The convex hull of the $n+1$ points $(0,0,0,...,0,0)$, $(0,0,0,...,0,1)$, $(0,0,0,...,1,1)$, ..., $(0,0,1,...,1,1)$, $(0,1,1,...,1,1)$, $(1,1,1,...,1,1)$ in $\mathbb R^n$ is an $n$-simplex. What is the combinatorial type of the intersection of this simplex with the hyperplane $x_1+...+x_n=\frac n2$? For example, for $n=2$ it is just the segment with vertices $(0,1)$ and $(\frac12,\frac12)$, while for $n=3$ it is the quadrangle with vertices $(0,\frac12,1)$, $(0,\frac34,\frac34)$, $(\frac14,\frac14,1)$ and $(\frac12,\frac12,\frac12)$ (not a square, neither a parallelogram or trapezoid, rather something of a kite). What is it for general $n$?
What I was able to figure out: it is the convex hull of the points $$ (\underbrace{0,...,0}_i,\underbrace{r,...,r}_{n-i-j},\underbrace{1,...,1}_j) $$ with $0\leqslant i,j\leqslant\frac{n-1}2$ and $r=\frac{\frac n2-j}{n-i-j}$, and one more point with $i=j=\frac n2$ for even $n$. The number of such points is $m^2$ for $n=2m-1$ and $m^2+1$ for $n=2m$.