Diameter of Reduced Spherical Convex Bodies
The intersection of two different non-opposite hemispheres of the unit sphere is called a lune. By we denote the distance of the centers of the semicircles bounding . By the thickness of a convex body we mean the minimal value of over all lunes . We call a convex body reduced provided for every convex body being a proper subset of . Our aim is to estimate the diameter of , where , in terms of its thickness.
Key words and phrases:spherical convex body, spherical geometry, hemisphere, lune, width, constant width, thickness, diameter
2010 Mathematics Subject Classification:52A55
Let be the unit sphere of the -dimensional Euclidean space . A great circle of is the intersection of with any two-dimensional linear subspace of . By a pair of antipodes of we mean any pair of points being the intersection of with a one-dimensional subspace of . Observe that if two different points are not antipodes, there is exactly one great circle passing through them. By the arc connecting them we understand the shorter part of the great circle through and . By the distance of and we mean the length of the arc . The notion of the diameter of any set not containing antipodes is taken with respect to this distance and denoted by .
A subset of is called convex if it does not contain any pair of antipodes of and if together with every two points it contains the arc connecting them. A closed convex set with non-empty interior is called a convex body. Its boundary is denoted by . If in there is no arc, we say that the body is strictly convex. Convexity on is considered in very many papers and monographs. For instance in , , , ,  and .
The set of points of in the distance at most , where , from a point is called a spherical disk, or shorter a disk, of radius and center . Disks of radius are called hemispheres. Two hemispheres whose centers are antipodes are called opposite hemispheres. The set of points of a great circle of which are at distance at most from a fixed point of this great circle is called a semicircle. We call the center of this semicircle.
Let be a convex body and . We say that a hemisphere supports at provided and is in the great circle bounding .
Since the intersection of every family of convex sets is also convex, for every set contained in an open hemisphere of there is the smallest convex set containing . We call it the convex hull of .
If non-opposite hemispheres and are different, then is called a lune. The semicircles bounding and contained in and , respectively, are denoted by and . The thickness of is defined as the distance of the centers of and .
After  recall a few notions. For any hemisphere supporting a convex body we find a hemisphere supporting such that the lune is of the minimum thickness (by compactness arguments at least one such a hemisphere exists). The thickness of the lune is called the width of determined by and it is denoted by . By the thickness of a convex body we understand the minimum of over all supporting hemispheres of . We say that a convex body is reduced if for every convex body different from we have . This definition is analogous to the definition of a reduced body in normed spaces (for a survey of results on reduced bodies see ). If for all hemispheres supporting the numbers are equal, we say that is of constant width. Spherical bodies of constant width are discussed in  and applied in .
Just bodies of constant width, and in particular disks, are simple examples of reduced bodies on . Also each of the four parts of a disk dissected by two orthogonal great circles through the center of this disk is a reduced body. It is called a quarter of a disk. There is a wide class of reduced odd-gons on (see ). In particular, the regular odd-gons of thickness at most are reduced.
2. Two lemmas
Let be a lune of thickness at most whose bounding semicircles are and . For every in such that and for every we have .
If and is the center of , then the distance between and any point of is the same, and thus the assertion is obvious. Consider the opposite case when , or but is not the center of . Clearly, the closest point to is unique. Observe that for the distance increases as the distance increases. This easily implies the assertion of our lemma. ∎
In a standard way, an extreme point of a convex body is defined as a point for which the set is convex (see ). The set of extreme points of is denoted by .
For every convex body of diameter at most we have .
In order to show the opposite inequality , thanks to , it is sufficient to show that for every . If , this is trivial. In the opposite case, at least one of these points does not belong to . If, say , then having in mind that is a convex body and we see that there are different from such that . From , we see that . Since also , the arc is a subset of .
Recall that by Theorem 3 of  we have for every hemisphere supporting . In particular, for the hemisphere supporting at all points of the arc . Thus by the assumption that we obtain for our particular . Hence we may apply Lemma 2.1 taking this in the part of there. We obtain .
If , from we conclude that . If , from we see that there are such that . Similarly to the preceding paragraph we show that and . By these two inequalities and by the preceding paragraph we get , which ends the proof. ∎
The assumption that is substantial in Lemma 2.2, as it follows from the example of a regular triangle of any diameter greater than . The weaker assumption that is not sufficient, which we see taking in the part of any isosceles triangle with and the arms longer than (so with the base shorter than ). The diameter of equals to the distance between the midpoint of the base and the opposite vertex of . Hence is over the length of each of the sides.
3. Diameter of reduced spherical bodies
For every reduced spherical body with we have . This value is attained if and only if is the quarter of disk of radius . If , then .
Assume that . In order to show the first statement, by Lemma 2.2 it is sufficient to show that the distance between any two different points of is at most . Since is reduced, according to the statement of Theorem 4 in  there exist lunes , where , of thickness with as the center of one of the two semicircles bounding (see Figure). Denote by the center of the other semicircle bounding .
If or , then , which by is at most . Otherwise is a non-degenerate spherical quadrangle with points , , , in its consecutive sides. Therefore, since , arcs and intersect at exactly one point. Denote it by . Observe that it may happen .
Let be the great circle orthogonal to the great circle containing and passing through . Since , we see that intersects . Let
FIGURE. Illustration to the proof of Theorem 3.1.
be the intersection point of them. Note that we do not exclude the case . From we see that . Thus from the right spherical triangle we conclude that . Moreover, from and we obtain . Consequently, from the formula for the right spherical triangle with hypotenuse and legs applied to the triangle (again see Figure) we obtain .
Observe that thanks to , the shown inequality becomes the equality only if . In this case, by Proposition 3.2 of , our body is a quarter of disk of radius .
Finally, we show the last statement of the theorem. Assume that . By Theorem 4.3 of , the body is of constant width .
There is an arc of length equal to . Clearly, . Take a lune from Theorem 5.2 of  such that is the center of a semicircle bounding . Denote by the center of the other semicircle bounding . By the third part of Lemma 3 in , we have for every different from . Hence for every . So also for every . In particular, . Since and , we obtain .
Let us show the opposite inequality.
From the inequalities obtained in the two preceding paragraphs we get the equality . Hence the last sentence of the theorem is proved. ∎
Proposition 3.5 of  implies that for every reduced spherical body with on we have . Here is a more precise form of this statement.
Let be a reduced body. Then if and only if . Moreover, if and only if .
We start with proving the first statement of our proposition. The function is increasing in the interval as a composition of the decreasing functions and . From we conclude that in the interval the function accepts only the values below . Thus by Theorem 3.1, if , then . The opposite implication results from the inequality for every spherical convex body , which in turn follows from Theorem 3 and Proposition 1 of .
Let us show the second part of our proposition.
Assume that . The inequality is impossible, by the first statement of our proposition. Also the inequality is impossible, because by (see Proposition 1 of ), it would imply in contradiction to the assumption at the beginning of this paragraph. So .
Now assume that . By the second statement of Theorem 3.1 we get . ∎
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