# A solution to some problems of Conway and Guy on monostable polyhedra

Zsolt Lángi  MTA-BME Morphodynamics Research Group and Department of Geometry,
Budapest University of Technology, Egry József utca 1., Budapest 1111, Hungary
<EMAIL_ADDRESS>
###### Abstract.

A convex polyhedron is called monostable if it can rest in stable position
only on one of its faces. The aim of this paper is to investigate three
questions of Conway, regarding monostable polyhedra, which first appeared i a
1969 paper of Goldberg and Guy (M. Goldberg and R.K. Guy, _Stability of
polyhedra (J.H. Conway and R.K. Guy)_ , SIAM Rev. 11 (1969), 78-82). In this
note we answer two of these problems and make a conjecture about the third
one. The main tool of our proof is a general theorem describing approximations
of smooth convex bodies by convex polyhedra in terms of their static
equilibrium points. As another application of this theorem, we prove the
existence of a convex polyhedron with only one stable and one unstable point.

###### Key words and phrases:

monostable, unistable, equilibrium point, Gömböc, polyhedral approximation

###### 1991 Mathematics Subject Classification:

52B10, 52B15, 70C20

The author is supported by the National Research, Development and Innovation
Office, NKFI, K-134199, the János Bolyai Research Scholarship of the Hungarian
Academy of Sciences, the ÚNKP-20-5 New National Excellence Program by the
Ministry of Innovation and Technology and the NRDI Fund (TKP2020 IES, Grant
No. TKP2020 BME-IKA-VIZ).

##  1\. Introduction

The study of static equilibrium points of convex bodies started with the work
of Archimedes [22], and has been continued throughout the history of science
in various disciplines: from geophysics and geology [16, 29] leading to
examination of the possible existence of water on Mars [30], to robotics and
manufacturing [31, 4] to biology and medicine [1, 11, 17]. In modern times,
the mathematical aspects of this concept was started by a problem of Conway
and Guy [5] in 1966 who conjectured that there is no homogeneous tetrahedron
which can stand in rest only on one of its faces when placed on a horizontal
plane, but there is a homogeneous convex polyhedron with the same property.
These two questions were answered by Goldberg and Guy in [20] in 1969,
respectively (for a more detailed proof of the first problem, see [7]), who
called the convex polyhedra satisfying this property _monostable_ or
_unistable_. In addition, in [20] Guy presented some problems regarding
monostable polyhedra, stating that three of them are due to Conway (for
similar statements in the literature, see e.g. [6, 8, 18]). These three
questions appear also in the problem collection of Croft, Falconer and Guy [6]
as Problem B12. The aim of our paper is to examine these problems.

######  Problem 1.

Can a monostable polyhedron in the Euclidean $3$-space $\mathbb{R}^{3}$ have
an $n$-fold axis of symmetry for $n>2$?

Before the next problem, recall that the _girth_ of a convex body in
$\mathbb{R}^{3}$ is the minimum perimeter of an orthogonal projection of the
body onto a plane [18].

######  Problem 2.

What is the smallest possible ratio of diameter to girth for a monostable
polyhedron?

######  Problem 3.

What is the set of convex bodies uniformly approximable by monostable
polyhedra, and does this contain the sphere?

It is worth noting that, according to Guy [20], Conway showed that no body of
revolution can be monostable, and also that the polyhedron constructed in [20]
has a $2$-fold rotational symmetry. Problems 1-3 appear also in the problem
collection of Klamkin [25], and Problem 1 and some other problems for
monostable polyhedra appear in a 1968 collection of geometry problems of
Shephard [28], who described these objects as ‘a remarkable class of convex
polyhedra’ whose properties ‘it would probably be very rewarding and
interesting to make a study of’.

Before stating our main result, we collect some related results and problems
from the literature. Here, we should first mention from [20] the problem of
finding the minimal dimension $d$ in which a $d$-simplex can be monostable.
This problem have been investigated by Dawson et al. [7, 10, 9, 8], who proved
that there is no monostable $d$-simplex if $d\leq 8$ and there is a monostable
$11$-simplex. With regard to Problem XVI in [28], asking about the minimum
number of faces of a monostable polyhedron in $\mathbb{R}^{3}$, the original
construction of Guy [20] with $19$ faces (attributed also to Conway) was
modified by Bezdek [3] to obtain a monostable polyhedron with $18$ faces,
while a computer-aided search by Reshetov [26] yields a monostable polyhedron
with $14$ faces. In [23], Heppes constructed a homogeneous tetrahedron in
$\mathbb{R}^{3}$ with the property that putting it on a horizontal plane with
a suitable face, it rolls twice before finding a stable position. Another
interesting convex body is found by Dumitrescu and Tóth [18], who constructed
a convex polyhedron $P$ with the property that after placing it on a
horizontal plane with a suitable face, it covers an arbitrarily large distance
while rolling until it finds a stable position. Finally, we remark that a
systematic study of the equilibrium properties of convex polyhedra was started
in [13].

Our main result is the following, where $d_{H}(\cdot,\cdot)$ and
$\mathbf{B}^{3}$ denotes the Hausdorff distance of convex bodies, and the
closed unit ball in $\mathbb{R}^{3}$ centered at the origin $o$, respectively.

######  Theorem 1.

For any $n\geq 3$, $n\in\mathbb{Z}$ and $\varepsilon>0$ there is a homogeneous
monostable polyhedron $P$ such that $P$ has an $n$-fold rotational symmetry
and $d_{H}(P,\mathbf{B}^{3})<\varepsilon$.

Theorem 1 answers Problem 1, and also the case of a sphere in Problem 3. In
addition, from it we may deduce Corollary 1 (for a lower bound without proof,
see also [18]). This solves Problem 2. Here, for any convex body
$K\subset\mathbb{R}^{3}$, we denote by $\operatorname{diam}(K)$ and $g(K)$ the
_diameter_ and the _girth_ of $K$, respectively.

######  Corollary 1.

The infimum of the quantities $\frac{\operatorname{diam}(P)}{g(P)}$ over the
family of all monostable polyhedra $P$ is $\frac{1}{\pi}$.

The proof of Theorem 1 is based on a general theorem on approximation of
smooth convex bodies by convex polytopes. Before stating it, we briefly
introduce some elementary concepts regarding their equilibrium properties. Let
$K\subset\mathbb{R}^{3}$ be a smooth convex body with the origin as its center
of mass, and let $\delta_{K}:\operatorname{bd}(K)\to\mathbb{R}$ be the
Euclidean distance function measured from $o$, where $\operatorname{bd}(K)$
denotes the boundary of $K$. The critical points of $\delta_{K}$ are called
_equilibrium points_ of $K$. To avoid degeneracy, it is usually assumed that
$\delta_{K}$ is a Morse function; i.e. it has finitely many critical points,
$\operatorname{bd}(K)$ is twice continuously differentiable at least in a
neighborhood of each critical point, and at each such point the Hessian of
$\delta_{K}$ is nondegenerate [35]. Depending on the number of negative
eigenvalues of the Hessian, we distinguish between _stable_ , _unstable_ and
_saddle-type_ equilibrium points, corresponding to the local minima, maxima
and saddle points of $\delta_{K}$, respectively. The Poincaré-Hopf Theorem
implies that under these conditions, the numbers $S$, $U$ and $H$ of the
stable, unstable and saddle points of $K$, respectively, satisfy the equation
$S-H+U=2$.

Answering a conjecture of Arnold, Domokos and Várkonyi [32] proved that there
is a homogeneous convex body with only one stable and one unstable point. They
called the body they constructed ‘Gömböc’ (for more information, see [21]). In
addition to the existence of Gömböc, in their paper [32] Domokos and Várkonyi
proved the existence of a convex body with $S$ stable and $U$ unstable
equilibrium points for any $S,U\geq 1$. This investigation was extended in
[15] to the combinatorial equivalence classes defined by the Morse-Smale
complexes of $\rho_{K}$, and in [12] for transitions between these classes.
Based on these results, for any $S,U\geq 1$ we define the set $(S,U)_{c}$ as
the family of smooth convex bodies $K$ having $S$ stable and $U$ unstable
equilibrium points, where $K$ has no degenerate equilibrium point, and at each
such point $\operatorname{bd}(K)$ has a positive Gaussian curvature. We define
the class $(S,U)_{p}$ analogously for convex polyhedra, where stable and
unstable points of a convex polyhedron are defined formally in Section 2. Our
theorem is the following, where we call a convex body _centered_ if its center
of mass is the origin $o$.

