Covering a reduced spherical body by a disk
UDC 514 In this paper, the following theorems are proved: (1) every spherical convex body $W$ of constant width $\Delta (W) \geq \dfrac{\pi}{2}$ may be covered by a disk of radius $\Delta(W) + \arcsin \!\left(\dfrac{2\sqrt{3}}{3} \cos \dfrac{\Delta(W)}{2}\right) - \dfrac{\pi}{2};$ (2) every reduced...
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| author | Musielak, M. Musielak, M. |
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In this paper, the following theorems are proved: (1) every spherical convex body $W$ of constant width $\Delta (W) \geq \dfrac{\pi}{2}$ may be covered by a disk of radius $\Delta(W) + \arcsin \!\left(\dfrac{2\sqrt{3}}{3} \cos \dfrac{\Delta(W)}{2}\right) - \dfrac{\pi}{2};$ (2) every reduced spherical convex body $R$ of thickness $\Delta(R)<\dfrac{\pi}{2}$ may be covered by a disk of radius $\arctan \!\left(\sqrt{2} \tan \dfrac{\Delta(R)}{2}\right)\!.$
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| doi_str_mv | 10.37863/umzh.v72i10.6029 |
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DOI: 10.37863/umzh.v72i10.6029
UDC 514
M. Musielak (Univ. Sci. and Technology, Bydgoszcz, Poland)
COVERING A REDUCED SPHERICAL BODY BY A DISK
ПОКРИТТЯ РЕДУКОВАНОГО СФЕРИЧНОГО ТIЛА ДИСКОМ
In this paper, the following theorems are proved: (1) every spherical convex body W of constant width \Delta (W ) \geq \pi
2
may
be covered by a disk of radius \Delta (W ) + \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\biggr)
- \pi
2
; (2) every reduced spherical convex body R of
thickness \Delta (R) <
\pi
2
may be covered by a disk of radius \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\biggl( \surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
\biggr)
.
У цiй статтi доведено такi теореми: (1) кожне редуковане сферичне тiло W сталої ширини \Delta (W ) \geq \pi
2
можна
покрити диском радiуса \Delta (W ) + \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\biggr)
- \pi
2
; (2) кожне редуковане сферичне тiло R товщини
\Delta (R) <
\pi
2
можна покрити диском радiуса \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\biggl( \surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
\biggr)
.
1. Introduction. The subject of reduced bodies in the Euclidean space Ed has been well researched,
see for instance [6]. However, it is a natural idea to investigate such bodies in non-Euclidean
geometries. There are number of articles describing reduced bodies on the sphere, see [3, 5, 7, 8].
The question arises which of the results achieved in the Euclidean space can be transferred to the
sphere. In particular, the main theorem of this paper presents the spherical version of the variant of
Jung theorem for reduced bodies in E2 given in [4].
Let S2 be the unit sphere of the three-dimensional Euclidean space E3. By a great circle of S2
we mean the intersection of S2 with any two-dimensional subspace of E3. The set of points of a
great circle of S2 in the distance at most
\pi
2
from a point c of this great circle is called a semicircle
with center c. Any pair of points obtained as the intersection of S2 with a one-dimensional subspace
of E3 is called a pair of antipodes. Note that if two different points a, b are not antipodes, there is
exactly one great circle containing them. The shorter part of this great circle is called the spherical
arc connecting a and b, or shortly arc. It is denoted ab. By the spherical distance | ab| , or shortly
distance, of these points we mean the length of the arc ab. If a, b are antipodes, we put | ab| = \pi . If
p is a point of S2 and F is a closed set containing at least two points, then we define \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(p, F ) as
\mathrm{m}\mathrm{i}\mathrm{n}q\in F | pq| .
A subset of S2 is called convex if it does not contain any pair of antipodes of S2 and if together
with every two points it contains the arc connecting them. By a spherical convex body we mean a
closed convex set with non-empty interior. If there is no arc in the boundary of a spherical convex
body, we call the body strictly convex.
Let \rho \in
\Bigl(
0,
\pi
2
\Bigr]
. By disk of radius \rho and center c we mean the set of points of S2 in the distance
at most \rho from a point c \in S2. The boundary of a disk is called a spherical circle. By a hemisphere
we mean any disk of radius
\pi
2
. The hemisphere with center p is denoted by H(p). If p, q is are
antipodes, then H(p) and H(q) are called opposite hemispheres. Let t be a boundary point of a
c\bigcirc M. MUSIELAK, 2020
1400 ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
COVERING A REDUCED SPHERICAL BODY BY A DISK 1401
convex body C \subset S2. We say that a hemisphere H supports C at t if C \subset H and t belongs to
the great circle bounding H. If the body C is supported at p by exactly one hemisphere, we call
p a smooth point of the boundary of C. If all boundary points of C are smooth, we say that C is
smooth. We say that e is an extreme point of C if C \setminus \{ e\} is a convex set. If the set A is contained
in an open hemisphere, we define conv(A) as a smallest convex set containing A (for details see
definition before Lemma 1 in [5]).
If hemispheres G and H are different and not opposite, then L = G\cap H is called a lune. The two
semicircles bounding L and contained in G and H, respectively, are denoted by G/H and H/G.
