The Geometry of the Orthological Triangles

Ion Pătrașcu Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

Peer Reviewers:

Prof. dr. Bencze Mihály Department of Mathematics, “Áprily Lajos” Highschool, Brașov, Romania

Prof. univ. dr. Temistocle Bîrsan “Gh. Asachi” Technical University, Iași, Romania

Lect. univ. dr. Claudiu-Ionuț Popîrlan Department of Computer Science, Faculty of Sciences, University of Craiova, Romania

Ion Pătrașcu Florentin Smarandache

THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

Pons Editions Brussels, 2020

To my grandchildren LUCAS and EVA-MARIE, with all my love – ION PĂTRAȘCU

ISBN 978-1-59973-653-2

DTP: Alexandre Lukacs Translation from Romanian by Oana Dumitru

Pons Editions Quai du Batelage, 5 1000 - Brussels

TABLE OF CONTENTS

FOREWORD ...... 21 AUTHORS’ NOTE ...... 23

1 INTRODUCTION ...... 25 1.1 ORTHOLOGICAL TRIANGLES: DEFINITION ...... 25 DEFINITION 1 ...... 26 OBSERVATION 1 ...... 26 EXERCISE 1...... 26 1.2. CHARACTERIZATION OF THE ORTHOLOGY RELATION ...... 26 THEOREM 1 (L. CARNOT, 1803) ...... 26 PROOF ...... 27 OBSERVATION 2 ...... 28 LEMMA 1 ...... 28 EXERCISE 2...... 28 THEOREM 2 ...... 29 PROOF ...... 29 THEOREM 3 ...... 30 PROOF ...... 30 OBSERVATION 3 ...... 31 1.3. THE ORTHOLOGICAL TRIANGLES THEOREM ...... 31 THEOREM 4 (J. STEINER, 1828 – THEOREM OF ORTHOLOGICAL TRIANGLES) ... 31 PROOF 1 ...... 31 PROOF 2 ...... 31 PROOF 3 ...... 32 PROOF 4 ...... 33 PROOF 5 ...... 34 REMARK 1 ...... 35 REMARK 2 ...... 35 PROOF 6 (ANALYTICAL) ...... 35 DEFINITION 2 ...... 37 PROBLEM 1 ...... 37 PROBLEM 2 ...... 37

Ion Pătrașcu, Florentin Smarandache 5 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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2 ORTHOLOGICAL REMARKABLE TRIANGLES ...... 39 2.1 A TRIANGLE AND ITS COMPLEMENTARY TRIANGLE ...... 39 DEFINITION 3 ...... 39 PROPOSITION 1 ...... 39 PROOF ...... 39 OBSERVATION 4 ...... 40 PROBLEM 3 ...... 40 2.2 A TRIANGLE AND ITS ANTI-COMPLEMENTARY TRIANGLE ...... 41 DEFINITION 4 ...... 41 PROPOSITION 2 ...... 41 PROOF ...... 41 OBSERVATION 5 ...... 42 2.3 A TRIANGLE AND ITS ORTHIC TRIANGLE ...... 42 DEFINITION 5 ...... 42 OBSERVATION 6 ...... 42 DEFINITION 6 ...... 43 OBSERVATION 7 ...... 43 PROPOSITION 3 ...... 43 PROOF ...... 44 REMARK 3 ...... 44 PROPOSITION 4 ...... 44 PROOF ...... 44 PROPOSITION 5 ...... 45 PROOF ...... 45 REMARK 4 ...... 46 PROPOSITION 6 ...... 46 PROOF ...... 46 REMARK 5 ...... 47 OBSERVATION 8 ...... 47 2.4 THE MEDIAN TRIANGLE AND THE ORTHIC TRIANGLE ...... 47 THEOREM 5 ...... 47 PROOF ...... 47 OBSERVATION 9 ...... 48 PROPOSITION 7 ...... 49 PROPOSITION 8 ...... 49 PROPOSITION 9 ...... 49 PROOF ...... 49 OBSERVATION 10 ...... 49 6 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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PROBLEM 4 ...... 49 PROBLEM 5 ...... 50 2.5 A TRIANGLE AND ITS CONTACT TRIANGLE ...... 50 DEFINITION 7 ...... 50 OBSERVATION 11 ...... 50 PROPOSITION 10 ...... 50 PROOF ...... 51 OBSERVATION 12 ...... 51 PROPOSITION 11 ...... 51 PROOF ...... 52 2.6 A TRIANGLE AND ITS TANGENTIAL TRIANGLE ...... 52 DEFINITION 8 ...... 52 OBSERVATION 13 ...... 52 PROPOSITION 12 ...... 52 OBSERVATION 14 ...... 52 PROPOSITION 13 ...... 53 PROOF ...... 53 NOTE 1 ...... 54 PROPOSITION 14 ...... 55 PROOF ...... 56 2.7 A TRIANGLE AND ITS COTANGENT TRIANGLE ...... 56 DEFINITION 9 ...... 56 OBSERVATION 15 ...... 57 PROPOSITION 15 ...... 57 PROOF ...... 57 DEFINITION 10 ...... 57 PROPOSITION 16 ...... 57 PROOF ...... 57 OBSERVATION 16 ...... 57 PROBLEM 6 ...... 58 DEFINITION 11 ...... 58 PROOF ...... 58 PROPOSITION 17 ...... 59 PROOF ...... 59 2.8 A TRIANGLE AND ITS ANTI-SUPPLEMENTARY TRIANGLE ...... 59 DEFINITION 12 ...... 59 OBSERVATION 17 ...... 60 PROPOSITION 18 ...... 60

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OBSERVATION 18 ...... 60 PROPOSITION 19 ...... 60 DEFINITION 13 ...... 60 OBSERVATION 19 ...... 61 PROPOSITION 20 ...... 61 PROOF ...... 61 PROBLEM 7 ...... 62 2.9 A TRIANGLE AND ITS I-CIRCUMPEDAL TRIANGLE ...... 62 DEFINITION 14 ...... 62 OBSERVATION 20 ...... 63 PROPOSITION 21 ...... 63 PROOF ...... 63 OBSERVATION 21 ...... 64 PROPOSITION 22 ...... 64 PROOF ...... 64 REMARK 6 ...... 65 2.10 A TRIANGLE AND ITS H-CIRCUMPEDAL TRIANGLE ...... 65 PROPOSITION 23 ...... 65 PROOF ...... 65 2.11 A TRIANGLE AND ITS O-CIRCUMPEDAL TRIANGLE ...... 65 DEFINITION 15 ...... 65 PROPOSITION 24 ...... 66 PROOF ...... 66 2.12 A TRIANGLE AND ITS IA-CIRCUMPEDAL TRIANGLE ...... 67 PROPOSITION 25 ...... 67 PROOF ...... 67 OBSERVATION 22 ...... 68 2.13 A TRIANGLE AND ITS EX-TANGENTIAL TRIANGLE ...... 68 DEFINITION 26 ...... 68 OBSERVATION 23 ...... 68 PROPOSITION 26 ...... 70 PROOF 1 ...... 70 DEFINITION 17 ...... 70 LEMMA 3 ...... 71 PROOF ...... 71 OBSERVATION 24 ...... 71 PROOF 2 ...... 72 LEMMA 4 ...... 72 8 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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PROOF OF LEMMA ...... 72 OBSERVATION 25 ...... 72 2.14 A TRIANGLE AND ITS PODAL TRIANGLE ...... 73 DEFINITION 18 ...... 73 OBSERVATION 26 ...... 73 REMARK 7 ...... 73 DEFINITION 19 ...... 74 PROPOSITION 28 ...... 74 DEFINITION 2 ...... 74 REMARK 8 ...... 74 THEOREM 6 ...... 74 PROOF ...... 74 OBSERVATION 28 ...... 75 PROPOSITION 29 ...... 75 PROPOSITION 30 ...... 75 PROOF ...... 75 OBSERVATION 29 ...... 76 DEFINITION 20 ...... 76 OBSERVATION 30 ...... 76 PROPOSITION 31 ...... 76 PROOF ...... 77 OBSERVATION 31 ...... 77 THEOREM 7 (THE CIRCLE OF SIX POINTS)...... 78 PROOF ...... 78 OBSERVATION 32 ...... 79 THEOREM 8 (THE RECIPROCAL OF THE THEOREM 7) ...... 79 PROOF ...... 79 PROPOSITION 32 ...... 79 PROOF ...... 80 2.15 A TRIANGLE AND ITS ANTIPODAL TRIANGLE ...... 81 DEFINITION 21 ...... 81 OBSERVATION 33 ...... 81 PROPOSITION 33 ...... 82 OBSERVATION 34 ...... 82 PROPOSITION 34 ...... 82 PROOF ...... 82 OBSERVATION 35 ...... 83 PROPOSITION 35 ...... 83

Ion Pătrașcu, Florentin Smarandache 9 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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PROOF ...... 83 REMARK 9 ...... 84 PROPOSITION 36 ...... 84 OBSERVATION 36 ...... 84 PROPOSITION 37 ...... 84 PROOF ...... 84 PROBLEM 8 ...... 85 2.16 A TRIANGLE AND ITS CYCLOCEVIAN TRIANGLE ...... 85 DEFINITION 22 ...... 85 OBSERVATION 37 ...... 86 DEFINITION 23 ...... 86 THEOREM 9 (TERQUEM – 1892) ...... 86 PROOF ...... 86 DEFINITION 24 ...... 87 OBSERVATION 38 ...... 87 THEOREM 10 ...... 87 PROOF ...... 87 DEFINITION 25 ...... 88 OBSERVATION 39 ...... 89 2.17 A TRIANGLE AND ITS THREE IMAGES TRIANGLE ...... 89 DEFINITION 26 ...... 89 OBSERVATION 40 ...... 89 PROPOSITION 28 ...... 90 PROOF ...... 90 THEOREM 11 (V. THÉBAULT – 1947)...... 90 PROOF ...... 90 PROPOSITION 39 ...... 91 2.18 A TRIANGLE AND ITS CARNOT TRIANGLE ...... 92 DEFINITION 27 ...... 92 DEFINITION 28 ...... 92 PROPOSITION 40 ...... 92 PROOF ...... 92 PROPOSITION 41 ...... 92 PROOF ...... 93 OBSERVATION 41 ...... 93 PROPOSITION 42 ...... 94 PROOF ...... 94 OBSERVATION 42 ...... 94 10 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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DEFINITION 29 ...... 94 REMARK 10 ...... 94 PROPOSITION 43 ...... 94 OBSERVATION 43 ...... 94 2.19 A TRIANGLE AND ITS FUHRMANN TRIANGLE ...... 95 DEFINITION 30 ...... 95 OBSERVATION 43 ...... 95 PROPOSITION 44 ...... 95 PROOF ...... 96 PROPOSITION 45 ...... 96 PROOF ...... 96 OBSERVATION 44 ...... 97 THEOREM 12 (HOUSEL'S LINE) ...... 97 PROOF 1 ...... 97 PROOF 2 ...... 98 PROPOSITION 46 ...... 98 PROOF ...... 99 PROPOSITION 47 ...... 99 PROOF ...... 100 PROPOSITION 48 ...... 100 PROOF ...... 100 PROPOSITION 49 ...... 101 PROOF ...... 102 THEOREM 13 (M. STEVANOVIC – 2002) ...... 102 PROOF ...... 102 PROPOSITION 50 ...... 103

3 ORTHOLOGICAL DEGENERATE TRIANGLES ...... 105 3.1 DEGENERATE TRIANGLES; THE ORTHOPOLE OF A LINE ...... 105 DEFINITION 31 ...... 105 PROPOSITION 51 ...... 105 THEOREM 14 (THE ORTHOPOLE THEOREM; SOONS – 1886) ...... 105 PROOF 1 (NICULAE BLAHA, 1949) ...... 106 PROOF 2 ...... 106 PROOF 3 (TRAIAN LALESCU – 1915)...... 107 DEFINITION 32 ...... 108 OBSERVATION 45 ...... 108

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3.2 SIMSON LINE ...... 108 THEOREM 15 (WALLACE, 1799) ...... 108 PROOF ...... 108 OBSERVATION 46 ...... 109 THEOREM 16 (THE RECIPROCAL OF THE SIMSON-WALLACE THEOREM) ...... 109 PROOF ...... 109 PROPOSITION 52 ...... 110 PROOF ...... 111 OBSERVATION 47 ...... 111 PROPOSITION 53 ...... 111 PROOF ...... 111 OBSERVATION 48 ...... 112 THEOREM 17 (J. STEINER) ...... 113 PROOF ...... 113 PROPOSITION 54 ...... 114 PROOF ...... 115 REMARK 11 ...... 115 THEOREM 18 ...... 115 PROOF ...... 115 REMARK 12 ...... 117 PROPOSITION 55 ...... 117 PROOF ...... 118 REMARK 13 ...... 119 PROPOSITION 56 ...... 119 PROOF ...... 119 OBSERVATION 48 ...... 120 PROPOSITION 57 ...... 120 PROOF ...... 121 PROPOSITION 58 ...... 121 PROOF ...... 121 OBSERVATION 49 ...... 121 PROPOSITION 59 ...... 121 PROOF ...... 122

12 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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4 S TRIANGLES OR ORTHOPOLAR TRIANGLES ...... 125 4.1 S TRIANGLES: DEFINITION, CONSTRUCTION, PROPERTIES ...... 125 DEFINITION 33 ...... 125 CONSTRUCTION OF S TRIANGLES ...... 125 OBSERVATION 50 ...... 126 PROPOSITION 60 ...... 127 THEOREM 19 (TRAIAN LALESCU, 1915) ...... 127 PROOF ...... 127 REMARK 14 ...... 129 4.2 THE RELATION OF EQUIVALENCE S IN THE SET OF TRIANGLES INSCRIBED IN THE SAME CIRCLE ...... 129 DEFINITION 34 ...... 129 PROPOSITION 61 ...... 129 PROOF ...... 130 PROPOSITION 62 ...... 130 PROOF ...... 131 REMARK 15 ...... 131 PROPOSITION 63 ...... 131 PROOF ...... 131 4.3 SIMULTANEOUSLY ORTHOLOGICAL AND ORTHOPOLAR TRIANGLES ... 131 LEMMA 5 ...... 131 PROOF ...... 132 THEOREM 19 ...... 132 PROOF ...... 132 PROPOSITION 64 ...... 133 PROPOSITION 65 ...... 133

5 ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER ...... 135 5.1 THEOREMS REGARDING THE ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER ...... 135 THEOREM 20 ...... 135 THEOREM 21 (N. DERGIADES, 2003) ...... 135 PROOF (ION PĂTRAȘCU) ...... 135 LEMMA 6 ...... 137 PROOF 1 ...... 137 OBSERVATION 51 ...... 138 PROOF 2 (ION PĂTRAȘCU) ...... 138

Ion Pătrașcu, Florentin Smarandache 13 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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PROOF 3 (MIHAI MICULIȚA) ...... 139 PROOF OF THEOREM 20 ...... 139 REMARK 16 ...... 139 THEOREM 22 ...... 139 PROOF ...... 140 OBSERVATION 52 ...... 141 PROPOSITION 66 ...... 141 PROOF ...... 142 PROPOSITION 67 ...... 142 PROOF ...... 142 5.2 RECIPROCAL POLAR TRIANGLES ...... 143 DEFINITION 34 ...... 143 THEOREM 23 ...... 143 PROOF ...... 143 REMARK 17 ...... 145 5.3 OTHER REMARKABLE ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER ...... 145 DEFINITION 35 ...... 146 OBSERVATION 53 ...... 146 PROPOSITION 68 ...... 146 PROOF ...... 146 REMARK 18 ...... 146 5.4. BIORTHOLOGICAL TRIANGLE ...... 147 DEFINITION 36 ...... 147 OBSERVATION 54 ...... 147 THEOREM 24 (A. PANTAZI, 1896 – 1948) ...... 147 PROOF 1 ...... 147 PROOF 2 ...... 148 REMARK 18 ...... 149 THEOREM 25 (C. COCEA, 1992) ...... 149 PROOF ...... 149 OBSERVATION 55 ...... 151 THEOREM 26 (MIHAI MICULIȚA) ...... 151 CONSEQUENCES ...... 151 PROOF ...... 151 THEOREM 27 (LEMOINE) ...... 153 PROOF ...... 153

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6 BILOGICAL TRIANGLES ...... 155 6.1 SONDAT’S THEOREM. PROOFS ...... 155 THEOREM 28 (P. SONDAT, 1894) ...... 155 PROOF 1 (V. THÉBAULT, 1952) ...... 155 PROOF 2 (ADAPTED AFTER THE PROOF GIVEN BY JEAN-LOUIS AYMÉ) ...... 158 6.2 REMARKABLE BILOGICAL TRIANGLES ...... 159 6.2.1 A TRIANGLE AND ITS FIRST BROCARD TRIANGLE ...... 159 DEFINITION 37 ...... 159 OBSERVATION 56 ...... 159 PROPOSITION 69 ...... 160 PROOF ...... 160 OBSERVATION 57 ...... 160 THEOREM 29 ...... 161 DEFINITION 38 ...... 162 PROPOSITION 70 ...... 162 PROOF ...... 162 DEFINITION 39 ...... 162 LEMMA 7 ...... 162 PROOF ...... 162 OBSERVATION 58 ...... 164 LEMMA 8 ...... 164 PROOF ...... 164 OBSERVATION 59 ...... 164 LEMMA 9 ...... 164 PROOF ...... 165 OBSERVATION 60 ...... 165 LEMMA 10 ...... 165 PROOF ...... 165 LEMMA 11 ...... 166 PROOF ...... 166 REMARK 20 ...... 166 6.2.2 A TRIANGLE AND ITS NEUBERG TRIANGLE ...... 167 DEFINITION 40 ...... 167 OBSERVATION 61 ...... 167 THEOREM 30 ...... 167 PROOF ...... 167 REMARK 21 ...... 169 DEFINITION 41 ...... 170

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THEOREM 31 ...... 171 PROOF ...... 171 REMARK 22 ...... 172 THEOREM 32 ...... 172 PROOF ...... 173 6.2.3 A TRIANGLE AND THE TRIANGLE THAT DETERMINES ON ITS SIDES THREE CONGRUENT ANTIPARALLELS ...... 173 THEOREM 33 (R. TUCKER) ...... 173 PROOF ...... 173 THEOREM 34 ...... 175 PROOF ...... 175 REMARK 23 ...... 176 PROPOSITION 71 ...... 176 PROOF ...... 176 6.2.4 A TRIANGLE AND THE TRIANGLE OF THE PROJECTIONS OF THE CENTER OF THE CIRCLE INSCRIBED ON ITS MEDIATORS ...... 178 THEOREM 35 ...... 178 PROOF ...... 178 REMARK 24 ...... 180 6.2.5 A TRIANGLE AND THE TRIANGLE OF THE PROJECTIONS OF THE CENTERS OF EX-INSCRIBED CIRCLES ON ITS MEDIATORS ...... 180 PROPOSITION 72 ...... 181 PROOF ...... 181 6.2.6 A TRIANGLE AND ITS NAPOLEON TRIANGLE ...... 181 DEFINITION 42 ...... 181 OBSERVATION 62 ...... 182 DEFINITION 43 ...... 182 DEFINITION 44 ...... 182 THEOREM 36 ...... 183 PROOF ...... 183 OBSERVATION 63 ...... 183 OBSERVATION 64 ...... 184 THEOREM 37 ...... 184 PROOF ...... 185 THEOREM 38 ...... 186 PROOF ...... 186 REMARK 25 ...... 188 THEOREM 39 ...... 188 16 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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7 ORTHOHOMOLOGICAL TRIANGLES ...... 191 7.1. ORTHOGONAL TRIANGLES ...... 191 DEFINITION 42 ...... 191 OBSERVATION 62 ...... 191 PROBLEM 9 ...... 192 SOLUTION ...... 192 PROBLEM 10 ...... 193 SOLUTION ...... 193 OBSERVATION 63 ...... 194 PROBLEM 11 ...... 194 SOLUTION ...... 195 7.2. SIMULTANEOUSLY ORTHOGONAL AND ORTHOLOGICAL TRIANGLES .. 195 PROPOSITION 73 (ION PĂTRAȘCU) ...... 195 PROOF ...... 196 PROPOSITION 74 (ION PĂTRAȘCU) ...... 197 PROOF ...... 197 OBSERVATION 64 ...... 198 7.3. ORTHOHOMOLOGICAL TRIANGLES ...... 198 DEFINITION 43 ...... 198 PROBLEM 12 ...... 198 LEMMA 12 ...... 199 DEFINITION 44 ...... 199 PROOF ...... 199 DEFINITION 45 ...... 200 OBSERVATION 65 ...... 200 SOLUTION OF PROBLEM 12 ...... 200 REMARK 24 ...... 201 THEOREM 40 ...... 201 PROOF ...... 201 THEOREM 41 ...... 202 PROOF ...... 202 THEOREM 42 (P. SONDAT) ...... 204 PROOF ...... 205 PROPOSITION 75 (ION PĂTRAȘCU) ...... 207 PROOF ...... 207 REMARK 25 ...... 207 THEOREM 43 ...... 207 PROOF ...... 208

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DEFINITION 46 ...... 210 OBSERVATION 66 ...... 210 7.4. METAPARALLEL TRIANGLES OR PARALLELOLOGIC TRIANGLES ...... 211 DEFINITION 47 ...... 211 THEOREM 44 ...... 211 PROOF ...... 211 OBSERVATION 67 ...... 212 REMARK 26 ...... 212 THEOREM 45 ...... 213

8 ANNEXES ...... 215 8.1 ANNEX 1: BARYCENTRIC COORDINATES ...... 215 8.1.1 BARYCENTRIC COORDINATES OF A POINT IN A PLANE ...... 215 DEFINITION 48 ...... 216 THEOREM 46 ...... 217 OBSERVATION 68 ...... 217 THEOREM 47 (THE POSITION VECTOR OF A POINT) ...... 217 OBSERVATION 69 ...... 217 THEOREM 48 ...... 217 THEOREM 49 (BARYCENTRIC COORDINATES OF A VECTOR) ...... 218 DEFINITION 49 ...... 218 OBSERVATION 70 ...... 218 THEOREM 50 (THE POSITION VECTOR OF A POINT THAT DIVIDES A SEGMENT INTO A GIVEN RATIO) ...... 218 THEOREM 51 (THE COLLINEARITY CONDITION OF TWO VECTORS) ...... 219 THEOREM 52 (THE CONDITION OF PERPENDICULARITY OF TWO VECTORS) .... 219 THEOREM 53 ...... 219 THEOREM 54 ...... 220 THEOREM 55 ...... 220 THEOREM 56 (THE THREE-POINT COLLINEARITY CONDITION) ...... 220 OBSERVATION 71 ...... 220 OBSERVATION 72 ...... 221 THEOREM 57 (THE CONDITION OF PARALLELISM OF TWO LINES) ...... 221 THEOREM 58 (THE CONDITION OF PERPENDICULARITY OF TWO LINES) ...... 221 THEOREM 59 ...... 221 THEOREM 60 (BARYCENTRIC COORDINATES OF A VECTOR PERPENDICULAR TO A GIVEN VECTOR) ...... 222 THEOREM 61 ...... 222 18 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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REMARK ...... 222 THEOREM 62 (THE CONDITION OF CONCURRENCY OF THREE LINES) ...... 222 THEOREM 63 ...... 222 THEOREM 64 ...... 223 THEOREM 65 ...... 223 THEOREM 66 ...... 223 THEOREM 67 ...... 223 8.1.2 THE BARYCENTRIC COORDINATES OF SOME IMPORTANT POINTS IN THE TRIANGLE GEOMETRY ...... 224 OBSERVATION 73 ...... 224 8.1.3 OTHER BARYCENTRIC COORDINATES AND USEFUL EQUATIONS ...... 224 8.1.4 APLICATIONS ...... 226 OBSERVATION 74 ...... 227 OBSERVATION 75 ...... 228 8.2 ANNEX 2: THE SIMILARITY OF TWO FIGURES ...... 229 8.2.1 PROPERTIES OF THE SIMILARITY ON A PLANE ...... 229 DEFINITION 50 ...... 229 PROPOSITION 76 ...... 229 PROOF ...... 230 REMARK 27 ...... 230 PROPERTY 77...... 230 PROOF ...... 230 REMARK 28 ...... 231 DEFINITION 51 ...... 231 OBSERVATION 76 ...... 231 DEFINITION 52 ...... 231 REMARK 29 ...... 231 PROPOSITION 78 ...... 231 PROOF ...... 232 REMARK 30 ...... 232 TEOREMA68 ...... 232 PROOF ...... 233 REMARK 31 ...... 233 THEOREM 69 ...... 233 PROOF ...... 233 REMARK 32 ...... 234 8.2.2 APPLICATIONS ...... 234

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8.3 ANNEX 3: THE MIQUEL’S POINT, TRIANGLE, CIRCLES ...... 237 8.3.1 DEFINITIONS AND THEOREMS ...... 237 THEOREM 70 (J. STEINER, 1827) ...... 237 PROOF ...... 237 OBSERVATION 77 ...... 238 THEOREM 71 (J. STEINER, 1827) ...... 238 PROOF ...... 238 REMARK 33 ...... 239 THEOREM 72 ...... 239 THEOREM 73 ...... 240 PROOF ...... 240 OBSERVATION 78 ...... 241 PROPOSITION 79 ...... 241 THEOREM 71 ...... 241 PROOF ...... 241 8.3.2 APPLICATIONS ...... 242

9 PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES ..... 249 9.1 PROPOSED PROBLEMS ...... 249 9.2 OPEN PROBLEMS ...... 267

10 SOLUTIONS, INDICATIONS, ANSWERS TO THE PROPOSED ORTHOLOGY PROBLEMS ...... 269

BIBLIOGRAPHY ...... 309

20 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

FOREWORD

Plants and trees grow perpendicular to the plane tangent to the soil surface, at the point of penetration into the soil; in vacuum, the bodies fall perpendicular to the surface of the Earth - in both cases, if the surface is horizontal. Starting from the property of two triangles to be orthological, the two authors have designed this work that seeks to provide an integrative image of elementary geometry by the prism of this "filter". Basically, the property of orthology is the skeleton of the present work, which establishes many connections of some theorems and geometric properties with it. The book "The Geometry of The Orthological Triangles" is divided into ten chapters. In the first seven, the topic is introduced and developed by connecting it with other beautiful properties of geometry, such as the homology of triangles. Chapter 8 includes three annexes intended to clarify to the readers some results used in the rest of the chapters. Chapter 9 is a collection of problems where orthological triangles usually appear; it is especially intended for students preparing for participation in different mathematics competitions. The last chapter contains solutions and answers to the problems in chapter 9. The work ends with a rich bibliography that has been consulted and used by the authors. It is noteworthy that this book highlights the contributions of Romanian mathematicians Traian Lalescu, Gheorghe Ţițeica, Cezar Coşniţa, Alexandru Pantazi et al. to this treasure that constitutes THE GEOMETRY! We congratulate the authors for the beauty and depth of the chosen topic, which can be explained by the passion of the distinguished teachers, complementary in the complex world of integral culture in the sense of A. Huxley - C. P. Snow: ‒ The geometer Ion Patrașcu - the classical teacher attracted by exciting topics such as orthogonality - an instrument of duality and lighthouse in knowledge / progress in Mathematics; ‒ The scientist Florentin Smarandache - renowned and unpredictable innovator in the Philosophy of Science, from Fundamentals to his well-

Ion Pătrașcu, Florentin Smarandache 21 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

FOREWORD known work "Hadron Models and related New Energy issues" (2007), also targeted in our pop-symphonic composition "LHC - Large Hadron Collider", uploaded on YouTube. The orthogonality is a universal geometric property, because it represents the local quintessence of the system {point , geodesic } in any Riemannn-dimensional space, in particular in an Euclidean space, for 𝑀𝑀 𝑔𝑔 = at least 2. I propose to the readers to try an "orthological reading" on the classical sphere, starting from the system { , } and advancing, through 𝑛𝑛 similarity, towards geodesic triangles. The topic can be taken further, in 𝑀𝑀 𝑔𝑔 post-university / doctoral studies, in the Riemannian context, under the empire of the incredible "isometric diving theorems" of brilliant John Forbes Nash, Jr. (MAGIC mechanism - Great ATTENTION! - of " Nash teleportation": from the -Dim Euclidean world to the Riemannian -Dim world, where is a polynomial of degree -I think- 2 or 𝑛𝑛 3 inn - see Professor web). 𝑘𝑘 𝑘𝑘 Prof. Dr. Valentin Boju Member of the "Cultural Merit" Order of Romania Officer rank, Category H, Scientific Research

22 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

AUTHORS’ NOTE

The idea of this book came up when writing our previous book, The Geometry of Homological Triangles (2012). We try to graft on the central theme - the orthology of triangles-, many results from the elementary geometry. In particular, we approach the connection between the orthological and homological triangles; also, we review the "S" triangles, highlighted for the first time by the great Romanian mathematician Traian Lalescu. The book is addressed to both those who have studied and love geometry, as well as to those who discover it now, through study and training, in order to obtain special results in school competitions. In this regard, we have sought to prove some properties and theorems in several ways: synthetic, vectorial, analytical. Basically, the book somewhat resembles a quality police novel, in which the pursuits are the orthological triangles, and, in their search, GEOMETRY is actually discovered. We adress many thanks to the distinguished professor Mihai Miculiţa from Oradea for drawing the geometrical figures in this book and contributed with interesting observations and mentions that enriched our work.

Prof. ION PĂTRAŞCU MA in Mathematics “Fraţii Buzeşti” National College Craiova, Romania [email protected]

Prof. FLORENTIN SMARANDACHE PhD in Mathematics University of New Mexico, United States of America [email protected]

Ion Pătrașcu, Florentin Smarandache 23 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

1.1 Orthological triangles: definition Let be a scalene triangle and – a point in its plane. We build from the lines , , , respectively perpendicular to 𝑨𝑨𝑨𝑨𝑨𝑨 𝑷𝑷 , and . On these lines, we consider the points , , 𝑷𝑷 𝒂𝒂𝟏𝟏 𝒃𝒃𝟏𝟏 𝒄𝒄𝟏𝟏 , such that they are not collinear (see Figure 1). About the 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏 𝑩𝑩𝟏𝟏 triangle we say that it is an orthological triangle in 𝑪𝑪𝟏𝟏 relation to the triangle , and about the point we say that it 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 is the orthology center of the triangle in relation to the 𝑨𝑨𝑨𝑨𝑨𝑨 𝑷𝑷 triangle . 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 𝑨𝑨𝑨𝑨𝑨𝑨 A

C1 P c1

B C

A1 b1

Figure 1

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INTRODUCTION

Definition 1 We say that the triangle is orthological in relation to the triangle if the perpendiculars taken from , , 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 respectively to , and are concurrent. 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏 𝑩𝑩𝟏𝟏 𝑪𝑪𝟏𝟏 About the concurrency point of the previously mentioned 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 𝑨𝑨𝑨𝑨 perpendiculars, we say that it is the orthology center of the triangle in relation to the triangle .

𝟏𝟏 𝟏𝟏 𝟏𝟏 Observation𝑨𝑨 𝑩𝑩 1 𝑪𝑪 𝑨𝑨𝑨𝑨𝑨𝑨 The presented construction leads to the conclusion that, being given a triangle and a point in its plane, we can build an infinity of triangles , such that to be orthological with , . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 𝐴𝐴𝑛𝑛𝐵𝐵𝑛𝑛𝐶𝐶𝑛𝑛 𝑛𝑛 𝑛𝑛 𝑛𝑛 Exercise𝐴𝐴 𝐵𝐵 𝐶𝐶 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑛𝑛 ∈ ℕ Being given a triangle , build the triangle to be orthological in relation to , such that its orthology center to be the vertex . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 1.2. Characterization𝐴𝐴𝐴𝐴𝐴𝐴 of the orthology relation 𝐴𝐴 A triangle can be considered orthological in relation to itself. Indeed, the perpendiculars taken from , , respectively to , , 𝐴𝐴𝐴𝐴𝐴𝐴 are also the altitudes of the triangle, therefore they are concurrent lines. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 The orthology center is the orthocenter of the triangle. 𝐴𝐴𝐴𝐴 We can say that the orthology relation is reflexive in the set of triangles. We establish in the following the conditions𝐻𝐻 that are necessary and sufficient for two triangles to be in an orthology relation. The following theorem plays an important role in this approach:

Theorem 1 (L. Carnot, 1803) If , , are points on the sides , , respectively of a given triangle , the perpendiculars raised at these points to 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 , respectively are concurrent if and only if the following 𝐴𝐴𝐴𝐴𝐴𝐴 relation takes place: 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 + + = 0. (1) 2 2 2 2 2 2 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵

26 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

Proof We consider that the perpendiculars raised in , , , to , respectively , are concurrent in a point (see Figure 2). 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 From Pythagoras’ Theorem applied in the triangles , , we have 𝐴𝐴𝐴𝐴 𝑃𝑃 = + and = + . 1 1 2 2 2 2 2 2 𝑃𝑃𝑃𝑃 𝐵𝐵 𝑃𝑃𝑃𝑃 𝐶𝐶 1 1 1 1 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 𝐴𝐴 𝐵𝐵 𝑃𝑃 𝑃𝑃 A 𝑃𝑃𝑃𝑃 𝐴𝐴 𝐶𝐶

C1 B1 P

B A1 C

Figure 2 By subtraction member by member, we find: = . (2) Similarly2 , 2we find the2 relations2 : 𝑃𝑃𝑃𝑃 − 𝑃𝑃𝑃𝑃 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 = , (3) 2 2 = 2 2 1 1 . (4) 𝑃𝑃𝑃𝑃2 − 𝑃𝑃𝑃𝑃2 𝐵𝐵 𝐶𝐶2 − 𝐵𝐵 𝐴𝐴2 By addition member1 by 1member of the relations (2), (3) and (4), we get the rela𝑃𝑃tion𝑃𝑃 (1).− 𝑃𝑃 𝑃𝑃 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 Reciprocally Let us assume that the relation (1) is true and let us prove the concurrency of the perpendiculars raised in , , , respectively to , and . Let be the intersection of the perpendicular in to with the perpendicular in 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑃𝑃 to CA; we denote by the projection of to . According to those 𝐴𝐴1 𝐵𝐵𝐵𝐵 previously demonstrated, we have the relation: 𝐵𝐵1 𝐶𝐶′1 𝑃𝑃 𝐴𝐴𝐴𝐴 + + = 0 (5) This2 relation 2and the 2relation 2(1) implies2 that: 2 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶′1𝐴𝐴 − 𝐶𝐶′1𝐵𝐵 = (6) 2 2 2 2 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐶𝐶′1𝐴𝐴 − 𝐶𝐶′1𝐵𝐵 Ion Pătrașcu, Florentin Smarandache 27 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

We prove that this relation is true if and only if = . Indeed, let us suppose that ( ) and ( ) (see Figure 3) and 𝐶𝐶1 𝐶𝐶′1 that the relation (6) takes place here. 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐵𝐵 ∈ 𝐴𝐴𝐶𝐶′1

A C1 B C'1

Figure 3 We obtain that: ( )( + ) = ( )( + ). (7) Because + = and + = , from (7) we have that 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐶𝐶1𝐴𝐴 𝐶𝐶1𝐵𝐵 𝐶𝐶′1𝐴𝐴 − 𝐶𝐶′1𝐵𝐵 𝐶𝐶′1𝐴𝐴 𝐶𝐶′1𝐵𝐵 = is absurd. 𝐶𝐶1𝐴𝐴 𝐶𝐶1𝐵𝐵 𝐴𝐴𝐴𝐴 𝐶𝐶′1𝐴𝐴 𝐶𝐶′1𝐵𝐵 𝐴𝐴𝐴𝐴 Similarly, we find that the the relation (6) cannot be satisfied in the 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐶𝐶′1𝐴𝐴 − 𝐶𝐶′1𝐵𝐵 hypothesis , separated by the points or . If, for example, , ( ), then relation (7) leads to = , which implies = and the 𝐶𝐶1 𝐶𝐶′1 𝐴𝐴 𝐵𝐵 𝐶𝐶1 𝐶𝐶′1 ∈ implication of the theorem is proved. 𝐴𝐴𝐴𝐴 𝐶𝐶1𝐴𝐴 𝐶𝐶′1𝐴𝐴 𝐶𝐶1 𝐶𝐶′1 Observation 2 a) The points , , from the theorem’s hypothesis can be collinear. b) If the points , , from Carnot’s theorem statement are 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 noncollinear, then relation (1) expresses a necessary and sufficient 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 condition that the triangle to be orthological in relation to the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 c) From the proof of the Carnot’s theorem, the following lemma was 𝐴𝐴𝐴𝐴𝐴𝐴 inferred:

Lemma 1 The geometric place of the points in plane with the property = , where and are two given fixed points and – a real constant2 , 𝑀𝑀 𝑀𝑀𝑀𝑀 − is a 2line perpendicular to . 𝑀𝑀𝑀𝑀 𝑘𝑘 𝐴𝐴 𝐵𝐵 𝑘𝑘

d) Carnot’s theorem𝐴𝐴𝐴𝐴 can be used to prove the concurrency of perpendiculars raised on the sides of a triangle.

