ON THE GERGONNE AND NAGEL POINTS FOR A HYPERBOLIC TRIANGLE

INTERNATIONAL JOURNAL OF GEOMETRY Vol 6 07 No - ON THE GERGONNE AND NAGEL POINTS FOR A HYPERBOLIC TRIANGLE PAUL ABLAGA Abstract In this note we prove the existence of the analogous points of the Gergonne Nagel points their adjoint points for the case of a non-degenerate hyperbolic triangle nd their barycentric coordinates with respect to the given triangle Unlike their classical counterparts the hyperbolic points we construct may as well be ideal or ultra-ideal depending on the triangle Introduction This article represent a continuation of our attempt started a few years ago to contribute to the developing of the plane hyperbolic geometry by using an analytic approach based on barycentric or areal coordinates While not new the barycentric coordinates have been introduced by Sommerville see [9] in 93 they have been seldom used in the last decades although they proved to be very useful especially in problems related to the geometry of the hyperbolic triangles Nevertheless in recent years there is a new interest in the Cayley-Klein geometry among which there is also the projective model of the hyperbolic geometry see the monographs [5] [7] Sommerville [9] introduced the barycentric coordinates as a special type of homogeneous coordinates in the projective plane by using a triangle a unit point a stard procedure in projective geometry Nevertheless later Coxeter see [4] showed that these coordinates are more naturally introduced by using a polarity of the projective plane If be denote by X 0 ; X ; X the barycentric coordinates of a point of the projective plane the polarity can be written in point coordinates as X; Y = X 0 Y 0 + cosh cx 0 Y + X Y 0 + cosh bx 0 Y + X Y 0 + + X Y + cosh ax Y + X Y + X Y ; which means that in particular the equation of the Absolute also in point coordinates is Keywords phrases: hyperbolic plane hyperbolic triangle 00Mathematics Subject Classi cation: 5M09 5M0 Received: 50707 Accepted: 00807

On the Gergonne Nagel points for a hyperbolic triangle 3 X X = X 0 + cosh cx 0 X + cosh bx 0 X + X + + cosh ax X + X = 0 On the other h when written in line coordinates the polarity reads: [ξ η] = [ sinh a ξ 0 η 0 + sinh a sinh b cos C ξ 0 η + ξ η 0 γ 3 sinh b ξ η + sinh a sinh c cos B ξ 0 η + ξ η 0 sinh c ξ η + + sinh b sinh c cos A ξ η + ξ η ] while the equation of the Absolute in line coordinates is [ξ ξ] = [ sinh a ξ0 + sinh a sinh b cos C ξ 0 ξ sinh b ξ η + γ 4 + sinh a sinh c cos B ξ 0 ξ sinh c ξ + sinh b sinh c cos A ξ ξ ] = 0 Here a b c are the hyperbolic lengths of the sides of the triangle while γ = + cosh a cosh b cosh c cosh a cosh b cosh c > 0 is the determinant of the polarity written in point coordinates For other version of barycenter coordinates see [0] As mentioned in the abstract the aim of this paper is to prove the existence of hyperbolic analogues of the Gergonne Nagel points of their adjoints For the Euclidean versions see [] [6] The points lines mentioned within the article can be seen in the Figure 6 The red ellipse represents the Absolute The Gergonne point of a hyperbolic triangle Theorem Let ABC be a non-degenerate triangle in the hyperbolic plane We connect the contact points of the incircle of the triangle with each side to the opposite vertex Then the three cevians intersect each other at a point that we will call the Gergonne point of the triangle Remark The incircle as it is well-known is always a proper circle not an equidistant or a horocycle Also as the cevians corresponding to the contact points of the incircle with the sides lie within the sides not on their extension we expect the Gergonne point if it exists to be a proper point as it should lie inside the triangle Proof of Theorem We shall prove that the three cevians do indeed intersect we will find the barycentric coordinates of the intersection point We start by finding the coordinates of the contact points between the incircle the sides of the triangle To this end it is enough to find the projections of the incenter on the sides As it is known the barycentric coordinates of the incenter I of the triangle ABC can be written as sin A sin B sin C We denote by K a the foot of the perpendicular from I to the side BC The equation of the line IK a is obviously of the form 5 ξ 0 X 0 + ξ X + ξ X = 0 We have to find the coefficients ξ i i = 0 Of course as the equation of the line is homogeneous we only need two conditions to find the coefficients