The thickness \Delta (L) of L \subset S2 is defined as the distance of the centers of G/H and H/G. By the
corners of L we understand the two points of the set (G/H) \cap (H/G).
For every hemisphere K supporting a convex body C \subset S2 we find hemispheres K\ast supporting
C such that the lunes K \cap K\ast are of the minimum thickness (by compactness arguments at least one
such a hemisphere K\ast exists). The thickness of the lune K\cap K\ast is called the width of C determined
by K and it is denoted by \mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h}K(C) (see [5]). If for all hemispheres K supporting C the numbers
\mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h}K(C) are equal, we say that C is of constant width (see [5]). By the thickness \Delta (C) of a
convex body C \subset S2 we understand the minimum width of C determined by K over all supporting
hemispheres K of C (see [5]).
After [5] we call a spherical convex body R \subset S2 reduced if \Delta (Z) < \Delta (R) for every convex
body Z \subset R different from R. Simple examples of reduced spherical convex bodies on S2 are
spherical bodies of constant width and, in particular, the disks on S2. Also each of the four parts of
a spherical disk on S2 dissected by two orthogonal great circles through the center of the disk is a
reduced spherical body. It is called a quarter of a spherical disk.
Remark. Most of the above notions can be defined also in higher dimensions, for details see, for
instance, [5].
For the convenience of the reader recall a few formulas of spherical geometry which are fre-
quently applied in this paper. Consider the right spherical triangle with hypotenuse C and legs A,B.
Denote by \alpha the angle opposite to A and by \beta the angle opposite to B. By [11] the following
formulas hold true:
\mathrm{t}\mathrm{a}\mathrm{n}A = \mathrm{c}\mathrm{o}\mathrm{s}\beta \mathrm{t}\mathrm{a}\mathrm{n}C, (1)
\mathrm{s}\mathrm{i}\mathrm{n}A = \mathrm{s}\mathrm{i}\mathrm{n}\alpha \mathrm{s}\mathrm{i}\mathrm{n}C, (2)
\mathrm{c}\mathrm{o}\mathrm{s}C = \mathrm{c}\mathrm{o}\mathrm{s}A \mathrm{c}\mathrm{o}\mathrm{s}B, (3)
\mathrm{c}\mathrm{o}\mathrm{s}C = \mathrm{c}\mathrm{o}\mathrm{t}\alpha \mathrm{c}\mathrm{o}\mathrm{t}\beta . (4)
After [2] we define the circumradius of a convex body C as the smallest \rho such that C can
be covered by a disk of radius \rho . Such a disk is unique and is called the disk circumscribed about
C . The boundary of this disk is called the circle circumscribed about C . In particular, the circle
circumsribed on a spherical triangle contains all vertices of this triangle, which is a consequence of
the spherical Ceva’s theorem (see Theorem 2 of [9]).
Determining the circumradius in the three special cases presented in the below Lemmas 1 – 3 is
useful in the further part of the paper.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
1402 M. MUSIELAK
Lemma 1. Let Q \subset S2 be a quarter of disk. The circumradius of Q is equal
\rho = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\biggl( \surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (Q)
2
\biggr)
.
Proof. Denote by c the center of the disk whose quarter is considered and by a, b the two
different extreme points of Q such that ca and cb are subsets of the boundary of Q. It is easily seen
that if a disk contains points a, b, c, then it contains Q. Therefore we are looking for the radius \rho
of the disk circumscribed on the triangle abc. Denote by o the center of this disk and by p a point
on ac such that the angle \angle cpo is right. Since o is in equal distances from a and c, the point p is in
the middle of ac and thus | cp| = \Delta (Q)
2
. Clearly, the angle \angle pco equals
\pi
4
and | oc| is equal to the
radius of our disk. Hence, considering the triangle cop, by (1) we have \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (Q)
2
= \mathrm{c}\mathrm{o}\mathrm{s}
\pi
4
\mathrm{t}\mathrm{a}\mathrm{n} \rho .
By evaluating \rho we obtain the thesis of our lemma.
Recall after [5] that Reuleaux triangle is the intersection of three disks of radius \sigma such that the
centers of these disks are pairwise distant by \sigma .
Lemma 2. The circumradius of a spherical Reuleaux triangle R is equal
\rho = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\biggl(
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
\Delta (R)
2
\biggr)
.
Proof. Denote by a, b, c the three points of bd(R) that are not smooth. It is easy to check
that any disk containing these three points contains R. Thus \rho is equal to the radius of the disk
circumscribed on the triangle abc. Denote by o the center of this disk and by p the middle of the
arc ab. Clearly, the angle \angle opa is right, the angle \angle aop is equal
\pi
3
and | ap| = \Delta (R)
2
. Therefore, if
we look at the triangle aop, by (2) we obtain \mathrm{s}\mathrm{i}\mathrm{n}
\Delta (R)
2
= \mathrm{s}\mathrm{i}\mathrm{n}
\pi
3
\mathrm{s}\mathrm{i}\mathrm{n} \rho . We easily evaluate \rho which
establishes the promised formula.