Exercise 2 Prove with the help of Carnot’s theorem that: a) The mediators of a triangle are concurrent;

28 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

b) The altitudes of a triangle are concurrent.

Theorem 2 Let and be two triangles in plane. is orthological in relation to the triangle if and only if the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 following relation is true: 𝐴𝐴𝐴𝐴𝐴𝐴 + + = + + (7) 2 2 2 2 2 2 1 1 1 1 1 1 Proof𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐵𝐵 We consider that is orthological in relation to ; we denote by the orthology point; let { } = , { } = , { } = 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 (see Figure 4); we prove′ the relation (7). ′ ′ 𝐴𝐴 𝐵𝐵𝐵𝐵 ∩ 𝑃𝑃𝑃𝑃1 𝐵𝐵 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵1𝑃𝑃 𝐶𝐶 𝐴𝐴𝐴𝐴 ∩ 1 𝑃𝑃𝑃𝑃 A1

B A' C P C' B'

B1 C1

Figure 4 According to Theorem 1, we have: + + = 0 (8) But′ :2 ′ 2 ′ =2 ′ 2 ′ (2Pythagoras’′ 2 theorem). 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 Similarly′ 2: ′ 2 2 2 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 29 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

= and = . ′ 2 ′ 2 2 2 ′ 2 ′ 2 2 2 By adding member1 by member1 the last three relation1 s, and1 taking into account𝐵𝐵 𝐶𝐶 (8)−,𝐵𝐵 we𝐴𝐴 deduce𝐵𝐵 𝐶𝐶 the− relation𝐵𝐵 𝐴𝐴 (7).𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 Reciprocally Let us consider the triangles , , and such that the relation (7) is satisfied; we prove that , , is orthological in relation to . 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Pythagoras’ theorem and the relation (7) lead to: 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 + + = 0. According′ 2 ′ to2 Theorem2 1, th2e perpendicular′ 2 ′ 2s in , , to , , 𝐴𝐴 𝐶𝐶 − 𝐴𝐴 𝐵𝐵 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 (which pass respectively through , , ) are concur′ ′ rent′ , therefore the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 triangle is orthological in relation to the triangle . 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 Another way to determine the orthology of two triangles is given by: 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Theorem 3 The triangle is orthological in relation to the triangle if and only if the following relation takes place for any point in 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 their plane: + + = 0. (9) 𝑀𝑀 1 1 1 Proof 𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐵𝐵𝐵𝐵��⃗ �𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐶𝐶𝐶𝐶��⃗ �𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐴𝐴𝐴𝐴��⃗ We denote: ( ) = + + , and we prove that ( ) has this value whatever . 𝐸𝐸 𝑀𝑀 𝑀𝑀��𝑀𝑀�⃗1 ∙ 𝐵𝐵𝐵𝐵��⃗ 𝑀𝑀��𝑀𝑀�⃗1 ∙ 𝐶𝐶𝐶𝐶��⃗ 𝑀𝑀��𝑀𝑀�⃗1 ∙ 𝐴𝐴𝐴𝐴��⃗ Let ( ) = + + , where is a point in the 𝐸𝐸 𝑀𝑀 𝑀𝑀 plane, different from . 𝐸𝐸 𝑁𝑁 𝑁𝑁��𝑁𝑁�⃗1 ∙ 𝐵𝐵𝐵𝐵��⃗ 𝑁𝑁��𝑁𝑁�⃗1 ∙ 𝐶𝐶𝐶𝐶��⃗ 𝑁𝑁��𝑁𝑁�⃗1 ∙ 𝐴𝐴𝐴𝐴��⃗ 𝑁𝑁 We have: 𝑀𝑀 ( ) ( ) = +

+ +1( 1 ) . 𝐸𝐸 𝑀𝑀 − 𝐸𝐸 𝑁𝑁 �𝑀𝑀��𝑀𝑀�⃗ − 𝑁𝑁��𝑁𝑁�⃗ � ∙ 𝐵𝐵𝐵𝐵��⃗ Therefore1 (1 ) ( ) =1 (1 + + ). ��𝑀𝑀��𝑀𝑀�⃗ −�𝑁𝑁��𝑁𝑁�⃗ � ∙ 𝐶𝐶𝐶𝐶��⃗ 𝑀𝑀��𝑀𝑀�⃗ − 𝑁𝑁��𝑁𝑁�⃗ ∙ 𝐴𝐴𝐴𝐴��⃗ Since + + = 0, it follows that ( ) ( ) = 0 = 0. 𝐸𝐸 𝑀𝑀 − 𝐸𝐸 𝑁𝑁 𝑀𝑀��𝑀𝑀��⃗ ∙ 𝐵𝐵𝐵𝐵��⃗ 𝐶𝐶𝐶𝐶��⃗ 𝐴𝐴𝐴𝐴��⃗ We proved that, if the relation (9) is true for a point in the plane, then it is ��⃗ ��⃗ ��⃗ �⃗ ��⃗ �⃗ true for any𝐵𝐵𝐵𝐵 other𝐶𝐶𝐶𝐶 point𝐴𝐴𝐴𝐴 in the plane. Let us𝐸𝐸 consider𝑀𝑀 − 𝐸𝐸 now𝑁𝑁 that𝑀𝑀𝑀𝑀 the∙ triangle is orthological in relation to the triangle , and that is the orthology center. Obviously, = = = 0 and hence 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 the relation (9) is true for the point , consequently it is also true for any other 𝑃𝑃��𝑃𝑃�⃗1 ∙ 𝐵𝐵𝐵𝐵��⃗ 𝑃𝑃��𝑃𝑃�⃗1 ∙ 𝐶𝐶𝐶𝐶��⃗ 𝑃𝑃��𝑃𝑃�⃗1 ∙ 𝐴𝐴𝐴𝐴��⃗ point from plane. 𝑃𝑃 𝑀𝑀

30 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Reciprocally If the relation (9) is satisfied, let us prove that the triangle is orthological in relation to the triangle . Let us denote by the intersection 𝐴𝐴1𝐵𝐵1𝐶𝐶1 of the given perpendicular from to with the perpendicular taken from 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 to . The relation (9) becomes in this case: = 0, which shows that 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵1 is perpendicular to and consequently the triangle is 𝐶𝐶𝐶𝐶 𝑀𝑀��𝑀𝑀�⃗1 ∙ 𝐴𝐴𝐴𝐴��⃗ orthological in relation to , the point being the orthology center. 𝑀𝑀𝑀𝑀1 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Observation 3 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 From Theorem 3, we note that, in order to prove that a triangle is orthological in relation to another triangle , it is sufficient to show that there 𝐴𝐴1𝐵𝐵1𝐶𝐶1 exists a point in their plane such that the relation (9) to be satisfied. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 1.3. The theorem of orthological triangles We noticed that the orthology relation, in the set of triangles in plane, is reflexive. The following theorem shows that the orthology relation is symmetrical.

Theorem 4 (J. Steiner, 1828 – Theorem of orthological triangles) If the triangle is orthological in relation to the triangle , then the triangle is orthological as well in relation to 𝐴𝐴1𝐵𝐵1𝐶𝐶1 . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Proof𝐴𝐴 𝐵𝐵 𝐶𝐶 1 It is based on Theorem 2. The relation (7), being symmetrical, we can write it like this: + + = + + (10) From2 Theorem2 2, 2it follows2 that the2 triangle2 is orthological in relation 𝐴𝐴𝐵𝐵1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Proof 2 𝐴𝐴 𝐵𝐵 𝐶𝐶 We use Theorem 3. Let the triangle be orthological in relation to the triangle ; then the relation (9) takes place, and we consider here = 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ; it follows that: + + = 0; now see relation (9), 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝐴𝐴1 𝐴𝐴��1��𝐶𝐶�⃗1 ∙ 𝐶𝐶𝐶𝐶��⃗ 𝐶𝐶��1��𝐴𝐴�⃗1 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐴𝐴��1��𝐵𝐵�⃗1

Ion Pătrașcu, Florentin Smarandache 31 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION where = , which shows that the triangle is orthological in relation to the triangle . 𝑀𝑀 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Proof 3 𝐴𝐴 𝐵𝐵 𝐶𝐶 Let be the orthology center of the triangle , in relation to the triangle , and be the point of intersection of perpendiculars taken from 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and respectively to and (see Figure 5). 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃1 We denote: = , = , = ; = , = , = . 𝐴𝐴 𝐵𝐵 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 From the hypothesis, we deduce that = 0, ( ) = 0 and 𝑃𝑃��𝐴𝐴��1⃗ 𝑎𝑎��1⃗ 𝑃𝑃��𝐵𝐵��1⃗ 𝑏𝑏��1⃗ 𝑃𝑃��𝐶𝐶��1⃗ 𝑐𝑐��1⃗ 𝑃𝑃��1��𝐴𝐴�⃗ 𝑎𝑎⃗ 𝑃𝑃��1��𝐵𝐵�⃗ 𝑏𝑏�⃗ 𝑃𝑃��1��𝐶𝐶�⃗ 𝑐𝑐⃗ = ( ) = 0. 1 1 𝑎𝑎��⃗ ∙ �𝑏𝑏�⃗ − 𝑐𝑐⃗� 𝑏𝑏��⃗ ∙ 𝑐𝑐⃗ − 𝑎𝑎⃗ Using1 1the 1obvious identity1 1: + ( ) + = 𝑎𝑎��⃗ ∙ �𝑏𝑏��⃗ − 𝑐𝑐⃗ � 𝑏𝑏�⃗ ∙ 𝑐𝑐⃗ − 𝑎𝑎⃗ + ( ) + (1 ), we deduce1 that 1 ( ) = 𝑎𝑎��⃗ ∙ �𝑏𝑏�⃗ − 𝑐𝑐⃗� 𝑏𝑏��⃗ ∙ 𝑐𝑐⃗ − 𝑎𝑎⃗ 𝑐𝑐��⃗ ∙ �𝑎𝑎⃗ − 𝑏𝑏�⃗� 0, ie. , which shows that the triangle is orthological in relation 𝑎𝑎⃗ ∙ �𝑏𝑏��1⃗ − 𝑐𝑐⃗1� 𝑏𝑏�⃗ ∙ 𝑐𝑐⃗1 − 𝑎𝑎⃗1 𝑐𝑐⃗ ∙ 𝑎𝑎⃗1 − 𝑐𝑐⃗1 𝑐𝑐⃗ ∙ 𝑎𝑎⃗1 − 𝑐𝑐⃗1 to , the orthology center being the point . 𝑃𝑃1𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 A1 𝑃𝑃 B

P1 P

B1 C1

Figure 5

32 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

Proof 4 We denote by the pedal triangle of orthology center of the triangle in relation to′ ′the′ triangle (see Figure 6). Also, we denote by , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 , the intersections with , , of perpendiculars taken from , ′′, 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 ′′respectively′′ to , and . 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 We have: ~ (because′ ′ and 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴 𝐵𝐵 as angles with′′ the sides′ respectively perpendicular′′ ).′ It follows that: ′′ ∆𝐴𝐴 𝐴𝐴𝐴𝐴 ∆𝐴𝐴 𝐶𝐶1𝑃𝑃 ∢𝐴𝐴 𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴 𝐶𝐶1𝑃𝑃 𝐴𝐴𝐴𝐴�𝐴𝐴 ≡ ′ 1 = . (1) 𝐶𝐶�𝑃𝑃𝐴𝐴′′ ′′ 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐵𝐵 ′ ′ 𝐴𝐴 𝐶𝐶1 𝐴𝐴~𝑃𝑃 (because and as angles ′′with sides′ respectively perpendicular′′ ). It′ follows that: ′′ ′ ∆𝐴𝐴 𝐴𝐴𝐴𝐴 ∆𝐴𝐴 𝐵𝐵1𝑃𝑃 ∢𝐴𝐴 𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴 𝐵𝐵1𝑃𝑃 𝐴𝐴𝐴𝐴�𝐴𝐴 ≡ 𝐵𝐵�1𝑃𝑃𝐴𝐴 = . (2) ′′ ′′ 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐶𝐶 ′ ′ 𝐴𝐴 𝐵𝐵1 𝐴𝐴 𝑃𝑃 A

B1 A'

B" C1

C" P P 1 C'

B A" C

Figure 6

Ion Pătrașcu, Florentin Smarandache 33 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

From relations (1) and (2), we obtain that: = (3). ′ ′′ 𝐴𝐴 𝐵𝐵1 𝐴𝐴 𝐵𝐵 ′ ′′ Similarly𝐴𝐴 𝐶𝐶1 𝐴𝐴 ,𝐶𝐶 we obtain that: = , (4) ′ ′′ 𝐵𝐵 𝐶𝐶1 𝐵𝐵 C ′ ′′ 𝐵𝐵 𝐴𝐴1 = 𝐵𝐵 𝐴𝐴. (5) ′ ′′ 𝐶𝐶 𝐴𝐴1 𝐶𝐶 𝐴𝐴 ′ ′′ Because𝐶𝐶 𝐵𝐵1 𝐶𝐶 𝐵𝐵 , , are concurrent in , we obtain from Ceva’s theorem that: ′ ′ ′ 𝐴𝐴1𝐴𝐴 𝐵𝐵1𝐵𝐵 𝐶𝐶1𝐶𝐶 𝑃𝑃 = 1. (6) ′ ′ ′ 𝐴𝐴 𝐵𝐵1 𝐵𝐵 𝐶𝐶1 𝐶𝐶 𝐴𝐴1 ′ ′ ′ The𝐴𝐴 𝐶𝐶1 ∙relations𝐵𝐵 𝐴𝐴1 ∙ 𝐶𝐶 𝐵𝐵 (3),1 (4), (5), (6) and Ceva’s theorem show that the cevians , , are concurrent, hence the triangle is orthological in relation′′ to′′ ′′ . 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Proof 5𝐴𝐴 𝐵𝐵 𝐶𝐶

C2 B2

C B P

O P 1 O1

B1 C1

Figure 7

34 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

Let be the orthology center of the triangle in relation to the triangle . We denote by , and the centers of the circles 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 circumscribed to the triangles , , (see Figure 7). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 The lines of the centers ; ; being the mediators of the 𝐵𝐵1𝑃𝑃𝐶𝐶1 𝐶𝐶1𝑃𝑃𝐴𝐴1 𝐴𝐴1𝑃𝑃𝐵𝐵1 segments , , respectively , are parallel with the sides of the triangle 𝐵𝐵2𝐶𝐶2 𝐶𝐶2𝐴𝐴2 𝐴𝐴2𝐵𝐵2 . The triangles and are homothetic (having parallel sides), 𝑃𝑃𝐴𝐴1 𝑃𝑃𝐵𝐵1 𝑃𝑃𝐶𝐶1 and the homothety center was denoted by . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 Perpendiculars from , and on the sides of the triangle are 𝑂𝑂 its mediators, and hence they are concurrent in the center of the circle 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 circumscribed to the triangle , which we denote by . Because the triangles and are homothetic, it follows that the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑂𝑂1 perpendiculars taken from ; , to , , respectively , will also 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 be concurrent (they are parallels with , and ) in a point , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 which shows that the triangle is orthological in relation to the triangle 𝐴𝐴2𝑂𝑂1 𝐵𝐵2𝑂𝑂1 𝐶𝐶2𝑂𝑂1 𝑃𝑃1 . 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 𝐴𝐴 𝐵𝐵Remark𝐶𝐶 1 The theorem of orthological triangles shows that, in the set of triangles in the plane, the relation of orthology is symmetrical.

Remark 2 Saying that the triangles and are orthological, it is obvious that we must assent that the order of the vertices of the two triangles was put in agreement. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 6 (analytical) We consider the triangles and such that: ( , ), ( , ), ( , ), (0, ), ( , 0), ( , 0) (see Figure 8). 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝑎𝑎1 𝑎𝑎2 The equations of the sides , , are: 𝐵𝐵1 𝑏𝑏1 𝑏𝑏2 𝐶𝐶1 𝑐𝑐1 𝑐𝑐2 𝐴𝐴 𝑎𝑎 𝐵𝐵 𝑏𝑏 𝐶𝐶 𝑐𝑐 : = 0, 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 : + 1 = 0, 𝐵𝐵𝐵𝐵 𝑦𝑦𝑥𝑥 𝑦𝑦 + 1 = 0 𝐴𝐴𝐴𝐴: 𝑏𝑏 𝑎𝑎 − . 𝑥𝑥 𝑦𝑦 The perpendiculars taken from , , to , , respectively have 𝐴𝐴𝐴𝐴 𝑐𝑐 𝑎𝑎 − the equations: 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 = 0, = ( ), = ( ). 𝑐𝑐 𝑏𝑏 The fact1 that these 2perpendiculars1 are concurrent2 in a1 point (see Figure 8) 𝑥𝑥 − 𝑎𝑎 𝑦𝑦 − 𝑏𝑏 𝑎𝑎 𝑥𝑥 − 𝑏𝑏 𝑦𝑦 − 𝑐𝑐 𝑎𝑎 𝑥𝑥 − 𝑐𝑐 is expressed by the condition: 𝑃𝑃

Ion Pătrașcu, Florentin Smarandache 35 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

1 0 = 0 (11) −𝑎𝑎1 �𝑐𝑐 −𝑎𝑎 𝑎𝑎𝑏𝑏2 − 𝑐𝑐𝑏𝑏1� 𝑏𝑏 −𝑎𝑎 𝑎𝑎𝑐𝑐2 − 𝑏𝑏𝑐𝑐1 B1(b1;b2) y

C (c ;c ) 1 1 2 A(0;a)

B(b;0) O C(c;0) x

A1(a1;a2) Figure 8 This condition can be written equivalently: ( ) + ( ) + ( ) = 0 (12)

The2 equations2 of the1 sides1 of the1 triangle1 are: 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 𝑏𝑏 𝑐𝑐1 − 𝑎𝑎 𝑐𝑐 𝑎𝑎 − 𝑏𝑏 1 1 1 : 1 = 0 or 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑥𝑥 𝑦𝑦 1 1 1 1 2 (𝐵𝐵 𝐶𝐶 �𝑏𝑏) 𝑏𝑏 ( � ) + = 0 1 2 ; 𝑐𝑐 𝑐𝑐 1 2 2 1 1 1 2 2 1 𝑏𝑏 −: 𝑐𝑐 𝑥𝑥 − 𝑏𝑏 1−=𝑐𝑐 0𝑦𝑦 or 𝑏𝑏 𝑐𝑐 − 𝑏𝑏 𝑐𝑐 𝑥𝑥 𝑦𝑦 1 1 1 1 2 𝐶𝐶( 𝐴𝐴 �𝑎𝑎) 𝑎𝑎 ( � ) + = 0 1 2 ; 𝑐𝑐 𝑐𝑐 1 2 2 1 1 1 2 2 1 𝑎𝑎 −: 𝑐𝑐 𝑥𝑥 − 𝑎𝑎 1−=𝑐𝑐 0𝑦𝑦 or 𝑎𝑎 𝑐𝑐 − 𝑎𝑎 𝑐𝑐 𝑥𝑥 𝑦𝑦 1 1 1 2 (𝐴𝐴 𝐵𝐵 � 𝑎𝑎) 𝑎𝑎 ( � ) + = 0. 𝑏𝑏1 𝑏𝑏2 𝑎𝑎2 − 𝑏𝑏2 𝑥𝑥 − 𝑎𝑎1 − 𝑏𝑏2 𝑦𝑦 𝑎𝑎1𝑏𝑏2 − 𝑎𝑎2𝑏𝑏1 36 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

INTRODUCTION

The slopes of these lines are: = , = , = . 𝑏𝑏2−𝑐𝑐2 𝑎𝑎2−𝑐𝑐2 𝑎𝑎2−𝑏𝑏2 1 1 1 1 1 1 𝑚𝑚The𝐵𝐵 𝐶𝐶perpendiculars𝑏𝑏1−𝑐𝑐1 𝑚𝑚𝐶𝐶 taken𝐴𝐴 𝑎𝑎from1−𝑐𝑐1 𝑚𝑚, 𝐴𝐴 ,𝐵𝐵 respectively𝑎𝑎1−𝑏𝑏2 to , and have the equations: 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 = , 𝑏𝑏1−𝑐𝑐1 𝑦𝑦 −= 𝑎𝑎 𝑏𝑏2−𝑐𝑐(2 𝑥𝑥 ), 𝑎𝑎1−𝑐𝑐1 𝑦𝑦 = − 𝑎𝑎2−𝑐𝑐2 (𝑥𝑥 − 𝑏𝑏). 𝑎𝑎1−𝑏𝑏1 𝑦𝑦The concurrency− 𝑎𝑎2−𝑐𝑐2 𝑥𝑥 − of𝑐𝑐 these lines is expressed by the condition: ( ) ( ) = 0 1 1 2 2 2 2 (13) 𝑏𝑏 − 𝑐𝑐 𝑏𝑏 − 𝑐𝑐 −𝑎𝑎(𝑏𝑏 − 𝑐𝑐 ) �𝑎𝑎1 − 𝑐𝑐1 𝑎𝑎2 − 𝑐𝑐2 −𝑏𝑏 𝑎𝑎1 − 𝑐𝑐1 � In this determinant, if from line 1 we subtract line 2 and add line 3 ( 𝑎𝑎1 − 𝑏𝑏1 𝑎𝑎2 − 𝑏𝑏2 −𝑐𝑐 𝑎𝑎1 − 𝑏𝑏1 + ), in the obtained determinant, taking into account condition (12), 𝐿𝐿1 → we1 find2 that 3the first line is null, therefore the determinant is null, and condition (13)𝐿𝐿 − is𝐿𝐿 satisfied𝐿𝐿 .

Definition 2 If two orthological triangles have different orthology centers, we will say that the line determined by them is the orthology axis of the triangles.

Problem 1 Show that the orthology relation in the set of triangles in the plane is not a transitive relation.

Problem 2 Let be a triangle, – a point in its interior, and , , – the centers of the circles circumscribed to the triangles , respectively 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 𝑂𝑂𝐴𝐴 𝑂𝑂𝐵𝐵 𝑂𝑂𝑐𝑐 . Prove that the triangles and are orthological. Specify 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 the orthology axis. 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝐴𝐴𝑂𝑂𝐵𝐵𝑂𝑂𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 37 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

2.1 A triangle and its complementary triangle

Definition 3 It is called complementary triangle or median triangle of a given triangle – the triangle determined by the sides of that triangle.

Proposition 1 A given triangle and its complementary triangle are orthological triangles. The orthology centers are respectively the orthocenter and the center of the circle circumscribed to the given triangle.

Proof We denote by the complementary triangle of the given triangle (see Figure 9). 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Because is a median line in the triangle , the perpendicular from to is also the altitude from of the triangle ; similarly, the 𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 perpendiculars from and to , respectively are altitudes. 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 The altitudes of the triangle , being concurrent in the orthocenter of 𝐵𝐵 𝐶𝐶 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 the triangle, it follows that is orthological in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 The perpendiculars from , , to , , are the mediators of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangle , consequently – the center of the circle circumscribed to the 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 triangle , is the orthology center of the complementary triangle in relation 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 39 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

C1 B1 O G

B Ha A1 C

Figure 9

Observation 4 a) The center of the circle circumscribed to the triangle , , is the orthocenter of the complementary triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 b) The triangle and its complementary triangle are homothetic by 𝐴𝐴1𝐵𝐵1𝐶𝐶1 homothety , . The homothety center is – the gravity center 𝐴𝐴𝐴𝐴𝐴𝐴 1 of the triangle . ℎ �𝐺𝐺 − 2� 𝐺𝐺 c) The points , , are collinear, and their line is called Euler line. 𝐴𝐴𝐴𝐴𝐴𝐴 Problem 3 𝐻𝐻 𝐺𝐺 𝑂𝑂 Let be a given triangle, its complementary triangle and the complementary triangle of the triangle . Prove that the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangle and the triangle are orthological. Determine the 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 orthology centers. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2

40 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

2.2 A triangle and its anti-complementary triangle

Definition 4 It is called anticomplementary triangle of a given triangle the triangle determined by the parallels taken through the vertices of the triangle at its opposite sides.

Proposition 2 A given triangle and its anticomplementary triangle are orthological triangles. The orthology centers are the center of the circumscribed circle and the orthocenter of the anticomplementary triangle.

Proof The proof of Proposition 2 is immediate if we take into account the fact that, for the anticomplementary triangle of the given triangle (see Figure 10), the latter is a complementary triangle. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 C1 A B1

H B 1 C

Figure 10

We are thus in the conditions of Proposition 1. However, we notice that the perpendiculars raised in , , , respectively to , and , are the mediators of the triangle , and the perpendiculars taken from , , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 to , and are altitudes in the anticomplementary triangle. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 41 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Observation 5 a) The center of the circumscribed circle of the anticomplementary triangle of a triangle is the orthocenter of the given triangle. b) The anticomplementary and complementary triangles of a given triangle are homothetic by homothety of center in the gravity center of the given triangle and ratio 4:1. c) The anticomplementary and complementary triangles of a given triangle are orthological triangles. The orthology centers are the orthocenter and the center of the circle circumscribed to the given triangle.

2.3 A triangle and its orthic triangle

Definition 5 It is called orthic triangle of a given (non-right) triangle – the triangle created by the feet of the altitudes of the given triangle.

a b

Figure 11

Observation 6 For a right triangle, the notion of orthic triangle is not defined.

42 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Definition 6 We say that the lines , are antiparallels in relation to the concurrent lines and if the angle created by the lines and 𝒄𝒄 𝒅𝒅 is congruent with the angle created by the lines and . 𝒂𝒂 𝒃𝒃 𝒂𝒂 𝒅𝒅 In Figure 11, the lines ( , ) are antiparallels in relation𝒃𝒃 to ( 𝒄𝒄, ); ( , ) ( , ). 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏 ∢ 𝑎𝑎 𝑑𝑑 ≡ ∢ 𝑏𝑏Observation𝑐𝑐 7 a) The lines ( , ), antiparallel with ( , ) form an inscribable quadrilateral with them. 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏 b) If , are antiparallels in relation to , , then the lines , are as well antiparallels in relation with the lines , . 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 Proposition 3 𝑐𝑐 𝑑𝑑 If the lines , are antiparallels in relation to the lines , and the line ’ is parallel with , then the lines ’, are antiparallels in relation to , . 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏

𝑐𝑐 𝑐𝑐 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏

a b c'

A' c B' A B

D C

Figure 12

Ion Pătrașcu, Florentin Smarandache 43 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof In Figure 12, we denote by , , , the vertices of the inscribable quadrilateral determined by the antiparallels , in relation to the lines , . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷 We denote by ’ and ’ – the intersections of the parallel ’ with 𝑐𝑐 𝑑𝑑 𝑎𝑎 𝑏𝑏 respectively , and we observe that the quadrilateral is inscribable; 𝐴𝐴 𝐵𝐵 𝑐𝑐 𝑎𝑎 consequently, , are antiparallels in relation to , . 𝑏𝑏 𝐴𝐴′𝐵𝐵′𝐶𝐶𝐶𝐶 Remark 3 𝑐𝑐′ 𝑑𝑑 𝑎𝑎 𝑏𝑏 If is a non-isosceles triangle ( ) and the line intersects respectively in and such that , we say about and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ≠ 𝐴𝐴𝐴𝐴 𝑎𝑎 𝐴𝐴𝐴𝐴 that they are antiparallels or that is an antiparallel to (see Figure 13). 𝐴𝐴𝐴𝐴 𝐴𝐴′ 𝐵𝐵′ ∢𝐴𝐴𝐴𝐴′𝐴𝐴′ ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 𝐵𝐵𝐵𝐵 𝑎𝑎 A 𝐵𝐵𝐵𝐵 a C'

B C

Figure 13

Proposition 4 The orthic triangle of a given non-isosceles triangle has the sides respectively antiparallels with the sides of this triangle.

Proof In Figure 14, we consider an obtuse triangle and its orthic triangle. Because = = 90 , it follows that the quadrilateral 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 is inscribable, and consequently0, is antiparallel with . ∢𝐵𝐵𝐵𝐵1𝐶𝐶 ∢𝐵𝐵𝐶𝐶1𝐶𝐶 Similarly, it is shown that and are antiparallels with and 𝐵𝐵𝐵𝐵𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 respectively . 𝐴𝐴1𝐶𝐶1 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴

44 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proposition 5 In a (non-isosceles) triangle, the tangent at a vertex to the circumscribed circle is antiparallel with the opposite side. A

C 1 H

B A1 C

Figure 14

Proof Let be the tangent to the circumscribed circle of the triangle (see Figure 15). We have (inscribed angles with the same measure). 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 The previous relation shows that and are antiparallels in relation to ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 and . 𝑃𝑃𝑃𝑃 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 A

B' P C' H

B C

Figure 15

Ion Pătrașcu, Florentin Smarandache 45 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Remark 4 If we take the projections of and on respectively , and we denote them by ’ respectively ’, we note that: 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 The tangent in to the circumscribed circle of the triangle is parallel with 𝐵𝐵 𝐶𝐶 the side ’ ’ of the orthic triangle ’ ’ ’ of this triangle. 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 Indeed, we established in Proposition 4 that ’ ’ is antiparallel with , since 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 ’ ’ ’ ; taking into account relation (4), we obtain that ’ ’ 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 which shows that ’ ’. ∢𝐴𝐴 𝐶𝐶 𝐵𝐵 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐴𝐴𝐴𝐴 𝐵𝐵 Proposition 6𝑃𝑃 𝑃𝑃 ∥ 𝐵𝐵 𝐶𝐶 A given triangle and its orthic triangle are orthological triangles. The orthology centers are the center of the circumscribed circle and the orthocenter of the given triangle.

Proof We consider the acute triangle , and let ’ ’ ’ be its orthic triangle (see Figure 16). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 A

C' H O

B A' C

Figure 16

Obviously, the altitudes ’ , ’ , ’ are concurrent in the orthocenter , hence ’ ’ ’ is orthological in relation to . 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐻𝐻 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

46 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

From the theorem of orthological triangles, it follows that is also orthological in relation to ’ ’ ’. Let us prove that the orthology center is , 𝐴𝐴𝐴𝐴𝐴𝐴 the center of the circle circumscribed to the triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 The perpendicular taken from to ’ ’, considering Proposition 4, is also 𝐴𝐴𝐴𝐴𝐴𝐴 perpendicular to the tangent in to the circumscribed circle, therefore it passes 𝐴𝐴 𝐵𝐵 𝐶𝐶 through . Similarly, the perpendiculars from and to ’ ’, respectively ’ ’ 𝐴𝐴 pass through the center of the circumscribed circle. The the𝑂𝑂 orem can similarly be proved if 𝐵𝐵 is an obtu𝐶𝐶 se𝐴𝐴 triangle𝐶𝐶 . 𝐴𝐴 𝐵𝐵

Remark 5 𝐴𝐴𝐴𝐴𝐴𝐴 The orthology centers and are isogonal conjugate points.

Observation 8 𝐻𝐻 𝑂𝑂 The triangles and ’ ’ ’ are homological. The homology center is , and the homology axis is the orthic axis (see [24]). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻 2.4 The median triangle and the orthic triangle

Theorem 5 In a given triangle, the median triangle, the orthic triangle and the triangle with the vertices in the midpoints of the segments determined by the orthocenter and the vertices of the given triangle are inscribed in the same circle (the circle of nine points).

