55

As I see it, the key intuition is passing from the equator orthogonal to a single vector to looking at a whole orthonormal basis.
Suppose we pick a random unit vector $(x_1,\dots,x_n)$. What we want to know is why $x_1$ is probably near zero, since this is equivalent to being near the equator relative to the first basis vector. But this feels intuitively ...

41

This question was explored here:
Lenhart, William J., and Sue H. Whitesides. "Reconfiguring closed polygonal chains in Euclidean $d$-space." Discrete & Computational Geometry 13, no. 1 (1995): 123-140; DOI: 10.1007/BF02574031, eudml.
From the Abstract: "It is shown that in three or more dimensions, reconfiguration is always possible, but that in ...

answered Aug 28 '19 at 11:32

Joseph O'Rourke

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35

James Wenk and I just finished a paper proving Zalgaller's sphere inspection conjecture for closed curves:
Shortest closed curve to inspect a sphere.
We show that in $R^3$ any closed curve $\gamma$ which inspects the unit sphere $S^2$, i.e., lies outside $S^2$ and contains $S^2$ within its convex hull, has length $L(\gamma)\geq 4\pi$. Equality holds only ...

28

That isn't a conjecture but a routine exercise assigned after the students learn about Bang's solution of the Tarski plank problem. The proof goes in 2 steps:
1) Consider all sums $\sum_j \varepsilon_i u_i$ with $\varepsilon_i=\pm 1$ and choose the longest one. Replacing some $u_j$ with $-u_j$ if necessary, we can assume WLOG that it is $y=\sum_i u_i$. ...

27

Visualize first, for comparison, the 2-dimensional unit sphere in 3-dimensional Euclidean space (something that I can visualize!), and imagine it cut, by circles of latitude (perpendicular to the $z$-axis), into narrow zones. Of course, the zones closer to the poles have smaller radii, and therefore smaller circumferences, than the zones near the equator. ...

answered Jun 27 '15 at 20:13

Andreas Blass

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26

Let me address the specific complaint of that review. The situation is the following. Our (bounded, open) convex set is denoted $K\subseteq\mathbb R^n$ with closure $\overline K$, and we consider the "distance to $p$" function $d_p:\partial K\to\mathbb R$ for $p\in\overline K$. Let $V\subseteq\overline K$ be the set of $p\in\overline K$ for which $d_p$ ...

25

In dimensions $3$ or greater, it is false. Take some non-coplanar points. You can connect them with a nonconvex polyhedron with arbitrarily small surface area by thickening a spanning tree, for example. The convex hull has surface area at least as great as the surface area of the convex hull of the points.
In dimension $2$, it is true. In two dimensions, ...

25

There has been a bunch of work along these lines, and I think the idea has been rediscovered several times. I suggest looking at the papers of Anna Romanowska, who refers to them as "barycentric algebras", to get an idea of what's known. Her book with Smith, "Modes", covers this as well as generalizations where $t$ is not required to be in $[0,1]$. Here ...

24

Take a very obtuse isosceles triangle and chop its acute angles.

answered Apr 25 '21 at 1:11

Anton Petrunin

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23

The numbers that are not of the form $a^2-b^2-c^2$ with $b$ and $c$ positive and $a>b+c$ are precisely the idoneal numbers apart from $7$, $28$, $112$, $15$, $60$, and $240$. As noted in the problem, the paper by Hertel and Richter shows that the numbers not of this form are necessarily idoneal numbers, and a quick calculation shows that $7$, $28$, $112$,...

23

Consider the sphere with equator 4.
Divide it into spherical cubes, the central projections from an inscribed cube.
Note that the exponential map from tangent plane to the sphere is short.
Note also that if one maps a unit cube centered at the origin by the exponential map it will cover the spherical cube.
It is easy to modify the map to get a short ...

answered Mar 16 '17 at 17:37

Anton Petrunin

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23

The following proposition answers OP's question regarding the upper bound of $$\tau(C) \Doteq f(C)/\lambda^2(C).$$
Let $B_n$ be the closed Euclidean unit ball of $\mathbb{R^n}$ centred at $0$, that is $$B_n = \{ (x_1,\dots, x_n) \in \mathbb{R^n} \, \vert \,\,
x_1^2 + \cdots + x_n^2 \le 1\},$$
and let $\tau_n = \tau(B_n)$.
Proposition.
The Euclidean unit ...

20

I have recently finished a paper called
The length, width, and inradius of space curves
where it is shown that the length $L$ of any closed curve $\gamma\colon[a,b]\to \mathbf{R}^3$ inspecting the unit sphere $\mathbf{S}^2$ must be at least
$$
6\sqrt{3}\approx 10.3923,
$$
which is almost $83$% of the conjectured lower bound $4\pi\approx 12.5664$ by ...

20

This fails already for $d=3$.
Consider a tetrahedron, e.g. the convex hull of the points $v_1,v_2,v_3,v_4$. Let $K$ be the closed subset consisting of $\sum_{i=1}^4 a_i v_i$ with $\sum_{i=1}^4 a_i=1$ and $a_3 +a_4 \leq 1/2-\epsilon$. Let $L$ be defined similarly, but $a_1+a_2 \leq 1/2-\epsilon$. Clearly the convex hull of $K$ and $L$ is this tetrahedron.
...

