# Sections and projections of nested convex bodies

I. González-García1, J. Jerónimo-Castro2
E. Morales-Amaya3, and D. J. Verdusco-Hernández4
1,2Facultad de Ingeniería
Universidad Autónoma de Querétaro, México
3,4Facultad de Matemáticas-Acapulco,
Universidad Autónoma de Guerrero, México

###### Abstract

One of the most important problems in Geometric Tomography is to establish
properties of a given convex body if we know some properties over its sections
or its projections. There are many interesting and deep results that provide
characterizations of the sphere and the ellipsoid in terms of the properties
of its sections or projections. Another kind of characterizations of the
ellipsoid is when we consider properties of the support cones. However, in
almost all the known characterizations, we have only a convex body and the
sections, projections, or support cones, are considered for this given body.
In this article we proved some results that characterizes the Euclidean ball
or the ellipsoid when the sections or projections are taken for a pair of
nested convex bodies, i.e., two convex bodies $K$, $L$ such that
$L\subset\text{int}\,K.$ We impose some relations between the corresponding
sections or projections and some apparently new characterizations of the ball
or the ellipsoid appear. We also deal with properties of support cones or
point source shadow boundaries when the apexes are taken in the boundary of
$K$.

##  1 Introduction

One of the most important problems in Geometric Tomography is to establish
properties of a given convex body if we know some properties over its sections
or its projections. A very interesting topic was initiated by C. A. Rogers
[25] who proved:

_Let $K\subset\mathbb{R}^{n}$ be a convex body, $n\geq 3$, and $p$ be a point
in $\mathbb{R}^{n}$. If all the $2$-dimensional sections, i.e., the
intersections of $K$ with planes of dimension 2, through $p$ have a centre of
symmetry, then $K$ has also a centre of symmetry._

Rogers also conjectured that if $p$ is not the centre of $K$ then it is an
ellipsoid. This conjecture was first proved by P. W. Aitchison, C. M. Petty,
and C. A. Rogers [1] and is known as the False Centre Theorem. A simpler proof
was given by L. Montejano and E. Morales-Amaya in [21]. The following is a
remarkable theorem due to S. Olovjanishnikov [24]:

_A convex body $K\subset\mathbb{R}^{n}$, $n\geq 3$, is a solid ellipsoid
provided all hyperplanar sections of $K$ that divide the volume of $K$ in a
given ratio $\mu\neq 1$ have a centre of symmetry._

We say that two sets $A$ and $B$ in $\mathbb{R}^{n}$ are affinely equivalent
if there is an invertible affine transformation
$f:\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}$ such that $A=f(B).$ A result
similar to the False Centre Theorem was proved by L. Montejano [20] which
combines the method of P. Mani [16] and some results of algebraic topology:

_Let $K\subset\mathbb{R}^{n}$ be a convex body, $n\geq 3$, and
$p\in\emph{int}\,K$. If all sections of $K$ by hyperplanes through $p$ are
affinely equivalent, then $K$ is either a solid ellipsoid or has centre of
symmetry at $p$._

Recently, L. Montejano proved a similar result [22] where he consider affinely
equivalent projections instead of affinely equivalent sections.

A different kind of characterizations of the ellipsoid is when we consider
properties of the support cones. Let $K\in\mathbb{R}^{n}$, $n\geq 2$, be a
convex body, i.e., a compact and convex set with non-empty interior. Given a
point $x\in\mathbb{R}^{n}\setminus K$ we denote the cone generated by $K$ with
apex $x$ by $\text{C}(K,x)$, i.e., $\text{C}(K,x):=\\{x+\lambda(y-x):y\in
K,\,\lambda\geq 0\\}.$ The boundary of $\text{C}(K,x)$ is denoted by
$\text{S}(K,x)$, in other words, $\text{S}(K,x)$ is the support cone of $K$
from the point $x$. We denote the _graze_ of $K$ from $x$ by $\Sigma(K,x)$,
i.e., $\Sigma(K,x):=\text{S}(K,x)\cap\partial K.$ In [17] A. Marchaud proved
the following:

_Let $K\subset\mathbb{R}^{3}$ be a convex body and $H$ be a plane which is
either disjoint from $K$ or meets $K$ at a single point. Then $K$ is a solid
ellipsoid if for every point $x\in H\setminus K$, the graze $\Sigma(K,x)$
contains a planar convex curve $\gamma$ such that
$\emph{conv}\,\gamma\cap\emph{int}\,K\neq\emptyset$_.

This result is also true if the apexes of the cones are points at infinity,
i.e., the grazes, in this case called _shadow boundaries_ are obtained by
intersections of $\partial K$ with circumscribed cylinders. In this case is
more convenient to think that the body is illuminated by light rays parallel
to a given direction $u\in\mathbb{S}^{n-1}$, and the shadow boundary is
precisely the border between the light part and the dark part of $\partial K$.
The shadow boundary of $K$ with respect the direction of illumination $u$ is
denoted by $\Lambda(K,u)$. The firs proof that a convex body is an ellipsoid
if and only if every shadow boundary lies in a hyperplane is due to H.
Blaschke [3]. However, if we change the plane $H$ by a surface $\Gamma$ which
encloses to $K$ it is not known whether if $K$ is an ellipsoid or not, even in
the case where $\Gamma$ is the boundary of a convex body that contains $K$ in
its interior. We suspect the following is true.

###### Conjecture 1

Let $L\subset\emph{int}\,K\subset\mathbb{R}^{n}$ be convex bodies such that
for every point $x\in\partial K$ it holds that $\Sigma(L,x)$ lies in a
hyperplane. Then $L$ is an ellipsoid.