######  Theorem 2.

Let $\varepsilon>0$, $S,U\geq 1$ be arbitrary, and let $G$ be any subgroup of
the orthogonal group $\mathcal{O}(3)$. Then for any centered, $G$-invariant
convex body $K\in(S,U)_{c}$, there is a centered $G$-invariant convex
polyhedron $P\in(S,U)_{p}$ such that $d_{H}(K,P)<\varepsilon$.

Here we note that the fact that any nondegenerate convex polyhedron can be
approximated arbitrarily well by a smooth convex body with the same number of
equilibrium points is regarded as ‘folklore’ (we use a simple argument to show
it in Section 2). On the other hand, it is shown in [14] that any sufficiently
fine approximation of a smooth convex body $K$ by a convex polyhedron $P$,
using an equidistant partition of the parameter range of the boundary of $K$,
has strictly more stable, unstable and saddle points in general than the
corresponding quantities for $K$.

Even though the convex body constructed in [32] is not $C^{2}$-class at its
two equilibrium points, in [15] it is shown that class $(1,1)_{c}$ is not
empty. Thus, Theorem 2 readily implies the existence of a polyhedron in class
$(1,1)_{p}$.

######  Corollary 2.

There is a convex polyhedron with a unique stable and a unique unstable point.

Furthermore, we remark that the elegant construction in the paper [18] of
Dumitrescu and Tóth yields an _inhomogeneous_ monostable convex polyhedron
arbitrarily close to a sphere. Nevertheless, we must add that dropping the
requirement of uniform density may significantly change the equilibrium
properties of a convex body. To show it we recall the construction of Conway
(see [8]) of an inhomogeneous monostable tetrahedron in $\mathbb{R}^{3}$, and
observe that spheres with inhomogeneous density, as also roly-poly toys, yield
trivial solutions to Arnold’s conjecture.

For completeness, we also recall the remarkable result of Zamfirescu [34]
stating that a typical convex body (in Baire category sense) has infinitely
many equilibrium points, and note that critical points of the distance
function from another point are examined in Riemannian manifolds, e.g. in [2,
19, 24].

In Section 2, we introduce our notation and collect the necessary tools for
proving Theorems 1 and 2, including more precise definitions for some of the
concepts mentioned in Section 1. In Section 3 we prove Theorem 1 and Corollary
1. In Section 4 we prove Theorem 2. Finally, in Section 5 we collect some
additional remarks and ask some open questions.

##  2\. Preliminaries

In the paper, for any $p,q\in\mathbb{R}^{3}$, we denote by $[p,q]$ the closed
segment with endpoints $p,q$, and by $|p|$ the Euclidean norm of $p$. We
denote the closed $3$-dimensional unit ball centered at the origin $o$ by
$\mathbf{B}^{3}$, and its boundary by $\mathbb{S}^{2}$. Furthermore, for any
set $S\subset\mathbb{R}^{3}$ we let $\operatorname{conv}(S)$ denote the convex
hull of $S$. By a convex body we mean a compact, convex set with nonempty
interior.

Let $K\subset\mathbb{R}^{3}$ be a convex body. The _center of mass_ $c(K)$ of
$K$ is defined by the fraction $c(K)=\frac{1}{\operatorname{vol}(K)}\int_{x\in
K}x\,dv$, where $v$ denotes $3$-dimensional Lebesgue measure. We remark that
the integral in this definition is called the _first moment_ of $K$, and note
that we clearly have $c(K)\in\operatorname{int}(K)$ for any convex body $K$.
If $q\in\operatorname{bd}(K)$ satisfies the property that the plane through
$q$ and orthogonal to the vector $q-c(K)$ supports $K$, then we say that $q$
is an _equilibrium point_ of $K$. Here, if $K$ is smooth, then the equilibrium
points of $K$ coincide with the critical points of the Euclidean distance
function measured from $c(K)$ and restricted to $\operatorname{bd}(K)$. We
remark that a convex body $K\subset\mathbb{R}^{3}$ is called _smooth_ if for
any boundary point $x$ of $K$ there is a unique supporting plane of $K$ at
$q$; this property coincides with the property that $\operatorname{bd}(K)$ is
a $C^{1}$-class submanifold of $\mathbb{R}^{3}$ (cf. [27]).

We define nondegenerate equilibrium points only in two special cases. If $K$
is smooth, $q\in\operatorname{bd}(K)$ is an equilibrium point of $K$ with a
$C^{2}$-class neighborhood in $\operatorname{bd}(K)$, and the Hessian of the
Euclidean distance function on $\operatorname{bd}(K)$, measured from $c(K)$,
is nondegenerate, we say that $q$ is _nondegenerate_. In this case $q$ is
called a _stable, saddle-type_ or _unstable_ point of $K$ if the number of the
negative eigenvalues of the Hessian at $q$ is $0,1$ or $2$, respectively [15].
Consider now the case that $K$ is a convex polyhedron in $\mathbb{R}^{3}$, and
$q\in\operatorname{bd}(K)$ is an equilibrium point of $K$. Then there is a
unique vertex, edge or face of $K$ that contains $q$ in its relative interior,
where by the relative interior of a vertex we mean the vertex itself. Let $F$
denote this vertex, edge or face, and let $H$ be the supporting plane of $K$
through $q$ that is perpendicular to $q-c(K)$. Observe that $F\subset K\cap
H$. We say that $q$ is _nondegenerate_ if $F=K\cap H$. In this case we call
$q$ a _stable, saddle-type_ or _unstable_ point of $K$ if the dimension of $F$
is $2,1$ or $0$, respectively [13]. In both the smooth and the polyhedral
cases $K$ is called nondegenerate if it has only finitely many equilibrium
points, and each such point is nondegenerate; note that the first condition is
automatically satisfied for convex polyhedra. We remark that in the above
definitions, we may replace the center of mass of $K$ by any fixed reference
point $c\in K$. In this case we write about equilibrium points _relative to
$c$_. We emphasize that in the paper, unless it is stated otherwise, if the
reference point is not specified, then it is meant to be the center of mass of
the body.

Let $K\in\mathbb{R}^{3}$ be a nondegenerate smooth convex body with $S$
stable, $H$ saddle-type and $U$ unstable equilibrium points. Using a standard
convolution technique, we may assume that $K$ has a $C^{\infty}$-class
boundary, and hence, by the Poincaré-Hopf Theorem, we have $S-H+U=2$ [15]. We
show that the same holds if $K$ is a nondegenerate convex polyhedron. Indeed,
let $\tau>0$ be sufficiently small, and set
$K(\tau)=(K\div(\tau\mathbf{B}^{3}))+(\tau\mathbf{B}^{3})$, where $\div$
denotes Minkowski difference and $+$ denotes Minkowski addition [27]. Then for
any $\tau>0$, $K(\tau)$ is a smooth nondegenerate convex body having the same
numbers of stable, saddle-type and unstable points relative to $c(K)$; hence,
we may apply the Poincaré-Hopf Theorem for $K(\tau)$ (here we note that by
Lemma 2 the same property holds relative to $c(K(\tau))$ as well). Thus, for
any nondegenerate convex body, the numbers of stable and unstable points
determine the number of saddle-type points. We define class $(S,U)_{c}$ as the
family of nondegenerate, smooth convex bodies $K\subseteq\mathbb{R}^{3}$ with
$S$ stable, $U$ unstable points with the additional assumption that at each
equilibrium point of $K$, the principal curvatures of $\operatorname{bd}(K)$
are positive. Similarly, by $(S,U)_{p}$ we mean the family of nondegenerate
convex polyhedra with $S$ stable and $U$ unstable points. Observe that if $K$
is nondegenerate, the point of $\operatorname{bd}(K)$ closest to or farthest
from $c(K)$ is necessarily a stable or unstable point, respectively, implying
that the numbers $S,U$ in the above symbol are necessarily positive.

For the following remark, see Lemma 7 from [15].

######  Remark 1.

Let $K\in(S,U)_{c}$ and for any equilibrium point $q$ of $K$, let $V_{q}$ be
an arbitrary compact neighborhood of $q$ containing no other equilibrium point
of $K$. Then $c(K)$ has an open neighborhood $U$ such that for any $x\in U$,
$K$ has $S$ stable and $U$ unstable points relative to $x$, and for any
equilibrium point $q$ of $K$ relative to $c(K)$, $V_{q}$ contains exactly one
equilibrium point of $K$ relative to $x$, and the type of this point is the
same as the type of $q$.

For Remark 2, see the paragraph in [14] after Definition 2.