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

This won't be a complete answer to Q2, but something of a starting point, at least for all two-dimensional convex $P$. Assuming for simplicity that the area of $P$ is 1, $P$ contains a parallelogram of area at least $1/2$ (triangular $P$-s are extreme in this respect). This is quite easy to prove, and for polygonal $P$, an algorithm can be produced to find ...

17

I don't understand your reference to the model (since the geometry of the hyperbolic plane does not depend on any model), but, in fact, the Beltrami-Klein model demonstrates that any qualitative statement about convex sets in the Euclidean plane holds in the Hyperbolic plane and vice versa, since the model maps convex sets to convex sets.
EDIT This has (...

answered Feb 2 '18 at 20:01

Igor Rivin

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16

Take iid Gaussian random variables $X_1,\ldots,X_d$ with mean $0$ and variance $1/d$. Normalizing the vector $X=(X_1,\ldots,X_d)$ will produce a random point on the unit sphere, but it's already close to having unit norm, so we will avoid this for the sake of intuition. For each unit vector $v$, there is an equator given by
$$\{x\in S^{d-1}:\langle x,v\...

answered Jun 27 '15 at 20:28

Dustin G. Mixon

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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

I think the identity you want is
$$2\inf_x f(x)=\inf_x(f^\ast(x)+f^\ast(-x))\mbox{.}$$
(I'm skipping a bunch of conditions required of $f$ to make this hold. We'll need convexity at least.)
Let's use $\oplus$ for infimal convolution and let $g(x)=f(-x)$.
By definition $(f\oplus g)(x)=\inf_y(f(x-y)+g(y))$.
Infimal convolution gives us $(f\oplus g)^\ast=f^...

15

[This is an attempt to explain the details in Anton Petrunin's answer to this question, since the comments suggest that a number of people have found it hard to understand as it was written.]
Let $C$ be the surface of the unit cube in $\mathbb{R}^d$ (for $d\geq 2$); when considered as a metric space, it is endowed with the distance-on-the-surface: so the ...

14

Take a path that joins the antipodes and concatenate it with its symmetric image. Get a centrally symmetric closed path on the boundary of the cube. If this path avoids one of the facets of the cube (and hence the antipodal facet as well), then we can project it to the boundary of the cube one dimension less and get a centrally symmetric path of at most the ...

14

Denote the diameter by $d$ and distance by $|x-y|$. Then there
are $y,z$ such that $d=|y-z|$ and we have by triangle inequality for every $x$:
$$d=|y-z|\leq |y-x|+|x-z|\leq 2f(x),$$
so we obtain your inequality. Notice that I did not use convexity, or any other of your assumptions, only the
triangle inequality.

answered Sep 7 '19 at 0:25

Alexandre Eremenko

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13

The baseball stitches curve suggested by Gjergji Zaimi appears in another paper of Zalgaller:
V. A. Zalgaller. Extremal problems on the convex hull of a space curve. Algebra i Analiz, 8(3):1–13, 1996.
Here Zalgaller also conjectures that this curve should be the minimizer.

13

As it was noticed by Abhinav Kumar, you only can hope for equality up to $\mathrm{GL}_n(\mathbb Z)$ transformations.
The following picture shows two $\mathrm{GL}_n(\mathbb Z)$-distinct plane figures which have the same number of integer points after any rescaling.
It still might be true that such bodies are $\mathrm{GL}_n(\mathbb Z)$-equidecomposable.
P.S....

answered Nov 20 '13 at 13:01

Anton Petrunin

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13

The Dehn functional of a polyhedron $P$ with edge lengths $\ell_i$ and exterior dihedral angles $\theta_i$ is $D(P) = \sum_i \ell_i \otimes_{\mathbb Q} (\theta_i\, \mathrm{mod}\, \mathbb{Q}\pi)$. It looks similar to the (twice) the total mean curvature $S(P) = \sum_i \ell_i \theta_i$. Both of them satisfy the "valuation property":
$$
F(P \cup Q) = F(P) + F(Q)...

12

This problem has been considered before:
The notes to chapter 4 of
Rahman/Schmeißer: Analytic Theory of Polynomials, Oxford University Press, 2002
mention this problem, state that conv(P) is a proper subset of hull(P) in general, and give two references:
1) J. L. Walsh: The location of Critical Points of Analytic and Harmonic Functions,
AMS Colloquium ...

12

This answer a different questions than was intended. The intended question remains unanswered!
Suppose pair of points $x,y\in K$ attains the diameter. Then $d(x,y)=d(-x,-y)=\operatorname{diam}(K)$. The points $x,-x,y,-y$ are coplanar. Look inside that plane. Consider the parallelogram spanned by these four points. By the parallelogram law, $$d(x,-x)^2+d(y,-...

12

I am posting this on behalf of Costin Vîlcu (with whom I've had the pleasure of coauthoring). —J.O'Rourke
The posted problem in $\mathbb{R}^3$ was answered in:
Y. G. Nikonorov and Y. V. Nikonorov,
The Intrinsic Diameter of the Surface of a Parallelepiped,
Discrete Comput. Geom. 40 (2008), 504-527.
I (Costin) know nothing about a higher ...

answered Apr 18 '14 at 17:52

Joseph O'Rourke

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12

The answer is yes for all dimensions.
An old theorem by Fenchel states that for a compact set $K$ in $\mathbb{R}^n$ every point in the convex hull can be either written as a convex combination of at most $n$ points or $K$ can be separated by a hyperplane. Since $M$ is path-connected, such a separation is not possible.
But there are also more modern ...

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