The area of reduced spherical polygons
TThe area of reduced spherical polygons Cen Liu † , Yanxun Chang † , Zhanjun Su ‡† Department of Mathematics, Beijing Jiaotong University, Beijing 100044, China ‡ School of Mathematical Sciences, Hebei Normal University, Shijiazhuang, 050024, [email protected], [email protected], [email protected]
Abstract:
We confirm two conjectures of Lassak on the area of reduced spherical poly-gons. The area of every reduced spherical non-regular n -gon is less than that of the reg-ular spherical n -gon of the same thickness. Moreover, the area of every reduced sphericalpolygon is less than that of the regular spherical odd-gons of the same thickness andwhose number of vertices tends to infinity. Keywords : reduced convex body, spherical polygon, thickness, area
Mathematics Subject Classification (2010) : 52A55
We focus on the reduced convex bodies introduced by Heil in [2]. Reduced convex bodiesare helpful for solving various extremal problems concerning the minimal width of convexbodies. Some basic properties of the reduced convex bodies in two-dimensional Euclideanspace E are introduced by Lassak in [3]. Lassak [7] demonstrates that in E the areaof every reduced non-regular n -gon is less than that of the regular n -gon of the samethickness.The notions about reduced convex bodies are extended to the d -dimensional unitsphere S d in [4]. Lassak [4, 5] discusses the properties of reduced convex bodies on S d ;he [6] further characterizes reduced convex polygons on S and proposes the followingconjectures:(1) The area of every reduced spherical polygon is less than that of the regular sphericalodd-gons of the same thickness and whose number of vertices tends to infinity .(2)
The area of every reduced spherical non-regular n -gon is less than that of theregular spherical n -gon of the same thickness .In Section 2, we present the necessary notions of reduced spherical convex bodies andreview some results in the literature. Several useful lemmas are established in Section 3.Then Section 4 aims to confirm the above two conjectures. Let S be the unit sphere in E centered at the origin. In this paper, all the notionsare discussed in S . A great circle is the intersection of S with any two-dimensionalsubspace of E . A pair of antipodes are the intersection of S with any one-dimensionalsubspace of E . Supported by NSFC under Grant 11971053 and Science Foundation of Hebei Normal UniversityL2020Z01. a r X i v : . [ m a t h . M G ] S e p learly, if different points a, b ∈ S are not antipodes, then there is exactly one greatcircle containing them; denote by arc ab , shortly ab , the shorter part of the great circlecontaining them. The spherical distance | ab | , or shortly distance of a and b is the lengthof ab .A set C ⊂ S is called convex , if it does not contain any pair of antipodes of S andfor arbitrary points a, b ∈ C , it is true that arc ab ⊆ C . The convex body is a closedconvex set with non-empty interior.By a spherical disk of radius r ∈ (0 , π/
2] and center k ∈ S we mean the set B = { p : | pk | ≤ r, p ∈ S } ; and the boundary bd( B ) is called a spherical circle . The sphericaldisk of radius π/ hemisphere . If hemispheres G and H are different and theircenters are not antipodes, then L = G ∩ H is called a lune of S . The parts of bd( G ) andbd( H ) contained in G ∩ H are denoted by G/H and
H/G , respectively. We define the thickness ∆( L ) of the lune L = G ∩ H as the distance of the centers of G/H and
