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While it is clear that the regular dodecahedron $D$ cannot be realized with all integer coordinates, it is easy to find a polytope, which is combinatorially equivalent (face lattice isomorphic) to $D$ …

While writing the code for this answer, I noticed that I not only could calculate the number of unfoldings of the $4$-cube, but also the number of the $n$-cube for more values of $n$. Basically, we c …

Too long for a comment: Stringham gave a talk about the content of his thesis here in the Seminar of Felix Klein in Göttingen on Monday, 1880/11/29, you can look at the scans here: Ueber reguläre Körp …

Too long for a comment: The $f$-vector of the polytopes in question appear to be: P_1 (1, 2, 1) P_2 (1, 4, 4, 1) P_3 (1, 5, 8, 5, 1) P_4 (1, 9, 18, 15, 6, 1) P_5 (1, 10, 27, 33, 21, 7, 1) P_6 (1, 16, …

Not all simple polytopes are incribable, e.g. the dual of the cyclic polytope $C_4(8)$ is simple and not inscribable, as shown recently in Combinatorial Inscribability Obstructions for Higher-Dimensio …

Alathea Jensen defines "self-polar": Self-polar polytopes are convex polytopes that are equal to an orthogonal transformation of their polar sets. and writes some interesting things about self-p …

A (trivial) necessary condition is that the diagonal of the inner one is not longer than the diagonal of the outer one. So if $(a,b,c)$ is supposed to fit in $(x,y,z)$, then we should have $$a^2+b^2+ …

If I understand correctly, your polytopes are a subset of the polytopes where every inequality has at most two variables with non-zero coefficient. Those are studied in On the Complexity of Polytope …

I guess it would be difficult to prove that the answero your question is "no", since proving that "no a priori reason exists" might be hard. More modestly, I can say that I don't really know a good r …

Which are the configrations $P\subset \mathbb{R}^2$ of points, such that the following property holds: Property M (for Monochromatic): Every two-coloring of $\mathbb{R}^2$ contains a monochromatic …

The answer is "yes". Let me give the outline of a proof of a stronger statement, that implies all of the cases you consider. Proposition: In all colorings of the plane with two colors there is an equ …

This is a well studied problem and there are a couple of algorithms, for example using linear programming. For an overview take a recent reference, for example: Fitting Voronoi Diagrams to Planar Tes …

It seems that this has been answered here by S. Mustonen: PointsInGrid.pdf There $f(n,m)$ is denoted by $L_n(m)$ and a formula would be $$L_n(m)=\frac{1}{2}[f(m,n+1)-2f(m,n)+f(m,n-1)]$$ for $$f(m,k) …

Here is one way to view this problem: Form the graph $G$ with vertices $\{1,\dots,n\}$ and edges $(i,j)$ for all $i,j$ such that $|i-j|\in S$. Then $F(n,S)$ is the size of a maximal independent set i …

Looking at the last paragraph of your question, it might seem reasonable to reformulate the question as: can we always select weights $\mathbf{w}$, such that the values $Q(V_i(\mathbf{w}))$ satisfy pr …