######  Remark 2.

Let $q$ be an equilibrium point of a centered convex body $K$ in $(S,U)_{c}$
for some $S,U\geq 1$. Let $|q|=\rho$, and let $\kappa_{1},\kappa_{2}$ denote
the principal curvatures of $\operatorname{bd}(K)$ at $q$. Then
$\kappa_{1},\kappa_{2}\neq\frac{1}{\rho}$. Furthermore,
$0\leq\kappa_{1},\kappa_{2}<\frac{1}{\rho}$ if and only if $q$ is a stable
point, $\kappa_{1},\kappa_{2}>\frac{1}{\rho}$ if and only if $q$ is an
unstable point, and
$0<\min\\{\kappa_{1},\kappa_{2}\\}<\frac{1}{\rho}<\max\\{\kappa_{1},\kappa_{2}\\}$
if and only if $q$ is a saddle-type equilibrium point.

######  Lemma 1.

The symmetry group of any nondegenerate convex body $K$ is finite.

###### Proof.

Let $K$ be a nondegenerate convex body with symmetry group $G$. Without loss
of generality, assume that $K$ is centered, i.e. $c(K)=o$. Since $c(K)$ is
clearly a fixed point of any symmetry in $G$, we have that $G$ is a subgroup
of the orthogonal group $\mathcal{O}(3)$. Clearly, $G$ is closed in
$\mathcal{O}(3)$, and thus, it is a Lie group embedded in $\mathcal{O}(3)$ by
Cartan’s Closed Subgroup Theorem. On the other hand, the Lie subgroups of
$\mathcal{O}(3)$ are well known, and in particular we have that if $G$ is
infinite, then it contains, up to conjugacy, $\mathcal{SO}(2)$ as a subgroup.
In other words, $K$ is rotationally symmetric. Thus, by nondegeneracy, $K$ has
exactly one stable and one unstable equilibrium point. But this property
contradicts Conway’s result mentioned in Section 1 that no rotationally
symmetric convex body is monostable. ∎

We finish Section 2 with two lemmas and two remarks, where $X\triangle Y$
denotes the symmetric difference of the sets $X,Y$.

######  Lemma 2.

Let $K(\tau)\subset\mathbb{R}^{3}$ be a $1$-parameter family of convex bodies,
where $\tau\in[0,\tau_{0}]$ for some $\tau_{0}>0$. For any
$\tau\in[0,\tau_{0}]$, let $c(\tau)$ denote the center of mass of $K(\tau)$,
and let $K=K(0)$ and $c=c(0)$. Assume that for some $C>0$ and $m>0$,
$\operatorname{vol}(K(\tau)\triangle K)\leq C\tau^{m}$ holds for any
sufficiently small value of $\tau$. Then there is some $C^{\prime}>0$ such
that $|c(\tau)-c|\leq C^{\prime}\tau^{m}$ holds for any sufficiently small
value of $\tau$.

###### Proof.

Without loss of generality, we may assume that $K(\tau)\subseteq
r\mathbf{B}^{3}$ for some suitable value of $r>0$ if $\tau$ is sufficiently
small. By definition, $c(\tau)=\frac{\int_{x\in
K(\tau)}x\,dv}{\operatorname{vol}(K(\tau))}$. On the other hand, by the
conditions, we have $|\operatorname{vol}(K(\tau))-\operatorname{vol}(K)|\leq
C\tau^{m}$, and $|\int_{x\in K(\tau)}x\,d\lambda-\int_{x\in K}x\,dv|\leq
rC\tau^{m}$ for all sufficiently small values of $\tau$. From these
inequalities and the fact that $\operatorname{vol}(K)>0$, the assertion
readily follows. ∎

######  Lemma 3.

Let $p\in\operatorname{int}(\mathbf{B}^{2})\subset\mathbb{R}^{2}$ and
$q\in\mathbb{S}^{1}$ such that $p$, $q$ and $o$ are not collinear, and let $L$
be a line through $p$ such that $L$ does not separate $o$ and $q$.
Furthermore, if $A$ denotes the convex angular region with $q\in A$ and
bounded by a half line of $L$ starting at $p$, and the half line starting at
$p$ and containing $o$, then assume that the angle of $A$ is obtuse. Then
there is a convex polygon $Q\subset\mathbf{B}^{2}$ with vertices
$o,x_{0}=q,x_{1},\ldots,x_{k}=p$ in cyclic order in $\operatorname{bd}(Q)$
such that $x_{i-1}x_{i}o\angle>\frac{\pi}{2}$ for all values of $i$, and $L$
supports $Q$.

Figure 1. The construction of the points $x_{i}^{\prime}$ in the proof of
Lemma 3. The dotted curves indicate arcs in the Thales circles of the segments
$[o,x_{i}^{\prime}]$.

We remark that the conditions in Lemma 3 imply that the Euclidean distance
function $x\mapsto|x|$, $x\in\mathbb{R}^{2}$ strictly decreases along the
curve $\bigcup_{i=1}^{k}[x_{i-1},x_{i}]$ from $q$ to $p$.

###### Proof.

Without loss of generality, we may assume that $q=(1,0)$ and the
$y$-coordinate of $p$ is positive. Set $poq\angle=\beta\in(0,\pi)$, and choose
an arbitrary positive integer $k$. For any $i=0,1,\ldots,k$, define the point
$x_{i}^{\prime}=\left(r_{i}\cos\frac{i\beta}{k},r_{i}\sin\frac{i\beta}{k}\right)$,
where $r_{i}=\cos^{i}\frac{\beta}{k}$. Then $x^{\prime}_{0}=q$, and
$x^{\prime}_{i}$ is on the Thales circle of the segment
$[0,x^{\prime}_{i-1}]$, and thus,
$x_{i-1}^{\prime}x_{i}^{\prime}o\angle=\frac{\pi}{2}$ (cf. Figure 1) for all
$i=1,2,\ldots,k$. Using elementary calculus, we obtain that
$\lim_{k\to\infty}\cos^{k}\frac{\beta}{k}=1$, which yields that there is some
value of $k$ such that $|x^{\prime}_{k}|>|p|$. Since $x^{\prime}_{k}$ and $p$
are on the same half line, we may decrease the values of $r_{i}$ for
$i=1,2,\ldots,k$ slightly such that for the points $x_{i}$ obtained in this
way the convex polygon $Q=\operatorname{conv}\\{o,x_{0},x_{1},\ldots,x_{k}\\}$
satisfies the required conditions apart from the one for $L$. Now, if $L$
supports $Q$, we are done. On the other hand, if $L$ does not support $Q$,
then we may take the polygon obtained as the intersection of $Q$ and the
closed half plane bounded by $L$ and containing $o$ in its interior. ∎

######  Remark 3.

Let $a,b>0$, where $a\neq b$, and let $E\subset\mathbb{R}^{2}$ be the ellipse
with equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\leq 1$. Then, for any
$\delta>0$ there is some $\varepsilon>0$ such that if $K\subset\mathbb{R}^{2}$
is a plane convex body satisfying $E\subseteq K\subseteq(1+\delta)E$, and the
vector $w$ is perpendicular to a supporting line of $K$ through
$w\in\operatorname{bd}(K)$, then the angle between $w$ and the $x$-axis or the
$y$-axis is at most $\delta$.

######  Remark 4.

Let $f,g$ be two real functions defined in a neighborhood of $a\in\mathbb{R}$.
If $f,g$ are both locally strictly increasing (resp., decreasing) at $a$, then
so are $\min\\{f,g\\}$ and $\max\\{f,g\\}$.

Finally, we remark that in the proof of Theorem 2, we use ideas also from [15,
12, 18].

##  3\. Proofs of Theorem 1 and Corollary 1

First, we show how Theorem 2 implies Theorem 1.

Let $n\geq 3$ be a positive integer and let $\varepsilon>0$ be a sufficiently
small fixed value. By Theorem 2, it is sufficient to construct a smooth convex
body $K\in(1,m)_{c}$ for some value of $m$ with $n$-fold rotational symmetry
and satisfying $d_{H}(K,\mathbf{B}^{3})\leq\varepsilon$. Let $P$ be a regular
$n$-gon inscribed in a fixed circle $C$ on $\mathbf{B}^{3}$ parallel to, but
not contained in the $(x,y)$-plane. Let the vertices of $P$ be $p_{i}$,
$i=1,2,\ldots,n$. Let
$Q(\varepsilon)=\operatorname{conv}\left(\mathbf{B}^{3}\cup\\{(1+\varepsilon)p_{1},\ldots,(1+\varepsilon)p_{n}\\}\right)$
(cf. Figure 2).

Figure 2. The construction of $Q(\varepsilon)$ with $\varepsilon=0.15$ and
$n=5$, where $q_{i}=(1+\varepsilon)p_{i}$.

