# On the area of ordinary hyperbolic reduced polygons

Ádám Sagmeister  Alfréd Rényi Institute of Mathematics, Reáltanoda u. 13-15,
H-1053 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

A convex body $R$ in the hyperbolic plane is reduced if any convex body
$K\subset R$ has a smaller minimal width than $R$. We examine the area of a
family of hyperbolic reduced $n$-gons, and prove that, within this family,
regular $n$-gons have minimal area.

###### Key words and phrases:

convex geometry, hyperbolic geometry, minimal width, thickness, reduced
bodies, reduced polygons

###### 2010 Mathematics Subject Classification:

Primary: 51M09, 51M10, 52A55

In loving memory of my grandmother.

##  1\. Introduction

The concept of reducedness was introduced by Heil [4] in 1978 motivated by
some volume minimizing problems. A convex body (i. e. a convex compact set of
non-empty interior) $K$ is called reduced if an arbitrary convex body
$L\subsetneq K$ strictly contained in $K$ has smaller minimal width than $K$.
Pál [20] proved in 1921 that for fixed minimal width, the regular triangle has
minimal area among convex bodies in the Euclidean plane. This result of his is
also known as the isominwidth inequality. The same problem in higher
dimensions remains open, as there are no reduced simplices in
${\mathbb{R}}^{n}$ for $n\geq 3$ (see [18, 19]), therefore there are no really
good candidates for the volume minimizing problems – so far the best one in
${\mathbb{R}}^{3}$ is the so-called Heil body, which has a smaller volume than
any rotationally symmetric body of the same minimal width. The problem can be
naturally generalized to other spaces, a natural approach is to examine the
problem in spaces of constant curvature. Bezdek and Blekherman [1] proved
that, if the minimal width is at most $\frac{\pi}{2}$, the regular triangle
minimizes the area in $S^{2}$. However, for spherical bodies of larger minimal
width, the minimizers of the isominwidth problem are polars of Reuleaux
triangles. Surprisingly enough, there is no solution of the isominwidth
problem in the hyperbolic space for arbitrary dimension (this is part of an
ongoing work joint with K. J. Böröczky and A. Freyer).

A reverse isominwidth problem is about finding the maximal volume if the
minimal width is fixed. Naturally, this problem does not have a maximizer in
general, but for reduced bodies we can ask for the convex body that maximizes
the volume. However, in ${\mathbb{R}}^{3}$, the diameter of a reduced body of
a given minimal width can be arbitrarily large, and hence the Euclidean
problem is only interesting on the plane. It is conjectured, that the unique
planar reduced bodies maximizing the area and of minimal width $w>0$ in
${\mathbb{R}}^{2}$ are the circular disk of radius $\frac{w}{2}$ and the
quarter of the disk of radius $w$. A big step towards the proof of this
conjecture was made by Lassak, who proved that among reduced $k$-gons, regular
ones maximize the area, and as a consequence, all reduced polygons have
smaller area than the circle. Following Lassak’s footsteps, the same
conclusion was derived in $S^{2}$ by Liu, Chan and Su [17]. Interestingly
enough, the characterization of hyperbolic reduced polygons is still unclear,
but clearly it must be different from the Euclidean and spherical ones; there
exist reduced rhombi on the hyperbolic plane, while Euclidean and spherical
reduced polygons are all odd-gons (see Lassak [8, 10]). However, the so-called
ordinary reduced polygons can be examined the same way (these are odd-gons
whose vertices have distance equal to the minimal width of the polygon from
the opposite sides such that the projection of the vertices to these sides are
in the relative interior of the sides). We have the following main result.

######  Theorem 1.1.

On the hyperbolic plane among ordinary reduced $n$-gons of minimal width $w$,
regular $n$-gons of the same minimal width have the smallest area.

##  2\. Preliminaries

We use the notation $H^{2}$ for the hyperbolic plane, which is equipped with
the geodesic metric. The geodesic distance of two points $x,y\in H^{2}$ will
be denoted as $d\left(x,y\right)$. In this section, we introduce hyperbolic
convexity. Many of the concepts are identical with their Euclidean analogues,
but as we will soon see, there are exceptions.