Lemma 3. The circumradius \rho of a spherical equilateral triangle T of thickness less than
\pi
2
equals \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\sqrt{}
9 + 8 \mathrm{t}\mathrm{a}\mathrm{n}2\Delta (T ) - 3
2 \mathrm{t}\mathrm{a}\mathrm{n}\Delta (T )
.
Proof. Denote by o the center of the disk circumscribed on our triangle, by d the center
of a side of this triangle and by a an endpoint of this side. Clearly, \angle doa =
\pi
3
, | oa| = \rho and
| od| = \Delta (T ) - \rho . Therefore by (1) we obtain \mathrm{t}\mathrm{a}\mathrm{n}(\Delta (T ) - \rho ) =
1
2
\mathrm{t}\mathrm{a}\mathrm{n} \rho . Using the subtraction
formula for the tangent function we can rewrite this equation as
\mathrm{t}\mathrm{a}\mathrm{n}\Delta (T ) - \mathrm{t}\mathrm{a}\mathrm{n} \rho
1 + \mathrm{t}\mathrm{a}\mathrm{n}\Delta (T ) \mathrm{t}\mathrm{a}\mathrm{n} \rho
=
\mathrm{t}\mathrm{a}\mathrm{n} \rho
2
. This
is equivalent to \mathrm{t}\mathrm{a}\mathrm{n}\Delta (T )(\mathrm{t}\mathrm{a}\mathrm{n} \rho )2 + 3 \mathrm{t}\mathrm{a}\mathrm{n} \rho - 2 \mathrm{t}\mathrm{a}\mathrm{n}\Delta (T ) = 0. Consequently, by \mathrm{t}\mathrm{a}\mathrm{n} \rho > 0 we get
\mathrm{t}\mathrm{a}\mathrm{n} \rho =
\sqrt{}
9 + 8 \mathrm{t}\mathrm{a}\mathrm{n}2\Delta (T ) - 3
2 \mathrm{t}\mathrm{a}\mathrm{n}\Delta (T )
, which ends the proof.
2. Covering a body of constant width over
\bfitpi
\bftwo
by a disk. For any set F on the sphere S2 we
define the set F\oplus as \{ p : F \subset H(p)\} .
In [3] and [12] there is used the notion of the polar set of the set F on the sphere, which is
defined as F o =
\bigcap
p\in F H(p).
Proposition 1. For every set F on the sphere we have F o = F\oplus .
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
COVERING A REDUCED SPHERICAL BODY BY A DISK 1403
Proof. For any point q we have
q \in F o \leftrightarrow \forall p\in F q \in H(p) \leftrightarrow \forall p\in F | pq| \leq \pi
2
\leftrightarrow \forall p\in F p \in H(q) \leftrightarrow F \subset H(q) \leftrightarrow q \in F\oplus .
However, applying F\oplus is more convenient in this paper.
We omit here a simple proof of the next lemma.
Lemma 4. If C is a spherical convex body, then C\oplus is also a spherical convex body.
By the way, observe that (C\oplus )\oplus = C for a spherical convex body C.
Proposition 2. If W is a spherical convex body of constant width, then W\oplus is a spherical
convex body of constant width \pi - \Delta (W ).
Proof. Consider a hemisphere H(a) supporting W\oplus . We intend to show that widthH(a)(W
\oplus ) =
= \pi - \Delta (W ). Since clearly a is a boundary point of W, by Theorem 7 of [7] there exist hemispheres
K and M supporting W such that the lune K \cap M is of thickness \Delta (W ) and a is the center
of K/M. Denote the center of the semicircle M/K by b and the centers of the hemispheres K
and M by a\prime and b\prime , respectively. Since a, b, a\prime , b\prime are in the same distance from both corners
of K \cap M, all these points lay on the same great circle. From the above we easily obtain that
| ab| + | a\prime b\prime | = | aa\prime | + | bb\prime | = \pi and thus | a\prime b\prime | = \pi - \Delta (W ). Since every point of W\oplus is in the
distance at most
\pi
2
from b, the body W\oplus is contained in H(b). Therefore, W\oplus is contained in the
lune H(a)\cap H(b), which is of thickness | a\prime b\prime | . It means that widthH(a)(W
\oplus ) is at most \pi - \Delta (W ).
Assume that widthH(a)(W
\oplus ) < \pi - \Delta (W ). Then there exists a point b such that the lune
H(a)\cap H(b) is of thickness less that \pi - \Delta (W ) and contains W\oplus . From | ab| +\Delta (H(a)\cap H(b)) = \pi
we conclude that | ab| > \Delta (W ). But this contradicts the fact that every body W of constant width
has diameter \Delta (W ). Hence, widthH(a)(W
\oplus ) = \pi - \Delta (W ), which ends the proof.
In [2] there was estimated the diameter of a spherical compact set by the function of its circum-
radius (see the second part of Theorem 2 in [2]). Recall this result on S2 in a different form: if d
is the diameter of a compact set and \sigma is its circumradius, then \mathrm{s}\mathrm{i}\mathrm{n}\sigma \leq 2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
d
2
. In particular,
it holds true for every spherical convex body W of constant width and in this case d = \Delta (W ). If
\Delta (W ) is at most
2\pi
3
, then
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
d
2
is at most 1. In this case our inequality is equivalent to the
statement that every spherical convex body of constant width at most
2\pi
3
can be covered by a disk
of radius \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
\Delta (W )
2
\Biggr)
. If \Delta (W ) is greater than
2\pi
3
, then
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
d
2
is greater than 1
and in this case our inequality does not estimate \sigma in a nontrivial way.
