This is a (slightly modified) crosspost.

**Subsequent edit** - coordinates are changed to obtain simpler expressions; the existing answer is not affected.

There is a family of convex polytopes: $P_n$ is $n$-dimensional and has $(m+1)^2$ vertices for $n=2m$ and $(m+1)^2+1$ vertex for $n=2m+1$.

I define $P_n$ to be the section of the simplex in $\mathbb R^{n+1}$ with vertices $$ (\underbrace{-1,...,-1}_i,\underbrace{1,...,1}_{n+1-i}), $$ $i=0,1,...,n+1$ with the "midway" hyperplane $x_0+...+x_n=0$. For example, $P_1$ is just the segment in $\mathbb R^2$ with vertices $(-1,1)$ and $(0,0)$,

while $P_2$ it is the quadrangle in $\mathbb R^3$ with vertices $(-1,0,1)$, $(-1,\frac12,\frac12)$, $(-\frac12,-\frac12,1)$ and $(0,0,0)$ (not a square, neither a parallelogram or trapezoid, rather something of a kite).

In general, $P_n$ is the convex hull of the points $$ (\underbrace{-1,...,-1}_i,\underbrace{r,...,r}_{n+1-i-j},\underbrace{1,...,1}_j) $$ with $0\leqslant i,j\leqslant\frac n2$ and $r=\frac{i-j}{n+1-i-j}$, and one more point $(-1,..,-1,1,...,1)$ with $i=j=\frac{n+1}2$ for odd $n$. The number of such points is $m^2$ for $n=2m$ and $m^2+1$ for $n=2m+1$.

In fact some experimenting suggests that $P_{2m+1}$ is a *pyramid* - namely, removing that extra point $(-1,...,-1,1,...,1)$ gives a single facet of $P_{2m+1}$. In fact the base of the perpendicular from $(-1,...,-1,1,...,1)$ to this facet is another vertex $(-1,...,-1,0,0,1,...,1)$. Whether this facet is combinatorially equivalent to $P_{2m}$ I don't know. It is certainly not isometric to $P_{2m}$.

Is there some nicer description? Say, a more tidy realization of the same combinatorial type? For example, can one easily find numbers of faces of all dimensions in $P_n$? What about the dual polytope?

**Update**

Using the Sage code from the @MoritzFirsching's answer one checks that $P_{2m}$ is in fact the Cartesian square of an $m$-simplex. How to prove this?

Cartesian productof two simplices?? But then it would have $m^2$ facets while you computed that it has only $2m+2$ facets... $\endgroup$ – მამუკა ჯიბლაძე Dec 16 '20 at 12:13`[(polytopes.simplex(n)*polytopes.simplex(n)).is_combinatorially_isomorphic(get_P(2*n)) for n in range(14)]`

evaluates to`True`

throughout! I was wrong about facets, it seems to fit. So it "only" remains to prove it $\endgroup$ – მამუკა ჯიბლაძე Dec 16 '20 at 15:583more comments