For a subset $X$ of the hyperbolic plane $H^{2}$, we say that $X$ is _convex_
, if for any pair of points $x$ and $y$, the unique geodesic segment
$\left[x,y\right]$ connecting $x$ and $y$ is a subset of $X$ (where
$\left[x,x\right]=\left\\{x\right\\}$). A _convex body_ is a convex compact
set of non-empty interior. It is clear, that similarly to Euclidean convexity,
the intersection of an arbitrary family of convex sets in the hyperbolic plane
is also convex, so we define the _convex hull_ of a set $X\subseteq H^{2}$ as
the intersection of all convex sets in $H^{2}$ containing $X$ as a subset, and
we will use the notation $\operatorname{\mathrm{conv}}\left(X\right)$ for the
convex hull of $X$. The convex body obtained as the convex hull of the finite
set $X=\left\\{x_{1},\ldots,x_{n}\right\\}$ is called a _polygon_ , and we use
the notation $\left[x_{1},\ldots,x_{n}\right]$ for
$\operatorname{\mathrm{conv}}\left(X\right)$. A point $x_{j}\in X$ is a
_vertex_ of the polygon $X$ if
$x_{j}\not\in\operatorname{\mathrm{conv}}\left(X\setminus\left\\{x_{j}\right\\}\right)$;
a _$k$ -gon_ in the hyperbolic plane is a polygon of $k$ vertices.

For convex bodies, width is an important concept. On the hyperbolic plane
there are many different notions of width (see Santaló [21], Fillmore [3],
Leichtweiss [16], Jerónimo-Castro–Jimenez-Lopez [6], G. Horváth [5],
Böröczky–Csépai–Sagmeister [2], Lassak [9]). We will use the width function
introduced by Lassak, but we note that it is identical with the extended
Leichtweiss width defined by Böröczky, Csépai and Sagmeister on supporting
lines. A hyperbolic line $\ell$ is called a _supporting line_ of the convex
body $K$ if $K\cap\ell\neq\emptyset$, and $K$ is contained in one of the
closed half-spaces bounded by $\ell$. The _width_ of the convex body $K$ with
respect to the supporting line $\ell$ is the distance of $\ell$ and
$\ell^{\prime}$, where $\ell^{\prime}$ is a (not necessarily unique) most
distant supporting line from $\ell$. It is known that this width function is
continuous, and its maximal value coincides with the diameter of the convex
body, which will be denoted as $\operatorname{\mathrm{diam}}\left(K\right)$.
The _minimal width_ (i.e. the minimal value of the width function on the set
of all supporting lines, also known as the _thickness_) of $K$ is denoted by
$w\left(K\right)$. This notion of minimal width is a monotonic function, that
is for arbitrary convex bodies $K,L$ such that $K\subseteq L$, we have
$w\left(K\right)\leq w\left(L\right)$. Hence, the concept of hyperbolic
reducedness makes perfect sense. A convex body $K$ is called _reduced_ , if
for any convex body $K^{\prime}\subsetneq K$,
$w\left(K^{\prime}\right)<w\left(K\right)$. Reduced bodies are well-studied
(see Heil [4], Lassak–Martini [12, 13, 14], Lassak–Musielak [15]), as they are
often extremizers of volume minimizing problems, and bodies of constant width
are also reduced.

If we consider the Poincaré disk model of the hyperbolic plane $H^{2}$,
hyperbolic lines are either diameters of the unit disk, or circular arcs
intersecting the unit circle orthogonally. The boundary points of the unit
disk are the _ideal points_ of the hyperbolic plane, and hence there is a
natural bijection between hyperbolic lines and pairs of ideal points. Besides
the identity, there are three types of orientation preserving isometries of
the hyperbolic plane depending the number of fixed points. If there is one
fixed point, the isometry is called an _elliptic isometry_ (or _rotation_). We
call an isometry with exactly one fixed ideal point, it is called a _parabolic
isometry_. Finally, isometries with exactly two fixed ideal points are called
_hyperbolic isometries_ , which maps the line corresponding to the two fixed
ideal points to itself.