Proof We denote by the orthic triangle, – the median triangle, and , , – the midpoints of the segments , , ( – the orthocenter 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 of the triangle , see Figure 17). 𝐴𝐴3 𝐵𝐵3 𝐶𝐶3 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 𝐻𝐻 We have that is midline in the triangle , therefore and 𝐴𝐴𝐴𝐴𝐴𝐴 = . Similarly2 3 , is midline in the triangle , therefore2 3 1 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 ∥ 𝐴𝐴𝐴𝐴 2 ;3 since , therefore2 3 ; we have that ; having2 3 𝐵𝐵 𝐶𝐶 2 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐶𝐶 𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴 𝐶𝐶 ∥ , it follows that the quadrilateral is parallelogram, hence 𝐵𝐵𝐵𝐵 𝑂𝑂𝐵𝐵2 ⊥ 𝐴𝐴𝐴𝐴 𝑂𝑂𝐵𝐵2 ∥ 𝐵𝐵𝐵𝐵 𝐴𝐴2𝐶𝐶3 ∥ 𝑂𝑂𝐵𝐵2 = . Because and ( ) = ( ), we obtain that the 𝑂𝑂𝐴𝐴2 ∥ 𝐵𝐵2𝐶𝐶3 𝑂𝑂𝐴𝐴2𝐶𝐶3𝐵𝐵2 quadrilateral is parallelogram. Denoting by the midpoint of the 𝐵𝐵2𝐶𝐶3 𝑂𝑂𝐴𝐴2 𝑂𝑂𝐴𝐴2 ∥ 𝐴𝐴3𝐻𝐻 𝑂𝑂𝐴𝐴2 𝐴𝐴3𝐻𝐻 segment , we have that , , are collinear and = . 𝑂𝑂𝐴𝐴2𝐻𝐻𝐴𝐴3 𝑂𝑂9 𝑂𝑂𝑂𝑂 𝐴𝐴3 𝑂𝑂9 𝐴𝐴2 𝑂𝑂9𝐴𝐴3 𝑂𝑂9𝐴𝐴2

Ion Pătrașcu, Florentin Smarandache 47 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Because is midline in the triangle , we have that: = , 1 therefore 𝑂𝑂9𝐴𝐴=3 . In the right triangle 𝐴𝐴𝐴𝐴𝐴𝐴 , is median𝑂𝑂9𝐴𝐴, 3therefore2 𝑂𝑂𝑂𝑂 1 = 𝑂𝑂9𝐴𝐴3= 2 𝑅𝑅 = . 𝐴𝐴3𝐴𝐴1𝐴𝐴2 𝐴𝐴1𝑂𝑂9 1 A 𝐴𝐴1𝑂𝑂9 𝑂𝑂9𝐴𝐴3 𝑂𝑂9𝐴𝐴2 2 𝑅𝑅

A C1 3

C2 B2

O O B1 9 H

B3 C3

B A2 A1 C

Figure 17

Similarly, it follows that the points , , are at distance from 1 and also from the points , , . The circle1 2 of points3 , , , , , 9, 𝐵𝐵 𝐵𝐵 𝐵𝐵 2 𝑅𝑅 𝑂𝑂 , , is also called Euler circle; as we saw, its radius is half the radius of 𝐶𝐶1 𝐶𝐶2 𝐶𝐶3 𝐴𝐴1 𝐴𝐴2 𝐴𝐴3 𝐵𝐵1 𝐵𝐵2 𝐵𝐵3 the circle circumscribed to the given triangle. 𝐶𝐶1 𝐶𝐶2 𝐶𝐶3 Observation 9 a) The median triangle and the triangle determined by the midpoints of the segments , , are congruent. b) The proof of the Theorem 5 can be made in the same way if the triangle 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 is obtuse. c) The Euler circle is homothetic to the circle circumscribed to the 𝐴𝐴𝐴𝐴𝐴𝐴 triangle by homothety of center and ratio . 1 d) From the previous Observation, it derives two useful properties related 𝐻𝐻 2 to the orthocenter of a triangle, namely:

48 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proposition 7 The symmetric point of the orthocenter of a triangle towards the sides of the triangle belongs to the circumscribed circle.

Proposition 8 The symmetric points of a triangle’s orthocenter towards the vertices of the median triangle belong to the circle circumscribed to the given triangle.

Proposition 9 The median triangle and the orthic triangle of a given triangle are orthological triangles. The orthology centers are the center of the circle of the nine points and the orthocenter of the given triangle.

Proof The segment is diameter in the circle of the nine points, having

= = 2 3(medians in rectangular triangles), and 1𝐴𝐴 𝐴𝐴 1 2 1= 290 ; we have that , therefore also2 1 3 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 2 𝐵𝐵𝐵𝐵 ∢𝐴𝐴 𝐵𝐵 𝐴𝐴 ≡ , and hence0 is mediator of the segment . Similarly, we show that ∢𝐴𝐴2𝐶𝐶1𝐴𝐴3 ∆𝐴𝐴2𝐵𝐵1𝐴𝐴3 ≡ ∆𝐴𝐴2𝐶𝐶1𝐴𝐴3 𝐴𝐴3𝐵𝐵1 ≡ the perpendicular from to passes through center of the circle of nine 𝐴𝐴3𝐶𝐶1 𝐴𝐴2𝐴𝐴3 𝐵𝐵1𝐶𝐶1 points. The fact that the orthocenter is the orthology center of the orthic 𝐵𝐵2 𝐴𝐴1𝐶𝐶1 𝑂𝑂9 triangle in relation to the median triangle is obvious. 𝐻𝐻 Observation 10 We can prove that the perpendicular from to passes through and, considering that is antiparallel to , therefore is parallel with the 𝐴𝐴2 𝐵𝐵1𝐶𝐶1 𝑂𝑂9 tangent in to the circle of the nine points, and consequently the perpendicular 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐵𝐵1𝐶𝐶1 from to , being perpendicular to the tangent in passes through the center 𝐴𝐴2 of the circle. 𝐴𝐴2 𝐵𝐵1𝐶𝐶1 𝐴𝐴2 9 𝑂𝑂 Problem 4 Show that the complementary triangle and the anticomplementary triangle of a given triangle are orthological triangles. Specify the orthology centers.

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Problem 5 Show that the orthic triangle and the anticomplementary triangle of a given triangle are orthological triangles. Specify the orthology centers.

2.5 A triangle and its contact triangle

Definition 7 We call a contact triangle of a given triangle – the triangle determined by the tangent (contact) points of the circle inscribed in the triangle with its sides.

Observation 11 In Figure 18, the contact triangle of the triangle was denoted by . is the center of the circle inscribed in the triangle (the bisectors’ intersection). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 𝐼𝐼 A 𝐴𝐴𝐴𝐴𝐴𝐴

Ba Cc I

B Ca C

Figure 18

Proposition 10 A given triangle and its contact triangle are orthological triangles. The contact centers of these triangles coincide with the center of the circle inscribed in the given triangle.

50 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof We use Figure 18. The tangents , taken from to the inscribed circle are equal, so the perpendicular taken from to is the bisector of the 𝐴𝐴𝐶𝐶𝑏𝑏 𝐴𝐴𝐶𝐶𝑐𝑐 𝐴𝐴 angle , which, obviously, passes through . The perpendicular taken from 𝐴𝐴 𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 to is radius of the inscribed circle; it contains the circle’s center , which 𝐵𝐵𝐵𝐵𝐵𝐵 𝐼𝐼 is the common center of the two orthologies between the considered triangles. 𝐶𝐶𝑎𝑎 𝐵𝐵𝐵𝐵 𝐼𝐼 Observation 12 a) The triangle and its contact triangle are bilogical triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 b) The triangles and are homological triangles, the center of

homology being Gergonne𝑎𝑎 𝑏𝑏 point𝑐𝑐 , and the axis of homology being Lemoine line 𝐴𝐴𝐴𝐴𝐴𝐴(see [24]).𝐶𝐶 𝐶𝐶 𝐶𝐶

Proposition 11 The contact triangle and the median triangle of a given triangle are orthological triangles. The orthology centers are the centers of the inscribed circles in the given triangle and in the median triangle of the given triangle. A

C1 B1

Cc I I1

B Ca A1 C

Figure 19

Ion Pătrașcu, Florentin Smarandache 51 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof We denote by the median triangle of the given triangle (see Figure 19); because and , it follows that the perpendicular 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 taken from to passes through , the center of the circle inscribed in the 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝐼𝐼𝐶𝐶𝑎𝑎 ⊥ 𝐵𝐵𝐵𝐵 triangle . The quadrilateral is parallelogram, the angle bisectors 𝐶𝐶𝑎𝑎 𝐵𝐵1𝐶𝐶1 𝐼𝐼 and are parallel; since , it follows that also 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐴𝐴𝐶𝐶1 (we denoted by the center of the circle inscribed in the median triangle). 𝐵𝐵1𝐴𝐴𝐶𝐶1 𝐵𝐵1𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 ⊥ 𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 𝐴𝐴1𝐼𝐼1 ⊥ 𝑏𝑏 𝑐𝑐 1 2.6𝐶𝐶 𝐶𝐶 A triangle and𝐼𝐼 its tangential triangle

Definition 8 The tangential triangle of a given triangle is the triangle formed by the tangents in , and to the circumscribed circle 𝑨𝑨𝑨𝑨𝑨𝑨 of the triangle . 𝑨𝑨 𝑩𝑩 𝑪𝑪 Observation 13 𝑨𝑨𝑨𝑨𝑨𝑨 a) In Figure 20, we denoted by the tangential triangle of the triangle . 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 b) The triangle is the contact triangle for , its tangential 𝐴𝐴𝐴𝐴𝐴𝐴 triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 c) The center of the circle circumscribed to the triangle , , is the center of the circle inscribed in its tangential triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 Proposition 12 A given triangle and its tangential triangle are orthological triangles. The common center of orthology is the center of the circle circumscribed to the given triangle.

Observation 14 a) Proof of this property derives from the proof of Proposition 10. b) From Proposition 11, the tangential triangle of a given triangle and the median triangle of the tangential triangle are orthological. c) The tangential triangle of a given triangle and this triangle are homological triangle. The center of homology is the symmedian center (Lemoine point in the given triangle, see [24]).

𝐾𝐾

52 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

B C

Figure 20

Proposition 13 The tangential triangle and the median triangle of a given triangle are orthological triangles. The orthology centers are the center of the circumscribed circle and the center of the nine points circle of the given triangle.

Proof Let and the tangential and median triangles of the given triangle (see Figure 21). Obviously, is mediator of segment and, 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝐴𝐴1𝐵𝐵1𝐶𝐶1 since , we have that , and passes through , the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑎𝑎𝐴𝐴1 𝐵𝐵𝐵𝐵 center of the circle circumscribed to the triangle . 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝑇𝑇𝑎𝑎𝐴𝐴1 ⊥ 𝐵𝐵1𝐶𝐶1 𝑇𝑇𝑎𝑎𝐴𝐴1 𝑂𝑂 Similarly, passes through and passes through O, therefore is 𝐴𝐴𝐴𝐴𝐴𝐴 orthology center. 𝑇𝑇𝑏𝑏𝐵𝐵1 𝑂𝑂 𝑇𝑇𝑐𝑐𝐶𝐶1 𝑂𝑂

Ion Pătrașcu, Florentin Smarandache 53 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

If we denote by the orthic triangle of the triangle , we have that the perpendicular taken from to passes through ; since 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 (both are antiparallels to ), we have that the perpendicular from to 𝐴𝐴1 𝐵𝐵2𝐶𝐶2 𝐴𝐴1 𝐵𝐵2𝐶𝐶2 ∥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 passes through . Similarly, it follows that the perpendiculars from and , 𝐵𝐵𝐵𝐵 𝑂𝑂9 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 respectively to and pass through . 𝑂𝑂9 𝐵𝐵1 𝐶𝐶1 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 9 Note 1 𝑇𝑇 𝑇𝑇 𝑇𝑇 𝑇𝑇 𝑂𝑂 In [1], Proposition 13 is presented as Cantor’s Theorem.

Tc B2

C1 B1

C2 O H O9

B A2 A1 C

Figure 21

54 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proposition 14 The tangential triangle of a given triangle and the median triangle of the orthic triangle of the triangle are orthological triangles. 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 The orthology centers are the orthocenter of and the center of the 𝐴𝐴𝐴𝐴𝐴𝐴 nine points circle of the triangle . 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑂𝑂9 𝐴𝐴𝐴𝐴𝐴𝐴 Tb

Tc B' M C' 1

M3 M2

B A' C

Figure 22

Ion Pătrașcu, Florentin Smarandache 55 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof We denote by the median triangle of the orthic triangle corresponding to the given acute triangle (see Figure 22). Because 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝐴𝐴′𝐵𝐵′𝐶𝐶′ and , it follows that ; similarly, 𝐴𝐴𝐴𝐴𝐴𝐴 and ′ ′ ′ ′ ; hence, the triangles and are 𝑂𝑂9𝑀𝑀1 ⊥ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 ∥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑂𝑂9𝑀𝑀1 ⊥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑂𝑂9𝑀𝑀2 ⊥ orthological, the orthology center being – the center of the circle of nine 𝑇𝑇𝑎𝑎𝑇𝑇𝑐𝑐 𝑂𝑂9𝑀𝑀3 ⊥ 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 points of triangle . The triangles and have the sides 𝑂𝑂9 respectively parallel. Ee denote by the orthocenter of the tangential triangle; 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 then will be perpendicular to , therefore is orthology center. 𝐻𝐻𝑇𝑇 𝑎𝑎 𝑇𝑇 2 3 𝑇𝑇 2.7 A𝑇𝑇 𝐻𝐻triangle and its cotangent𝑀𝑀 𝑀𝑀 triangle 𝐻𝐻

Definition 9 It is called cotangent triangle of a given triangle the triangle determined by the contacts of the ex-inscribed circles with the sides of the triangle.

Jc I Jb

B Ja C

Figure 23

56 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Observation 15 In Figure 23, is cotangent triangle of the triangle .

𝑎𝑎 𝑏𝑏 𝑐𝑐 Proposition 15𝐽𝐽 𝐽𝐽 𝐽𝐽 𝐴𝐴𝐴𝐴𝐴𝐴 A triangle and its cotangent triangle are orthological triangles.

Proof It is calculated without difficulty: = , = , = , = , = , = . 𝐴𝐴𝐽𝐽𝑏𝑏 𝑝𝑝 − 𝑐𝑐 𝐴𝐴𝐽𝐽𝑐𝑐 𝑝𝑝 − 𝑏𝑏 𝐵𝐵𝐽𝐽𝑐𝑐 𝑝𝑝 − We note that: 𝑎𝑎 𝐵𝐵𝐽𝐽𝑎𝑎 𝑝𝑝 − 𝑐𝑐 𝐶𝐶𝐽𝐽𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝐶𝐶𝐽𝐽𝑏𝑏 𝑝𝑝 − 𝑎𝑎 + + = + + . According2 2 to the2 Theorem2 2, the2 triangles2 and are orthological. 𝐴𝐴𝐽𝐽𝑐𝑐 𝐵𝐵𝐽𝐽𝑎𝑎 𝐶𝐶𝐽𝐽𝑏𝑏 𝐽𝐽𝑏𝑏𝐴𝐴 𝐽𝐽𝑐𝑐𝐵𝐵 𝐽𝐽𝑎𝑎𝐶𝐶 𝑎𝑎 𝑏𝑏 𝑐𝑐 Definition 10 𝐴𝐴𝐴𝐴𝐴𝐴 𝐽𝐽 𝐽𝐽 𝐽𝐽 It is called Bevan point of the triangle – the intersection of perpendiculars taken from the centers of ex-inscribed circles , 𝑨𝑨𝑨𝑨𝑨𝑨 , respectively to , and . 𝑰𝑰𝒂𝒂 𝒃𝒃 𝒄𝒄 𝑰𝑰Proposition𝑰𝑰 16 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 𝑨𝑨𝑨𝑨 The cotangent triangle of a given triangle and the given triangle have as orthology center the Bevan point.

Proof From Proposition 15, the perpendiculars taken from , , to , , are concurrent. The points , , being respectively the contacts of the ex- 𝐽𝐽𝑎𝑎 𝐽𝐽𝑏𝑏 𝐽𝐽𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 inscribed circles with the sides , , , from the uniqueness of the 𝐽𝐽𝑎𝑎 𝐽𝐽𝑏𝑏 𝐽𝐽𝑐𝑐 perpendicular in a point on a line, we have that perpendiculars taken in to 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 , in to and in to pass respectively through , , , therefore 𝐽𝐽𝑎𝑎 the Bevan𝑏𝑏 point is orthology𝑐𝑐 center of the cotangent triangle𝑎𝑎 in 𝑏𝑏relation𝑐𝑐 to the 𝐵𝐵𝐵𝐵given triangle𝐽𝐽 𝐶𝐶𝐶𝐶. 𝐽𝐽 𝐴𝐴𝐴𝐴 𝐼𝐼 𝐼𝐼 𝐼𝐼

Observation 16 The triangle and its cotangent triangle are homological triangles. The homology center is the Nagel point (see [24]). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐽𝐽𝑎𝑎𝐽𝐽𝑏𝑏𝐽𝐽𝑐𝑐

Ion Pătrașcu, Florentin Smarandache 57 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Problem 6 Let be a triangle and ( ), ( ), ( ), such that = , = and = . Prove that the triangles and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ 𝐴𝐴𝐴𝐴 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 are orthological. What can you say about the triangle ? 𝐴𝐴𝐵𝐵1 𝐵𝐵𝐴𝐴1 𝐵𝐵𝐶𝐶1 𝐶𝐶𝐵𝐵1 𝐴𝐴𝐶𝐶1 𝐶𝐶𝐴𝐴1 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 1 1 1 𝐴𝐴 𝐵𝐵Definition𝐶𝐶 11 𝐴𝐴 𝐵𝐵 𝐶𝐶 We call adjoint A-cotangent triangle of the triangle the triangle that has as vertices the projections of the center of the 𝑨𝑨𝑨𝑨𝑨𝑨 A-ex-inscribed circle, , on the sides of the triangle . It can be defined similarly the adjoint cotangent triangles 𝑰𝑰𝒂𝒂 𝑨𝑨𝑨𝑨𝑨𝑨 corresponding to the vertices and .

Proof 𝑩𝑩 𝑪𝑪 We denote by the adjoint A-cotangent triangle of the triangle (see Figure 24). Obviously′ ′ , the perpendiculars taken in the contacts of the A- 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 ex-inscribed circle on the sides , , pass through .

A 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎

B Ja C

J'b

J'c

Figure 24

58 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proposition 17 A given triangle and its adjoint cotangent triangle are orthological triangles. The common center of orthology is the center of the corresponding ex-inscribed circle.

Proof We denote by the adjoint A-cotangent triangle of the triangle (see Figure 25). Obviously′ ′ , the perpendiculars taken in the contacts of the - 𝐽𝐽𝑎𝑎𝐽𝐽𝑏𝑏𝐽𝐽𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 ex-inscribed circle to the sides , , pass through - the center of the 𝐴𝐴 -ex-inscribed circle; hence, the triangle is orthological in relation to 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎 . Because = (tangents taken from′ ′ to the -ex-inscribed circle), 𝐴𝐴 𝐽𝐽𝑎𝑎𝐽𝐽𝑏𝑏𝐽𝐽𝑐𝑐 it follows that the′ the perpendicular′ from to is the bisector of the angle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐽𝐽𝑏𝑏 𝐴𝐴𝐽𝐽𝑐𝑐 𝐴𝐴 𝐴𝐴 , therefore it passes through ; similarly, the′ perpendiculars′ taken from 𝐴𝐴 𝐽𝐽𝑏𝑏𝐽𝐽𝑐𝑐 and to , respectively to , are exterior bisectors corresponding to the 𝐵𝐵𝐵𝐵𝐵𝐵 𝐼𝐼𝑎𝑎 𝐵𝐵 angles and′ of the triangle ′ , thereforethey pass through . 𝐶𝐶 𝐽𝐽𝑎𝑎𝐽𝐽𝑐𝑐 𝐽𝐽𝑎𝑎𝐽𝐽𝑏𝑏 𝑎𝑎 𝐵𝐵 𝐶𝐶 A 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼

M2 M3 M

B M1 C

Figure 25

2.8 A triangle and its anti-supplementary triangle

Definition 12 It is called antisupplementary triangle of a given triangle the triangle determined by exterior bisectors of this triangle.

Ion Pătrașcu, Florentin Smarandache 59 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 17 a) The antisupplementary triangle of the triangle is the triangle with the vertices in the centers of the circles ex-inscribed to the 𝐴𝐴𝐴𝐴𝐴𝐴 triangle . 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 b) The triangle is the orthic triangle of its triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 𝑏𝑏 𝑐𝑐 Proposition 18 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 𝐼𝐼 𝐼𝐼 A triangle and its antisupplementary triangle are orthological triangles. The orthology centers are the center of the inscribed circle and Bevan point of the given triangle. The proof of this Proposition derives from Propositions 6 and 16.

Observation 18 a) The center of the circle inscribed in the triangle , the point , is the orthocenter of the antisupplementary triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 b) The Bevan point of the triangle is the center of the circle 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 circumscribed to the antisupplementary triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 c) The orthology axis of a triangle and of its antisupplementary𝑎𝑎 𝑏𝑏 𝑐𝑐 triangle is Euler line of the antisupplementary triangle. 𝐼𝐼 𝐼𝐼 𝐼𝐼

Proposition 19 If is a given triangle, is the center of its inscribed circle, and its antisupplementary triangle, then the pairs of triangles ( , ), 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 ( , ), ( , ) have the same orthology center. Their 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐼𝐼𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 orthology𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑐𝑐center𝑎𝑎 is𝑎𝑎 I.𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼𝐼𝐼 𝐼𝐼 Proof of this property is obvious, because the altitudes of the antisupplementary triangle are the bisectors of the given triangle.

Definition 13 It is called pedal triangle of a point from the plane of the triangle – the triangle that has as vertices the intersections 𝑴𝑴 of the cevians , , , respectively with , and . 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 𝑨𝑨𝑨𝑨

60 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 19 a) In Figure 25, the triangle is the pedal triangle of the point in relation to the triangle . We say about the triangle that 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝑀𝑀 it is the -pedal triangle of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1𝑀𝑀2𝑀𝑀3 b) The orthic triangle of the triangle is its -pedal triangle. 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 Proposition 20 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 The antisupplementary triangle of the triangle is orthological with the -pedal triangle of the triangle ( is the center of the circle inscribed 𝐴𝐴𝐴𝐴𝐴𝐴 in the triangle ). 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 We denote by the -pedal triangle of the triangle (see Figure 26).

𝐼𝐼1𝐼𝐼2𝐼𝐼3 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 Ib

I2 I3 I

B I1 C

Figure 26

Ion Pătrașcu, Florentin Smarandache 61 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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The triangle is the orthic triangle of its antisupplementary triangle , therefore the orthology center of the triangle in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 is the orthocenter of , therefore . 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐼𝐼1𝐼𝐼2𝐼𝐼3 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 From the orthological triangles theorem, it follows that the perpendiculars 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐼𝐼 taken from , , respectively to , , are concurrent as well in the second orthology center. 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 𝐼𝐼𝑐𝑐 𝐼𝐼2𝐼𝐼3 𝐼𝐼1𝐼𝐼 𝐼𝐼1𝐼𝐼2 Problem 7 Let be an isosceles triangle with = and its antisupplementary triangle. Show that the triangle is orthological in 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 relation to -pedal triangle of the triangle . 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝑎𝑎 2.9 A triangle𝐼𝐼 and its I-circumpedal𝐴𝐴𝐴𝐴𝐴𝐴 triangle

Definition 14 It is called circumpedal triangle (or metaharmonic triangle) of a point from the plane of a triangle – the triangle with the vertices in the intersections of the semi-lines ( , ( , ( 𝑴𝑴 𝑨𝑨𝑨𝑨𝑨𝑨 with the circumscribed circle of the triangle . 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 A𝑨𝑨𝑨𝑨𝑨𝑨

M2 M

B C

Figure 27

62 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Observation 20 In Figure 27, we denoted by the circumpedal triangle of the point in relation to the triangle . We will say about that it is a - 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝑀𝑀 circumpedal triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1𝑀𝑀2𝑀𝑀3 𝑀𝑀 Proposition 21 A given triangle and its -circumpedal triangle are orthological. The orthology centers are and – the center of the circle circumcribed to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 the triangle ABC. 𝐼𝐼 𝑂𝑂 Proof

Let 1 2 3 the -circumpedal triangle of the triangle (see Figure 28). We denote { } = ; we observe that ( ) = + 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ 1 + 𝐴𝐴 𝐼𝐼2=𝐼𝐼3 ∩ 𝐴𝐴𝐼𝐼1 + + 𝑚𝑚=∢90𝐼𝐼1𝐴𝐴. 𝐼𝐼2 2 𝑚𝑚�𝐴𝐴�𝐼𝐼3� 1 1 1 0 2 𝑚𝑚�𝐶𝐶�𝐼𝐼2� 2 𝑚𝑚�𝐶𝐶�𝐼𝐼1� 2 � 𝑚𝑚 � 𝐴𝐴 ̂ �A 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐶𝐶̂��

I2 A' I3

B C

Figure 28

Consequently: ; similarly, it follows that and , therefore the triangles and 1 2 3 are orthological, and the orthology 𝐴𝐴𝐼𝐼1 ⊥ 𝐼𝐼2𝐼𝐼3 𝐵𝐵𝐼𝐼2 ⊥ 𝐼𝐼1𝐼𝐼3 𝐶𝐶𝐼𝐼3 ⊥ center is . 𝐼𝐼1𝐼𝐼2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐼𝐼 Ion Pătrașcu, Florentin Smarandache 63 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Because 1 is the midpoint of the arc , it means that the perpendicular from 1 to is the mediator of , therefore it passes through – the center of the 𝐼𝐼 𝐵𝐵𝐵𝐵 circle circumscribed to the triangle . This is the second orthology center of 𝐼𝐼 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝑂𝑂 the considered triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 Observation 21 a) The center of the circle inscribed in the triangle is the orthocenter of the -circumpedal triangle of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 b) The line is the Euler line of the -circumpedal triangle. 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 Proposition 𝑂𝑂22𝑂𝑂 𝐼𝐼 The -circumpedal triangle of the triangle and the contact triangle of the triangle are orthological. The orthology centers are the 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 points𝑎𝑎 𝑏𝑏 𝑐𝑐 and (the orthocenter of the contact triangle). 𝐶𝐶 𝐶𝐶 𝐶𝐶 ′ 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 𝐼𝐼 𝐻𝐻

Cb I3 Cc H' I

B Ca C

Figure 29

64 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

We have , and , therefore the - circumpedal triangle 1 2 3 and the contact triangle are orthological. 𝐴𝐴𝐼𝐼1 ⊥ 𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 𝐵𝐵𝐼𝐼2 ⊥ 𝐶𝐶𝑐𝑐𝐶𝐶𝑎𝑎 𝐶𝐶𝐼𝐼3 ⊥ 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏 𝐼𝐼 Because the contact triangle and the -circumpedal triangle are homothetic, 𝐼𝐼 𝐼𝐼 𝐼𝐼 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 it follows that the second orthology center of the triangles and 1 2 3 is 𝐼𝐼 the orthocenter of the contact triangle. 𝑎𝑎 𝑏𝑏 𝑐𝑐 ′ 𝐶𝐶 𝐶𝐶 𝐶𝐶 𝐼𝐼 𝐼𝐼 𝐼𝐼 Remark 6 𝐻𝐻 1. The center of homothety of the triangles and is the isogonal of the Nagel point of the triangle (see [15], p. 290). 𝐼𝐼1𝐼𝐼2𝐼𝐼3 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 2. The′ isogonal of the Nagel point of the triangle is to be found on the 𝑁𝑁 𝐴𝐴𝐴𝐴𝐴𝐴 line (see [24] and [15], p. 291). 𝐴𝐴𝐴𝐴𝐴𝐴 3. The points , , and are collinear. 𝑂𝑂𝑂𝑂 ′ ′ 𝑁𝑁 𝐻𝐻 𝐼𝐼 𝑂𝑂 2.10 A triangle and its H-circumpedal triangle

Proposition 23 A non-right given triangle and its -circumpedal triangle are orthological. The orthology centers are and (the orthocenter and the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 center of the circle circumscribed to the triangle ). 𝐻𝐻 𝑂𝑂 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 The -circumpedal triangle of the triangle is the homothetic of the orthic triangle of the triangle by homothety of center and ratio 2 (see 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 Proposition 6). Because the orthic triangle of the triangle is orthological 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 with it (see Proposition 6), it follows that the the perpendiculars taken from , 𝐴𝐴𝐴𝐴𝐴𝐴 , on the sides of the -circumpedal triangle (parallels to the sides of the 𝐴𝐴 orthic triangle) will be concurrent in . The other orthology center is obviously 𝐵𝐵 𝐶𝐶 𝐻𝐻 the orthocenter of the triangle . 𝑂𝑂 2.11 A triangle𝐻𝐻 and its O-circumpedal𝐴𝐴𝐴𝐴𝐴𝐴 triangle

Definition 15 The symmetric of the orthocenter of the triangle with respect to the center of its circumscribed circle is called 𝑯𝑯 𝑨𝑨𝑨𝑨𝑨𝑨 Longchamps point, , of the triangle . 𝑶𝑶 𝑳𝑳 𝑨𝑨𝑨𝑨𝑨𝑨

Ion Pătrașcu, Florentin Smarandache 65 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proposition 24 A non-right triangle and its -circumpedal triangle are orthological triangles. The orthology centers are the orthocenter and the Longchamps 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 point of the triangle . 𝐻𝐻 Proof𝐿𝐿 𝐴𝐴𝐴𝐴𝐴𝐴 Let be the -circumpedal triangle of the acute triangle in Figure 30; this is the symmetric of the triangle with respect to , hence 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝑂𝑂 , and . The perpendiculars taken from , , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂2𝑂𝑂3 ∥ respectively on , , are the altitudes of the triangle , 𝐵𝐵𝐵𝐵 𝑂𝑂3𝑂𝑂1 ∥ 𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2 ∥ 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 consequently is the orthology center. 𝑂𝑂2𝑂𝑂3 𝑂𝑂3𝑂𝑂1 𝑂𝑂1𝑂𝑂2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 A

O3 O2

L O H

B C

Figure 30

For reasons of symmetry, it follows that the other orthology center will be the symmetric of with respect to , ie. the Longchamps point of the triangle . 𝐻𝐻 𝑂𝑂 𝐿𝐿 𝐴𝐴𝐴𝐴𝐴𝐴 66 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

2.12 A triangle and its -circumpedal triangle

𝒂𝒂 Proposition 25 𝑰𝑰 The -circumpedal triangle of a given triangle and the triangle are orthological triangles. 𝐼𝐼𝑎𝑎 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴Proof A

O B C

Figure 31

In Figure 31, we denoted by 1 1 1 the -circumpedal triangle of the center of the -ex-inscribed circle of the triangle (with > ). Because 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐼𝐼𝑎𝑎 1 is the midpoint of the arc , it follows that the the perpendicular taken 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴̂ 𝐶𝐶̂ from 1 to is mediator of , therefore it passes through , the center 𝐴𝐴 𝐵𝐵𝐵𝐵 of the circle circumscribed to the triangle . 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 67 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

We prove that the perpendiculars taken from 1 to and from 1 to

pass through as well. The quadrilateral 1 is inscribed in the 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐶𝐶 circumscribed circle, therefore: 𝐴𝐴𝐴𝐴 𝑂𝑂 𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵 1 = 1 = ( + ); 1 = ( ) + 1 . 1 𝑚𝑚But𝐵𝐵� 𝐴𝐴𝐴𝐴1 𝑚𝑚=𝐵𝐵�𝐵𝐵𝐵𝐵 12 𝑚𝑚= 𝐴𝐴�( )𝐶𝐶� 𝑚𝑚𝐵𝐵�1𝐶𝐶𝐶𝐶 = 𝑚𝑚(𝐶𝐶�) 𝑚𝑚 𝐵𝐵�(𝐶𝐶𝐶𝐶+ ). 1 It follows𝑚𝑚𝐵𝐵�𝐶𝐶𝐶𝐶 that 𝑚𝑚𝐵𝐵𝐵𝐵𝐵𝐵�1 =𝑚𝑚 𝐴𝐴�( − 𝑚𝑚𝐵𝐵�) 𝐴𝐴𝐴𝐴and 𝑚𝑚1 𝐴𝐴� =− 2(𝑚𝑚) +𝐴𝐴� 𝐶𝐶� ( 1 1 ) = ( + ). 𝑚𝑚𝐵𝐵�𝐶𝐶𝐶𝐶 2 𝑚𝑚 𝐴𝐴� − 𝐶𝐶� 𝐵𝐵�𝐶𝐶𝐶𝐶 𝑚𝑚 𝐶𝐶� 2 𝑚𝑚 𝐴𝐴� − 1 Hence: therefore = 𝐶𝐶� 2 𝑚𝑚 𝐴𝐴� 1 𝐶𝐶� 1 , 1 1 and, consequently, the perpendicular taken from 1 to passes through ; similarly, we show 𝐵𝐵�𝐴𝐴𝐴𝐴 ≡ 𝐵𝐵�𝐶𝐶𝐶𝐶 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐶𝐶 that 1 = 1 , therefore the perpendicular taken from 1 to passes 𝐵𝐵 𝐴𝐴𝐴𝐴 𝑂𝑂 through as well. 𝐶𝐶 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 We showed that the triangle 1 1 1 is orthological with and the 𝑂𝑂 orthology center is . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 Observation 22 𝑂𝑂 In the previous proof, we showed that: The intersection of the exterior bisector of the angle of a triangle with the circle circumscribed to it, is the midpoint of the high arc subtended by the angle’s opposite side.