Another characterization of ellipsoids with respect to the properties of
support cones is the following due to P. Gruber and T. Ódor [10]:

_Let $K\subset\mathbb{R}^{n}$, $n\geq 3$, be a convex body whose boundary is a
hypersurface of class $C^{4}$ with positive Gauss curvature. If for every
exterior point $x$ of $K$ which is sufficiently close to $K$, the cone
$\emph{C}(K,x)$ has an axis of symmetry, then $K$ is an ellipsoid._

In this article we proved some results that characterizes the Euclidean ball
or the ellipsoid when the sections or projections are taken for a pair of
nested convex bodies, i.e., two convex bodies $K$, $L$ such that
$L\subset\text{int}\,K.$ We impose some relations between the corresponding
sections or projections and some apparently new characterizations of the ball
or the ellipsoid appear. We also deal with properties of support cones or
point source shadow boundaries when the apexes are taken in the boundary of
$K$.

##  2 By properties of sections or orthogonal projections

We give first the definition of a very important convex body. For the given
convex body $K$, the floating body of $K$ is defined as follows: for a given
positive number $\delta<\text{vol}_{n}(K)$, the _floating body_ denoted as
$K_{\delta}$ is defined as the intersection of all the closed half-spaces
which cut off from $K$ a cap with volume $\delta$. For instance, if $K$ is a
Euclidean disc with unit area in the plane and $\delta<1/2$, then $K_{\delta}$
is a disc concentric with $K$. Notice that the midpoint of every chord of $K$
and tangent to $K_{\delta}$ belongs to $K_{\delta}.$ This property holds for
every convex body and its floating body $K_{\delta}$ (see for instance [27]).

Figure 1: The floating body

The following result is related to the notions of floating body and a special
set of hyperplanes called an $n$-cycle of hyperplanes (see for instance [14]):
Let $\delta:\mathbb{S}^{n}\longrightarrow\mathbb{R}$ be a continuous function
such that $\delta(-x)=-\delta(x)$. We denote by $\mathcal{P}$ the following
set of hyperplanes in $\mathbb{R}^{n+1}$:

$\mathcal{P}=\\{H_{y}=\\{x\in\mathbb{R}^{n+1}|\langle
x,y\rangle=\delta(y)\\}\\}_{y\in\mathbb{S}^{n}},$

and we say that $\mathcal{P}$ an $n$-cycle of hyperplanes. If for a given
convex body $K$, it holds that $H_{p}\cap H_{q}\cap\text{int}\,K\neq\emptyset$
for every two hyperplanes $H_{p},H_{q}\in\mathcal{P}$, we say that
$\mathcal{P}$ is an $n$-cycle of planes for $K$. In [15], D. Larman, L.
Montejano, and E. Morales proved some results which characterizes the
ellipsoid (in $\mathbb{R}^{3}$) in terms of the properties of the sections,
cut by the members of a $2$-cycle of planes, in a given convex body.

We have the following interesting lemma.

###### Lemma 1

Let $K\subset\mathbb{R}^{n+1}$ a convex body and let $\ell\subset K$ be any
$(n-1)$-dimensional section of $K$. If $\mathcal{P}$ is an $n$-cycle of
hyperplanes for $K$ then there exists $\Pi\in\mathcal{P}$ such that
$\ell\subset\Pi$.

_Proof._ For every unitary vector $u\in\mathbb{S}^{n}$, we may consider that
$\delta(u)$ is the distance with sign from $H_{u}$ to the origin $O$. Recall
that $H_{u}$ is the plane in $\mathcal{P}$ that is orthogonal to $u$. Let
$\mathcal{P}_{\ell}$ be the subset of $\mathcal{P}$ such that for every
$H_{y}\in\mathcal{P}_{\ell}$ it holds that $y$ is orthogonal to
$\text{aff}\,\ell$. Denote by
$S(\ell)=\\{\upsilon\in\mathbb{S}^{n}:\upsilon\bot\text{aff}\,\ell\\}.$ Let
$\delta_{\ell}$ be the restriction of $\delta$ to the set $S(\ell)$. Let
$O^{\prime}$ be the centroid of $\ell$, and let
$\sigma:S(\ell)\longrightarrow\mathbb{R}$ be the function such that
$\sigma(x)$ is the distance with sign from $O^{\prime}$ to $H_{x}$. Since
$\delta_{\ell}$ is a continuous function we have that $\sigma$ is also a
continuous function. Moreover, $\sigma(-x)=-\sigma(x)$, i.e., $\sigma$ is an
odd function. By the Borsuk-Ulam theorem (see for instance [18]), there exists
$x_{0}\in S(\ell)$ such that $\sigma(x_{0})=0$, this means that the plane
$H_{x_{0}}$ contains to $\ell$. $\Box$

The following is a more convenient reformulation, four our purposes, of the
theorem of Olovjanishnikov [24].

Theorem O. Let $K$ and $L$ be convex bodies in $\mathbb{R}^{n}$, with
$L\subset\text{int}\,K$, such that every hyper-section of $K$ tangent to $L$
is centrally symmetric and its centre belongs to $\partial L$, then $K$ and
$L$ are homothetic and concentric ellipsoids.

###### Theorem 1

Let $K,L\subset\mathbb{R}^{n}$, $n\geq 3$, be two convex bodies with
$L\subset\emph{int}\,K$. Suppose that one of the following properties is
satisfied:

  1. (a)

There is an $(n-1$)-cycle of planes $\mathcal{P}$ for $L$ such that for every
$H\in\mathcal{P}$ it holds that $H\cap L$ is a floating body of $H\cap K$.

  2. (b)

There is a sphere $\mathcal{S}$ contained in $\emph{int}\,L$, such that for
every supporting hyperplane $H$ of $\mathcal{S}$ we have that $H\cap L$ is a
floating body of $H\cap K$.