##  3\. Ordinary reduced polygons

Lassak proved that hyperpolic odd-gons of thickness $w$ are reduced if all
vertices are of distance $w$ from the opposite sides, and the orthogonal
projections of these vertices onto the opposite sides are in the relative
interior of these sides (see [9]). Such polygons are called _ordinary reduced
polygons_ , since this property characterizes reducedness both in
${\mathbb{R}}^{2}$ and in $S^{2}$ (see Lassak [8, 10]), but not in the
hyperbolic plane. In an ongoing work with Ansgar Freyer and Károly Jr.
Böröczky we show that, for each $w>0$ there are reduced rhombi, whose
diameters are unbounded. The characterization of hyperbolic reduced polygons
is therefore unclear, so we focus on ordinary reduced polygons in this paper.
For the diameter of an ordinary reduced polygon of thickness $w$, we have the
following by Lassak [9].

######  Theorem 3.1.

Let $R\subset H^{2}$ be an ordinary reduced polygon of thickness $w$ and
diameter $d$. Then,

$w<d<\operatorname{\mathrm{arcosh}}\left(\cosh
w\sqrt{1+\frac{\sqrt{2}}{2}\sinh w}\right).$

As a consequence, for each $n$ we can expect an $n$-gon of extremal area among
ordinary reduced $n$-gons of thickness $w$ by Blaschke’s Selection Theorem. In
the remainder of the section we will discuss the area of hyperbolic reduced
$n$-gons based on the arguments of Lassak [7] and Liu–Chang–Su [17].

From now on, $R$ denotes an ordinary reduced $n$-gon in $H^{2}$ whose vertices
are $v_{1},\ldots,v_{n}$ in cyclic order with respect to the positive
orientation. For each $i$, let $t_{i}$ be the orthogonal projection of $v_{i}$
on the line through $v_{i+\frac{n-1}{2}}$ and $v_{i+\frac{n+1}{2}}$, where the
indices are taken mod $n$. By definition, $t_{i}$ is in the relative interior
of $\left[v_{i+\frac{n-1}{2}},v_{i+\frac{n+1}{2}}\right]$, and hence the
geodesic segments $\left[v_{i},t_{i}\right]$ and
$\left[v_{i+\frac{n+1}{2}},t_{i+\frac{n+1}{2}}\right]$ intersect in a point
$p_{i}$. Let $B_{i}$ be the union of the two triangles:

$B_{i}=\left[v_{i},p_{i},t_{i+\frac{n+1}{2}}\right]\cup\left[v_{i+\frac{n+1}{2}},p_{i},t_{i}\right];$

we will call $B_{i}$ a _butterfly_. We observe that these butterflies cover
the polygon.

######  Lemma 3.2.

Let $R\subset H^{2}$ be an ordinary reduced $n$-gon, and $B_{i}$ be its
$i^{\mathrm{th}}$ butterfly. Then,

$R=\bigcup_{i=1}^{n}B_{i}.$

###### Proof.

Let us denote the line containing $v_{i}$ and $t_{i}$ by $\ell_{i}$. Since
$t_{i}$ is always in the relative interior of
$\left[v_{i+\frac{n-1}{2}},v_{i+\frac{n+1}{2}}\right]$, we have readily
$B_{i}\subset R$ by the convexity of $R$, and hence

$R\supseteq\bigcup_{i=1}^{n}B_{i}.$

Now we want to see that every point of $R$ is covered by some butterfly
$B_{i}$. To prove this, we observe that $B_{i}$ is the union of the chords
obtained by continuously rotating the line $\ell_{i}$ around $p_{i}$ into
$\ell_{i+\frac{n+1}{2}}$, and then these lines are intersected by $R$. Then we
apply a hyperbolic isometry that preserves the line $\ell_{i+\frac{n+1}{2}}$
such that the image of $p_{i}$ is $p_{i+\frac{n+1}{2}}$. After successively
mapping the lines
$\ell_{1},\ell_{1+\frac{n+1}{2}},\ell_{2},\ell_{2+\frac{n+1}{2}},\ldots,\ell_{\frac{n+1}{2}},\ell_{1}$
by using alternating elliptic and hyperbolic isometries, the composition of
these isometries will be a rotation around $p_{1}$ by $\pi$, and hence

$R\subseteq\bigcup_{i=1}^{n}B_{i}.$

∎

We introduce a few additional notations for some angles of the butterflies.
Let

$\varphi=\angle\left(v_{i},p_{i},t_{i+\frac{n+1}{2}}\right)=\angle\left(t_{i},p_{i},v_{i+\frac{n+1}{2}}\right)$

and

$\alpha_{i}=\angle\left(t_{i},v_{i+\frac{n+1}{2}},p_{i}\right).$

The following lemma shows that the two triangles involved in the butterfly
$B_{i}$ are congruent.