2.13 A triangle and its ex-tangential triangle

Definition 26 Let a non-right given triangle and its ex-inscribed circles of centers , , . The exterior common tangents to the ex- 𝑨𝑨𝑨𝑨𝑨𝑨 inscribed circles (which does not contain the sides of the triangle 𝑰𝑰𝒂𝒂 𝑰𝑰𝒃𝒃 𝑰𝑰𝒄𝒄 ) determine a triangle called ex-tangential triangle of the triangle . 𝑨𝑨𝑨𝑨𝑨𝑨 𝑬𝑬𝒂𝒂𝑬𝑬𝒃𝒃𝑬𝑬𝒄𝒄 Observation 23 𝑨𝑨𝑨𝑨𝑨𝑨 a) The ex-tangential triangle of the triangle is represented in Figure 32. 𝐸𝐸𝑎𝑎𝐸𝐸𝑏𝑏𝐸𝐸𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 b) If is a right triangle, then the ex-tangential triangle is not defined for this triangle. Indeed, let a triangle with = 90 . 𝐴𝐴𝐴𝐴𝐴𝐴 0 𝐴𝐴𝐴𝐴𝐴𝐴 𝑚𝑚�𝐴𝐴̂�

68 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Because the common tangent to -ex-inscribed and -ex- inscribed circles is perpendicular to the interior common tangent 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 taken from the symmetry center with respect to the line , the 𝐴𝐴𝐴𝐴 exterior common tangent to the ex-inscribed circles ( ), ( ) will be 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏 perpendicular to . 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 Similarly, the exterior common tangent to the circles ( ), ( ) will be 𝐵𝐵𝐵𝐵 perpendicular to . 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 Since the exterior common tangent taken through and from the 𝐵𝐵𝐵𝐵 ex-inscribed circles are parallel, the triangle ex-tangential is not 𝐸𝐸𝑏𝑏 𝐸𝐸𝑐𝑐 defined.

B1 D C1

E c Ib A Ic B' C' I B C A' D1

Figure 32

Ion Pătrașcu, Florentin Smarandache 69 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proposition 26 A non-right given triangle and its ex-tangential triangle are orthological triangles. The orthology center of the given triangle and its ex-tangential triangle is the center of the circle circumscribed to the given triangle.

Eb I c A Ib I B C

Figure 33

Proof 1 For the proof, we first make the following mentions:

Definition 17 Two cevians of a triangle are called isogonal if they are symmetric in relation to the bisector of the triangle with which they have the common vertex.

70 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Lemma 3 The altitude from a vertex of the triangle and the radius of the circumscribed circle corresponding to that vertex are isogonal cevians.

Proof Let be an altitude in the triangle and – the center of the circumscribed circle (see Figure 34). We have: = 90 , 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 = 2 ( ). 0 𝑚𝑚�𝐷𝐷�𝐷𝐷𝐷𝐷� − 𝑚𝑚�𝐶𝐶̂� 𝑚𝑚�𝐴𝐴�Therefore𝐴𝐴𝐴𝐴� 𝑚𝑚 𝐶𝐶̂ = 180 2 = . 1 0 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 2 ∙ � − 𝑚𝑚 � 𝐶𝐶 ̂ � �A 𝑚𝑚�𝐷𝐷�𝐷𝐷𝐷𝐷�

E O

B D C

Figure 34

Or (Solution by Mihai Miculița):  1      m( EOA) =. m( AOB) = m( ACB) ⇒≡ EOA ACB  2   ⇒≡OAE CAD OE⊥ AB   ⇒≡AEO ADC AD⊥ BC  (Two triangles which have two angles respectively congruent, have the third congruent angle as well).

Observation 24 a) Obviously: = . Writing in (right-angled): sin = , we find sin = , therefore = 2 – sinus theorem. 𝐴𝐴𝐴𝐴 ∢𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝐶𝐶 𝐶𝐶 ∆𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂 𝐶𝐶 2𝑅𝑅 sin 𝐶𝐶 𝑅𝑅 Ion Pătrașcu, Florentin Smarandache 71 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

b) Lemma 3 is proved similarly in the case of the obtuse or right triangles.

We prove now Proposition 26. We denote: { 1} = and { 1} = . For reasons of symmetry, the line is axis of symmetry of the 𝐵𝐵 𝐴𝐴𝐴𝐴 ∩ 𝐸𝐸𝑏𝑏𝐸𝐸𝑐𝑐 𝐶𝐶 figure formed by the -ex-inscribed and -exinscribed circles and from their 𝐴𝐴𝐴𝐴 ∩ 𝐸𝐸𝑏𝑏𝐸𝐸𝑐𝑐 𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 exterior and interior common tangents. It follows that the triangle is 𝐵𝐵 𝐶𝐶 congruent with the triangle . We take , . We denote 𝐴𝐴𝐴𝐴𝐴𝐴 { 1} = , and we have that 1 = 1; taking into account 𝐴𝐴𝐶𝐶1𝐵𝐵1 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵1𝐶𝐶1 𝐷𝐷 ∈ 𝐵𝐵1𝐶𝐶1 Lemma 1 and the indicated congruence of triangles, it follows that passes 𝐷𝐷 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 ∢𝐷𝐷𝐷𝐷𝐵𝐵 ∢𝐵𝐵𝐵𝐵𝐷𝐷 through , the circumscribed center of the triangle . Similarly, we show1 that the perpendicular taken from to passes through and that the perpendicular𝐴𝐴𝐷𝐷 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 from to passes through . 𝐵𝐵 𝐸𝐸𝑎𝑎𝐸𝐸𝑐𝑐 𝑂𝑂 𝑎𝑎 𝑏𝑏 Proof𝐶𝐶 𝐸𝐸2 𝐸𝐸 𝑂𝑂 We use the following lemma:

Lemma 4 The ex-tangential triangle of the non-right triangle and the orthic triangle of the triangle are homothetic triangles. ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 Proof𝐴𝐴 of𝐵𝐵 Lemma𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 We use Figure 32; from the congruence of triangles and , it follows that . On the other hand, is antiparallel with , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐶𝐶1𝐵𝐵1 therefore . ′ ′ ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴𝐶𝐶1𝐵𝐵1 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 The previous′ ′ relations lead to , which implies that ∢𝐴𝐴𝐵𝐵 𝐶𝐶 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ . 1 1 ′ ′ ∢𝐴𝐴𝐵𝐵 𝐶𝐶 ≡ ∢𝐴𝐴𝐶𝐶 𝐵𝐵 𝑏𝑏 Similarly𝑐𝑐 , it is shown that and . The ex-tangential 𝐸𝐸 𝐸𝐸 ∥ 𝐵𝐵 𝐶𝐶 ′ ′ ′ ′ triangle and the orthic triangle𝑎𝑎, having𝑏𝑏 respectively𝑎𝑎 𝑐𝑐 parallel sides, are hence homothetic. 𝐸𝐸 𝐸𝐸 ∥ 𝐴𝐴 𝐵𝐵 𝐸𝐸 𝐸𝐸 ∥ 𝐵𝐵 𝐶𝐶

Observation 25 The homothety center of the orthic triangle is called Clawson point.

We are able now to complete Proof 2 of Proposition 26. We observed (Proposition 6) that the non-right triangle ABC and its orthic triangle are orthological. The perpendiculars taken from , , to the sides of the orthic triangle are concurrent in the center of the circumscribed circle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 72 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Because the sides of the orthic triangle are parallels with those of the ex- tangential triangle , it follows that the triangle and its ex-tangential triangle are orthological, the orthology center being . 𝐸𝐸𝑎𝑎𝐸𝐸𝑏𝑏𝐸𝐸𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 2.14 A triangle and its podal triangle 𝑂𝑂

Definition 18 It is called podal triangle of a point from the plane of a given triangle – the triangle determined by the orthogonal projections of the point on the sides of the triangle.

Observation 26 a) In Figure 35, the podal triangle of the point is . b) The podal triangle of the orthocenter of a non-right′ ′ ′triangle is the 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthic triangle of that triangle.

Remark 7 For the points that belongs to the circumscribed circle of a triangle , the podal triangle is not defined because the projections of the point on sides are 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 collinear points. The line of these projections is called Simson line. 𝑀𝑀 A

B' C' P

B A' C

Figure 35

Ion Pătrașcu, Florentin Smarandache 73 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Definition 19 The circumscribed circle of the podal triangle of a point is called podal circle.

Proposition 28 The isogonal cevians of concurrent cevians in a triangle are concurrent.

Definition 2 The concurrency points and of cevians in a triangle and of their isogonal are called isogonal′ points or isogonal conjugate 𝑷𝑷 𝑷𝑷 points.

Remark 8 The center of the circle circumscribed to a triangle and its orthocenter are isogonal points.

Theorem 6 The podal triangle of a point from the interior of a triangle and the given triangle are orthological triangles. The orthology centers are the given point and its isogonal point.

Proof A

C' B' P

B A' C

Figure 36

74 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Let be the podal triangle of in relation to the triangle (see Figure 36′). ′Obviously′ , the triangles and are orthological, and is 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 orthology center. From the theorem of′ orthological′ ′ triangles, it follows that the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 perpendiculars taken from , , to , respectively are concurrent in a point . It remains to be shown′ ′ that′ ′ the points and′ ′ are 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 isogonal points. ′ ′ 𝑃𝑃 𝑃𝑃 𝑃𝑃 Because the quadrilateral is inscribable, it follows that . These congruent angles′ are′ the complements of the angles ′ ′ 𝐴𝐴𝐶𝐶 𝑃𝑃𝐵𝐵 ∢𝐴𝐴𝑃𝑃 𝐶𝐶 ≡ respectively′ ′ . The congruence of these last angles show that the cevians ∢𝐴𝐴𝐵𝐵 𝐶𝐶 𝑃𝑃𝑃𝑃𝑃𝑃 and are′ isogonal. 𝑃𝑃 𝐴𝐴𝐴𝐴 Similarly′ , it follows that and are isogonal cevians and and 𝑃𝑃𝑃𝑃 𝑃𝑃 𝐴𝐴 are isogonal cevians, consequently ′ and – the orthology centers, are′ 𝑃𝑃𝑃𝑃 𝑃𝑃 𝐵𝐵 𝑃𝑃𝑃𝑃 𝑃𝑃 𝐶𝐶 isogonal conjugate points. ′ 𝑃𝑃 𝑃𝑃 Observation 28 a) Theorem 6 is true also for the case when the point is located in the exterior of triangle . 𝑃𝑃 b) Theorem 6 generalizes Propositions 1, 2, 6, 10. 𝐴𝐴𝐴𝐴𝐴𝐴 c) From Theorem 6, it follows that:

Proposition 29 A given triangle and the podal of the center of its ex-inscribed circle are orthological triangles. The common orthology center is the center of the ex- inscribed circle.

Proposition 30 The podal triangle of a point from the interior of the given triangle and the complementary triangle of are orthological triangles. 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴Proof 𝐴𝐴𝐴𝐴𝐴𝐴 Let be the podal of the point and the triangle of the complementar′ ′ y′ triangle (see Figure 37). 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Having , the perpendicular from to will be also 𝐴𝐴𝐴𝐴𝐴𝐴 perpendicular to and will pass through . The point′ is orthology center 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐵𝐵𝐵𝐵 of the triangle 1 1 in relation to . ′𝐵𝐵′ 𝐶𝐶′ 𝑃𝑃 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1

Ion Pătrașcu, Florentin Smarandache 75 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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C1 B1

B' P

B A1 A' C

Figure 37

Observation 29 The previous Proposition is true for any point from the exterior of the triangle which does not belong to its circumscribed circle. 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴Definition 20 The symmetric of a median of a triangle with respect to the bisector of the triangle with the origin at the same vertex of the triangle is called symmedian.

Observation 30 a) The symmedians of a triangle are concurrent. Their concurrency point is called the symmedian center of the triangle or Lemoine point of the triangle. b) The symmedian center and the gravity center are isogonal conjugate points.

Proposition 31 The podal triangle of the symmedian center and the median triangle of a median triangle of a given triangle are orthological triangles. The orthology centers are the symmedian center and the gravity center of the given triangle.

76 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proof

B1 C' K G

B A' A1 C

Figure 38

Let be the podal of the symmedian center and the median triangle of′ ′ ′ (see Figure 38). The cevians and being isogonal, we 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐾𝐾 𝐴𝐴1𝐵𝐵1𝐶𝐶1 have that: ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 . (1) The quadrilateral este inscribable, therefore: ∢𝐴𝐴1𝐴𝐴𝐴𝐴 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 . ′ ′ (2) 𝐴𝐴𝐴𝐴 𝐾𝐾𝐵𝐵 Because′ ′ ′ + = 90 , (3) ∢𝐶𝐶 𝐾𝐾𝐴𝐴 ≡ ∢𝐶𝐶 𝐵𝐵 𝐴𝐴 from (1) and (2) we obtain that′ : 0 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝐶𝐶�𝐾𝐾𝐴𝐴� + = 90 . (4) This relation shows′ ′that 0 , therefore the perpendicular from �1 � 𝑚𝑚�𝐴𝐴 𝐴𝐴𝐴𝐴� 𝑚𝑚�𝐶𝐶 𝐵𝐵 𝐴𝐴� ′ ′ to is the median . Similarly1 , it follows that and 1 ′ ′ 𝐴𝐴𝐴𝐴 ⊥ {𝐵𝐵 𝐶𝐶} = ′ ′ 𝐴𝐴 , therefore the gravity1 center 1 is the orthology1 ′ 𝐵𝐵′ 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴 𝐶𝐶 𝐶𝐶𝐶𝐶 ⊥ center of the triangle in relation to 1 . 1 1 𝐴𝐴 𝐵𝐵 𝐺𝐺 𝐴𝐴𝐴𝐴′ ′∩′𝐵𝐵𝐵𝐵 ∩ 𝐶𝐶𝐶𝐶 1 1 1 Observation 31 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 Proposition 31 can be proved also using Theorem 6. Indeed, the podal of and are orthological triangles, therefore the perpendiculars from , , on the 𝐾𝐾 sides of the triangle pass through the isogonal of , videlicet through the 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐾𝐾 Ion Pătrașcu, Florentin Smarandache 77 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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gravity center . These perpendiculars, being the medians of the triangle , pass through , , . The uniqueness of the perpendicular taken from a point to a line 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 show that 1 is1 orthology1 center of the triangle and . 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ ′ ′ 1 1 1 Theorem𝐺𝐺 7 (The circle of six points) 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 If the points , are isogonal conjugate points in the interior of the triangle , and their podal triangles, then 𝑃𝑃1 𝑃𝑃2 these triangles have the same podal circle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 Proof From Theorem 6, it follows that: . (1) If we denote = , since is inscribable, it follows that: 𝐶𝐶𝑃𝑃1 ⊥ 𝐴𝐴2𝐵𝐵2 = . (2) 𝑚𝑚�𝑃𝑃1�𝐵𝐵1𝐴𝐴1� 𝑥𝑥 𝑃𝑃𝐴𝐴1𝐶𝐶𝐵𝐵1 Taking (1) into consideration, we have that = 90 . But 𝑚𝑚�𝑃𝑃�1𝐶𝐶𝐴𝐴1� 𝑥𝑥 = 90 as well, therefore the points , , , 0 (3) are 𝑚𝑚�𝐵𝐵�2𝐴𝐴2𝐶𝐶� − 𝑥𝑥 concyclic (the lines0 and are antiparallels (see Figure 39). 𝑚𝑚�𝐴𝐴�1𝐵𝐵1𝐶𝐶� − 𝑥𝑥 𝐴𝐴1 𝐴𝐴2 𝐵𝐵1 𝐵𝐵2 A𝐴𝐴1𝐵𝐵1 𝐴𝐴2𝐵𝐵2

A4 B1 A3

C2 B2

P2 C1 P P1

B A1 A2 C

Figure 39

78 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Because the mediators of the segments , pass through the midpoint of the segment , it follows that is the center of the circle on 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝑃𝑃 which the points from (3) are found. Similarly, we show that the points , , 𝑃𝑃1𝑃𝑃2 𝑃𝑃 , are concyclic, and that is the center of their circle (4). 𝐵𝐵1 𝐵𝐵2 From (3) and (4), we have that the points , , , , , are at the 𝐶𝐶1 𝐶𝐶2 𝑃𝑃 same distance from , therefore they are concyclic. 𝐴𝐴1 𝐴𝐴2 𝐵𝐵1 𝐵𝐵2 𝐶𝐶1 𝐶𝐶2 Observation 32 𝑃𝑃 The circle of projections on the sides of a triangle of two isogonal conjugate points from its interior is called the circle of six points.

Theorem 8 (The reciprocal of the Theorem 7) If , are two distinct points in the interior of the triangle and their podal circles coincide, then , are isogonal conjugate 𝑃𝑃1 𝑃𝑃2 𝐴𝐴𝐴𝐴𝐴𝐴 points. 𝑃𝑃1 𝑃𝑃2 Proof We use Figure 39; if , , , , , are concyclic, then the center of this circle will be – the midpoint of the segment . 𝐴𝐴1 𝐴𝐴2 𝐵𝐵1 𝐵𝐵2 𝐶𝐶1 𝐶𝐶2 Let , be the midpoints of segment respectively . From the 𝑃𝑃 𝑃𝑃1𝑃𝑃2 conciclicity of the six points, we have that = . We observe that: 𝐴𝐴3 𝐴𝐴4 𝑃𝑃1𝐴𝐴 𝑃𝑃2𝐴𝐴 (S.S.S.) - is midline in , therefore = 𝑃𝑃𝑃𝑃1 𝑃𝑃𝑃𝑃2 1, and3 1 is2 median in the right3 triangle , hence1 2 = 3, so 1∆𝑃𝑃𝐶𝐶 𝐴𝐴 ≡ ∆𝑃𝑃𝐴𝐴 𝑃𝑃 𝑃𝑃𝐴𝐴 ∆𝐴𝐴𝑃𝑃 𝑃𝑃 1 𝑃𝑃𝐴𝐴 2 = 2; similarly4 , it follows that 2=2 . The 2congruence4 2 of 2 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝑃𝑃 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐴𝐴 2 𝑃𝑃 𝐴𝐴 triangles implies that , and from here, taking into account 𝑃𝑃𝐴𝐴3 𝐵𝐵2𝐴𝐴4 𝐶𝐶1𝐴𝐴3 𝑃𝑃𝐴𝐴4 that , we obtain that . ∢𝐶𝐶1𝐴𝐴3𝑃𝑃 ≡ ∢𝐵𝐵2𝐴𝐴4𝑃𝑃 These latter angles are exterior angles to the isosceles triangles , ∢𝑃𝑃1𝐴𝐴3𝑃𝑃 ≡ ∢𝑃𝑃2𝐴𝐴1𝑃𝑃 ≡ ∢𝑃𝑃1𝐴𝐴𝑃𝑃2 ∢𝑃𝑃1𝐴𝐴3𝐶𝐶1 ≡ ∢𝑃𝑃2𝐴𝐴4𝐵𝐵2 respectively , hence , and thus it follows that the 𝐶𝐶1𝐴𝐴3𝐴𝐴 cevians and are isogonal. Similarly, it is proved that and are 𝐵𝐵2𝐴𝐴4𝐴𝐴 ∢𝑃𝑃1𝐴𝐴𝐶𝐶1 ≡ ∢𝑃𝑃2𝐴𝐴𝐵𝐵2 isogonal, and that and are isogonal cevians, hence and are 𝑃𝑃1𝐴𝐴 𝑃𝑃2𝐴𝐴 𝑃𝑃1𝐵𝐵 𝑃𝑃2𝐵𝐵 isogonal conjugate points. 𝑃𝑃1𝐶𝐶 𝑃𝑃2𝐶𝐶 𝑃𝑃1 𝑃𝑃2 Proposition 32 Let be a point in the interior of the triangle , and its podal triangle; the podal circle of intersects a second time ( ), ( ), 𝑃𝑃1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 respectively ( ) in the points , , respectively . 𝑃𝑃1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 Ion Pătrașcu, Florentin Smarandache 79 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Then the triangles and are orthological. The orthology centers are the points and , where is the isogonal of . 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 1 2 2 1 Proof 𝑃𝑃 𝑃𝑃 𝑃𝑃 𝑃𝑃

C2 B2

P2 O1 C1 P1

B A1 A2 C

Figure 40

We denote by the center of the podal circle of the point (see Figure 40). The mediators of the segments , , are obviously concurrent 𝑂𝑂1 𝑃𝑃1 in . We denote by the symmetric of the point with respect to . The 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 symmetric of the line with respect to will be and is 𝑂𝑂1 𝑃𝑃2 𝑃𝑃1 𝑂𝑂1 perpendicular to ; similarly, the symmetric of the line and with 𝑃𝑃1𝐴𝐴1 𝑂𝑂1 𝑃𝑃2𝐴𝐴2 𝑃𝑃2𝐴𝐴2 respect to will be perpendicular in and to respectively ; they 𝐵𝐵𝐵𝐵 𝑃𝑃1𝐵𝐵1 𝑃𝑃2𝐶𝐶1 also contain the point . The points and have the same podal circle; 𝑂𝑂1 𝐵𝐵2 𝐶𝐶2 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 applying Theorem 8, it follows that these points are isogonal. 𝑃𝑃2 𝑃𝑃1 𝑃𝑃2 Applying now Theorem 6, we obtain that the point – the isogonal of , is orthology center of triangles and ; from here, we have as well 𝑃𝑃2 𝑃𝑃1 that is orthology center of the triangle in relation to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑃𝑃1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2

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2.15 A triangle and its antipodal triangle

Definition 21 It is called an antipodal triangle of the point from the plane of the triangle – the triangle formed by the perpendiculars in 𝑷𝑷 , , , respectively , , . 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨Observation𝑩𝑩 𝑪𝑪 33 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 a. In Figure 41, the antipodal triangle of the point is represented. This point is called the antipodal point′ of′ the′ triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ 𝑃𝑃′ ′ B'𝐴𝐴 𝐵𝐵 𝐶𝐶

A C'

B C

Figure 41

b. The antipodal triangle of the center of the circle circumscribed to a triangle is its tangential triangle. c. The antipodal triangle of the orthocenter of a triangle is the anticomplementary triangle of this triangle. d. The antipodal triangle of the center of the circle inscribed in a triangle is the antisupplementary triangle of this triangle. e. For the points that belong to the sides of the given triangle, the antipodal triangle is not defined.

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Proposition 33 A triangle and its antipodal triangle are orthological triangles.

The proof of this Proposition is obvious, because perpendiculars from , , to , and are concurrent in (see Figure 38). ′ ′ ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵Observation𝐶𝐶 𝐴𝐴 𝐶𝐶 𝐶𝐶34𝐴𝐴 𝑃𝑃 The orthology centers of the triangles and its antipodal triangle are the point and its isogonal conjugate in the antipodal triangle, as it follows′ ′ from′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Theorem 6. ′ 𝑃𝑃 𝑃𝑃 Proposition 34 The antipodal triangle of a point is orthological with the -pedal triangle in the triangle . 𝑃𝑃 𝑃𝑃 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 In Figure 42, let the antipodal triangle of and the -pedal triangle in . ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 B'

A C' B1 C1 P

B A1 C

Figure 42

82 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

It is obvious that the perpendiculars taken from , , , respectively , ′ ′ and are concurrent in , therefore 1 1is orthological1 with the ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 antipode , the orthology center being . 1 1 1 𝐶𝐶 𝐴𝐴 𝐴𝐴′ 𝐵𝐵′ ′ 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 Observation𝐴𝐴 𝐵𝐵 𝐶𝐶 35 𝑃𝑃 This Proposition generalizes the Proposition 20.

Proposition 35 The antipodal triangle of the orthocenter of a non-right triangle and its orthic triangle are orthological triangles. The orthology center is the orthocenter of the given triangle, point that coincides with the center of the circle circumscribed to the antipodal triangle.

Proof The antipodal triangle of the orthocenter of the triangle is the anticomplementary triangle of ; the perpendiculars taken from , , 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 to , and are the mediators of the antipodal triangle and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 they ′are′ concurrent′ ′ in′ ′ – the orthocenter of the triangle . ′ ′ ′ 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐶𝐶 C' 𝐻𝐻 A 𝐴𝐴𝐴𝐴𝐴𝐴 B'

C 1 H

B A1 C

Figure 43

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On the other hand, the perpendicular taken from to ( is the orthic triangle of , see Figure 43) is radius in the′ circumscribed circle of 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the triangle (Proposition 5), therefore it passes through the center of the 𝐴𝐴𝐴𝐴𝐴𝐴 circle circumscribed′ ′ ′ to the triangle , which we showed that it is . 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ ′ ′ Remark 9 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻 Starting from Theorem 6 and referring to Figure 33 that we "complete" with the antipodal triangle of the point , we observe that the antipodal triangle of has parallel sides with the podal triangle′ of , therefore it is homothetic with it. ′ 𝑃𝑃 𝑃𝑃 We formulate this way: 𝑃𝑃 Proposition 36 The antipodal triangle of a point from interior of a given triangle is homothetic with the podal triangle of the isogonal conjugate point of . 𝑃𝑃 ′ Observation 36 𝑃𝑃 𝑃𝑃 The antipodal triangle of a point from the interior of a given triangle and the podal triangle of its isogonal being homothetic are orthological triangles. ′ 𝑃𝑃 Proposition 37 𝑃𝑃 The antipodal triangle of a point from the interior of the triangle is orthological with the -circumpedal triangle of the triangle . 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 In Figure 44, we denoted by the antipode of and by the -circumpedal triangle of . ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Because , , are symmetrics of , , with respect to , we will 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 have that has parallel sides to the sides of the triangle . Since 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 and its antipode are orthological, the perpendiculars from , , to , , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 are concurrent, but these lines are perpendicular also to′ ′ , ′ , , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 therefore and are orthological triangles. 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 For reasons′ ′ ′of symmetry, the orthology center of triangles and 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1 will be the symmetric of the point with respect to . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ′ (′Indeed′ , the triangles ′and are congruent with , it 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 ′ 𝑃𝑃 𝑂𝑂 ′ follows that .) 1 1 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝑃𝑃 𝑂𝑂 𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴 𝑃𝑃 𝐴𝐴1𝑃𝑃 ⊥ 𝐵𝐵 𝐶𝐶

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C' A B'

C1 B1

P' O P

B C

Figure 44

Problem 8 Let be a triangle in which the angles are smaller than 120 ; iin this triangle there is a point (called isogon center), such that 0 = 𝐴𝐴𝐴𝐴𝐴𝐴 = = 120 . Prove that the antipodal triangle of the point 𝑇𝑇 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� is equilateral. 0 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝐶𝐶𝐶𝐶𝐶𝐶�� 2.1𝑇𝑇 6 A triangle and its cyclocevian triangle

Definition 22 Let be the concurrency point of cevians , , from triangle . The circumscribed circle of the′ triangle′ ′ 𝑷𝑷 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 intersects the second time the sides of the triangle in′ the′ ′ 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨 𝑩𝑩 𝑪𝑪 points , , . The triangle is called the cyclocevian 𝑨𝑨𝑨𝑨𝑨𝑨 triangle ′′of the′′ triangle′′ corresponding′′ ′′ ′′ to the point . 𝑨𝑨 𝑩𝑩 𝑪𝑪 𝑨𝑨 𝑩𝑩 𝑪𝑪 𝑨𝑨𝑨𝑨𝑨𝑨 𝑷𝑷

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Observation 37 In Figure 45, the triangle is the cyclocevian triangle of the triangle , corresponding to the point , the′′ intersection′′ ′′ of cevians , , . We can say 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 that the circumscribed circle of the -pedal triangle of a point′ intersect′ ′ the sides of 𝑃𝑃 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 the triangle in the vertices of the cyclocevian triangle of the point – we call the 𝑃𝑃 triangle the -cyclocevian triangle of the triangle . ′′ ′′ ′′ 𝑃𝑃 Definition𝐴𝐴 𝐵𝐵 𝐶𝐶 23 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 The triangles and are called homological triangles if the lines , , are′ concurrent′ ′ . The concurrency point is 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨 𝑩𝑩 𝑪𝑪 called the homology′ ′ center′ . 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 Theorem 9 (Terquem – 1892) The triangle and its -cyclocevian triangle are homological triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 Proof For proof, we use Figure 45. Since , , are concurrent in , we have from Ceva’s theorem: ′ ′ ′ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑃𝑃 = . (1) ′ ′ ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 ∙ 𝐵𝐵 𝐶𝐶 ∙ 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐶𝐶 ∙ 𝐵𝐵 A𝐴𝐴 ∙ 𝐶𝐶 𝐵𝐵

B' C" B" C' P

B A' A" C

Figure 45

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ORTHOLOGICAL REMARKABLE TRIANGLES

Considering the powers of vertices of triangle over the circumscribed circle of the triangle , we have: 𝐴𝐴𝐴𝐴𝐴𝐴 = ′ ′ ′, (2) 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ ′′ = ′ ′′, (3) 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 ′ ′′ = ′ ′′. (4) 𝐵𝐵𝐵𝐵 ∙ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 ∙ 𝐵𝐵𝐵𝐵 Multiplyi′ ′′ng these′ last′′ three relations and taking into account the relation (1), 𝐶𝐶𝐶𝐶we get∙ 𝐶𝐶𝐶𝐶: 𝐶𝐶𝐶𝐶 ∙ 𝐶𝐶𝐶𝐶 = , or equivalent: ′′ ′′ ′′ ′′ ′′ ′′ = 1. (5) 𝐴𝐴𝐴𝐴′′ ∙ 𝐵𝐵𝐵𝐵′′ ∙′′𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵𝐵𝐵 ∙ 𝐶𝐶𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 The′′ relation′′ ′′(5) and Ceva’s theorem show that the cevians , , 𝐴𝐴 𝐶𝐶 ∙ 𝐵𝐵 𝐴𝐴 ∙ 𝐶𝐶 𝐵𝐵 are concurrent, consequently the triangles and are homological′′ ′′ ′′. ′′ ′′ ′′ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 Definition 24 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 The concurrency point of lines , , is called the cyclocevian of the point . ′′ ′′ ′′ 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 Observation 38 𝑷𝑷 a) The orthocenter and the gravity center of a triangle are cyclocevian points because the orthic triangle and the median triangle 𝐻𝐻 𝐺𝐺 are inscribed in the circle of nine points. b) The median triangle is the -cyclocevian triangle of the triangle . c) The orthic triangle is the -cyclocevian triangle. 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 Theorem 10 𝐺𝐺 In the triangle , let be -pedal triangle and -cyclocevian triangle. If the triangles and are 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑃𝑃 𝐴𝐴2𝐵𝐵2𝐶𝐶2 orthological, and , are its orthological centers, then: 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 i. and are isogonal conjugate points; 𝑄𝑄1 𝑄𝑄2 ii. The triangles and are orthological; 𝑄𝑄1 𝑄𝑄2 iii. The orthology centers of the triangles and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 are the points and . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 1 2 Proof 𝑄𝑄 𝑄𝑄 i. Let be the orthology center of the triangles and (the triangle is the podal triangle of the point ). We denote by the 𝑄𝑄1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑄𝑄1 𝑄𝑄2

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ORTHOLOGICAL REMARKABLE TRIANGLES second orthology center of triangles and . According to Theorem 6, we have and – isogonal points. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 ii. If we denote by the podal triangle of and we take into 𝑄𝑄1 𝑄𝑄2 account Theorem 9, it follows′ ′ that′ the points , , , , , are 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝑄𝑄2 concyclic. If two circles have three points in common,′ then they′ coincide′ , it 𝐴𝐴1 𝐴𝐴2 𝐵𝐵1 𝐵𝐵2 𝐶𝐶1 𝐶𝐶2 follows that = , = , = , therefore the podal triangle of is the -cyclocevian′ triangle′, videlicet ′ . This triangle, being podal triangle 𝐴𝐴2 𝐴𝐴2 𝐵𝐵2 𝐵𝐵2 𝐶𝐶2 𝐶𝐶2 𝑄𝑄2 of , is orthological with (Theorem 6), the orthology center being . 𝑃𝑃 𝐴𝐴2𝐵𝐵2𝐶𝐶2 iii. Applying Theorem 6, we also have that the perpendiculars taken from 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄𝑎𝑎 , , to , , are concurrent in the isogonal of the point , therefore in the point . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵2𝐶𝐶2 𝐶𝐶2𝐴𝐴2 𝐴𝐴2𝐵𝐵2 𝑄𝑄2 1 𝑄𝑄 A

C1 B1 B Q 1 P 2 Q2

B A1 A2 C

Figure 46

Definition 25 About two triangles that are simultaneously orthological and homological triangles we say that they are bilogical triangles.

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Observation 39 The triangles and from the previous theorem are bilogical triangles. The homology derives from Theorem 9. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 2.17 A triangle and its three images triangle

Definition 26 The triangle having as vertices the symmetrics of vertices of a given triangle to its opposite sides is called the triangle of the three images of the given triangle.

Observation 40 a) In Figure 47, the triangle is the triangle of the three images of the triangle . ′′ ′′ ′′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 b) The triangle of the three images is not possible in the case of a right 𝐴𝐴𝐴𝐴𝐴𝐴 triangle. B"

A C" B'

C' H

B A' C

Figure 47

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Proposition 28 A given non-right triangle and the triangle of its three images are bilogical triangles.

Proof Obviously, , , (see Figure 43) are concurrent in , the orthocenter of the triangle′′ ′′ , ′′point which is the homology center of triangles 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐻𝐻 and . The perpendiculars taken from , , to , , 𝐴𝐴𝐴𝐴𝐴𝐴 are “altitude′′s”′′ in ′′ , therefore the orthocenter is also′′ an′′ orthology′′ center of 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 triangles and . ′′ ′′ ′′𝐴𝐴𝐴𝐴𝐴𝐴 Theorem𝐴𝐴 𝐵𝐵 11𝐶𝐶 (V. Th𝐴𝐴𝐴𝐴𝐴𝐴ébault – 1947) The triangle of the three images of a given triangle is homothetic with the podal triangle of the center of the circle of the nine points corresponding to the given triangle.

