Research ArticleA New Hyperbolic Area Formula ofa Hyperbolic Triangle and Its Applications

Hui Bao and Xingdi Chen

Department of Mathematics Huaqiao University Quanzhou Fujian 362021 China

Correspondence should be addressed to Xingdi Chen chxtthqueducn

Received 14 May 2014 Accepted 25 July 2014 Published 13 August 2014

Academic Editor Ming-Sheng Liu

Copyright copy 2014 H Bao and X ChenThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study some characterizations of hyperbolic geometry in the Poincare diskWefirst obtain the hyperbolic area and length formulaof Euclidean disk and a circle represented by its Euclidean center and radius Replacing interior angles with vertices coordinateswe also obtain a new hyperbolic area formula of a hyperbolic triangle As its application we give the hyperbolic area of a Lambertquadrilateral and some geometric characterizations of Lambert quadrilaterals and Saccheri quadrilaterals

1 Introduction

Hyperbolic geometry was created in the first half of thenineteenth century in order to prove the dependence ofEuclidrsquos fifth postulate on the first four ones It was exten-sively studied by Nikolai Lobachevskii and Johann BolyaiBecause of this hyperbolic geometry is also known as Bolyai-Lobachevskian geometry The basis necessary for an analyticstudy of hyperbolic geometry was laid by Leonhard EulerGaspard Monge and Friedrich Gauss in their investigationof curved surfaces Later two of the most famous analyticmodels of the hyperbolic geometry were built which areknown as the Klein model and the Poincare model in thename of their inventors [1] One can refer to [2ndash4] for morehistory about hyperbolic geometry

For a complex number 119911 we assume that 119911 = 119909 + 119894119910where 119909 119910 isin R For convenience we denote by 119911

119860= 119909

119860+ 119894119910

119860

the coordinate of the point 119860 We denote the complex planethe unit disk and the upper half plane by R2 B2 and H2respectively We denote by 119878

1(119888 119903) the Euclidean circle with

center 119888 and radius 119903 Let 119869[119886 119887] be the hyperbolic geodesicsegment joining 119886 and 119887 when 119886 119887 isin B2 or H2 or thehyperbolic geodesic ray with one of the two points 119886 and119887 is on 120597B2 or 120597H2 Let Ω sub R2 be a simply connecteddomain of hyperbolic type and let 120588

Ω(119911)|119889119911| be the hyperbolic

metric of Ω with the Gaussian curvature minus1 In particular

the hyperbolic metric density inH2 andB2 with the Gaussiancurvature minus1 is given by

119908H2 =1

119910

119908B2 =2

1 minus |119911|2 (1)

respectivelyDefine the hyperbolic area of a measurable subset 119864 of Ω

as

119860hyp (119864) = ∬

119864

120588Ω(119911)

2|119889119911|

2 (2)

Particularly if 119864 is a measurable subset in B2 then

119860hyp (119864) = ∬

119864

(

2

1 minus |119911|2)

2

119889119909 119889119910 (3)

The integrand at (3) is replaced with 11199102 when 119864 sube H2 We

also denote the Euclidean area of a measurable subset 119864 sub Ω

by 119860euc(119864) For any rectifiable curve 120574 in Ω the hyperboliclength of 120574 is

119897hyp (120574) = int

120574

120588Ω (119911) |119889119911| (4)

Hindawi Publishing CorporationJournal of MathematicsVolume 2014 Article ID 838497 8 pageshttpdxdoiorg1011552014838497

2 Journal of Mathematics

IfΩ isB2 orH2 then the integrands in (4) are 2(1minus|119911|2) and1119910 respectively The hyperbolic distance between two points1199111 119911

2isin Ω is defined by

119889hyp (1199111 1199112) = inf120574119897hyp (120574) (5)

where the infimum is taken over all rectifiable curves in Ω

joining 1199111and 119911

2 We say that a curve 120574 [0 1] rarr Ω is a

hyperbolic geodesic joining 120574(0) and 120574(1) if for all 119905 isin (0 1) itfollows

119889hyp (120574 (0) 120574 (1)) = 119889hyp (120574 (0) 120574 (119905)) + 119889hyp (120574 (119905) 120574 (1))

(6)

If there exists a polygon enclosed by 119899 hyperbolicgeodesics then we call it a hyperbolic 119899 polygon Particularlyif 119899 takes 3 or 4 then we call it a hyperbolic triangle andhyperbolic quadrilateral respectively The interior angle of ahyperbolic polygon denotes the intersectional angle of thetangents of two geodesic arcs at the vertex If there exists ahyperbolic quadrilateral with angles (1205872 1205872 120601 1205872) 0 le

120601 lt 1205872 then it is said to be a Lambert quadrilateral [5p 156] If there exists a hyperbolic quadrilateral with angles(1205872 1205872 120601 120601) 0 le 120601 lt 1205872 then it is called a Saccheriquadrilateral [5 p 156]

Given two nonempty subsets 119860 119861 of 119866 let 119889120588(119860 119861)

denote the hyperbolic distance between them defined as

119889120588 (119860 119861) = inf

1199111isin1198601199112isin119861

120588119866(119911

1 119911

2) (7)

where 120588119866(1199111 119911

2) stands for the hyperbolic distance between

two points 1199111and 119911

2

We also need the following three explicit formulas

cosh 120588H2 (1199111 1199112) = 1 +

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

2

211991011199102

(8)

120588B2 (1199111 1199112) = 2arth10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

radic10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

2+ (1 minus

10038161003816100381610038161199111

1003816100381610038161003816

2) (1 minus

10038161003816100381610038161199112

1003816100381610038161003816

2)

(9)

for all 1199111 119911

2isin B2 In particular for 119911 isin B2

120588B2 (0 119911) = log 1 + |119911|

1 minus |119911|

= 2arth |119911| (10)

The above basic facts can be found in our standardreferences [5 6] and in many other sources on hyper-bolic geometry such as [7 8] Far-reaching and specializedadvanced texts discussing hyperbolic geometry include [910]

Hyperbolic triangles and Lambert quadrilaterals are fun-damental geometric quantities for the study of hyperbolicgeometry theory and its applications Curien andWerner [11]constructed random triangulations of the Poincare disc byhyperbolic triangles Demirel [12] Yang and Fang [13 14]gave characterizations of Mobius transformations by use ofhyperbolic triangles or Lambert quadrilaterals Pambuccian

[15] showed that mappings preserving the area equalityof hyperbolic triangles are motions One can see [16ndash18]for more characterizations about quasiconformal mappingsand harmonic quasiconformal mappings in the sense ofhyperbolic metrics

Further study of geometric properties of hyperbolictriangles Lambert quadrilaterals and hyperbolic polygonsalso raises onersquos interest Rostamzadeh and Taherian [19]considered the Klein model of the real hyperbolic plane andgave a new definition of its defect and area A hyperbolic areaformula and the radius of the inscribed circle of a hyperbolictriangle in the Poincare model can be found in [5 p 150 152]Recently Vuorinen andWang [20] obtained sharp bounds forthe product and the sum of two hyperbolic distances betweentwo opposite sides of hyperbolic Lambert quadrilaterals inthe unit disk Kanesaka and Nakamura [21] gave an explicithyperbolic area formula for a 120579-acute triangle One can referto [22ndash24] for distortion estimates of hyperbolic areas ofmeasurable subsets under quasiconformal mappings

By the hyperbolic radius of a disc the following is shown

TheoremA (see [5]) (1)The area of a hyperbolic disc of radius119903 is 4120587sinh2(1199032)

(2) The length of a hyperbolic circle of radius 119903 is 2120587 sinh 119903

Instead of a hyperbolic radius we use the Euclideancenter and radius of Euclidean disc inH2 to give its hyperbolicarea formula as follows

Theorem 1 For Euclidean disc1198631= 119911 |119911minus(119886+119887119894)| lt 119877

1 sube

H2 the hyperbolic area of1198631is

119860hyp (1198631) = 2120587(

119887

radic1198872minus 119877

2

1

minus 1)

119897hyp (1205971198631) =

21205871198771

radic1198872minus 119877

2

1

(11)

where 119887 gt 1198771gt 0

Area formula for a hyperbolic polygon can be determinedby its interior angles [5 p 150]

Theorem B For any hyperbolic triangle 119879(119860 119861 119862) withinterior angles 120572 120573 120574

119860hyp (119879 (119860 119861 119862)) = 120587 minus (120572 + 120573 + 120574) (12)

In this paper instead of interior angles we will usethe coordinates of vertexes of a hyperbolic triangle to giveanother hyperbolic area formula We first study a hyperbolictriangle with a vertex at infinity

Theorem 2 Let 1199111 119911

2isin H2 with 119911

1= 119909

1+ 119894119910

1 119911

2= 119909

2+

1198941199102(119909

1= 119909

2)

Journal of Mathematics 3

(1) If (1199091minus 119888)(119909

2minus 119888) gt 0

119860hyp (119879 (1199111 119911

2infin)) =

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

minus arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(13)

(2) If (1199091minus 119888)(119909

2minus 119888) lt 0

119860hyp (119879 (1199111 119911

2infin))

= 120587 minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(14)

where 119888 = (|1199111|2minus |119911

2|2)2(119909

1minus 119909

2)

For a general hyperbolic triangle inH2 its hyperbolic areacan be represented by four hyperbolic triangles with a vertexat infinity (see Remark 6)

As some applications of the hyperbolic area formula ofa hyperbolic triangle we first give an explicit formula forthe hyperbolic area of a Lambert quadrilateral in Section 3Moreover we obtain the length of two hyperbolic diagonallines in a Lambert quadrilateral in B2 as follows

