Covering a reduced spherical polygon by a disk

Abstract

A spherical convex body R is called reduced if any convex body properly contained in R has thickness smaller than the thickness $\Delta (R)$ of R. We show that for every reduced spherical convex polygon V there exists its boundary point such that the disk of radius $\Delta (V)$ centered at this point contains V.

1 Introduction

In Ref. [2] it is proved that for every reduced polygon in the Euclidan plane there exists a boundary point of this polygon, such that the ball centered at this point and with radius equal to the thickness of the polygon covers it. In this paper we present a proof of an analogous theorem for sphere, but we use here a different approach.

At first recall some basic notions.

By $S^2$ we denote the unit sphere in the three-dimensional Euclidean space $E^{3}$. By a great circle of $S^2$ we mean the intersection of $S^2$ with any two-dimensional subspace of $E^{3}$. The intersection of $S^2$ with a one-dimensional subspace of $E^{3}$ is a pair of points, which is called a pair of antipodes. Observe that if two different points a, b are not antipodes, there is exactly one great circle containing them. This great circle consists of two parts laying between a and b. The shorter of these parts is called the arc connecting a and b, or shortly arc. We denote it by ab. By the spherical distance (or shortly distance) of a and b we mean the length of the arc ab and we denote it by |ab|. For a, b being a pair of antipodes, we put $|ab|=\pi$.

A subset of $S^2$ is called convex if it does not contain any pair of antipodes of $S^2$ and if together with every two points it contains the arc connecting them. A closed convex set with non-empty interior is called a spherical convex body.

Let $\rho \in \left( 0, \frac{\pi }{2}\right]$. By the disk $B_\rho (c)$ of radius $\rho$ and center $c\in S^2$ we mean the set of points of $S^2$ at the distance at most $\rho$ from c. The boundary of a disk is called a circle. We denote by ${\overline{B}}_\rho (c)$ the circle bounding $B_\rho (c)$.

Let a and b be different points of ${\overline{B}}_\rho (c)$ such that a, b, c do not lay on a great circle. Observe that a and b divide ${\overline{B}}_\rho (c)$ into two parts. The smaller of these parts is called a small arc of ${\overline{B}}_\rho (c)$. We denote it by ${\widehat{ab}}$. We say that ${\widehat{ab}}$ is of radius $\rho$ and its center is c. Points a and b are called the endpoints of this small arc.

For any point p the disk $B_{\frac{\pi }{2}}(p)$ is called a hemisphere and is denoted by H(p). If p, q are antipodes, then H(p) and H(q) are called opposite hemispheres. By a semicircle we mean the intersection of a hemisphere with a great circle different from the boundary of this hemisphere. For any boundary point s of a convex body $C \subset S^2$ we say that a hemisphere H supports C at s if $C \subset H$ and s belongs to the great circle bounding H. If F is a subset of an open hemisphere, we define $\text {conv} (F)$ as the smallest convex set containing F. We say that e is an extreme point of C if $C\setminus \{ e\}$ is a convex set. We denote the set of all extreme points of C by E(C).

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 $L \subset S^2$ is defined as the distance of the midpoints of G/H and H/G. The arc connecting these midpoints is called the string of the lune L.

For every hemisphere K supporting a convex body $C\subset S^2$ we find the hemispheres $K^*$ supporting C such that the lunes $K\cap K^*$ are of the minimum thickness (by compactness arguments at least one such a hemisphere $K^*$ exists). The thickness of the lune $K \cap K^*$ is called the width of C determined by K and it is denoted by ${\textrm{width}}_K (C)$ (see [4]). If for all hemispheres K supporting C the numbers ${\textrm{width}}_K (C)$ are equal, we say that C is of constant width (see [4]). By the thickness $\Delta (C)$ of C we understand the minimum width of C determined by K over all supporting hemispheres K of C (see [4]).