4 Paul A Blaga We start with the condition of perpendicularity The equation of the side BC is X 0 = 0 therefore we have to impose the perpendicularity condition to the lines of coefficients given by the vectors ξ [ξ 0 ξ ξ ] η [ 0 0] As it is known this condition is simply [ξ η] = 0 or in our particular case ξ 0 sinh a + ξ sinh a sinh b cos C + ξ sinh a sinh c cos B = 0 or ξ 0 sinh a + ξ sinh b cos C + ξ sinh c cos B = 0 or finally using the hyperbolic sine theorem 6 ξ 0 sin A + ξ sin B cos C + ξ sin C cos B = 0 The second equation we need is based on the fact that the point I belongs to the line IK a hence 7 ξ 0 sin A + ξ sin B + ξ sin C = 0 After solving the system formed by the equations 6 7 performing some manipulations we get for IK a the equation 8 IK a : cos B cos C X 0 tan B sin A X + tan B sin A X = 0 To find the coordinates of K a we intersect IK a to the side BC of equation X 0 = 0 we get 9 K a = K a 0 tan C tan B Now we have all we need to write the equation of the cevian AK a An easy computation shows that this equation is 0 AK a : tan B X tan C X = 0 The other two contact points of the incenter are obtained in the same manner they are: K b = K b tan A 0 tan C K c = K c tan A tan B 0 therefore the two other analogous cevians are given by 3 BK b : tan A X 0 tan C X = 0 4 CK c : tan A X 0 tan B X = 0 If we intersect the lines AK a BK b which are obviously concurrent we obtain the point 5 Ge = Ge tan A tan B tan C it is easy to check that CK c also passes through this point The point Ge defined by 5 is called the Gergonne point of the triangle ABC It

On the Gergonne Nagel points for a hyperbolic triangle 5 is always a proper point of the hyperbolic plane it lies in the interior of the triangle 3 The adjoint Gergonne points The adjoint Gergonne points are constructed analogously to their Euclidean counterparts Namely we construct the perpendiculars from the center of an excycle to the sides of the triangle consider the cevians passing through the feet of these perpendiculars prove that they intersect each other at the same point Unlike the case of the Gergonne point the adjoint points may be ideal or ultra-ideal Let us consider thus the excenter I a given by the barycentric coordinates 6 I a = I a sin A sin B sin C We start by constructing the perpendicular from I a to the side BC We shall denote by K aa the foot of this perpendicular Proceeding as before we get for the line I a K aa the equation 7 I a K aa : sin B sin Ccos B cos CX 0 sin A cos C cos BX + + sin A sin B cos CX = 0 To intersect this line to the side BC we put X 0 = 0 we get after some trigonometric manipulations 8 K aa = K aa 0 tan C tan B We denote now the foot of the perpendicular from I a to CA by K ab We search again the equation of the line I a K ab under de form 5 It is easy to check that the equation of the line is 9 I a K ab : sin B sin C + cos AX 0 + sin A sin Ccos A + cos CX + + sin A sin B cos CX = 0 thus the we get for K ab : 0 K ab = K ab tan A 0 cot C If we denote by K ac the foot of the perpendicular from I a to AB then we get in the same manner K ac = K ac tan A cot B 0 We connect now the vertices of the triangle to the points K aa K ab K ac AK aa should be of the form ξ X + ξ X = 0 Using the fact that the line passes through K aa we get after some computation AK aa : tan B X tan C X = 0 Analogously we get 3 BK ab : cot A X 0 + tan C X = 0