Theorem 3 Let 119876(V119886 V

119887 V

119888 V

119889) be the Lambert quadrilateral

in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 Then

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052

120588B2 (V119887 V119889) = 2 arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(15)

where 1199051= |V

119887minus V

119886| 119905

2= |V

119889minus V

119886|

Some geometric characterizations of Saccheri quadrilat-erals are studied and a relation about twohyperbolic distancesbetween two pairs of opposite sides is given in Section 4 (seeTheorem 14)

2 Hyperbolic Areas of HyperbolicTriangles in H2

Proof of Theorem 1 Let 1198632= 119911 |119911| lt 119877

2 be Euclidean disc

in B2 By direct calculation the hyperbolic area of1198632is given

by

119860hyp (1198632) = ∬

1198632

(

2

1 minus |119911|2)

2

119889119909 119889119910

= int

2120587

0

int

1198772

0

119903(

2

1 minus 1199032)

2

119889119903 119889120579 =

41205871198772

2

1 minus 1198772

2

(16)

Let 119891(119911) = 1198912∘ 119891

1(119911) which maps 119863

1= 119911 |119911 minus (119886 +

119887119894)| lt 1198771 sube H2 onto the disk with center 0 and radius

(119887 minus radic1198872minus 119877

2

1)119877

1 where

1198911 (119911) =

119911 minus (119886 + 2119887119894)

119911 minus (119886 minus 2119887119894)

1198912 (119911) = 119905 sdot

119911 minus 119904

119911 minus 119905

119904 = 1 minus

4119887 (2119887 minus radic1198872minus 119877

2

1)

31198872+ 119877

2

1

119905 = 1 minus

4119887 (2119887 + radic1198872minus 119877

2

1)

31198872+ 119877

2

1

(17)

Then we have

119860hyp (1198631) = 2120587(

119887

radic1198872minus 119877

2

1

minus 1)

119897hyp (1205971198631) =

21205871198771

radic1198872minus 119877

2

1

(18)

Lemma 4 Let 120574 be a hyperbolic geodesic in H2 with twoendpoints 119860 = 119894119905 119861 = 119905119890

119894120579(0 le 120579 le 120587 120579 = 1205872 119905 gt 0)

Then the hyperbolic area of the hyperbolic triangle 119879(119860 119861infin)

is given as follows

119860hyp (119879 (119860 119861infin)) =

1003816100381610038161003816100381610038161003816

120579 minus

120587

2

1003816100381610038161003816100381610038161003816

(19)

ProofCase 1 If 0 le 120579 lt 1205872 then119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

119905 cos 120579

0

119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= arcsin (cos 120579) = arcsin(sin(1205872

minus 120579))

=

120587

2

minus 120579

(20)

Particularly when 120579 = 0 then 119860hyp(119879(119860 119861infin)) = 1205872

Case 2 If 1205872 lt 120579 le 120587 we have119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

0

119905 cos 120579119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= minus arcsin (cos 120579) = minus arcsin(sin(1205872

minus 120579))

= 120579 minus

120587

2

(21)

In particular when 120579 = 120587 119860hyp(119879(119860 119861infin)) = 1205872

4 Journal of Mathematics

Lemma 5 Let 120574 be a hyperbolic geodesic with two endpoints119860 = 119888 + 119905119890

1198941205791 119861 = 119888 + 119905119890

1198941205792(119888 isin R 119905 gt 0 0 le 120579

1le 120587 0 le 120579

2le

120587 1205791

= 1205792) in H2 then the hyperbolic area of the hyperbolic

triangle 119879(119860 119861infin) is as follows

119860hyp (119879 (119860 119861infin)) =10038161003816100381610038161205792minus 120579

1

1003816100381610038161003816 (22)

Proof Let 119862 = 119888 + 119894119905 Without loss of generality we assume1205791lt 120579

2

(1) If 0 le 1205791lt 120579

2le 1205872 by Lemma 4 we have

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119861 119862infin))

= (

120587

2

minus 1205791) minus (

120587

2

minus 1205792) = 120579

2minus 120579

1

(23)

(2) If 1205872 le 1205791lt 120579

2le 120587 using Lemma 4 again we get

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119861 119862infin)) minus 119860hyp (119879 (119860 119862infin))

= (1205792minus

120587

2

) minus (1205791minus

120587

2

) = 1205792minus 120579

1

(24)

(3) If 0 le 1205791lt 1205872 lt 120579

2le 120587 Lemma 4 also implies

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) + 119860hyp (119879 (119861 119862infin))

= (1205792minus

120587

2

) + (

120587

2

minus 1205791) = 120579

2minus 120579

1

(25)

Similarly if 1205792lt 120579

1we have the same conclusion

119860hyp (119879 (119860 119861infin)) = 1205791minus 120579

2 (26)

Thus the proof of Lemma 5 is completed

Theorem 2 is an application of Lemma 5 For the proofof Theorem 2 we also need a sharp formula of the circleorthogonal to the boundary ofH2 [25]

Lemma A Let 1199111 119911

2isin H2 with 119909

1= 119909

2 Then 119878

1(119888 119903) is

orthogonal to 120597H2 where

119888 =

10038161003816100381610038161199111

1003816100381610038161003816

2minus10038161003816100381610038161199112

1003816100381610038161003816

2

2 (1199091minus 119909

2)

119903 = radic1199102

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(27)

Proof of Theorem 2 Wemay assume that 1199091gt 119909

2 Let 1198781(119888 119903)

be the circle which is through the two points 1199111 119911

2 where 119888

and 119903 are represented as in (27) Then we have 1199111= 119888 + 119903119890

1198941205791

1199112= 119888 + 119903119890

1198941205792

Case 1 ((1199091minus 119888)(119909

2minus 119888) gt 0) If 119888 lt 119909

2lt 119909

1 then by Lemma 5

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199102

1199092minus 119888

minus arctan1199101

1199091minus 119888

(28)

If 1199092lt 119909

1lt 119888 so we have

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199101

119888 minus 1199091

minus arctan1199102

119888 minus 1199092

(29)

Case 2 ((1199091minus119888)(119909

2minus119888) lt 0)This case implies that119909

2lt 119888 lt 119909

1

then

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= 120587 minus arctan

1199101

1199091minus 119888

minus arctan1199102

119888 minus 1199092

(30)

Remark 6 For a general hyperbolic triangle119879(119860 119861 119862) inH2we can give its hyperbolic area formula by Theorem 2 Giventhree points 119860(119909

119860 119910

119860) 119861(119909

119861 119910

119861) and 119862(119909

119862 119910

119862) in H2 with

(119909119860minus 119909

119861)(119909

119861minus 119909

119862)(119909

119862minus 119909

119860) = 0 we may assume that

119909119862lt 119909

119860lt 119909

119861 Let 1198781(119888 119903) be the circlewhich is through119861 and

119862 and orthogonal to 120597H2 Let 119863(119909119863 119910

119863) be the point which

is the intersection of 1198781(119888 119903) and the hyperbolic geodesicthrough point 119860 and orthogonal to 120597H2 Then we have

119909119860= 119909

119863 119910

119863= radic119903

2minus (119909

119860minus 119888)

2 (31)

where 119888 = (|119911119861|2

minus |119911119862|2)2(119909

119861minus 119909

119862) and 119903 =

radic1199102

119861+ ((119910

2

119862+ (119909

119861minus 119909

119862)2minus 119910

2

119861)2(119909

119861minus 119909

119862))

2Thus it follows

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863))

(32)

where

119860hyp (119879 (119860 119862119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119862119863infin))

10038161003816100381610038161003816

119860hyp (119879 (119860 119861119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119861infin)) minus 119860hyp (119879 (119861119863infin))

10038161003816100381610038161003816

(33)

then byTheorem 2 we can get the result of 119860hyp(119879(119860 119861 119862))

Example 7 Let 119860 = 119894 119861 = 1 + 2119894 119862 = minus1 + 3119894 by calculationwe have

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863)) asymp 0372

(34)

where119863 = radic301198942

Remark 8 There are lots of triangles which have the samehyperbolic area but different Euclidean areas in H2 For

Journal of Mathematics 5

instance 119879(119860 119861 119862) and 119879(119863 119861 119864) are two hyperbolic trian-gles inH2 with 119860 = minus1 119861 = 0 119862 = 1119863 = minus2 119864 = 2 and

119860hyp (119879 (119860 119861 119862)) = 119860hyp (119879 (119863 119861 119864)) = 120587

120587

4

= 119860euc (119879 (119860 119861 119862)) = 119860euc (119879 (119863 119861 119864)) = 120587

(35)

Remark 9 For any hyperbolic 119899-polygons 119875(1198601 119860

2 119860

119899)

(119899 ge 3) inH2 we always can divide it into several hyperbolictriangles 119879

1 119879

2 119879

119894(119894 ge 1) then we have

119860hyp (119875 (1198601 119860

2 119860

119899)) =

119894

sum

119896=1

119860hyp (119879119896) (36)

Remark 10 For any hyperbolic 119899-polygons119875(1198601 119860

2 119860

119899)

(119899 ge 3) in B2 since the hyperbolic area is invariant underMobius transformation sowe canmap the119875(119860

1 119860

2 119860

119899)

intoH2 by a Mobius transformation therefore we obtain thehyperbolic area of 119875(119860

1 119860

2 119860

119899) in B2

3 Hyperbolic Areas of Lambert Quadrilaterals

As an application of the explicit hyperbolic area formulaof a hyperbolic triangle we obtain the hyperbolic area of aLambert quadrilateral

Theorem 11 Let 119876(119860 119861 119862119863) be the Lambert quadrilateralwith angles (1205872 1205872 120601 1205872) 0 le 120601 lt 1205872 in H2 and 119860 =

1198880+ 119894119903

0 119861 = 119888

0+ 119894119903 then

(1) if 119903 le 119889

119860hyp (119876 (119860 119861 119862119863))

= arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(37)

(2) if 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1198881minus 119909

2

(38)

where 119889 = radic(1198881minus 119888

0)2+ 119903

2

1 119862 = 119909

2+ 119894119910

2119863 = 119909

1+ 119894119910

1 1198881and

1199031are given at (39)

Proof Without loss of generality we assume that 1198880lt 119888

1 0 lt

1199030lt 119903 and 119862 isin 119878

1(1198880 119903) cap 119878

1(1198881 1199031) 119863 isin 119878

1(1198880 1199030) cap 119878

1(1198881 1199031)

(see Figure 1) which implies 1198880+ 119903

0gt 119888

1minus 119903

1 1198881+ 119903

1gt 119888

0+ 119903

By (27) we have

1198881=

1199092

1+ 119910

2

1minus 119909

2

2minus 119910

2

2

2 (1199091minus 119909

2)

1199031= radic119910

2

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(39)

C0C1O

Y

A

B E

D

C

X

Figure 1

When 119903 le 119889 we get

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

(40)

where 119864 = 1199091+ 119894radic119903

2minus (119909

1minus 119888

0)2 So the following three

relations119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

10038161003816100381610038161198881minus 119888

0

1003816100381610038161003816

2= (119909

1minus 119888

0)2+ 119910

2

1+ (119909

1minus 119888

1)2+ 119910

2

1

(41)

imply that

119860hyp (119876 (119860 119861 119862119863)) = arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(42)

When 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0 it follows

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

(43)

and then we get

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1199092minus 119888

1

(44)

The proof of the case that 1198880gt 119888

1is similar to the case that

1198881gt 119888

0 So the proof of Theorem 11 is complete

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Journal of Mathematics

IfΩ isB2 orH2 then the integrands in (4) are 2(1minus|119911|2) and1119910 respectively The hyperbolic distance between two points1199111 119911

2isin Ω is defined by

119889hyp (1199111 1199112) = inf120574119897hyp (120574) (5)

where the infimum is taken over all rectifiable curves in Ω

joining 1199111and 119911

2 We say that a curve 120574 [0 1] rarr Ω is a

hyperbolic geodesic joining 120574(0) and 120574(1) if for all 119905 isin (0 1) itfollows

119889hyp (120574 (0) 120574 (1)) = 119889hyp (120574 (0) 120574 (119905)) + 119889hyp (120574 (119905) 120574 (1))

(6)

If there exists a polygon enclosed by 119899 hyperbolicgeodesics then we call it a hyperbolic 119899 polygon Particularlyif 119899 takes 3 or 4 then we call it a hyperbolic triangle andhyperbolic quadrilateral respectively The interior angle of ahyperbolic polygon denotes the intersectional angle of thetangents of two geodesic arcs at the vertex If there exists ahyperbolic quadrilateral with angles (1205872 1205872 120601 1205872) 0 le

120601 lt 1205872 then it is said to be a Lambert quadrilateral [5p 156] If there exists a hyperbolic quadrilateral with angles(1205872 1205872 120601 120601) 0 le 120601 lt 1205872 then it is called a Saccheriquadrilateral [5 p 156]

Given two nonempty subsets 119860 119861 of 119866 let 119889120588(119860 119861)

denote the hyperbolic distance between them defined as

119889120588 (119860 119861) = inf

1199111isin1198601199112isin119861

120588119866(119911

1 119911

2) (7)

where 120588119866(1199111 119911

2) stands for the hyperbolic distance between

two points 1199111and 119911

2

We also need the following three explicit formulas

cosh 120588H2 (1199111 1199112) = 1 +

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

2

211991011199102

(8)

120588B2 (1199111 1199112) = 2arth10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

radic10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

2+ (1 minus

10038161003816100381610038161199111

1003816100381610038161003816

2) (1 minus

10038161003816100381610038161199112

1003816100381610038161003816

2)

(9)

for all 1199111 119911

2isin B2 In particular for 119911 isin B2

120588B2 (0 119911) = log 1 + |119911|

1 minus |119911|

= 2arth |119911| (10)

The above basic facts can be found in our standardreferences [5 6] and in many other sources on hyper-bolic geometry such as [7 8] Far-reaching and specializedadvanced texts discussing hyperbolic geometry include [910]

Hyperbolic triangles and Lambert quadrilaterals are fun-damental geometric quantities for the study of hyperbolicgeometry theory and its applications Curien andWerner [11]constructed random triangulations of the Poincare disc byhyperbolic triangles Demirel [12] Yang and Fang [13 14]gave characterizations of Mobius transformations by use ofhyperbolic triangles or Lambert quadrilaterals Pambuccian

[15] showed that mappings preserving the area equalityof hyperbolic triangles are motions One can see [16ndash18]for more characterizations about quasiconformal mappingsand harmonic quasiconformal mappings in the sense ofhyperbolic metrics

Further study of geometric properties of hyperbolictriangles Lambert quadrilaterals and hyperbolic polygonsalso raises onersquos interest Rostamzadeh and Taherian [19]considered the Klein model of the real hyperbolic plane andgave a new definition of its defect and area A hyperbolic areaformula and the radius of the inscribed circle of a hyperbolictriangle in the Poincare model can be found in [5 p 150 152]Recently Vuorinen andWang [20] obtained sharp bounds forthe product and the sum of two hyperbolic distances betweentwo opposite sides of hyperbolic Lambert quadrilaterals inthe unit disk Kanesaka and Nakamura [21] gave an explicithyperbolic area formula for a 120579-acute triangle One can referto [22ndash24] for distortion estimates of hyperbolic areas ofmeasurable subsets under quasiconformal mappings

By the hyperbolic radius of a disc the following is shown

TheoremA (see [5]) (1)The area of a hyperbolic disc of radius119903 is 4120587sinh2(1199032)

(2) The length of a hyperbolic circle of radius 119903 is 2120587 sinh 119903

Instead of a hyperbolic radius we use the Euclideancenter and radius of Euclidean disc inH2 to give its hyperbolicarea formula as follows

Theorem 1 For Euclidean disc1198631= 119911 |119911minus(119886+119887119894)| lt 119877

1 sube

H2 the hyperbolic area of1198631is

119860hyp (1198631) = 2120587(

119887

radic1198872minus 119877

2

1

minus 1)

119897hyp (1205971198631) =

21205871198771

radic1198872minus 119877

2

1

(11)

where 119887 gt 1198771gt 0

Area formula for a hyperbolic polygon can be determinedby its interior angles [5 p 150]

Theorem B For any hyperbolic triangle 119879(119860 119861 119862) withinterior angles 120572 120573 120574

119860hyp (119879 (119860 119861 119862)) = 120587 minus (120572 + 120573 + 120574) (12)

In this paper instead of interior angles we will usethe coordinates of vertexes of a hyperbolic triangle to giveanother hyperbolic area formula We first study a hyperbolictriangle with a vertex at infinity

Theorem 2 Let 1199111 119911

2isin H2 with 119911

1= 119909

1+ 119894119910

1 119911

2= 119909

2+

1198941199102(119909

1= 119909

2)

Journal of Mathematics 3

(1) If (1199091minus 119888)(119909

2minus 119888) gt 0

119860hyp (119879 (1199111 119911

2infin)) =

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

minus arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(13)

(2) If (1199091minus 119888)(119909

2minus 119888) lt 0

119860hyp (119879 (1199111 119911

2infin))

= 120587 minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(14)

where 119888 = (|1199111|2minus |119911

2|2)2(119909

1minus 119909

2)

For a general hyperbolic triangle inH2 its hyperbolic areacan be represented by four hyperbolic triangles with a vertexat infinity (see Remark 6)

As some applications of the hyperbolic area formula ofa hyperbolic triangle we first give an explicit formula forthe hyperbolic area of a Lambert quadrilateral in Section 3Moreover we obtain the length of two hyperbolic diagonallines in a Lambert quadrilateral in B2 as follows

Theorem 3 Let 119876(V119886 V

119887 V

119888 V

119889) be the Lambert quadrilateral

in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 Then

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052

120588B2 (V119887 V119889) = 2 arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(15)

where 1199051= |V

119887minus V

119886| 119905

2= |V

119889minus V

119886|

Some geometric characterizations of Saccheri quadrilat-erals are studied and a relation about twohyperbolic distancesbetween two pairs of opposite sides is given in Section 4 (seeTheorem 14)

2 Hyperbolic Areas of HyperbolicTriangles in H2

Proof of Theorem 1 Let 1198632= 119911 |119911| lt 119877

2 be Euclidean disc

in B2 By direct calculation the hyperbolic area of1198632is given

by

119860hyp (1198632) = ∬

1198632

(

2

1 minus |119911|2)

2

119889119909 119889119910

= int

2120587

0

int

1198772

0

119903(

2

1 minus 1199032)

2

119889119903 119889120579 =

41205871198772

2

1 minus 1198772

2

(16)

Let 119891(119911) = 1198912∘ 119891

1(119911) which maps 119863

1= 119911 |119911 minus (119886 +

119887119894)| lt 1198771 sube H2 onto the disk with center 0 and radius

(119887 minus radic1198872minus 119877

2

1)119877

1 where

1198911 (119911) =

119911 minus (119886 + 2119887119894)

119911 minus (119886 minus 2119887119894)

1198912 (119911) = 119905 sdot

119911 minus 119904

119911 minus 119905

119904 = 1 minus

4119887 (2119887 minus radic1198872minus 119877

2

1)

31198872+ 119877

2

1

119905 = 1 minus

4119887 (2119887 + radic1198872minus 119877

2

1)

31198872+ 119877

2

1

(17)

Then we have

119860hyp (1198631) = 2120587(

119887

radic1198872minus 119877

2

1

minus 1)

119897hyp (1205971198631) =

21205871198771

radic1198872minus 119877

2

1

(18)