Then $K$ and $L$ are homothetic and concentric ellipsoids.

_Proof._ First we suppose that (a) holds. Let $x$ be any point in $\partial L$
and let $\ell$ be any supporting $(n-2)$-dimensional plane through $x$. By
Lemma 1 we have that there exists a hyperplane $H\in\mathcal{P}$ such that
$H\cap L$ is a floating body of $H\cap K$. Since $\ell\subset H$ is a
supporting $(n-2)$-dimensional plane of $H\cap L$ through $x$, we have that
$x$ is the centroid of $\ell\cap K$ (see for instance [27]). Because $\ell$ is
any $(n-2)$-dimensional supporting plane of $L$ through $x$, we have that $x$
is the centre of symmetry of the section of $K$ tangent to $L$ at $x$ (see
Theorem 5.6.19 in [7]). By Theorem O we have that $K$ and $L$ are homothetic
and concentric ellipsoids.

Now, suppose that (b) holds. Let $x$ be any point in $\partial L$ and let
$\ell$ be any supporting $(n-2)$-dimensional plane through $x$. Consider one
of the two supporting hyperplanes of $\mathcal{S}$ that contain to $\ell$, to
say $H$. From this point we proceed exactly as in the proof of (a). Hence $K$
and $L$ are homothetic and concentric ellipsoids. $\Box$

With respect to projections we have the following conjecture.

###### Conjecture 2

Let $K,L\subset\mathbb{R}^{n}$, $n\geq 3$, be two convex bodies with
$L\subset\emph{int}\,K$ a strictly convex body (i.e., with no segments on its
boundary). Suppose for every $u\in\mathbb{S}^{n-1}$ the orthogonal projection
of $L$, in direction $u$, is a floating body of the orthogonal projection of
$K$. Then $K$ and $L$ are homothetic and concentric ellipsoids.

As evidence of the veracity of this conjecture we have a proof for the case
$n=3$.

_Proof for $n=3$._ Let $x\in\partial L$ be any point of $L$, i.e., there
exists a support plane $H$ of $L$ such that $H\cap L:=\\{x\\}.$ Let
$u\in\mathbb{S}^{2}$ be any vector parallel to $H$ and denote by $\pi_{u}$ the
orthogonal projection onto the subspace $u^{\bot}$. The projection of
$\pi_{u}(H)$ is a supporting line of $\pi_{u}(L)$ through the point
$x^{\prime}:=\pi_{u}(x)$ (see Fig. 2).

Figure 2: The point $x$ projects onto the midpoint of
$[a^{\prime},b^{\prime}]$

Since $\pi_{u}(L)$ is the floating body of $\pi_{u}(K)$ we have that
$x^{\prime}$ is the midpoint of the segment
$[a^{\prime},b^{\prime}]:=H\cap\pi_{u}(K)$ (see for instance [27]). This is
true for any vector $u$ parallel to $H$, we then have that $x$ is the centre
of symmetry for $H\cap K$. Now, since this last is true for every point
$x\in\partial L$, we apply Theorem O and conclude that $K$ and $L$ are
homothetic and concentric ellipsoids. $\Box$

The proof of the general case of this conjecture depends upon the following
conjecture.

###### Conjecture 3

Let $K\mathbb{\subset}\mathbb{R}^{n}$, $n\geq 3$, be a convex body and let $q$
be a point in its interior. If for every $u\in\mathbb{S}^{n-1}$ we have that
$\pi_{u}(q)$ is the centroid of $\pi_{u}(K)$, then $q$ is the centre of
symmetry of $K$.

We were not able to prove it or to find this in the literature, neither we
have a counterexample.

###### Remark 1

Let $K$ and $L$ be convex bodies in $\mathbb{R}^{3}$, with
$L\subset\emph{int}\,K$. If for every hyperplane $H$, the projection of $L$
onto $H$ is the floating body of the projection of $K$, then $K$ and $L$ are
homothetic and concentric ellipsoids and then $L$ is a floating body of $K$.
However, the converse is not true, i.e., if $L$ is the floating body of $K$,
then it is not necessarily true that every projection of $L$ is the floating
body of the projection of $K$.

###### Remark 2

Let $K\subset\mathbb{R}^{n}$ be a convex body with boundary of class $C^{1}$.
A billiard ball is a point in $K$ which moves along a straight line in the
interior of $K$ until it hits $\partial K$ where it is reflected in the usual
way: the angle of reflection, with respect to the normal line at the point, is
equal to the angle of incidence and the incidence line, the normal line, and
the reflection line, are coplanar. A convex body $L$ in the interior of $K$ is
a convex caustic if any trajectory which touches $L$ once, touches it again
after each reflection. In [8], P. Gruber proved the following: _Let
$K\subset\mathbb{R}^{n}$ be a convex body with boundary of class $C^{1}$, let
$L$ be a convex body in $\text{int}\,K$, and let $r\in\\{2,3,\ldots,n-1\\}$.
Then $L$ is a caustic of $K$ if and only if for every $r$-dimensional plane
$H$, the orthogonal projection of $L$ onto $H$ is a caustic of the orthogonal
projection of $K$._ In some sense, Conjecture 2 is similar to Gruber’s
theorem. Indeed, in [9] Gruber proved that only ellipsoids have convex
caustics.

