39

There are many points in the interior of this polyhedron, constructed (independently) by
Raimund Seidel and Bill Thurston, that see no vertices.
Interior regions are cubical spaces with "beams" from the indentations passing above
and below, left and right, fore and aft. Standing in one of these cubical cells,
you are surrounded by these beams and can see ...

answered Jun 8 '14 at 14:35

Joseph O'Rourke

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39

Euclidean case
Using the formula for the tan of the half solid angle that Robin Houston quotes, and expressing everything in terms of edge lengths by using the cosine law to convert the dot products, I end up with the following linear relationship:
$$\frac{1}{P_i} = \alpha_E \cot\left(\frac{\Omega_i}{2}\right)+\beta_E$$
where:
$$\alpha_E = \frac{12 V}{...

37

there is no doubt the answer to your question is "no"; for a wonderful and scholarly recent book on the whole story, see Euler's Gem: The Polyhedron Formula and the Birth of Topology by David Richeson.
They all missed it. The ancient Greeks -- mathematical luminaries
such as Phythagoras, Theaetetus, Plato, Euclid, and
Archimedes, who where infatuated ...

answered Jan 14 '14 at 10:18

Carlo Beenakker

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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 ...

31

Today almost nobody shares anymore old Leibnitz' optimistic idea There is no ignorabimus in mathematics (in Hilbert's words). We know that there are true facts in mathematics that will never be proved, either because our maths society (if not the whole mankind) will extinguish before, or because the shortest proof has by far more symbols than there are ...

30

This is not an answer to the question, but an experimental observation that suggests a sharper conjecture: it’s only written as an answer because I’d like to flesh it out a bit more than there’s room for in a comment.
The image of $e^{i\Omega}$ under the Möbius transformation $z\mapsto\frac{i-zi}{1+z}$ is $\tan(\frac\Omega2)$, so an equivalent statement of ...

27

This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvability of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.
...

23

Theorem. Let $c(g)$ be the minimum number of cubes such that the boundary of some configuration of $c(g)$ cubes is a genus $g$ surface. Then $c(g)/g \to 2$ as $g \to \infty$.
Proof. We write $\chi(X)$ for the compactly supported Euler characteristic
of $X$, i.e.,
$\chi(X) = \sum (-1)^i \dim H^i_c(X, \mathbb{Q})$.
Note this is not a homotopy invariant:...

21

The most direct answer I know for this is the following:
Any maximal subgroup of a finite solvable group has index equal to a power of a prime. The unsolvability of $A_{5}$ is provable, therefore, by exhibiting a maximal subgroup of index 6 or 10 in $A_{5}$. These are both easy to see in terms of the geometry of the icosahedron.
The icosahedron has 12 ...

21

I have a simpler example and I see that its idea is similar to the above one.
Cut the vertices of a cube to form 8 small triangles and suppose the triangles are rigid but the faces are not. Then rotate the triangles, 4 of them clockwise and the rest counter-clockwise, in an alternating manner.
The images are drawn using Geogebra 3D.

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

This is exercise 21.3, of Mathematical Omnibus: Thirty Lectures on Classic Mathematics, by D.B. Fuks and Serge Tabachnikov. The solution is given on page 448:
Assume that $P$ is circumscribed. Consider a face $A_1$, $A_2$,...
$A_n$ and let $O$ be its tangency point with the sphere. Clearly, the
sum of angles $A_i O A_{i+1}$ is $2\pi$. We shall sum up ...

answered Jun 23 '14 at 11:54

Carlo Beenakker

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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.

17

I think this statement does not hold. The following tetrahedron should give a counterexample:
$A=(-0.5, 0, 0)$, $B=(1,0,0)$, $C=(0,\varepsilon^2, \varepsilon)$, $D=(0,\varepsilon^2, -\varepsilon)$, $1>>\varepsilon>0$.
Here $BCD$ has largest area, but the spherical angles at $C$ and $D$ should be significantly larger than the one at $A$. (I have ...

