# Diameter, width and thickness in the hyperbolic plane

Ákos G.Horváth  Á. G.Horváth
Department of Geometry
Mathematical Institute
Budapest University of Technology and Economics
H-1521 Budapest
Hungary<EMAIL_ADDRESS>
###### Abstract.

This paper contains a new concept to measure the width and thickness of a
convex body in the hyperbolic plane. We compare the known concepts with the
new one and prove some results on bodies of constant width, constant diameter
and given thickness.

###### Key words and phrases:

constant width property, diameter of a convex set, hyperbolic plane, thickness
of a convex set, width function

###### 2010 Mathematics Subject Classification:

51M09,51M10

##  1\. Introduction

In hyperbolic geometry there are several concepts to measure the breadth of a
convex set. In the first section of this paper we introduce a new idea and
compare it with four known ones. Correspondingly, we define three classes of
bodies, bodies of constant with, bodies of constant diameter and bodies having
the constant shadow property, respectively. In Euclidean space more or less
these classes are agree but in the hyperbolic plane we need to differentiate
them. Among others we find convex compact bodies of constant width which size
essential in the fulfilment of this property (see Statement 7) and others
where the size dependence is non-essential (see Statementst:circle). We prove
that the property of constant diameter follows to the fulfilment of constant
shadow property (see Theorem 2), and both of them are stronger as the property
of constant width (see Theorem 1). In the last part of this paper, we
introduce the thickness of a constant body and prove a variant of Blaschke’s
theorem on the larger circle inscribed to a plane-convex body of given
thickness.

###  1.1. The hyperbolic concepts of width (breadth) introduced earlier

####  1.1.1. $\mathrm{width}_{1}(\cdot)$:

Santaló in his paper [14] developed the following approach. For a convex set
$K$, let $z$ be a point in the boundary of $K$. Let $L_{z}$ be the supporting
line of $K$ passing through $z$ and let $L_{z}^{\bot}$ be the orthogonal line
to $L_{z}$ at the point $z$. Choose another supporting line $L_{v}$ of $K$
that is orthogonal to $L_{z}^{\bot}$, then the _breadth of $K$ corresponding
to $z$_ is the hyperbolic distance between $L_{v}$ and $L_{z}$, i.e., the
length of the segment $L_{z}^{\bot}$ joining the lines $L_{v}$ and $L_{z}$.
Under this approach, the set $K$ is said to have constant width $\lambda$ if
the breadth of $K$ with respect to every point in its boundary is $\lambda$.
The problem with this approach that there is no chance to define ”opposite
points” on the boundary of $K$, and the Euclidean concept of affine diameter
cannot be interpretable.

####  1.1.2. $\mathrm{width}_{2}(\cdot)$:

Historically the next development is due to J. Filmore in [7] who introduced
the following notion. If a set of the hyperbolic plane holds the property that
with every pair $x,y$ of its points it contains the intersection of all
paracyclic (horocyclic) domains which contain $x$ and $y$, then we say that
the set is a h-convex set of the plane. Of course every h-convex set is also
convex in the standard hyperbolic meaning but for example a convex polygon is
not h-convex. Let $K$ be an h-convex set with smooth boundary and fix a point
$O$ in the plane. Consider a half-line through the point $O$ and consider that
pair of paracycles which tangents to $K$ and go through on the ideal point of
the given half-line. The width in the direction of this half-line is the
length of the segment which endpoints are the intersection points of the above
paracycles with the line containing the given half-line. The width in this
approach depends on the point $O$ and cannot measure the sets which convex but
not h-convex one.

####  1.1.3. $\mathrm{width}_{3}(\cdot)$:

K. Leichtweiss introduced a concept in [12] which don’t use touching lines or
paracycles to measure the width of a h-convex set. His idea is to start with
strip regions which are homogeneous in the sense that they admit one-parameter
subgroups of motions, which in the hyperbolic case is the group of
translations correspondingly to a given line. (More precisely, regard those
translations which are the product of two reflections in lines which are
orthogonal to the given line. These translations form a subgroup of motions of
the hyperbolic plane. The strips which invariant against the elements of this
subgroup must be bounded by geodesic parallel curves of constant curvature,
which are the so-called hypercycles (distance curves) of the given line.) Thus
it is quite natural to define the width a horocyclic convex curve $C$ in a
certain direction by means of the width of the supporting strip region of this
direction. In this approach the condition of h-convexity also needs because
the curvature of the boundary points of the strip region is less than the
curvature of paracycles but again it is non-zero. Consequently, this method
cannot be applied for polygons, too. On the other hand it is a strong
connection with the notions of spherical geometry and we can define
appropriate signed support function $h(\theta)$ for a h-convex set like in the
classical convex geometry. The width $w(\theta)$ in a direction holds the
equality $w(\theta)=h(\theta)+h(\theta+\pi)$. Observe that these functions
depends on the origin and so the ”width of the h-convex set in a direction”
doesn’t an inner property of the set, like the earlier concept of Filmore.
Advantage of this building up that in the case of h-convex sets we can say
about ”opposite points” and ”affine diameters” as in the Euclidean situation.

####  1.1.4. $\mathrm{width}_{4}(\cdot)$:

Recent approach in the literature is in the paper of J. Jeronimo-Castro and F.
G. Jimenez-Lopez [6]. The authors first introduce the notion of opposite
points for an arbitrary compact convex body $K$ of the hyperbolic plane. We
translate the definition to the model-free language of the hyperbolic plane.
Transform with an isometry the point $z$ of the boundary of $K$ to a given
(and fixed) point $O$ of the plane such a way that a given (and fixed) line
$l$ of the plane through $O$ will be a supporting (tangent) line $L_{z}$ of
the image. In the second step search a point $w$ on the boundary of $K$ for
which a support line $L_{w}$ at $w$ has equal alternate angle with the segment
$[z,w]$ as the angle between $[z,w]$ and $L_{z}$. By definition such a pair
$z$ and $w$ of points are _opposite points_ with respect to $K$ and the
segment $[z,w]$ is an _isometric diameter_. The existence at least one
opposite pair to a given $z$ can be seen immediately from the Poincare’s disk
model. Similarly, it can be seen easily that if $K$ smooth and h-convex $z$
uniquely determines its opposite pair $w$. Let denote by $L_{z,w}$ the line of
the segment $[z,w]$. Since $L_{z,w}$ has equal alternate angles with $L_{z}$
and $L_{w}$ these latter are ultraparallel lines (non intersected and non-
parallel) implying that, there is an unique distance of it. This distance is
the width of the body corresponding to the isometric diameter $[z,w]$. Note
that in this approach there is no such width function which for every
direction uniquely determines a real number. It is possible, that for the same
boundary point $z$ we find two opposite points $w^{\prime}$ and
$w^{\prime\prime}$ for which the corresponding width are distinct to each
other. But immediately can be defined the notion of _curve of constant width_
with the property that all widths let be the same value with respect to all
pairs of opposite points.

Our method gives a synthesization of the above concepts preserving the
important properties of those ones.

###  1.2. Comparison of the known concepts of width

First we observe that we can compare the Santaló’s breadth
$\mathrm{width}_{1}(z)$ with $\mathrm{width}_{4}(z)$ of Jeronimo-Castro and
Jimenez-Lopez. Both of them corresponds the value of width to the points of
the boundary of $K$. Concentrating to the method of [6], if the boundary of
$K$ smooth and horocyclic convex, a point on the boundary has an unique
opposite pair and the width with respect to this point can be defined as the
distance of the opposite (in this sence) tangent lines.

######  Statement 1.

Let $K$ be a convex compact body of the plane with smooth and h-convex
boundary. Then we have

(1) $\mathrm{width}_{4}(z)\geq\mathrm{width}_{1}(z).$

Equality holds at that point $z$ where the normal is a double-normal.

###### Proof.

The breadth of [14] can be seen as the distance of the tangent at the points
$z$ and $v$, while the width of [6] can be found as the distance of the lines
$L_{z}$ and $L_{w}$ (see Fig. 1). From synthetic absolute geometry (see in
[1]) we know that this latter quantity attend at the perpendicular to $L_{z}$
from the midpoint $M$ of the segment $[z,w]$.