Proof Let be the orthocenter of triangle , the median triangle of and the -podal triangle in (see Figure 48). We denote by 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the symmetric of to , videlicet the intersection of semi-line ( with 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑂𝑂9 𝐴𝐴𝐴𝐴𝐴𝐴 the circumscribed circle of the triangle . Because is the midpoint ′of the 𝐻𝐻1 𝐻𝐻 𝐵𝐵𝐵𝐵 𝐻𝐻𝐴𝐴 segment , it follows that in the right trapeze we have midline, 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂9 therefore 2 = + . ′ 𝑂𝑂𝑂𝑂 𝐻𝐻𝐴𝐴 𝐴𝐴1𝑂𝑂 𝑂𝑂9𝐴𝐴2 Taking into account that 2′ = , we have: 4 = 2 + 2 , 𝑂𝑂9𝐴𝐴2 𝑂𝑂𝐴𝐴1 𝐻𝐻𝐴𝐴 namely: 4 = + = . ′ 𝑂𝑂𝐴𝐴1 𝐴𝐴𝐴𝐴 𝑂𝑂9𝐴𝐴2 𝑂𝑂𝐴𝐴1 𝐻𝐻𝐴𝐴 On the other hand, 4 ′′= ( -′′ the gravity center), we get from the last 𝑂𝑂9𝐴𝐴2 𝐻𝐻𝐻𝐻1 𝐴𝐴 𝐻𝐻1 𝐻𝐻𝐴𝐴 relation that: = = . Because , , are collinear and , 𝑂𝑂9𝐺𝐺 𝐺𝐺𝐺𝐺 𝐺𝐺 𝑂𝑂9𝐴𝐴2 𝑂𝑂9𝐺𝐺 1 ′ we have that the′′ triangles and 9 are similar, therefore the9 points2 , 𝐻𝐻𝐴𝐴 𝐻𝐻𝐻𝐻 4 𝐻𝐻 𝑂𝑂 𝐺𝐺 𝑂𝑂 𝐴𝐴 ∥ 𝐻𝐻𝐴𝐴 , are collinear. ′′ 𝑂𝑂9𝐺𝐺𝐴𝐴2 𝐻𝐻𝐻𝐻𝐴𝐴 𝐺𝐺 Consequently′′ , we have that: = . Similarly, we find that , , and 𝐴𝐴2 𝐴𝐴 𝐺𝐺𝐴𝐴2 1 ′′ ′′ , , are collinear and: 𝐺𝐺=𝐴𝐴 4= . The obtained relations𝐺𝐺 𝐵𝐵 show2 𝐵𝐵 that ′′ 𝐺𝐺𝐵𝐵2 𝐺𝐺𝐶𝐶2 1 the triangles2 (the podal′′ of ′′) and the triangle of the three 𝐺𝐺 𝐶𝐶 𝐶𝐶 𝐺𝐺𝐵𝐵 𝐺𝐺𝐶𝐶 4 ′′ ′′ ′′ images are homothetic2 2 2 by homothety 9 ; . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 1𝐴𝐴 𝐵𝐵 𝐶𝐶

ℎ �𝐺𝐺 4�

90 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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C' H C1 B1 O 9 G O

B A' A 2 A1 C

Figure 48

Proposition 39 The triangle of the three images of a given triangle and the podal triangle of the center of the circle of the nine points are orthological triangles.

The proof of this Proposition follows from Theorem 11 and from the fact that if two triangles are homothetic, they are orthological.

Ion Pătrașcu, Florentin Smarandache 91 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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2.18 A triangle and its Carnot triangle

Definition 27 We call Carnot circles of the given non-right triangle , of orthocenter , the circles circumscribed to triangles , , 𝑨𝑨𝑨𝑨𝑨𝑨 . 𝑯𝑯 𝑩𝑩𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪𝑪𝑪 𝑨𝑨Definition𝑨𝑨𝑨𝑨 28 The triangle determined by the centers of the Carnot circles of the triangle is called the Carnot triangle of the 𝑶𝑶𝒂𝒂𝑶𝑶𝒃𝒃𝑶𝑶𝒄𝒄 triangle . 𝑨𝑨𝑨𝑨𝑨𝑨 Proposition𝑨𝑨𝑨𝑨𝑨𝑨 40 A given triangle and its Carnot triangle are orthological triangles. The orthology centers are the orthocenter and the center of the circle circumscribed to the given triangle.

Proof In Figure 49, we consider an acute triangle of orthocenter . Because is mediator of segment (common chord in Carnot circles ( ), ( )) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 and , it follows that . 𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐴𝐴𝐴𝐴 𝑂𝑂𝑏𝑏 𝑂𝑂𝑐𝑐 Similarly, and , hence the triangles and 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵𝐵𝐵 𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 ∥ 𝐵𝐵𝐵𝐵 have respectively𝑎𝑎 𝑏𝑏parallel sides𝑎𝑎, therefore𝑐𝑐 are homothetic and, consequently𝑎𝑎 𝑏𝑏 𝑐𝑐, orthological. 𝑂𝑂 𝑂𝑂 ∥ 𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 ∥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 𝑂𝑂 The orthology center of the triangle in relation to is , because the perpendiculars taken from , , to the sides of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐻𝐻 are altitudes of the triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 The perpendiculars from , , to , , are the mediators of the 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 triangle , therefore the second orthology center is , the center of the circle 𝑂𝑂𝑎𝑎 𝑂𝑂𝑏𝑏 𝑂𝑂𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 circumscribed to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 Proposition 41 𝐴𝐴𝐴𝐴𝐴𝐴 The Carnot circles of a triangle are congruent with the circle circumscribed to the triangle.

92 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Oc Ob

H O 9 O

B M1 C

Oa H1

Figure 49

Proof We denote by the symmetric of to ; according to Proposition 8, this belongs to the circle circumscribed to the triangle , hence the 𝐻𝐻1 𝐻𝐻 𝐵𝐵𝐵𝐵 symmetric of the triangle to is the triangle , and consequently 𝐴𝐴𝐴𝐴𝐴𝐴 the symmetric of Carnot circle ( ) to is the circumscribed circle of the 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐵𝐵𝐻𝐻1𝐶𝐶 triangle , therefore these circles are congruent. 𝑂𝑂𝑎𝑎 𝐵𝐵𝐵𝐵 Observation𝐴𝐴𝐴𝐴𝐴𝐴 41 Propositions 40, 41 are true also if is an obtuse triangle.

𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 93 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proposition 42 The Carnot triangle is congruent with the triangle .

𝑎𝑎 𝑏𝑏 𝑐𝑐 Proof 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 Let be the midpoint of ( ); we have = ; since is the 1 midpoint 1of ( ), it follows that the quadrilateral1 is parallelogram1 . 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑂𝑂𝑀𝑀 2 𝐴𝐴𝐴𝐴 𝑀𝑀 The parallelogram’s center is the midpoint of segment , consequently 𝑂𝑂𝑂𝑂𝑎𝑎 𝐴𝐴𝐴𝐴𝑂𝑂𝑎𝑎𝑂𝑂 passes through . We notice that (S.A.S.), hence 𝑂𝑂9 𝑂𝑂𝑂𝑂 ( ) = ( ), similarly we find that ( ) = ( ) and ( ) = ( ). 𝐴𝐴𝑂𝑂𝑎𝑎 𝑂𝑂9 ∆𝑂𝑂𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏 ≡ ∆𝐻𝐻𝐻𝐻𝐻𝐻 The triangles and are congruent; also we have that its Carnot 𝐴𝐴𝐴𝐴 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏 𝐵𝐵𝐵𝐵 𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐶𝐶𝐶𝐶 𝑂𝑂𝑎𝑎𝑂𝑂𝑐𝑐 triangle is the homothetic of triangle by homothety ( , 1) or 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 equivalently the Carnot triangle is symmetric to of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 ℎ 𝑂𝑂9 − 9 Observation 42 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 Proposition 42 can be proved similarly in the case of the obtuse triangle.

Definition 29 A quartet of points (a quadruple) such that any of them is the orthocenter of the triangle determined by the other three points is called an orthocentric quadruple.

Remark 10 The quadruple formed by the vertices of a non-right triangle and its orthocenter is a non-right orthocentric quadruple.

Proposition 43 If is a non-right triangle with – orthocenter, and we denote by the center of the circumscribed circle, and is Carnot triangle, then 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝑂𝑂 the triangles and , and , and are 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 orthological. The proof derives from Proposition 40. The orthology centers 𝐵𝐵𝐵𝐵𝐵𝐵 𝑂𝑂𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐶𝐶𝐶𝐶𝐶𝐶 𝑂𝑂𝑂𝑂𝑎𝑎𝑂𝑂𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏 of each pair of triangles in the hypothesis are the points and .

Observation 43 𝑂𝑂 𝐻𝐻 The triangles in the pairs mentioned in the previous Proposition are symmetrical to – the center of the circle of the nine points of the triangle .

𝑂𝑂9 𝐴𝐴𝐴𝐴𝐴𝐴 94 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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2.19 A triangle and its Fuhrmann triangle

Definition 30 It is called Fuhrmann triangle of the triangle the triangle whose vertices are the symmetrics of the means of small arcs , 𝑨𝑨𝑨𝑨𝑨𝑨 , of the circumscribed circle to the sides , , 𝑩𝑩𝑩𝑩� respectively . 𝑪𝑪𝑪𝑪� 𝑨𝑨𝑨𝑨� 𝑩𝑩𝑩𝑩 𝑪𝑪𝑪𝑪 Observation 43𝑨𝑨𝑨𝑨 In Figure 46, the Fuhrmann triangle is denoted by . The circle circumscribed to Fuhrmann triangle is called Fuhrmann circle. 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐 Proposition 44 A given triangle and its Fuhrmann triangle are orthological triangle.

A C'

B' Fa Fb O

Fc B C

Figure 50

Ion Pătrașcu, Florentin Smarandache 95 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proof In Figure 50, we denoted by , , the midpoints of the small arcs , , . The lines , , ′ are′ the′ mediator of the sides of the triangle 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵� , hence they are′ concurrent′ ′ in – the center of the circumscribed circle, 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐹𝐹𝑎𝑎 𝐵𝐵 𝐹𝐹𝑏𝑏 𝐶𝐶 𝐹𝐹𝑐𝑐 therefore� � the triangles and are orthological, and is orthology 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 center. We will denote the other orthology center by . We further prove some 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 properties that will help us define starting from the Fuhrmann triangle 𝑃𝑃 of the triangle . 𝑃𝑃 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐 Proposition𝐴𝐴𝐴𝐴𝐴𝐴 45 In a triangle , the lines determined by the orthocenter and by the vertices of the Fuhrmann triangle are respectively perpendicular to the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 bisectors , , . A" 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 A

Fa H

B C

H1 A'

Figure 51

Proof In Figure 51, we considered an acute triangle where the bisector intersects the circumscribed circle of the triangle in , and intersects the 𝐴𝐴𝐴𝐴 same circle in . We denote by the diameter of ′ in the circumscribed 1 𝐴𝐴 𝐴𝐴𝐴𝐴 ′′ ′ 𝐻𝐻 𝐴𝐴 𝐴𝐴 96 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES circle. We have that (the latter being mediator of BC), therefore the chords and are congruent.′ ′′ On the other hand, being the symmetric 1 𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴 𝐴𝐴 of with′ respect′′ to ′ and the symmetric of with respect to BC, we have 𝐴𝐴 𝐴𝐴 𝐻𝐻 𝐴𝐴 𝐻𝐻1 that the quadrilateral ′ is isosceles trapezoid, consequently = 𝐻𝐻 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐹𝐹𝑎𝑎 . From = and from′ previous relation, we obtain that = 𝐻𝐻1𝐴𝐴𝐹𝐹𝑎𝑎𝐴𝐴 𝐻𝐻𝐹𝐹𝑎𝑎 ;′ this equality′′ (together′ with ) leads to the ascertainment ′′that 𝐻𝐻1𝐴𝐴 𝐴𝐴𝐴𝐴 𝐻𝐻1𝐴𝐴 𝐴𝐴𝐴𝐴 is a parallelogram. Because ′′ , it follows that . 𝐻𝐻𝐹𝐹𝑎𝑎 𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴 𝐹𝐹𝑎𝑎 Similarly′′ , we prove the other perpendicularities′′ . ′ 𝐴𝐴𝐴𝐴𝐹𝐹𝑎𝑎𝐴𝐴 𝐴𝐴 𝐴𝐴 ⊥ 𝐴𝐴𝐴𝐴 𝐻𝐻𝐹𝐹𝑎𝑎 ⊥ 𝐴𝐴𝐴𝐴 Observation 44 The parallelism can be deduced thus: is antiparallel with ; ′′ ′′ ′ is antiparallel with , 𝑎𝑎hence, and are parallel (see Proposition1 3). 𝐴𝐴𝐴𝐴 ∥ 𝐻𝐻′𝐹𝐹 ′′ 𝐴𝐴𝐴𝐴 𝐻𝐻 𝐴𝐴 𝑎𝑎 1 𝑎𝑎 𝐻𝐻𝐹𝐹Theorem 12 (Housel's𝐻𝐻 𝐴𝐴 line) 𝐴𝐴𝐴𝐴 𝐻𝐻𝐹𝐹 In a triangle, the center of the inscribed circle, the gravity center and the Nagel point are collinear and = 2 .

Proof 1 𝐺𝐺𝐺𝐺 𝐼𝐼𝐼𝐼 We denote by the projection of on and by the foot of cevian . The point is the projection on of the center of the′ -ex-inscribed circle, 𝐶𝐶𝑎𝑎 𝐼𝐼 𝐵𝐵𝐵𝐵 𝐼𝐼𝑎𝑎 𝐴𝐴𝐴𝐴 also we denote′ by the foot of the altitude in (see Figure 52). 𝐼𝐼𝑎𝑎 𝐵𝐵𝐵𝐵 𝐼𝐼𝑎𝑎 𝐴𝐴 We prove that the′ triangles and are similar ( is the midpoint 𝐴𝐴 𝐴𝐴 of ). ′ ′ 𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎 𝐼𝐼𝐶𝐶𝑎𝑎𝐴𝐴1 𝐶𝐶𝐶𝐶1 Because the points and are isotomic, we have: = = . 𝐵𝐵𝐵𝐵 We calculate: ′ ′ 𝐶𝐶𝑎𝑎 𝐼𝐼𝑎𝑎 𝐵𝐵𝐵𝐵𝑎𝑎 𝐶𝐶𝐼𝐼𝑎𝑎 𝑝𝑝 − 𝑏𝑏 = = cos ( ). ′ ′ ′ ′ ( ) Because𝑎𝑎 cos = 𝑎𝑎 we obtain = . We have: = 𝐼𝐼 𝐴𝐴 𝑎𝑎 − 𝐵𝐵𝐴𝐴 − 𝐼𝐼 𝐶𝐶2 2 𝑎𝑎 2− 𝑐𝑐 𝐵𝐵 − 𝑝𝑝 − 𝑏𝑏 𝑎𝑎 +𝑐𝑐 −𝑏𝑏 ′ ′ 𝑝𝑝 𝑏𝑏−𝑐𝑐 ( ), = = , 2𝑎𝑎= . It follows that𝑎𝑎 : 𝑎𝑎= = . 𝑎𝑎 1 𝑐𝑐 𝐵𝐵 𝐼𝐼 𝐴𝐴 ′ ′ ′ 𝐶𝐶 𝐴𝐴 1 𝑆𝑆 ′ 2𝑆𝑆 𝐼𝐼𝑎𝑎𝐴𝐴 𝐴𝐴𝐴𝐴 2𝑝𝑝 2 𝑏𝑏The− 𝑐𝑐 indicated𝐼𝐼𝐼𝐼𝑎𝑎 𝑟𝑟 triangles𝑝𝑝 𝐴𝐴𝐴𝐴 are similar,𝑎𝑎 and consequently𝐴𝐴1𝐶𝐶𝑎𝑎 : 𝐼𝐼𝐼𝐼𝑎𝑎 𝑎𝑎 . Applying Menelaus’s theorem in the triangle for transverses - - 𝐼𝐼𝐼𝐼1 ∥ 𝐴𝐴𝐴𝐴 we find that = ( - the foot of Nagel cevian ′). We denote { } = ′ 𝐴𝐴𝐼𝐼𝑎𝑎𝐶𝐶 𝐵𝐵 𝑁𝑁 𝐼𝐼𝑏𝑏 𝐴𝐴𝐴𝐴 𝑝𝑝 ′ ′ ′ , we have𝐴𝐴𝐼𝐼:𝑎𝑎 𝑎𝑎=𝐼𝐼𝑏𝑏 = , which shows that 𝐵𝐵𝐵𝐵 divides the median𝐺𝐺 𝐼𝐼𝐼𝐼 by∩ ′ 𝐼𝐼𝐺𝐺 𝐼𝐼𝐴𝐴1 1 ′ ′ ratio𝐴𝐴𝐴𝐴1 , hence 𝐺𝐺 =𝑁𝑁 –𝐴𝐴𝐴𝐴 the gravity2 center of the triangle𝐺𝐺 and 2 =𝐴𝐴𝐴𝐴1 . 2 ′ 1 𝐺𝐺 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼 𝐺𝐺𝐺𝐺

Ion Pătrașcu, Florentin Smarandache 97 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

I G N

B A' I a A 1 I'a C

Figure 52

Proof 2

Let be a point in the plane of the triangle ; because = , we ′ 𝐵𝐵𝐼𝐼𝑎𝑎 𝑝𝑝−𝑐𝑐 ′ 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎𝐶𝐶 𝑝𝑝−𝑏𝑏 have: = 𝑝𝑝−𝑐𝑐 . We obtain: = + . Since = , 𝑃𝑃��𝑃𝑃�⃗+𝑝𝑝−𝑏𝑏∙�𝑃𝑃��𝑃𝑃�⃗ ′ ′ 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 𝐴𝐴𝐴𝐴 𝑄𝑄 𝑎𝑎 𝑝𝑝−𝑐𝑐 𝑎𝑎 ′ 𝑃𝑃��𝐼𝐼��⃗ 1+𝑝𝑝−𝑏𝑏 𝑃𝑃��𝐼𝐼��⃗ 𝑎𝑎 ∙ 𝑃𝑃��𝑃𝑃�⃗ 𝑎𝑎 𝑃𝑃��𝑃𝑃�⃗ 𝑁𝑁𝐼𝐼𝑎𝑎 𝑝𝑝−𝑎𝑎 = 𝑎𝑎 it follows that ��′⃗, therefore the position vector of Nagel point is: 𝑃𝑃��𝑃𝑃�⃗+𝑝𝑝−𝑎𝑎∙𝑃𝑃𝑃𝑃𝑎𝑎 𝑎𝑎 = 𝑃𝑃��+𝑃𝑃��⃗ 1+𝑝𝑝−+𝑎𝑎 . 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 = 𝑃𝑃��𝑃𝑃��⃗Considering𝑝𝑝 ∙ 𝑃𝑃��𝑃𝑃�⃗ in 𝑝𝑝this∙ 𝑃𝑃��relation�𝑃𝑃�⃗ 𝑝𝑝 ∙ 𝑃𝑃��𝑃𝑃�⃗ (the gravity center) and taking into account that + + = 0, we get: 𝑃𝑃 𝐺𝐺 = ��⃗ ��+�⃗ ��⃗+ �⃗ . (1) 1𝐺𝐺𝐺𝐺 𝐺𝐺𝐺𝐺 𝐺𝐺𝐺𝐺 𝐺𝐺�It�� 𝐺𝐺��is⃗ known− 𝑝𝑝 � 𝑎𝑎that�𝐺𝐺��𝐺𝐺�⃗ the𝑏𝑏 position�𝐺𝐺��𝐺𝐺�⃗ 𝑐𝑐 𝐺𝐺��vector��𝐺𝐺�⃗� of the center of the inscribed circle is: = + + , taking = , we have: 1 𝐼𝐼 𝑃𝑃��𝑃𝑃⃗ = 2𝑟𝑟 �𝑎𝑎𝑃𝑃��𝑃𝑃�⃗ + 𝑏𝑏�𝑃𝑃��𝑃𝑃�⃗ + 𝑐𝑐𝑃𝑃��𝑃𝑃�⃗�. 𝑃𝑃 𝐺𝐺 (2) 1 2 = 𝐺𝐺�The��𝐺𝐺⃗ relations2𝑝𝑝 �𝑎𝑎�𝐺𝐺��𝐺𝐺� ⃗(1) and𝑏𝑏�𝐺𝐺��𝐺𝐺� ⃗(2) show𝑐𝑐𝐺𝐺��𝐺𝐺�⃗� that , , are collinear, and that .

Proposition 46 𝐼𝐼 𝐺𝐺 𝐻𝐻 𝐼𝐼𝐼𝐼 𝐺𝐺𝐺𝐺 The Fuhrmann circle of the triangle has as diameter the segment determined by the orthocenter and by Nagel point. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻𝐻𝐻

98 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof

Fa H I N

B A1 C

Figure 53

In Figure 53, we considered an acute triangle of orthocenter . From Theorem 11, it follows that and 2 = ( , the midpoint of BC). 𝐻𝐻 We build – the intersection of the line with ; because in the triangle 𝐼𝐼𝐼𝐼1 ∥ 𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼1 𝐴𝐴𝐴𝐴 𝐴𝐴1 we have′ and 2 = , it follows that is midline in the 𝑁𝑁 𝑁𝑁𝑁𝑁1 𝐴𝐴𝐴𝐴 triangle′ , therefore = ; having = , we obtain that the 𝑁𝑁 𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼1 ∥ 𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼1 𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼1 quadrilateral′ is parallelogram′ , consequently′ . We proved 𝑁𝑁 𝑁𝑁𝑁𝑁 𝑁𝑁 𝐴𝐴1 𝐴𝐴1𝑁𝑁 𝐴𝐴 𝐴𝐴1 𝐴𝐴1𝐹𝐹𝑎𝑎 that , it follo′ ′ws therefore that = 90 , ie. belongs to the 𝑁𝑁𝐹𝐹𝑎𝑎𝑁𝑁 𝐴𝐴 𝑁𝑁𝑁𝑁𝑎𝑎 ∥ 𝐴𝐴𝐴𝐴 circle of diameter . Similarly, it is shown that , 0 are on the circle of 𝐻𝐻𝐻𝐻𝑎𝑎 ⊥ 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐻𝐻�𝐹𝐹𝑎𝑎𝑁𝑁� 𝐹𝐹𝑎𝑎 diameter , and similarly the Proposition can be proved in the case of the 𝐻𝐻𝐻𝐻 𝐹𝐹𝑏𝑏 𝐹𝐹𝑐𝑐 obtuse triangle. 𝐻𝐻𝐻𝐻 Proposition 47 The measures of the angles of the Fuhrmann triangle of the triangle are 90 , 90 , 90 . 𝐴𝐴𝐴𝐴𝐴𝐴 0 𝐴𝐴 0 𝐵𝐵 0 𝐶𝐶 − 2 − 2 − 2

Ion Pătrașcu, Florentin Smarandache 99 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof is perpendicular to the bisector , and is perpendicular to the bisector . Because = 90 + , and the angle is its 𝐻𝐻𝐻𝐻𝑎𝑎 𝐴𝐴𝐴𝐴 𝐻𝐻𝐹𝐹𝑏𝑏 0 𝐶𝐶 supplement;𝐵𝐵𝐵𝐵 it has the measure𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴 90� . On2 the other hand, 𝐹𝐹𝑎𝑎�𝑁𝑁𝐹𝐹𝑏𝑏 , 0 𝐶𝐶 therefore = 90 . Similarly− 2 , it follows that 𝐹𝐹𝑎𝑎�𝐻𝐻𝐹𝐹𝑏𝑏 ≡=𝐹𝐹𝑎𝑎90�𝐹𝐹𝑐𝑐𝐹𝐹𝑏𝑏 0 𝐶𝐶 0 and 𝑚𝑚�𝐹𝐹𝑎𝑎�𝐹𝐹𝑐𝑐𝐹𝐹=𝑏𝑏�90 −. 2 𝑚𝑚�𝐹𝐹𝑏𝑏�𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏� − 𝐴𝐴 0 𝐵𝐵 2 𝑚𝑚�𝐹𝐹𝑐𝑐�𝐹𝐹𝑏𝑏𝐹𝐹𝑎𝑎� − 2 Proposition 48 The second orthology center, , of the triangle and of the Fuhrmann triangle, , is the intersection of the circles: ( ; ), 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 ( ; ), ( ; ). 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐 𝒞𝒞 𝐹𝐹𝑎𝑎 𝐹𝐹𝑎𝑎𝐵𝐵 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑐𝑐 𝒞𝒞 𝐹𝐹Proof𝐹𝐹 𝐶𝐶 𝒞𝒞 𝐹𝐹 𝐹𝐹 𝐴𝐴

Fa P F T c

Fb B A1 C

Figure 54

100 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

We denote by the intersection of the perpendicular taken from to with the line (see Figure 54). Since = 90 , it follows that 𝑇𝑇 𝐶𝐶 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏 0 𝐴𝐴 = 𝐹𝐹,𝑎𝑎 𝐹𝐹a𝑐𝑐lso, = ; we𝑚𝑚 obtain�𝐹𝐹�𝑏𝑏𝐹𝐹𝑎𝑎 𝐹𝐹𝑐𝑐that� the point− 2 is on the 𝐴𝐴 𝐴𝐴 circumscribed circle of the triangle . 𝑚𝑚�𝐹𝐹�𝑐𝑐𝐵𝐵𝐵𝐵� 2 𝑚𝑚�𝐹𝐹�𝑎𝑎𝐵𝐵𝐵𝐵� 2 𝑇𝑇 If is the center of orthology of triangles and , we note that 𝐴𝐴𝐴𝐴𝐴𝐴 (angles with sides respectively perpendicular), also, we have 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑎𝑎 (the quadrilateral is inscribable). The angles and ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝑇𝑇𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏 are also congruent (sides respectively perpendicular); we obtain that: ∢𝑇𝑇𝐹𝐹𝑎𝑎𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐹𝐹𝑎𝑎𝑇𝑇 𝐵𝐵𝐵𝐵𝐵𝐵 , and then that or that: = 𝐹𝐹𝑏𝑏𝐹𝐹𝑎𝑎𝐴𝐴1 𝑎𝑎 . This𝑏𝑏 last𝑎𝑎 1 relation shows that the point is on𝑎𝑎 1the circle of center ∢1 𝑇𝑇𝐹𝐹 𝐵𝐵 ≡ ∢𝐹𝐹 𝐹𝐹 𝐴𝐴 ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐵𝐵𝐹𝐹 𝐴𝐴 ∢𝐵𝐵𝐵𝐵𝐵𝐵 which𝑎𝑎 passes through and . In the same way, we prove that belongs to the𝑎𝑎 2 ∢𝐵𝐵𝐹𝐹 𝐶𝐶 𝑃𝑃 𝐹𝐹 circles: ( , ), ( , ). 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑐𝑐 Proposition𝒞𝒞 𝐹𝐹 𝐹𝐹 𝐶𝐶49 𝒞𝒞 𝐹𝐹 𝐹𝐹 𝐴𝐴 The symmetrics of vertices of an acute given triangle with respect to the sides of its -circumpedal triangle are the vertices of Fuhrmann triangle of its -circumpedal triangle. 𝐻𝐻 𝐻𝐻 A B'

T Z V

C' X H

B U C

Figure 55

Ion Pătrașcu, Florentin Smarandache 101 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Proof In Figure 55, we denote by the -circumpedal triangle of and by , , the symmetrics of the points′ ′ ′ , , to , , respectively . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 We noticed that the -circumpedal triangle of ′ ′ is′ h′omothetic with ′the′ 𝑋𝑋 𝑌𝑌 𝑍𝑍 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 orthic triangle of . Also, the triangle and its orthic triangle are 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 orthological, and , the center of the circumscribed circle, is orthology center. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 Because the perpendicular taken from to passes through , it means 𝑂𝑂 that it is also the mediator of the side ; hence′ , ′ is the midpoint of the arc 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 and is vertex of the Fuhrmann triangle′ ′ of the triangle . 𝐵𝐵 𝐶𝐶 𝐴𝐴 ′ Similar′ proof for the other vertices. ′ ′ ′ 𝐵𝐵�𝐶𝐶 𝑋𝑋 𝐴𝐴 𝐵𝐵 𝐶𝐶 Theorem 13 (M. Stevanovic – 2002) In a triangle acute, the orthocenter of Fuhrmann triangle coincides with the center of the circle inscribed in the given triangle.

Proof We use Figure 51, where we note that is Fuhrmann triangle of the triangle ( -circumpedal of ). It is sufficient to prove that H, the 𝑋𝑋𝑋𝑋𝑋𝑋 orthocenter′ of′ ′ and the center of the circle inscribed in , is the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 orthocenter of . We will prove that , showing that ′ ′ ′ = 0. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 We have: 𝑋𝑋𝑋𝑋𝑋𝑋 𝑋𝑋𝑋𝑋 ⊥ 𝑌𝑌𝑌𝑌 𝑋𝑋��𝑋𝑋��⃗ ∙ 𝑌𝑌��𝑌𝑌�⃗ = , (1) = + + . (2) 𝑋𝑋��𝑋𝑋��⃗ 𝐴𝐴��𝐴𝐴��⃗ − 𝐴𝐴��𝐴𝐴�⃗ Because and are the symmetrics of , respectively , with respect to 𝑌𝑌��𝑌𝑌�⃗ 𝑌𝑌��𝑌𝑌�⃗ 𝐵𝐵𝐵𝐵��⃗ 𝐶𝐶𝐶𝐶��⃗ , respectively , with the parallelogram rule, we have: = + 𝑌𝑌 𝑍𝑍 𝐵𝐵 𝐶𝐶 ′ ′, = + ′ ; ′replacing these relations in (2), it results: ′ 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐵𝐵𝐵𝐵��⃗ �𝐵𝐵��𝐴𝐴��⃗ ′ = +′ +′ . (3) 𝐵𝐵��𝐶𝐶��⃗ 𝐶𝐶𝐶𝐶��⃗ �𝐶𝐶��𝐴𝐴��⃗ 𝐶𝐶��𝐵𝐵��⃗ But + +′ =′ 0, therefore′ :′ 𝑌𝑌��𝑌𝑌�⃗ 𝐵𝐵𝐵𝐵��⃗ 𝐶𝐶��𝐴𝐴��⃗ 𝐶𝐶��𝐵𝐵��⃗ − 𝐵𝐵��𝐴𝐴��⃗ − 𝐵𝐵��𝐶𝐶��⃗ = + ′ . ′ (4) 𝐵𝐵𝐵𝐵��⃗ 𝐶𝐶��𝐴𝐴��⃗ 𝐴𝐴��𝐵𝐵�⃗ Because: ′ ′ 𝑌𝑌��𝑌𝑌�⃗ 𝐶𝐶��𝐵𝐵��⃗ 𝐶𝐶��𝐵𝐵�⃗ + + + = 0, (5) we get: =′ +′ ′ . We′ evaluate: = , 𝐵𝐵𝐵𝐵��⃗ 𝐶𝐶��𝐵𝐵��⃗ 𝐵𝐵��𝐶𝐶��⃗ 𝐶𝐶��𝐵𝐵�⃗ �⃗ it follows that ′ =′ + . ′ ′ 𝑌𝑌��𝑌𝑌�⃗ 𝐶𝐶𝐶𝐶��⃗ 𝐶𝐶��𝐵𝐵��⃗ 𝑋𝑋��𝑋𝑋��⃗ ∙ 𝑌𝑌��𝑌𝑌�⃗ �𝐴𝐴��𝐴𝐴�⃗ − 𝐴𝐴��𝐴𝐴�⃗� ∙ �𝐶𝐶𝐶𝐶��⃗ − 𝐶𝐶��𝐵𝐵��⃗� But , so = 0 and ′ ′ . ′ ′ �𝑋𝑋��𝑋𝑋�⃗ ∙ 𝑌𝑌��𝑌𝑌�⃗ 𝐴𝐴��𝐴𝐴��⃗ ∙ 𝐶𝐶𝐶𝐶��⃗ 𝐴𝐴��𝐴𝐴��⃗ ∙ 𝐶𝐶��𝐵𝐵��⃗ − 𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐶𝐶𝐶𝐶��⃗ − 𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐶𝐶��𝐵𝐵��⃗ ′ ′ 𝐴𝐴𝐴𝐴 ⊥ 𝐶𝐶𝐶𝐶 �𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐶𝐶𝐶𝐶��⃗ 𝐴𝐴𝐴𝐴 ⊥ 𝐶𝐶 𝐵𝐵

102 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL REMARKABLE TRIANGLES

Therefore = 0; so: = ′ ′ + . (6) 𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐶𝐶��𝐵𝐵��⃗ We denote by { }′=′ and { } = , we have: 𝑋𝑋��𝑋𝑋��⃗ ∙ 𝑌𝑌��𝑌𝑌�⃗ 𝐴𝐴��𝐴𝐴��⃗ ∙ 𝐶𝐶��𝐵𝐵��⃗ 𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐵𝐵𝐵𝐵��⃗ = cos , = cos ′ ′ , 𝑈𝑈 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 𝑉𝑉 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵 𝐶𝐶 = cos , = cos . 𝐴𝐴��𝐴𝐴�⃗ ∙ 𝐵𝐵𝐵𝐵��⃗ 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵𝐵𝐵 ∙ 𝐴𝐴�𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵𝐵𝐵 ∙ 𝐴𝐴�𝐴𝐴𝐴𝐴 We observe′ ′ that ′ ′ ′(sides′ respectively′ ′ perpendicular′ ). The 𝐴𝐴��𝐴𝐴��⃗ ∙ 𝐶𝐶��𝐵𝐵��⃗ 𝐴𝐴𝐴𝐴 ∙ 𝐶𝐶 𝐵𝐵 ∙ 𝐴𝐴𝐴𝐴�𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 ∙ 𝐶𝐶 𝐵𝐵 ∙ �𝐴𝐴�𝐴𝐴𝐶𝐶 � point is the symmetric of with respect′ to , therefore ; ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴𝐴𝐴𝐶𝐶 similarly′ , it is obtained that and from these two relations′: 𝐵𝐵 𝐻𝐻 𝐴𝐴𝐴𝐴 ∢𝐻𝐻𝐻𝐻𝐻𝐻 ≡ ∢𝐶𝐶𝐶𝐶𝐵𝐵 = 2 . ′ ∢𝐻𝐻𝐻𝐻𝐻𝐻 ≡ ∢𝐵𝐵𝐵𝐵𝐶𝐶 ′Sinus′ theorem in the triangles and provides the relations = ∢𝐵𝐵 𝐴𝐴𝐶𝐶 𝐴𝐴̂ 2 sin 2 , = 2 sin 2 ( – the′ radius′ of the circumscribed circle).′ We′ 𝐴𝐴𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 show that = is equivalent to 2 sin = 𝑅𝑅 𝐴𝐴 𝐵𝐵𝐵𝐵 𝑅𝑅 𝐴𝐴 𝑅𝑅 2 sin 2 , videlicet with =′ 2′ cos . From = 2 and 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 ∙ 𝑅𝑅 𝐴𝐴 𝐴𝐴𝐴𝐴 ∙ , it follows that = . On the other hand, ′= ′ (because = ̂ 𝑅𝑅′ ′ 𝐴𝐴 ′𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴 ∢𝐵𝐵′ 𝐴𝐴𝐶𝐶 𝐴𝐴 𝐴𝐴𝐴𝐴 ⊥ and = ). Since ̂ = and = cos = cos , it 𝐵𝐵 𝐶𝐶 ∢𝑇𝑇𝑇𝑇𝐶𝐶 𝐴𝐴 1 𝐴𝐴𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 follows′ that ′ = 2′ cos . The angles of lines ( ,′ ) and ( , ) 𝐴𝐴𝐵𝐵 𝐴𝐴𝐵𝐵 𝐴𝐴𝐶𝐶 𝐴𝐴𝐴𝐴 2 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐶𝐶 𝐴𝐴 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴 are supplementar, therefore we get from (6) that =′ 0′ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 ∢ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐵𝐵 ∢ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 Similarly, we prove that , so is the orthocenter of the Fuhrmann 𝑋𝑋��𝑋𝑋��⃗ ∙ 𝑌𝑌��𝑌𝑌�⃗ triangle . 𝑌𝑌𝑌𝑌 ⊥ 𝑋𝑋𝑋𝑋 𝐻𝐻 Proposition𝑋𝑋𝑋𝑋𝑋𝑋 50 The Fuhrmann triangle and the Carnot triangle corresponding to a given triangle are orthological.

