On the non-generic Tzitzeica-Johnson s Configuration

Similar documents
Unit 1: Introduction to Proof

Ion Patrascu, Florentin Smarandache Theorems with Parallels Taken through a Triangle s Vertices and Constructions Performed only with the Ruler

right angle an angle whose measure is exactly 90ᴼ

Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg

DISCOVERING GEOMETRY Over 6000 years ago, geometry consisted primarily of practical rules for measuring land and for

Chapter 3. Betweenness (ordering) A system satisfying the incidence and betweenness axioms is an ordered incidence plane (p. 118).

Triangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z.

Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg ( )

Geometry Arcs and Chords. Geometry Mr. Austin

Triangle Geometry. Often we can use one letter (capitalised) to name an angle.

Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions

Geometry GENERAL GEOMETRY

Geometry Problem booklet

Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg ( )

triangles in neutral geometry three theorems of measurement

Geometry Arcs and Chords. Geometry Mr. Peebles Spring 2013

Question 1 (3 points) Find the midpoint of the line segment connecting the pair of points (3, -10) and (3, 6).

TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions

Math 1230, Notes 2. Aug. 28, Math 1230, Notes 2 Aug. 28, / 17

UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).

UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).

10. Show that the conclusion of the. 11. Prove the above Theorem. [Th 6.4.7, p 148] 4. Prove the above Theorem. [Th 6.5.3, p152]

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in

3.2. Parallel Lines and Transversals

3.2. Parallel Lines and Transversals

1) If AB is congruent to AC, then B is congruent to C.

(RC3) Constructing the point which is the intersection of two existing, non-parallel lines.

Ion Patrascu, Florentin Smarandache A Sufficient Condition for the Circle of the 6 Points to Become Euler s Circle

Mathematics 3210 Spring Semester, 2005 Homework notes, part 8 April 15, 2005

North Carolina Math 2 Transition Edition Unit 4 Assessment: Similarity and Congruence

INVERSION IN THE PLANE BERKELEY MATH CIRCLE

Page 1 of 15. Website: Mobile:

Exercise 2.1. Identify the error or errors in the proof that all triangles are isosceles.

chapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?

Lesson 9.1 Skills Practice

Neutral Geometry. October 25, c 2009 Charles Delman

Sample Question Paper Mathematics First Term (SA - I) Class IX. Time: 3 to 3 ½ hours

Name: Date: Period: 1. In the diagram below,. [G.CO.6] 2. The diagram below shows a pair of congruent triangles, with and. [G.CO.

2012 GCSE Maths Tutor All Rights Reserved

New Jersey Center for Teaching and Learning. Progressive Mathematics Initiative

SOME EQUIVALENT CONDITIONS FOR CONIC SECTIONS

Exercises for Unit V (Introduction to non Euclidean geometry)

Additional Mathematics Lines and circles

Cylindrical Tzitzeica Curves Implies Forced Harmonic Oscillators

Solutions and scoring for 1st Midterm Exam Geometry 515 Fall Oct 2 Mon 4:00-5:20 pm

SOLUTIONS FOR. MTH 338 Final, June A 3 5 note card allowed. Closed book. No calculator. You may also use your list of the first 15 SMGS axioms.

Postulates, Definitions, and Theorems (Chapter 4)

Foundations of Neutral Geometry

Congruence. Chapter The Three Points Theorem

7. m JHI = ( ) and m GHI = ( ) and m JHG = 65. Find m JHI and m GHI.


5-1 Perpendicular and Angle Bisectors

The equation of Euler s line yields a Tzitzeica surface

EXERCISE 10.1 EXERCISE 10.2

3. In the Poincaré Plane let l = 0 L and. 4. Prove the above Theorem. [Th 7.1.4, p173] 5. Show that in the Poincaré Plane there is

4 Arithmetic of Segments Hilbert s Road from Geometry

It is known that the length of the tangents drawn from an external point to a circle is equal.

CME Project, Geometry 2009 Correlated to: Kentucky Core Content for Mathematics Assessment 4.1 (High School, Grade 11)

Geometry: Introduction, Circle Geometry (Grade 12)

Midterm Review Packet. Geometry: Midterm Multiple Choice Practice

TOPIC 4 Line and Angle Relationships. Good Luck To. DIRECTIONS: Answer each question and show all work in the space provided.

Circles in Neutral Geometry

Gossard s Perspector and Projective Consequences

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)

Common Core State Standards for Mathematics - High School

Ch 5 Practice Exam. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.

2005 Palm Harbor February Invitational Geometry Answer Key

0811ge. Geometry Regents Exam

Triangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y?

