthogonal projection of $\Delta$.