6 Paul A Blaga 4 CK ac : cot A X 0 + tan B X = 0 We intersect first the lines 3 We get the point 5 J a = J a tan A cot B cot C On the other h clearly J a is also on the line CK ac hence the three lines are concurrent We have to mention nevertheless that the point J a is not necessarily an ordinary point of the hyperbolic plane In exactly the same manner if we use the excenter I b sin A sin B sin C we get the contact points 6 K ba = K ba 0 tan B cot C 7 K bb = K bb tan C 0 tan A 8 K bc = K bc cot A tan B 0 the intersection point 9 J b = J b cot A tan B cot C while finally if we use the excenter I c sin A sin B sin C we get the contact points 30 K ca = K ca 0 cot B tan C 3 K cb = K cb cot A 0 tan C 3 K cc = K cc tan B tan A 0 the intersection point 33 J b = J b cot A cot B tan C Thus we have the following theorem: Theorem Let ABC a nondegenerate hyperbolic triangle Then the lines connecting the vertices of the triangle to the projections of an excenter of the triangle on the opposite sides of the vertices are concurrent at a point which may be ordinary ideal or ultra-infinite The three points J a J b J c obtained in this way are called the adjoint Gergonne points of the triangle ABC Remark 3 Unlike the Gergonne points the adjoint Gergonne points may as well be ideal or ultra-ideal

On the Gergonne Nagel points for a hyperbolic triangle 7 4 The Nagel point of a hyperbolic triangle Theorem 3 Let ABC be a nondegenerate hyperbolic triangle We consider the cevians connecting each vertex to the contact point of the opposite excycle with the oposite side Then the three cevians are concurrent at a point we shall call the Nagel point of the triangle ABC Proof With the notations introduced before we consider the lines AK aa BK bb CK cc We claim that the three lines are concurrent We have already the equation of AK aa see We get in the same way 34 BK bb : tan A X 0 tan C X = 0 35 CK cc : tan A X 0 tan B X = 0 By intersecting the lines 33 we get the point 36 N cot A cot B cot C Clearly N is also on the line CK cc so the three lines are concurrent 5 The adjoint Nagel points of a hyperbolic triangle In the Euclidean geometry of the triangle an adjoint Nagel point associated to a side let s say BC is defined in the following way We consider the lines that connect the extremities of the side B C in our case to the contact points of the excircles inscribed into the angles adjacent to this side lying on the extensions of the lines opposite to the considered vertices More precisely we connect B to K cb C to K bc We also consider the line connecting the third vertex of the triangle A in our case to the contact point of the opposite side BC in our case with the incircle We shall call these three cevians the the adjoint Nagel cevians associated to the side a In a similar manner are defined the adjoint Nagel cevians associated to the two other sides of the triangle In the Euclidean geometry the three adjoint Nagel cevians of each side are concurrent at one point We get thus the three adjoint Nagel points of the triangle N a N b N c Our intention is to prove that an analogous result holds for the hyperbolic plane More specifically we have the following result: Theorem 4 Let ABC be a nondegenerate hyperbolic triangle Then the three adjoint Nagel cevians associated to a side of the triangle are concurrent at a point We get thus three points N a N b N c that we shall call the the adjoint Nagel points of the triangle ABC They are either proper ideal or ultra-ideal points of the hyperbolic plane Proof As we said we shall make the computations for the side BC the vertex A For the other two situations they are completely analogous We start with the line BK cb Clearly the equation of the line should be of the form ξ 0 X 0 + ξ X = 0

8 Paul A Blaga As the line passes through K cb we must have ξ 0 cot A ξ tan C = 0 or ξ 0 = tan A tan C ξ As such the equation of BK cb is: In a similar manner we get BK cb : tan A X 0 + cot C X = 0 CK bc : tan A X 0 + cot B X = 0 On the other h we saw already that the equation of AK a is AK a : tan C X tan B X = 0 As such the intersection between BK cb CK bc is the point N a given by N a = N a cot A tan B tan C it is easily checked that N a belongs to the line AK a as well We get analogously for the other two adjoint Nagel points: N b = N b tan A cot B tan C N c = N c tan A tan B cot C respectively 6 The expressions of the barycentric coordinates in terms of the sides lengths We found so far the expressions of the coordinates of the Gergonne Nagel points their associates in terms of the angles of the triangle ABC we discovered that they are identical to those from the Euclidean geometry We shall find now their expressions in terms of the lengths of the sides Although they are not identical to their Euclidean versions as we shall see the differences are minimal they agree to a general rule according to which many formulas can be adapted to hyperbolic geometry by simple replacing some lengths with the hyperbolic sine of these lengths We have the following theorem: Theorem 5 Let ABC be a nondegenerated hyperbolic triangle Then the barycentric coordinated of the Gergonne Nagel points of their adjoints written in terms of the lengths of the sides the semiperimeter s