BUILT. for. Regents Pathways

Activity Sheet 1: Constructions

Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch

Chapter - 7. (Triangles) Triangle - A closed figure formed by three intersecting lines is called a triangle. A

Y. D. Chai and Young Soo Lee

Prentice Hall Intermediate Algebra, 5th Edition 2009 (Martin-Gay)

Prentice Hall PreCalculus, 3rd Edition 2007, (Blitzer)

Alhazen s Hyperbolic Billiard Problem

Euclidean Geometry Proofs

The Inversion Transformation

0811ge. Geometry Regents Exam BC, AT = 5, TB = 7, and AV = 10.

MATHEMATICS CLASS - IX MATHEMATICS CLASS - IX

GEO REVIEW TEST #1. 1. In which quadrilateral are the diagonals always congruent?

arxiv: v1 [math.ho] 29 Nov 2017

CO-ORDINATE GEOMETRY. 1. Find the points on the y axis whose distances from the points (6, 7) and (4,-3) are in the. ratio 1:2.

Chapter 10. Properties of Circles

North Carolina Standard Course of Study North Carolina Math 2

Euclidean Geometry Rediscovered

Name: Class: Date: 5. If the diagonals of a rhombus have lengths 6 and 8, then the perimeter of the rhombus is 28. a. True b.

2. In ABC, the measure of angle B is twice the measure of angle A. Angle C measures three times the measure of angle A. If AC = 26, find AB.

21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.

The analysis method for construction problems in the dynamic geometry

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:

CIRCLES MODULE - 3 OBJECTIVES EXPECTED BACKGROUND KNOWLEDGE. Circles. Geometry. Notes

THE FIVE GROUPS OF AXIOMS.

Regent College. Maths Department. Core Mathematics 4. Vectors

Some New Facts in Discrete Asymptotic Analysis

Generalization of the Wallace-Simson line

9 th CBSE Mega Test - II

Pearson Geometry Postulates And Theorems

Maths Module 4: Geometry. Teacher s Guide

Transcription:

An. Şt. Univ. Ovidius Constanţa Vol. 20(2), 2012, 21 26 On the non-generic Tzitzeica-Johnson s Configuration Wladimir G. Boskoff, Şerban Bărcănescu and Alexandru Bobe Abstract Working in the context of Hilbert s axiomatic system, we prove that the Tzitzeica-Johnson configuration has sense in neutral geometry. The elements of a Tzitzeica-Johnson configuration allow us to analyse the Euclidean or the non-euclidean character of the metric of the plane. 1 Introduction The statement of the analogue of Tzitzeica-Johnson theorem in the neutral geometry is described by the following theorem as it appears in [2]. Theorem 1. Let H, O 1, O 2, O 3 four points in the neutral plane such that O 1 H = = O 2 H = O 3 H and H Int( O 1 O 2 O 3 ). Consider A = S O1O 2 H, B = S O3O 1 H, C = S O2O 3 H. Then there exists a point O in the neutral plane such that OA = OB = OC. Therefore in the case when H Int( O 1 O 2 O 3 ), called the generic case, it exists a Tzitzeica-Johnson configuration in the neutral plane and it was denoted by [H, O 1 O 2 O 3 O, ABC] (see [2]). In the Euclidean plane, the generic Tzitzeica-Johnson s configuration has the property that [OA], [OB], [OC], [HO 1 ], [HO 2 ], [HO 3 ] all are congruent segments. Key Words: Neutral plane, axiom of parallels, Tzitzeica problem, Johnson problem. 2010 Mathematics Subject Classification: 51F05, 51M05, 51M09, 51M10. Received: August, 2011. Accepted: February, 2012. 21

22 Wladimir G. Boskoff, Şerban Bărcănescu and Alexandru Bobe In the neutral plane we have that [OA], [OB], [OC] are congruent and separately that [HO 1 ], [HO 2 ], [HO 3 ] are congruent. It is natural to compare [OA] and [HO 1 ] in order to establish certain properties of the neutral plane itself. To formulate the next result we use the comparison of the angles and of the segments as introduced in [10]. By the notation R we mean the right angle in the plane. By the sign R we mean any of the relations >, =, <. The result which allows to highlight the importance of a generic Tzitzeica- Johnson s configuration in the neutral plane in establishing the type of the metric is the following (see [2]). Theorem 2. Let [H, O 1 O 2 O 3 O, ABC] be a generic Tzitzeica-Johnson s configuration in the neutral plane. Let Σ be the sum of the angles of O 1 O 2 O 3. If [AO] R [HO 1 ] then Σ R 2R. At this point we introduce a new definition. We call non-generic Tzitzeica- Johnson configuration, the geometric figure in the neutral plane such that the point H doesn t lie in the interior of O 1 O 2 O 3. Our goal becomes to obtain an equivalent theorem in the case of non-generic Tzitzeica-Johnson s configuration. We have to analyze two possibilities. 2 Undecidability in the case H ( O 1 O 2 O 3 ) Suppose now that H lies on the boundary of O 1 O 2 O 3. Denote this case by [H, (O 1 O 2 O 3 ) O, ABC] Without loss of generality (see Figure 1) we can assume that H lies on the segment [O 1 O 3 ], such that HO 1 = HO 2 = HO 3. Then S O1O 3 H = H = B, S O1O 2 H = A S O2O 3 H = C. Denote {T } = AH O 1 O 2 and {K} = CH O 2 O 3. Remark that AT O 2 HT O 2, which yields O 2 A = O 2 H (= O 2 B ). Similarly from CKO 2 HKO 2 we get CO 2 = O 2 H (= O 2 B ). Therefore O 2 is equidistant from A, B = H, and C. This shows that when H lies on the boundary, the Tzitzeica-Johnson configuration has the same structure in both Euclidean and non-euclidean cases. In fact, in this case, one can not decide if the geometry is Euclidean or hyperbolic.