According to Lemma 2 the example of a spherical Reuleaux triangle shows that the estimate can
not be improved for bodies of constant width at most
\pi
2
. The following theorem describes the case
of bodies of constant width at least
\pi
2
and in particular gives an improvement of the estimate recalled
from [2] for convex bodies of constant width greater than
\pi
2
.
Theorem 1. Every spherical body W of constant width at least
\pi
2
is contained in a disk of
radius \Delta (W ) + \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
- \pi
2
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
1404 M. MUSIELAK
Proof. Observe that for \Delta (W ) =
\pi
2
we have
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
+\Delta (W ) - \pi
2
= \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
\Delta (W )
2
\Biggr)
.
Therefore by the above recalled result of Dekster the thesis of the theorem holds true for bodies of
constant width
\pi
2
.
Assume now that \Delta (W ) >
\pi
2
. By Proposition 2 the body W\oplus is of thickness \pi - \Delta (W ). Hence
by the above recalled result of Dekster, W\oplus is contained in a disk of radius
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{s}\mathrm{i}\mathrm{n}
\pi - \Delta (W )
2
\Biggr)
= \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
.
Denote the center of this disk by o and let p be a boundary point of W. By Proposition 3 of [7] W
is smooth, and therefore there exists exactly one hemisphere supporting W at p. Denote its center by
a and notice that a is a boundary point of W\oplus . Moreover, H(p) is a supporting hemisphere of W\oplus
and it supports W\oplus at a. By the proof of the first part of Theorem 1 of [5], there exists a unique
point of W\oplus closest to p. Denote it by b. Again by Theorem 1 of [5] b is the center of one of the
two semicircles bounding the lune of thickness widthH(p)(W
\oplus ) = \Delta (W\oplus ) = \pi - \Delta (W ). Thus,
| ab| = \Delta (W\oplus ) and clearly b \in ap. Hence
| op| \leq | ob| + | bp| = | ob| + | ap| - | ab| = | ob| + \pi
2
- (\pi - \Delta (W )) = | ob| +\Delta (W ) - \pi
2
.
Since | ob| is at most \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
, the distance between o and p is at most
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
+\Delta (W ) - \pi
2
, which ends the proof.
Observe that in general we can not improve the estimate from Theorem 1. By proof of this
theorem we see that the value \Delta (W )+\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}
\Biggl(
2
\surd
3
3
\mathrm{c}\mathrm{o}\mathrm{s}
\Delta (W )
2
\Biggr)
- \pi
2
is attained for every W such
that W\oplus is a Reuleaux triangle.
3. Covering reduced bodies of thickness at most
\bfitpi
\bftwo
by a disk. The main theorem of this
paper is analogous to Theorem of [4]. However, we are not able to present a similar proof as in [4]
due to the lack of the notion parallelism on the sphere. For this reason the proof of our main theorem
is based on a different idea.
Lemma 5. Let c be a positive number less than
1
4
and a be a number from the interval\Biggl(
1
2
,
1
2
+
\sqrt{}
1
4
- c
\Biggr)
. The function f(x) =
\surd
1 - x -
\sqrt{}
1 - x - c
x
satisfies
f(x) \leq \mathrm{m}\mathrm{a}\mathrm{x}
\biggl(
f
\biggl(
1
2
\biggr)
, f(a)
\biggr)
for every x \in
\biggl[
1
2
, a
\biggr]
.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
COVERING A REDUCED SPHERICAL BODY BY A DISK 1405
Proof. Notice that if x > 0, then 1 - x - c
x
\geq 0 is equivalent to x2 - x+ c \leq 0. This inequality
is satisfied for x \in
\biggl[
1
2
-
\sqrt{}
1
4
- c,
1
2
+
\sqrt{}
1
4
- c
\biggr]
. In particular, we conclude that
\sqrt{}
1 - x - c
x
= 0
for x =
1
2
+
\sqrt{}
1
4
- c and f(x) is well defined in the interval
\biggl(
1
2
,
1
2
+
\sqrt{}
1
4
- c
\biggr)
.
In order to prove the thesis we check the sign of the first derivative of f(x). We have
f \prime (x) = - 1
2
\surd
1 - x
- 1
2
\sqrt{}
1 - x - c
x
\Bigl(
- 1 +
c
x2
\Bigr)
=
=
1
2
\sqrt{}
1 - x - c
x
\biggl(
1 - c
x2
-
\sqrt{}
1 - c
x(1 - x)
\biggr)
.
Put g(x) = 1 - c
x2
-
\sqrt{}
1 - c
x(1 - x)
. Clearly, for every x the sign of f \prime (x) is the same as the
sign of g(x). We obtain
g(x) = 0 \leftrightarrow 1 - c
x2
=
\sqrt{}
1 - c
x(1 - x)
\leftrightarrow 1 - 2c
x2
+
c2
x4
=
= 1 - c
x(1 - x)
\leftrightarrow 1
1 - x
- 2
x
+
c
x3
= 0 \leftrightarrow 3x3 - 2x2 + c(1 - x) = 0.