######  Lemma 3.3.

The two triangles $\left[v_{i},p_{i},t_{i+\frac{n+1}{2}}\right]$ and
$\left[v_{i+\frac{n+1}{2}},p_{i},t_{i}\right]$ defining $B_{i}$ are congruent.

###### Proof.

It is enough to show that
$\angle\left(p_{i},t_{i},v_{i+\frac{n+1}{2}}\right)=\alpha_{i}$. To verify
this, we show that the triangles
$\left[v_{i},t_{i},v_{i+\frac{n+1}{2}}\right]$ and
$\left[v_{i+\frac{n+1}{2}},v_{i},t_{i+\frac{n+1}{2}}\right]$ are congruent.

These right triangles share a hypotenuse
$\left[v_{i},v_{i+\frac{n+1}{2}}\right]$, and furthermore

$d\left(v_{i},t_{i}\right)=d\left(v_{i+\frac{n+1}{2}},t_{i+\frac{n+1}{2}}\right)=w.$

Thus, by the hyperbolic law of sines we get

$\sin\angle\left(t_{i},v_{i+\frac{n+1}{2}},v_{i}\right)=\frac{\sinh w}{\sinh
d\left(v_{i},v_{i+\frac{n+1}{2}}\right)}=\sin\angle\left(v_{i},t_{i+\frac{n+1}{2}},v_{i+\frac{n+1}{2}}\right).$

Since the angle sum of a hyperbolic triangle is always less than $\pi$, we
deduce

$\angle\left(t_{i},v_{i+\frac{n+1}{2}},v_{i}\right)=\angle\left(v_{i},t_{i+\frac{n+1}{2}},v_{i+\frac{n+1}{2}}\right).$

We recall the following Napier analogy: for a hyperbolic triangle of sides
$a,b,c$ and opposite angles $\alpha,\beta,\gamma$ we have

$\cot\frac{\gamma}{2}=\tan\frac{\alpha+\beta}{2}\cdot\frac{\cosh\frac{a+b}{2}}{\cosh\frac{a-b}{2}}.$

This implies the equality of the angles

$\angle\left(v_{i+\frac{n+1}{2},v_{i},t_{i}}\right)=\angle\left(t_{i+\frac{n+1}{2}},v_{i+\frac{n+1}{2}},v_{i}\right),$

so the triangles $\left[v_{i},t_{i},v_{i+\frac{n+1}{2}}\right]$ and
$\left[v_{i+\frac{n+1}{2}},v_{i},t_{i+\frac{n+1}{2}}\right]$ are indeed
congruent. Finally, we get

$\displaystyle\angle\left(t_{i},v_{i},t_{i+\frac{n+1}{2}}\right)=\angle\left(v_{i+\frac{n+1}{2}},v_{i},t_{i+\frac{n+1}{2}}\right)-\angle\left(v_{i+\frac{n+1}{2}},v_{i},t_{i}\right)=$
$\displaystyle=\angle\left(t_{i},v_{i+\frac{n+1}{2}},v_{i}\right)-\angle\left(t_{i+\frac{n+1}{2}},v_{i+\frac{n+1}{2}},v_{i}\right)=\alpha_{i}.$

∎

As a consequence, we can derive the following area formula. Let
$F_{w}\left(x\right)=f_{w}\left(g_{w}\left(x\right)\right)$, where

$f_{w}\left(x\right)=\arcsin\frac{x\sqrt{1-\tanh^{2}w}}{\tanh w-x}\text{ and
}g_{w}\left(x\right)=\frac{1+\cos x-\sqrt{\left(1+\cos
x\right)^{2}-4\tanh^{2}w\cos x}}{2\tanh w}.$

######  Theorem 3.4.

Let $R\subset H^{2}$ be an ordinary reduced $n$-gon of thickness $w$ described
as above. Then, its area is

$\operatorname{\mathrm{area}}\left(R\right)=\left(n-2\right)\pi-2\sum_{i=1}^{n}F_{w}\left(\varphi_{i}\right).$

###### Proof.

We know from Lemma 3.3 and the Gauss–Bonnet formula that the area of $R$ is

$\operatorname{\mathrm{area}}\left(R\right)=\left(n-2\right)\pi-2\sum_{i=1}^{n}\alpha_{i},$

so we only have to calculate $\alpha_{i}$. Let
$b_{i}=d\left(p_{i},t_{i}\right)$ and
$c_{i}=d\left(p_{i},v_{i+\frac{n+1}{2}}\right)$. From the law of sines, we
have