H/G .We recall some notions in [4]. We say that a hemisphere H supports a convex body C ⊆ S at point p if C ⊆ H and p ∈ bd( H ) ∩ C . For any convex body C ⊆ S andany hemisphere K supporting C , we define the the width of C determined by K as theminimum thickness of a lune K ∩ K ∗ over all hemispheres K ∗ (cid:54) = K supporting C and wedenote it by width K ( C ); the thickness of C is defined by∆( C ) = min { width K ( C ) : K is a supporting hemisphere of C } . The thickness of C is nothing else but the minimum thickness of a lune containing C .A convex body C ⊂ S is said to be reduced if ∆( R ) < ∆( C ) for each convex body R being a proper subset of C .We recall some definitions given in [6]. Let p be a point in a hemisphere differentfrom its center and let l be the great circle bounding this hemisphere. The projection of p on l is the point t such that | pt | = min {| pc | : c ∈ l } . If C is a subset of a convex set of S , then the intersection of all convex sets containing C is called a convex hull of C . Theconvex hull of k ≥ S such that each of them does not belong to the convexhull of the remaining points is called a spherically convex k -gon. If V is a sphericallyconvex k -gon, we denote by v , . . . , v k the vertices of V in the counterclockwise order. Aspherically convex polygon with sides of equal length and interior angles of equal measureis called a regular spherical polygon . Lemma 2.1 [9, Theorem 41.2]
Let V be a convex n -gon in the unit sphere with angles γ , γ , . . . , γ n , then area( V ) = γ + . . . + γ n − ( n − π . For a convex odd-gon V = v v · · · v n , by the opposite side to the vertex v i we meanthe side v i +( n − / v i +( n +1) / , the indices are taken modulo n . Lemma 2.2 [6, Theorem 3.2]
Every reduced spherical polygon is an odd-gon of thicknessat most π . A spherically convex odd-gon V with ∆( V ) < π is reduced if and only if theprojection of every its vertices on the great circle containing the opposite side belongs tothe relative interior of this side and the distance of this vertex from this side is ∆( V ) . Lemma 2.3 [6, Corollary 3.3]
Every regular spherical odd-gon of thickness at most π is reduced. Working lemmas
We recall a few formulas of spherical geometry in [8] which are the basic method for theresearch. Consider the right spherical triangle with hypotenuse c and legs a, b , we use A, B and C to represent the corresponding angles of edges a, b and c , respectively. Thencos B = cos b sin A, (1)cos A = tan b cot c, (2)sin b = sin c sin B. (3)In a reduced spherical polygon V = v v · · · v n , according to Lemma 2.2, we give somerelated notations. Denote by t i the projection of v i in the opposite side v i +( n − / v i +( n +1) / .Denote by o i the intersection of v i t i and v i +( n +1) / t i +( n +1) / ; put α i = ∠ v i +1 v i t i , β i = ∠ t i v i v i +( n +1) / , and ϕ i = ∠ v i o i t i +( n +1) / = ∠ t i o i v i +( n +1) / , where i ∈ { , , . . . , n } . Forexample, Fig. 1 presents some notations in a reduced spherical pentagon.Fig. 1: Some notations Lemma 3.1 [6, Corollary 3.9] If V = v v · · · v n is a reduced spherical polygon with ∆( V ) < π , then β i ≤ α i . Actually, we can gain the following lemma by Corollary 3 . Lemma 3.2
For every reduced spherical polygon V = v v · · · v n with ∆( V ) < π , thespherical triangles v i o i t i +( n +1) / and v i +( n +1) / o i t i are congruent, where i ∈ { , , . . . , n } . Proof
Lemma 2.2 shows that | v i t i | = | v i +( n +1) / t i +( n +1) / | = ∆( V ), where i ∈ { , . . . , n } .From this and | v i v i +( n +1) / | = | v i +( n +1) / v i | , we find that the right spherical triangles v i t i v i +( n +1) / and v i +( n +1) / t i +( n +1) / v i are congruent. Then we acquire α i + β i = ∠ t i +( n +1) / v i v i +( n +1) / = ∠ v i v i +( n +1) / t i (4)3nd β i = ∠ t i v i v i +( n +1) / = ∠ v i v i +( n +1) / t i +( n +1) / . (5)From (4) and (5), we obtain that α i = ∠ t i +( n +1) / v i o i = ∠ t i v i +( n +1) / o i . Applying thisand ∠ v i o i t i +( n +1) / = ∠ t i o i v i +( n +1) / = ϕ i , it turns out that the right spherical triangles v i o i t i +( n +1) / and v i +( n +1) / o i t i are congruent. (cid:50) Lemma 3.3
All the vertices of a regular spherical polygon are contained in a sphericalcircle.