Figure 1. Comparison of Santaló width function to the width of Jerónimo-Castro
and Jimenez-Lopez

In the figure it is the length of the segment $[T_{z},T_{w}]$. Since the
points $w$ and $z$ are in distinct half-planes of the line $T_{z}T_{w}$. the
boundary of $K$ intersects the segment $[T_{z}T_{w}]$ in two points, one of
them (say $N$) is on the segment $[M,T_{w}]$. Note that the lines $L_{w}$ and
$L_{z}^{\bot}$ either non-intersecting or intersect each other in a point $L$.
In the second case, the quadrangle with vertices $L$,$z$,$T_{z}$ and $T_{w}$
is a Saccheri quadrangle implying that $T_{w}Lz_{\angle}$ is acute angle. From
this follows that the touching point $v$ – which needed to the determination
of the Santalo’s breadth – is on the arc $wNz$. This also true in the first
case, hence $L_{v}$ intersects the segment $[N,T_{w}]$ in a point $T$. This
provides the following inequalities:

(2)
$\mathrm{width}_{4}(z)=|[T_{z},T_{w}]|\geq|[T,T_{z}]|\geq|[K,z]|=\mathrm{dist}(L_{z},L_{v})=\mathrm{width}_{1}(z).$

(The second inequality follows from the fact that the distance of two lines is
less or equal to the length of another transversalis of them.) Clearly in the
case of equality we have $z=T_{z}$, $w=v=T_{w}$ and the isometric diameter
$[z,w]$ is a double normal of the boundary of $K$. ∎

Our second note that there is a simple comparison of $\mathrm{width}_{2}$ and
$\mathrm{width}_{3}$. As we can see on Fig. 2 these functions correspond to a
direction of the plane defined by a half-line starting from a given point $O$
of the plane. Here we assume that the origin $O$ is in the interior of the
body $K$. (We can consider that the origin is an arbitrary point of the plane,
but more comfortable if the getting support functions have only positive
values, respectively. In the case of Fillmore’s paper this follows from the
fact that the domain of the investigated paracycles contains $K$ and
Leichtweiss directly assumed that the origin is in the interior of $K$.)

######  Statement 2.

Let $K$ be a horocycle-convex compact body of the plane with the origin in its
interior. Then we have

(3) $\mathrm{width}_{2}(\theta)\geq\mathrm{width}_{3}(\theta).$

Moreover, equality holds if and only if the touching points $z$ and $v$ are
the endpoints of a double-normal of the boundary of $K$.

Figure 2. Connection between the width function of Fillmore and the width
function of Leichtweiss

###### Proof.

Let denote by $h_{2}(\cdot)$ and $h_{3}(\cdot)$ the corresponding support
functions. If $\theta$ any parameter determining the direction of the half-
line $\overrightarrow{OP}$ then
$\mathrm{width}_{2}(\theta)=h_{2}(\theta)+h_{2}(\theta+\pi)=|OP|+|OQ|$. In
Fig.2 $p$ is that paracycle which touches $K$. One of its axis is
$\overrightarrow{OP}$. The angle $\theta$ also corresponds to a value of the
support function $h_{3}(\theta)$ of paper [12]. Since it should be the
distance of a hypercycle of the leading line $l$, which is perpendicular to
the direction $\overrightarrow{OP}$ we have to consider the hypercycle $h$
which goes through the point $P$. However, $h$ doesn’t touch $K$, because its
curvature is less than the curvature of the paracycle $p$. This means that
$h_{3}(\theta)\leq h_{2}(\theta)$ consequently
$\mathrm{width}_{3}(\theta)\leq\mathrm{width}_{2}(\theta)$ as we stated.
Equality holds for a $\theta$ if and only if $P=z$ and $Q=v$ meaning that the
normals at the touching points go through the origin $O$, respectively. Hence
the segment $[z,v]$ is a double-normal of the boundary of $K$ (and so it is
also an isometric-diameter). ∎

##  2\. A new width function of the hyperbolic plane

In this section, we propose a new definition of the width function. We
eliminate that dependence of the width function on the origin, which is
present in the concept $\mathrm{width}_{3}$. Using again hypercycles we define
such strips which leading-lines don’t go through a fix origin. Denote by
$\mathcal{I}$ the set of ideal points. (To name ”ideal point” in our
terminology means a point on the boundary of a disk model. In a model-
independent approach an ideal points is such a new object which corresponds to
a congruence class of hyperbolic parallel half-lines.)

######  Definition 1.

Let $O$ be a point in the interior of the convex compact body $K$. Let
$\overrightarrow{OX}$ be a half-line with ideal point $X$ and consider a line
$YX$ where $Y\in\mathcal{I}$, $YX$ is parallel to $\overrightarrow{OX}$ and
intersects $K$. Then there are two hypercycles corresponding to the leading-
line $YX$, which touch $K$ and the strip between them contains $K$. The
distance of these distance-curves are $d_{K}^{+}(YX)$ and $d_{K}^{-}(YX)$,
respectively. Let denote by $d_{K}(YX):=d_{K}^{+}(YX)+d_{K}^{-}(YX)$. The
_width of $K$ in the direction of $X$_ is

(4) $\mathrm{width}_{K}(X):=\sup\\{d_{K}(YX):Y\in\mathcal{I}\\}.$

It follows by a standard compactness argument, that for an ideal point $X$
there is at least one $Y^{\star}$ such that
$\mathrm{width}_{K}(X)=d_{K}(Y^{\star}X)$. So the supremum in (4) is a
maximum.

######  Remark 1.

In Euclidean geometry the width in a direction means the needed size of a gate
on which we can push across the body along the given direction (without any
rotation). In hyperbolic geometry, we imagine this gate at an ideal point. We
can push the body along several distinct half-lines corresponding to this
point. The gate has to be enough big to that, the passage through the gate
should be successful regardless of which route we used. This explains why we
chose supremum instead of infimum in our definition.

######  Statement 3.

For every ideal point $X$ holds the inequality:
$\mathrm{width}_{K}(X)\geq\mathrm{width}_{3}(X)$. Moreover, for every
$X\in\mathcal{I}$ we can chose a point $z(X)$ on the boundary of $K$ for which
holds the inequality $\mathrm{width}_{K}(X)\geq\mathrm{width}_{1}(z(X))$. If
$K$ is strictly convex and smooth then there is a homeomorphism
$\varphi:\mathcal{I}\longrightarrow\mathrm{bd}(K)$ $\varphi(X)=z(X)$ such that
$\mathrm{width}_{K}(X)\geq\mathrm{width}_{1}(\varphi(X))$.

###### Proof.

It is clear, that if $O^{\prime}$ any point of the interior of $K$ then all
half-line $\overrightarrow{O^{\prime}X}$ is a subset of such a line $YX$ which
is parallel to $\overrightarrow{OX}$. Hence
$\mathrm{width}_{K}(X)\geq\mathrm{width}_{3}(\overrightarrow{OX})$ holds with
respect to any pair of $O$ and $\overrightarrow{OX}$.

Figure 3. Comparisation of the width function of Santaló with
$\mathrm{width}_{K}(\cdot)$

For a given direction $\overrightarrow{OX}$ let $Y$ such that $YX$ touches $K$
at the point $v$. One of the corresponding hypercycles is the line $YX$ and
the other touches the boundary of $K$ at the point $z$. Now the distance of
the point $z$ to $YX$ is $d_{K}(YX)$ which also represents
$\mathrm{width}_{1}(z)$ (see Fig.3). Hence if we correspond to the direction
$\overrightarrow{OX}$ the above point $z(X)$ on the boundary of $K$ we get
that $\mathrm{width}_{1}(z(X))\leq\mathrm{width}_{K}(X)$.

To prove the last statement consider a Cayley-Klein model. In this model the
smooth and strictly convex body has the same property. Obviously we can assume
that $K$ contains the origin $O$ of the model. For a point $X$ of the model
circle we can associate two points, which are the tangent points of the
tangent lines of $K$ through $X$. We start from that point $X_{0}$ of the
model circle for which
$\overrightarrow{OX_{0})$isthepositivehalfoftheaxis$x$ofanEuclideancoordinatesystem.Let$\varphi(X_{0}}$
that tangent point for which the line $\varphi(X_{0})X_{0}$ has positive
directional tangent. Hence the triangle $O\varphi(X_{0}){X_{0}}_{\triangle}$
has positive (counterclockwise) direction and by definition we choose that
tangent point as the $\varphi(X)$ image of $X$ for which this property holds.
Clearly our conditions on the boundary of $K$ guarantees that $\varphi$ is a
homeomorphism.

∎

We have also an immediate connection between $\mathrm{width}_{4}(z)$ and our
function $\mathrm{width}_{K}(X)$.

######  Statement 4.

Let $K$ be a h-convex compact body. For every point $X\in\mathcal{I}$ there is
a point $Y\in\mathcal{I}$ such that the supporting strip of the leading line
$YX$ touches $K$ in the endpoints $z,w$ of an isometric diagonal. Conversely,
for every isometric diameter $[z,w]$ there is a line $YX$ with
$X,Y\in\mathcal{I}$ for which the respective touching points of the supporting
strip of $YX$ are $z$ and $w$. Moreover for corresponding pairs
$X,Y\in\mathcal{I}$ and $z,w\in\mathrm{bd}(K)$ holds
$\mathrm{width}_{K}(X)\geq
d_{K}(YX)\geq\mathrm{width}_{4}(z)=\mathrm{width}_{4}(w)$, with equality in
the second inequality if and only if the isometric diameter $[z,w]$ is a
double-normal.

###### Proof.

Let $Y\in\mathcal{I}$ such that holds the equality
$d_{K}^{+}(YX)=d_{K}^{-}(YX)$. (The continuity property of the distance
function implies the existence of such line $YX$.) Denote by $z$ and $w$ a
pair of touching points lying in the distinct bounding hypercycles. Then $z$
and $w$ are opposite points in the sense of [6]. In fact, the triangles
$zT_{z}M_{\triangle}$ and $wT_{w}M_{\triangle}$ in the left hand figure of
Fig. 4 are congruent so the chord $[z,w]$ has equal alternate angle with the
corresponding tangent lines $t_{z}$ and $t_{w}$. Since these lines disjoint to
this strip that their distance is greater than the width of it. Equality holds
if and only if the isometric diameter $[z,w]$ is a double-normal of $K$.

