61

Here are a couple pictures. If you'd like to do more, I created this with a simple povray script. Feel free to e-mail me and I'll send it to you.
With four reflections.
With 8 levels of reflections.
One with 24 reflections.
And one with 32 reflections, and each mirror having a slightly different tint, a red sphere, and a slightly wider viewing ...

53

Nima Arkani-Hamed had a series of talks at JHU roughly 6 months ago which I attended related to this topic. He discussed it at Stony Brook a little bit over a week ago (pointed out by Emilio Pisanty in the comments above in which he used the term "amplituhedron", but my understanding of this comes mostly from his earlier talks.
Update: I managed to track ...

39

These are the so-called Regge symmetries, described by T. Regge in a 1970-ish paper. For a bit on it, with references, see the paper
Philip P. Boalch, MR 2342290 Regge and Okamoto symmetries, Comm. Math. Phys. 276 (2007), no. 1, 117--130.\

answered Jan 15 '17 at 21:13

Igor Rivin

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36

The 600-cell can be tiled by five 24-cells in exactly ten different ways. These are written explicitly in table 2 of "Parity proofs of the Bell-Kochen-Specker theorem based on the 600-cell", where you can also see an application of this fact to giving a proof of the Kochen-Specker theorem, ruling out the existence of noncontextual hidden variable theories in ...

answered Dec 15 '17 at 0:48

Gjergji Zaimi

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26

If you fix $m$, this is known as the $m$-dimensional multiplication problem. In 2010 Koukoulopoulos showed that as $n\rightarrow \infty$ $$P(m,n)=\left|\lbrace a_1\cdots a_m\ :\ a_i\leq n \text{ for all } \ i\rbrace\right|\asymp \frac{n^{m+1}}{(\log n)^{c_m}(\log\log n)^{3/2}}$$ where $$c_{m}=\int_{1}^{\frac{k}{\log(m+1)}}\log x\text{d}x=\frac{\log(m+1)+m\...

23

Edit: The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it.
This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Serre tree associated to $SL_2$ (really, a Bruhat-Tits building associated to $SL_2(F)$, where $F$ is a local field), hence is virtually free.
...

21

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-Dimensional Polytopes by Doolittle, Labbé, Lange, Sinn, Spreer and Ziegler
In dimension $3$, there is a combinatorial criterion by Rivin describing inscribabilty ...

answered Oct 2 '20 at 12:36

Moritz Firsching

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20

To address the scissors congruence question at the end of the post: the Regge symmetries produce tetrahedra which are scissors congruent. This is proved in Section 6 (Theorem 9 and Corollary 10) of
J. Roberts. Classical $6j$-symbols and the tetrahedron. Geom. & Top. 3 (1999), pp. 21-66. (link to paper on arXiv)
The argument is indirect, proving that ...

19

Using $k$ half-spaces, the polytope has at most $k$ facets. For a fixed number of facets, the number of vertices is maximized, for example, by the dual polytope of the cyclic polytope with $k$ vertices. More generally, by the dual polytope of any neighborly polytope with $k$ vertices.
This maximum number of vertices is equal to
$${k-\lceil n/2 \rceil \choose ...

19

The answer to the third question is no. This is a rather counter-intuitive discovery of Micha Perles from the sixties. See this paper of Ziegler, for a simpler construction and other pertinent information. However, for polytopes in dimension $3$, the answer is yes, as mentioned in the same paper of Ziegler.

18

Edit: a preprint concerning this problem can now be found on the arXiv: http://arxiv.org/abs/1407.0683
Let me give an exhaustive answer. Croft closes his paper with a list of the unsolved cases:
Here $\kappa$ denotes the maximal edge length of the inner polyhedron. The ratio of the volumes can be easily computed when $\kappa$ is known.
Let's fill in the ...

answered Jan 8 '14 at 22:57

Moritz Firsching

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18

Did you ever find any answer to this?
I find it intriguing that figuring out which shapes of holes a given solid object can pass through is widely considered to be a suitable puzzle for 2 year olds, yet we still don't know the smallest possible hole even for a regular simplex.
Just playing I found a hole shape which I believe brings the upper bound down from ...

18

It seems that here is an example for rational exponents.
Let $n=2209=47^2>3^7>2^{11}$, and $N=4\,385\,664=2^7\cdot 3^6\cdot 47=2048^{7/11}\cdot 2187^{6/7}\cdot 2209^{1/2}\cdot 1^{1/154}$ with $7/11+6/7+1/2+1/154=2$. If $N=ab$ with $a,b\leq 47^2$, then 47 divides one of $a$ and $b$ (say, $a=47k$); since $a,b\leq 47^2$ we have $47\geq k\geq 40$ (the ...

18

I implemented the ideas in the paper using Mathematica. I pushed it a bit further to actually generate the images below. You can download this Mathematica notebook to see the code and detailed explanation.
You might notice Dali's original in the middle of the third row from the bottom.

18

Let me give a geometric interpretation for the case of tetrahedra of volume zero. The statement becomes as follows:
Given four positive numbers $a,b,c,d$ that satisfy the quadrangle inequalities, denote
$$
s=\frac{a+b+c+d}2, \quad a'=s-a \text{ etc.}
$$
Take a quadrilateral with side lengths $a,b,c,d$ (in this cyclic order). If its diagonals have lengths $x,...