Lemma 4 Let 120574 be a hyperbolic geodesic in H2 with twoendpoints 119860 = 119894119905 119861 = 119905119890

119894120579(0 le 120579 le 120587 120579 = 1205872 119905 gt 0)

Then the hyperbolic area of the hyperbolic triangle 119879(119860 119861infin)

is given as follows

119860hyp (119879 (119860 119861infin)) =

1003816100381610038161003816100381610038161003816

120579 minus

120587

2

1003816100381610038161003816100381610038161003816

(19)

ProofCase 1 If 0 le 120579 lt 1205872 then119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

119905 cos 120579

0

119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= arcsin (cos 120579) = arcsin(sin(1205872

minus 120579))

=

120587

2

minus 120579

(20)

Particularly when 120579 = 0 then 119860hyp(119879(119860 119861infin)) = 1205872

Case 2 If 1205872 lt 120579 le 120587 we have119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

0

119905 cos 120579119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= minus arcsin (cos 120579) = minus arcsin(sin(1205872

minus 120579))

= 120579 minus

120587

2

(21)

In particular when 120579 = 120587 119860hyp(119879(119860 119861infin)) = 1205872

4 Journal of Mathematics

Lemma 5 Let 120574 be a hyperbolic geodesic with two endpoints119860 = 119888 + 119905119890

1198941205791 119861 = 119888 + 119905119890

1198941205792(119888 isin R 119905 gt 0 0 le 120579

1le 120587 0 le 120579

2le

120587 1205791

= 1205792) in H2 then the hyperbolic area of the hyperbolic

triangle 119879(119860 119861infin) is as follows

119860hyp (119879 (119860 119861infin)) =10038161003816100381610038161205792minus 120579

1

1003816100381610038161003816 (22)

Proof Let 119862 = 119888 + 119894119905 Without loss of generality we assume1205791lt 120579

2

(1) If 0 le 1205791lt 120579

2le 1205872 by Lemma 4 we have

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119861 119862infin))

= (

120587

2

minus 1205791) minus (

120587

2

minus 1205792) = 120579

2minus 120579

1

(23)

(2) If 1205872 le 1205791lt 120579

2le 120587 using Lemma 4 again we get

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119861 119862infin)) minus 119860hyp (119879 (119860 119862infin))

= (1205792minus

120587

2

) minus (1205791minus

120587

2

) = 1205792minus 120579

1

(24)

(3) If 0 le 1205791lt 1205872 lt 120579

2le 120587 Lemma 4 also implies

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) + 119860hyp (119879 (119861 119862infin))

= (1205792minus

120587

2

) + (

120587

2

minus 1205791) = 120579

2minus 120579

1

(25)

Similarly if 1205792lt 120579

1we have the same conclusion

119860hyp (119879 (119860 119861infin)) = 1205791minus 120579

2 (26)

Thus the proof of Lemma 5 is completed

Theorem 2 is an application of Lemma 5 For the proofof Theorem 2 we also need a sharp formula of the circleorthogonal to the boundary ofH2 [25]

Lemma A Let 1199111 119911

2isin H2 with 119909

1= 119909

2 Then 119878

1(119888 119903) is

orthogonal to 120597H2 where

119888 =

10038161003816100381610038161199111

1003816100381610038161003816

2minus10038161003816100381610038161199112

1003816100381610038161003816

2

2 (1199091minus 119909

2)

119903 = radic1199102

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(27)

Proof of Theorem 2 Wemay assume that 1199091gt 119909

2 Let 1198781(119888 119903)

be the circle which is through the two points 1199111 119911

2 where 119888

and 119903 are represented as in (27) Then we have 1199111= 119888 + 119903119890

1198941205791

1199112= 119888 + 119903119890

1198941205792

Case 1 ((1199091minus 119888)(119909

2minus 119888) gt 0) If 119888 lt 119909

2lt 119909

1 then by Lemma 5

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199102

1199092minus 119888

minus arctan1199101

1199091minus 119888

(28)

If 1199092lt 119909

1lt 119888 so we have

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199101

119888 minus 1199091

minus arctan1199102

119888 minus 1199092

(29)

Case 2 ((1199091minus119888)(119909

2minus119888) lt 0)This case implies that119909

2lt 119888 lt 119909

1

then

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= 120587 minus arctan

1199101

1199091minus 119888

minus arctan1199102

119888 minus 1199092

(30)

Remark 6 For a general hyperbolic triangle119879(119860 119861 119862) inH2we can give its hyperbolic area formula by Theorem 2 Giventhree points 119860(119909

119860 119910

119860) 119861(119909

119861 119910

119861) and 119862(119909

119862 119910

119862) in H2 with

(119909119860minus 119909

119861)(119909

119861minus 119909

119862)(119909

119862minus 119909

119860) = 0 we may assume that

119909119862lt 119909

119860lt 119909

119861 Let 1198781(119888 119903) be the circlewhich is through119861 and

119862 and orthogonal to 120597H2 Let 119863(119909119863 119910

119863) be the point which

is the intersection of 1198781(119888 119903) and the hyperbolic geodesicthrough point 119860 and orthogonal to 120597H2 Then we have

119909119860= 119909

119863 119910

119863= radic119903

2minus (119909

119860minus 119888)

2 (31)

where 119888 = (|119911119861|2

minus |119911119862|2)2(119909

119861minus 119909

119862) and 119903 =

radic1199102

119861+ ((119910

2

119862+ (119909

119861minus 119909

119862)2minus 119910

2

119861)2(119909

119861minus 119909

119862))

2Thus it follows

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863))

(32)

where

119860hyp (119879 (119860 119862119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119862119863infin))

10038161003816100381610038161003816

119860hyp (119879 (119860 119861119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119861infin)) minus 119860hyp (119879 (119861119863infin))

10038161003816100381610038161003816

(33)

then byTheorem 2 we can get the result of 119860hyp(119879(119860 119861 119862))

Example 7 Let 119860 = 119894 119861 = 1 + 2119894 119862 = minus1 + 3119894 by calculationwe have

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863)) asymp 0372

(34)

where119863 = radic301198942

Remark 8 There are lots of triangles which have the samehyperbolic area but different Euclidean areas in H2 For

Journal of Mathematics 5

instance 119879(119860 119861 119862) and 119879(119863 119861 119864) are two hyperbolic trian-gles inH2 with 119860 = minus1 119861 = 0 119862 = 1119863 = minus2 119864 = 2 and

119860hyp (119879 (119860 119861 119862)) = 119860hyp (119879 (119863 119861 119864)) = 120587

120587

4

= 119860euc (119879 (119860 119861 119862)) = 119860euc (119879 (119863 119861 119864)) = 120587

(35)

Remark 9 For any hyperbolic 119899-polygons 119875(1198601 119860

2 119860

119899)

(119899 ge 3) inH2 we always can divide it into several hyperbolictriangles 119879

1 119879

2 119879

119894(119894 ge 1) then we have

119860hyp (119875 (1198601 119860

2 119860

119899)) =

119894

sum

119896=1

119860hyp (119879119896) (36)

Remark 10 For any hyperbolic 119899-polygons119875(1198601 119860

2 119860

119899)

(119899 ge 3) in B2 since the hyperbolic area is invariant underMobius transformation sowe canmap the119875(119860

1 119860

2 119860

119899)

intoH2 by a Mobius transformation therefore we obtain thehyperbolic area of 119875(119860

1 119860

2 119860

119899) in B2

3 Hyperbolic Areas of Lambert Quadrilaterals

As an application of the explicit hyperbolic area formulaof a hyperbolic triangle we obtain the hyperbolic area of aLambert quadrilateral

Theorem 11 Let 119876(119860 119861 119862119863) be the Lambert quadrilateralwith angles (1205872 1205872 120601 1205872) 0 le 120601 lt 1205872 in H2 and 119860 =

1198880+ 119894119903

0 119861 = 119888

0+ 119894119903 then

(1) if 119903 le 119889

119860hyp (119876 (119860 119861 119862119863))

= arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(37)

(2) if 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1198881minus 119909

2

(38)

where 119889 = radic(1198881minus 119888

0)2+ 119903

2

1 119862 = 119909

2+ 119894119910

2119863 = 119909

1+ 119894119910

1 1198881and

1199031are given at (39)

Proof Without loss of generality we assume that 1198880lt 119888

1 0 lt

1199030lt 119903 and 119862 isin 119878

1(1198880 119903) cap 119878

1(1198881 1199031) 119863 isin 119878

1(1198880 1199030) cap 119878

1(1198881 1199031)

(see Figure 1) which implies 1198880+ 119903

0gt 119888

1minus 119903

1 1198881+ 119903

1gt 119888

0+ 119903

By (27) we have

1198881=

1199092

1+ 119910

2

1minus 119909

2

2minus 119910

2

2

2 (1199091minus 119909

2)

1199031= radic119910

2

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(39)

C0C1O

Y

A

B E

D

C

X

Figure 1

When 119903 le 119889 we get

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

(40)

where 119864 = 1199091+ 119894radic119903

2minus (119909

1minus 119888

0)2 So the following three

relations119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

10038161003816100381610038161198881minus 119888

0

1003816100381610038161003816

2= (119909

1minus 119888

0)2+ 119910

2

1+ (119909

1minus 119888

1)2+ 119910

2

1

(41)

imply that

119860hyp (119876 (119860 119861 119862119863)) = arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(42)

When 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0 it follows

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

(43)

and then we get

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1199092minus 119888

1

(44)

The proof of the case that 1198880gt 119888

1is similar to the case that

1198881gt 119888

0 So the proof of Theorem 11 is complete

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Mathematics 3

(1) If (1199091minus 119888)(119909

2minus 119888) gt 0

119860hyp (119879 (1199111 119911

2infin)) =

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

minus arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(13)