Then $Q(\varepsilon)$ is the union of $\mathbf{B}^{3}$ and $n$ cones $C_{i}$,
$i=1,2,\ldots,n$, with spherical circles centered at the points $p_{i}$ as
directrixes. By symmetry, the center of mass $c$ of $Q(\varepsilon)$ is on the
$z$-axis, and by the Thales Theorem and Lemma 2, its distance from $o$ is of
magnitude $O(\varepsilon^{2})$. Thus, the points $(1+\varepsilon)p_{i}$ are
equilibrium points of $Q(\varepsilon)$ if $\varepsilon$ is sufficiently small.
Furthermore, we have $c\neq o$. On one hand, from this we have that there are
exactly two equilibrium points of $Q(\varepsilon)$ on $\mathbb{S}^{2}$, namely
the points $(0,0,1)$ and $(0,0,-1)$, and exactly one of these points is
stable, and the other one is unstable. On the other hand, this also implies
that $Q(\varepsilon)$ has exactly one equilibrium point on each cone $C_{i}$
apart from its vertex; this point is a saddle point in the relative interior
of a generating segment of $C_{i}$ (cf. Figure 3).

Figure 3. An illustration for the proof of Theorem 1: equilibrium points on a
conic part of $\operatorname{bd}(Q(\varepsilon)$.

Now, we set
$Q^{\prime}(\varepsilon)=(Q(\varepsilon)\div(\tau\mathbf{B}^{3}))+(\tau\mathbf{B}^{3})$,
where $\tau>0$ is negligible compared to $\varepsilon$. Then
$Q^{\prime}(\varepsilon)$ is a smooth convex body which has $1$ stable, $n$
saddle-type and $(n+1)$ unstable points by Lemma 2. To guarantee that the body
has positive principal curvatures at each equilibrium point, we may replace
the generating segments of the cones by circular arcs of radius $R>0$, where
$\frac{1}{R}$ is negligible compared to $\tau$. The obtained convex body
$K(\varepsilon)\in(1,n+1)_{c}$ satisfies the required conditions.

Finally, we prove Corollary 1. Clearly,
$\operatorname{diam}(\mathbf{B}^{3})=2$ and $g(\mathbf{B}^{3})=2\pi$, and
hence, the first statement follows from the continuity of diameter and girth
with respect to Hausdorff distance. On the other hand, it is well known that
for any convex body $K$ in $\mathbb{R}^{3}$,
$\frac{\operatorname{diam}(K)}{g(K)}\geq\frac{1}{\pi}$; here we include the
proof only for completeness. Let $K\subset\mathbb{R}^{3}$ be a convex body,
and let $w(\cdot)$ and $\operatorname{perim}(\cdot)$ denote mean width and
perimeter, respectively. Then, for any projection $M$ of $K$, we have
$w(M)\leq\operatorname{diam}(M)\leq\operatorname{diam}(K)$. On the other hand,
it is well known that $w(M)=\frac{\operatorname{perim}(M)}{\pi}$, implying
that
$\frac{\operatorname{diam}(K)}{\operatorname{perim}(M)}\geq\frac{1}{\pi}$.
From this, we readily obtain
$\frac{\operatorname{diam}(K)}{g(K)}\geq\frac{1}{\pi}$.

##  4\. Proof of Theorem 2

First, observe that by Lemma 1, $G$ is finite.

We construct $P$ by truncating $K$ with finitely many suitably chosen planes;
or more precisely by taking its intersection with finitely many suitably
chosen closed half spaces. We carry out the construction of $P$ in three
steps.

In Step 1, we replace some small regions of $\operatorname{bd}(K)$ by
polyhedral regions disjoint from all equilibrium points of $K$. These
polyhedral regions will serve as ‘controlling regions’; that is, after
constructing a polyhedron with $S$ stable and $U$ unstable points relative to
$o$, we modify these regions to move back the center of mass of the
constructed polyhedron to $o$. In Step 2, we truncate a neighborhood of each
equilibrium point to replace it by a polyhedral surface in such a way that
each polyhedral surface contains exactly one equilibrium point relative to
$o$, and the type of this point is the same as the type of the corresponding
equilibrium point of $K$. Finally, in Step 3 we truncate the remaining part of
$\operatorname{bd}(K)$ such that no new equilibrium point is created. We
describe these steps in three separate subsections.

In the proof, we denote by $\mathcal{E}$ the set of the equilibrium points of
$K$, and for any point $q\in\operatorname{bd}(K)$, we denote by $H_{q}$ the
unique supporting plane of $K$ at $q$. Observe that by the definition of
$(S,U)_{c}$, $H_{q}\cap K=\\{q\\}$ for any $q\in\mathcal{E}$, and set
$X=\operatorname{bd}(K)\setminus\mathcal{E}$. Finally, by $F$ we denote the
set of the fixed points of $G$, and note that $F$ is a linear subspace of
$\mathbb{R}^{3}$ that contains the center of mass of any $G$-invariant convex
body.

###  4.1. Step 1: Truncating some small regions disjoint from all equilibrium
points

We distinguish two cases depending on $\dim(F)$.

Case 1, if $F=\mathbb{R}^{3}$.
By Carathéodory’s theorem, there are points $z_{1},z_{2},z_{3},z_{4}\in X$
such that $o\in\operatorname{conv}\\{z_{1},z_{2},z_{3},z_{4}\\}$. Since $X$ is
open in $\operatorname{bd}K$, we may choose these points to satisfy
$o\in\operatorname{int}\operatorname{conv}\\{z_{1},z_{2},z_{3},z_{4}\\}$. By
the definition of $(S,U)_{c}$, we have that $H_{z_{i}}$ is disjoint from
$H_{q}$ for any $1\leq i\leq 4$ and $q\in\mathcal{E}$. We show that the
$z_{i}$s can be chosen such that the planes $H_{z_{i}}$ are pairwise distinct.
Suppose for contradiction that, say, three of these planes coincide. Without
loss of generality, assume that $H_{z_{1}}=H_{z_{2}}=H_{z_{3}}$, and denote
this common plane by $H$. Then there are points
$z_{1}^{\prime},z_{2}^{\prime},z_{3}^{\prime}\in\mathrm{relbd}(K\cap H)$ such
that
$\operatorname{conv}\\{z_{1},z_{2},z_{3}\\}\subseteq\operatorname{conv}\\{z_{1}^{\prime},z_{2}^{\prime},z_{3}^{\prime}\\}$.
Now we may replace $z_{2}^{\prime}$ and $z_{3}^{\prime}$ by two points
$z_{2}^{\prime\prime},z_{3}^{\prime\prime}\notin H$ such that
$z_{i}^{\prime\prime}$ is sufficiently close to $z_{i}^{\prime}$ for $i=2,3$.
Then we have
$o\in\operatorname{int}\operatorname{conv}\\{z_{1}^{\prime},z_{2}^{\prime\prime},z_{3}^{\prime\prime},z_{4}\\}$,
where no supporting plane of $K$ contains three of the points. If a supporting
plane of $K$ contains two of these points, we may repeat the above procedure,
and finally obtain some points $w_{1},\ldots,w_{4}\in X$ such that
$o\in\operatorname{int}\operatorname{conv}\\{w_{1},\ldots,w_{4}\\}$, and the
sets $H_{w_{i}}\cap K$ are pairwise disjoint.

Let $\delta>0$, and let us truncate $K$ by planes $H_{1},\ldots,H_{4}$ such
that for all $i$s $H_{i}$ is parallel to $H_{z_{i}}$ and it is at the distance
$\delta$ from it in the direction of $o$. We denote the truncated convex body
by $K^{\prime}$ and its center of mass by $c^{\prime}$. By Remark 1, if
$\delta$ is sufficiently small, then $K$ has $S$ stable and $U$ unstable
equilibrium points relative to $c^{\prime}$, and each such equilibrium point
is contained in
$\operatorname{bd}(K^{\prime})\setminus(\bigcup_{i=1}^{4}H_{i})$. Furthermore,
if $\delta$ is sufficiently small, then for any point $q\in
H_{i}\cap\operatorname{bd}(K)$ and any plane $H$ supporting $K^{\prime}$ at
$q$, $q$ is not perpendicular to $H$. Finally, since
$o\in\operatorname{int}\operatorname{conv}\\{w_{1},\ldots,w_{4}\\}$, we may
choose points $w_{i}^{\prime}\in\operatorname{relint}(H_{i}\cap K^{\prime})$
such that
$c^{\prime}\operatorname{int}\operatorname{conv}\\{w_{1}^{\prime},\ldots,w_{4}^{\prime}\\}$.
For any $w_{i}^{\prime}$, choose some convex $n_{i}$-gon
$P_{i}\subset\operatorname{relint}(H_{i}\cap K^{\prime})$ such that the center
of mass of $P_{i}$ is $w_{i}^{\prime}$ (cf. Figure 4). Now we obtain the body
$K^{\prime\prime}$ by truncating $K^{\prime}$ by $n_{i}$ planes almost
parallel to $H_{i}$ such that for each $i$, every side of $P_{i}$ is contained
in one of the truncating planes, and we have $P_{i}=H_{i}\cap
K^{\prime\prime}$. We choose the truncating planes such that the center of
mass $c^{\prime\prime}$ of $K^{\prime\prime}$ satisfies
$c^{\prime\prime}\in\operatorname{int}\operatorname{conv}\\{w^{\prime}_{1},\ldots,w^{\prime}_{4}\\}$,
and $K^{\prime\prime}$ has $S$ stable and $U$ unstable points on the smooth
part of its boundary, and no equilibrium point on the non-smooth part. Now, we
set $K_{1}=K^{\prime\prime}-c^{\prime\prime}$,
$w_{i}^{\prime\prime}=w_{i}^{\prime}-c^{\prime\prime}$ and
$P_{i}^{\prime}=P_{i}-c^{\prime\prime}$ for all $i$s, and for some
sufficiently small $\bar{\tau}>0$ we define four $1$-parameter families of
polyhedral cones
$C_{i}(\tau_{i})=\operatorname{conv}(P_{i}^{\prime}\cup\\{(1+\tau_{i})w_{i}^{\prime\prime}\\})$,
$\tau_{i}\in[0,\bar{\tau}]$, $i=1,2,3,4$. Furthermore, for later use, we set
$K_{0}=K-c^{\prime\prime}$, and call the set
$X_{1}=K_{1}\cap\operatorname{bd}(K_{0})$ the _non-truncated part of
$\operatorname{bd}(K_{1})$_.