For V (x) = 3x3 - 2x2 + c(1 - x) we have V (0) = c > 0 and V
\biggl(
1
2
\biggr)
=
1
2
\biggl(
c - 1
4
\biggr)
< 0.
Thus V (x) has three zeros: one is less than 0, one is in the interval
\biggl(
0,
1
2
\biggr)
, and one is greater than
1
2
. Hence, g(x) has exactly one zero in the interval
\Biggl(
1
2
,
1
2
+
\sqrt{}
1
4
- c
\Biggr)
. Denote it by x0. Due to
the continuity of the function g(x) in the interval
\Biggl(
1
2
,
1
2
+
\sqrt{}
1
4
- c
\Biggr)
, it has constant sign in the
interval
\biggl(
1
2
, x0
\biggr)
and constant sign in the interval
\Biggl(
x0,
1
2
+
\sqrt{}
1
4
- c
\Biggr)
. Notice that
g
\biggl(
1
2
\biggr)
= 1 - 4c -
\surd
1 - 4c = (1 - 4c)
\bigl( \surd
1 - 4c - 1
\bigr)
< 0
and
g
\Biggl(
1
2
+
\sqrt{}
1
4
- c
\Biggr)
=
\sqrt{}
1
2
-
\sqrt{}
1
4
- c > 0.
Hence, g(x) < 0 for x \in
\biggl(
1
2
, x0
\biggr)
and g(x) > 0 for x \in (x0, a). Therefore, f(x) is decreasing
in
\biggl(
1
2
, x0
\biggr)
and increasing in (x0, a). The thesis of our lemma is an immediate consequence of this
statement.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
1406 M. MUSIELAK
Lemma 6. Let a, b be points on a spherical circle. Consider any point t such that a, t, b lay
on this circle in this order according to the positive orientation. The measure of the angle \angle atb is
the greatest if t is equidistant from a and b.
Proof. Denote by o the center of our circle and by \rho its radius. Denote by t\prime the point
on our circle laying in equal distances from a and b, and on the same side of the great circle
containing ab as t. Put \alpha =
1
2
| \angle aot| ] and \beta =
1
2
| \angle bot| =, and let k be the midpoint of the arc
at. Observe that | \angle aot\prime | = | \angle bot\prime | = \alpha + \beta . Since | \angle tok| = \alpha , applying (4) we easily obtain that
| \angle ota| = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}\alpha ). Analogously | \angle otb| = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}\beta ) and
| \angle ot\prime a| = | \angle ot\prime b| = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}
\biggl(
\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}
\alpha + \beta
2
\biggr)
.
Our aim is to show that | \angle at\prime b| \geq | \angle atb| .
This inequality is equivalent to
\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}
\biggl(
\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}
\alpha + \beta
2
\biggr)
\geq \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}\alpha ) + \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}\beta )
2
.
In order to show this it is sufficient to show that the function f(x) = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{t}\mathrm{a}\mathrm{n}x) is concave
in the interval
\Bigl(
0,
\pi
2
\Bigr)
. The reader may check that
f \prime \prime (x) =
2 \mathrm{c}\mathrm{o}\mathrm{s} \rho \mathrm{s}\mathrm{i}\mathrm{n}x \mathrm{c}\mathrm{o}\mathrm{s}x(\mathrm{c}\mathrm{o}\mathrm{s}2 \rho - 1)
(\mathrm{c}\mathrm{o}\mathrm{s}2 x+ \mathrm{c}\mathrm{o}\mathrm{s}2 \rho \mathrm{s}\mathrm{i}\mathrm{n}2 x)2
.
It is easily seen that f \prime \prime (x) < 0 for x \in
\Bigl(
0,
\pi
2
\Bigr)
and therefore the function f(x) is concave in
the interval
\Bigl(
0,
\pi
2
\Bigr)
, which completes the proof.
Theorem 2. Every reduced spherical body R of thickness at most
\pi
2
is contained in a disk of
radius \rho = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\biggl( \surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
\biggr)
.
Proof. Note that every boundary point of R belongs to an arc whose ends are extreme points
of R. Therefore if a disk contains all extreme points of R, then it contains all boundary points of R
and so the whole body R. Thus, it is sufficient to show that all extreme points of R are in a disk of
radius \rho . Moreover, according to the spherical Helly’s theorem (see [1, 10]), it is even sufficient to
show that every three extreme points of R are contained in a disk of radius \rho .
Let e1, e2, e3 be any three different extreme points of R. By Theorem 4 of [5] there exist lunes
L1, L2, L3 such that ei is the center of one of the semicircles bounding Li for i = 1, 2, 3. Denote
by fi the center of the other semicircle bounding Li for i = 1, 2, 3.