Figure 4. Illustration of the proof of Statement4

Conversely, we prove that every isometric diameter of a h-convex body can be
get with the above method. Precisely, if $[z,w]$ is an isometric diameter
there is a line $YX$ for which $\mathrm{dist}(z,XY)=\mathrm{dist}(w,XY)$. See
the picture on the right hand side of Fig. 4. This shows the isometric
diameter $[z,w]$ in the Poincare’s disk model. Let $X$ and $Y$ be the
respective intersections of the Euclidean tangent line at the point $w$ and
the model circle. Then the Euclidean line $XY$ represents an hypercycle of the
hyperbolic plane corresponding to the leading hyperbolic line with ideal
points $X$ and $Y$. Using an inversion to the absolute (to the circle of ideal
points) we get from this hypercycle its opposite pair which has the same
distance from the hyperbolic line $YX$. Inversion sends the Euclidean line
$YX$ to a circle which goes through the origin, the intersections with the
model circle are $X$ and $Y$ and tangents $K$ at $z$. The only question is
that this strip contains or not the body $K$. If $K$ is a h-convex body then
its curvature at the point $z$ is greater than the curvature of the hypercycle
$YzX$ and the strip contains the body, as we stated. ∎

Note that if $K$ is a polygon the two endpoints of a double-normal typically
cannot be considered to the touching pair of points of a strip which is
bounded by hypercycles of the same leading line.

We call a strip _symmetric to its leading line_ if the boundary hypercycles
have the same distance from the leading line.

######  Corollary 1.

Let $K$ be a h-convex compact body. Assume that the strip with leading line
$YX$ touches $K$ at the points $z$ and $w$, respectively. From the proof of
Lemma 4 we can see easily that the following properties are equivalent to each
other:

  * •

$[z,w]$ is a double normal of $K$

  * •

$[z,w]$ is orthogonal to $YX$.

When this orthogonality property holds the chord $[z,w]$ is an isometric
diameter. If $[z,w]$ isn’t orthogonal to $YX$ then it is an isometric diameter
if and only if the strip symmetric with respect to the leading line $YX$.

######  Statement 5.

Let $AB$ be a segment of length $d$ and take its figure in a projective model
on the horizontal diameter in symmetric position (see Fig.5). The
corresponding ideal points of the line $AB$ are $X_{{AB}}$ and $X_{{BA}}$,
respectively. Denote by $X_{1},X_{2}\in\mathcal{I}$ those ideal points for
which $X_{1}BA_{\angle}=X_{2}AB_{\angle}=\pi/2$. We say that the ideal points
$X^{\prime}$ and $X^{\prime\prime}$ holds the inequality
$X^{\prime}<X^{\prime\prime}$, if there are on the upper half of the model and
in counterclockwise direction the point $X^{\prime}$ precedes the point
$X^{\prime\prime}$. Finally, denote by $\alpha$ the angle $XAB_{\angle}$ for
$X<X_{1}$ and by $\beta$ the angle $XBA_{\angle}$ if $X>X_{2}$. Then the width
function of the segment $AB$ is

(5) $\mathrm{width}_{AB}(X)=\left\\{\begin{array}[]{ccc}0&\mbox{ if
}&X=X_{AB}\\\ \sinh^{-1}\left(\frac{\sin\alpha\sinh d}{\cosh d-\cos\alpha\sinh
d}\right)&\mbox{ if }&X_{AB}<X\leq X_{1}\\\ d&\mbox{ if }&X_{1}<X\leq X_{2}\\\
\sinh^{-1}\left(\frac{\sin\beta\sinh d}{\cosh d-\cos\beta\sinh
d}\right)&\mbox{ if }&X_{2}<X<X_{BA}\\\ 0&\mbox{ if
}&X=X_{BA}.\end{array}\right.$

The width function can be extend symmetrically to the other directions of the
plane.

###### Proof.

Let $K$ be the closed segment $AB$. Let $X\in\mathcal{I}$ arbitrary ideal
point and take those lines $YX$ which belong to the sector bounding by $AX$
and $BX$. The triangle $ABX_{\triangle}$ has one ideal vertex at $X$ so it can
be determined by the length $d=AB$ and the angle $\alpha=BAX_{\angle}$. Assume
that $BAX_{\angle}\leq ABX_{\angle}$. A line $YX$ should take into
consideration if its angle $XMB_{\angle}$ is greater than $\alpha$ and is less
or equal to $XB{X_{AB}}_{\angle}$ (see Fig. 5).

Figure 5.

Clearly, for fixed point $X$ we have to determine the sum of the distances
$d_{A}+d_{B}$ of the points $A$ and $B$ from the line $YX$ and have to take
the supremum of these values varying the length $t=|AM|$ between the bounds
$0$ and $d$. To this we introduce some notation. Let $X^{\star}$ be the upper
ideal point of the bisector of the segment $AB$. Then the parallel angle
$\alpha^{\star}$ holds the equality

$\cos\alpha^{\star}=\tanh\frac{d}{2}.$

We have two symmetries in the calculation, the first one is with respect to
the line of $AB$ and the second one is with respect to the bisector of $AB$,
meaning that we can assume that $\alpha\leq\alpha^{\star}$. Denote by $t$ the
length of the segment $AM$ and consider a perpendicular from $M$ to $AX$. Its
length is $m$. This segment divides the angle $AMX_{\angle}$ into two parts;
one of them say $\gamma$ is the second acute angle of a right triangle, the
other one is the parallel angle $\mu$ corresponding to the distance $m$. Hence
we have

$\cos\mu=\tanh m\quad\quad\sin\mu=\frac{1}{\cosh m}.$

From the right triangle $AMT_{\triangle}$ we get the equalities

$\cos\gamma=\frac{\tanh m}{\tanh
t}\quad\quad\sin\gamma=\frac{\sqrt{\tanh^{2}t-\tanh^{2}m}}{\tanh t}.$

By the further notation of the figure, we get

$\sin\delta=\sin(\pi-\delta)=\frac{\tanh m}{\cosh m\tanh t}+\frac{\tanh
m\sqrt{\tanh^{2}t-\tanh^{2}m}}{\tanh t}.$

Finally, from the right triangle $AMT_{\triangle}$ we get

$\sinh m=\sin\alpha\sinh t\quad\cosh
m=\sqrt{1+\sin^{2}\alpha\sinh^{2}t}\quad\tanh m=\frac{\sin\alpha\sinh
t}{\sqrt{1+\sin^{2}\alpha\sinh^{2}t}}.$

Now we can deduce the following formula

$\sin\delta=\frac{\sin\alpha(\cosh t+\cos\alpha\sinh
t)}{1+\sin^{2}\alpha\sinh^{2}t}=\frac{\sin\alpha}{\cosh t-\cos\alpha\sinh t}.$

Our function is

(6) $f(t):=d_{A}+d_{B}=\sinh^{-1}\left(\sin\delta\sinh
t\right)+\sinh^{-1}\left(\sin\delta\sinh(d-t)\right)=$
$\sinh^{-1}\frac{\sin\alpha\sinh t}{\cosh t-\cos\alpha\sinh
t}+\sinh^{-1}\frac{\sin\alpha\sinh(d-t)}{\cosh t-\cos\alpha\sinh t}.$

The derivative of $f(t)$ is

$f^{\prime}(t)=\frac{\sin\alpha}{\left(\cosh t-\cos\alpha\sinh
t\right)\sqrt{\left(\cosh t-\cos\alpha\sinh
t\right)^{2}+\sin^{2}\alpha\sinh^{2}t}}+$ $\frac{\sin\alpha\left(-\cosh
d+\cos\alpha\sinh d\right)}{\left(\cosh t-\cos\alpha\sinh
t\right)\sqrt{\left(\cosh t-\cos\alpha\sinh
t\right)^{2}+\sin^{2}\alpha\sinh^{2}(d-t)}}.$

Since $\cosh t-\cos\alpha\sinh t>0$ holds for all positive $t$ and also we
assumed that $\sin\alpha>0$, the condition $f^{\prime}(t)=0$ is equivalent to
the following equality:

$\cosh d-\cos\alpha\sinh d=\frac{\sqrt{\left(\cosh t-\cos\alpha\sinh
t\right)^{2}+\sin^{2}\alpha\sinh^{2}(d-t)}}{\sqrt{\left(\cosh
t-\cos\alpha\sinh t\right)^{2}+\sin^{2}\alpha\sinh^{2}t}}.$

Introducing the notation $x(t):=\cosh t-\cos\alpha\sinh t$ we can write

$x^{2}(d)-1=\frac{\sin^{2}\alpha\left(\sinh^{2}(d-t)-\sinh^{2}t\right)}{x^{2}(t)+\sin^{2}\alpha\sinh^{2}t}.$

This leads to

$\left(\sinh^{2}d(1+\cos^{2}\alpha)-\cos\alpha\sinh(2d)\right)\left(\cosh
2t-\cos\alpha\sinh 2t\right)=$
$\sin^{2}\alpha\left(\sinh^{2}d\cosh(2t)-\sinh(2t)\sinh d\cosh d\right),$

from which straightforward calculation shows the following equivalent form:

$0=\left(\cosh d-\cos\alpha\sinh
d\right)\left((1+\cos^{2}\alpha)\sinh(2t)-2\cos\alpha\cosh(2t)\right).$

Since the first term is always positive we get that the only zero is attend at

$\tanh(2t)=\frac{2\cos\alpha}{1+\cos^{2}\alpha}.$

From this $\tanh t=\cos\alpha$, therefore the function $f(t)$ has one maximum
at the point $t=\tanh^{-1}\left(\cos\alpha\right)$. It can be seen that at
this value $\delta=\pi/2$, so the triangle $AXM_{\triangle}$ is a right one.
Of course, the fix value $d$ sometimes less than that $t$ where the maximum is
attend, in this case from that $t$’s which we allowed, $t=d$ produces the
larger widthness. ∎

##  3\. Bodies of constant width

One of the most interesting question with respect to the width function is the
characterization problem of bodies of constant width. In this section we
investigate this question. First we determine the with function of an
important body, the with function of a symmetric bounded hypercycle domain.