##  3 By properties of the support cones

Let $K\in\mathbb{R}^{n}$, $n\geq 2$, be a convex body, i.e., a compact and
convex set with non-empty interior. We say that a hyperplane $\ell$ is a
support hyperplane of $K$ if $\ell\cap K\neq\emptyset$ and $K$ is contained in
one of the half-spaces bounded by $\ell$. In the Euclidean plane we have the
following: for every $t\in[0,2\pi]$ denote by $\ell(t)$ the support line of
$K$ with outward normal vector $u(t)=(\cos t,\sin t),$ and let $p(t)$ denote
the distance with the sign from the origin $O$ to $\ell(t).$ A well known
result in the geometry of convex sets is that a body is determined in a unique
way if the support function is given (see for instance [28]). Particularly, a
convex body is a disc centred at the origin if its support function is
constant.

In [13], J. Jerónimo-Castro and T. McAllister proved the following: _Let
$\mathcal{C}\subset\mathbb{R}^{n}$, $n\geq 3$, be a convex solid cone. If for
every $y\in\mathbb{R}^{n}$ the intersection of
$(\partial\mathcal{C})\cap(y-\partial\mathcal{C})$, when non empty, lies in a
hyperplane, then $K$ is an elliptic cone, i.e., every $(n-1)$-dimensional
bounded section of $K$ is an $(n-1)$-dimensional ellipsoid._ The following
theorem is similar to this mentioned theorem, however, there is more order in
the position of the apexes of the cones.

###### Theorem 2

Let $K$ be a convex body contained in the interior of the unit ball of
$\mathbb{R}^{n}$, $n\geq 2$, such that for every unit vector
$x\in\mathbb{S}^{n-1}$ we have that $S(K,x)\cap S(K,-x)$ is contained in the
subspace $x^{\bot}.$ Then $K$ is a Euclidean ball.

_Proof._ We first note that the origin $O$ is contained in the interior of
$K$. Suppose this is not the case, then there exists a hyperplane $H$ which
separates $O$ from $K$ (it could happen that $O\in H$). Let
$z\in\mathbb{S}^{n-1}$ be the vector orthogonal to $H$ which is contained in
the opposite half-space bounded by $H$ that $K$. Clearly, the intersection
$S(K,z)\cap S(K,-z)$ cannot be contained in $z^{\bot}$. This contradiction
shows that $O$ must be in the interior of $K$.

Now we will prove that the central projections of $K$ onto $x^{\bot}$, from
the points $x$ and $-x$, coincide for every $x\in\mathbb{S}^{n-1}.$ In order
to do this, let $L$ be any $2$-dimensional plane through the points $x$ and
$-x$, and consider the section $K_{L}:=K\cap L$. Consider the support lines of
$K_{L}$, from $x$ and $-x$, that touch $K_{L}$ in the same half-plane bounded
by the line through $x$ and $-x$. The intersection point $q$ of these support
lines must be, by hypothesis, in the hyperplane $x^{\bot}$. This argument
proves the assertion.

Now, we continue the proof by induction.

  1. 1.

For $n=2$. Let $x_{0}\in\mathbb{S}^{1}$ be such that the angle $\beta$ between
one of the support lines of $K$ from $x_{0}$ and the line generated by $x_{0}$
is an irrational multiple of $\pi$ (as shown in Fig. 3). The existence of such
$x_{0}$ is guaranteed by the following argument: if for every
$x\in\mathbb{S}^{1}$ the associated angle $\beta$ between the corresponding
support line and the vector $x$ is equal to a constant value $\beta_{0}$, then
the support function of $K$ is also constant and consequently, $K$ is a disc
(see [28]). Otherwise, there is an interval $[\beta_{1},\beta_{2}]$ such that
$\beta$ takes every value in this interval in a continuous way. Obviously,
there is a value in this interval such that $\beta$ is an irrational multiple
of $\pi$. Now, let $x_{1}$ be the intersection between this support line from
$x_{0}$ and $\mathbb{S}^{1}$. Denote by $K_{x_{0}}:=C(K,x_{0})\cap
x_{0}^{\bot}$, i.e., $K_{x_{0}}$ is the central projection of $K$ onto
$x_{0}^{\bot}$ from the point $x_{0}$. By hypothesis $K_{-x_{0}}=K_{x_{0}}$,
hence the angle between the corresponding support line of $K$ from $-x_{0}$
and the line generated by $-x_{0}$ is also $\beta$.

Figure 3:

Since triangle $\triangle x_{0}Ox_{1}$ is isosceles, we have that
$\measuredangle Ox_{1}x_{0}=\beta$. Let $z$ be the intersection between the
segment $[x_{0},x_{1}]$ and the line $x_{1}^{\bot}.$ By the hypothesis of the
theorem we have that the angle between the segments $[x_{1},-x_{1}]$ and
$[-x_{1},z]$ is equal to $\beta$. Now, let $x_{2}$ be the point in
$\mathbb{S}^{1}$ (different of $-x_{1}$) that is collinear with $-x_{1}$ and
$z$. Since $\measuredangle x_{2}Ox_{1}$ is a central angle which intersects
the arc $\widehat{x_{1}x_{2}}$ we have that $\measuredangle
x_{2}Ox_{1}=2\beta$. Repeating this process we obtain a sequence of points
$x_{3},$ $x_{4},\ldots$ such that for every natural number $i$ it holds that
$\measuredangle x_{i+1}Ox_{i}=2\beta$, and since $2\beta$ is an irrational
multiple of $\pi$ we obtain a sequence of points $\\{x_{i}\\}_{i=1}^{\infty}$
which is dense in the circle $\mathbb{S}^{1}$. Moreover, since the distance
from every support segment $[-x_{i},x_{i+1}]$ to $O$ is equal to the distance
from $[x_{0},x_{1}]$ to $O$, we have that the support function of $K$ is
constant and hence $K$ is a disc.

  2. 2.

We suppose the theorem is proved for every number $n\leq r$, for some natural
number $r$.

  3. 3.