Proof of this property is immediate if we observe that and belong to the mediator of the side . The orthology center of the Carnot triangle in 𝑂𝑂𝑎𝑎 𝐹𝐹𝑎𝑎 relation to the Fuhrmann triangle is – the center of the circle circumscribed 𝐵𝐵𝐵𝐵 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 to the triangle . 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐 𝑂𝑂

𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 103 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

In this section, we will define the concept of orthopole of a line in relation to a triangle, and we will establish connections between this notion and the orthological triangles.

3.1 Degenerate triangles; the orthopole of a line

Definition 31 We say that a triplet of distinct collinear points, joined together by segments, is a degenerate triangle of sides ( ), ( ), ( ) and of vertices , , . We will admit that any two parallel lines are "concurrent"𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 in a point𝑪𝑪𝑪𝑪 "thrown to infinity";𝑨𝑨 𝑩𝑩 𝑪𝑪consequently, we can formulate:

Proposition 51 Two degenerate triangles and are orthological. An important case is when we consider a triangle – a scalene 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangle, and a triangle – a degenerate triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Theorem 14 (The Orthopole𝐴𝐴 𝐵𝐵 𝐶𝐶 Theorem; Soons – 1886)

If is a given triangle, is a certain line, and 1, 1, 1 are the orthogonal projections of vertices , , on , then the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴 𝐵𝐵 𝐶𝐶 perpendiculars taken from 1, 1, 1 respectively to , and 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑 are concurrent in a point called the orthopole of the line in 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 relation to the triangle . 𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 105 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

Proof 1 (Niculae Blaha, 1949) We consider , , – the vertices of a degenerate triangle. The lines , , , being perpendicular to , are concurrent to infinity; 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 consequently, the triangles and are orthological. By the theorem 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑑𝑑 of orthological triangles, we have that the perpendiculars taken from , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 respectively to , and are concurrent as well. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 Proof 2 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 We denote by , , the orthogonal projections of the points , , respectively on , ′ ′ and′ (see Figure 56). 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 A

C' B' O

d B1 A1 C1

C A'

Figure 56 We have that: = = + ′ 2 ′ 2 = 2 2 = 2 + 2 2 2 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵𝐵𝐵1 𝐵𝐵1𝐴𝐴1 − 𝐶𝐶𝐶𝐶1 − 𝐶𝐶1𝐴𝐴1 ′ 2 ′ 2 = 2 2 = 2 + 2 2 2 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶𝐶𝐶1 𝐵𝐵1𝐶𝐶1 − 𝐴𝐴𝐴𝐴1 − 𝐵𝐵1𝐴𝐴1 From the′ 2 three′ previous2 relations2 ,2 we obtain2 that: 2 2 2 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐴𝐴𝐴𝐴1 𝐴𝐴1𝐶𝐶1 − 𝐵𝐵𝐵𝐵1 − 𝐵𝐵1𝐶𝐶1 + + = 0. According′ 2 ′ to2 Carnot’s′ 2 theorem′ 2 , ′it 2follows′ 2that the lines , , 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 are concurrent. ′ ′ ′ 𝐴𝐴 𝐴𝐴1 𝐵𝐵 𝐵𝐵1 𝐶𝐶 𝐶𝐶1

106 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

Proof 3 (Traian Lalescu – 1915) The points and are equally distant from the midpoint of the side because the perpendicular taken from to is mediator of the side . 𝐵𝐵1 𝐶𝐶1 𝑀𝑀𝑎𝑎 Similarly, and are equally distant from – the midpoint of , and 𝐵𝐵𝐵𝐵 𝑀𝑀𝑎𝑎 𝑑𝑑 𝐵𝐵1𝐶𝐶𝑖𝑖0 and are equally distant from (see Figure 57). 𝐴𝐴1 𝐶𝐶1 𝑀𝑀𝑏𝑏 𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝑀𝑀𝑐𝑐 A1 A

Mc Mb

B Ma C

Figure 57 We consider the circles ( , ), ( , ), ( , ). The first two circles, having in common the point , still have another common 𝒞𝒞 𝑀𝑀𝑎𝑎 𝑀𝑀𝑎𝑎𝐵𝐵1 𝒞𝒞 𝑀𝑀𝑏𝑏 𝑀𝑀𝑏𝑏𝐶𝐶1 𝒞𝒞 𝑀𝑀𝑐𝑐 𝑀𝑀𝑐𝑐𝐴𝐴1 point, and their common chord is perpendicular to (the center line), since 𝐶𝐶1 is parallel with , which means that the common chord is perpendicular 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏 to . 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏 𝐴𝐴𝐴𝐴 Similarly, the circles of centers and have in common the chord that 𝐴𝐴𝐴𝐴 have the extremity which, being perpendicular to , is also 𝑀𝑀𝑏𝑏 𝑀𝑀𝑐𝑐 perpendicular to . 𝐴𝐴1 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 Finally, the circles of centers and have in common a chord that has 𝐵𝐵𝐵𝐵 the extremity , and it is perpendicular to . 𝑀𝑀𝑎𝑎 𝑀𝑀𝑐𝑐 𝐵𝐵1 𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 107 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

The three circles, having two by two a common chord and their centers being noncollinear – according to a theorem, it follows that these chords are concurrent. Their point of concurrency, namely the radical center of the considered circles, is the orthopole of the line in relation to the triangle .

Definition 32 𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴 If is a triangle; , , are the projections of its vertices on a line ; and is the orthopole of the line , – then the 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏 𝑩𝑩𝟏𝟏 𝑪𝑪𝟏𝟏 circumscribed circles of triangles , , are 𝒅𝒅 𝑶𝑶 𝒅𝒅 called orthopolar circles of the triangle . 𝑶𝑶𝑶𝑶𝟏𝟏𝑩𝑩𝟏𝟏 𝑶𝑶𝑶𝑶𝟏𝟏𝑪𝑪𝟏𝟏 𝑶𝑶𝑶𝑶𝟏𝟏𝑨𝑨𝟏𝟏 Observation 45 𝑨𝑨𝑨𝑨𝑨𝑨 From this Definition and from Proof 3 of the previous theorem, it follows that the centers of the orthopolar circles are midpoints of the sides of the given triangle.

3.2 Simson line

Theorem 15 (Wallace, 1799) The projections of a point belonging to the circumscribed circle of a triangle on the sides of that triangle are collinear points.

Proof Let be a point on the circumscribed circle of the triangle and the orthogonal projections of , , respectively (see Figure 58). 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 From the inscribable quadrilateral , it follows that: 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 , and from this relation we obtain that: 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝑀𝑀𝑀𝑀𝑀𝑀 ≡ , (1) ∢𝑀𝑀𝑀𝑀𝐶𝐶1 being the complement of the previous angles. ∢𝐶𝐶𝐶𝐶𝐴𝐴1 ≡ ∢𝐶𝐶1𝑀𝑀𝑀𝑀 The quadrilateral and are inscribable, hence: , (2) 𝑀𝑀𝐵𝐵1𝐴𝐴1𝐶𝐶 𝑀𝑀𝐵𝐵1𝐴𝐴𝐶𝐶1 . (3) ∢𝐴𝐴1𝐵𝐵1𝐶𝐶 ≡ ∢𝐴𝐴1𝑀𝑀𝑀𝑀 The relations (1), (2) and (3) implies , consequently the ∢𝐶𝐶1𝐵𝐵𝐵𝐵 ≡ ∢𝐶𝐶1𝑀𝑀𝑀𝑀 points , and are collinear. ∢𝐴𝐴1𝐵𝐵1𝐶𝐶 ≡ ∢𝐴𝐴𝐴𝐴1𝐶𝐶1 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶

108 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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C1 A M

B A1 C

Figure 58

Observation 46 a) It is called the Simson line of a point to the triangle the line of the points , , . 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 b) If is the Simson line of the triangle in relation to the point 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 , we can consider as a degenerate triangle. This triangle is 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 orthological with the triangle , the point being an orthology 𝑀𝑀 𝐴𝐴1𝐵𝐵1𝐶𝐶1 center, and the other orthology center being “thrown to infinity”. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 Theorem 16 (The reciprocal of the Simson-Wallace theorem)

Let be a given triangle and 1 1 1 a degenerate triangle, with 1 , 1 , 1 . The orthology center of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 triangle 1 1 1 in relation to is a point that belongs to the 𝐴𝐴 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵 ∈ 𝐶𝐶𝐶𝐶 𝐶𝐶 ∈ 𝐴𝐴𝐴𝐴 circle circumscribed to the triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 Because is orthological in relation to (the orthology center is “thrown to infinity”), it follows that is orthological in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Let be the orthology center (see Figure 59). 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Because is a degenerate triangle, we have: 𝑀𝑀 . (1) 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∢𝐴𝐴1𝐵𝐵1𝐶𝐶 ≡ ∢𝐶𝐶1𝐵𝐵1𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 109 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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On the other hand, from the inscribable quadrilaterals and , we note that: 𝑀𝑀𝐵𝐵1𝐴𝐴𝐶𝐶1 , (2) 𝑀𝑀𝐵𝐵1𝐴𝐴1𝐶𝐶 . (3) ∢𝐴𝐴1𝐵𝐵1𝐶𝐶 ≡ ∢𝐴𝐴1𝑀𝑀𝑀𝑀 From relations (1), (2) and (3), we get: ∢𝐶𝐶1𝐵𝐵1𝐴𝐴 ≡ ∢𝐴𝐴𝐴𝐴𝐶𝐶1 . (4)

∢𝐴𝐴1𝑀𝑀𝑀𝑀 ≡ ∢𝐶𝐶1𝑀𝑀𝑀𝑀

C1 A M

B A1 C

Figure 59 The angles from the relation (4) are complements of the angles and , and consequently: ∢𝑀𝑀𝑀𝑀𝐴𝐴1 . (5) 1 The∢𝑀𝑀 relation𝑀𝑀𝐶𝐶 (5) shows that the quadrilateral is inscribable, therefore ∢𝑀𝑀𝑀𝑀𝑀𝑀1 ≡ ∢𝑀𝑀𝑀𝑀𝐶𝐶1 belongs to the circle circumscribed to the triangle . 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀 Proposition 52 𝐴𝐴𝐴𝐴𝐴𝐴 In the triangle , the point belongs to the circle circumscribed to the triangle. We denote by ’ the second intersection of the perpendicular 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 taken from to on the circle ( ). Then, the Simson line 𝑀𝑀 of the point is parallel with ’. 𝑀𝑀𝐴𝐴1 𝑀𝑀 𝐵𝐵𝐵𝐵 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝑀𝑀 𝐴𝐴𝐴𝐴

110 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proof The quadrilateral is inscribable ( and are feet of perpendiculars taken from to respectively , see Figure 60); it follows 𝑀𝑀𝐵𝐵1𝐴𝐴1𝐶𝐶 𝐵𝐵1 𝐶𝐶1 that (1). 𝑀𝑀 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 But 1 1 1 ’ (2) (having the same measure ’ . ∢𝐵𝐵 𝐴𝐴 𝑀𝑀 ≡ ∢𝐵𝐵 𝐶𝐶𝐶𝐶 1 It follows1 1that ’ and hence ’. ∢𝐵𝐵 𝐴𝐴 𝑀𝑀 ≡ ∢𝐴𝐴𝐴𝐴 𝑀𝑀 2 𝑚𝑚�𝐴𝐴�𝐴𝐴 � 1 1 1 1 ∢𝐴𝐴𝐴𝐴 𝑀𝑀 ≡ ∢ 𝐵𝐵 C𝐴𝐴1 𝑀𝑀 𝐴𝐴 𝐵𝐵 ∥ 𝐴𝐴𝐴𝐴

A M

B A1 C

Figure 60

Observation 47 Similarly, it is proved that the Simson line of the point ’ is parallel with .

Proposition 53 𝑀𝑀 𝐴𝐴𝐴𝐴 The Simson lines of two diametrically opposed points in the circumscribed circle of the triangle are perpendicular.

Proof 𝐴𝐴𝐴𝐴𝐴𝐴 Let and ’ two diametrically opposed points in the circumscribed circle of the triangle (see Figure 61). 𝑀𝑀 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 111 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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We denote by the second intersection with the circle of the perpendicular taken to , and with the second intersection with the circle of the 𝑀𝑀1 perpendicular taken to ′ . 𝑀𝑀𝐴𝐴1 𝐵𝐵𝐵𝐵 𝑀𝑀1 From Proposition′ ′ 52, it follows that the Simson line of the point is 𝑀𝑀 𝐴𝐴1 𝐵𝐵𝐵𝐵 parallel with the line , and the Simson line of is parallel with the line 𝑀𝑀 , because the lines and are perpendicular,′ we obtain that the 𝐴𝐴𝑀𝑀1 𝑀𝑀 mentioned′ Simson lines are perpendicular.′ 𝑀𝑀𝐴𝐴1 𝐴𝐴𝑀𝑀1 𝐴𝐴𝐴𝐴1 Indeed, the quadrilateral is a rectangle because the points , are diametrically opposed; it follows′ ′ that and , their orthogonal′ 𝑀𝑀𝑀𝑀1𝑀𝑀 𝑀𝑀1 𝑀𝑀 𝑀𝑀 projections, are equally distanced from the center of the ′circumscribed circle 𝐴𝐴1 𝐴𝐴1 and hence the chords and are parallel; being equally distant from ′ ′ 𝑂𝑂 , they are congruent. The1 points1 , , are collinear and consequently 𝑀𝑀𝑀𝑀 𝑀𝑀 𝑀𝑀 ′ . 1 1 𝑂𝑂 ′ 𝑀𝑀 𝑂𝑂 𝑀𝑀 𝑀𝑀1𝐴𝐴 ⊥ 𝑀𝑀1𝐴𝐴 A

M1 M'

B A1 Ma A'1 C

M M'1

Figure 61

Observation 48 The Simson lines of the the extremities of a diameter are isotomic transverse. Indeed, the points and and analogs are isotomic points.

𝐴𝐴1 𝐴𝐴′1 112 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 17 (J. Steiner) The Simson line of a point that belongs to the circumscribed circle of the triangle contains the midpoint of the segment 𝑀𝑀 determined by and by the orthocenter of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 𝑀𝑀 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 We build the symmetric of the circle circumscribed to the triangle with respect to the side . We denote by the projection of the point on 𝐴𝐴𝐴𝐴𝐴𝐴 and with the intersection of with the circumscribed circle of the triangle 𝐵𝐵𝐵𝐵 𝐴𝐴1 𝑀𝑀 𝐵𝐵𝐵𝐵 ; also, ′we denote by the intersection of chord ( ) with the circle 𝑀𝑀 𝑀𝑀𝐴𝐴1 symmetric with the circle circumscribed′′ to the triangle (see′ Figure 62). 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝑀𝑀𝑀𝑀 A 𝐴𝐴𝐴𝐴𝐴𝐴 M' H M"

M0 B A 1 C

Figure 62 Also, we denote by the second intersection of the altitude with the circumscribed circle of the triangle . From Proposition 52, we have that the 𝐻𝐻1 𝐴𝐴𝐴𝐴 Simson line of the point is the parallel taken through with . 𝐴𝐴𝐴𝐴𝐴𝐴 ′ 𝑀𝑀 𝐴𝐴1 𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 113 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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On the other hand, we have that the quadrilaterals and are isosceles trapezoids; the first – because ; and′ the second′′ – 𝐴𝐴𝐻𝐻1𝑀𝑀𝑀𝑀 𝐻𝐻1𝑀𝑀𝑀𝑀 𝐻𝐻 because , and is the symmetric of with′ respect to (see 𝐴𝐴𝐻𝐻1 ∥ 𝑀𝑀𝑀𝑀 Proposition 7), and, ′′also, is the symmetric of with respect to due to 𝐻𝐻𝐻𝐻1 ∥ 𝑀𝑀𝑀𝑀 𝐻𝐻1 𝐻𝐻 𝐵𝐵𝐵𝐵 the performed construction. ′′ 𝑀𝑀 𝑀𝑀 𝐵𝐵𝐵𝐵 From the considered isosceles trapezoids, we obtain that , and then the Simson line of the point is parallel with , because′′ ′is the 𝑀𝑀𝑀𝑀 ∥ 𝐴𝐴𝑀𝑀 midpoint of the segment [ ]; it follows that the Simson′′ line of the point 𝑀𝑀 𝐻𝐻𝐻𝐻 𝐴𝐴1 is middle line in the triangle′′ and consequently passes through the 𝑀𝑀𝑀𝑀 𝑀𝑀 midpoint of the segment . ′′ 𝑀𝑀𝑀𝑀 𝐻𝐻 Proposition 54 𝑀𝑀𝑀𝑀 The midpoint of the segment determined by the point that belongs to the circle circumscribed to the triangle and by the orthocenter of the 𝑀𝑀 triangle belongs to the circle of nine points of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 M 𝐴𝐴𝐴𝐴𝐴𝐴 A

O O H 9

B C

Figure 63

114 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proof Let be the center of the circumscribed circle, and let – the center of the circle of nine points (the midpoint of the segment ( ), see Figure 63). If is 𝑂𝑂 𝑂𝑂9 the midpoint of the segment [ ], then in the triangle , is middle 𝑂𝑂𝑂𝑂 𝑃𝑃 line, therefore = ; consequently, = , which shows that9 the point 𝑂𝑂𝑂𝑂 𝐻𝐻𝐻𝐻 𝑅𝑅 𝐻𝐻𝐻𝐻𝐻𝐻 𝑃𝑃𝑂𝑂 belongs to the circle9 of nine points. 9 𝑃𝑃𝑂𝑂 2 𝑃𝑃𝑂𝑂 2 𝑃𝑃 Remark 11 This Proposition shows that the circle of nine points is homothetic to the circle circumscribed to the triangle by homothety of center and ratio . 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 2 Theorem 18

Let be a given triangle, and 1 1 1, 2 2 2 – degenerate triangles, with , , , {1, 2}; the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthology centers of the latter in relation to are , . We 𝐴𝐴𝑖𝑖 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵𝑖𝑖 ∈ 𝐶𝐶𝐶𝐶 𝐶𝐶𝑖𝑖 ∈ 𝐴𝐴𝐴𝐴 𝑖𝑖 ∈ denote by { } = ; then is the orthopole of the line 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1 𝑀𝑀2 in relation to the triangle . 𝑄𝑄 𝐴𝐴1𝐵𝐵1 ∩ 𝐴𝐴2𝐵𝐵2 𝑄𝑄 1 2 Proof𝑀𝑀 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 We denote by and the projections of points and on the line (see Figure 64). We′ prove ′that the quadrilateral is inscribable. 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝑀𝑀1𝑀𝑀2 Indeed, from the inscribable quadrilateral ′ ′, we note that: 𝐴𝐴1𝐵𝐵 𝐶𝐶 𝐴𝐴2 . ′ (1) 𝑀𝑀2𝐴𝐴2𝐶𝐶𝐶𝐶 From the′ inscribable quadrilateral , we have: ∢𝑀𝑀2𝐶𝐶 𝐴𝐴2 ≡ ∢𝑀𝑀2𝐶𝐶𝐶𝐶2 . (2) 𝐵𝐵𝐵𝐵1𝑀𝑀2𝐶𝐶 The inscribable quadrilateral′ leads to: ∢𝑀𝑀2𝐶𝐶𝐶𝐶 ≡ ∢𝐵𝐵𝐵𝐵1𝐵𝐵 . ′ (3) 𝐵𝐵 𝐵𝐵𝐵𝐵1𝑀𝑀1 From the′ relations′ (1), (2) and (3), we obtain that: , ∢𝐵𝐵𝐵𝐵1𝐵𝐵 ≡ ∢𝐵𝐵 𝐴𝐴1𝐵𝐵 therefore the points , , , are concyclic. ′ ′ ∢𝑀𝑀2𝐶𝐶 𝐴𝐴2 ≡ ∢𝐵𝐵 𝐴𝐴1𝐵𝐵 We denote by the′ orthopole′ of the line in relation to the triangle 𝐵𝐵 𝐴𝐴1 𝐴𝐴2 𝐶𝐶 . Having ′ and , it follows that = 180 𝑄𝑄 𝑀𝑀1𝑀𝑀2 . We prove that′ ′ is on the circle′ ′ of the points , , , ′ .′ It′ is sufficient0 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝑄𝑄 ⊥ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝑄𝑄 ⊥ 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐵𝐵�𝑄𝑄 𝐶𝐶 � − to show that ′ = 180 . ′ ′ 𝐴𝐴 𝑄𝑄 𝐵𝐵 𝐴𝐴1 𝐴𝐴2 𝐶𝐶 We have: ′ ′ = 1800 + . 𝑚𝑚𝐵𝐵�𝐴𝐴2𝐶𝐶 − 𝐴𝐴 ′ ′ 0 ′ ′ 𝑚𝑚𝐵𝐵�𝐴𝐴2𝐶𝐶 − �𝑚𝑚𝐵𝐵�𝐴𝐴2𝐵𝐵 𝑚𝑚𝐶𝐶�𝐴𝐴2𝐶𝐶�

Ion Pătrașcu, Florentin Smarandache 115 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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O C2 B2 Q

B A1 A2 C B C1 2

C' M2 B' M1

Figure 64

From the inscribable quadrilateral , we note that . On the other hand, and are on the ′circumscribed circle′ of the 𝐵𝐵𝐴𝐴2𝑀𝑀2𝐵𝐵 𝐵𝐵�𝐴𝐴2𝐵𝐵 ≡ triangle′ , so we have: , therefore . 𝐵𝐵�𝑀𝑀2𝐵𝐵 𝑀𝑀1 𝑀𝑀2 Similarly, we find that ′ . We obtain that ′ + = 𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝐵𝐵 𝑀𝑀2𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵𝑀𝑀1 ∢𝐵𝐵 𝐴𝐴2𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵𝑀𝑀1 and, consequently, ′( ) = 180 . ′ ′ 𝐶𝐶�𝐴𝐴2𝐶𝐶 ≡ 𝑀𝑀�1𝐴𝐴𝐴𝐴 ∢𝐵𝐵 𝐴𝐴2𝐵𝐵 ∢𝐶𝐶 𝐴𝐴2𝐶𝐶 Let us show that . ′It is sufficient′ 0 to prove that and ∢𝐴𝐴 𝑚𝑚 ∢𝐵𝐵 𝐴𝐴2𝐶𝐶 − 𝐴𝐴 . Because ′ , it is necessary to prove that ′ ′ . 𝑄𝑄 ≡ 𝑄𝑄 𝑄𝑄 ∈ 𝐴𝐴1𝐶𝐶1 𝑄𝑄 ∈ But ′ . Also, (because ′ 𝐴𝐴2𝐵𝐵2 𝑄𝑄 ∈ 𝐴𝐴1𝐶𝐶1 ∢𝑄𝑄 𝐴𝐴1𝐴𝐴2 ≡ ∢𝐶𝐶1𝐴𝐴1𝐵𝐵 ′ ), ′ ′ (being perpendicular′ to ). It ′ follows that ∢𝑄𝑄 𝐴𝐴1𝐴𝐴2 ≡ ∢𝐶𝐶2𝐶𝐶 𝑄𝑄 𝐵𝐵𝑀𝑀1 ∥ 𝐴𝐴2𝐶𝐶 ∢𝐵𝐵𝑀𝑀1𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵𝑀𝑀2 ≡ ′ . ′ ′ ∢𝐴𝐴2𝐶𝐶 𝑀𝑀2 𝑀𝑀1𝐶𝐶1 ∥ 𝐶𝐶 𝑄𝑄 𝐴𝐴𝐴𝐴 Because′ ′ the quadrilateral is inscribable, . ∢𝐴𝐴2𝐶𝐶 𝑄𝑄 ≡ ∢𝐶𝐶2𝑀𝑀1𝐵𝐵 Thus, we have that , as such is on the Simson line of 𝐵𝐵𝐶𝐶1𝑀𝑀1𝐴𝐴1 ∢𝐶𝐶1𝑀𝑀𝑀𝑀 ≡ ∢𝐶𝐶1𝐴𝐴1𝐵𝐵 the point . ′ ′ ∢𝑄𝑄 𝐴𝐴1𝐴𝐴2 ≡ ∢𝐵𝐵𝐴𝐴1𝐶𝐶1 𝑄𝑄 Similarly1 , we show that , therefore . 𝑀𝑀 ′ ′ 𝑄𝑄 ∈ 𝐴𝐴2𝐵𝐵2 𝑄𝑄 ≡ 𝑄𝑄

116 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

Remark 12 1. The theorem shows that the orthopole of the line (which intersect the circumscribed circle of the triangle in and ) is the intersection of Simson 𝑀𝑀1𝑀𝑀2 lines of the points and in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1 𝑀𝑀2 2. From the theorem, it follows that, if a line "rotates" around a point from 𝑀𝑀1 𝑀𝑀2 𝐴𝐴𝐴𝐴𝐴𝐴 the circle to the circle circumscribed to the triangle , then the orthopole of the 𝑑𝑑 𝑀𝑀 line belongs to the Simson line of the point in relation to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 3. Also, from this theorem, we obtain that: The𝑑𝑑 orthopole of a diameter of the circumscribed𝑀𝑀 circle of a triangle in𝐴𝐴𝐴𝐴𝐴𝐴 relation to this triangle belongs to the circle of nine points of the triangle.

Proposition 55 The orthopolar circles of a triangle relative to a tangent taken to the circumscribed circle of the triangle are tangent to the sides of the triangle. The tangent points are collinear, and their lines contain the orthopole of the tangent. A

B1 B A C Q 1

B' A' M C' d

Figure 65

Ion Pătrașcu, Florentin Smarandache 117 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

Proof Let the tangent point with the circle of tangent (see Figure 65). We denote by , , the projections of vertices of the triangle on the line and 𝑀𝑀 𝑑𝑑 by , , ′ the′ projections′ of on the sides of the triangle . Also, we 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑 denote by the orthopole of tangent , and by – the center of the orthopolar 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 circle circumscribed to the triangle .We prove that belongs to this 𝑄𝑄 𝑑𝑑 𝑄𝑄𝑎𝑎 circle. ′ ′ 𝐵𝐵 𝑄𝑄𝐶𝐶 𝐴𝐴1 We have = 180 . (1) We prove as well′ that′ 0 = 180 . (2) 𝑚𝑚�𝐵𝐵�𝑄𝑄𝐶𝐶 � − 𝐴𝐴 The quadrilateral and′ ′ are0 inscribable; from here, we note 𝑚𝑚�𝐵𝐵�𝐴𝐴1𝐶𝐶 � − 𝐴𝐴 that: ′ ′ 𝐴𝐴1𝑀𝑀𝐵𝐵 𝐵𝐵 𝐴𝐴1𝑀𝑀𝐶𝐶 𝐶𝐶 , (3) ′ ′ . (4) ∢𝐵𝐵 𝐴𝐴1𝐵𝐵 ≡ ∢𝐵𝐵 𝑀𝑀𝑀𝑀 Since′ is tangent′ to the circle circumscribed to the triangle , we have: ∢𝐶𝐶 𝐴𝐴1𝐶𝐶 ≡ ∢𝐶𝐶 𝑀𝑀𝑀𝑀 , (5) 𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴 ′ . (6) ∢𝐵𝐵 𝑀𝑀𝑀𝑀 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 But′: + = . (7) ∢𝐶𝐶 𝑀𝑀𝑀𝑀 ≡ ∢𝐶𝐶𝐶𝐶𝐶𝐶 Also: ( ) = 180 ( ) + ( ), therefore we ∢𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝐶𝐶𝐶𝐶𝐶𝐶 ∢𝐴𝐴 obtain the relation′ (2),′ which, together0 with′ (1), prove the concyclicity′ of points 𝑚𝑚 ∢𝐵𝐵 𝐴𝐴1𝐶𝐶 − 𝑚𝑚 ∢𝐵𝐵 𝐴𝐴1𝐵𝐵 𝑚𝑚 ∢𝐶𝐶 𝐴𝐴1𝐶𝐶 , , , . ′ We prove′ that the orthopolar circle ( ) is tangent in to the . 𝐵𝐵 𝑄𝑄 𝐴𝐴1 𝐶𝐶 From the inscribable quadrilateral , we have the relation (4), which, 𝑄𝑄𝑎𝑎 𝐴𝐴1 𝐵𝐵𝐵𝐵 together with MBC (consequence′ of the fact that is tangent to 𝐴𝐴1𝑀𝑀𝐶𝐶 𝐶𝐶 the circumscribed circle′ ) and with M (the′ ′quadrilateral ∢𝐶𝐶𝐶𝐶𝐶𝐶 ≡ ∢ 𝐵𝐵 𝐶𝐶 is inscribable) lead to , which′ shows that is ∢𝑀𝑀𝑀𝑀𝐴𝐴1 ≡ ∢𝐴𝐴1𝐵𝐵 tangent′ to the orthopolar circle ( ). ′Similarly′, we show that and are 𝐴𝐴1𝑀𝑀𝑀𝑀 𝐵𝐵 ∢𝐴𝐴1𝐵𝐵 𝑀𝑀 ≡ ∢𝐶𝐶 𝐴𝐴1𝐶𝐶 𝐵𝐵𝐵𝐵 contact points with respectively of the orthopolar circles ( ) and ( ). 𝑄𝑄𝑎𝑎 𝐵𝐵1 𝐶𝐶1 The points , , belong to the Simson line of the point . We prove that 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝑄𝑄𝑏𝑏 𝑄𝑄𝑐𝑐 belongs to this Simson line. It is satisfactory to prove that: 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑀𝑀 Q . (8) 𝑄𝑄 From the inscribable quadrilateral , we note that: ∢𝐵𝐵1𝐴𝐴1𝐶𝐶 ≡ ∢ 𝐴𝐴1𝐵𝐵 . (9) 𝐴𝐴1𝑀𝑀𝑀𝑀𝐵𝐵1 The line is tangent to the the circumscribed circle, therefore: ∢𝐵𝐵1𝐴𝐴1𝐶𝐶 ≡ ∢𝐵𝐵1𝑀𝑀𝑀𝑀 M . (10) 𝑑𝑑 ′ From (8) and (9), it1 follows that: . (11) ∢𝑀𝑀𝑀𝑀𝑀𝑀 ≡ ∢ 𝐵𝐵 𝐴𝐴 ′ Since and are parallel, we find1 that: ′ 𝐵𝐵 𝐴𝐴 ∥ 𝑀𝑀𝑀𝑀 1. (12) ′ 𝐵𝐵 𝑄𝑄 𝑀𝑀𝐵𝐵 ∢𝑄𝑄𝐵𝐵 𝐴𝐴1 ≡ ∢𝐵𝐵1𝑀𝑀𝑀𝑀 118 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

The line is tangent in to the orthopolar circle ( ), consequently: Q . (13) 𝐵𝐵𝐵𝐵 𝐴𝐴1 𝑄𝑄𝑐𝑐 The relations′ (9), (12) and (13) lead to the relation (8). ∢𝑄𝑄𝐵𝐵 𝐴𝐴1 ≡ ∢ 𝐴𝐴1𝐵𝐵 Remark 13 This Proposition can be considered a particular case of Theorem 18. Indeed, if we consider the tangent in to the “limit” position of a secant with tending to be confused with , then also the projections , on are to be 𝑀𝑀 𝑀𝑀1𝑀𝑀2 𝑀𝑀1 confused, and the circle circumscribed to the quadrilateral becomes 𝑀𝑀2 𝐴𝐴1 𝐴𝐴2 𝐵𝐵𝐵𝐵 tangent in to ; also it has been observed that belongs to this circle and, since 𝐵𝐵′𝐴𝐴1𝐴𝐴2𝐶𝐶′ is found at the intersection of Simson lines of the points , , it will be situated 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝑄𝑄 on the Simson line corresponding to the point of tangent . 𝑄𝑄 𝑀𝑀1 𝑀𝑀2 Proposition 56 𝑀𝑀 A given triangle and the triangle formed by the centers of the orthopolar circles corresponding to the orthopole of a secant to the circle are orthological triangles.