######  Statement 6.

The width function of the symmetric hypercycle domain $ABCD$ is a combination
of the width functions of the ”diagonal chords” $AC$ and $BD$. Using the
notation of Fig.6 we have on the arc $X_{0}\leq X\leq X^{\prime}$ the
following function

(7)
$\mathrm{width}_{K}(X)=\max\\{\mathrm{width}_{AC}(X),\mathrm{width}_{BD}(X)\\}=\left\\{\begin{array}[]{ccc}\sinh^{-1}\frac{\tanh
h(\cosh a+\sinh a)}{\sqrt{\cosh^{2}a\cosh^{2}h-1}}&\mbox{ if }&X_{0}<X\leq
X_{1}\\\ d&\mbox{ if }&X_{1}<X\leq X_{2}\\\
\tanh^{-1}\cos(\gamma+\delta)&\mbox{ if
}&X_{2}<X<X^{\prime}\end{array}\right..$

The complete function has rotational symmetry of order 4.

###### Proof.

Let $K$ be a convex, compact body and let $X$ be a direction. Then we have

$\mathrm{width}_{K}(X)=\sup\\{\mathrm{width}_{AB}(X)\quad:\quad A,B\in K\\}.$

From this we can deduce the width function of a symmetric hypercycle domain
$D$. Let its height $2h$ meaning that its leading line has distance $h$ from
the points of the hypercycle arcs, respectively (see the left picture in
Fig.6).

Figure 6.

The chord $FG$ on the leading line has length $2a$. Clearly $AC$ and $BD$ are
the longest chords in the domain, they common length is
$d=2\cosh^{-1}\left(\cosh a\cosh h\right)$. Observe that any hypercycle domain
which contains these two segments also contains those segments which one
endpoint is $AB$ and the other one is in $CD$. Hence we have to concentrate
those segments which ends lying in the hypercycle arcs. From those ideal
points which are on the dashed arcs on Fig.6 the value of the width is $d$ by
Lemma5. The question is the following: What can we say about the widthness
from the points on the bold arcs of the model circle? Clearly, we have
$\mathrm{width}_{D}(X_{0})\geq 2h$. On the other hand we can calculate the
width of the segment $BD$ from this direction, too. In fact, Lemma5 says, that

$\mathrm{width}_{BD}(X)=\sinh^{-1}\left(\frac{\sin\alpha\sinh d}{\cosh
d-\cos\alpha\sinh d}\right),$

where $\alpha+\beta$ is the parallel-angle of the distance $h$ and
$\sin\beta=\frac{\sinh a}{\sinh d/2}$. Since $\cos(\alpha+\beta)=\tanh h$ we
get

$\tanh
h=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos\alpha\sqrt{1-\frac{\sinh^{2}a}{\sinh^{2}\frac{d}{2}}}-\sin\alpha\frac{\sinh
a}{\sinh\frac{d}{2}}.$

By the hyperbolic theorem of cosine $\cosh\frac{d}{2}=\cosh a\cosh h$ and we
get that

$\sin\alpha=\frac{\tanh h(\cosh a-\sinh a)}{\sinh\frac{d}{2}},$

which gives that

$\cos\alpha=\frac{\cosh a\sinh^{2}h+\sinh a}{\cosh h\sinh\frac{d}{2}}.$

We can use formula (5) to determine the width:

$\mathrm{width}_{BD}(X_{0})=\sinh^{-1}\frac{\sin\alpha\sinh d}{\cosh
d-\cos\alpha\sinh d}.$

Long but straightforward calculation shows that

$\mathrm{width}_{BD}(X_{0})=\sinh^{-1}\frac{\tanh h(\cosh a+\sinh
a)}{\sinh\frac{d}{2}}=\sinh^{-1}\frac{\tanh h(\cosh a+\sinh
a)}{\sqrt{\cosh^{2}a\cosh^{2}h-1}}.$

For fixed value $h$ we can use the above formula to determine the width of the
chord $B(t)D(t)$ as a function of $a(t)$. The least possible value of $a(t)$
attend at that $a^{\prime}$ when the corresponding
${B^{\prime}D^{\prime}X_{0}}_{\angle}$ is the parallel angle of the value
$d^{\prime}$. From this we get

$\frac{\tanh h(\cosh a^{\prime}-\sinh
a^{\prime})}{\sinh\frac{d^{\prime}}{2}}=\sin\alpha^{\prime}=\frac{1}{\cosh
d^{\prime}}=\frac{1}{\cosh^{2}\frac{d^{\prime}}{2}+\sinh^{2}\frac{d^{\prime}}{2}}.$

If $a(t)\leq a^{\prime}$ from (5) we get that

$d(t)=\mathrm{width}_{B(t)D(t)}(X_{0})\leq\sinh^{-1}\frac{\tanh h(\cosh
a^{\prime}+\sinh
a^{\prime})}{\sinh\frac{d^{\prime}}{2}}=\sinh^{-1}\left(\frac{1}{\cosh
d^{\prime}}+2\frac{\sinh a^{\prime}}{\sinh\frac{d^{\prime}}{2}}\right).$

The width a chord from $X_{0}$ which connects two points of the opposite
hypercycle is equal to the width from $X_{0}$ of such a chord which goes
through $O$. So we get

$\mathrm{width}_{K}(X_{0})=\mathrm{width}_{BD}(X_{0})=\sup\left\\{\mathrm{width}_{B(t)D(t)}(X_{0}):0\leq
a(t)\leq a\right\\}.$

The same true for the other points of the arc $X_{0}X_{1}$.

Assume that $X_{2}<X<X_{3}$ holds for the point $X$ (see the right picture in
Fig. 6). We can observe that from these points the width of $K$ is always
equal to the width of one of the segments $AC$ and $BD$. On the figure, we
drew that situation when $\mathrm{width}_{BD}(X)=BT$ implying the equality

$\tanh(\mathrm{width}_{BD}(X))=\tanh BT=\cos(\gamma+\delta),$

where we used the notation of the figure. This proves the statement. ∎

Figure 7. Bodies of constant width

######  Definition 2.

We say that the compact convex body $K$ has _constant width_ if there is a
$\lambda\in\mathbb{R}^{+}$ that for every ideal point $X$
$\mathrm{width}_{K}(X)=\lambda$ holds.

######  Remark 2.

There is an interesting consequence of the Statement 6. When the arcs
$\\{X:X_{8}\leq X\leq X_{1}\\}$ and $\\{X:X_{2}\leq X\leq X_{3}\\}$ degenerate
to a point the width function of the hypercycle domain has constant value. In
this case, $\tanh(d/2)=\cos\pi/4=\sqrt{2}/{2}$, from which
$\cosh^{2}d/2-1=\sinh^{2}d/2=(1/2)\cosh^{2}d/2$ implying that $\cosh
d/2=\sqrt{2}$ and the unique value of the width function is
$d=2\cosh^{-1}(\sqrt{2})=2\ln(\sqrt{2}+1)$. Since $\sqrt{2}=\cosh a\cosh h$
and also $\sinh h=\sqrt{2}/{2}\sinh d/2=\sqrt{2}/{2}$ we get that
$a=\cosh^{-1}\sqrt{\frac{4}{3}}$ and $h=\cosh^{-1}\sqrt{\frac{3}{2}}$. This
phenomenon exclude the characterization of bodies of constant width, because
each convex body belonging to a circular disk of diameter $d=2\ln(\sqrt{2}+1)$
and containing the endpoints of a pair of orthogonal diameters is a convex
body of constant width. (E.g. see in Fig. 7 the circle, the quadrangle with
equal angles and sides, and the body which bounded by the dotted curve,
respectively.)

The following statement generalises this observation.