$\sin\alpha_{i}=\frac{\sinh b_{i}}{\sinh c_{i}},$

so it renains to express $b_{i}$ and $c_{i}$ using $\varphi$ and $w$. By Lemma
3.3, we have $c_{i}=w-b_{i}$, and then for the right triangle
$\left[v_{i+\frac{n+1}{2}},p_{i},t_{i}\right]$ we have

$\cos\varphi_{i}=\frac{\tanh b_{i}}{\tanh c_{i}}=\frac{\tanh
b_{i}}{\tanh\left(w-b_{i}\right)}=\frac{\tanh b_{i}\left(1-\tanh w\tanh
b_{i}\right)}{\tanh w-\tanh b_{i}}.$

This leads to a quadratic equation for $\tanh b_{i}$, whose solutions are

$\tanh
b_{i}=\frac{1+\cos\varphi_{i}\pm\sqrt{\left(1+\cos\varphi_{i}\right)^{2}-4\tanh^{2}w\cos\varphi_{i}}}{2\tanh
w}.$

However, it is easy to see that

$\frac{1+\cos\varphi_{i}+\sqrt{\left(1+\cos\varphi_{i}\right)^{2}-4\tanh^{2}w\cos\varphi_{i}}}{2\tanh
w}>\tanh w,$

so from $b_{i}<w$ we get

$\tanh
b_{i}=\frac{1+\cos\varphi_{i}-\sqrt{\left(1+\cos\varphi_{i}\right)^{2}-4\tanh^{2}w\cos\varphi_{i}}}{2\tanh
w}=g_{w}\left(\varphi_{i}\right)$

as the only solution. Using the identity

$\sinh\left(\operatorname{\mathrm{artanh}}\left(x\right)\right)=\frac{x}{\sqrt{1-x^{2}}},$

we get

$\sinh
b_{i}=\frac{g_{w}\left(\varphi_{i}\right)}{\sqrt{1-g_{w}\left(\varphi_{i}\right)^{2}}}$

and

$\displaystyle\sinh
c_{i}=\sinh\left(\operatorname{\mathrm{artanh}}\left(\frac{\tanh w-\tanh
b_{i}}{1-\tanh w\tanh
b_{i}}\right)\right)=\sinh\left(\operatorname{\mathrm{artanh}}\left(\frac{\tanh
w-g_{w}\left(\varphi_{i}\right)}{1-\tanh w\cdot
g_{w}\left(\varphi_{i}\right)}\right)\right)=$
$\displaystyle=\frac{\frac{\tanh w-g_{w}\left(\varphi_{i}\right)}{1-\tanh
w\cdot g_{w}\left(\varphi_{i}\right)}}{\sqrt{1-\left(\frac{\tanh
w-g_{w}\left(\varphi_{i}\right)}{1-\tanh w\cdot
g_{w}\left(\varphi_{i}\right)}\right)^{2}}}=\frac{\tanh
w-g_{w}\left(\varphi_{i}\right)}{\sqrt{\left(1-\tanh^{2}w\right)\left(1-g_{w}\left(\varphi_{i}\right)^{2}\right)}},$

and hence

$\alpha_{i}=\arcsin\left(\frac{\frac{g_{w}\left(\varphi_{i}\right)}{\sqrt{1-g_{w}\left(\varphi_{i}\right)^{2}}}}{\frac{\tanh
w-g_{w}\left(\varphi_{i}\right)}{\sqrt{\left(1-\tanh^{2}w\right)\left(1-g_{w}\left(\varphi_{i}\right)^{2}\right)}}}\right)=f_{w}\left(g_{w}\left(\varphi_{i}\right)\right)=F_{w}\left(\varphi_{i}\right).$

∎

Now we can show that contrary with the Euclidean and spherical case, regular
$n$-gons minimize the area among ordinary reduced $n$-gons.

######  Theorem 3.5.

Let $R\subset H^{2}$ be an ordinary reduced $n$-gon of thickness $w$, and let
$\widetilde{R}$ be a regular $n$-gon of thickness $w$. Then,

$\operatorname{\mathrm{area}}\left(R\right)\geq\operatorname{\mathrm{area}}\left(\widetilde{R}\right)$

with equality if and only if $R$ and $\widetilde{R}$ are congruent.