ON THE NON-GENERIC TZITZEICA-JOHNSON S CONFIGURATION 23 3 The case when H lies in the exterior of O 1 O 2 O 3 : Analysis Now we analyze the Tzitzeica-Johnson configuration when H lies in the exterior of O 1 O 2 O 3. For this discussion we refer to Figure 2. Suppose that [HO 1 is between [HO 2 and [HO 3 as in Figure 4. Let A = S O1O 2 H, B = S O1O 3 H, and C = S O2O 3 H. The rays [HA, [HB and [HC bisect O 2 HO 1, O 1 HO 3, and O 2 HO 3, respectively. Consider now the perpendicular bisectors of [AB] and [AC]. One of them is perpendicular on [AB], the other one is slant. In the Hilbert axiomatic sistem corresponding to Euclidean geometry is easy to see that these perpendicular bisectors meet. On the other hand, in the axiomatic framework corresponding to neutral geometry there is no criterion to decide if the two lines intersect. So, in the case of a non-generic Tzitzeica-Johnson configuration if there is not O such that OA = OB = OC it results that our geometry is hyperbolic. Let us denote such configuration by [H, ext (O 1 O 2 O 3 ) O, ABC]. Let us analyze the Tzitzeica-Johnson configuration in the case when the perpendicular bisectors of [AB] and [AC] meet in a point O in neutral geometry. Let us denote this case by [H, ext (O 1 O 2 O 3 ) O, ABC]. First, let us describe the construction of these perpendicular bisectors. Let L be the midpoint of [AB]. Then O 1 L is perpendicular on AB since O 1 LB and O 1 LA are congruent (case SSS). Similarly, denote by D the midpoint of [AC]. In the same way, O 2 D AC by the same congruence case in O 2 DA O 2 DC. Thus, the perpendicular bisector of [AB] and [AC] are O 1 L and O 2 D, respectively. By our assumption, let O be their point of intersection. Then the perpendicular bisector of [BC] passes also through O. The construction of the perpendicular bisector of [BC] is similar: the midpoint Y of [BC] determines the line O 3 Y which is the perpendicular bisector of [BC]. Furthermore, O [O 3 Y. At this point, let us focus on O 2 O 3 O. We denote by x the equal angles HO 1 O 2 = HO 2 O 1 = AO 1 O 2 = AO 2 O 1. Then remark that OO 2 O 3 has also measure x, since [O 2 O 3 bisects HO 2 C and [O 2 O bisects CO 2 A. By analogy, denoting by y the measure of equal angles HO 1 O 3 = HO 3 O 1 = BO 1 O 3 = BO 3 O 1, we obtain that O 2 O 3 O has measure y. For O 2 OO 3, the reasoning is different. Denote by z the measure of AO 1 O = BO 1 O, and by u the measure of AOO 1 = BOO 1. The reasoning made to measure O 2 OO 3 still holds true due to the fact that [OO 2 bisects COA and [OO 3 bisects COB. Thus O 2 OO 3 has measure u. The sum of the angles around O 1 yields 2x+2y+2z = 4R (we use the standard notation with R for the measure of a right angle), thus x + y + z = 2R. The sum of angles in O 2 OO 3 is x + y + u.