######  Statement 7.

Let $k\geq 2$ an integer number. Assume that the radius $r$ of the
circumscribed circle of a regular $n$-polygon $P$ holds

$\tanh r\geq\tanh
r_{\min}:=\left\\{\begin{array}[]{ccc}\frac{-1+\sqrt{1+4\sin\frac{\pi}{2k-1}\cot\frac{(k-1)\pi}{2k-1}}}{2\cos\frac{(k-1)\pi}{2k-1}}&\mbox{
if }&n=2k-1,\\\ &&\\\
\frac{1}{\sqrt{\cos^{2}\frac{(k-1)\pi}{2k}+\tan^{2}\frac{(k-1)\pi}{2k}}}&\mbox{
if }&n=2k.\end{array}\right.$

Then $P$ is a body of constant width $d$, where

$\sinh\frac{d}{2}=\left\\{\begin{array}[]{ccc}\frac{\sinh
r}{\sin\frac{(k-1)\pi}{2k-1}}&\mbox{ if }&n=2k-1,\\\ &&\\\ \frac{\sinh
r}{\sin\frac{(k-1)\pi}{2k}}&\mbox{ if }&n=2k.\end{array}\right.$

###### Proof.

If $n=2k$ is an even number then we can use the train of though of Remark 2.

Figure 8. Regular $2k$-polygon of constant width

Put the polygon into the center of a projective model with a large radius
similarly as you can see on Fig. 8. Consider those pairs of support lines of
$P$ which go through the opposite vertices of a diameter. This $2k$ bands
intersect the boundary of the model those ideal points from which at least one
diameter has width $d$. We determine the radius of the model such a way that
the getting arcs cover the circle but not overlap. The common points are
$X_{i}$ $i=1,\cdots 2k$. The endpoints of one support line are
$X_{i}X_{i+n-1}$ where we take the indices modulo $2k$. The central angle of
such a line is $\frac{(k-1)}{k}\pi$ thus the parallel angle corresponds to
$d/2$ is $\frac{(k-1)}{2k}\pi$. Hence

$\tanh\frac{d}{2}=\cos\frac{(k-1)\pi}{2k}$

from which we get that

$\tanh d=\frac{2\cos\frac{(k-1)\pi}{2k}}{1+\cos^{2}\frac{(k-1)\pi}{2k}}$

The corresponding radius $r$ of the circumscribed circle holds:

$\sinh
r=\frac{\sinh\frac{d}{2}}{\sin\frac{(k-1)\pi}{2k}}=\frac{\frac{\tanh\frac{d}{2}}{\sqrt{1-\tanh^{2}\frac{d}{2}}}}{\sin\frac{(k-1)\pi}{2k}}=\frac{\cos\frac{(k-1)\pi}{2k}}{\sin^{2}\frac{(k-1)\pi}{2k}},$

or

$\tanh
r=\frac{\cos\frac{(k-1)\pi}{2k}}{\sqrt{\sin^{4}\frac{(k-1)\pi}{2k}+\cos^{2}\frac{(k-1)\pi}{2k}}}=\frac{\cos\frac{(k-1)\pi}{2k}}{\sqrt{1-\sin^{2}\frac{(k-1)\pi}{2k}\cos^{2}\frac{(k-1)\pi}{2k}}}=\frac{1}{\sqrt{\cos^{2}\frac{(k-1)\pi}{2k}+\tan^{2}\frac{(k-1)\pi}{2k}}}.$

The odd case is more complicated. A diametral chord doesn’t contain the
origin. It determines a set of ideal points from which the widthness of the
chord is $d$. In the model these sets are the union of two non-congruent arcs,
the longest one is in that half-plane of the chord which contains the origin.
We guarantee the constant widthness property if the rotated copies ($n$-times
by the angle $2\pi/n$) of these sets cover the whole boundary of the model. If
the vertices of the polygon are $P_{1},\ldots,P_{2k-1}$, then the diametral
chords are the segments $P_{i}P_{i+k-1}$ respectively. The diametral chords
are rotated copies of each other by the integer multipliers of the angle
$2\pi/(2k-1)$ around the origin. (See Fig. 9.)

Figure 9. Regular $2k-1$-polygon of constant width

Denote by $X_{1}X_{2}\cup X_{2k-1}X_{2k}$ and $X_{3}X_{4}\cup
X_{2k+1}X_{2k+2}$ the sets of ideal points from which the diametral chords
$P_{1}P_{k}$ and $P_{2}P_{k+1}$ can be seen in their common real length $d$.
When $X_{2k-1}X_{2k}$ and $X_{2k+1}X_{2k+2}$ overlap the rotational symmetry
guarantees that the polygon is a body of constant width $d$, since for every
ideal points one of the diametral chord can be seen in its real length $d$.
Thus we have to determine the central angle $X_{2k-1}{OX_{2k}}_{\angle}$. The
minimal value of $r$ occurs when $X_{2k-1}{OX_{2k}}_{\angle}=2\pi/(2k-1)$.

Since $P_{3}OQ_{\angle}=\frac{(k-1)\pi}{2k-1}$ and the radius of the model is
$1$, the Euclidean distance of the diametral chords to the origin is
$OT=\cos\frac{(k-1)\pi}{2k-1}\tanh r$. The pole of the chord containing the
segment $P_{1}P_{k}$ is $Q$ so $OQ=(\cos\frac{(k-1)\pi}{2k-1}\tanh r)^{-1}$.
Let denote by $\alpha=\frac{(k-1)\pi}{2k-1}$ and $\beta:=OQ{P_{k}}_{\angle}$,
respectively. Since $Q$, $X_{2}$ and $X_{2k-1}$ are collinear points we have
$QX_{2}\cdot QX_{2k-1}=OQ^{2}$ where $QX_{2}$ and $QX_{2k-1}$ are the roots of
the polynomial equation of second order:

$0=x^{2}-2OQ\cdot x\cos\beta+OQ^{2}-1.$

Hence the two roots are $x_{1,2}=OQ\cos\beta\pm\sqrt{1-\sin^{2}\beta OQ^{2}}$,
so $QX_{2k-1}=OQ\cos\beta+\sqrt{1-\sin^{2}\beta OQ^{2}}$. Now if the half of
the central angle of the arc $X_{2k-1}X_{2k}$ is equal to the half of the
central angle $P_{1}{OP_{2}}_{\angle}=\frac{2\pi}{2k-1}=\frac{\alpha}{k-1}$,
then we get the equality:

$\sin\frac{\alpha}{k-1}=\sin(\pi-\frac{\alpha}{k-1})=QX_{2k-1}\sin\beta=\sin\beta\left(OQ\cos\beta+\sqrt{1-\sin^{2}\beta
OQ^{2}}\right).$

From this equality we can determine again $OQ$ and get

$0=OQ^{2}-2\sin\frac{\alpha}{k-1}\cot\beta
OQ+\frac{\sin^{2}\frac{\alpha}{k-1}}{\sin^{2}\beta}-1,$

from which

$OQ=\sin\frac{\alpha}{k-1}\cot\beta\pm\cos\frac{\alpha}{k-1}.$

From the Euclidean right triangle ${TQP_{k}}_{\triangle}$ we have that

$\cot\beta=\frac{OQ-OT}{\sin\alpha\tanh r}=\frac{\frac{1}{\cos\alpha\tanh
r}-\cos\alpha\tanh r}{\sin\alpha\tanh
r}=\frac{1-\cos^{2}\alpha\tanh^{2}r}{\cos\alpha\sin\alpha\tanh^{2}r},$

and get the equality

$\frac{1}{\tanh
r\cos\alpha}=\sin\frac{\alpha}{k-1}\frac{1-\cos^{2}\alpha\tanh^{2}r}{\cos\alpha\sin\alpha\tanh^{2}r}\pm\cos\frac{\alpha}{k-1}.$

From this equation we get either the equality $(\star)$:

$(\star)\qquad 0=\tanh^{2}r\sin\frac{(k-2)\alpha}{k-1}\cos\alpha-\tanh
r\sin\alpha+\sin\frac{\alpha}{k-1}$

or the equality $(\star\star)$:

$(\star\star)\qquad 0=\tanh^{2}r\sin\frac{k\alpha}{k-1}\cos\alpha+\tanh
r\sin\alpha+\sin\frac{\alpha}{k-1}.$

The corresponding solutions are

$(\star)\qquad\tanh
r=\frac{\sin\alpha\pm\sqrt{\sin^{2}\alpha-4\sin\frac{(k-2)\alpha}{k-1}\sin\frac{\alpha}{k-1}\cos\alpha}}{2\sin\frac{(k-2)\alpha}{k-1}\cos\alpha}$

and

$(\star\star)\qquad\tanh
r=\frac{-\sin\alpha\pm\sqrt{\sin^{2}\alpha+4\sin\frac{k\alpha}{k-1}\sin\frac{\alpha}{k-1}\cos\alpha}}{2\sin\frac{k\alpha}{k-1}\cos\alpha},$

respectively. Substituting $\alpha=\frac{(k-1)\pi}{2k-1}$ we get

$(\star)\qquad\tanh
r=\frac{\sin\frac{(k-1)\pi}{2k-1}\pm\sqrt{\sin^{2}\frac{(k-1)\pi}{2k-1}-4\sin\frac{(k-2)\pi}{2k-1}\sin\frac{\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}}{2\sin\frac{(k-2)\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}$

and

$(\star\star)\qquad\tanh
r=\frac{-\sin\frac{(k-1)\pi}{2k-1}\pm\sqrt{\sin^{2}\frac{(k-1)\pi}{2k-1}+4\sin\frac{k\pi}{2k-1}\sin\frac{\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}}{2\sin\frac{k\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}.$

