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What is this 4D shape with Radial equilateral symmetry property?

From your very definition of the vertex set, i.e. $$\mathcal{Q}_d=\{\mathbf{q}=\langle q_1,q_2,...,q_d\rangle : q_i = +1, q_j = -1, q_{\not = i \not = j} = 0, \: \forall \: i,j \in [1,d] \; | \; i \...
1 vote
Accepted

Question on convex polyhedra - how would I approach this?

The $n$-gonal prism has $n+2$ faces, so that settles the problem for $n\ge5$. The tetrahedron has $4$ faces and it is trivial to show that no polyhedron can have fewer faces. So the polyhedra you want ...
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3 votes
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Are there any other solids like the pseudo-rhombicuboctahedron (Miller's solid)?

Given the definitions of local and global uniformity, the problem has long been solved – at least for the convex case. The Johnson solids are the strictly convex polyhedra with regular polygon faces ...
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1 vote
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What is this 4D shape with Radial equilateral symmetry property?

Well, the four vertices (1,-1,0,0,0), (1,0,-1,0,0), (1,0,0,-1,0), (1,0,0,0,-1) are all the same distance apart and hence form a tetrahedron. Your figure has 10 of these; besides, it has some other ...
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3 votes

3D picture of the 38-sided Engel space-filling polyhedron

Now there is a computed 3D model of it with animation, interactive rotation, and space-filling arrangement of a few polyhedrons: https://community.wolfram.com/groups/-/m/t/2617634
1 vote

How to Prove that hyperplane is facet defining?

The meaning of a "facet-defining" hyperplane of $\mathbb{R}^n$ for a polyhedron of dimension $d$ is not so obvious when $d<n$. Here I assume it is facet-defining iff the intersection of ...
5 votes
Accepted

38-sided space-filling polyhedron

The images of $\left(\frac{427}{6984}, \frac{761}{6984}, \frac{1421}{6984}\right)$ under $I4_132$ indeed generate this polyhedron as their Voronoi cells. I used this description of $I4_132$ and wrote ...
5 votes

Cutting seven cuboids and forming a cube

Yes, it is possible, since the two sets of polyhedra both have Dehn invariant $0$; a result of Sydler (1965) shows that any two collections of polyhedra with the same Dehn invariant can be cut into ...
0 votes

Why isn't the Euler characteristic equal to 2 for this polyhedron?

Who needs a dodecahedron? The same thing happens with an octahedron. Say you have a regular octahedron and eight regular tetrahedra while edges are congruent with those of the octahedron. If you affix ...
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