(2) If (1199091minus 119888)(119909

2minus 119888) lt 0

119860hyp (119879 (1199111 119911

2infin))

= 120587 minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199101

1003816100381610038161003816119888 minus 119909

1

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

minus

100381610038161003816100381610038161003816100381610038161003816

arctan1199102

1003816100381610038161003816119888 minus 119909

2

1003816100381610038161003816

100381610038161003816100381610038161003816100381610038161003816

(14)

where 119888 = (|1199111|2minus |119911

2|2)2(119909

1minus 119909

2)

For a general hyperbolic triangle inH2 its hyperbolic areacan be represented by four hyperbolic triangles with a vertexat infinity (see Remark 6)

As some applications of the hyperbolic area formula ofa hyperbolic triangle we first give an explicit formula forthe hyperbolic area of a Lambert quadrilateral in Section 3Moreover we obtain the length of two hyperbolic diagonallines in a Lambert quadrilateral in B2 as follows

Theorem 3 Let 119876(V119886 V

119887 V

119888 V

119889) be the Lambert quadrilateral

in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 Then

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052

120588B2 (V119887 V119889) = 2 arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(15)

where 1199051= |V

119887minus V

119886| 119905

2= |V

119889minus V

119886|

Some geometric characterizations of Saccheri quadrilat-erals are studied and a relation about twohyperbolic distancesbetween two pairs of opposite sides is given in Section 4 (seeTheorem 14)

2 Hyperbolic Areas of HyperbolicTriangles in H2

Proof of Theorem 1 Let 1198632= 119911 |119911| lt 119877

2 be Euclidean disc

in B2 By direct calculation the hyperbolic area of1198632is given

by

119860hyp (1198632) = ∬

1198632

(

2

1 minus |119911|2)

2

119889119909 119889119910

= int

2120587

0

int

1198772

0

119903(

2

1 minus 1199032)

2

119889119903 119889120579 =

41205871198772

2

1 minus 1198772

2

(16)

Let 119891(119911) = 1198912∘ 119891

1(119911) which maps 119863

1= 119911 |119911 minus (119886 +

119887119894)| lt 1198771 sube H2 onto the disk with center 0 and radius

(119887 minus radic1198872minus 119877

2

1)119877

1 where

1198911 (119911) =

119911 minus (119886 + 2119887119894)

119911 minus (119886 minus 2119887119894)

1198912 (119911) = 119905 sdot

119911 minus 119904

119911 minus 119905

119904 = 1 minus

4119887 (2119887 minus radic1198872minus 119877

2

1)

31198872+ 119877

2

1

119905 = 1 minus

4119887 (2119887 + radic1198872minus 119877

2

1)

31198872+ 119877

2

1

(17)

Then we have

119860hyp (1198631) = 2120587(

119887

radic1198872minus 119877

2

1

minus 1)

119897hyp (1205971198631) =

21205871198771

radic1198872minus 119877

2

1

(18)

Lemma 4 Let 120574 be a hyperbolic geodesic in H2 with twoendpoints 119860 = 119894119905 119861 = 119905119890

119894120579(0 le 120579 le 120587 120579 = 1205872 119905 gt 0)

Then the hyperbolic area of the hyperbolic triangle 119879(119860 119861infin)

is given as follows

119860hyp (119879 (119860 119861infin)) =

1003816100381610038161003816100381610038161003816

120579 minus

120587

2

1003816100381610038161003816100381610038161003816

(19)

ProofCase 1 If 0 le 120579 lt 1205872 then119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

119905 cos 120579

0

119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= arcsin (cos 120579) = arcsin(sin(1205872

minus 120579))

=

120587

2

minus 120579

(20)

Particularly when 120579 = 0 then 119860hyp(119879(119860 119861infin)) = 1205872

Case 2 If 1205872 lt 120579 le 120587 we have119860hyp (119879 (119860 119861infin))

= ∬

119879(119860119861infin)

1

1199102119889119909 119889119910 = int

0

119905 cos 120579119889119909int

infin

radic1199052minus1199092

1

1199102119889119910

= minus arcsin (cos 120579) = minus arcsin(sin(1205872

minus 120579))

= 120579 minus

120587

2

(21)

In particular when 120579 = 120587 119860hyp(119879(119860 119861infin)) = 1205872

4 Journal of Mathematics

Lemma 5 Let 120574 be a hyperbolic geodesic with two endpoints119860 = 119888 + 119905119890

1198941205791 119861 = 119888 + 119905119890

1198941205792(119888 isin R 119905 gt 0 0 le 120579

1le 120587 0 le 120579

2le

120587 1205791

= 1205792) in H2 then the hyperbolic area of the hyperbolic

triangle 119879(119860 119861infin) is as follows

119860hyp (119879 (119860 119861infin)) =10038161003816100381610038161205792minus 120579

1

1003816100381610038161003816 (22)

Proof Let 119862 = 119888 + 119894119905 Without loss of generality we assume1205791lt 120579

2

(1) If 0 le 1205791lt 120579

2le 1205872 by Lemma 4 we have

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119861 119862infin))

= (

120587

2

minus 1205791) minus (

120587

2

minus 1205792) = 120579

2minus 120579

1

(23)

(2) If 1205872 le 1205791lt 120579

2le 120587 using Lemma 4 again we get

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119861 119862infin)) minus 119860hyp (119879 (119860 119862infin))

= (1205792minus

120587

2

) minus (1205791minus

120587

2

) = 1205792minus 120579

1

(24)

(3) If 0 le 1205791lt 1205872 lt 120579

2le 120587 Lemma 4 also implies

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) + 119860hyp (119879 (119861 119862infin))

= (1205792minus

120587

2

) + (

120587

2

minus 1205791) = 120579

2minus 120579

1

(25)

Similarly if 1205792lt 120579

1we have the same conclusion

119860hyp (119879 (119860 119861infin)) = 1205791minus 120579

2 (26)

Thus the proof of Lemma 5 is completed

Theorem 2 is an application of Lemma 5 For the proofof Theorem 2 we also need a sharp formula of the circleorthogonal to the boundary ofH2 [25]

Lemma A Let 1199111 119911

2isin H2 with 119909

1= 119909

2 Then 119878

1(119888 119903) is

orthogonal to 120597H2 where

119888 =

10038161003816100381610038161199111

1003816100381610038161003816

2minus10038161003816100381610038161199112

1003816100381610038161003816

2

2 (1199091minus 119909

2)

119903 = radic1199102

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(27)

Proof of Theorem 2 Wemay assume that 1199091gt 119909

2 Let 1198781(119888 119903)

be the circle which is through the two points 1199111 119911

2 where 119888

and 119903 are represented as in (27) Then we have 1199111= 119888 + 119903119890

1198941205791

1199112= 119888 + 119903119890

1198941205792

Case 1 ((1199091minus 119888)(119909

2minus 119888) gt 0) If 119888 lt 119909

2lt 119909

1 then by Lemma 5

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199102

1199092minus 119888

minus arctan1199101

1199091minus 119888

(28)

If 1199092lt 119909

1lt 119888 so we have

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199101

119888 minus 1199091

minus arctan1199102

119888 minus 1199092

(29)

Case 2 ((1199091minus119888)(119909

2minus119888) lt 0)This case implies that119909

2lt 119888 lt 119909

1

then

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= 120587 minus arctan

1199101

1199091minus 119888

minus arctan1199102

119888 minus 1199092

(30)

Remark 6 For a general hyperbolic triangle119879(119860 119861 119862) inH2we can give its hyperbolic area formula by Theorem 2 Giventhree points 119860(119909

119860 119910

119860) 119861(119909

119861 119910

119861) and 119862(119909

119862 119910

119862) in H2 with

(119909119860minus 119909

119861)(119909

119861minus 119909

119862)(119909

119862minus 119909

119860) = 0 we may assume that

119909119862lt 119909

119860lt 119909

119861 Let 1198781(119888 119903) be the circlewhich is through119861 and

119862 and orthogonal to 120597H2 Let 119863(119909119863 119910

119863) be the point which

is the intersection of 1198781(119888 119903) and the hyperbolic geodesicthrough point 119860 and orthogonal to 120597H2 Then we have

119909119860= 119909

119863 119910

119863= radic119903

2minus (119909

119860minus 119888)

2 (31)

where 119888 = (|119911119861|2

minus |119911119862|2)2(119909

119861minus 119909

119862) and 119903 =

radic1199102

119861+ ((119910

2

119862+ (119909

119861minus 119909

119862)2minus 119910

2

119861)2(119909

119861minus 119909

119862))

2Thus it follows

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863))

(32)

where

119860hyp (119879 (119860 119862119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119862119863infin))

10038161003816100381610038161003816

119860hyp (119879 (119860 119861119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119861infin)) minus 119860hyp (119879 (119861119863infin))

10038161003816100381610038161003816

(33)

then byTheorem 2 we can get the result of 119860hyp(119879(119860 119861 119862))

Example 7 Let 119860 = 119894 119861 = 1 + 2119894 119862 = minus1 + 3119894 by calculationwe have

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863)) asymp 0372

(34)

where119863 = radic301198942

Remark 8 There are lots of triangles which have the samehyperbolic area but different Euclidean areas in H2 For

Journal of Mathematics 5

instance 119879(119860 119861 119862) and 119879(119863 119861 119864) are two hyperbolic trian-gles inH2 with 119860 = minus1 119861 = 0 119862 = 1119863 = minus2 119864 = 2 and

119860hyp (119879 (119860 119861 119862)) = 119860hyp (119879 (119863 119861 119864)) = 120587

120587

4

= 119860euc (119879 (119860 119861 119862)) = 119860euc (119879 (119863 119861 119864)) = 120587

(35)