are given by: 37 38 39 40 4 4 43 44 On the Gergonne Nagel points for a hyperbolic triangle 9 Ge = Ge sinhs a sinhs b J a = J a sinh s sinhs c sinhs b J b = J b sinhs c sinh s sinhs b sinhs a J c = J c sinhs b sinhs a sinh s N = N sinhs a sinhs b sinhs b N a = N a sinh s sinhs c sinhs b N b = N b sinhs c sinh s sinhs a N c = N c sinhs b sinhs a sinh s Proof Clearly all we have to do is to find expressions for tan A analogues in terms of the lengths of the sides In the Euclidean case the formula is well-known: tan A = s bs c ss a the where s is the semiperimeter of the triangle ABC In the hyperbolic case Nestorovich see [7] gives the following formulas: whence sin A = sinhs b sinhs c sinh b sinh c cos A sinh s sinhs a = sinh b sinh c tan A = sinhs b sinhs c sinh s sinhs a As such we have Ge = Ge tan A tan B tan C = sinhs b sinhs c sinhs c sinhs a = Ge sinh s sinhs a sinh s sinhs b sinhs a sinhs b sinh s sinhs c or Ge = Ge sinhs a sinhs b For the first adjoint Gergonne point we get sinhs b

0 Paul A Blaga Nb Ic Kca Na Kcb Jc Kac Kbc A Kcc Kb Kc Jb I Kbb Ge N B Kaa C Ka Ja Kab Ia Kba Nc Ib Figure The Gergonne Nagel points their adjoints J a = J a tan A cot B cot C sinhs b sinhs c sinh s sinhs b = J a sinh s sinhs a sinhs c sinhs a sinh s sinhs c sinhs a sinhs b or J a = J a sinh s sinhs c sinhs b

On the Gergonne Nagel points for a hyperbolic triangle analogously J b = J b sinhs c sinh s sinhs a J c = J c sinhs b sinhs a sinh s As it follows easily from the theorems 4 the Nagel point is the isotomic conjugate of the Gergonne point each adjoint Nagel point is isotomic conjugate to the corresponding adjoint Gergonne point see [3] Therefore we have at once: N = N sinhs a sinhs b sinhs b N a = N a sinh s sinhs c sinhs b N b = N b sinhs c sinh s sinhs a N c = N c sinhs b sinhs a sinh s References [] Andrica D Nguyen KL A note on the Nagel Gergonne points Creative Math & Inf 7008 pp 7-36 [] Blaga PA Barycentric trilinear coordinates in the hyperbolic plane Automat Comput Appl Math Volume 03 Number pp 49-58 [3] Blaga PA The isotomic transformation in the hyperbo ic plane Stud Univ Babeş- Bolyai Math 5904 No 4 pp 53-5 [4] Coxeter H Non-Euclidean Geometry 5th Edition University of Toronto Press 965 [5] Kowol G Projektive Geometrie und Cayley-Klein Geometrien der Ebene Birkhäuser 009 [6] Mihalescu C The Geometry of Remarkable Elements XYZ Press 06 [7] Nestorovich NM Geometrical Constructions in the Lobachevskian Plane in Russian GITTL 95 [8] Onishchik A Sulanke R Projective Cayley-Klein Geometries Springer 006 [9] Sommerville D Metrical coordinates in non-euclidean geometry Proceedings of the Edinburgh Mathematical Society 393 pp 6-5 [0] Ungar A Barycentric Calculus in Euclidean Hyperbolic Geometry World Scientific 00 FACULTY OF MATHEMATICS AND COMPUTER SCIENCE BABEŞ-BOLYAI UNIVERSITY CLUJ-NAPOCA ROMANIA E-mail address: pablaga@csubbclujro