Figure 4. An illustration for Step 1. Panel (a): The points $w_{i}$ and the
planes supporting $K$ at these points. Panel (b): the polygon $P_{1}$
constructed on the intersection $H_{1}\cap K^{\prime}$.

If $\bar{\tau}$ is sufficiently small, then
$K_{1}\cup\bigcup_{i=1}^{4}C_{i}(\tau_{i})$ is convex for all values of the
parameters $\tau_{i}$. Furthermore, note that the center of mass of
$C_{i}(\tau_{i})$ lies on the line through $[o,w^{\prime\prime}_{i}]$ for any
value of $i$. This yields that the first moment of
$\bigcup_{i=1}^{4}C_{i}(\tau_{i})$ is
$\sum_{i=1}^{4}(\alpha_{i}\tau_{i}+\beta_{i}\tau_{i}^{2})w_{i}^{\prime\prime}$
for some suitable constants $\alpha_{i},\beta_{i}>0$, which implies that it is
surjective in a neighborhood of $o$. Thus, since $K_{1}$ is centered, after we
replace the non-truncated part of $\operatorname{bd}(K_{1})$ by a polyhedral
surface in Steps 2 and 3, we may choose values of the $\tau_{i}$s in such a
way that the sum of the first moment of $\bigcup_{i=1}^{4}C_{i}(\tau_{i})$ and
of the first moment of the polyhedron $P$ obtained after Step 3 is equal $o$.
This makes the polyhedron $P\cup\bigcup_{i=1}^{4}C_{i}(\tau_{i})$ centered.
Finally, we observe that by choosing sufficiently small values of $\delta$ and
$\bar{\tau}$, for all values of the parameters, no point of $C_{i}(\tau_{i})$
is an equilibrium point of $K_{1}$ relative to $o$.

Case 2, if $F\neq\mathbb{R}^{3}$.
In this case $F$ is a plane or a line through $o$, or $F=\\{o\\}$. Consider
the case that $F$ is a plane. Then, by the properties of isometries, the orbit
of any point $p$ under $G$ consists of $p$ and its reflection about $F$. Let
$K_{F}=F\cap K$, and observe that since $K$ is symmetric about $F$, for any
$q\in\operatorname{bd}(K)$ $H_{q}$ is either disjoint from $K_{F}$ or
$q\in\mathrm{relbd}(K_{F})$. Thus, we may apply the argument in Case 1 for
$K_{F}$, and obtain some points $z_{1},z_{2},z_{3}\in X\cap K_{F}$ such that
$o\in\operatorname{relint}\operatorname{conv}\\{z_{1},z_{2},z_{3}\\}$ and the
planes $H_{z_{i}}$ are pairwise disjoint. But then there are some points
$z_{3}^{\prime}$ and $z_{3}^{\prime\prime}$, sufficiently close to $H_{z_{3}}$
such that $z_{3}^{\prime\prime}$ is the reflected copy of $z_{3}^{\prime}$
about $F$,
$o\in\operatorname{int}\operatorname{conv}\\{z_{1},z_{2},z_{3}^{\prime},z_{3}^{\prime\prime}\\}$,
and the supporting planes at these points are pairwise disjoint. Clearly, the
set $\\{z_{1},z_{2},z_{3}^{\prime},z_{3}^{\prime\prime}\\}$ is $G$-invariant.
From now on, we may apply the argument in Case 1.

If $F$ is a line, we may apply a similar argument. Indeed, let $F\cap
K=[p,p^{\prime}]$. Then, by symmetry, $p$ and $p^{\prime}$ are equilibrium
points with respect to $o$, and thus, the planes $H$ and $H^{\prime}$,
containing $p$ and $p^{\prime}$, and perpendicular to $[o,p]$ and
$[o,p^{\prime}]$, respectively, support $K$. Thus, we may choose a point
$w_{1}$, with $\\{w_{1},w_{2},\ldots,s_{k}\\}$ as its orbit, such that $w_{1}$
is sufficiently close to $H$, and the orbit of $H_{1}\cap K$, where $H_{1}$ is
the supporting plane of $K$, contains mutually disjoint sets. Choosing points
$w_{1}^{\prime},w_{2}^{\prime},\ldots,w_{k}^{\prime}$ similarly and
sufficiently close to $H^{\prime}$, we have
$o\in\operatorname{int}\operatorname{conv}\\{w_{1},\ldots,w_{k},w_{1}^{\prime},\ldots,w_{k}^{\prime}\\}$,
and we may proceed as in Case 1.

Finally, if $F=\\{o\\}$, then any $G$-invariant convex body (and in particular
the convex polyhedron constructed in Steps 2 and 3) is centered. Thus, in this
case we may skip Step 1.

Based on the existence of the families $C_{i}(\tau_{i})$, in Steps 2 and 3 all
equilibrium points are meant to be _relative to $o$_. We denote by
$\mathcal{E}_{1}$ the set of the equilibrium points of $K_{1}$.

###  4.2. Step 2: Truncating small neighborhoods of equilibrium points

In this step we take all points $q\in\mathcal{E}_{1}$, and truncate
neighborhoods of them in $\operatorname{bd}(K_{1})$ simultaneously for all
points in the orbit of $q$. Here we observe that the orbit of an equilibrium
point consists of equilibrium points. We carry out the truncations in such a
way that the regions truncated in Step 1 or Step 2 are pairwise disjoint. We
denote the convex body obtained in this step by $K_{2}$, and set
$X_{2}=\operatorname{bd}(K_{1})\cap K_{2}$. We construct $K_{2}$ in such a way
that for any point $p\in X_{2}$ there is no supporting plane $H$ of $K_{2}$
through $p$ which contains an equilibrium point of $K_{2}$.

Consider some $q\in\mathcal{E}_{1}$. Without loss of generality, we may assume
that $q=(0,0,\rho)$ for some $\rho>0$, and denote by $e_{x},e_{y},$ and
$e_{z}$ the vectors of the standard orthonormal basis. With a little abuse of
notation, for any $p\in\operatorname{bd}(K_{0})$, we denote by $H_{p}$ the
unique supporting plane of $K_{0}$ at $p$.

Case 1, the stabilizer of $q$ in $G$ is the identity; i.e. $q$ not fixed under
any element of $G$ other than the identity.

_Subcase 1.1_ , $q$ is a stable point of $K_{1}$.
In this case we truncate $K_{1}$ by a plane $H_{q}^{\prime}$ parallel to, and
sufficiently close to $H_{q}$. Then we truncate $K_{1}$ by finitely many
additional planes such that any point of
$H_{q}^{\prime}\cap\operatorname{bd}(K_{1})$ is truncated by at least one of
them, and for any point $p$ of the non-truncated part $X_{2}$ of
$\operatorname{bd}(K_{1})$ there is no supporting plane $H$ of $K_{2}$ through
$p$ which contains an equilibrium point of $K_{2}$ relative to $o$.

_Subcase 1.2_ , $q$ is a saddle-type equilibrium point.
Note that by Remark 2, $q$ is not an umbilic point of
$\operatorname{bd}(K_{1})$, and its principal curvatures
$\kappa_{1}<\kappa_{2}$ satisfy the inequalities
$0<\kappa_{1}<\frac{1}{\rho}<\kappa_{2}$.