Proof
Let V = v v · · · v n be a regular spherical polygon. Denote by γ the length of theside of V and by θ the interior angle of V . Let D be a spherical circle passing through v , v and v , and denote by o be the center of the spherical disk whose boundary is D .Connect o with v i , where i ∈ { , . . . , n } .In the spherical triangles ov v and ov v , we have | ov | = | ov | = | ov | . By thedefinition of regular spherical polygon, we have | v v | = | v v | = γ , then ov v and ov v are congruent. Since | ov | = | ov | = | ov | , we have ∠ ov i v i +1 = ∠ ov i +1 v i = θ , where i = 1 , ov v and ov v . From | ov | = | ov | , ∠ ov v = ∠ ov v = θ ,and | v v | = | v v | , we obtain that ov v and ov v are congruent. Hence we have | ov | = | ov | = | ov | = | ov | .Similarly, the spherical triangles ov v and ov j v j +1 are congruent, where j ∈ { , . . . , n } and the indices are taken modulo n . Hence | ov | = | ov | = | ov j | = | ov j +1 | with j ∈ { , . . . , n } . Consequently, we have | ov | = | ov | = · · · = | ov n | , and thus v , v , · · · , v n are all contained in D . (cid:50) Fact 3.4
For every reduced spherical polygon V = v v · · · v n with ∆( V ) < π , we have < ϕ i < π , where i ∈ { , . . . , n } . Proof
Consider the right spherical triangle v i t i v i +( n +1) / for every i ∈ { , . . . , n } . From(1), we obtain cos( α i + β i ) = cos | v i t i | sin β i . As | v i t i | = ∆( V ) < π , we get α i + β i < π .Applying this and Lemma 3.1, it follows that 2 β i ≤ α i + β i < π .It is obvious that the area of the spherical triangle o i v i v i +( n +1) / is nonnegative.Hence by Lemma 2.3 and (5), we get that β i + β i + ( π − ϕ i ) − π ≥
0. Therefore, ϕ i ≤ β i < π . The proof is complete. (cid:50) By a rotation of a set C ⊆ S around a point p ∈ S , we mean the rotation of C around the straight line through p and the center of E . For any two points a, b ∈ bd( C ), ab is called a chord of C ⊆ S . Lemma 3.5
For any reduced spherical polygon V = v v · · · v n with ∆( V ) < π , we have (cid:80) ni =1 ϕ i ≥ π . Proof
Lemma 2.2 shows that V is an odd-gon. We use a similar technic as thatof Lemma in [7] to show the statement. Let B i = v i o i t i +( n +1) / ∪ v i +( n +1) / o i t i , then B i ⊆ V , where i ∈ { , . . . , n } . Thus we have B ∪ · · · ∪ B n ⊆ V .We intend to show V ⊆ B ∪ · · · ∪ B n . We present every B i as the union of chords of V which pass through o i . All the chords of successively B , B n +1) / , . . . , B ( n +1) / are4n big circles which step by step rotate changing the centers of rotation; those centerssuccessively are o , o n +1) / , . . . , o ( n +1) / . We assume that all the above chords in B i are oriented with the origins in v i t i +( n +1) / . For any point p ∈ V , we assume that p is inthe left hand side of v t . When we start from v t , after total rotation by ϕ + · · · + ϕ n ,we arrive at t v which has the opposite direction. Now p is in the right hand side of theoriented chord v t . Since the described changes of v t are continuous, there is a positionsuch that the chord contains p . Hence p ∈ B ∪ · · · ∪ B n and then V ⊆ B ∪ · · · ∪ B n Consequently, V = B ∪ · · · ∪ B n . Claim 1: If V is a