We will prove the theorem for $n=r+1$. Consider any point
$x_{0}\in\mathbb{S}^{r}$ and let $\upsilon\in\mathbb{S}^{r}$ be any vector
orthogonal to $x_{0}.$ Denote by $K_{\upsilon}:=K\cap\upsilon^{\bot}$ and
$S_{\upsilon}:=\mathbb{S}^{r}\cap\upsilon^{\bot}$. Since $K_{\upsilon}$ have
the hypothesis of the theorem from every point $x\in S_{\upsilon}$ we obtain
that $K_{\upsilon}$ is a $r$-dimensional ball for every $\upsilon$ orthogonal
to $x_{0}.$ We conclude that $K$ is a $(r+1)$-dimensional ball.

Therefore, the proof is complete and $K$ is a ball. $\Box$

There are convex bodies in the plane, besides discs, which are contained in
the interior of the unit circle $\mathbb{S}^{1}$ and such that all its
projections onto $\mathbb{S}^{1}$ from points in $\mathbb{S}^{1}$ are
congruent. These bodies indeed, are bodies which have the unit disc as one of
is isoptic curves, i.e., curves under which the body is seen under a constant
angle (see for instance [6], [5]). However, as we will see in Theorem 3, if we
consider projections over lines through the origin then the only convex body
with the property of congruent projections is the disc. Moreover, the
$(2n+1)$-dimensional version of this result, for every natural number $n$,
still being true. Myroshnychenko had proved [23] that two polytopes $K$ and
$L$ are the same polytope if for any $x\in\mathbb{S}^{n-1}$ the projections of
$K$ and $L$ onto $\mathbb{S}^{n-1}$ are congruent.

We enunciate the following problem.

Problem 1. It would be interesting to know whether or not a convex body
$K\subset\mathbb{R}^{n}$ is a ball (or at least have centre of symmetry) if
for every unit vector $x\in\mathbb{S}^{n-1}$ the sets $C(K,x)\cap x^{\bot}$
and $C(K,-x)\cap x^{\bot}$ are congruent. Even the case when both sets differ
by a translation over the hyperplane $x^{\bot}$ seems to be interesting
enough. The problem when we only know that $C(K,x)\cap x^{\bot}$ and
$C(K,-x)\cap x^{\bot}$ have the same $(n-1)$-dimensional volume looks more
difficult.

In the case of the Euclidean space with odd dimension we have the following.

###### Theorem 3

Let $K$ be a convex body contained in the interior of the unit ball of
$\mathbb{R}^{2k+1}$, $k\geq 1$, and let $L$ be a $2k$-dimensional convex body.
Suppose that for every unit vector $x\in\mathbb{S}^{2k}$ we have that
$C(K,x)\cap x^{\bot}$ is a convex body congruent to $L$. Then $K$ is a
Euclidean ball.

For the proof of this theorem we need to introduce an important notion: the
concept of field of congruent convex bodies. Let $A\subset\mathbb{R}^{n}$ be
an $n$-dimensional convex body, a _field of bodies congruent to_ $A$ is a
continuous function $A(u)$ defined for $u$ in the unit sphere
$\mathbb{S}^{n}=\\{x\in\mathbb{R}^{n+1}:\|x\|=1\\}$, where $A(u)$ is a
congruent copy of $A$ lying in a hyperplane of $\mathbb{R}^{n+1}$
perpendicular to $u$; here $A(u)$ is meant to be continuous in the Hausdorff
metric. If in addition $A(u)=A(-u)$, for each $u$, we say $A(u)$ is a complete
turning of $A$ in $\mathbb{R}^{n+1}$. Clearly if all the $n$-dimensional
sections of an $(n+1)$-dimensional convex body $K\subset\mathbb{R}^{n+1}$,
through a fixed inner point are congruent, they give rise to a complete
turning of some $n$-dimensional body $M\subset\mathbb{R}^{n}$ in
$\mathbb{R}^{n+1}$.

In [16] P. Mani proved that if $n$ is even, fields of convex bodies tangent to
$\mathbb{S}^{n}$ can be constructed only if $A$ is an $n$-dimensional ball. In
[4], G. R. Burton used the methods of H. Hadwiger [11] and P. Mani [16] to
prove that a 3-dimensional convex body $M\subset\mathbb{R}^{4}$ can be
completely turned in $\mathbb{R}^{4}$ if and only if $M$ is centrally
symmetric. In [20], L. Montejano generalized Burton’s result, i.e., _convex
bodies $M\subset\mathbb{R}^{n+1}$, $n\geq 2$, can be completely turned in
$\mathbb{R}^{n+1}$ if and only if $M$ is centrally symmetric_. Montejano used
his result to generalize a theorem of R. Schneider (see [26]) related to a
famous problem of Banach.

Let $I_{n+1}$ be the group of isometries of $\mathbb{R}^{n+1}$. Let us denote
by $\Omega$ the space of all convex bodies in $\mathbb{R}^{n+1}$ with the
topology induced by the Hausdorff metric. We consider the (continuous) map
$\tau:\Omega\rightarrow\mathbb{R}^{n+1}$ which assigns to each $C\in\Omega$
the centre of its circumsphere. Let $C_{1}$ and $C_{2}\in\Omega$, we say that
$C_{1}$ is congruent to $C_{2}$ if there is $g\in I_{n+1}$ such that
$g(C_{1})=C_{2}$.