Proof The center of the circle ( ) is the center of the circle circumscribed to the quadrilateral (see Figure 64). The perpendicular from to 𝑄𝑄𝑎𝑎 passes through ′the midpoint′ of ( ). This perpendicular is parallel with ′ ′ 𝐵𝐵 𝐴𝐴1𝐴𝐴2𝐶𝐶 𝑄𝑄𝑎𝑎 𝐵𝐵 𝐶𝐶 and with , and hence passes through′ ′ the midpoint of the side ( ). If′ 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 we denote by′ the midpoint of the chord to the circle circumscribed to 𝐶𝐶𝐶𝐶 𝑀𝑀𝑎𝑎 𝐵𝐵𝐵𝐵 the triangle , we have , so . The mediator of 𝑃𝑃 𝑀𝑀1𝑀𝑀2 ( ) passes through and through (it is parallel with and it is 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄𝑄𝑄 ⊥ 𝑀𝑀1𝑀𝑀2 𝑄𝑄𝑄𝑄 ⊥ 𝑄𝑄𝑎𝑎𝑀𝑀𝑎𝑎 middle line in the trapezoid ), hence is parallel with . The 𝐴𝐴1𝐴𝐴2 𝑃𝑃 𝑄𝑄𝑎𝑎 𝑀𝑀1𝐴𝐴1 quadrilateral is parallelogram. Since , it follows that the 𝑀𝑀1𝐴𝐴1𝐴𝐴2𝑀𝑀2 𝑄𝑄𝑎𝑎𝑃𝑃 𝑂𝑂𝑂𝑂𝑎𝑎 perpendicular taken from to passes through . 𝑄𝑄𝑎𝑎𝑃𝑃𝑃𝑃𝑀𝑀𝑎𝑎 𝑂𝑂𝑀𝑀𝑎𝑎 ⊥ 𝐵𝐵𝐵𝐵 Similarly, we show that the perpendiculars from and from to , 𝑄𝑄𝑎𝑎 𝐵𝐵𝐵𝐵 𝑃𝑃 respectively , pass through the midpoint of the chord [ ], point that 𝑄𝑄𝑏𝑏 𝑄𝑄𝑐𝑐 𝐴𝐴𝐴𝐴 is orthology center of the triangle in relation to . 𝐴𝐴𝐴𝐴 𝑃𝑃 𝑀𝑀1𝑀𝑀2 We show that is parallel and congruent with ; similarly, it follows 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 that and are parallel and congruent with , and it is obtained that 𝑄𝑄𝑎𝑎𝑀𝑀𝑎𝑎 𝑂𝑂𝑂𝑂 the triangle is congruent with and have sides parallel to its 𝑄𝑄𝑏𝑏𝑀𝑀𝑏𝑏 𝑄𝑄𝑐𝑐𝑀𝑀𝑐𝑐 𝑂𝑂𝑂𝑂 sides, so actually the triangle is the translation of the median triangle 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 by vector translation . 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 �𝑂𝑂��𝑂𝑂�⃗ Ion Pătrașcu, Florentin Smarandache 119 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

The triangles and are orthological, and the orthology center is the orthocenter of the triangle ; it follows that is also the orthology 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 center of the triangle in relation to the triangle . 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝑎𝑎 𝑏𝑏 𝑐𝑐 Observation 48 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄 𝑄𝑄 𝑄𝑄 The orthology centers of the triangles and are the orthocenters of these triangles. Indeed, the perpendiculars taken from , , to , 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 respectively are concurrent in , but is parallel with , and is 𝑄𝑄𝑎𝑎 𝑄𝑄𝑏𝑏 𝑄𝑄𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 parallel with , hence the perpendicular from to is perpendicular to 𝐴𝐴𝐴𝐴 𝑃𝑃 𝐵𝐵𝐵𝐵 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 , so belongs to the altitude from of the triangle ; similarly, we 𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝑄𝑄𝑎𝑎 𝐵𝐵𝐵𝐵 obtain that belongs also to the altitude from of the same triangle, therefore 𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 𝑃𝑃 𝑄𝑄𝑎𝑎 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 is the orthocenter of the triangle 𝑃𝑃 𝑄𝑄𝑏𝑏 𝑃𝑃 𝑄𝑄𝑎𝑎𝑄𝑄𝑏𝑏𝑄𝑄𝑐𝑐 A N

B A1 C

B' N1 1 A'

Figure 66

Proposition 57 Let and ’ two triangles inscribed in the same circle such that the points and ’ are diametrically-opposed. The Simson lines of a point 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵𝐵𝐵 that belongs to the circle in relation to the triangles and ’ are 𝐴𝐴 𝐴𝐴 orthogonal, and their intersection point is the orthogonal projection on 𝑁𝑁 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵𝐵𝐵 of the point . 𝐵𝐵𝐵𝐵 𝑁𝑁 120 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

Proof The Simson line of the point , denoted in Figure 66, in relation to the triangle , is parallel with (Proposition 52). By we denoted the 𝑁𝑁 𝐴𝐴1𝐶𝐶1 intersection of the perpendicular taken from to with the circle. Also, the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝑁𝑁1 𝑁𝑁1 Simson line of in relation to the triangle , denoted , is parallel with 𝑁𝑁 𝐵𝐵𝐵𝐵 . Because the angle is right, it follows′ that the Simson′ lines are also 𝑁𝑁 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴1𝐵𝐵1 perpendicular′ . They obviously′ pass through the projection of to . 𝐴𝐴 𝑁𝑁1 𝐴𝐴𝑁𝑁1𝐴𝐴 1 Proposition 58 𝐴𝐴 𝑁𝑁 𝐵𝐵𝐵𝐵 In a triangle, the orthopole of the diameter of the circumscribed circle is the symmetric of projection of a vertex of the triangle on this diameter, in relation to the side of the median triangle opposite to that vertex.

Proof Let be the projection of vertex on on the diameter (see Figure 67). The ′point belongs obviously to the circle circumscribed to the triangle 𝐴𝐴 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 (we denoted′ by the median triangle of ). 𝐴𝐴 This circle has as diameter the radius of the circumscribed circle and it 𝐴𝐴𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 is the symmetric with respect to of the circle of nine points of the triangle 𝐴𝐴𝐴𝐴 . 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 We know that the orthopole of belongs to the latter circle, on the other 𝐴𝐴𝐴𝐴𝐴𝐴 hand belongs to the perpendicular taken from to , therefore is the 𝑄𝑄 𝑑𝑑 ′ symmetric of with respect to . 𝑏𝑏 𝑐𝑐 𝑄𝑄 ′ 𝐴𝐴 𝑀𝑀 𝑀𝑀 𝑄𝑄 𝑏𝑏 𝑐𝑐 Observation𝐴𝐴 49 𝑀𝑀 𝑀𝑀 This Proposition can be as well formulated this way: The projections of vertices of a triangle on a diameter of the circumscribed circle are the vertices of a degenerate orthological triangle with the median triangle of the given triangle and having as orthology center the orthopole of diameter in relation to the given triangle.

Proposition 59 Let be a triangle inscribed in the circle of center and a line that passes through . We denote by , , the symmetrics of vertices of 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑑𝑑 the triangle to , and by , , – the symmetrics of points , , 𝑂𝑂 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 respectively to , and . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 121 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

d A A'

Mc Mb O

B Ma C

Figure 67 Then: i. The triangle and the triangle are symmetrical to the orthopole of the line in relation to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 ii. The triangles and are orthological. Their orthology 𝑄𝑄 𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴 centers are the orthocenters of these triangles, and , and the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 line passes through the orthopole of the line . 𝐻𝐻 𝐻𝐻2 2 Proof 𝐻𝐻𝐻𝐻 𝑄𝑄 𝑑𝑑 Let be the projection of to the line that passes through . This point is on the circle′ with center in , the midpoint of (see Figure 68). This circle 𝐴𝐴 𝐴𝐴 𝑑𝑑 𝑂𝑂 is the homothetic of the circle circumscribed to the triangle by homothety 𝑂𝑂1 𝑂𝑂𝑂𝑂 of center and ratio . The symmetric of with respect to the line is , and 1 𝐴𝐴𝐴𝐴𝐴𝐴 the symmetric of with respect to is , as we proved in the previous 𝐴𝐴 2 𝐴𝐴 𝑑𝑑 𝐴𝐴1 Proposition. The symmetric′ of with respect to is , and , 𝐴𝐴 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝑄𝑄 hence , , are collinear, and , are symmetrical with respect′ to . 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐴𝐴2 𝐴𝐴 𝑄𝑄 ∥ 𝐴𝐴1𝐴𝐴2 𝐴𝐴 𝑄𝑄 𝐴𝐴2 𝐴𝐴1 𝐴𝐴2 𝑄𝑄 122 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL DEGENERATE TRIANGLES

C2 B2 A'

Mc Mb H2 O Q H A1

B Ma C

A2 d B1

Figure 68 Similarly, we prove that , and , are symmetrical with respect to . The triangles and have parallel and congruent homologous 𝐵𝐵1 𝐵𝐵2 𝐶𝐶1 𝐶𝐶2 sides. It is obvious that the perpendiculars taken from , , to , 𝑄𝑄 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 respectively (altitudes of the triangle) will be perpendicular to , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 respectively and concurrent in . 𝐴𝐴𝐴𝐴 𝐵𝐵2𝐶𝐶2 𝐶𝐶2𝐴𝐴2 Similarly, is the orthocenter of the triangle in relation to ; 𝐴𝐴2𝐵𝐵2 𝐻𝐻 moreover, and are symmetric in relation to , the orthopole of the line 𝐻𝐻2 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 in relation to the triangle . 𝐻𝐻 𝐻𝐻2 𝑄𝑄 𝑑𝑑

𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 123 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

𝒮𝒮

TRIANGLES OR ORTHOPOLAR TRIANGLES 𝓢𝓢

The triangles were introduced in geometry by the illustrious Romanian mathematician Traian Lalescu. In this chapter, we will present this notion and some𝒮𝒮 theorems related to it, and establish connections with the orthological triangles.

4.1 triangles: definition, construction, properties

Definition𝓢𝓢 33 We say that the triangle is a triangle in relation to the triangle if these triangles are inscribed in the same circle 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 𝓢𝓢 and if the Simson line of the vertex of the triangle (in 𝑨𝑨𝑨𝑨𝑨𝑨 relation to the triangle ) is perpendicular to the opposite side 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 of that vertex of the triangle . 𝑨𝑨𝑨𝑨𝑨𝑨 𝟏𝟏 𝟏𝟏 𝟏𝟏 Construction of triangles 𝑨𝑨 𝑩𝑩 𝑪𝑪

Being given a triangle𝓢𝓢 inscribed in a circle ( ), we show how another triangle can be built in order to be triangle in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 We present two ways of accomplishing the construction (see Figure 69). 𝐴𝐴𝐴𝐴𝐴𝐴 I. 1. We fix a point on the circle ( ). 2. We build the Simson line - - . 𝐴𝐴1 𝑂𝑂 3. We build the chord ( ) i′n the′ circle′ ( ), perpendicular to 𝐴𝐴 𝐵𝐵 𝐶𝐶 the Simson line . 𝐵𝐵1𝐶𝐶1 𝑂𝑂 The triangle is triangle′ ′ in relation to the triangle . 𝐴𝐴 𝐵𝐵 II. 1. Let us fix the points and on the circumscribed circle of 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴 the triangle . 𝐵𝐵1 𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 125 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

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2. We build the chord ( ), perpendicular to . 3. We build the chord ( ′′ ), perpendicular to . 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 The triangle is triangle ′′in relation to the triangle . 𝐴𝐴 𝐴𝐴1 𝐵𝐵𝐵𝐵 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 C'𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴

A1 A

B A' C

B1 A"

Figure 69 Figure 69 was made to illustrate both constructions above. The construction II is based on the result of Proposition 52.

Observation 50 1. From the construction I, it follows that, being fixed the point on the circumscribed circle of the triangle , we can build an infinity of triangles 𝐴𝐴 to be triangles in relation to . These triangles have the side of 𝐴𝐴𝐴𝐴𝐴𝐴 fixed direction (that of the perpendicular to the Simson line of the point ). 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 We can formulate: 𝐴𝐴1

126 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

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Proposition 60 If the triangle is triangle in relation to the triangle , then any triangle ’ ’ ’ , where ’ ’ is a chord parallel with in the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴 circumscribed circle of the triangle , is triangle in relation to . 𝐴𝐴 1𝐵𝐵 1𝐶𝐶 1 𝐵𝐵 1𝐶𝐶 1 𝐵𝐵𝐵𝐵

2. Both constructions provide an infinity𝐴𝐴𝐴𝐴𝐴𝐴 of 𝒮𝒮 triangles in relation to ABC𝐴𝐴𝐴𝐴𝐴𝐴.

Theorem 19 (Traian Lalescu, 1915) 𝒮𝒮 If the triangle is triangle in relation to the triangle , then: 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴 1. The algebraic sum of measures of the arcs , , considered on the circumscribed circle of the triangle , on 𝐴𝐴𝐴𝐴�1 𝐵𝐵𝐵𝐵�1 𝐶𝐶𝐶𝐶�1 which a positive sense of traversing was set, equals zero. 𝐴𝐴𝐴𝐴𝐴𝐴 2. The Simson lines of the vertices of triangle in relation to the triangle are respectively perpendiculars on the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 opposite sides of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 3. The Simson lines of vertices of the triangle in relation 𝐴𝐴1𝐵𝐵1𝐶𝐶1 to the triangle are concurrent. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 4. The triangle is a triangle in relation to the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 . 𝐴𝐴𝐴𝐴𝐴𝐴 𝒮𝒮 5. The six Simson lines of vertices of the triangles and 𝐴𝐴1𝐵𝐵1𝐶𝐶1 are concurrent in the midpoint of the segment determined by 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the orthocenters of these triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 Proof 1. We refer to Figure 70. Suppose that the trigonometric direction of traversing the arcs was fixed on the circle circumscribed to the triangle . The Simson line of point is perpendicular to . We denote: { } = 𝐴𝐴𝐴𝐴𝐴𝐴 ; we have: (as angles with perpendicular sides), where 𝐴𝐴1 𝐵𝐵1𝐶𝐶1 𝑋𝑋 𝐵𝐵𝐵𝐵 ∩ is the intersection with the circle′′ of the perpendicular to . We have: ′′ 1 1 �1 �1 𝐵𝐵 𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶 ≡ 𝐴𝐴𝐴𝐴 𝐴𝐴 ′ 𝐴𝐴 = + , 1 1 𝐴𝐴 𝐴𝐴 𝐵𝐵𝐵𝐵 𝑚𝑚�𝐶𝐶𝐶𝐶�𝐶𝐶1� =2 �𝑚𝑚�𝐵𝐵𝐵𝐵�1� . 𝑚𝑚�𝐶𝐶�1𝐶𝐶�� 1 ′′ = + 𝑚𝑚From�𝐴𝐴�𝐴𝐴 𝐴𝐴1� 2 𝑚𝑚�𝐴𝐴�1𝐴𝐴� , it follows that: + + = 0. 𝑚𝑚�𝐴𝐴�1𝐴𝐴� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶𝐶𝐶�1� 𝑚𝑚�𝐴𝐴�1𝐴𝐴� 𝑚𝑚�𝐵𝐵�1𝐵𝐵� 𝑚𝑚�𝐶𝐶𝐶𝐶�1� Ion Pătrașcu, Florentin Smarandache 127 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

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2. We prove the Simson line of vertex in relation to the triangle is perpendicular to . 𝐵𝐵1 𝐴𝐴𝐴𝐴𝐴𝐴 We take the perpendicular from to and we denote by its 𝐴𝐴1𝐶𝐶1 intersection with the circle. We join with and we denote by { } = ′′ 𝐵𝐵 𝐴𝐴1𝐶𝐶1 𝐵𝐵 . We have: ′′ 𝐵𝐵 𝐵𝐵1 𝑌𝑌 𝐴𝐴𝐴𝐴 ∩ 1 1 = , 𝐴𝐴 𝐶𝐶 1 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴1� =2 �𝑚𝑚�𝐴𝐴�1𝐴𝐴� −. 𝑚𝑚�𝐶𝐶𝐶𝐶�1�� ′′ 1 1 + 1 + = 0 𝑚𝑚Since�𝐵𝐵�𝐵𝐵 𝐵𝐵 � 2 𝑚𝑚�𝐵𝐵�𝐵𝐵� , it derives that: . �1 �1 �1 𝑚𝑚�𝐴𝐴 𝐴𝐴� ′′𝑚𝑚�𝐵𝐵 𝐵𝐵� 𝑚𝑚�𝐶𝐶 𝐶𝐶� 1 1 ∢𝐴𝐴𝐴𝐴 𝐴𝐴 ≡ ∢ C'𝐵𝐵𝐵𝐵 𝐵𝐵

A A1

M C1 P B"

H B'

B N X A' C

B1 Y

Figure 70 These acute angles, having , they also have , therefore the Simson line of vertex ′′ is parallel with and as such′′ it is 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴1𝑌𝑌 𝐵𝐵1𝐵𝐵 ⊥ 𝐴𝐴𝐴𝐴 perpendicular to . ′′ 𝐵𝐵1 𝐵𝐵𝐵𝐵 Similarly, it is proved that the Simson line of the vertex is perpendicular 𝐴𝐴1𝐶𝐶1 to . 𝐶𝐶1 𝐴𝐴1𝐵𝐵1

128 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Remark 14 Basically the condition + + = 0(mod 3600) is necessary and sufficient as the triangle to be triangle in relation to the 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶�𝐶𝐶1� triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 3. We denote by , , respectively the midpoints of segments , , 𝐴𝐴𝐴𝐴𝐴𝐴 , where is the orthocenter of . From Steiner’s theorem, it follows that the 𝑀𝑀 𝑁𝑁 𝑃𝑃 𝐻𝐻𝐴𝐴1 𝐻𝐻𝐵𝐵1 Simson lines of vertices , , pass respectively through , respectively , 𝐻𝐻𝐶𝐶1 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 and, on the other hand, these Simson lines are perpendicular to , 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑀𝑀 𝑁𝑁 𝑃𝑃 respectively , lines respectively parallel to , and . Consequently, 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 the Simson lines of vertices , , are altitudes in the triangle , therefore 𝐴𝐴1𝐵𝐵1 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀 they are concurrent. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑀𝑀𝑀𝑀𝑀𝑀 4. It derives from + + = 0(mod 3600). 5. The triangle is the homothetic of the triangle by homothety of 𝑚𝑚�𝐴𝐴�1𝐴𝐴� 𝑚𝑚�𝐵𝐵�1𝐵𝐵� 𝑚𝑚�𝐶𝐶�1𝐶𝐶� center and ratio . It follows that its orthocenter is the midpoint1 1 1 of the segment 𝑀𝑀1 𝑀𝑀𝑀𝑀 𝐴𝐴 𝐵𝐵 𝐶𝐶 determined by the homothety center and the homologue point , the orthocenter 𝐻𝐻 2 of the triangle . We saw that the Simson lines of the vertices of triangle 𝐻𝐻 𝐻𝐻1 pass through this point. The relation between the triangles and 𝐴𝐴1𝐵𝐵1𝐶𝐶1 being symmetrical, we have that the Simson lines of the vertices of the triangle 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 in relation to are concurrent in the same point, the midpoint of the segment 𝐴𝐴𝐴𝐴𝐴𝐴 [ ]. 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐻𝐻𝐻𝐻1 We write: 1. and read: the triangle is in “ ” relation with the triangle ABC. ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1𝒮𝒮∆𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝒮𝒮 2. - the set of triangles inscribed in a given circle.

𝒯𝒯∆ 4.2 The relation of equivalence in the set of triangles inscribed in the same circle 𝓢𝓢 Definition 34 We say that the triangle is in relation with the triangle if the triangle is triangle in relation to the triangle 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 𝓢𝓢 . 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 𝓢𝓢 𝑨𝑨𝑨𝑨𝑨𝑨Proposition 61 The relation in the set is an equivalence relation.

𝒮𝒮 𝒯𝒯∆ Ion Pătrașcu, Florentin Smarandache 129 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

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Proof We must prove that the relation has the properties: reflexivity, symmetry and transitivity. 𝒮𝒮 Reflexivity Whatever the triangle from the set , we have: . Indeed, the Simson line of vertex in relation to is the altitude from of 𝐴𝐴𝐴𝐴𝐴𝐴 𝒯𝒯∆ ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1𝒮𝒮∆𝐴𝐴𝐴𝐴𝐴𝐴 the triangle , this being perpendicular to ; we obtain that is 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 triangle in relation to itself. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝒮𝒮 Another proof: + + = 0 (mod 360 ). Therefore: . 0 𝑚𝑚�𝐴𝐴𝐴𝐴� � 𝑚𝑚�𝐵𝐵𝐵𝐵� � 𝑚𝑚�𝐶𝐶𝐶𝐶� � Symmetry ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴𝐴𝐴𝐴𝐴 If , it has been shown in T. Lalescu’s theorem that , therefore the “ ” relation is symmetrical. ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1𝒮𝒮∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴𝐴𝐴𝐴𝐴Transitivity𝒮𝒮∆𝐴𝐴1𝐵𝐵1𝐶𝐶 1 𝒮𝒮 In the set , we consider the triangles , , such that: and . 𝒯𝒯∆ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 Let us prove that: . ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1𝒮𝒮∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 From , we have that: ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 + + = 0 (mod 360 ). ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 From , we have that: 0 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶�𝐶𝐶1� + + = 0 (mod 360 ). 𝐴𝐴1𝐵𝐵1𝐶𝐶1𝒮𝒮∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 Adding member by member the previous relations0, we obtain that: 𝑚𝑚�𝐴𝐴�1𝐴𝐴2� 𝑚𝑚�𝐵𝐵�1𝐵𝐵2� 𝑚𝑚�𝐶𝐶�1𝐶𝐶2� + + = 0 (mod 360 ), hence: . 0 𝑚𝑚�𝐴𝐴�𝐴𝐴2� 𝑚𝑚�𝐵𝐵�𝐵𝐵2� 𝑚𝑚�𝐶𝐶�𝐶𝐶2� If is a fixed triangle from , we define the set: ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 = / . 𝐴𝐴𝐴𝐴𝐴𝐴 𝒯𝒯∆ The∆𝐴𝐴𝐴𝐴𝐴𝐴 set ′ is′ a ′modulo “ ” equivalence′ ′ ′ class of set . 𝒮𝒮 �∆𝐴𝐴 𝐵𝐵 𝐶𝐶 ∈ 𝒯𝒯∆ ∆𝐴𝐴𝐴𝐴𝐴𝐴𝒮𝒮∆𝐴𝐴 𝐵𝐵 𝐶𝐶 � ∆𝐴𝐴𝐴𝐴𝐴𝐴 𝒮𝒮 𝒮𝒮 𝒯𝒯∆ Proposition 62 The Simson line of a point that belongs to the circle circumscribed to the triangles of the same equivalence class has the fixed direction in relation to these triangles.

130 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

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Proof Let be a point on the circumscribed circle of the triangle , and – a line determining together with a triangle in relation to . Because the 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 Simson line of in relation to , or any other triangle from , is 𝑀𝑀 𝒮𝒮 𝐴𝐴𝐴𝐴𝐴𝐴 perpendicular to , it will have a fixed direction. 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝒮𝒮∆𝐴𝐴𝐴𝐴𝐴𝐴 Remark 15 𝑑𝑑 The denomination of triangles of the orthopolar triangles is justified by the fact that two triangles in relation to a line have the same orthopole. 𝒮𝒮 Indeed, let be a triangle and – a line. The projections of vertices of 𝒮𝒮 triangle on are , , ; and the perpendiculars taken from , , to 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 , respectively are concurrent with , the orthopole of line in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 . This point is the intersection of the Simson lines of the intersection points 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑂𝑂 𝑑𝑑 with the circle (of line ) circumscribed to the triangle . Basically, these 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 perpendiculars are Simson lines of the triangle in relation to the triangle that 𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴 has the line d as one of its sides. 𝒮𝒮 𝒮𝒮 Proposition 63 In two triangles, the orthopoles of one’s sides in relation to the other coincide with the midpoint of the segment determined by its orthocenters. 𝒮𝒮 Proof The orthopole of a line in relation to a scalene triangle is the intersection point of Simson lines, of the intersection points of line with the circle. From 𝑂𝑂 𝑑𝑑 T. Lalescu’s theorem, the six Simson lines of the vertices of a triangle in relation to the other triangle are concurrent lines in the midpoint𝑑𝑑 of the segment determined by its orthocenters.

4.3 Simultaneously orthological and orthopolar triangles

Lemma 5 If the triangles and are inscribed in the same circle, with , , , then: and are antiparallels in 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 relation to and ; , and – antiparallels in relation to and 𝐴𝐴𝐴𝐴1 ∥ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵1 ∥ 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶1 ∥ 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 ; and – antiparallels in relation to and . 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵

Ion Pătrașcu, Florentin Smarandache 131 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

𝒮𝒮

Proof and form the inscribed quadrilateral , therefore they are antiparallels in relation to and ; since and , we have 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵1𝐶𝐶𝐶𝐶1 that and are antiparallels in relation to and . Similarly, the other 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝐵𝐵𝐵𝐵1 ∥ 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶1 ∥ 𝐴𝐴𝐴𝐴 requirements are proved. 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 Theorem 19 If the triangles and inscribed in the same circle have , , , then the triangles are 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 simultaneously orthological and orthopolar. 𝐴𝐴𝐴𝐴1 ∥ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵1 ∥ 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶1 ∥ 𝐴𝐴𝐴𝐴 Proof First of all, we prove that the triangles and are orthological. Indeed, the perpendicular from to , which is antiparallel with , passes 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 through – the center of the circle circumscribed to the triangle 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵 (Proposition 4). Similarly, the perpendiculars from and to 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 respectively pass through , therefore is an orthology center. We prove 𝐵𝐵 𝐶𝐶 𝐶𝐶1𝐴𝐴1 that the triangles and are triangles. We show that: 𝐴𝐴1𝐵𝐵1 𝑂𝑂 𝑂𝑂 + + 1 1 1 = 0 (mod360 ). (1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝒮𝒮 0 �1 �1 �1 𝑚𝑚�𝐴𝐴𝐴𝐴 � 𝑚𝑚�𝐵𝐵𝐵𝐵 � 𝑚𝑚 � 𝐶𝐶 A𝐶𝐶 � A1

B C

Figure 71

132 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

TRIANGLES OR ORTHOPOLAR TRIANGLES

𝒮𝒮

We use Figure 71; we have that the quadrilateral is an isosceles trapezoid, therefore: 𝐴𝐴𝐴𝐴1𝐶𝐶𝐶𝐶 = 360 4 2 . (2) The quadrilateral0 is also an isosceles trapezoid; we have: 𝑚𝑚�𝐴𝐴�𝐴𝐴1� − 𝑚𝑚�𝐶𝐶̂� − 𝑚𝑚�𝐴𝐴̂� = 360 4 2 . (3) 𝐵𝐵𝐵𝐵1𝐶𝐶𝐶𝐶 The quadrilateral0 is an isosceles trapezoid; it follows that: 𝑚𝑚�𝐵𝐵�𝐵𝐵1� − 𝑚𝑚�𝐶𝐶̂� − 𝑚𝑚�𝐵𝐵�� = 360 4 2 . (4) 𝐶𝐶𝐶𝐶1𝐴𝐴𝐴𝐴 Taking into account0 that, in the relation (1), the arcs must be transversed in 𝑚𝑚�𝐶𝐶�𝐶𝐶1� − 𝑚𝑚�𝐴𝐴̂� − 𝑚𝑚�𝐶𝐶̂� the same direction, we have that: = 4 + 2 . Then: 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐶𝐶̂� 𝑚𝑚�𝐴𝐴̂� + + = 4 + 2 + 360 4 2 + 360 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶�𝐶𝐶1� 4 2 . 0 0 𝑚𝑚�𝐶𝐶̂� 𝑚𝑚�𝐴𝐴̂� − 𝑚𝑚�𝐶𝐶̂� − 𝑚𝑚�𝐵𝐵�� It follows that: − 𝑚𝑚�𝐴𝐴̂� − 𝑚𝑚�𝐶𝐶̂� + + = 720 2 + + . Therefore: 0 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶�𝐶𝐶1� − �𝑚𝑚�𝐴𝐴̂� 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐶𝐶̂�� + + = 360 , and, consequently, the relation (1) is true0; therefore the triangles and 𝑚𝑚�𝐴𝐴�𝐴𝐴1� 𝑚𝑚�𝐵𝐵�𝐵𝐵1� 𝑚𝑚�𝐶𝐶�𝐶𝐶1� are triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 𝐴𝐴 𝐵𝐵Proposition𝐶𝐶 𝒮𝒮 64 The median triangle and the orthic triangle of a given non-right triangle are simultaneously orthopolar and orthological triangles. Proof of this property follows as a consequence of Theorem 5. The fact that the median triangle and the orthic triangle are orthological was established in Proposition 9.

Proposition 65 A given triangle and the triangle determined by the intesections of the exterior bisectors of the angles , , with the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴0𝐵𝐵0𝐶𝐶0 circumscribed circle of the triangle are simultaneously orthopolar and 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthological triangles. Proof of this property derives from the fact that the 𝐴𝐴𝐴𝐴𝐴𝐴 triangle and the triangle are respectively the orthic triangle and the median triangle in the antisupplementary triangle of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴0𝐵𝐵0𝐶𝐶0 triangle . 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 133 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

In this chapter, we will prove some important theorems regarding the orthological triangles with common orthology center, and we will address some issues related to the biorthological triangles.

5.1 Theorems regarding the orthological triangles with the same orthology center

Theorem 20 Two orthological triangles with common orthology center are homological triangles. In the proof of this theorem, we will use:

Theorem 21 (N. Dergiades, 2003) Let ( , ), ( , ), ( , ) be three circles that pass respectively through vertices and , and , and of a 𝒞𝒞1 𝑂𝑂1 𝑅𝑅1 𝒞𝒞2 𝑂𝑂2 𝑅𝑅2 𝒞𝒞3 𝑂𝑂3 𝑅𝑅3 triangle . We denote by , , the second intersection point 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 respectively between ( ) and ( ), ( ) and ( ), ( ) and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷 𝐸𝐸 𝐹𝐹 ( ). The perpendiculars taken from the points , , to , 𝒞𝒞2 𝒞𝒞3 𝒞𝒞3 𝒞𝒞1 𝒞𝒞1 respectively intersect the sides , , in the points , , 𝒞𝒞1 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 . The points , and are collinear. 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑋𝑋 𝑌𝑌 Proof𝑍𝑍 (Ion Pătrașcu)𝑋𝑋 𝑌𝑌 𝑍𝑍 Let be the median triangle of the triangle (see Figure 72). In the proof,1 w2e use3 Menelaus’s reciprocal theorem. We 𝐴𝐴have𝐵𝐵 :𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 ( ) ( ) ( ) = = = ( ) ( ) ( ). 𝑋𝑋𝑋𝑋 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑋𝑋𝑋𝑋𝑋𝑋 𝐷𝐷𝐷𝐷∙sin 𝑋𝑋𝑋𝑋𝑋𝑋 𝐷𝐷𝐷𝐷∙cos 𝐴𝐴𝐴𝐴𝐴𝐴 𝑋𝑋𝑋𝑋 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑋𝑋𝑋𝑋𝑋𝑋 𝐷𝐷𝐷𝐷∙sin 𝑋𝑋𝑋𝑋𝑋𝑋 𝐷𝐷𝐷𝐷∙cos 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 135 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Similarly, we find: ( ) ( ) = = ( ), ( ). 𝑌𝑌𝑌𝑌 𝐸𝐸𝐸𝐸 cos 𝐵𝐵𝐵𝐵𝐵𝐵 𝑍𝑍𝑍𝑍 𝐹𝐹𝐹𝐹 cos 𝐶𝐶𝐶𝐶𝐶𝐶 From𝑌𝑌𝑌𝑌 𝐸𝐸𝐸𝐸the∙ inscribedcos 𝐵𝐵𝐵𝐵𝐵𝐵 quadrilaterals𝑍𝑍𝑍𝑍 𝐹𝐹𝐹𝐹 ∙ cos 𝐶𝐶𝐶𝐶𝐶𝐶 , and , we note that: = , = , = . 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 Z ∢𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝐶𝐶𝐶𝐶𝐶𝐶 ∢𝐶𝐶𝐶𝐶𝐶𝐶 ∢𝐴𝐴𝐴𝐴𝐴𝐴

O3 C1 B1 D P E F

X B A1 C

Figure 72

136 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Consequently: = . (1) 𝑋𝑋𝑋𝑋 𝑌𝑌𝑌𝑌 𝑍𝑍𝑍𝑍 𝐷𝐷𝐷𝐷 𝐸𝐸𝐸𝐸 𝐹𝐹𝐹𝐹 On the other hand: 𝑋𝑋𝑋𝑋 ∙ 𝑌𝑌𝑌𝑌 ∙ 𝑍𝑍𝑍𝑍 𝐷𝐷𝐷𝐷 ∙ 𝐸𝐸𝐸𝐸 ∙ 𝐹𝐹𝐹𝐹 = 2 sin( ), = 2 sin( ), 𝐷𝐷𝐷𝐷 𝑅𝑅3 𝐵𝐵𝐵𝐵𝐵𝐵 = 2 sin( ), 𝐸𝐸𝐸𝐸 𝑅𝑅3 𝐴𝐴𝐴𝐴𝐴𝐴 = 2 sin( ), 𝐷𝐷𝐷𝐷 𝑅𝑅2 𝐶𝐶𝐶𝐶𝐶𝐶 = 2 sin( ), 𝐹𝐹𝐹𝐹 𝑅𝑅2 𝐴𝐴𝐴𝐴𝐴𝐴 = 2 sin( ). 𝐹𝐹𝐹𝐹 𝑅𝑅1 𝐵𝐵𝐵𝐵𝐵𝐵 Coming back1 to the relation (1), we obtain: 𝐸𝐸𝐸𝐸 𝑅𝑅 𝐶𝐶𝐶𝐶𝐶𝐶( ) ( ) ( ) = ( ) ( ) ( ). (2) 𝑋𝑋𝑋𝑋 𝑌𝑌𝑌𝑌 𝑍𝑍𝑍𝑍 sin 𝐵𝐵𝐵𝐵𝐵𝐵 sin 𝐶𝐶𝐶𝐶𝐶𝐶 sin 𝐴𝐴𝐴𝐴𝐴𝐴 According𝑋𝑋𝑋𝑋 ∙ 𝑌𝑌𝑌𝑌 ∙ 𝑍𝑍𝑍𝑍 to Carnot’sin 𝐶𝐶𝐶𝐶𝐶𝐶 s∙ tsinheorem,𝐴𝐴𝐴𝐴𝐴𝐴 ∙ thesin 𝐵𝐵common𝐵𝐵𝐵𝐵 chords of circles , , are concurrent, which means that = { } ( is the radical center of 𝒞𝒞1 𝒞𝒞2 𝒞𝒞3 these circles). 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 ∩ 𝐶𝐶𝐶𝐶 𝑃𝑃 𝑃𝑃 The cevians , , being concurrent, we can write the trigonometric form of Ceva’s theorem, from where we find: ( ) (𝐴𝐴𝐴𝐴) 𝐵𝐵𝐵𝐵 ( 𝐶𝐶𝐶𝐶) = 1 ( ) ( ) ( ) . sin 𝐵𝐵𝐵𝐵𝐵𝐵 sin 𝐶𝐶𝐶𝐶𝐶𝐶 sin 𝐴𝐴𝐴𝐴𝐴𝐴 Bysin 𝐶𝐶this𝐶𝐶𝐶𝐶 relation∙ sin 𝐴𝐴𝐴𝐴𝐴𝐴, from∙ sin 𝐵𝐵(2)𝐵𝐵𝐵𝐵 we obtain that: = 1, 𝑋𝑋𝑋𝑋 𝑌𝑌𝑌𝑌 𝑍𝑍𝑍𝑍 which shows the collinearity of the points , , . 𝑋𝑋𝑋𝑋 ∙ 𝑌𝑌𝑌𝑌 ∙ 𝑍𝑍𝑍𝑍 To prove Theorem 20, we need: 𝑋𝑋 𝑌𝑌 𝑍𝑍 Lemma 6 Let and be two orthological triangles with the same orthology center, O. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴′𝐵𝐵′𝐶𝐶′ If and are the orthogonal projections of vertices and on the support lines of the sides [ ] and respectively [ ], then the points , 𝐸𝐸 𝐹𝐹 𝐵𝐵 𝐶𝐶 , and are four concyclic′ points′ . ′ ′ 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐸𝐸Proof 𝐹𝐹1 Let be the common orthology center of the given triangle; we denote by , the projections of and on respectively on ; also, we denote: 𝑂𝑂 { } = , { } = ′ ′ (see Figure 73). ′ ′ 𝐸𝐸 ′′𝐹𝐹 ′ ′ ′′𝐵𝐵 𝐶𝐶 𝐴𝐴′ 𝐶𝐶′ 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐸𝐸𝐸𝐸 ∩ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐹𝐹𝐹𝐹 ∩ 𝐴𝐴 𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 137 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