non-regular reduced spherical polygon, then (cid:80) ni =1 ϕ i ≥ π .From Lemmas 2.1 and 3.2, the area of V is S V = 2( α + · · · + α n ) − ( n − π . Thearea of B i is S B i = 2( ϕ i + α i − π ). From V = B ∪ · · · ∪ B n , we have S V ≤ (cid:80) ni =1 S B i ,that is 2 n (cid:88) i =1 α i − ( n − π ≤ n (cid:88) i =1 ( ϕ i + α i − π , (6)and then (cid:80) ni =1 ϕ i ≥ π . Claim 2: If V is a regular spherical polygon, then (cid:80) ni =1 ϕ i = π and ϕ i = πn , where i ∈ { , , . . . , n } .In this case, by Lemma 3.3, we have o = · · · = o n = o (the notation o is describedin Lemma 3.3). Then, B i = v i ot i +( n +1) / ∪ v i +( n +1) / ot i , where i ∈ { , , . . . , n } . Clearly,for arbitrary i, j ∈ { , , . . . , n } , B i and B j are congruent, and their interiors satisfyint( B i ) ∩ int( B j ) = ∅ . Thus we have S V = (cid:80) ni =1 S B i and then (6) becomes2 n (cid:88) i =1 α i − ( n − π = 2 n (cid:88) i =1 ( ϕ i + α i − π . Hence we obtain that (cid:80) ni =1 ϕ i = π and ϕ i = πn , where i ∈ { , , . . . , n } . (cid:50) In the following lemmas, we investigate the monotonicity and concavity of two kindsof functions, respectively, which are needed in Section 4.
Lemma 3.6
Let f ( x ) = arccos x √ λ λ − x and f ( x ) = arccos x (1+ λx ) λ − x . Then f ( x ) f ( x ) is adecreasing function of x , where λ ∈ (0 , + ∞ ) and x ∈ (0 , − √ λ λ ) . Proof
Set f ( x ) = f ( x ) f ( x ) . Let us show that f (cid:48) ( x ) <
0. The derivative of f ( x ) is f (cid:48) ( x ) = λ · h ( x )( λ − x ) √ x √− λ x − λx + λ f ( x ) , where h ( x ) = − (cid:112) (1 + λ )(1 + x ) f ( x ) + ( − x + 2 λx + 1) f ( x ) . From x < − √ λ λ < λ , and if h ( x ) <
0, then f (cid:48) ( x ) <
0. The first derivative of h ( x ) is h (cid:48) ( x ) = − x √ λ √ x f ( x ) + 2( λ − x ) f ( x ) . The second derivative of h ( x ) is h (cid:48)(cid:48) ( x ) = − λ √ λ √− λ x − λx + λ · λ − x − x (1 + x )( λ − x ) − √ λ (1 + x ) / f ( x ) − f ( x ) . x ∈ (0 , − √ λ λ ), it follows that λ − x > λ − x − x >
0. Also we have f ( x ) > f ( x ) >
0. Hence h (cid:48)(cid:48) ( x ) < h (cid:48) ( x ) > h (cid:48) ( − √ λ λ ) = 0. Thusfrom h (cid:48) ( x ) >
0, we get h ( x ) < h ( − √ λ λ ) = 0. Therefore, f (cid:48) ( x ) < (cid:50) Lemma 3.7
Let F ( x ) = arcsin g ( x ) √ λ λ − g ( x ) and g ( x ) = − (1+cos x )+ √ (1+cos x ) +4 λ cos x λ ,where λ ∈ (0 , + ∞ ) and x ∈ (0 , π ) . Then F (cid:48) ( x ) < and F (cid:48)(cid:48) ( x ) < . Proof
For convenience, set r ( x ) = (cid:112) (1 + cos x ) + 4 λ cos x . We find the first derivativeof F ( x ) is F (cid:48) ( x ) = − λ √ λ sin xr ( x ) √ − cos x (cid:112) λ + cos x − r ( x ) . Therefore, we obtain F (cid:48) ( x ) <
0. The second derivative of F ( x ) is F (cid:48)(cid:48) ( x ) = λ √ λ + 2 λ sin x (cid:0) − x ) − λ + 2(1 + cos x ) r ( x ) (cid:1) (1 − cos x ) r ( x ) (cid:112) λ + cos x − r ( x ) . We can check that − x ) − λ + 2(1 + cos x ) r ( x ) < − cos x ) r ( x ) > x ∈ (0 , π ). Hence F (cid:48)(cid:48) ( x ) < F ( x ) is a concave function of x . (cid:50) This section aims to prove the conjectures mentioned in the introduction. For ease ofnotations, we use ω to replace the thickness of a reduced spherical polygon in this part.Here