24 Wladimir G. Boskoff, Şerban Bărcănescu and Alexandru Bobe Denote by R one of the symbols >, =, <. angles of the triangle O 2 OO 3. Denote by Σ the sum of the Theorem 3. In the case of the non-generic Tzitzeica-Johnson configuration [H, ext (O 1 O 2 O 3 ) O, ABC] we have O 1 H R OA if and only if ΣR2R. Proof: In order to complete the proof let us observe that urz if and only if O 1 H R OA, that is if and only if ΣR2R. One more comment can be made. Since the sum of the angles of a triangle in Hilbert s neutral plane is less than or equal 2R (Legendre s theorem) we obtain both in the case of a generic configuration and in the case of a nongeneric configuration that the metric can be Euclidean of Hyperbolic only. Except the case when H ( O 1 O 2 O 3 ), one can decide the type of the metric knowing the type of the configuration and looking at the comparison between the segments O 1 H and OA. All these are syntetised in the following. Theorem 4. (Main Theorem) Consider a Tzitzeica-Johnson configuration. a) [H, O 1 O 2 O 3 O, ABC]: AO R HO 1 ΣR2R, where Σ is the sum of angles of the triangle O 1 O 2 O 3 ; the metric can be Euclidean or hyperbolic. b) [H, (O 1 O 2 O 3 ) O, ABC]: the metric can not be decided between Euclidean and hyperbolic. c) [H, ext (O 1 O 2 O 3 ) O, ABC]: the metric is hyperbolic. d) [H, ext (O 1 O 2 O 3 ) O, ABC]: O 1 H R OA ΣR2R, where Σ is the sum of angles of the triangle O 2 OO 3 ; the metric can be Euclidean or hyperbolic. This complets the study of the Tzitzeica-Johnson configuration according to Barbilian s point of view. References [1] D. Barbilian - Pagini inedite, volume 2, Editors G. Barbilian and V. G. Vodă, Editura Albatros, 1984. [2] W. Boskoff, S. Barcanescu - On the Tzitzeica-Johnson configuration, Journal of Geometry, 96(2009), 57-61. [3] W. Boskoff and P. Horja - The characterization of some special Barbilian spaces using the Tzitzeica construction, Stud. Cerc. Mat. 46 (1994), No. 5, 503-514. [4] A. Emch - Remarks on the foregoing circle theorem, American Mathematical Monthly, 23(1916), 162-164. [5] M. J. Greenberg - Aristotle s axiom in the foundations of geometry J. Geom. 33(1988), 53-57.

ON THE NON-GENERIC TZITZEICA-JOHNSON S CONFIGURATION 25 [6] M. J. Greenberg - Euclidean and Non-Euclidean geometries, Development and history, Third edition, W. H. Freeman & Co., 1993. [7] R. Hartshorne - Non-Euclidean III.36 Amer. Math. Monthly 110(2003), 495-502. [8] D. Hilbert - The Foundations of Geometry, Open Court Publishing Company, La Salle, 1962 (reprint of the translation by E. J. Townsend, 1902). [9] R. A. Johnson - A Circle Theorem, Amer. Math. Monthly, 23(1916), 161-162. [10] B. Kerekjarto Les fondements de la geometrie, vol. 1, Akad. Kiado, Budapest, 1955. [11] T. Lalescu - Geometry of the Triangle (in Romanian: Geometria triunghiului), Ed. Apollo, 1993 (original French edition in 1937). [12] N. N. Mihăileanu - Complementary Lessons in Geometry, (in Romanian: Lecţii complementare de geometrie), Editura didactică şi pedagogică, Bucureşti, 1976. [13] N. N. Mihăileanu et al. - A Problem Book of Synthetic and Projective Geometry (in Romanian: Culegere de probleme de geometrie sintetică şi proiectivă), Editura didactică şi pedagogică, 1971. [14] V. Pambuccian - Zum Stufenaufbau des Parallelenaxioms J. Geom. 51(1994), 79-88. [15] V. Pambuccian - Constructive axiomatizations of plane absolute, Euclidean and hyperbolic geometry Math. Log. Q. 47(2001), 129-135. [16] G. Pólya - Mathematical discovery. On understanding, learning, and teaching problem solving, Reprint in one volume. With a foreword by Peter Hilton. With a bibliography by Gerald Alexanderson. With an index by Jean Pedersen. John Wiley & Sons, Inc., New York, 1981, xxv+220 pp. [17] G. Tzitzeica On a Euler s Theorem(romanian), Gazeta Matematică, 13(1907), 293-294. [18] G. Ţiţeica - Problems of Geometry (in Romanian: Probleme de geometrie, Sixth Edition, with foreword by Gh. D. Simionescu, Editura Tehnică, 1962, reprinted 1981.

26 Wladimir G. Boskoff, Şerban Bărcănescu and Alexandru Bobe [19] V. Gh. Vodă - The Charm of Old-Fashioned Geometry (in Romanian:Vraja geometriei demodate), Editura Albatros, 1983. Wladimir G. BOSKOFF, Department of Mathematics, Ovidius University of Constanta, Bdul Mamaia 124, 900527 Constanta, Romania. Email: boskoff@univ-ovidius.ro Şerban BĂRCĂNESCU, Institute of Mathematics of the Romanian Academy, Calea Grivitei 21, Bucharest, Romania. Email: serban.barcanescu@imar.ro Alexandru BOBE, Department of Mathematics, Ovidius University of Constanta, Bdul Mamaia 124, 900527 Constanta, Romania. Email: alexb@univ-ovidius.ro