After [4] we call a spherical convex body $R \subset S^2$ reduced if $\Delta (Z) < \Delta (R)$ for every convex body $Z \subset R$ different from R. Simple examples of reduced spherical convex bodies on $S^2$ are spherical bodies of constant width and, in particular, the disks on $S^2$. From the paper [5] we see which spherical polygons are reduced. Theorem 3.1 of this paper says that all reduced spherical polygons are of thickness at most $\frac{\pi }{2}$.

Most of above notions are similar to ones in the Euclidan space $E^d$, where we have a wide knowledge about reduced bodies, see for instance [9]. Since this subject is well explored in the Euclidan plane, it is natural to investigate reduced bodies on the sphere. There are many papers about reduced bodies on $S^d$, see [3, 4, 7, 10, 11]. There are also some papers considering reduced bodies in finite-dimensional normal spaces, see for instance [8]. See also the monograph [12], which examines reduced bodies and bodies of constant width in various spaces, in particural on the sphere.

In Refs. [1] and [13] there are solutions of some of spherical reduced bodies problems, which earlier were solved for the plane. In this paper we give one more analogous property.

2 A few lemmas

For the proof of our theorem we need some lemmas.

Lemma 1

Let $F\subset S^2$ and $\rho \in \left( 0, \frac{\pi }{2}\right]$. If $conv (F)\ne S^2$ and $p \in conv (F)$, then $\bigcap _{f\in F}B_\rho (f) \subset B_\rho (p)$.

Proof

In the first case assume that there are extreme points $e_1,e_2$ of $\text {conv} (F)$ such that $p\in e_1e_2$. Clearly $e_1,e_2 \in F$. Let $x\in \bigcap _{f\in F}B_\rho (f)$. In particular $x\in B_\rho (e_1)$ and $x\in B_\rho (e_2)$, which means that $|xe_1|\le \rho$ and $|xe_2|\le \rho$. Let $x'$ be such a point that $p\in xx'$ and $|px|=|px'|$. It is easy to observe that $|x'e_1|=|xe_2|$. Therefore by the triangle inequality for the triangle $xx'e_1$ we get

$$\begin{aligned} |px| = \frac{1}{2} |xx'|\le \frac{1}{2} \left( |xe_1| +|x'e_1|\right) =\frac{1}{2} \left( |xe_1| +|xe_2| \right) \le \max \{ |xe_1|, |xe_2|\} \le \rho . \end{aligned}$$

It means that $x\in B_\rho (p)$ and so $\bigcap _{f\in F}B_\rho (f) \subset B_\rho (p)$, which ends the proof in the first case.

Consider now the second case when p does not belong to any arc with endpoints at extreme points of $\text {conv} (F)$. There exist an extreme point e and a boundary point b of $\text {conv} (F)$ such that $p\in eb$. There also exist extreme points $e_1,e_2$ of $\text {conv} (F)$ such that $b\in e_1e_2$. By the same reason as in the first case we have $B_\rho (e_1)\cap B_\rho (e_2)\subset B_\rho (b)$ and $B_\rho (e)\cap B_\rho (b)\subset B_\rho (p)$. Since $e,e_1,e_2\in F$, we conclude that

$$\begin{aligned} \bigcap _{f\in F}B_\rho (f)\subset B_\rho (e)\cap B_\rho (e_1)\cap B_\rho (e_2)\subset B_\rho (e)\cap B_\rho (b)\subset B_\rho (p). \end{aligned}$$

This ends the proof in the second case. $\square$

For any $F\subset S^2$ we define the set $F_\rho ^{\circ }$ as $\left\{ p: F\subset B_\rho (p)\right\}$.

Lemma 2

If C is a spherical convex body and $\rho \in \left( 0, \frac{\pi }{2}\right]$, then $C_\rho ^{\circ }= \bigcap _{e\in E(C)}B_\rho (e)$.

Proof

We have:

$$\begin{aligned}{} & {} x\in C_\rho ^{\circ }\Leftrightarrow C \subset B_\rho (x) \Leftrightarrow \forall _{p\in C}\ p\in B_\rho (x) \Leftrightarrow \\{} & {} \quad \Leftrightarrow \forall _{p\in C}\ |px| \le \rho \Leftrightarrow \forall _{p\in C}\ x\in B_\rho (p) \Leftrightarrow x \in \bigcap _{p\in C} B_\rho (p). \end{aligned}$$

Thus in order to end the proof it is sufficient to show that $\bigcap _{p\in C} B_\rho (p)=\bigcap _{e\in E(C)} B_\rho (e)$.