First, let us consider the case when two points from amongst e1, e2, e3, say e1 and e2, lay in L3
on the same side of the great circle containing e3 and f3. Since e1f1 and e3f3 intersect (see the proof
of Lemma 2 in [7]), the distance from e1 to e3f3 is less or equal than the distance from e1 to the point
of intersection of e1f1 with e3f3, and so it is less or equal \Delta (R). For the same reason the distance
from e2 to e3f3 is at most \Delta (R). Denote by c the corner of L3 laying on the same side of the great
circle containing e3 and f3 as e1 and e2. Denote by k the point of e3c such that | e3k| = \Delta (R) and
by l the point of f3c such that | f3l| = \Delta (R). Since e1, e2, e3 \in conv \{ e3, f3, k, l\} , it is sufficient to
show that conv \{ e3, f3, k, l\} may be covered by a disk of radius \rho . The triangle e3f3k is contained
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
COVERING A REDUCED SPHERICAL BODY BY A DISK 1407
in a quarter of a disk of thickness \Delta (R), so by Lemma 1 it can be covered by a disk of radius \rho .
What is more such a disk is unique thanks to Lemma 1 For the same reason the triangle e3f3l can
be covered by a disk of radius \rho and such a disk is unique. It is easily seen that the disk is the same
for both triangles. This disk covers conv \{ e3, f3, k, l\} , which ends proof in this case.
Assume now that for any i, j, k such that \{ i, j, k\} = \{ 1, 2, 3\} the points ei and ej lay in Lk on
the different sides of the great circle containing ek and fk. Let us stay with this assumption up to
the end of the proof.
Let us consider the case when the triangle e1e2e3 is obtuse or right. Without losing the generality
we can assume that the angle \angle e1e3e2 is obtuse or right. Let k be the point on the same side of the
great circle containing e1 and e2 as e3, such that | e1k| = | e2k| and the angle \angle e1ke2 is right. By
Theorem 8 of [7] | e1e2| is at least \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{c}\mathrm{o}\mathrm{s}2\Delta (R)). Therefore | e1k| and | e2k| are at most \Delta (R),
because otherwise by (3) the lenght | e1e2| is greater that \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{c}\mathrm{o}\mathrm{s}2\Delta (R)). Thus, by Lemma 1 the
circumradius of the triangle e1ke2 is at most \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
\biggl( \surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
\biggr)
.
By Lemma 6 we easily obtain that for any point t from the circle circumscribed on e1ke2 laying
on the same side of the great circle containing e1 and e2 as k, the angle \angle e1te2 is at least
\pi
2
. Thus
since \angle e1e3e2 is obtuse or right, e3 must lay inside this circumscribed circle, which ends the proof
in this case.
The last case is when the triangle e1e2e3 is acute. Clearly, all heights of this triangle have lengths
less that \Delta (R). Let g be the point closest to e2 such that e2 is on the arc e1g and g satisfies at least
one of following conditions: \angle e1e3g is right or \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(g, e1e3) = \Delta (R) or \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(e1, ge3) = \Delta (R). If
\angle e1e3g is right, then | e1e3| and | e3g| are at most \Delta (R). Therefore the triangle e1ge3 is contained in
a right triangle with legs of length \Delta (R), and so in the quarter of a disk of thickness \Delta (R). Hence,
by Lemma 1, e1, e2, e3 can be covered by a disk of radius \rho .
Otherwise, the angle \angle e1e3g is acute. Since | \angle e1ge3| < | \angle e1e2e3| , the triangle e1ge3 is acute.
One height of this triangle, say this from the vertex g, is of length \Delta (R). Moreover, the height
of this triangle from vertex e1 is of length at most \Delta (R). Let j be the point closest to e3 such
that e3 is on the arc e1j and j satisfies at least one of following conditions: \angle e1gj is right or
\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(j, e1g) = \Delta (R) or \mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}(e1, gj) = \Delta (R). If \angle e1gj is right
\biggl(
which by the way is possible only
if \Delta (R) =
\pi
2
\biggr)
, then we apply the same argument as in the preceding case. Otherwise the triangle
e1gj is acute and has two heights of length \Delta (R), say from vertices g, j. Observe that from the
construction of this triangle we obtain that the third height is of length at most \Delta (R). Since the
points e1, e2, e3 are contained in the triangle e1gj, it remains to prove that this triangle may be
covered by a disk of radius \rho .
Observe that since the triangle e1gj has two equal heights, it is an isosceles triangle. Denote
the angle at e1 by 2\alpha , the center of the circle circumscribed on e1gj by o and the radius of this
circle by \sigma . Denote also the point at the middle of je1 by k and the point on e1g closest to j by h.
Clearly, the triangles hje1 and oke1 are right. Put B = | je1| and notice that | ke1| =
B
2
. Since e1gj
is isosceles, we have | \angle oe1k| = \alpha . The length of the arc hj is \Delta (R).