Without loss of generality, we may assume that the sectional curvature of
$\operatorname{bd}(K_{1})$ in the $(x,z)$-plane is $\kappa_{1}$, and in the
$(y,z)$-plane it is $\kappa_{2}$. For any $\tau>0$, let $K_{1}(\tau)$ denote
the set of points of $K_{1}$ with $z$-coordinates at least $\rho-\tau$, and
observe that by the fact that $\kappa_{2}>\kappa_{1}>0$, for any
$\varepsilon>0$ there is some $\tau>0$ such that $K_{1}(\tau)$ is contained in
the neighborhood of $q$ of radius $\varepsilon$. For any
$\\{i,j\\}\subset\\{x,y,z\\}$, let $H_{ij}$ denote the $(i,j)$ coordinate
plane, and $\operatorname{proj}_{ij}$ denote the orthogonal projection of
$\mathbb{R}^{3}$ onto $H_{ij}$.

For any $\eta>0$, let $C(\eta)$ be the set of the points of the circular disk
$y^{2}+(z-\rho+\eta)^{2}\leq{\eta}^{2}$ in $H_{yz}$ whose $z$-coordinates are
at least $\rho-\tau$. Then, since $\operatorname{bd}(K_{1})$ is $C^{2}$-class
in a neighborhood of $q$, we have that for any $\eta_{1},\eta_{2}>0$
satisfying $\frac{1}{\rho}<\frac{1}{\eta_{1}}<\kappa_{2}<\frac{1}{\eta_{2}}$,
if $\tau$ is sufficiently small, then
$C(\eta_{2})\subseteq\operatorname{proj}_{yz}(K_{1}(\tau))\subseteq
C(\eta_{1})$ holds. Since $\operatorname{proj}_{yz}(K_{1})$ is convex,
$\mathrm{relbd}(\operatorname{proj}_{yz}(K_{1}))$ has exactly two points with
their $z$-coordinates equal to $\rho-\tau$. Let these points be
$q^{-}=(0,\sigma^{-},\rho-\tau)$ and $q^{+}=(0,\sigma^{+},\rho-\tau)$ such
that $\sigma^{-}<0<\sigma^{+}$ (cf. Figure 5). Then there are some supporting
lines $L_{-},L_{+}$ of $\operatorname{proj}_{yz}(K_{1})$ passing through
$q^{-}$ and $q^{+}$, respectively.

Figure 5. An illustration for Step 2 in the proof of Theorem 2.

Clearly, for $i\in\\{-,+\\}$, $L^{i}$ is the orthogonal projection of some
supporting plane $H^{i}$ of $K_{1}$ onto $H_{yz}$. Let $r^{i}$ be a point of
$L^{i}$, on the open half line starting at $q^{i}$ such that the line through
$[o,q^{i}]$ do not separate $q$ and $r_{i}$. Then, for any fixed values of
$\eta_{1}$ and $\eta_{2}$ and sufficiently small value of $\tau$, the angles
$oq^{i}r^{i}\angle$ are obtuse. Now we choose some sufficiently small value of
$\zeta>0$, and define ${q^{i}}^{\prime}=\zeta e_{z}+q^{i}$,
${r^{i}}^{\prime}=\zeta e_{z}+r^{i}$, ${L^{i}}^{\prime}=\zeta e_{z}+L^{i}$,
${H^{i}}^{\prime}=\zeta e_{z}+H^{i}$ and $q^{\prime}=-\zeta e_{z}+q$. Then we
may assume that $|{q^{i}}^{\prime}|<|q^{\prime}|$, the angles
$o{q^{i}}^{\prime}{r^{i}}^{\prime}\angle$ are obtuse, and the planes
${H^{i}}^{\prime}$ are disjoint from $K_{1}$.

Thus, by Lemma 3, for $i\in\\{-,+\\}$, there is a polygonal curve $\Gamma_{i}$
in $H_{yz}$, connecting $q^{\prime}$ to ${q^{i}}^{\prime}$ such that the
Euclidean distance measured from the points of $\Gamma_{i}$ to $o$ is strictly
decreasing as we move from $q^{\prime}$ to ${q^{i}}^{\prime}$ (see the remark
after Lemma 3), $\Gamma_{i}$ is contained in
$\mathrm{relbd}(\operatorname{conv}(\Gamma_{i}\cup\\{o\\}))$, and the latter
set is supported by ${L_{i}}^{\prime}$ in $H_{yz}$. Consider the closed,
convex set $C_{H}\subset H_{yz}$ bounded by $\Gamma_{-}\cup\Gamma_{+}$, the
half line of ${L^{+}}^{\prime}$ starting at ${q^{+}}^{\prime}$ and not
containing ${r^{+}}^{\prime}$, and the half line of ${L^{-}}^{\prime}$
starting at ${q^{-}}^{\prime}$ and not containing ${r^{-}}^{\prime}$, and set
$C=\operatorname{proj}_{yz}^{-1}(C_{H})\subset\mathbb{R}^{3}$. By the previous
consideration, $C$ is an infinite convex cylinder with the properties that
$o\in\operatorname{int}(C)$, $K_{1}\setminus C\subseteq K_{1}(\tau)$, and the
equilibrium points of $C$ relative to $o$ are $q$ and two stable points on
${L_{+}}^{\prime}$ and ${L_{-}}^{\prime}$, respectively. To construct $K_{2}$,
we truncate $K_{1}$ by $C$, and show that, apart from the saddle point
$q^{\prime}$, no new equilibrium point is created by this truncation.

Observe that by our construction, any new equilibrium point is a point of
$\operatorname{bd}(K_{1})\cap\operatorname{bd}(C)$. Suppose that there is some
equilibrium point $q\in\operatorname{bd}(K_{1})\cap\operatorname{bd}(C)$ of
$K_{1}\cap C$. To reach a contradiction, we identify $H_{xy}$ with
$\mathbb{R}^{2}$, and parametrize $\operatorname{bd}(K_{1})$ in a neighborhood
of $q$ as the graph of a function $f:\mathbb{R}^{2}\to\mathbb{R}$ and
$\operatorname{bd}(C)$ in a neighborhood of $q^{\prime}$ as the graph of a
function $g:\mathbb{R}^{2}\to\mathbb{R}$. Note that by the nondegeneracy of
$q$, for some value of $\phi>0$, $o$ has a neighborhood
$U\subset\mathbb{R}^{2}$ such that for any $w=(x_{0},y_{0})\in U$ whose angle
with the $x$-axis is at most $\phi$, $|(x,y,f(x,y)|$ is locally strictly
increasing at $w$ as a function of $x$ if $x_{0}>0$ and locally strictly
decreasing if $x_{0}<0$; furthermore, if the angle of $w$ with the $y$-axis is
at most $\phi$, then $|(x,y,f(x,y)|$ is locally strictly decreasing as a
function of $y$ if $y_{0}>0$ and locally strictly increasing if $y_{0}<0$.
Note that by Remark 4, the same property holds for
$\min\\{|(x,y,f(x,y))|,|(x,y,g(x,y))|\\}$ as well. Observe that since $q$ is
an equilibrium point of $K_{1}\cap C$, it is an equilibrium point of the
section of $K_{1}\cap C$ with the plane through $q$ parallel to $H_{xy}$.
Thus, by Remark 3, if $\tau>0$ is chosen small enough, then the angle of
$\operatorname{proj}_{xy}(q)$ with the $x$-axis or the $y$-axis is at most
$\phi$. But this contradicts our previous observation that at such a point
$\min\\{|(x,y,f(x,y))|,|(x,y,g(x,y))|\\}$ is locally strictly increasing or
decreasing parallel to the $x$\- or the $y$-axis.

Finally, to exclude the possibility that a support plane of $K_{1}\cap C$
through a point in $\operatorname{bd}(K_{1})\cap\operatorname{bd}(C_{1})$
contains $q^{\prime}$, we truncate all points of
$\operatorname{bd}(K_{1})\cap\operatorname{bd}(C)$ by planes, not containing
$q^{\prime}$, whose intersections with $K_{1}\cap C$ do not contain
equilibrium point.