_Proof of Theorem 3._ We denote by $A(x)$ the set $C(K,x)\cap x^{\perp}$. The
map $\alpha:\mathbb{S}^{2k}\rightarrow\Omega$ which assigns to each $x$ the
set $A(x)$, is continuous. Indeed, if $x\in\mathbb{S}^{2k}$ and the sequence
of points $\\{x_{r}\\}\subset\mathbb{S}^{2k}$ is such that $x_{r}\rightarrow
x$, when $r\rightarrow\infty$, then

$S(K,x_{r})\rightarrow S(K,x),\textrm{ }\textrm{ }\textrm{when}\textrm{
}\textrm{ }r\rightarrow\infty.$

On the other side, $x^{\perp}_{r}\rightarrow x^{\perp}$, thus
$A(x_{r})\rightarrow A(x)$. Now we define the set $B(x)$ as
$A(x)+(x-\tau(A(x)))$. Since $B(x)\subset x^{\perp}+x$, the family of convex
bodies $\\{B(x):x\in\mathbb{S}^{2k}\\}$ varies continuously and since all the
bodies $B(x)$ are congruent to $L$, we have found a field of bodies congruent
to $L$. By Mani’s theorem $L$ is a sphere. Now we apply a theorem due to G.
Bianchi and P. Gruber (see [2]) which affirm that if from any point $x$ in the
boundary of a convex body $M$, for a convex body $N\subset\text{int}\,M$ it
holds that $S(N,x)$ is an ellipsoidal cone, then $N$ is an ellipsoid. Now, is
very easy to see that only Euclidean balls have the property stated in the
theorem. Therefore, we conclude that $K$ is a Euclidean ball in
$\mathbb{R}^{2k+1}$. $\Box$

The following theorem is the $2$-dimensional case of Theorem 3, however, its
proof cannot be obtained from the proof of Theorem 3. For this reason, we
include here in a separate statement.

###### Theorem 4

Let $K$ be a convex body contained in the interior of the unit disc of
$\mathbb{R}^{2}$ such that for every unit vector $x\in\mathbb{S}^{1}$ we have
that $C(K,x)\cap x^{\bot}$ is a segment of constant length. Then $K$ is a
Euclidean disc.

_Proof._ Using a similar argument as the used in the proof of Theorem 2, we
obtain that the origin $O$ is contained in the interior of $K$. Suppose first
that $K$ is seen under a constant angle $\varphi$ from every point
$x_{0}\in\mathbb{S}^{1}$. Since the length of every segment
$K_{x_{0}}=C(K,x_{0})\cap x_{0}^{\bot}$ is also constant, we have that the two
angles formed by the support lines of $K$ from $x_{0}$ with the segment
$[x_{0},O]$ are $\alpha$ and $\theta$, for two fixed values $\alpha$ and
$\theta$ such that $\alpha+\theta=\varphi.$ Let $x_{1}\in\mathbb{S}^{1}$ be
the point such that the segment $[x_{1},x_{0}]$ is supporting $K$, as shown in
Fig. 4. Let $b,a$ be points in the line $x_{0}^{\bot}$ such that
$[b,a]=K_{x_{0}}$, and suppose with out loss of generality that
$\measuredangle bx_{0}O=\theta$ and $\measuredangle ax_{0}O=\alpha$. Since the
triangle $\triangle x_{0}Ox_{1}$ is isosceles, we have that $x_{0}^{\bot}$ and
$x_{1}^{\bot}$ make equal angles with $[x_{1},x_{0}]$. If
$[b^{\prime},a^{\prime}]$ is the projection of $K$ over the line
$x_{1}^{\bot}$ from the point $x_{1}$ (as shown in Fig. 4), we have that
$\measuredangle a^{\prime}x_{1}O=\alpha$. Let $x_{2}$ be the point where the
line $x_{1}a^{\prime}$ intersects again to $\mathbb{S}^{1}.$ Continuing this
process, we obtain a sequence of points, $x_{0},$ $x_{1}$, $x_{2}$, $\ldots$,
probably a finite sequence, such that the support function of $K$ takes only
the values $\sin\alpha$ and $\sin\theta.$ No matter what is the initial point
$x_{0}$, the associated sequence of points generates only the values
$\sin\alpha$ and $\sin\theta$ for the support function. Since the support
function of a convex body is a continuous function, we must have that
$\theta=\alpha$ and hence $K$ is a disc.

Now, suppose the angle $\varphi$ is not the same for every
$x\in\mathbb{S}^{1}$, then there is an interval $[\varphi_{1},\varphi_{2}]$,
such that $\varphi$ takes every value in this interval in a continuous way.
Consider a point $x_{0}\in\mathbb{S}^{1}$ such that $K$ is seen under an angle
$\varphi$ from $x_{0}$, with $\frac{\varphi}{\pi}$ an irrational number.
Again, suppose that $\measuredangle bx_{0}O=\theta$ and $\measuredangle
ax_{0}O=\alpha$. This time the sequence of points $x_{0}$, $x_{1},$ $x_{2}$,
$\ldots$, is an infinite sequence, an moreover, the set of points
$\\{x_{0},x_{1},x_{2},\ldots\\}$ is dense in the circle $\mathbb{S}^{1}$. If
follows that the support function of $K$ takes only the values $\sin\alpha$
and $\sin\theta$, and by the continuity of $p$, we must have that
$\theta=\alpha.$ We conclude again, that $K$ is a disc centred at $O$. $\Box$

Figure 4:

###### Remark 3

It would be interesting to know whether or not $K\subset\mathbb{R}^{n}$ is a
ball if we only know that the $(n-1)$-dimensional volume of $C(K,x)\cap
x^{\bot}$, for every $x\in\mathbb{S}^{n-1}$, is equal to a given constant
number $\lambda$.

Although the following result can be obtained by a theorem of G. Bianchi and
P. Gruber, we provide a direct proof here which uses only geometric arguments.
We believe this results goes towards finding of a proof for Conjecture 5.