Among these solutions there is only one which have to take into consideration.
In $(\star)$ (by the inequality $\tanh r<1$) we need that

$\left(\sin\alpha-2\sin\frac{(k-2)\alpha}{k-1}\cos\alpha\right)<\mp\sqrt{\sin^{2}\alpha-4\sin\frac{(k-2)\alpha}{k-1}\sin\frac{\alpha}{k-1}\cos\alpha},$

from which the inequality:

$-4\sin\alpha\sin\frac{(k-2)\alpha}{k-1}\sin\frac{\alpha}{k-1}\cos\alpha+4\sin^{2}\frac{(k-2)\alpha}{k-1}\cos^{2}\alpha<-4\sin\frac{(k-2)\alpha}{k-1}\sin\frac{\alpha}{k-1}\cos\alpha$

holds. Simplifying it we get

$\sin\frac{(k-2)\alpha}{k-1}\cos\alpha<\sin\alpha\sin\frac{\alpha}{k-1}-\sin\frac{\alpha}{k-1}=\left(\sin\alpha-1\right)\sin\frac{\alpha}{k-1},$

which gives a contradiction, because the right hand side is negative and the
left hand side is positive. Similarly $\tanh r>0$ implies that in
$(\star\star)$ we have to take into consideration only the sign plus. Hence we
have that

$\tanh
r=\frac{-\sin\frac{(k-1)\pi}{2k-1}+\sqrt{\sin^{2}\frac{(k-1)\pi}{2k-1}+4\sin\frac{k\pi}{2k-1}\sin\frac{\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}}{2\sin\frac{k\pi}{2k-1}\cos\frac{(k-1)\pi}{2k-1}}=\frac{-1+\sqrt{1+4\sin\frac{\pi}{2k-1}\cot\frac{(k-1)\pi}{2k-1}}}{2\cos\frac{(k-1)\pi}{2k-1}},$

proving the last statement. ∎

######  Remark 3.

For $k=2$ we have

$\tanh
r_{\min}:=\left\\{\begin{array}[]{ccc}\frac{-1+\sqrt{1+4\sin\frac{\pi}{3}\cot\frac{\pi}{3}}}{2\cos\frac{\pi}{3}}=\sqrt{3}-1&\mbox{
if }&n=3,\\\ &&\\\
\frac{1}{\sqrt{\cos^{2}\frac{\pi}{4}+\tan^{2}\frac{\pi}{4}}}=\sqrt{\frac{2}{3}}&\mbox{
if }&n=4,\end{array}\right.$

and the corresponding values of $d_{\min}$ are

$\sinh
d_{\min}=\left\\{\begin{array}[]{ccc}\frac{2\sqrt{3}-2}{\sqrt{6\sqrt{3}-9}}=1.2408&\mbox{
if }&n=3,\\\ &&\\\ 2&\mbox{ if }&n=4.\end{array}\right.$

Observe that the respective functions $\tanh r_{\min}(2k-1)$ and $\tanh
r_{\min}(2k)$ of $k$ are strictly decreasing tends to zero in the variable
$k$. The question is that there is or not a value of $k$ when the difference
of the two series change their sign or for every pairs of successive integers
$2k-1$ and $2k$ this difference is negative. This question can be answered by
computer easily. The first six values of the two functions can be found in the
table below showing that the merged series is monotone decreasing up to the
index $9$.

n | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12
---|---|---|---|---|---|---|---|---|---|---
$\tanh r$ | 0.7321 | 0.8165 | 0.5309 | 0.5547 | 0.4080 | 0.4091 | 0.3286 | 0.3233 | 0.2739 | 0.2673
Table 1. The sequence of the radiuses.

An interesting phenomenon of constant width property that it depends also on
the size of a body, not only on the combinatorial and symmetry properties of
it. In spite of that there are bodies of constant width which size is non-
essential in this point of view.

######  Statement 8.

The hyperbolic circle and the hyperbolic Reuleaux polygons are convex bodies
of constant width independently on their sides.

###### Proof.

Consider the projective or Cayley-Klein disk model of the hyperbolic plane and
consider a hyperbolic circle which center is the centre of the model disk.
This is an Euclidean circle in the model. Every leading line $YX$ is a chord
of the model and the corresponding hypercycles are such half-ellipses which
are tangent to the model circle at the points $Y$ and $X$, respectively; and
touching the given circle (see the left hand side of Fig. 10).

Figure 10. Hyperbolic circles and Realeaux polygons are bodies of constant
width

From the rotational symmetry of the figure we can see that for all line $YX$
we get the same width $d_{K}(YX)$ which is equal to the hyperbolic diameter of
the circle. From this the statement is obvious for the circle.

Consider the case of the Realeaux polygons. From the definition (see e.g. [14]
p.410) follows that the maximal distance $d$ of its points reach at such a
pair of points that one of them is a vertex of the polygon and the other lies
on the opposite arc of the boundary. Let now $YX$ a leading line and let be
the corresponding hypercyles $h^{+}$ and $h^{-}$ with the touching points $z$
and $w$, respectively (see the right figure in Fig. 10). If $z$ is a vertex
and $w$ lies on the opposite arc then there is a normal line at $w$ which also
perpendicular to $YX$ and go through $z$. This means that $d_{K}(YX)=d$. On
the other hand if $x$ and $y$ are two smooth points of contact of a strip
corresponding to a leading line $ZW$ then the distinct normals at $x$ and $y$
are orthogonal to the line $ZW$. But in a Realeaux polygon every normal goes
through the opposite vertex hence every two normals are intersect each other.
This is a contradiction because there is no common perpendicular of
intersecting lines. Hence $\mathrm{width}_{K}(X)=d$ for all $X\in\mathcal{I}$
implying that Realeaux-polygons are bodies of constant width. ∎

##  4\. Diametral chords and the bodies of constant diameter.

On the usual way we define the _diameter_ of a convex body $K$ as the greater
distance between its two points.

######  Statement 9.

The diameter of the compact, convex body $K$ is

$\mathrm{diam}(K)=\sup\\{\mathrm{width}_{K}(X):\quad X\in\mathcal{I}\\}.$

The supremum is a maximum, so there is at least one point
$X^{\star}\in\mathcal{I}$ such that
$\mathrm{diam}(K)=\mathrm{width}_{K}(X^{\star})$.

###### Proof.

Let $z$, $w$ be such points of the boundary of $K$ for which
$\mathrm{diam}(K)=d(z,w)$. Clearly, those lines $t^{+}$ and $t^{-}$ which are
orthogonal to the chord $[z,w]$ at $z$ and $w$ are supporting lines of $K$. If
$YX^{\star}$ a line which orthogonally intersects $[z,w]$ and $h^{+}$ and
$h^{-}$ are the distance curves corresponding to $YX^{\star}$ through $z$ and
$w$, respectively, then $d_{K}(YX^{\star})=d(z,w)$. Hence
$\mathrm{width}_{K}(X^{\star})\geq d_{K}(YX^{\star})=d(z,w)=\mathrm{diam}(K)$.
Assume now that $\mathrm{width}_{K}(X^{\star})>\mathrm{diam}(K)$. Then there
is an $Y^{\star}\in\mathcal{I}$ such that
$\mathrm{diam}(K)<d_{K}(Y^{\star}X^{\star})$. Let $\\{z^{\star},w^{\star}\\}$
be a touching pair of points in the intersection of the boundary of $K$ and
that strip which distance is $d_{K}(Y^{\star}X^{\star})$. Obviously,
$d(z^{\star},w^{\star})\geq d_{K}(Y^{\star}X^{\star})$ yielding a
contradiction. Hence in fact,
$\mathrm{diam}(K)=\sup\\{\mathrm{width}_{K}(X):\quad X\in\mathcal{I}\\}$, as
we stated. ∎

The above proof shows that the endpoints of a diametral chord are such points
of $\mathrm{bd}K$ in which the normals are double-normals. This fact is
trivial in all such spaces where orthogonality is defined by metric property.
There are three papers of P.V.Araújo on _metric width_ of the bodies of the
hyperbolic plane (see [2],[3], [4]) in which the concept of curves of constant
width was seriously investigated. The author said that a _plane-convex body is
of constant with $\delta$_ if its diameter is equal to $\delta$ and each point
of the boundary is an endpoint of such a chord which length is equal to
$\delta$. We call such a chord to _diametral chords_ in this paper. In another
terminology the _constant diameter_ property was rather used. We use also this
latter phrase. The most important properties of constant diameter property
(was proved first by Araújo) we now collect in a statement:

######  Statement 10 ([2],[3]).

A body of constant diameter of the hyperbolic plane holds the properties:

  * •

Every diametral chord is a double normal of the boundary of the body, cutting
the boundary curve orthogonally at both ends (more precisely, the
perpendicular line at each extremity of each diametral chord is a line of
support of the body).

  * •

Every chord of that is orthogonal to the boundary at one end is a diametral
chord, and therefore it is a double normal.

  * •

Any two distinct diametral chords must intersect each other.

Araújo also introduced the concepts of _track function_ and _intersection
function_ , respectively. He fixed a diametral chord and denoted by $\theta$
the oriented angle of another diametral chord with the fixed one (see the
oriented angle $CMB$ in Fig. 11).