Remark 9 For any hyperbolic 119899-polygons 119875(1198601 119860

2 119860

119899)

(119899 ge 3) inH2 we always can divide it into several hyperbolictriangles 119879

1 119879

2 119879

119894(119894 ge 1) then we have

119860hyp (119875 (1198601 119860

2 119860

119899)) =

119894

sum

119896=1

119860hyp (119879119896) (36)

Remark 10 For any hyperbolic 119899-polygons119875(1198601 119860

2 119860

119899)

(119899 ge 3) in B2 since the hyperbolic area is invariant underMobius transformation sowe canmap the119875(119860

1 119860

2 119860

119899)

intoH2 by a Mobius transformation therefore we obtain thehyperbolic area of 119875(119860

1 119860

2 119860

119899) in B2

3 Hyperbolic Areas of Lambert Quadrilaterals

As an application of the explicit hyperbolic area formulaof a hyperbolic triangle we obtain the hyperbolic area of aLambert quadrilateral

Theorem 11 Let 119876(119860 119861 119862119863) be the Lambert quadrilateralwith angles (1205872 1205872 120601 1205872) 0 le 120601 lt 1205872 in H2 and 119860 =

1198880+ 119894119903

0 119861 = 119888

0+ 119894119903 then

(1) if 119903 le 119889

119860hyp (119876 (119860 119861 119862119863))

= arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(37)

(2) if 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1198881minus 119909

2

(38)

where 119889 = radic(1198881minus 119888

0)2+ 119903

2

1 119862 = 119909

2+ 119894119910

2119863 = 119909

1+ 119894119910

1 1198881and

1199031are given at (39)

Proof Without loss of generality we assume that 1198880lt 119888

1 0 lt

1199030lt 119903 and 119862 isin 119878

1(1198880 119903) cap 119878

1(1198881 1199031) 119863 isin 119878

1(1198880 1199030) cap 119878

1(1198881 1199031)

(see Figure 1) which implies 1198880+ 119903

0gt 119888

1minus 119903

1 1198881+ 119903

1gt 119888

0+ 119903

By (27) we have

1198881=

1199092

1+ 119910

2

1minus 119909

2

2minus 119910

2

2

2 (1199091minus 119909

2)

1199031= radic119910

2

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(39)

C0C1O

Y

A

B E

D

C

X

Figure 1

When 119903 le 119889 we get

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

(40)

where 119864 = 1199091+ 119894radic119903

2minus (119909

1minus 119888

0)2 So the following three

relations119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

10038161003816100381610038161198881minus 119888

0

1003816100381610038161003816

2= (119909

1minus 119888

0)2+ 119910

2

1+ (119909

1minus 119888

1)2+ 119910

2

1

(41)

imply that

119860hyp (119876 (119860 119861 119862119863)) = arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(42)

When 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0 it follows

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

(43)

and then we get

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1199092minus 119888

1

(44)

The proof of the case that 1198880gt 119888

1is similar to the case that

1198881gt 119888

0 So the proof of Theorem 11 is complete

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Journal of Mathematics

Lemma 5 Let 120574 be a hyperbolic geodesic with two endpoints119860 = 119888 + 119905119890

1198941205791 119861 = 119888 + 119905119890

1198941205792(119888 isin R 119905 gt 0 0 le 120579

1le 120587 0 le 120579

2le

120587 1205791

= 1205792) in H2 then the hyperbolic area of the hyperbolic

triangle 119879(119860 119861infin) is as follows

119860hyp (119879 (119860 119861infin)) =10038161003816100381610038161205792minus 120579

1

1003816100381610038161003816 (22)

Proof Let 119862 = 119888 + 119894119905 Without loss of generality we assume1205791lt 120579

2

(1) If 0 le 1205791lt 120579

2le 1205872 by Lemma 4 we have

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119861 119862infin))

= (

120587

2

minus 1205791) minus (

120587

2

minus 1205792) = 120579

2minus 120579

1

(23)

(2) If 1205872 le 1205791lt 120579

2le 120587 using Lemma 4 again we get

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119861 119862infin)) minus 119860hyp (119879 (119860 119862infin))

= (1205792minus

120587

2

) minus (1205791minus

120587

2

) = 1205792minus 120579

1

(24)

(3) If 0 le 1205791lt 1205872 lt 120579

2le 120587 Lemma 4 also implies

119860hyp (119879 (119860 119861infin))

= 119860hyp (119879 (119860 119862infin)) + 119860hyp (119879 (119861 119862infin))

= (1205792minus

120587

2

) + (

120587

2

minus 1205791) = 120579

2minus 120579

1

(25)

Similarly if 1205792lt 120579

1we have the same conclusion

119860hyp (119879 (119860 119861infin)) = 1205791minus 120579

2 (26)

Thus the proof of Lemma 5 is completed

Theorem 2 is an application of Lemma 5 For the proofof Theorem 2 we also need a sharp formula of the circleorthogonal to the boundary ofH2 [25]

Lemma A Let 1199111 119911

2isin H2 with 119909

1= 119909

2 Then 119878

1(119888 119903) is

orthogonal to 120597H2 where

119888 =

10038161003816100381610038161199111

1003816100381610038161003816

2minus10038161003816100381610038161199112

1003816100381610038161003816

2

2 (1199091minus 119909

2)

119903 = radic1199102

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(27)

Proof of Theorem 2 Wemay assume that 1199091gt 119909

2 Let 1198781(119888 119903)

be the circle which is through the two points 1199111 119911

2 where 119888

and 119903 are represented as in (27) Then we have 1199111= 119888 + 119903119890

1198941205791

1199112= 119888 + 119903119890

1198941205792

Case 1 ((1199091minus 119888)(119909

2minus 119888) gt 0) If 119888 lt 119909

2lt 119909

1 then by Lemma 5

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199102

1199092minus 119888

minus arctan1199101

1199091minus 119888

(28)

If 1199092lt 119909

1lt 119888 so we have

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= arctan

1199101

119888 minus 1199091

minus arctan1199102

119888 minus 1199092

(29)

Case 2 ((1199091minus119888)(119909

2minus119888) lt 0)This case implies that119909

2lt 119888 lt 119909

1

then

119860hyp (119879 (1199111 119911

2infin))

=10038161003816100381610038161205791minus 120579

2

1003816100381610038161003816= 120587 minus arctan

1199101

1199091minus 119888

minus arctan1199102

119888 minus 1199092

(30)

Remark 6 For a general hyperbolic triangle119879(119860 119861 119862) inH2we can give its hyperbolic area formula by Theorem 2 Giventhree points 119860(119909

119860 119910

119860) 119861(119909

119861 119910

119861) and 119862(119909

119862 119910

119862) in H2 with

(119909119860minus 119909

119861)(119909

119861minus 119909

119862)(119909

119862minus 119909

119860) = 0 we may assume that

119909119862lt 119909

119860lt 119909

119861 Let 1198781(119888 119903) be the circlewhich is through119861 and

119862 and orthogonal to 120597H2 Let 119863(119909119863 119910

119863) be the point which

is the intersection of 1198781(119888 119903) and the hyperbolic geodesicthrough point 119860 and orthogonal to 120597H2 Then we have

119909119860= 119909

119863 119910

119863= radic119903

2minus (119909

119860minus 119888)

2 (31)

where 119888 = (|119911119861|2

minus |119911119862|2)2(119909

119861minus 119909

119862) and 119903 =

radic1199102

119861+ ((119910

2

119862+ (119909

119861minus 119909

119862)2minus 119910

2

119861)2(119909

119861minus 119909

119862))

2Thus it follows

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863))

(32)

where

119860hyp (119879 (119860 119862119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119862infin)) minus 119860hyp (119879 (119862119863infin))

10038161003816100381610038161003816

119860hyp (119879 (119860 119861119863))

=

10038161003816100381610038161003816119860hyp (119879 (119860 119861infin)) minus 119860hyp (119879 (119861119863infin))

10038161003816100381610038161003816

(33)

then byTheorem 2 we can get the result of 119860hyp(119879(119860 119861 119862))

Example 7 Let 119860 = 119894 119861 = 1 + 2119894 119862 = minus1 + 3119894 by calculationwe have

119860hyp (119879 (119860 119861 119862))

= 119860hyp (119879 (119860 119862119863)) + 119860hyp (119879 (119860 119861119863)) asymp 0372

(34)

where119863 = radic301198942

Remark 8 There are lots of triangles which have the samehyperbolic area but different Euclidean areas in H2 For

Journal of Mathematics 5

instance 119879(119860 119861 119862) and 119879(119863 119861 119864) are two hyperbolic trian-gles inH2 with 119860 = minus1 119861 = 0 119862 = 1119863 = minus2 119864 = 2 and

119860hyp (119879 (119860 119861 119862)) = 119860hyp (119879 (119863 119861 119864)) = 120587

120587

4

= 119860euc (119879 (119860 119861 119862)) = 119860euc (119879 (119863 119861 119864)) = 120587

(35)

Remark 9 For any hyperbolic 119899-polygons 119875(1198601 119860

2 119860

119899)

(119899 ge 3) inH2 we always can divide it into several hyperbolictriangles 119879

1 119879

2 119879

119894(119894 ge 1) then we have

119860hyp (119875 (1198601 119860

2 119860

119899)) =

119894

sum

119896=1

119860hyp (119879119896) (36)

Remark 10 For any hyperbolic 119899-polygons119875(1198601 119860

2 119860

119899)

(119899 ge 3) in B2 since the hyperbolic area is invariant underMobius transformation sowe canmap the119875(119860

1 119860

2 119860

119899)

intoH2 by a Mobius transformation therefore we obtain thehyperbolic area of 119875(119860

1 119860

2 119860

119899) in B2

3 Hyperbolic Areas of Lambert Quadrilaterals

As an application of the explicit hyperbolic area formulaof a hyperbolic triangle we obtain the hyperbolic area of aLambert quadrilateral