_Subcase 1.3_ , $q$ is an unstable point.
In this case both principal curvatures $\kappa_{1},\kappa_{2}$ of
$\operatorname{bd}(K_{1})$ at $q$ satisfy
$\kappa_{1},\kappa_{2}>\frac{1}{\rho}>0$, and thus, there is some constant
$\max\left\\{\frac{1}{\kappa_{1}},\frac{1}{\kappa_{2}}\right\\}<\eta<\rho$
such that the ball $\frac{\rho-\eta}{\rho}q+\eta\mathbf{B}^{3}$ contains a
neighborhood of $q$ in $\operatorname{bd}(K_{1})$. We parametrize
$\operatorname{bd}(K_{1})$ in a neighborhood of $q$ as the graph
$\\{z=f(x,y)\\}$ of a function $(x,y)\mapsto f(x,y)$, and note that by
nondegeneracy, there is some $r_{0}>0$ such that the function
$|(r\cos(\phi),r\sin(\phi),f(r\cos(\phi),r\sin(\phi)))|$ is a strictly
decreasing function of $r$ on $[0,r_{0}]$ for all values of $\phi$.

For any $\tau>0$, let $K_{1}(\tau)$ denote the set of points of $K_{1}$ with
$z$-coordinates at least $\rho-\tau$, let $H_{\tau}$ denote the plane with
equation $\\{z=\rho-\tau\\}$. Let $\tau>0$ be sufficiently small. Then there
is a circle $C_{0}$ centered at $(0,0,\rho-\tau)$ which is contained in
$H_{\tau}\cap\operatorname{int}(\eta\mathbf{B}^{3})$, and is disjoint from
$K_{1}$. Let $H$ be a plane supporting $K_{1}$ at a point of $H_{\tau}$ such
that its angle $\alpha$ with $H_{\tau}$ is minimal among these supporting
planes. Let $H^{\prime}$ be the translate of $H$ touching $C_{0}$ such that
$H$ strictly separates $o$ and $H^{\prime}$, and let the intersection point of
$H^{\prime}$ and the $z$-axis be $r$. Consider the infinite cone $C$ with apex
$r$ and base $C_{0}$, and observe that it contains $K_{1}\setminus
K_{1}(\tau)$ in its interior. Now, let $q^{\prime}=q-\zeta e_{z}$ for some
suitably small $\zeta>0$, and let $\Gamma$ be a polygonal curve connecting
$q^{\prime}$ to a point $p\in C_{0}$ such that the plane of $o,p,q^{\prime}$
contains $\Gamma$,
$\Gamma\subset\mathrm{relbd}(\operatorname{conv}(\Gamma\cup\\{o\\}))$, and the
Euclidean distance function is strictly decreasing along $\Gamma$ from
$q^{\prime}$ to $p$. Let $L_{p}$ denote the closed half line in the line of
$[r,p]$ starting at $p$ and not containing $r$, and let
$\Gamma^{\prime}=\Gamma\cup L_{p}$. Let $m\geq 3$ be arbitrary, and for any
$i=0,1,\ldots,m-1$, let $\Gamma^{\prime}_{i}$ denote the rotated copy of
$\Gamma^{\prime}$ around the $z$-axis, with angle $\frac{2\pi i}{m}$. Let
$P^{\prime}=\operatorname{conv}\bigcup_{i=0}^{m-1}\Gamma^{\prime}_{i}$. Then
$P^{\prime}$ is a convex polyhedral domain such that $K_{1}\setminus
P^{\prime}\subseteq K_{1}(\tau)$, and if $m$ is sufficiently large, then at
any boundary point of $P^{\prime}$ with $z$-coordinate greater than
$\rho-\tau$, $|(x,y,g(x,y))|$ is strictly locally increasing in a neighborhood
of $(0,0)$ as a function of $\sqrt{x^{2}+y^{2}}$, where
$\operatorname{bd}(P^{\prime})$ is given as the graph of the function
$z=g(x,y)$. Thus, by Remark 4 and following the idea at the end of Subcase 1.2
in Step 2, we may truncate a neighborhood of $q$ in $\operatorname{bd}(K_{1})$
by a convex polyhedral region $P^{\prime}$ such that the only equilibrium
point of the truncated body on $\operatorname{bd}(P^{\prime})$ is the unstable
point $q^{\prime}$, and the truncated body has no non-truncated boundary point
where some supporting plane contains an equilibrium point.

The procedure discussed in Subcases 1.1-1.3 for $q$ is applied for any
equilibrium point in the orbit of $q$ in an analogous way.

Case 2, if the stabilizer of $q$ in $G$ is not the identity.
In this case the procedure described in Case 1 is carried out in such a way
that the truncating polyhedral domain is invariant under any element of $G$
fixing $q$.

Summing up, to construct $K_{2}$ in Step 2 we truncated a neighborhood of each
equilibrium point of $K_{1}$ by a polyhedral region in such a way that each
region contains exactly one equilibrium point relative to $o$, and no plane
supporting $K_{2}$ at any point of $X_{2}=\operatorname{bd}(K_{1})\cap K_{2}$
contains an equilibrium point of $K_{2}$ relative to $o$. In addition, $K_{2}$
is $G$-invariant.

###  4.3. Step 3: Truncating the remaining part of the boundary

In this step let $Y=X_{1}\cap X_{2}$. Furthermore, for any plane $H$ let
$o_{H}$ denote the orthogonal projection of $o$ onto $H$, and let
$\mathcal{H}$ denote the family of planes $H$ with the property that $o_{H}\in
K_{2}$. Note that $\mathcal{H}$ consists of

  * •

all planes through $o$, and,

  * •

for any $p\in K_{2}\setminus\\{o\\}$, the (unique) plane passing through $p$
and perpendicular to $[o,p]$.

Observe that $Y$ is compact, and by our construction, for any plane $H$
supporting $K_{2}$ at some point $p\in Y$, we have $o_{H}\notin H\cap K_{2}$;
or equivalently, we have $\\{o_{H}:H\in\mathcal{H}\\}\cap Y=\emptyset$. Thus,
by compactness, there is some $\delta>0$ such that for any $H\in\mathcal{H}$
and $p\in Y$, the distance of $o_{H}$ and $p$ is at least $\delta>0$. Now, for
any $p\in Y$, let $H_{p}$ denote the unique closed half space whose boundary
supports $K_{0}$ at $p$ and which satisfies $\operatorname{int}(H_{p})\cap
K_{1}=\emptyset$. Let $u_{p}$ denote the outer unit normal vector of $K$ at
$p$, and for any $\zeta>0$, set $H_{p}(\zeta)=H_{p}-\zeta u_{p}$, and
$Y(\zeta)=K_{2}\cap\left(\bigcup_{p\in Y}H_{p}(\zeta)\right)$ (cf. Figure 6).
Clearly, $Y(\zeta)$ tends to $Y$ with respect to Hausdorff distance as
$\zeta\to 0^{+}$. Thus, there is some sufficiently small $\zeta_{0}>0$ such
that for any $H\in\mathcal{H}$, $o_{H}\notin Y(\zeta_{0})$.

Figure 6. An illustration for Step 3 in the proof of Theorem 2.

Now, for any $p\in Y$, set $U(p)=Y\cap\operatorname{int}(H_{p}(\zeta_{0}))$.
Then $U(p)$ is an open neighborhood of $p$ in $Y$. Thus, by the compactness of
$Y$, there are finitely many points $p_{1},\ldots,p_{m}$ such that
$\bigcup_{i=1}^{m}U(p_{i})=Y$. Then, clearly
$P=K_{2}\cap\left(\bigcap_{i=1}^{m}\left(\mathbb{R}^{3}\setminus\operatorname{int}(H_{p_{i}}(\zeta_{0}))\right)\right)$
is a convex polytope contained in $K_{2}$. Furthermore, since $\zeta_{0}>0$
can be arbitrarily small, $P$ can be arbitrarily close to $K_{2}$.

We show that no point $q\in\operatorname{bd}(P)$, contained in some
$\operatorname{bd}(H_{p_{i}}(\zeta_{0}))$ is an equilibrium point of $P$.
Indeed, if $q$ was such a point, then the plane $H$ through $q$ and
perpendicular to $[o,q]$ is contained in $\mathcal{H}$. On the other hand,
$q\in\operatorname{bd}(P)\cap\operatorname{bd}(H_{p_{i}}(\zeta_{0}))\subset
Y(\zeta_{0})$, which is impossible by our choice of $\zeta_{0}$.

Finally, we may choose the points $p_{1},p_{2},\ldots,p_{m}$ in such a way
that the set $\\{p_{1},\ldots,p_{m}\\}$ is invariant under the act of any
element of $G$.

##  5\. Remarks and open questions

First, we remark that by using truncations instead of conic extensions in the
proof of Theorem 1, we readily obtain Theorem 3. Here, a _mono-unstable_
convex body is meant to be a nondegenerate convex body with a unique unstable
point.

######  Theorem 3.

For any $n\geq 3$, $n\in\mathbb{Z}$ and $\varepsilon>0$ there is a homogeneous
mono-unstable polyhedron $P$ such that $P$ has an $n$-fold rotational symmetry
and $d_{H}(P,\mathbf{B}^{3})<\varepsilon$.