Figure 11. The properties of bodies with constant diameter

When a point from the right end of the fixed chord goes along the boundary
then the corresponding diametral chord change its position and angle,
respectively. Using the notation of the figure we can see that the boundary
curve can be determined by the distance of the point of intersection $M$ to
the right end $C$ (which is the value $x(\theta)$ of the track function) and
the distance $MB$ which is the value of the intersection function $f(\theta)$.
It was proven in [4] that the original chord $AC$ can be chosen such a way
that the following properties hold:

  * •

Both the track function and the intersection function are continuous on each
interval $(0;\pi)$ and $(\pi;2\pi)$ and for $\theta\in(0,\pi)$ hold that
$x(\theta+\pi)=x(\theta)$, $0\leq f(\theta)\leq d$ and
$f(\theta)+f(\theta+\pi)=d$.

  * •

The diametral chord $AC$ can be chosen in such away that the continuity
property of these functions extended to the interval $(0,2\pi)$ giving that
they are can be extended periodically such that the above properties remain
valid on $\mathbb{R}$.

  * •

For $C^{3}$ boundary between the track and intersection function there is the
connection $f^{\prime}(\theta)=-x^{\prime}(\theta)\cos\theta$ implying that
the intersection function determines the track function up to the choice of a
constant.

There is a nice recent result of M. Lassak in [11] which connects the
spherical concepts of constant width and constant diameter. He proved that a
convex body on the two-dimensional sphere of diameter $\delta$ is less than
$\pi/2$ is of constant diameter $\delta$ if and only if it is of constant
width $\delta$. As we saw in Remark 2 this is not true in the hyperbolic
plane, the quadrangle of this remark is a body of constant width but it has
only two diametral chords $AC$ and $BD$, respectively. We now prove the
reversal of this statement.

######  Theorem 1.

If the body $K$ is a convex body of constant diameter $d$ then it is a body of
constant width $d$.

###### Proof.

Since the diameter is the maximum value of the width function of $K$ we have
to prove that there is no such direction from which the width of $K$ is less
than the diameter $d$. Let $AC$ be a diametral chord with endpoints $A$, $C$
and length $d$. Let $YX$ be any line which intersects $AC$ orthogonally. Since
$AC$ is a double normal
$d\geq\mathrm{width}_{K}(X)\geq\mathrm{width}_{AC}(X)=d$. Let
$\mathcal{X}(AC)$ be the set of ideal points from which the width of $AC$ is
equal to $d$. We should prove that $\cup\\{\mathcal{X}(AC)\,:\,AC\mbox{ is a
diametral chord }\\}=\mathcal{I}$. Observe that the tangent lines at the
smooth points of the boundary continuously change their position such that the
function $\theta(s)$ parameterized by the arc-lengths is strictly increasing
and at a corner points each support line determines a diametral chord which is
orthogonal to it and one end is the given corner point. From this latter
follows that for each $\theta$ there is an unique support line of the body
such that the intersection points of the support line with a great circle
containing the body produce a continuous bijection between them. This means
that the union of the ideal points of the support lines already cover the
whole $\mathcal{X}$ implying that it is even more true for
$\cup\\{\mathcal{I}(AC)\,:\,AC\mbox{ is a diametral chord }\\}$. ∎

We saw that without further condition a body of constant width is not a body
of constant diameter. To give a positive result in this a direction we
introduce a new concept.

######  Definition 3.

We say that a convex, compact body has the _constant shadow property_ if there
is a positive number $d$ such that $d_{K}(YX)=d$ for all line $YX$ which
intersects the body $K$.

Of course a body which has the constant shadow property with the number $d$
then it is a body of constant width with the same $d$. A circle is a body with
this property, hence this class is not empty. We prove the following:

######  Theorem 2.

The convex, compact body $K$ of the hyperbolic plane has the constant shadow
property with the number $d$ then it is a body of constant diameter $d$.

###### Proof.

We prove that a body of constant shadow property is a body of constant
diameter. Let $d=d_{K}(YX)$, where $YX\cap K\neq\emptyset$. Then the diameter
of $K$ is less or equal to $d$. Really, if $\mathrm{diam}K=d^{\star}>d$ then
there is a chord $PQ$ for which $|PQ|=d^{\star}$. Let $YX$ is a line which
orthogonally intersects the segment $PQ$. Obviously $d_{K}(YX)\geq
d^{\star}>d$ which is a contradiction.

Consider a point $P$ on the boundary of $K$. Let $t$ be the supporting line of
$K$ at $P$; $PQ$ be a chord of $K$ which orthogonal to $t$ and $YX$ be an
intersecting line of $K$ perpendicular to $PQ$. Since $d_{K}(YX)=d$ we have
two possibilities. If $PQ$ is a double-normal (at $Q$ it is orthogonal to a
supporting line of $K$ through $Q$) then its length is $d$ and it is a chord
from $P$ with length $d$. If $PQ$ isn’t a double normal
$d=d_{K}(YX)<|PQ^{\star}|$, where $Q^{\star}$ is a common point of the another
branch of the hypercycle domain of minimal width containing $K$ with
fundamental line $YX$. This is impossible because $PQ^{\star}$ is longest
chord as the diameters of $K$. ∎

######  Corollary 2.

For a convex compact body of the plane equivalent the following two properties

  * •

It has the constant shadow property with number $d$.

  * •

All diameter of $K$ is a double-normal of $K$ in the general meaning. (We
extend the concept of normal to a non-smooth point of the boundary such that
it is a line which goes through the point and orthogonal to a supporting line
of $K$ at this point.)

Theorem 2 and the above equivalence implies the following theorem:

######  Theorem 3.

If $K$ is a plane convex compact body of constant shadow property then every
normal of $K$ is a double-normal.

Since by Theorem 3 in [6] the property that every normal of $K$ is a double-
normal is equivalent to the property that $K$ is a body of constant width with
respect to the function $\mathrm{widht}_{4}(\cdot)$, we can deduce that the
property of constant shadow knows this property of constant width of Jeronimo-
Castro paper.

Further analogises of this theorem well-known in Euclidean geometry (see [15]
and the references therein). in the hyperbolic plane this statement has been
proved for the function $\mathrm{widht}_{3}(\cdot)$ in [12] (Theorem 3.7), and
without proof was used for $\mathrm{widht}_{1}(\cdot)$ in [14] (Section 7).

##  5\. The thickness (or width) of a hyperbolic domain

Usual the width of a convex body $K$ is the minimal width of it with respect
to the directions of the plane. In the hyperbolic plane this property can be
understood in two ways:

  * •

In the first possibility we collect the possible values of the width function
and from these values we create a common one for example taking their infimum
(this leads to the usual definition of width of $K$;

  * •

in the second possibility we consider the intersecting lines to an independent
”directions” (see the concept of constant shadow property), and from these
values create the infimum.

These two methods lead to the same result, if in the first one we change the
concept of width with respect to a direction to something-like thickness of
the body changing the supremum to infimum in this definition.

######  Definition 4.

Let the _thickness_ of the body $K$ in the direction $X$ is

$\mathrm{thi}_{K}(X):=\inf\\{d_{K}(YX)\,:\quad YX\cap K\neq\emptyset\\}$

and let _the thickness of $K$_ ($\mathrm{thi}(K)$) be the infimum of the
values $\mathrm{thi}_{K}(X)$ take into consideration all direction
$X\in\mathcal{I}$.

By compactness argument we can see easily that there is at least one
$X^{\star}\in\mathcal{I}$, such that
$\mathrm{thi}_{K}(X^{\star})=\mathrm{thi}(K)$.

Note that the concept of the thickness for a convex hyperbolic body differs
from the concept of the usual width. The quadrangle $ABCD$ in Fig. 7 is a body
of constant width with $d=2\ln(\sqrt{2}+1)$ but its thickness is less or equal
to $2h=2\ln\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)<d$ (see Remark
2). Unfortunately, it is also possible that a chord which represents the
thickness of the body isn’t a double normal. Consider Fig. 12., where we can
see a plane-convex body which is the union of two regular triangles (forming a
quadrangle with equal sides), two quoters of circle and two quoters of an
ellipse. We denoted by dotted lines the bounding hypercycles of the hypercycle
domain. It can be seen easily that the thickness of this body is represented
by a strip containing the points $u,z$, and $v,w$, respectively.

Figure 12. A chord which represents the smaller strip distance isn’t a double-
normal.

Now the respective chords joining the corresponding touching points are
$[u,v]$, $[u,w]$, $[z,v]$ and $[z,w]$ but non of them could be double-normal.
On the other hand if the ellipse and the hypercycle at $w$ have a common
tangent then the chords $[w,w]^{\prime}$ and $[z,z]^{\prime}$ are double-
normals, respectively. This shows that there are chords which are double-
normals and also are isometric diameters (their endpoints are opposite points
in the meaning of [6]) but their lengths smaller than the thickness of $K$.
This phenomenon is differ from the standard idea of the geometry of a convex
body. From Statement 4 we can see that the reason is the definition of
convexity and for h-convex bodies this phenomenon isn’t perceptible. An
interesting question that a shorter isometric diameter of a h-convex body is
or isn’t a double normal, and what is the connection with the thickness of the
body. Presently we don’t know the answer.