###### Theorem 5

Let $K,L\subset\mathbb{R}^{n},$ $n\geq 3$, be two convex bodies with
$L\subset\emph{int}\,K$. Suppose that for every $x\in\partial K$ the grazes
$\Sigma(L,x)$ are $(n-1)$-dimensional ellipsoids. Then $L$ is an ellipsoid.

_Proof._ A convex body $L\subset\mathbb{R}^{n}$, $n\geq 3$, is an ellipsoid if
and only if there is a point $q\in\text{int}\,L$ and a fixed integer number
$2\leq k\leq n-1$, such that all its $k$-dimensional sections through $q$ are
ellipsoids. In particular, the above statement holds for $k=2$ and since the
graze of $H\cap L$ from a point $x$, for every $3$-dimensional plane $H$, is
precisely $H\cap\Sigma(L,x)$, we have that it is only need to prove the
theorem for $n=3.$ For the point $q$ we may consider any point in the interior
of $L$.

In order to prove that $L$ is an ellipsoid we are going to prove that all the
shadow boundaries of $L$ are planar, then we apply Blaschke’s theorem
(mentioned at the introduction) and the conclusion follows. Since ellipses are
strictly convex it follows immediately that $L$ is strictly convex. For
$x\in\partial K$ we denote by $\mathcal{E}_{x}$ the ellipse $\Sigma(L,x)$.
Consider a unit vector $u\in\mathbb{S}^{2}$ and denote by $F_{u}$ the family
of all grazes of $L$ contained in planes parallel to $u$. Let
$x_{1}\in\partial K$ such that $\mathcal{E}_{x_{1}}\in F_{u}$ and let $\ell$
be the line of affine symmetry of $\mathcal{E}_{x_{1}}$ corresponding to $u$
(see Fig. 5). Let $\Delta_{1}$ be the plane determined by $x_{1}$ and $\ell$.
Consider any point $c\in\ell\setminus L$ and let $a$, $b$, the contact points
of the supporting lines of $\mathcal{E}_{x_{1}}$ through $c$. Since $\ell$ is
the line of affine symmetry of $\mathcal{E}_{x_{1}}$ corresponding to $u$, the
segment $[a,b]$ is parallel to $u$ and its midpoint $d$ belongs to $\ell$. We
denote by $y$ the point in $L(x,c)\cap\partial K$ ($L(x,c)$ denotes the line
through $x$ and $y$), different of $x$, and denote by $\Omega$ the plane where
$\mathcal{E}_{y}$ is contained. Since the planes $\text{aff}\\{a,x_{1},y\\}$
and $\text{aff}\\{b,x_{1},y\\}$ are supporting planes of $L$, which make
contact with $L$ at $a$ and $b$, respectively, we have that $a$ and $b$ belong
to $\mathcal{E}_{y}$. Thus $\Omega$ is parallel to $u$, i.e.,
$\mathcal{E}_{y}\in F_{u}$. Since $\Omega\cap\text{aff}\\{a,x_{1},y\\}$ and
$\Omega\cap\text{aff}\\{b,x_{1},y\\}$ are supporting lines of
$\mathcal{E}_{y}$ passing through a point $z\in L(x_{1},y)$, the line of
affine symmetry of $\mathcal{E}_{y}$ corresponding to $u$ is the line $L(z,d)$
which is contained in $\Delta_{1}$.

Figure 5:

Now, since $\mathcal{E}_{x_{1}}$ is an ellipse and the segment $[n,m]$ is the
affine diameter of $\mathcal{E}_{x_{1}}$ contained in $\ell$, we have that the
tangent lines to $\mathcal{E}_{x_{1}}$ at the points $n$ and $m$, are parallel
to $u$. It follows that $n,m\in\Delta_{1}\cap\partial L$ are in the shadow
boundary of $L$ with respect to the direction of illumination $u$, i.e., in
$\Lambda(L,u)$. When we vary the point $c$ along the line $\ell$, we obtain
that for every point $e$ of the arc $\widehat{x_{2}w}$ in
$\Delta_{1}\cap\partial K$ (in clockwise sense, with respect to Fig. 5), it
holds that $\mathcal{E}_{e}\in F_{u}$ and consequently the corresponding line
of affine symmetry of $\mathcal{E}_{e}$, with respect to $u$, intersects the
boundary of $L$ in points that belong to $\Lambda(L,u)$. Again, we have that
these points belong to $\Delta_{1}$. In this way we have that there is an arc
in $\Delta_{1}\cap\partial L$ that belong to $\Lambda(L,u)$.

To see that all the points in $\Delta_{1}\cap\partial L$ belong to
$\Lambda(L,u)$, we need to prove that for every point $g$ in the arc
$\widehat{wx_{2}}$ (in clockwise sense) in $\Delta_{1}\cap\partial K$, it
holds that $\mathcal{E}_{g}\in F_{u}$. In order to do this, we consider the
sequence of points $x_{2},$ $x_{3}$, $\ldots$ in $\Delta_{1}\cap\partial K$
(in counter-clockwise sense) such that every segment $[x_{i-1},x_{i}]$ is
tangent to $\Delta_{1}\cap L$. Let $r$ be the minimum integer number such that
$x_{r}$ is in the arc $\widehat{x_{1}w}$ in clockwise sense (see Fig. 6). We
have that for all the points $h$ in the arc $\widehat{x_{2}x_{3}}$, in
counter-clockwise sense, it holds that $\mathcal{E}_{h}\in F_{u}$. We also
have the same property for the points in the arc $\widehat{x_{3}x_{4}}$, and
so on, until we arrive to the arc $\widehat{x_{r-1}x_{r}}$. The union of all
these arcs cover $\Delta_{1}\cap\partial K$, hence, $\Delta_{1}\cap\partial
L=\Lambda(L,u)$ (recall that $L$ is estrictly convex). This happens for every
direction $u$, therefore, by Blaschke’s theorem we conclude that $K$ is an
ellipsoid. $\Box$

Figure 6:

###### Theorem 6

Let $K,L\subset\mathbb{R}^{n},n\geq 3$, be two convex bodies with
$L\subset\emph{int}K$. Suppose that for every $x\in\partial K$ the sets
$S(L,x)\cap L$ and $S(L,x)\cap K\setminus\\{x\\}$ are contained in parallel
hyperplanes. Then $L$ and $K$ are homothetic and concentric ellipsoids.