we have ω ∈ (0 , π/ λ = tan ω , then λ ∈ (0 , + ∞ ). Denote by S the area of areduced spherical polygon.Let us define several functions which are needed in the following theorems. Set f ( x ) = arcsin x √ λ λ − x , f ( x ) = arccos x (1 + λx ) λ − x ,f ( x ) = π − f ( x ) = arccos x √ λ λ − x , where x ∈ (cid:0) , ( − √ λ ) /λ (cid:1) . Set g ( ϕ ) = − (1 + cos ϕ ) + (cid:112) (1 + cos ϕ ) + 4 λ cos ϕ λ , where ϕ ∈ (0 , π/ g ( ϕ ) ∈ (cid:0) , ( − √ λ ) /λ (cid:1) . Lemma 4.1
For a reduced spherical polygon V = v · · · v n with ω ∈ (0 , π ) , the area is S = 2 (cid:80) ni =1 f ( y i ) − ( n − π , where y i = g ( ϕ i ) . Proof
For every i ∈ { , , . . . , n } , we focus on the right spherical triangle o i t i v i +( n +1) / .Put | o i t i | = b i and | o i v i +( n +1) / | = c i . By Lemma 3.2 we have | o i t i | + | o i v i +( n +1) / | = b i + c i = ω . Here we have b i < ω . From (2), we obtaincos ϕ i = tan b i tan c i = tan b i tan( ω − b i ) = tan b i (1 + tan ω tan b i )tan ω − tan b i . (7)6y a simple calculation, we get that tan b i = g ( ϕ i ), for simplicity, we denote g ( ϕ i ) by y i .Hence b i = arctan y i . Fact 3.4 shows that 0 < ϕ i < π . From this and b i < ω , it followsthat y i ∈ (0 , − √ λ λ ).Hence (7) becomes cos ϕ i = y i (1+ λy i ) λ − y i and then ϕ i = f ( y i ) = arccos y i (1+ λy i ) λ − y i . More-over, we have tan c i = tan b i cos ϕ i = λ − y i λy i and thus c i = arctan λ − y i λy i .From (3), we obtainsin α i = sin b i sin c i = sin arctan y i sin arctan λ − y i λy i = y i √ λ λ − y i , and thus α i = f ( y i ) = arcsin y i √ λ λ − y i . Then Lemmas 2.1 and 3.2 imply that the area of V is S = 2 n (cid:88) i =1 α i − ( n − π = 2 n (cid:88) i =1 f ( y i ) − ( n − π. (8) (cid:50) Theorem 4.2
The regular spherical n -gon has the maximum area among all regularspherical k -gons of fixed thickness, with odd numbers k, n and ≤ k ≤ n . Proof
Let V = v v · · · v k be a regular spherical odd-gon. Lemma 2.3 shows that V isreduced, then we use the same notations as that in Lemma 4.1. By Claim 2 in Lemma3.5, we have ϕ = · · · = ϕ k = πk . Thus y = · · · = y k = g ( πk ), where g ( πk ) = − (1 + cos πk ) + (cid:112) (1 + cos πk ) + 4 λ cos πk λ , and g ( πk ) ∈ (0 , − √ λ λ ). For simplicity, we denote ϕ i and y i by ϕ and y , respectively,for every i ∈ { , , . . . , k } . From the proof process of Lemma 4.1, we can easily see that k = πϕ = πf ( y ) , where f ( y ) = ϕ = arccos y (1+ λy ) λ − y .From k = πf ( y ) and f ( y ) + f ( y ) = π , it follows that (8) becomes S = − π f ( y ) f ( y ) + 2 π .By Lemma 3.6, we get that f ( y ) f ( y ) is a decreasing function of y . Since y = g ( πk ),one can check that y is an increasing function of k . Consequently, f ( y ) f ( y ) is a decreasingfunction of k .Because S = − π f ( y ) f ( y ) +2 π , the above analysis implies that S is an increasing functionof k . This completes the proof. (cid:50) Corollary 4.3
The area of the regular spherical odd-gon with thickness ω ∈ (0 , π ) is − cos ω ) π when the number of vertices tends to infinity. Proof
Let V = v v · · · v n be a regular spherical odd-gon. Since λ = tan ω , we getcos ω λ (cid:114) λ (cid:115) √ λ + 1 + 12 √ λ + 1 .