Theorem 11 Let 119876(119860 119861 119862119863) be the Lambert quadrilateralwith angles (1205872 1205872 120601 1205872) 0 le 120601 lt 1205872 in H2 and 119860 =

1198880+ 119894119903

0 119861 = 119888

0+ 119894119903 then

(1) if 119903 le 119889

119860hyp (119876 (119860 119861 119862119863))

= arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(37)

(2) if 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1198881minus 119909

2

(38)

where 119889 = radic(1198881minus 119888

0)2+ 119903

2

1 119862 = 119909

2+ 119894119910

2119863 = 119909

1+ 119894119910

1 1198881and

1199031are given at (39)

Proof Without loss of generality we assume that 1198880lt 119888

1 0 lt

1199030lt 119903 and 119862 isin 119878

1(1198880 119903) cap 119878

1(1198881 1199031) 119863 isin 119878

1(1198880 1199030) cap 119878

1(1198881 1199031)

(see Figure 1) which implies 1198880+ 119903

0gt 119888

1minus 119903

1 1198881+ 119903

1gt 119888

0+ 119903

By (27) we have

1198881=

1199092

1+ 119910

2

1minus 119909

2

2minus 119910

2

2

2 (1199091minus 119909

2)

1199031= radic119910

2

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(39)

C0C1O

Y

A

B E

D

C

X

Figure 1

When 119903 le 119889 we get

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

(40)

where 119864 = 1199091+ 119894radic119903

2minus (119909

1minus 119888

0)2 So the following three

relations119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

10038161003816100381610038161198881minus 119888

0

1003816100381610038161003816

2= (119909

1minus 119888

0)2+ 119910

2

1+ (119909

1minus 119888

1)2+ 119910

2

1

(41)

imply that

119860hyp (119876 (119860 119861 119862119863)) = arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(42)

When 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0 it follows

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

(43)

and then we get

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1199092minus 119888

1

(44)

The proof of the case that 1198880gt 119888

1is similar to the case that

1198881gt 119888

0 So the proof of Theorem 11 is complete

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Mathematics 5

instance 119879(119860 119861 119862) and 119879(119863 119861 119864) are two hyperbolic trian-gles inH2 with 119860 = minus1 119861 = 0 119862 = 1119863 = minus2 119864 = 2 and

119860hyp (119879 (119860 119861 119862)) = 119860hyp (119879 (119863 119861 119864)) = 120587

120587

4

= 119860euc (119879 (119860 119861 119862)) = 119860euc (119879 (119863 119861 119864)) = 120587

(35)

Remark 9 For any hyperbolic 119899-polygons 119875(1198601 119860

2 119860

119899)

(119899 ge 3) inH2 we always can divide it into several hyperbolictriangles 119879

1 119879

2 119879

119894(119894 ge 1) then we have

119860hyp (119875 (1198601 119860

2 119860

119899)) =

119894

sum

119896=1

119860hyp (119879119896) (36)

Remark 10 For any hyperbolic 119899-polygons119875(1198601 119860

2 119860

119899)

(119899 ge 3) in B2 since the hyperbolic area is invariant underMobius transformation sowe canmap the119875(119860

1 119860

2 119860

119899)

intoH2 by a Mobius transformation therefore we obtain thehyperbolic area of 119875(119860

1 119860

2 119860

119899) in B2

3 Hyperbolic Areas of Lambert Quadrilaterals

As an application of the explicit hyperbolic area formulaof a hyperbolic triangle we obtain the hyperbolic area of aLambert quadrilateral

Theorem 11 Let 119876(119860 119861 119862119863) be the Lambert quadrilateralwith angles (1205872 1205872 120601 1205872) 0 le 120601 lt 1205872 in H2 and 119860 =

1198880+ 119894119903

0 119861 = 119888

0+ 119894119903 then

(1) if 119903 le 119889

119860hyp (119876 (119860 119861 119862119863))

= arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(37)

(2) if 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1198881minus 119909

2

(38)

where 119889 = radic(1198881minus 119888

0)2+ 119903

2

1 119862 = 119909

2+ 119894119910

2119863 = 119909

1+ 119894119910

1 1198881and

1199031are given at (39)

Proof Without loss of generality we assume that 1198880lt 119888

1 0 lt

1199030lt 119903 and 119862 isin 119878

1(1198880 119903) cap 119878

1(1198881 1199031) 119863 isin 119878

1(1198880 1199030) cap 119878

1(1198881 1199031)

(see Figure 1) which implies 1198880+ 119903

0gt 119888

1minus 119903

1 1198881+ 119903

1gt 119888

0+ 119903

By (27) we have

1198881=

1199092

1+ 119910

2

1minus 119909

2

2minus 119910

2

2

2 (1199091minus 119909

2)

1199031= radic119910

2

1+ (

1199102

2+ (119909

1minus 119909

2)2minus 119910

2

1

2 (1199091minus 119909

2)

)

2

(39)

C0C1O

Y

A

B E

D

C

X

Figure 1

When 119903 le 119889 we get

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

(40)

where 119864 = 1199091+ 119894radic119903

2minus (119909

1minus 119888

0)2 So the following three

relations119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

10038161003816100381610038161198881minus 119888

0

1003816100381610038161003816

2= (119909

1minus 119888

0)2+ 119910

2

1+ (119909

1minus 119888

1)2+ 119910

2

1

(41)

imply that

119860hyp (119876 (119860 119861 119862119863)) = arctan1199102

1198881minus 119909

2

+ arctan1199102

1199092minus 119888

0

minus

120587

2

(42)

When 119889 lt 119903 lt 1198881+ 119903

1minus 119888

0 it follows

119860hyp (119876 (119860 119861 119862119863))

= 119860hyp (119876 (119860 119861 119864119863)) + 119860hyp (119879 (119864 119862119863))

119860hyp (119876 (119860 119861 119864119863))

= 119860hyp (119879 (119860119863infin)) minus 119860hyp (119879 (119861 119864infin))

119860hyp (119879 (119864 119862119863))

= 119860hyp (119879 (119863 119862infin)) minus 119860hyp (119879 (119864 119862infin))

(43)

and then we get

119860hyp (119876 (119860 119861 119862119863))

=

120587

2

+ arctan1199102

1199092minus 119888

0

minus arctan1199102

1199092minus 119888

1

(44)

The proof of the case that 1198880gt 119888

1is similar to the case that

1198881gt 119888

0 So the proof of Theorem 11 is complete

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Journal of Mathematics

Y

Vd

Va

Vc

Vb

X

Figure 2

In particular when 119903 = 1198881+ 119903

1minus 119888

0 we have

119860hyp (119876 (119860 119861 119862119863)) =

120587

2

(45)

Remark 12 Since the Saccheri quadrilateral can be dividedinto twoLambert quadrilaterals therefore the hyperbolic areaof a Saccheri quadrilateral is twice of a Lambert quadrilateral

4 Hyperbolic GeometricCharacterization of Lambert Quadrilateralsand Saccheri Quadrilaterals

Lemma B (see [25]) Let 119886 isin R2 be a constant with |119886| gt 1Then 119878

1(119886 119903) is orthogonal to 1198781(0 1) for |119886|2 = 1 + 119903

2 Given1199111= 119909

1+ 119894119910

1 119911

2= 119909

2+ 119894119910

2isin B2 such that 0 119911

1 and 119911

2are

noncollinear the orthogonal circle 1198781(119886 119903) contains 1199111and 119911

2if

119886 = 119894

1199112(1 +

10038161003816100381610038161199111

1003816100381610038161003816

2) minus 119911

1(1 +

10038161003816100381610038161199112

1003816100381610038161003816

2)

2 (11990921199101minus 119909

11199102)

119903 =

10038161003816100381610038161199111minus 119911

2

1003816100381610038161003816

100381610038161003816100381610038161199111

10038161003816100381610038161199112

1003816100381610038161003816

2minus 119911

2

10038161003816100381610038161003816

210038161003816100381610038161199112

1003816100381610038161003816

100381610038161003816100381611990911199102minus 119909

21199101

1003816100381610038161003816

(46)

and 1198781(119886 119903) cap 119878

1(0 1) = 119911 isin R2

119911 = 119886|119886| exp(plusmn119894120579) 120579 =

arccos(1|a|)

The following result in Lemma 13 appeared in [20] forcompleteness we give its proof as follows

Lemma 13 Let 119876(V119886 V

119887 V

119888 V

119889) be a Lambert quadrilateral

(see Figure 2) in B2 with V119888= 119905119890

119894120579 0 lt 120579 lt 1205872 and let

1198891= 119889

120588(119869 [V

119886 V

119889] 119869 [V

119887 V

119888])

1198892= 119889

120588(119869 [V

119886 V

119887] 119869 [V

119888 V

119889])

(47)

then

1198891= arth (119871119903) 119889

2= arth (1198711199031015840) (48)

where 119871 = tanh 120588B2(V119886 V119888) 119903 = cos 120579 1199031015840 = radic1 minus 1199032= sin 120579

Proof Assume that V119886= 0 V

119887is on the real axis V

119889is on the

imaginary axis and V119888= 119905119890

119894120579 0 lt 119905 le 1 and 0 lt 120579 lt 1205872 Weuse (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(49)

Utilizing the relation (46) we have

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(50)

where 119887 119903119887denote the center and radius of the circle through

the geodesic 120574(V119887 V

119888) and 119889 and 119903

119889denote the center and

radius of the circle through the geodesic 120574(V119888 V

119889) Write119875(119887+

119903119887cos120595 119903

119887sin120595) isin 120574(V

119887 V

119888) 1205872 lt 120595 lt 120587 and 119876(0 119910) isin

120574(V119886 V

119889) 0 le 119910 le |V

119889| = |119889| minus 119903

119889 Let 119891(120595 119910) = 120588B2(119875 119876)

Then the relation (9) implies that

119891 (120595 119910) = 120588B2 (119875 119876)