To verify that the definition was chosen appropriately we prove a version of
Blaschke’s theorem in the hyperbolic plane. This theorem says that every
convex compact body in the Euclidean plane of thickness $1$ contains a circle
of radius $1/3$ (see example 17 in [5] or p.287 in [13]). For h-convex bodies
we can prove an analogous of this result only . Since in hyperbolic plane
there is no similarity, our result depends not only the thickness but also the
diameter of the body.

######  Theorem 4.

Let $K$ be a h-convex body of the hyperbolic plane with diameter $d$ and
thickness $t$. Then it contains a circle of radius $r$ for which

(8) $\tanh r=\frac{\sinh t\cosh t}{3\cosh^{2}d+\sinh^{2}t}.$

We note that if we use a first order approximation of the hyperbolic functions
this formula leads to the Euclidean one, namely we get $r=t/3$.

###### Proof.

Consider a boundary point $P$ of $K$ and a point $Q\in K$. The half-line
$\overrightarrow{PQ}$ intersects the boundary of $K$ at the point $Q^{\star}$.
Let $K_{P}$ be the set containing those points $Q$ of $K$ for which
$\tanh|PQ|\leq\left(1-\frac{1}{3\cosh^{2}d}\right)\tanh|PQ^{\star}|$. We prove
that $K_{P}$ is a convex set. Observe that a shrinking map which any point $X$
of the plane sends to that point $X^{\prime}$ of the half-line $PX$ for which
$\tanh|PX^{\prime}|=\lambda\tanh|PX|$ ($0<\lambda\leq 1$) is a collinearity
which maps to a line to another line. In fact, if the line $l$ given for which
the point $P$ doesn’t belong to $l$, then we can consider that point $T$ of it
for which $PT$ is perpendicular to $l$. Let another point of the line $X$, and
consider the triangles $PTX_{\triangle}$ and
$PT^{\prime}X^{\prime}_{\triangle}$, respectively. Let denote the angle
$TPX_{\angle}$ by $\alpha$. Then $\cos\alpha=\tanh PT/\tanh PX$ since the
first triangle is a right one. But $\tanh PT/\tanh PX=\tanh PT^{\prime}/\tanh
PX^{\prime}$ so the second triangle is a right triangle. Hence the points
$X^{\prime}$ are on the line through the point $T^{\prime}$ and orthogonal to
the line $PT$. Hence this mapping takes a convex set into a convex set.
Consider three points $A,B,C$ on the boundary of $K$ and the sets $K_{A}$,
$K_{B}$ and $K_{C}$, respectively. We prove that $K_{A}\cap K_{B}\cap K_{C}$
is non-empty. Consider the median point $M$ of the triangle $ABC_{\triangle}$.
(If $M_{A},M_{B}$ and $M_{C}$ are the middle point of the sides $BC$, $AC$ and
$AB$, respectively, then (by Menelaos theorem) the medians $MM_{A}$, $MM_{B}$
and $MM_{C}$ intersect each other in a point $M$.) Since the median point $M$
holds the equality

$\frac{\sinh AM}{\sinh MM_{A}}=2\cosh\frac{a}{2}=\frac{\cosh BM+\cosh
CM}{\cosh MM_{A}}$

(see Theorem 4.3 in [8], or Theorem 1.2 in [9] and the manuscript [10]
containing the proofs of these statements), we get

$\frac{\tanh AM}{\tanh MM_{A}}=\frac{\sinh AM\cosh MM_{A}}{\sinh MM_{A}\cosh
AM}=\frac{\cosh BM+\cosh CM}{\cosh AM}$

On the other hand we have that

$\frac{\tanh AM}{\tanh AM_{A}}=\frac{\tanh AM+\tanh^{2}AM\tanh MM_{A}}{\tanh
AM+\tanh MM_{A}}=\frac{\frac{\tanh AM}{\tanh MM_{A}}+\tanh^{2}AM}{\frac{\tanh
AM}{\tanh MM_{A}}+1}=1-\frac{1-\tanh^{2}AM}{\frac{\tanh AM}{\tanh MM_{A}}+1}=$
$=1-\frac{1}{\cosh^{2}AM\left(\frac{\tanh AM}{\tanh
MM_{A}}+1\right)}=1-\frac{1}{\cosh AM\left(\cosh AM+\cosh BM+\cosh
CM\right)}\leq 1-\frac{1}{3\cosh^{2}d}.$

Since Helly’s theorem on convex, compact sets of the hyperbolic plane is also
valid we get that there is a point $O$ in the body $K$ that for each chord
$PP^{\prime}$ through $O$ has the property $\tanh
PO\leq(1-\frac{1}{3\cosh^{2}d})\tanh PP^{\prime}$.

We prove that the circle with center $O$ and radius $r$ in Eq. 8 lies in $K$.
Let $P$ be a boundary point of $K$. Since $K$ is a h-convex compact body there
is a strip symmetric to its leading line such that one of its bounding
hypercycle tangents $K$ at $P$ (see Statement 4 and Corollary 1). The other
side of the strip supports $K$ at $Q$. Now consider the chord $QR$ which goes
through $O$, and a line $l$ through $O$ which perpendicular to the generator
$YX$ of the strip.

Figure 13. The proof of Theorem 4.

$l$ intersects the bounding hypercycles and the line $YX$ at the points
$L_{P}$,$L_{Q}$ and $T$ respectively (see Fig. 13). $S$ and $V$ the
intersection of $QR$ with the hypercycle (is going through $P$) and the line
$YX$, respectively. The orthogonal projections of the points $Q$ and $S$ to
the line $YX$ are $U$ and $W$. Using the hyperbolic theorem of sines we get

$\sin OVT_{\angle}=\frac{\sinh QU}{\sinh QV}=\frac{\sinh OT}{\sinh
OV}=\frac{\sinh SW}{\sinh SV}$

Since $L_{Q}O=OT+UQ$ and $L_{Q}L_{P}=SW+UQ$ we get

$\frac{\tanh L_{Q}O}{\tanh
L_{Q}L_{P}}=\frac{\tanh(OT+UQ)}{\tanh(SW+UQ)}=\frac{(\tanh OT+\tanh
UQ)(1+\tanh SW\tanh UQ)}{(\tanh SW+\tanh UQ)(1+\tanh OT\tanh UQ)}=$
$=\frac{(\sinh OT\cosh UQ+\sinh UQ\cosh OT)(\cosh SW\cosh UQ+\sinh SW\sinh
UQ)}{(\sinh SW\cosh UQ+\sinh UQ\cosh SW)(\cosh OT\cosh UQ+\sinh OT\sinh UQ)}=$
$=\frac{(\sinh OV\cosh UQ+\sinh QV\cosh OT)(\cosh SW\cosh UQ+\sinh SV\sinh
QV)}{(\sinh SV\cosh UQ+\sinh QV\cosh SW)(\cosh OT\cosh UQ+\sinh OV\sinh
QV)}\leq$ $\frac{\max\\{\cosh UV,\cosh VT,\cosh VW\\}}{\min\\{\cosh UV,\cosh
VT,\cosh VW\\}}\frac{\tanh QO}{\tanh QS}\leq\frac{\max\\{\cosh UV,\cosh
VW\\}}{\min\\{\cosh UV,\cosh VW\\}}\frac{\tanh QO}{\tanh QR}=\frac{\tanh
QO}{\tanh QR}.$

(The last equality follows from the fact, that the strip symmetric to its
leading line, hence $UV=VW$.) Hence we get

$\tanh L_{Q}O\leq\frac{\tanh QO}{\tanh QR}\tanh L_{Q}L_{P}\leq c\tanh
L_{Q}L_{P},$

where

$c:=\min\frac{\tanh QO}{\tanh QR}=\left(1-\frac{1}{3\cosh^{2}d}\right).$

This implies

$\tanh OP\geq\tanh OL_{P}=\tanh(L_{Q}L_{P}-L_{Q}O)=\frac{\tanh
L_{Q}L_{P}-\tanh L_{Q}O}{1-\tanh L_{Q}L_{P}\tanh L_{Q}O}\geq\frac{(1-c)\tanh
L_{Q}L_{P}}{1-c\tanh^{2}L_{Q}L_{P}}\geq$ $\geq\frac{(1-c)\tanh
t}{1-c\tanh^{2}t}=\frac{\frac{1}{3\cosh^{2}d}}{1-(1-\frac{1}{3\cosh^{2}d})\tanh^{2}t}\tanh
t=\frac{\tanh t}{3\cosh^{2}d(1-\tanh^{2}t)+\tanh^{2}t}=$ $\frac{\tanh
t\cosh^{2}t}{3\cosh^{2}d+\sinh^{2}t}=\frac{\sinh t\cosh
t}{3\cosh^{2}d+\sinh^{2}t},$

because $\tanh(x-y)$ is a strictly decreasing function of $y$ and
$f(x)=\frac{(1-c)\tanh x}{1-c\tanh^{2}x}$ is a strictly increasing functions
of $x$. Since the distance of an arbitrary boundary point from $O$ is at least
the required value, the statement is proved. ∎

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