Since $E(C) \subset C$, it is obvious that $\bigcap _{p\in C} B_\rho (p)\subset \bigcap _{e\in E(C)} B_\rho (e)$. Let us show that $\bigcap _{p\in C} B_\rho (p)\supset \bigcap _{e\in E(C)} B_\rho (e)$. Take any point $p\in C$. From $p \in \text {conv} (E(C))$ and Lemma 1 we conclude that $\bigcap _{e\in E(C)} B_\rho (e) \subset B_\rho (p)$. This inclusion holds true for every $p\in C$, therefore $\bigcap _{e\in E(C)} B_\rho (e) \subset \bigcap _{p\in C}B_\rho (p)$ which ends the proof. $\square$

The following lemma is a special case of Lemma 3 from [6].

Lemma 3

If the endpoints of a small arc ${\widehat{ab}}$ of radius $\rho$ are contained in $B_\rho (c)$, then ${\widehat{ab}}\subset B_\rho (c)$.

Next lemma describes the shape of the boundary of a spherical convex body. We denote the boundary of C in the common way by $\textrm{bd} (C)$ and its diameter by $\textrm{diam}(C)$.

Lemma 4

Let F be the intersection of a finite number of different disks of radius $\rho$ and F has the non-empty interior. Then $bd (F)$ consists of small arcs of radius $\rho$ with different centers.

Proof

We apply induction. For two disks the thesis of lemma is obvious. Assume that the intersection F of a finite family ${\mathcal {D}}$ of disks of radius $\rho$ has non-empty interior and its boundary consists of small arcs of radius $\rho$ with different centers. Intersect F with a disk B of radius $\rho$ whose center is different from the centers of the disks from ${\mathcal {D}}$. Also assume that $F\cap B$ has non-empty interior. If $F\subset B$, then $F\cap B= F$ and by the assumption we already know that the inductive thesis holds true. Otherwise the circle bounding B intersects the interior of F. Observe that if points a, b belong to $\textrm{bd}(B) \cap F$, then they belong to every disk from ${\mathcal {D}}$. Therefore by Lemma 3 it follows that ${\widehat{ab}}\subset \textrm{bd}(B)$. Hence ${\widehat{ab}}$ is a subset of every disk from ${\mathcal {D}}$ and thus it is a subset of F. As a consequence, $\textrm{bd}(B) \cap F$ is a small arc. Without losing the generality we may assume that a and b are the endpoints of this arc.

We conclude that the boundary of $B\cap F$ consists of the small arc ${\widehat{ab}}$ and small arcs being parts of boundaries of some disks from ${\mathcal {D}}$. Therefore also in this case the inductive thesis holds true. $\square$

We define $E^*(V)$ as $\left\{ e\in E(V): |\textrm{bd}(V_r^{\circ }) \cap {\overline{B}}_r(e)|>1\right\}$. From Lemmas 2 and 4 we immediately obtain the following corollary.

Corollary 1

If V is a spherical convex polygon, then $V_\rho ^{\circ }= \bigcap _{e\in E^{*}(V)}B_\rho (e)$. For every point $e\in E^*(V)$ the intersection of $\textrm{bd}(V_\rho ^{\circ })$ and ${\overline{B}}_\rho (e)$ is a small arc of ${\overline{B}}_\rho (e)$.

Applying this tool we prove the following theorem.

3 The covering by a disk

Theorem

Every spherical reduced polygon V is contained in a disk of radius $\Delta (V)$ centered at a boundary point of V.

Proof

Recall that $V_{\Delta (V)}^{\circ }$ is the set of centers of disks of radius $\Delta (V)$ that contain V. So it is sufficient to show that $V_{\Delta (V)}^{\circ }$ has a common point with the boundary of V.