= 2arth1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

radic1003816100381610038161003816119911119875minus 119911

119876

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816119911119875

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816119911119876

1003816100381610038161003816

2)

= 2arthradic1198872+ 119903

2

119887+ 2119887119903

119887cos120595 minus 2119910119903

119887sin120595 + 119910

2

1 + 1199102(119887

2+ 119903

2

119887+ 2119887119903

119887cos120595) minus 2119910119903

119887sin120595

(51)

From the relation (7) we know that 1198891= inf

120595119910120588B2(119875 119876) The

function 119891(120595 119910) reaches the minimum value

119891 (120587 0) = 2arth (119887 minus 119903119887) = arth2119905 cos 120579

1 + 1199052 (52)

Then

1198891= 120588B2 (0 V119887) = arth2119905 cos 120579

1 + 1199052

= arth (119871119903) (53)

where 119903 = cos 120579 Similarly we have

1198892= 120588B2 (0 V119889) = arth2119905 sin 120579

1 + 1199052

= arth (1198711199031015840) (54)

where 1199031015840 = radic1 minus 1199032= sin 120579

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Mathematics 7

Proof of Theorem 3 Assume that V119886= 0 and V

119888= 119905119890

119894120579 0 lt 119905 le

1 0 lt 120579 lt 1205872 and let V119887be on the real axis and V

119889on the

imaginary axis (see Figure 2) We use (10) to obtain

120588B2 (0 V119888) = log1 +

1003816100381610038161003816V119888

1003816100381610038161003816

1 minus1003816100381610038161003816V119888

1003816100381610038161003816

= log 1 + 119905

1 minus 119905

119871 = tanh 120588B2 (V119886 V119888) =2119905

1 + 1199052

(55)

which implies

120588B2 (V119886 V119888) = arth 2119905

1 + 1199052 (56)

By the relation (46) we have that the geodesic 120574(V119887 V

119888) is on

the circle 1198781(119887 119903119887) where

119903119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

119887 =

1 + 1199052

2119905 cos 120579

(57)

Similarly the geodesic 120574(V119888 V

119889) is on the circle 119878

1(119889 119903

119889)

where

119889 = 119894

1 + 1199052

2119905 sin 120579 119903

119889=

radic(1 + 1199052)2minus 4119905

2sin21205792119905 sin 120579

(58)

Hence 1199051= |V

119887minus V

119886| = 119887 minus 119903

119887and 119905

2= |V

119889minus V

119886| = |119889| minus 119903

119889

Utilizing the relation (9) we obtain

120588B2 (V119887 V119889) = 2arth1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

radic1003816100381610038161003816V119887minus V

119889

1003816100381610038161003816

2+ (1 minus

1003816100381610038161003816V119887

1003816100381610038161003816

2) (1 minus

1003816100381610038161003816V119889

1003816100381610038161003816

2)

= 2arthradic1199052

1+ 119905

2

2

1 + 1199052

11199052

2

(59)

This completes the proof of Theorem 3

Theorem 14 Let 119878(V1015840119889 V

119889 V

119888 V1015840

119888) be a Saccheri quadrilateral in

B2 with V119888= 119905119890

119894120579 V1015840119888= 119905119890

minus119894120579 0 lt 120579 lt 1205872 and

1198893= 119889

120588(119869 [V

119888 V1015840

119888] 119869 [V

119889 V1015840

119889])

1198894= 119889

120588(119869 [V

119888 V

119889] 119869 [V1015840

119888 V1015840

119889])

(60)

then

tanh21198893+ tanh2 (1198894

2

) = 1198712 (61)

where 119871 = tanh 120588B2(0 V119888)

Proof Assume that V119888= 119905119890

119894120579 0 lt 119905 le 1 0 lt 120579 lt 1205872 andV119886= 0 and let V

119887be on the real axis and V

119889on the imaginary

axis (see Figure 3) Then by the relation (46) we have that thecircle through points V

119888and V1015840

119888which is orthogonal to B2 is

1198781(119887 119903

119887) where

119887 =

1 + 1199052

2119905 cos 120579 119903

119887=

radic(1 + 1199052)2minus 4119905

2cos21205792119905 cos 120579

(62)

Y

Vd

Va

Vc

Vb

X

V998400

dV998400

c

Figure 3

It follows from the relations (10) and (48) that

1198893= 120588B2 (0 V119887) = arth (119871 cos 120579)

1198894= 2120588B2 (0 V119889) = 2arth (119871 sin 120579)

(63)

where 119871 = tanh 120588B2(0 V119888) = 2119905(1 + 1199052) isin (0 1] Thus

tanh21198893+ tanh2 (1198894

2

) = 1198712 (64)

This completes the proof of Theorem 14

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by the Natural Science Founda-tion of Fujian Province of China (2014J01013) NCETFJFund (2012FJ-NCET-ZR05) and the Promotion Program forYoung and Middle-Aged Teacher in Science and TechnologyResearch of Huaqiao University (ZQN-YX110)

References

[1] A F Beardon and C Pommerenke ldquoThe Poincare metric ofplane domainsrdquoThe Journal of the LondonMathematical SocietyII vol 18 no 3 pp 475ndash483 1978

[2] B Iversen Hyperbolic Geometry vol 25 of London Math-ematical Society Student Texts Cambridge University PressCambridge UK 1992

[3] J Milnor ldquoHyperbolic geometry the first 150 yearsrdquo AmericanMathematical Society vol 6 no 1 pp 9ndash24 1982

[4] S Kulczycki Non-Euclidean Geometry Courier Dover Mine-ola NY USA 2012

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Mathematics

[5] A F Beardon The Geometry of Discrete Groups vol 91 ofGraduate Texts in Mathematics Springer New York NY USA1983

[6] M VuorinenConformal Geometry and Quasiregular Mappingsvol 1319 of Lecture Notes in Mathematics Springer BerlinGermany 1988

[7] J W Anderson Hyperbolic Geometry Springer UndergraduateMathematics Series Springer London UK 2nd edition 2005

[8] L Keen and N Lakic Hyperbolic Geometry from a Local View-point London Mathematical Society Student Texts 68 Cam-bridge University Press Cambridge UK 2007

[9] A Marden Outer Circles An introduction to hyperbolic 3-manifolds Cambridge University Press Cambridge UK 2007

[10] J G Ratcliffe Foundations of Hyperbolic Manifolds GraduateTexts inMathmatics SpringerNewYorkNYUSA 2nd edition2006

[11] N Curien andWWerner ldquoTheMarkovian hyperbolic triangu-lationrdquo Journal of the EuropeanMathematical Society vol 15 no4 pp 1309ndash1341 2013

[12] O Demirel ldquoA characterization of Mobius transformations byuse of hyperbolic trianglesrdquo Journal of Mathematical Analysisand Applications vol 398 no 2 pp 457ndash461 2013

[13] A N Fang and S H Yang ldquoCorrigendum to lsquoa new char-acteristic of Mobius transformations in hyperbolic geometryrsquo[Journal of Mathematical Analysis and Applications vol 319no 2 pp 660ndash664 2006]rdquo Journal ofMathematical Analysis andApplications vol 376 pp 383ndash384 2011

[14] S Yang and A Fang ldquoA new characteristic of Mobius trans-formations in hyperbolic geometryrdquo Journal of MathematicalAnalysis and Applications vol 319 no 2 pp 660ndash664 2006

[15] V Pambuccian ldquoMappings preserving the area equality ofhyperbolic triangles are motionsrdquo Archiv der Mathematik vol95 no 3 pp 293ndash300 2010

[16] X Chen and A Fang ldquoA class of harmonic quasiconformalmappings with strongly hyperbolically convex rangesrdquoComplexVariables and Elliptic Equations vol 58 no 7 pp 1013ndash10222013

[17] X D Chen and A N Fang ldquoA Schwarz-Pick inequality for har-monic quasiconformal mappings and its applicationsrdquo Journalof Mathematical Analysis and Applications vol 369 no 1 pp22ndash28 2010

[18] M Knezevic and M Mateljevic ldquoOn the quasi-isometries ofharmonic quasiconformal mappingsrdquo Journal of MathematicalAnalysis and Applications vol 334 no 1 pp 404ndash413 2007

[19] M Rostamzadeh and S Taherian ldquoDefect and area in Beltrami-Klein model of hyperbolic geometryrdquo Results in Mathematicsvol 63 no 1-2 pp 229ndash239 2013

[20] M Vuorinen andGWang ldquoHyperbolic Lambert quadrilateralsand quasiconformalmappingsrdquoAnnales Academiaelig ScientiarumFennicaelig Mathematica vol 38 no 2 pp 433ndash453 2013

[21] N Kanesaka and H Nakamura ldquoOn hyperbolic area of themoduli of 120579-acute trianglesrdquoMathematical Journal of OkayamaUniversity vol 55 pp 191ndash200 2013

[22] F W Gehring and E Reich ldquoArea distortion under quasicon-formal mappingsrdquo Annales Academiae Scientiarum Fennicae Avol 388 pp 1ndash15 1966

[23] J A Kelingos ldquoDistortion of hyperbolic area under quasicon-formal mappingsrdquo Duke Mathematical Journal vol 41 pp 127ndash139 1974

[24] R M Porter and L F Resendis ldquoQuasiconformally explodablesetsrdquo Complex Variables Theory and Application vol 36 no 4pp 379ndash392 1998

[25] R Klen and M Vuorinen ldquoApollonian circles and hyperbolicgeometryrdquo Journal of Analysis vol 19 pp 41ndash60 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of