_Proof._ By the same argument given at the beginning of the proof of Theorem
5, we have that it is only need to prove the theorem for the case of dimension
$n=3$. We will show that the body $L$ is strictly convex. Suppose, to the
contrary, there exists a line segment $[a,b]\subset\partial L$. By virtue that
$L\subset K$ there exist points $x,y\in\partial K$ such that
$L(a,b)\cap\partial K=\\{x,y\\}$. By hypothesis there exist a plane $\Pi$ such
that $S(L,x)\cap L\subset\Pi$. Since $[a,b]\subset S(L,x)\cap L$ it follows
that $[a,b]\subset\Pi$. Consider a line $\tau\subset\Pi$ passing through $x$
such that $\tau\cap L=\emptyset$ and a point $q\in\text{int}\,L\setminus\Pi$.
Now we rotate the plane $\text{aff}\\{q,\tau\\}$ with respect to the line
$\tau$ until we get a plane $\Gamma$, $\Gamma\neq\Pi$, which is a supporting
plane of $L$ and makes contact with $L$ at the point $s$. By the choice of
$\tau$ and $\Gamma$, $s$ is not in $\Pi$. On the other hand, $s$ belongs to
the set $S(L,x)\cap L$, that is, $s\in\Pi$. This contradiction shows that $L$
is strictly convex.

Now, we are going to prove that for every supporting plane $\Pi$ of $K$ the
section $\Pi\cap K$ is centrally symmetric with centre at $\Pi\cap L$. Then by
Theorem O will follow that $L$ and $K$ are homothetic and concentric
ellipsoids.

Let $\Pi$ be any supporting plane of $L$. In order to prove that $\Pi\cap K$
is centrally symmetric we will use the following well known characterization
of central symmetry due to P. Hammer [12]:

_Let $K\subset\mathbb{R}^{n}$, $n\geq 2$, be a convex body and
$q\in\emph{int}\,K$. If for every line $l$ passing through $q$ there exist
parallel supporting hyperplanes of $K$ at the points $l\cap\partial K$, then
$K$ is centrally symmetric with centre at $q$._

Let $l\subset\Pi$ be a line passing through the point $p=\Pi\cap L$, we are
going to prove that there are parallel supporting lines of $\Pi\cap K$ at the
extreme points of the segment $l\cap\Pi\cap K$. We denote such points by $x,y$
and by $\Omega(K,x)$ the intersection $S(L,x)\cap K$ (See Fig. 7). By
hypothesis, there exists two parallel planes $\Delta_{K}$ and $\Delta_{L}$
such that $\Sigma(L,x)\subset\Delta_{L}$ and $\Omega(K,x)\subset\Delta_{K}$.
Since $\Sigma(L,x)$ is a strictly convex curve and $\Omega(K,x)$ is homothetic
to $\Sigma(L,x)$, with respect to the center of homothety $x$, we have that
the line $\ell_{y}=\Pi\cap\Delta_{K}$ is a supporting line of $\Omega(K,x)$ at
the point $y$ and intersects $\Omega(K,x)$ only at the point $y$. We have that
$\ell_{y}$ must be also a supporting line of the section $\Pi\cap K$.
Otherwise, let $z\in\Pi\cap K$ such that $z\in\ell_{y}$ and $z\neq y$, this
imply that $z\in\ell_{y}\cap\Omega(K,x)$, which contradicts that $\ell_{y}$
touches $\Omega(K,x)$ only at $y$. Finally, we observe that $\ell_{y}$ is
parallel to the line $\tau_{y}=\Pi\cap\Delta_{L}$.

Figure 7:

Let $\Delta_{L}^{\prime}$ be the plane that contains to $\Sigma(L,y)$ and let
$\Delta_{K}^{\prime}$ be the plane parallel to $\Delta_{L}^{\prime}$ through
$x$. By the same argument as before, we have that there exists a supporting
line $\ell_{x}$ of $\Pi\cap K$ at $x$ which is given by the intersection
$\Pi\cap\Delta_{K}^{\prime}$ and which is parallel to
$\tau_{x}=\Pi\cap\Delta_{L}^{\prime}$. If $\tau_{x}\neq\tau_{y}$ then there is
another point of intersection between $\Sigma(L,x)$ and $\Sigma(L,y)$, $q\neq
p$. Since $\Sigma(L,x)$ is a strictly convex curve, we have that
$(p,q)\in\text{int}\,L$, however, since the lines $L(y,q)$ and $L(x,q)$ are
supporting lines of $L$ through $q$, we have that the plane
$\text{aff}\,\\{y,q,x\\}$ is a supporting plane of $L$ through $q$. This
implies that $[p,q]\in\partial L$, which contradicts that
$(p,q)\in\text{int}\,L$. Hence, $\tau_{x}=\tau_{y}$, i.e., $\ell_{x}$ is
parallel to $\ell_{y}$. Since the line $l$ through $p$ was taken arbitrarily,
we conclude by Hammer’s theorem that $\Pi\cap K$ has centre of symmetry at
$p$. Therefore, as we said before, $K$ and $L$ are homothetic and concentric
ellipsoids. $\Box$

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