###### Proof.

Using the notations introduced earlier, Lemma 3.4 says that

$\operatorname{\mathrm{area}}\left(R\right)=\left(n-2\right)\pi-2\sum_{i=1}^{n}F_{w}\left(\varphi_{i}\right)$

and

$\operatorname{\mathrm{area}}\left(\widetilde{R}\right)=\left(n-2\right)\pi-2nF_{w}\left(\frac{\pi}{n}\right).$

Let us show that $F_{w}$ is concave on the
$\left(0,\max_{i}\varphi_{i}\right]$ interval. For the sake of simplicity, we
introduce

$r_{w}\left(x\right)=\sqrt{\left(1+\cos\varphi_{i}\right)^{2}-4\tanh^{2}w\cos\varphi_{i}}.$

We can calculate that the second derivative is

$F_{w}^{\prime\prime}\left(x\right)=\frac{\sqrt{1-\tanh^{2}w}}{\left(\tanh
w\right)\left(\tanh^{2}w\left(g_{w}\left(x\right)+1\right)-2\tanh
wg_{w}\left(x\right)\right)r_{w}^{2}\left(x\right)}h_{w}\left(x\right),$

where

$\displaystyle h_{w}\left(x\right)=\sin x\left(\left(\tanh
w-g_{w}\left(x\right)+1\right)\left(\sin
x-r_{w}^{\prime}\left(x\right)\right)-g_{w}^{\prime}\left(x\right)\left(2\tanh^{2}w-r_{w}\left(x\right)-\cos
x-1\right)\right)$ $\displaystyle-\cos x\left(\tanh
w-g_{w}\left(x\right)+1\right).$

First we observe that

$\frac{\sqrt{1-\tanh^{2}w}}{\left(\tanh
w\right)\left(\tanh^{2}w\left(g_{w}\left(x\right)+1\right)-2\tanh
wg_{w}\left(x\right)\right)r_{w}^{2}\left(x\right)}>0,$

so the concavity of $F$ only depends on the sign of $h_{w}$. We also notice
that for $0<x<\max_{i}\varphi_{i}<\frac{\pi}{2}$ both $\sin x$ and $\cos x$
are positive, $\tan w>g_{w}\left(x\right)$ by the argument in the proof of
Theorem 3.4, and then we also have the following:

$\sin x-r_{w}^{\prime}\left(x\right)=\frac{\sin
x}{r_{w}\left(x\right)}\left(r_{w}\left(x\right)+1+\cos\left(x\right)-2\tanh^{2}w\right)=\frac{2\tanh
w\sin x}{r_{w}\left(x\right)}\left(g_{w}\left(x\right)-\tanh w\right)<0.$

On the other hand,

$g^{\prime}\left(x\right)=\frac{-\sin x+r_{w}^{\prime}\left(x\right)}{2\tanh
w}=-\frac{\sin x}{r_{w}\left(x\right)}\left(g_{w}\left(x\right)-\tanh
w\right),$

so

$h_{w}\left(x\right)=\frac{2\sin^{2}x\tanh
w}{r_{w}\left(x\right)}\left(g_{w}\left(x\right)-\tanh w\right)\left(2\tanh
w-2g_{w}\left(x\right)+1\right)-\cos x\left(\tanh
w-g_{w}\left(x\right)+1\right),$

which is negative, and hence $-F_{w}$ is strictly convex in the interval
$\left(0,\max_{i}\varphi_{i}\right]$. Finally, Jensen’s inequality implies the
statement of the theorem. ∎

Since the limit of the areas of regular $n$-gons is the area of the circle of
the same width as $n$ tends to infinity, this might sound like a contradiction
if we consider Lassak’s conjecture about maximizing the area of reduced convex
bodies. He conjectures that on the Euclidean plane, the circle and the quarter
of the disk have maximal area. However, on one hand, hyperbolic width function
does not tend to the Euclidean one as $w$ approaches 0, on the other hand, the
quarter of the disk has a strictly greater area than the circle of the same
thickness in $H^{2}$.

Acknowledgement:  The author wants to thank Károly J. Böröczky and Ansgar
Freyer for their useful comments on the manuscript.

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