We ask the following.

######  Question 1.

What are the positive integers $n\geq 2$ such that class $(1,1)_{p}$ contains
a convex polyhedron with an $n$-fold axis of symmetry?

In light of the words of Shephard in [28] from Section 1 about monostable
polyhedra, we remark that a consequence of Theorem 2 is that to study the
metric properties of nondegenerate polyhedra, in particular monostable
polyhedra, it is sufficient to study the metric properties of their smooth
counterparts, which seem to be much more tractable.

Next, we conjecture that Theorem 2 remains true if we omit the condition that
the principal curvatures of $\operatorname{bd}(K)$ at every equilibrium point
of $K$ are strictly positive.

Finally, to propose a conjecture for the first part of Problem 3, we recall
the following concept from [33], where the function
$\rho_{K}:\mathbb{S}^{2}\to\mathbb{R}$, $\rho_{K}(x)=\max\\{\lambda:\lambda
x\in K\\}$ is called the _gauge function_ of the convex body $K$.

######  Definition 1.

Let $K\in\mathbb{R}^{3}$ be a centered convex body, and for any simple, closed
curve $\Gamma\subset\mathbb{S}^{2}$, let $\Gamma^{+}$ and $\Gamma^{-}$ denote
the two compact, connected domains in $\mathbb{S}^{2}$ bounded by $\Gamma$.
Then the quantities

$F(K)=\sup_{\Gamma}\sup_{p_{1}\in\Gamma^{+},p_{2}\in\Gamma^{-}}\frac{\min\\{\rho_{K}(s):s\in\Gamma\\}}{\max\\{\rho_{K}(p_{1}),\rho_{K}(p_{2})\\}}$

and

$T(K)=\sup_{\Gamma}\sup_{p_{1}\in\Gamma^{+},p_{2}\in\Gamma^{-}}\frac{\min\\{\rho_{K}(p_{1}),\rho_{K}(p_{2})\\}}{\max\\{\rho_{K}(s):s\in\Gamma\\}}$

are called the _flatness_ and the _thinness_ of $K$, respectively.

Domokos and Várkonyi in [33] proved that for any nondegenerate, centered
smooth convex body $K$, $F(K)=1$ if and only if $K$ is monostable, and
$T(K)=1$ if and only if $K$ is mono-unstable.

Recall that a nondegenerate convex body is _mono-monostatic_ if it has a
unique stable and a unique unstable point [32]. We conjecture the following.

######  Conjecture 1.

For any centered convex body $K\subset\mathbb{R}^{3}$, $K$ can be uniformly
approximated by monostable convex polyhedra if and only if $F(K)=1$, by mono-
unstable convex polyhedra if and only if $T(K)=1$, and by mono-monostatic
convex polyhedra if and only if $F(K)=T(K)=1$.

_Acknowledgements_. The author expresses his gratitude to Gábor Domokos and,
more generally, the members of the Morphodynamics Research Group with whom he
had many fruitful discussions on the subject, and also to an unknown referee
for many helpful comments.

## References

  * [1] A. Abramson et al., _An ingestible self-orienting system for oral delivery of macromolecules_ , Science 363 (6427) (2019), 611–615.
  * [2] I. Bárány, J. Itoh, C. Vîlcu and T. Zamfirescu, _Every point is critical_ , Adv. Math. 235 (2013), 390-397.
  * [3] A. Bezdek, _On stability of polyhedra_ , in: Workshop on Discrete Geometry, Sept. 13-16 2011, Fields Institute, Canada, 2490-2491.
  * [4] K.F. Bohringer, B.R. Donald, L.E. Kavraki and F. Lamiraux, _Part orientation with one or two stable equilibria using programmable vector fields_ , IEEE Transactions on Robotics Automation. 16(2) (2000), 157-170.
  * [5] J.H. Conway and R.K. Guy, _Stability of polyhedra_ , SIAM Rev. 8(3) (1966), 381.
  * [6] H.T. Croft, K. Falconer and R.K. Guy, _Unsolved Problems in Geometry_ , Springer, New York, NY, 1991.
  * [7] R. Dawson, _Monostatic simplexes_ , Amer. Math. Monthly 92 (1985), 541-546.
  * [8] R. Dawson and W. Finbow, _What shape is a loaded die?_ , Math. Intelligencer 22 (1999), 32-37.
  * [9] R. Dawson and W. Finbow, _Monostatic Simplexes III_ , Geom. Dedicata 84 (2001), 101-113.
  * [10] R. Dawson, W. Finbow and P. Mak, _Monostatic simplexes II_ , Geom. Dedicata 70 (1998), 209-219.
  * [11] G. Domokos, _The Gömböc Pill_ , Math. Intelligencer 41 (2019), 9–11.
  * [12] G. Domokos, P. Holmes and Z. Lángi, _A genealogy of convex solids via local and global bifurcations of gradient vector fields_ , J. Nonlinear Sci. 26 (2016), 1789-1815.
  * [13] G. Domokos, F. Kovács, Z. Lángi, K. Regős and P.T. Varga, _Balancing polyhedra_ , Ars Math. Contemp. 19 (2020), 95-124.
  * [14] G. Domokos, Z. Lángi and T. Szabó, _On the equilibria of finely discretized curves and surfaces_ , Monatsh. Math. 168 (2012) 321-345.
  * [15] G. Domokos, Z. Lángi and T. Szabó, _A topological classification of convex bodies_ , Geom. Dedicata 182 (2016), 95-116.
  * [16] G. Domokos, A.Á. Sipos, T. Szabó and P.L. Várkonyi, _Pebbles, shapes and equilibria_ , Math. Geosciences 42 (1) (2010), 29-47.
  * [17] G. Domokos and P. Várkonyi, _Geometry and self-righting of turtles_ , Proc. R. Soc. B. 275 (2007), 11-17.
  * [18] A. Dumitrescu and C.D. Tóth, _On the cover of the rolling stone_ , Proceedings of the 31st Annual ACM-SIAM Symposium on Discrete Algorithms (2020), 2575-2586.
  * [19] F. Galaz-García and L. Guijarro, _Every point in a Riemannian manifold is critical_ , Calc. Var. 54 (2015), 2079–2084.
  * [20] M. Goldberg and R.K. Guy, _Stability of polyhedra (J.H. Conway and R.K. Guy)_ , SIAM Rev. 11 (1969), 78-82.
  * [21] http://www.gomboc.eu/en/site.php?nyelv=2
  * [22] T.I. Heath (Ed.), _The Works of Archimedes_ , Cambridge University Press, 1897.
  * [23] A. Heppes, _A double-tipping tetrahedron_ , SIAM Rev. 9 (1967), 599-600.
  * [24] J. Itoh, C. Vîlcu and T. Zamfirescu, _With respect to whom are you critical?_ , Adv. Math. 369 (2020), 107187.
  * [25] M.S. Klamkin, _Problems in Applied Mathematics_ , SIAM, Philadelphia, USA, 1990.
  * [26] A. Reshetov, _A unistable polyhedron with $14$ faces (English summary)_, Internat. J. Comput. Geom. Appl. 24 (2014), 3959.
  * [27] R. Schneider, _Convex bodies: the Brunn-Minkowski theory_ , Encyclopedia of Mathematics and its Applications 44, Cambridge University Press, Cambridge, 1993.
  * [28] G.C. Shephard, _Twenty problems on convex polyhedra part II_ , Math. Gaz. 52 (1968), 359-367.
  * [29] T. Szabó and G. Domokos, _A new classification system for pebble and crystal shapes based on static equilibrium points_ , Central European Geology, 53(1) (2010), 1-19.
  * [30] T. Szabó, G. Domokos, J.P. Grotzinger and D.J. Jerolmack, _Reconstructing the transport history of pebbles on Mars_ , Nature Comm. 6 (2015), paper 8366.
  * [31] P.L. Várkonyi, _Estimating Part Pose Statistics With Application to Industrial Parts Feeding and Shape Design: New Metrics, Algorithms, Simulation Experiments and Datasets_ , IEEE Transactions on Automation Science and Engineering 11(3) (2014), 658 - 667.
  * [32] P.L. Várkonyi and G. Domokos, _Static equilibria of rigid bodies: dice, pebbles and the Poincaré-Hopf Theorem_ , J. Nonlinear Sci. 16 (2006), 255-281.
  * [33] P.L. Várkonyi and G. Domokos, _Mono-monostatic bodies: the answer to Arnold’s question_ , Math. Intelligencer 28 (2006), 34-38.
  * [34] T. Zamfirescu, _How do convex bodies sit?_ , Mathematika 42 (1995), 179-181.
  * [35] A. Zomorodian, _Topology for computing_ , Cambridge University Press, 2005.