7y Theorem 4.2, we get that the area of V is S = − π f ( y ) f ( y ) + 2 π , where y = g ( πn ) = − (1+cos πn )+ √ (1+cos πn ) +4 λ cos πn λ . When n tends to infinity, y tends to t = − √ λ +1 λ . Thenfrom lim y → t f ( y ) = 0 and lim y → t f ( y ) = 0, by using L (cid:48) Hospital rule, we getlim y → t f ( y ) f ( y ) = lim y → t f (cid:48) ( y ) f (cid:48) ( y ) = lim y → t √ λ (cid:112) y − y + 2 λy + 1 = (cid:115) √ λ + 1 + 12 √ λ + 1 . Consequently, lim n → + ∞ S = 2(1 − cos ω ) π . (cid:50) The next theorem shows that the second conjecture mentioned in the introduction istrue.
Theorem 4.4
The area of every reduced spherical non-regular n -gon is less than that ofthe regular spherical n -gon of the same thickness. Proof
Let V = v · · · v n be a reduced spherical odd-gon. By Lemma 4.1, the area of V is S = 2 (cid:80) ni =1 F ( ϕ i ) − ( n − π , where F ( ϕ i ) = f ( g ( ϕ i )) = arcsin g ( ϕ i ) √ λ λ − g ( ϕ i ) and by Fact3.4, we have ϕ i ∈ (0 , π ).By Lemma 3.7, we obtain that F ( x ) is a concave function of x . Thus from Jensen’sinequality [1], we have F ( ϕ ) + · · · + F ( ϕ n ) n ≤ F ( ϕ + · · · + ϕ n n ) , the equality holds when ϕ = · · · = ϕ n . Then the area of V satisfies S = 2 n ( F ( ϕ ) + · · · + F ( ϕ n ) n ) − ( n − π ≤ nF ( ϕ + · · · + ϕ n n ) − ( n − π. Case 1. If V is a regular spherical polygon, then by Claim 2 in Lemma 3.5 we have ϕ + · · · + ϕ n = π and ϕ = · · · = ϕ n = πn . In this case, S = 2 nF ( πn ) − ( n − π . Case 2. If V is a non-regular spherical polygon, then by Claim 1 in Lemma 3.5 wehave ϕ + · · · + ϕ n ≥ π .By Lemma 3.7, we obtain that F ( x ) is a decreasing function of x . Since ϕ + · · · + ϕ n ≥ π , it follows that F ( ϕ + ··· + ϕ n n ) ≤ F ( πn ). In this case, we have S ≤ nF ( ϕ + ··· + ϕ n n ) − ( n − π ≤ nF ( πn ) − ( n − π .The above two cases show that the area of V always satisfies S ≤ nF ( πn ) − ( n − π ,which is exactly the area of the regular spherical n -gon. This completes the proof. (cid:50) By Theorems 4.2, 4.4 and Corollary 4.3 we obtain the following corollary which showsthat the first conjecture is true.
Corollary 4.5
The area of every reduced spherical polygon V is less than − cos ∆( V )2 ) π ,which is the limit value for the area of the regular spherical odd-gons whose number ofvertices tends to infinity. eferences ∼∼