18

According to Jeff Miller's Earliest Known Uses of
Some of the Words of Mathematics, the first known occurrence is in Schoute’s Mehrdimensionale Geometrie of 1902.

17

For $n=3$ this question was
asked in 1996 by
James Propp,
conjecturing that the answer is Yes.
(I got the reference from this
page, which concerns the very special case of a rectangular box
in ${\bf R}^3$; this is already a nontrivial problem, as Jim Propp noted,
and the cuboid page
reports calculations claimed to prove it in that case and to locate
the ...

answered Apr 17 '14 at 16:07

Noam D. Elkies

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17

Consider the polytope in $\mathbb{R}^3$ with $8$ vertices at coordinates $(\pm 1, \pm 2, 1), (\pm 2, \pm 1, -1)$. Geometrically this looks like a cube where the top face is stretched in the direction of the $y$-axis and the bottom face is stretched in the direction of the $x$-axis, but it still has the face structure of a cube and in particular has $6$ ...

16

W. Wernick has tabulated 139 triangle construction problems using a
list of sixteen points associated with the triangle. In each case three points are given and the goal is to construct a triangle for which the three 'special' points listed are the points given. There are still some open problems in Wernick's list. Notation:
$A$, $B$, $C$ Three vertices,
...

16

Yes. Let $\mathbb{R}^{alg}$ be the field of real algebraic numbers. This is a real closed field which means that, for any statement of first order logic, using the symbols $0$, $1$, $+$, $\times$, $=$, $<$, that statement is true in $\mathbb{R}^{alg}$ if and only if it is true in $\mathbb{R}$.
The statement "The volume of $\mathrm{Hull}(x_1, x_2, \ldots, ...

answered Jul 27 '13 at 2:36

David E Speyer

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16

I claim that this is false for $n=6$. I find it convenient to shift the variables to $w_{ij} = 2 v_{ij} -1$. So the inequalities are
$$-1 \leq w_{ij} \leq 1 \quad (1)$$
$$w_{ij} + w_{ji}=0 \quad (2)$$
$$-1 \leq w_{ij} + w_{jk} + w_{k i} \leq 1. \quad (3)$$
Consider the point $x_{12} = x_{34} = x_{56} = 1$, $x_{23} = x_{45} = x_{61} = -1$ and $x_{ij}=0$ in ...

answered Aug 11 '14 at 16:30

David E Speyer

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16

The number of integer points in $P_n$ is the number of forests on $[n]$; see Section 3 of Stanley's Decompositions of rational convex polytopes. In fact you can see there a simple description of its entire Ehrhart polynomial in terms of forests. (See also Section 9.3 of Beck and Robins's book Computing the continuous discretely.)
The toric variety ...

15

Chromatic Number of the Plane or Hadwiger–Nelson problem asks for the minimum number of colors required to color the plane such that no two points at distance $1$ from each other have the same color.
It is only known that
$$5 ≤ \chi ≤ 7.$$

15

What you call the derived polygon is a construction that has appeared many times in literature. I believe the first occurrence was in a 1878 paper by Darboux, "Sur un problème de géométrie élémentaire", but it has also been a topic of many problems and articles in the American Mathematical Monthly, which is where I've learned about it. :-)
Not only is the ...

answered Nov 25 '13 at 5:23

Gjergji Zaimi

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15

As Ian Agol said, the group is virtually free.
Now, let us note that all your quaternions are congruent to $1$ modulo $\sqrt 5$. This implies your group is torsion free (a kind of Minkowski theorem), and hence is free.
[ Proof : Let $\Gamma$ be the group of quaternions of norm $1$ in $\mathbf Z[\frac{1}{2},\sqrt{5}]$, and let $\Gamma_0$ be the subgroup of ...

15

You can find Frank Lutz's lists of simplicial spheres (and other manifolds) here:
Let me shamelessly self-advertise my list of simplicial 4-polytopes with up to $10$ vertices and various families of neighborly polytopes here. I give realization with rational coordinates (inscribed, if possible). You can extract the lists of facets easily from the ...

answered Sep 14 '15 at 21:41

Moritz Firsching

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15

An example
Yes, here is a list of rational coordinates lying on the unit sphere, the convex hull of which is combinatorially equivalent to a regular dodecahedron. This polyhedron is invariant under reflections in three orthogonal hyperplanes (having a symmetry group $C_2 \times C_2 \times C_2$ of order 8, much smaller than the order-120 symmetry group of a ...

answered Jul 25 '20 at 19:18

Adam P. Goucher

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15

Philosophical questions deserve philosophical answers, so I am afraid no amount of references and specific results will probably satisfy you. Let me try to explain it in a somewhat generic way.
Think about it this way - why care about sequences like $\{n!\}$, Fibonacci or Catalan numbers? The honest answer is "because they come up all the time". Now, ...

14

The problem is NP hard. Here is a proof sketch.
The problem is to determine if there is a point $y$ with $\|y\|=1$ outside of the convex hull of given points $x_1,\dots, x_n$. Note that such point exists if and only if there is hyperplane at distance less than $1$ from the origin such that all points $x_1, \dots, x_n$ and $0$ lie on one side of the ...

14

Your question is essentially about extension complexity. In general, the extension complexity of a polytope $P$ is the minimum number of facets over all polytopes $Q$ which project to $P$. You are interested in the extension complexity of polygons. Fiorini, Rothvoß, and Tiwary proved that regular $n$-gons have extension complexity $O(\log n)$. For lower ...

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