From (1) for the triangle oke1 we obtain \mathrm{t}\mathrm{a}\mathrm{n}
B
2
= \mathrm{c}\mathrm{o}\mathrm{s}\alpha \mathrm{t}\mathrm{a}\mathrm{n}\sigma . Formula (2) for the triangle hje1
gives \mathrm{s}\mathrm{i}\mathrm{n}\Delta (R) = \mathrm{s}\mathrm{i}\mathrm{n} 2\alpha \mathrm{s}\mathrm{i}\mathrm{n}B and in different form \mathrm{s}\mathrm{i}\mathrm{n}B \mathrm{c}\mathrm{o}\mathrm{s}\alpha =
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
2 \mathrm{s}\mathrm{i}\mathrm{n}\alpha
. Using formulas from
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
1408 M. MUSIELAK
last two sentences and the trigonometric formula \mathrm{t}\mathrm{a}\mathrm{n}
B
2
=
1 - \mathrm{c}\mathrm{o}\mathrm{s}B
\mathrm{s}\mathrm{i}\mathrm{n}B
, we obtain
\mathrm{t}\mathrm{a}\mathrm{n}\sigma =
\mathrm{t}\mathrm{a}\mathrm{n}
B
2
\mathrm{c}\mathrm{o}\mathrm{s}\alpha
=
1 - \mathrm{c}\mathrm{o}\mathrm{s}B
\mathrm{s}\mathrm{i}\mathrm{n}B \mathrm{c}\mathrm{o}\mathrm{s}\alpha
=
1 - \mathrm{c}\mathrm{o}\mathrm{s}B
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
2 \mathrm{s}\mathrm{i}\mathrm{n}\alpha
=
=
2 \mathrm{s}\mathrm{i}\mathrm{n}\alpha
\Bigl(
1 -
\sqrt{}
1 - \mathrm{s}\mathrm{i}\mathrm{n}2B
\Bigr)
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
=
2 \mathrm{s}\mathrm{i}\mathrm{n}\alpha
\Biggl(
1 -
\sqrt{}
1 - \mathrm{s}\mathrm{i}\mathrm{n}2\Delta (R)
\mathrm{s}\mathrm{i}\mathrm{n}2 2\alpha
\Biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
=
=
2
\Biggl(
\mathrm{s}\mathrm{i}\mathrm{n}\alpha -
\sqrt{}
\mathrm{s}\mathrm{i}\mathrm{n}2 \alpha - \mathrm{s}\mathrm{i}\mathrm{n}2 \alpha \mathrm{s}\mathrm{i}\mathrm{n}2\Delta (R)
\mathrm{s}\mathrm{i}\mathrm{n}2 2\alpha
\Biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
=
=
2
\Biggl(
\surd
1 - \mathrm{c}\mathrm{o}\mathrm{s}2 \alpha -
\sqrt{}
1 - \mathrm{c}\mathrm{o}\mathrm{s}2 \alpha - \mathrm{s}\mathrm{i}\mathrm{n}2\Delta (R)
4 \mathrm{c}\mathrm{o}\mathrm{s}2 \alpha
\Biggr)
\mathrm{s}\mathrm{i}\mathrm{n}\Delta (R)
.
The greatest possible value of \alpha is
\pi
4
and the smallest is this value for which the third height is
of length \Delta (R). In the first case the triangle e1jg is right and by Lemma 1 we have \mathrm{t}\mathrm{a}\mathrm{n}\sigma 1 =
=
\surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
. By the formula \mathrm{t}\mathrm{a}\mathrm{n}\Delta (R) =
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
1 - \mathrm{t}\mathrm{a}\mathrm{n}2
\Delta (R)
2
we easily obtain that in this case
\mathrm{t}\mathrm{a}\mathrm{n}\sigma 1 =
\surd
2
\sqrt{}
4 + 4 \mathrm{t}\mathrm{a}\mathrm{n}2\Delta (R) - 2
2 \mathrm{t}\mathrm{a}\mathrm{n}\Delta (R)
=
\sqrt{}
2 + 2 \mathrm{t}\mathrm{a}\mathrm{n}2\Delta (R) -
\surd
2
\mathrm{t}\mathrm{a}\mathrm{n}\Delta (R)
.
In the second case the triangle e1jg is equilateral and by Lemma 3 in this case \mathrm{t}\mathrm{a}\mathrm{n}\sigma 2 =
=
\sqrt{}
9 + 8 \mathrm{t}\mathrm{a}\mathrm{n}2\Delta (R) - 3
2 \mathrm{t}\mathrm{a}\mathrm{n}\Delta (R)
. For shortness denote \mathrm{t}\mathrm{a}\mathrm{n}\Delta (R) = t. We have
\mathrm{t}\mathrm{a}\mathrm{n}\sigma 2 =
\surd
9 + 8t2 - 3
2t
=
1
2t
\Bigl( \sqrt{}
9 + 8t2 -
\sqrt{}
8 + 8t2 - 3 +
\sqrt{}
8 + 8t2
\Bigr)
=
=
1
2t
\biggl(
1\surd
9 + 8t2 +
\surd
8 + 8t2
- 3 +
\sqrt{}
8 + 8t2
\biggr)
<
1
2t
\biggl(
1\surd
9 +
\surd
8
- 3 +
\sqrt{}
8 + 8t2
\biggr)
=
=
1
2t
\Bigl(
3 - 2
\surd
2 - 3 +
\sqrt{}
8 + 8t2
\Bigr)
=
\surd
2 + 2t2 -
\surd
2
t
= \mathrm{t}\mathrm{a}\mathrm{n}\sigma 1.