Assume the opposite that $V_{\Delta (V)}^{\circ }$ is a subset of the interior of V.

By order we always mean the order according to the positive orientation.

Let $e_1, e_2, \ldots , e_m$ be all points of $E^*(V)$ laying on the boundary of V in that order. Put $e_{m+1}=e_1$. For $i=1,2, \ldots , m$ denote by $p_{i+1}$ the point of $V_{\Delta (V)}^{\circ }$ which belongs to ${\overline{B}}_{\Delta (V)}(e_i)\cap {\overline{B}}_{\Delta (V)}(e_{i+1})$. Put $p_{m+1}=p_1$. According to Corollary 1 we see that the boundary of $V_{\Delta (V)}^{\circ }$ consists only of small arcs $\widehat{p_ip_{i+1}}$ of ${\overline{B}}_r(e_i)$.

Since for every $i=1,2,\ldots , m$ the point $e_i$ is an extreme point of V, by Theorem 4 of Ref. [4] there exists a hempishere $L_i$ of thickness $\Delta (V)$ containing V, whose string has an endpoint at $e_i$. Denote the other end of this string by $b_i$.

Since $|e_1b_1| = \Delta (V)$, we see that $b_1$ lays in the boundary of $B_{\Delta (V)}(e_1)$. But from our assumption that $V_{\Delta (V)}^{\circ }$ is a subset of the interior of V we conclude that $b_1\notin \widehat{p_1p_2}$. Therefore $b_1$ lays either on the same side of the great circle containing $e_1p_2$ as $p_3$, or on the same side of the great circle containing $e_1p_1$ as $e_2$. Without losing the generality we can assume that the first of these possibilites takes place.

Let v, u be points of $\textrm{bd}(V)$ such that |vu| is equal to $\textrm{diam}(V)$. By Proposition 4.1 of Ref. [5] we know that v and u are vertices of V and $\{ v,u\}$ = $\{ v_i, v_j\}$ for some i, j which differ by $\frac{n-1}{2}$ or $\frac{n+1}{2}$ modulo n. Let t and s be points of $\textrm{bd}(V)$ such that vt and su are strings of lunes of thickness $\Delta (V)$ containing V. Observe that exactly one of the following statements holds true: t, the midpoint of tu and u lay on the boundary of V in this order, or s, the midpoint of sv and v lay on the boundary of V in this order. We may assume without losing the generality, that the first of these statements holds true.

Clearly, there exists i such that points $e_i$, v and $e_{i+1}$ lay on the boundary of V in this order. We do not lose the generality assuming that $i=1$, i.e. that $e_1$, v and $e_2$ are in this order. We do not exclude the possibility that $e_1=v$, but we do not allow the possibility that $e_2=v$.

By Corollary 3.8 from Ref. [5] the string of every lune of thickness $\Delta (V)$ containing V halves perimeter of V. Thus strings of every pair of lunes of thickness $\Delta (C)$ containing C intersect. Hence we conclude that $b_1$, t and $b_2$ lay on the boundary of V in this order. Observe that there exists $k\ge 2$ such that t does not belong to $B_{\Delta (V)}(e_k)$, otherwise $t\in V_{\Delta (V)}^{\circ }$. Clearly, such $e_k$ lays on the opossite side of the great circle containing $e_1p_2$ than t.

Since $|te_k|>\Delta (V)$, there exists a point x such that $x\in te_k$ and $|tx|=\Delta (V)$. The triangle vxt is isosceles with $|tv|=|tx|$. Consider the great circle containing t and the midpoint m of the arc vx. Clearly it is the set of all points of $S^2$ equally distanced from v and x. But since $|\angle mtu| = |\angle mtv| + |\angle vtu|= |\angle mtv| + \frac{\pi }{2}< \pi$, we conclude that u lays on the same side of the great circle containing mt as v. Therefore $|ux|>|uv| = \textrm{diam}(V)$. But this is impossible, because $x\in V$ and there does not exist an arc inside V longer than $\textrm{diam}(V)$. The obtained contradiction ends the proof. $\square$