17

Here's the abstract of K.A. Post, "Triangle in a triangle: On a problem of Steinhaus", Geom Dedicata (1993) 45: 115; this paper was cited in the one given in the comment by Nemo.
A necessary and sufficient condition on the sides $p, q, r$ of a triangle $PQR$ and the sides $a, b, c$ of a triangle $ABC$ in order that $ABC$ contains a congruent copy of $PQR$ ...

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

May I suggest this article by Joseph Malkevitch, an AMS Feature Column.
Note particularly his point below (screen snapshot--not searchable)
that polyhedra were simply not viewed,
at the time, in terms of vertices, edges, and faces:
This point is also made in Imre Lakotos's
Proofs and Refutations (Cambridge Univ. Press, 1976).
Addendum. Joe ...

answered Apr 18 '14 at 18:34

Joseph O'Rourke

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16

Here goes the elementary proof of the claim by Robert Houston that the quadraples $(P_1^{-1},P_2^{-1},P_3^{-1},P_4^{-1})$ and $(\cot \frac{\Omega_1}2,\cot \frac{\Omega_2}2,\cot \frac{\Omega_3}2,\cot \frac{\Omega_4}2)$ are affinely equivalent. In the spherical case the first quadraple should be replaced to $\{\cot\frac{P_i}2\}$, in the hyperbolic case to ...

answered Aug 27 '19 at 21:53

Fedor Petrov

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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

Not an answer, just a long comment. This comment slowly evolved into a partial numerical solution: the regular pentagon is the local maximum.
I've used Maple to produce the formula for $\frac{S'}{S}$ and the result is a rational function on $\mathbb{R}^{10}$ with both numerator and denominator of degree 12. The numerator has 11495 monomial terms while the ...

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 ...

15

This is very similar to the cryo-electron microscopy problem: You want to image a certain macromolecule, and the scale of the macromolecule requires the use of an electron microscope. Unfortunately, such an imaging process is harmful to the specimen. In the 80s, they realized you can protect the specimen by freezing it first, but a single exposure will undo ...

answered May 31 '15 at 14:16

Dustin G. Mixon

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14

There is nothing particularly mysterious here. Roughly, fix the lattice (incidence relations of all faces) and the edge lengths. Then there are several realizations of this structure, typically a finite number of them, but not always (see here). The space of realizations is a semi-algebraic set, but the degrees of the polynomials are typically very large, ...

14

Here are a few trivial lemmas. I won't use anything about the rolling motion, just that the distance is defined by gluing pentagons edge-to-edge:
The $dd$-circle of radius $k$, which I'll call $C_k$, is a closed polygonal curve. Let $D_k$ be the closed $dd$-disk of radius $k$; note that $D_k$ may contain holes and $C_k$ is not in general simple! (This ...

14

Here one can find an algebra-geometric proof of the trigonometric relations. The main point is that given a (say, spherical) tetrahedron $T$ one can construct a rational elliptic surface $X_T$ with the property that surfaces corresponding to a tetrahedron and its dual are isomorphic. The cross-ratios in the question are some invariants of $X_T$ computed in ...

13

[Corrected (two typos noted by V.T.) and expanded
(alternative coordinates)]
Also not an answer, but this time a simpler algebraic formulation.
All indices are cyclic mod $5$. Let the vertices be $P_i$ ($i \bmod 5$),
and let $A_i$ be the area of triangle $P_{i-1} P_i P_{i+1}$.
Then $S$ is the larger root of the quadratic equation
$$
S^2 - S \sum_{i\,\bmod\...

answered Dec 9 '13 at 22:43

Noam D. Elkies

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13

Once I was lecturing to high school students about this theorem. My proof began with the words: "suppose that the net is drawn on a surface of a rubber ball...".
One student asked: "Did rubber exist at the time of Euler?"
I think the moral of this story is that the very statement of the question was foreign to
ancient Greek mathematics. Did they have a ...

answered Jan 16 '14 at 15:46

Alexandre Eremenko

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13

If you have no information about the vertices and allow nonconvexity there could be problems first you will only get information about the surface of the 3d shape but even there you will have problems. If you remove a tiny cube near the center of one of the faces of the cube the surface will not be the same but I don't think it will make a difference in the ...

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