If we put \mathrm{c}\mathrm{o}\mathrm{s}2 \alpha = x and
\mathrm{s}\mathrm{i}\mathrm{n}2\Delta (R)
4
= c, then by Lemma 5 we conclude that \mathrm{t}\mathrm{a}\mathrm{n}\sigma has the
greatest value \mathrm{m}\mathrm{a}\mathrm{x} (\mathrm{t}\mathrm{a}\mathrm{n}\sigma 1, \mathrm{t}\mathrm{a}\mathrm{n}\sigma 2) =
\surd
2 \mathrm{t}\mathrm{a}\mathrm{n}
\Delta (R)
2
, from which we obtain the thesis of theorem.
Note that by Lemma 1 in general we can not improve the estimate from Theorem 2.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
COVERING A REDUCED SPHERICAL BODY BY A DISK 1409
References
1. L. Danzer, B. Grünbaum, V. Klee, Helly’s theorem and its relatives, Proc. Sympos. Pure Math., Vol. VII, Convexity,
(1963), p. 99 – 180.
2. B. V. Dekster, The Jung theorem for spherical and hyperbolic spaces, Acta Math. Hungar., 67, № 4, 315–331 (1995).
3. H. Han, T. Nishimura, Self-dual Wulff shapes and spherical convex bodies of constant width
\pi
2
, J. Math. Soc. Japan,
69, № 4, 1475 – 1484 (2017).
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5. M. Lassak, Width of spherical convex bodies, Aequat. Math., 89, № 3, 555 – 567 (2015).
6. M. Lassak, H. Martini, Reduced convex bodies in Euclidean space — a survey, Expo. Math., 29, 204 – 219 (2011).
7. M. Lassak, M. Musielak, Reduced spherical convex bodies, Bull. Pol. Acad. Sci. (to appear) (see also
arXiv:1607.00132v1).
8. M. Lassak, M. Musielak, Spherical bodies of constant width, Aequat. Math. (to appear).
9. L. A. Masal’tsev, Incidence theorems in spaces of constant curvature, J. Math. Sci., 72, 3201 – 3206 (1994).
10. J. Molnár, Über einen Übertragung des Hellyschen Satzes in sphärische Räume, Acta Math. Acad. Sci. Hung., 8,
315 – 318 (1957).
11. D. A. Murray, Spherical trigonometry, Longmans Green and Co, London etc. (1900).
12. T. Nishimura, Y. Sakemi, Topological aspect of Wulff shapes, J. Math. Soc. Japan, 66, 89 – 109 (2014).
Received 21.09.17,
after revision — 05.01.18
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 10
|
| id | umjimathkievua-article-6029 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:25:31Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/f8/9efa1d967819f5202e4324806eac8ff8.pdf |
| spelling | umjimathkievua-article-60292025-03-31T08:49:43Z Covering a reduced spherical body by a disk Covering a reduced spherical body by a disk Musielak, M. Musielak, M. spherical convex body spherical geometry hemisphere lune width thickness disk spherical convex body spherical geometry hemisphere lune width thickness disk UDC 514 In this paper, the following theorems are proved: (1) every spherical convex body $W$ of constant width $\Delta (W) \geq \dfrac{\pi}{2}$ may be covered by a disk of radius $\Delta(W) + \arcsin \!\left(\dfrac{2\sqrt{3}}{3} \cos \dfrac{\Delta(W)}{2}\right) - \dfrac{\pi}{2};$ (2) every reduced spherical convex body $R$ of thickness $\Delta(R)&lt;\dfrac{\pi}{2}$ may be covered by a disk of radius $\arctan \!\left(\sqrt{2} \tan \dfrac{\Delta(R)}{2}\right)\!.$ &nbsp; UDC 514 Покриття редукованого сферичного тiла диском У цій статті доведено такі теореми: (1) кожне редуковане сферичне тіло $W$ сталої ширини $\Delta (W) \geq \dfrac{\pi}{2}$ можна покрити диском радіуса $\Delta(W) + \arcsin \!\left(\dfrac{2\sqrt{3}}{3} \cos \dfrac{\Delta(W)}{2}\right) - \dfrac{\pi}{2};$ (2) кожне редуковане сферичне тіло $R$ товщини $\Delta(R)&lt;\dfrac{\pi}{2}$ можна покрити диском радіуса $\arctan \!\left(\sqrt{2} \tan \dfrac{\Delta(R)}{2}\right)\!.$ Institute of Mathematics, NAS of Ukraine 2020-10-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6029 10.37863/umzh.v72i10.6029 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 10 (2020); 1400 - 1409 Український математичний журнал; Том 72 № 10 (2020); 1400 - 1409 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6029/8762 |
| spellingShingle | Musielak, M. Musielak, M. Covering a reduced spherical body by a disk |
| title | Covering a reduced spherical body by a disk |
| title_alt | Covering a reduced spherical body by a disk |
| title_full | Covering a reduced spherical body by a disk |
| title_fullStr | Covering a reduced spherical body by a disk |
| title_full_unstemmed | Covering a reduced spherical body by a disk |
| title_short | Covering a reduced spherical body by a disk |
| title_sort | covering a reduced spherical body by a disk |
| topic_facet | spherical convex body spherical geometry hemisphere lune width thickness disk spherical convex body spherical geometry hemisphere lune width thickness disk |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6029 |
| work_keys_str_mv | AT musielakm coveringareducedsphericalbodybyadisk AT musielakm coveringareducedsphericalbodybyadisk |