C" D B"

C' O

F E B A" C

Figure 73 is the orthocenter in the triangle . It consequently follows that . Because is antiparal′lel′′ with′′ , it follows from 𝑂𝑂 𝐴𝐴 𝐵𝐵 𝐶𝐶 Proposition′ ′′ ′′4 that is antiparallel with , which shows′′ ′′ that the quadrilateral 𝐴𝐴 𝑂𝑂 ⊥ 𝐵𝐵 𝐶𝐶 𝐸𝐸𝐸𝐸 𝐵𝐵 𝐶𝐶 is inscribable. 𝐸𝐸𝐸𝐸 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵Observation 51 Similarly, if is the projection of on , it follows that the points , , , and , , , are concyclic. 𝐷𝐷 𝐴𝐴 𝐵𝐵′𝐶𝐶′ 𝐴𝐴 𝐷𝐷 𝐹𝐹 𝐶𝐶 Proof𝐴𝐴 𝐷𝐷 𝐸𝐸 2 𝐵𝐵(Ion Pătrașcu) We denote: { } = . The quadrilaterals , are inscribable. The power′′ of′ the point over the circles circumscribed′′ ′ ′ ′′to these 𝐴𝐴 𝐴𝐴 𝑂𝑂 ∩ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐹𝐹𝐹𝐹 quadrilaterals provides the relations: = and = 𝑂𝑂 . We get from here that: ′′= ′ , accordingly the′′ points′ , 𝑂𝑂𝐴𝐴 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐴𝐴 ∙ 𝑂𝑂𝑂𝑂 , , are concyclic. 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝐵𝐵 𝐶𝐶 𝐹𝐹 𝐸𝐸 138 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Proof 3 (Mihai Miculița) Denoting { } = = ( ) (see Figure 73), we have: ′′ ′ ′ inscribable𝐵𝐵𝐵𝐵 𝐴𝐴 ′ ′′ 𝐴𝐴 𝑂𝑂 ∩ 𝐵𝐵𝐵𝐵 𝑝𝑝𝑟𝑟 𝐴𝐴 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴 𝐶𝐶 inscribable ′ ′ 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 − ⇒ ∢OEF ≡ ∢O𝐴𝐴 F 𝐴𝐴 𝑂𝑂 ⊥ 𝐵𝐵𝐵𝐵 � ⇒ � ′′ ′ inscribable ′ ′ ′ 𝐴𝐴 𝐴𝐴 𝐶𝐶𝐶𝐶 − . ⇒ ∢O𝐴𝐴 F ≡ ∢BCO 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴 𝐵𝐵 ⇒Proof∢OEF of T≡heorem∢𝐵𝐵𝐵𝐵𝐵𝐵 20⇒ 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 − We use the configuration from Figure 73. The quadrilaterals , , being inscribable, we observe that their circumscribed circles satisfy the 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 hypotheses in Theorem 21. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 Applying this theorem, it follows that the lines and , and , and intersect respectively in the points , , , which are collinear. 𝐵𝐵′𝐶𝐶′ 𝐵𝐵𝐵𝐵 𝐶𝐶′𝐴𝐴′ 𝐶𝐶𝐶𝐶 Using Desargues’s theorem (see [24]), it follows that , and are 𝐴𝐴′𝐵𝐵′ 𝐴𝐴𝐴𝐴 𝑋𝑋 𝑌𝑌 𝑍𝑍 concurrent and, hence, the triangles and ’ ’ ’ are homological. 𝐴𝐴𝐴𝐴′ 𝐵𝐵𝐵𝐵′ 𝐶𝐶𝐶𝐶′ Remark 16 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 a) The triangles (formed by the centers of the circles circumscribed to the quadrilaterals , respectively ) 𝑂𝑂1𝑂𝑂2𝑂𝑂3 and are orthological. The orthology centers are the points – the 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 radical center of the circles of centers , , ; and – the center 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 of the circle circumscribed to . 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝑂𝑂 b) The triangles and (formed by the projections of vertices 𝐴𝐴𝐴𝐴𝐴𝐴 of triangles on the sides of ’ ’ ’) are orthological. The 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝐷𝐷𝐷𝐷𝐷𝐷 orthology centers are: the center of the circle circumscribed to the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 triangle ; and – the center of the radical circle of the circles of centers , , . Indeed, the perpendiculars taken from , , 𝐷𝐷𝐷𝐷𝐷𝐷 𝑃𝑃 to , respectively are mediators of these segments, therefore 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 they are concurrent in the center of the circle circumscribed to the 𝐸𝐸𝐸𝐸 𝐹𝐹𝐹𝐹 𝐷𝐷𝐷𝐷 triangle , and the perpendiculars taken from , , to the sides of the triangle , being the common chords , , , they 𝐷𝐷𝐷𝐷𝐷𝐷 𝐷𝐷 𝐸𝐸 𝐹𝐹 will be concurrent in the point . 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 Theorem 22 𝑃𝑃 If is a point in the interior of the given triangle , is its podal triangle, and the points ’, ’, ’ are such that: 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 139 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

i. , ’ are collinear vectors; , ’ are collinear vectors; , ’ are collinear vectors; 𝑂𝑂��𝐴𝐴��1⃗ �𝑂𝑂��𝑂𝑂��⃗ 𝑂𝑂��𝐵𝐵��1⃗ �𝑂𝑂��𝑂𝑂��⃗ ii. ’ = ’ = ’, so the triangles 𝑂𝑂��𝐶𝐶��1⃗ �𝑂𝑂��𝑂𝑂��⃗ and ’ ’ ’ are orthological with – common orthology 𝑂𝑂��𝐴𝐴��1⃗ ∙ �𝑂𝑂��𝑂𝑂��⃗ 𝑂𝑂��𝐵𝐵��1⃗ ∙ �𝑂𝑂��𝑂𝑂��⃗ 𝑂𝑂��𝐶𝐶��1⃗ ∙ �𝑂𝑂��𝑂𝑂��⃗ 𝐴𝐴𝐴𝐴𝐴𝐴 center. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 Proof We build the circle of diameter and we denote by its second point of intersection with . We obtain (see Figure′ 74) that (1) and 𝐴𝐴𝐶𝐶 𝐷𝐷 = (2). Because = (3),′ from relations i) and 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐷𝐷 ⊥ 𝐴𝐴𝐴𝐴 𝑂𝑂𝐶𝐶1 ∙ ii), ′we have = . This′ relation shows′ that the points , , 𝑂𝑂𝐶𝐶 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐶𝐶1 ∙ 𝑂𝑂𝐶𝐶 𝑂𝑂𝐵𝐵1 ∙ 𝑂𝑂𝐵𝐵 , are concyclic. ′ ′ 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐵𝐵1 ∙ 𝑂𝑂𝐵𝐵 𝐵𝐵1 𝐵𝐵 𝐴𝐴 𝐷𝐷 A B' D B1 C' C1 P

B A1 C

Figure 74 The angle is right; it follows that is also right, hence (4). The relations′ (1) and (4) show that the points′ , , are collinear′ , 𝐴𝐴𝐵𝐵1𝐵𝐵 ∢𝐴𝐴𝐴𝐴𝐵𝐵 𝐵𝐵 𝐷𝐷 ⊥ and . Similarly, we deduce that and′ ′ , therefore 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐷𝐷 𝐶𝐶 the triangles′ ′ and are orthological, of ′common′ center′ .′ 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵 𝐶𝐶 ′ ′ ′ 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴 𝐶𝐶 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴 𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 140 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Observation 52 The reciprocal of Theorem 22 is also true, and its proof is made without difficulty.

Proposition 66 If and ’ ’ ’ are two orthological triangles of common orthology center , and of homology with the homology axis , then , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 are respectively perpendiculars to ’, ’, ’. 𝑂𝑂 𝑋𝑋𝑋𝑋𝑋𝑋 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 Y 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶

B Y' 1 C' C 1 O X' F E X B A1 C

Figure 75

Ion Pătrașcu, Florentin Smarandache 141 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Proof We denote by , , the orthogonal projections of on the sides , respectively ,′ and′ by ′ , , – the orthogonal projections of on , 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝑂𝑂 𝐵𝐵𝐵𝐵 , respectively . ′ ′ 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝑂𝑂 𝐵𝐵 𝐶𝐶 ′ Applying′ Lemma′ 6′, it follows that the points , , , are concyclic 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 (see Figure 75). ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐸𝐸 𝐷𝐷 Considering the power of the point to the circle of the previous points, we write: 𝑂𝑂 = . (1) The′ points′ and ′ are′ on the circle of diameter ; considering the power 𝑂𝑂𝐷𝐷 ∙ 𝑂𝑂𝐴𝐴 𝑂𝑂𝐸𝐸 ∙ 𝑂𝑂𝐵𝐵 of the point over this′ circle, it follows that: ′ 𝐹𝐹 𝐸𝐸 𝐶𝐶𝐵𝐵 = . (2) 𝑂𝑂 Also from Lemma′ 6,′ we note that the points , , , are concyclic; the 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐸𝐸 ∙ 𝑂𝑂𝐵𝐵 power of over the circle of these points implies that: 𝐴𝐴 𝐶𝐶 𝐹𝐹 𝐷𝐷 = . (3) 𝑂𝑂 From the relations (1), (2), (3), we obtain: 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 ’ ’ = . (4) The relation (4) shows that the points , ’, ’, are concyclic, therefore: 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 ’ ’ . (5) 𝐴𝐴 𝐴𝐴 𝐷𝐷 𝐷𝐷 The points , , , ’ belong to the circle of diameter ( ), therefore we ∢𝐷𝐷𝐷𝐷 𝑂𝑂 ≡ ∢𝐴𝐴 𝐴𝐴𝐴𝐴 have: 𝑂𝑂 𝐷𝐷 𝑋𝑋 𝐷𝐷 𝑂𝑂𝑂𝑂 ’ . (6) The relations (5) and (6) lead to: ∢𝐷𝐷𝐷𝐷𝐷𝐷 ≡ ∢𝐷𝐷 𝐷𝐷𝐷𝐷 ’ ’ . (7) This relation, together with ’ ’ and with the reciprocal theorem of ∢𝐷𝐷 𝐷𝐷𝐷𝐷 ≡ ∢𝐴𝐴 𝐴𝐴𝐴𝐴 the angles with the perpendicular sides, we obtain that ’ . Similarly, we 𝑂𝑂𝑂𝑂 ⊥ 𝐵𝐵 𝐶𝐶 prove that ’ and ’ . 𝐴𝐴𝐴𝐴 ⊥ 𝑂𝑂𝑂𝑂 Proposition𝐵𝐵𝐵𝐵 ⊥ 67𝑂𝑂 𝑂𝑂 𝐶𝐶𝐶𝐶 ⊥ 𝑂𝑂𝑂𝑂 If the triangles and ’ ’ ’ are orthological, of common center , and homological with and – respectively their center of homology and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 the axis of homology; then: . 𝑃𝑃 𝑋𝑋𝑋𝑋 Proof 𝑂𝑂𝑂𝑂 ⊥ 𝑋𝑋𝑋𝑋 From the previous Proposition, we remember that ’, ’, ’. 𝑂𝑂𝑂𝑂 ⊥ 𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂 ⊥ 𝐵𝐵𝐵𝐵 𝑂𝑂𝑂𝑂 ⊥ 𝐶𝐶𝐶𝐶 142 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

We denote { ’} = ’, { ’} = ’, { ’} = ’ (see Figure 75). The quadrilateral ’ ’ ’ is inscriptable; considering the power of 𝑋𝑋 𝑂𝑂𝑂𝑂 ∩ 𝐴𝐴𝐴𝐴 𝑌𝑌 𝑂𝑂𝑂𝑂 ∩ 𝐵𝐵𝐵𝐵 𝑍𝑍 𝑂𝑂𝑂𝑂 ∩ 𝐶𝐶𝐶𝐶 point over its circumscribed circle, we write: 𝑋𝑋𝑋𝑋 𝑂𝑂 𝐴𝐴 ’ = ’ ’. (1) 𝑂𝑂 On the other hand: 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝐴𝐴 ’ = ’. The points ’, , ’, are located on the circle of diameter ’; writing 𝑂𝑂𝐴𝐴1 ∙ 𝑂𝑂𝐴𝐴 𝑂𝑂𝐵𝐵1 ∙ 𝑂𝑂𝐵𝐵 the power of over this circle, it turns out that: 𝐵𝐵 𝐵𝐵1 𝑌𝑌 𝑌𝑌 𝑌𝑌𝑌𝑌 ’ = ’ . (2) 𝑂𝑂 The relations (1), (2), (3) lead to: 𝑂𝑂𝐵𝐵1 ∙ 𝑂𝑂𝐵𝐵 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑌𝑌 ’ = ’ , (3) relation showing that the points ’, , , ’ are concyclic, hence ’ ’ is 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ∙ 𝑂𝑂𝑌𝑌 antiparallel with in relation to and . Also, ’ ’ is antiparallel with the 𝑋𝑋 𝑋𝑋 𝑌𝑌 𝑌𝑌 𝑋𝑋 𝑌𝑌 tangent taken in to the circle of diameter . Consequently, the tangent to 𝑋𝑋𝑋𝑋 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 𝑋𝑋 𝑌𝑌 the circle is parallel with , and since is perpendicular to the tangent, we 𝑂𝑂 𝑂𝑂𝑂𝑂 have that . 𝑋𝑋𝑋𝑋 𝑂𝑂𝑂𝑂 5.2 Reciprocal𝑂𝑂𝑂𝑂 ⊥ 𝑋𝑋𝑋𝑋 polar triangles

Definition 34 Two triangles are called reciprocal polar in relation to a given circle if the sides of one are the other’s vertices polars with respect to a circle. The circle against which the two triangles are reciprocal polars is called director circle.

Theorem 23 Two reciprocal polar triangles in relation to a given circle are orthological triangles, having the center of the circle as common orthology center.

Proof Let and ’ ’ ’ be two reciprocal polar triangles in relation to the director circle of center (see Figure 76). Because the polar of , videlicet 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 ’ ’, is perpendicular to the line determined by the point and by the circle’s 𝑂𝑂 𝐴𝐴 center , we have that ’ ’, similarly ’ ’ and ’ ’. 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝑂𝑂 𝑂𝑂𝑂𝑂 ⊥ 𝐵𝐵 𝐶𝐶 𝑂𝑂𝑂𝑂 ⊥ 𝐴𝐴 𝐶𝐶 𝑂𝑂𝑂𝑂 ⊥ 𝐴𝐴 𝐵𝐵 Ion Pătrașcu, Florentin Smarandache 143 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Also, the polar of ’ with respect to the circle, videlicet , is perpendicular to ’ , and similarly ’ and ’ , therefore and ’ ’ ’ are 𝐴𝐴 𝐵𝐵𝐵𝐵 orthological triangles with the point – common orthology center. 𝐴𝐴 𝑂𝑂 𝐵𝐵 𝑂𝑂 ⊥ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝑂𝑂 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 The points , , are the inverses of points , , – with respect to the 𝑂𝑂 inversion circle, with as center. 𝐴𝐴0 𝐵𝐵0 𝐶𝐶0 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 B'

A0 C'

C0 B0 B C

Figure 76

144 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Remark 17 A triangle and its tangential triangle are reciprocal polar triangles in relation to the circumscribed circle of the triangle , therefore they are 𝐴𝐴𝐴𝐴𝐴𝐴 orthological triangles with the common orthology center in the circle circumscribed 𝐴𝐴𝐴𝐴𝐴𝐴 to the triangle .

𝐴𝐴𝐴𝐴𝐴𝐴 5.3 Other remarkable orthological triangles with the same orthology center In the Second Chapter, we presented a few pairs of orthological triangles with only one orthology center, namely a given triangle and its contact triangle; the given triangle and its tangent triangle. In the following, we will investigate another pair of orthological triangles that have the same orthology center.

Oc O

B C

Figure 77

Ion Pătrașcu, Florentin Smarandache 145 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Definition 35 Being given a non-right triangle , the triangle with the vertices in the centers of the circles circumscribed to the triangles 𝑨𝑨𝑨𝑨𝑨𝑨 , , , where is the center of the circle circumscribed to the triangle – is called Coșniță triangle. 𝑶𝑶𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶𝑶𝑶 𝑶𝑶𝑶𝑶𝑶𝑶 𝑶𝑶 Observation 53 𝑨𝑨𝑨𝑨𝑨𝑨 In Figure 77, the Coșniță triangle of the triangle was denoted .

𝑎𝑎 𝑏𝑏 𝑐𝑐 Proposition 68 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 𝑂𝑂 A non-right given triangle and the Coșniță triangle are orthological triangles with the common orthology center – the center of the circle circumscribed to the given triangle. 𝑂𝑂 Proof Obviously, the perpendiculars taken from , , to , , are the mediators of the triangle , therefore are concurrent in . On the other hand, 𝑂𝑂𝑎𝑎 𝑂𝑂𝑏𝑏 𝑂𝑂𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 is a common chord in the circles circumscribed to triangles and , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 hence . Similarly, it follows that the perpendiculars from and 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 respectively to and pass through the point . 𝐴𝐴𝐴𝐴 ⊥ 𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 𝐵𝐵 𝐶𝐶 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 Remark 18𝑂𝑂 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝑂𝑂 1. In [24], we proved that the triangles and are homological (with the help of Ceva’s theorem); this result can now be obtained 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 using Theorem 20. We recall that the homology center is called Coșniță point and it is the isogonal conjugate of the center of the circle of nine points; also, the homology axis is called Coșniță line. 2. If the triangle is acute; is the podal triangle of the center of the circumscribed circle ; and is Coșniță triangle, – it can 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 be shown that: 𝑂𝑂 𝑂𝑂𝑎𝑎𝑂𝑂𝑏𝑏𝑂𝑂𝑐𝑐 = = = . 1 2 1 𝑎𝑎 1 𝑏𝑏 1 𝑐𝑐 𝑂𝑂𝐴𝐴 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐵𝐵 ∙ 𝑂𝑂𝑂𝑂 𝑂𝑂𝐶𝐶 ∙ 𝑂𝑂𝑂𝑂 2 𝑅𝑅 If we consider ’, ’, ’ belonging respectively to the lines , , , such that ’ = ’ = ’, then, from Theorem 22, we obtain 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝑂𝑂𝑎𝑎 𝐵𝐵1𝑂𝑂𝑏𝑏 𝐶𝐶1𝑂𝑂𝑐𝑐 that the triangles and ’ ’ ’ are orthological. From Theorem 20, we see that 𝑂𝑂��𝐴𝐴��1⃗ ∙ 𝑂𝑂��𝑂𝑂��⃗ 𝑂𝑂��𝐵𝐵��1⃗ ∙ 𝑂𝑂��𝑂𝑂��⃗ 𝑂𝑂��𝐶𝐶��1⃗ ∙ 𝑂𝑂��𝑂𝑂��⃗ these triangles are homological. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶

146 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

In [21], we called the homology center of these triangles – the generalized Coșniță point, and the homology axis of these triangles is called – the generalized Coșniță line. We call the triangle ’ ’ ’ – the generalized Coșniță triangle. Considering Proposition 68, we can state: 𝐴𝐴 𝐵𝐵 𝐶𝐶 The line determined by the center of the circumscribed circle of a triangle and the generalized Coșniță point is perpendicular to the generalized Coșniță line.

5.4. Biorthological triangle

Definition 36 If the triangle is orthological with the triangle and with the triangle , we say that the triangles and 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 are biorthological. 𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏𝑨𝑨𝟏𝟏 𝑨𝑨𝑨𝑨𝑨𝑨 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝑨𝑨Observation𝑩𝑩 𝑪𝑪 54 In Figure 78, the triangles and are biorthological. We denoted by the orthology center of the triangle in relation to the triangle and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 with – the orthology center of the triangle in relation to the triangle . 𝑂𝑂1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 2 1 1 1 Theorem𝑂𝑂 24 (A. Pantazi, 1896 – 1948)𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 If the triangle is simultaneously orthological in relation to the triangles and , then the triangle is 𝐴𝐴𝐴𝐴𝐴𝐴 orthological in the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1𝐴𝐴1 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Proof 1 𝐶𝐶 𝐴𝐴 𝐵𝐵 The triangles and being orthological, we have: + + = 0. (1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 The2 triangles2 2 and 2 being2 ort2hological, we can write: 𝐴𝐴𝐵𝐵1 − 𝐴𝐴𝐶𝐶1 𝐵𝐵𝐶𝐶1 − 𝐵𝐵𝐴𝐴1 𝐶𝐶𝐴𝐴1 − 𝐶𝐶𝐵𝐵1 + + = 0. (2) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1𝐴𝐴1 Adding2 member2 by2 member2 the relations2 2 (1) and (2), we obtain: 𝐴𝐴𝐶𝐶1 − 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐴𝐴1 − 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐵𝐵1 − 𝐶𝐶𝐶𝐶1 + + = 0. (3) The2 relation2 (3) expresses2 2 the necessary2 2 and sufficient condition for the 𝐴𝐴𝐵𝐵1 − 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐶𝐶1 − 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐴𝐴1 − 𝐶𝐶𝐶𝐶1 triangles and to be orthological.

𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶1𝐴𝐴1𝐵𝐵1

Ion Pătrașcu, Florentin Smarandache 147 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Proof 2 Let the triangle simultaneously orthological with the triangles and . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Using Theorem 3, we have: 𝐵𝐵1𝐶𝐶1𝐴𝐴1 + + = 0, (4) + + = 0, (5) 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐵𝐵��1��𝐶𝐶��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐶𝐶��1��𝐴𝐴��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐴𝐴��1��𝐵𝐵��1⃗ Adding member by member the relations (4) and (5), we obtain: 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐶𝐶��1��𝐴𝐴��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐴𝐴��𝐵𝐵��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐵𝐵��1��𝐶𝐶��1⃗ + + + +

+ 1 1 +1 1 = 0, 1 1 1 1 (6) 𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐵𝐵��𝐶𝐶��⃗ 𝐶𝐶��𝐴𝐴��⃗� 𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐶𝐶��𝐴𝐴��⃗ 𝐴𝐴��𝐵𝐵��⃗� whatever – a point in plane. �𝑀𝑀��𝑀𝑀�⃗ ∙ �𝐴𝐴��1��𝐵𝐵��1⃗ 𝐵𝐵��1��𝐶𝐶��1⃗� 𝑀𝑀 A

B C

Figure 78

148 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Because + = , + = and + = (Chasles relation), we have: 𝐵𝐵��1��𝐶𝐶��1⃗ 𝐶𝐶��1��𝐴𝐴��1⃗ 𝐵𝐵��1��𝐴𝐴��1⃗ 𝐶𝐶��1��𝐴𝐴��1⃗ 𝐴𝐴��1��𝐵𝐵��1⃗ 𝐶𝐶��1��𝐵𝐵��1⃗ 𝐴𝐴��1��𝐵𝐵��1⃗ 𝐵𝐵��1��𝐶𝐶��1⃗ + + = 0, (7) 𝐴𝐴��1��𝐶𝐶��1⃗ whatever – a point in plane. 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐵𝐵��1��𝐴𝐴��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐶𝐶��1��𝐵𝐵��1⃗ 𝑀𝑀��𝑀𝑀�⃗ ∙ 𝐴𝐴��1��𝐶𝐶��1⃗ This relation shows that the triangles and are orthological. 𝑀𝑀 1 1 1 Remark 18 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵 The Pantazi’s theorem can be formulated as follows: If two triangles are biorthological, then they are triorthological.

Theorem 25 (C. Cocea, 1992) (i) Two inversely oriented equilateral triangles and are three times orthological and namely in the 𝐴𝐴𝐴𝐴𝐴𝐴 orders: 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ; ; . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 (ii) If� we denote �by� , , � �the orthology� centers 𝐴𝐴1 1 1 1 1 1 1 1 1 corresponding𝐵𝐵 𝐶𝐶 to 𝐵𝐵the 𝐶𝐶terns𝐴𝐴 above𝐶𝐶, then𝐴𝐴 the𝐵𝐵 triangle 𝑂𝑂′1 𝑂𝑂′2 𝑂𝑂′3 is equilateral and congruent with the triangle . 𝑂𝑂′1𝑂𝑂′2𝑂𝑂′3 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 i) Denoting by = ( ), = ( ) and by { } = , to prove that is′ orthological with′ , it is sufficient to show′ that:′ 𝐴𝐴 𝑝𝑝𝑟𝑟𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝑝𝑝𝑟𝑟𝐴𝐴1𝐶𝐶1 𝐵𝐵 𝑂𝑂1 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 (see Figure 79). ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 We have: 𝑂𝑂1𝐶𝐶 ⊥ 𝐴𝐴1𝐵𝐵1 - inscribable ′ 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵1𝐶𝐶1 ′ ′ ′ � ⟹ 𝑂𝑂1𝐴𝐴 𝐶𝐶1𝐵𝐵 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴1𝐶𝐶1 (1) 𝐵𝐵𝐵𝐵𝐵𝐵� ≡ 𝐴𝐴1�𝐶𝐶1𝐵𝐵1 [ ] [ ] = { } ′ ⟹ 𝐴𝐴1�𝐶𝐶1𝐵𝐵1 ≡ 𝐵𝐵�𝑂𝑂1𝐴𝐴 � ⟹ 𝐵𝐵𝐵𝐵𝐵𝐵� ≡ 𝐵𝐵�𝑂𝑂1𝐴𝐴 ′ ′ ′ 𝐴𝐴𝐴𝐴 ∩- inscribable𝐵𝐵𝐵𝐵 𝑂𝑂1 ⇒ 𝐵𝐵�𝑂𝑂1𝐴𝐴 ≡ 𝐵𝐵�𝑂𝑂 1𝐴𝐴 (2, 3) 1 1 { } = 𝑂𝑂 ∈ 𝐴𝐴𝐴𝐴𝐴𝐴 Finally⟹ 𝐴𝐴𝐴𝐴𝑂𝑂, d𝐶𝐶enoting: ⟹ � 1 , we obtain that: ′ 𝐴𝐴𝑂𝑂 𝐶𝐶 ≡ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶 𝑂𝑂1𝐶𝐶 ∩ 𝐴𝐴1𝐵𝐵1

Ion Pătrașcu, Florentin Smarandache 149 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

( ) = = 60 ( ) = (1) 0 𝑚𝑚�𝐵𝐵�𝑂𝑂1𝐴𝐴� = 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� = 60 ′ ′ 3 1 = 180 + 0� ⇒ 𝑚𝑚 𝐶𝐶 𝑂𝑂= 𝐵𝐵 𝑚𝑚�𝐴𝐴�𝑂𝑂1𝐶𝐶� 𝑚𝑚�𝐴𝐴𝐴𝐴𝐴𝐴�� = 180 (600 + 60 ) = 60 = − �𝑚𝑚�𝐵𝐵�𝑂𝑂1𝐴𝐴� 𝑚𝑚�𝐴𝐴�𝑂𝑂1𝐶𝐶�� 0 0 0 0 inscribable 1�1 1 − ′ ′ 𝑚𝑚�𝐵𝐵 𝐴𝐴 𝐶𝐶 � ⟹ ′ ′ 𝑂𝑂1𝐵𝐵 𝐴𝐴1𝐶𝐶 − ⟹ 𝐶𝐶�𝑂𝑂1𝐵𝐵 ≡ 𝐵𝐵1�𝐴𝐴1𝐶𝐶1 ⟹( ) ′ . ⟹ 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴1𝐶𝐶1 Similarly, it is shown that the triangles′ and are orthological, ⟹ 𝐶𝐶𝐶𝐶 𝑂𝑂1𝐶𝐶 ⊥ 𝐴𝐴1𝐵𝐵1 of center ( belongs to the circle circumscribed to the triangle ), and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1𝐴𝐴1 that the triangles and are orthological, of orthology center , a 𝑂𝑂2 𝑂𝑂2 𝐴𝐴𝐴𝐴𝐴𝐴 point that belongs to the circle circumscribed to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶1𝐴𝐴1𝐵𝐵1 𝑂𝑂3 A'" 𝐴𝐴𝐴𝐴𝐴𝐴 A O 3 B" O2

C' C B

O1 A"

A1 B'

B'" A' C" C 1 B1

Figure 79

150 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

ii) Because (they are perpendicular to ) and is inscribable quadrilateral (these points belong to the circle circumscribed to the 𝐶𝐶𝐶𝐶3 ∥ 𝐴𝐴𝑂𝑂2 𝐴𝐴1𝐶𝐶1 𝐴𝐴1𝑂𝑂2𝐶𝐶1𝑂𝑂3 triangle ), we have that is an isosceles trapezoid, therefore = . Similarly, is an isosceles trapezoid, hence = ; 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝑂𝑂2𝐶𝐶𝑂𝑂3 and is an isosceles trapezoid, hence = . But the triangle 𝑂𝑂2𝑂𝑂3 𝐴𝐴𝐴𝐴 𝐴𝐴𝑂𝑂1𝐵𝐵𝑂𝑂3 𝑂𝑂1𝑂𝑂3 𝐴𝐴𝐴𝐴 is equilateral, thus is an equilateral triangle, congruent with . 𝐵𝐵𝑂𝑂2𝐶𝐶𝑂𝑂1 𝑂𝑂1𝑂𝑂2 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 1 2 3 Observation 55𝑂𝑂 𝑂𝑂 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 Similarly, it can be shown that, if we denote by , , the orthology centers corresponding to the terns below: 𝑂𝑂′1 𝑂𝑂′2 𝑂𝑂′3 ; ; , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1𝐴𝐴1 𝐶𝐶1𝐴𝐴1𝐵𝐵1 then� the triangle� � � � is equilateral� and congruent with the triangle . It can𝐴𝐴𝐴𝐴𝐴𝐴 be proved𝐴𝐴𝐴𝐴𝐴𝐴 without difficulty𝐴𝐴𝐴𝐴𝐴𝐴 that the triangle is biorthological in 𝑂𝑂′1𝑂𝑂′2𝑂𝑂′3 𝐴𝐴1𝐵𝐵1𝐶𝐶1 relation to the triangle . 𝑂𝑂1𝑂𝑂2𝑂𝑂3 1 2 3 Theorem 26 (Mihai𝑂𝑂′ 𝑂𝑂Miculița)′ 𝑂𝑂′ Two certain similar triangles and , inversely oriented, are orthological. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Consequences 1). During the proof, we showed that the quadrilateral is inscribable, it follows that . Therefore, the orthology center 𝑃𝑃𝐵𝐵1𝐴𝐴1𝐶𝐶1 of the triangle 1 1 with respect to the triangle is to be found on 1 𝑃𝑃 ∈ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑃𝑃 the circumscribed circle of the triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 1 1 1 𝐴𝐴𝐴𝐴𝐴𝐴 2). If and are two inversely oriented equilateral triangles, 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 then we have: ~ ~ ~ , so the point i) of 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Theorem 25 is a particular case of the above Theorem. ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∆𝐵𝐵1𝐶𝐶1𝐴𝐴1 ∆𝐶𝐶1𝐴𝐴1𝐵𝐵1 Proof Denoting by = ( ), = ( ) and { } = , and by { } = ′ ′ ′ ′ , the proof𝐵𝐵𝐵𝐵 is1 now reduced𝐴𝐴𝐴𝐴 to1 show only: ′ = ( 𝐴𝐴) 𝑝𝑝𝑟𝑟 𝐴𝐴 𝐶𝐶 𝑝𝑝𝑟𝑟 𝐶𝐶 𝑃𝑃 𝐴𝐴𝐴𝐴 ∩ 𝐶𝐶𝐶𝐶 1 (see Figure 80). 𝐵𝐵 ′ 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵 𝑃𝑃 We have𝐴𝐴:𝐴𝐴 1 𝐵𝐵 𝑝𝑝𝑟𝑟 𝐵𝐵= ( )

′ = ( ) ′ 𝐴𝐴 𝑝𝑝𝑟𝑟𝐵𝐵𝐵𝐵 𝐴𝐴1 ⇔ 𝐴𝐴1𝐴𝐴 ⊥ 𝐵𝐵𝐵𝐵 ′ ′ ′ ′ � ⟹ 𝐶𝐶�𝐴𝐴 𝑃𝑃 ≡ 𝐵𝐵�𝐶𝐶 𝑃𝑃 ⟹ 𝐶𝐶 𝑝𝑝𝑟𝑟𝐴𝐴𝐴𝐴 𝐴𝐴1 ⇔ 𝐶𝐶1𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 151 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

{ } = inscribable′ ′ =′ ′ 𝑃𝑃 𝐴𝐴𝐴𝐴 ∩ 𝐶𝐶𝐶𝐶 ⇒ 𝐴𝐴�1𝑃𝑃𝑃𝑃1 𝐶𝐶�𝑃𝑃𝐴𝐴 ′ ′ ~ ′=⟹′ ⟹ � �� ⟹ 𝐴𝐴 𝐵𝐵𝐶𝐶 𝑃𝑃 − = ⟹ 𝐶𝐶 𝑃𝑃𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ⟹ 𝐴𝐴𝐴𝐴𝐴𝐴� 𝐴𝐴1�𝐵𝐵1𝐶𝐶1 = inscribable = ⟹ 𝐴𝐴�1𝑃𝑃𝑃𝑃1 𝐴𝐴1�𝐵𝐵1𝐶𝐶1 ⟹ [ ] [ ] = { } = 𝐴𝐴�1𝑃𝑃𝑃𝑃1 𝐴𝐴1�𝐵𝐵1𝐶𝐶1 ⟹ 𝑃𝑃𝐵𝐵1𝐴𝐴1𝐶𝐶1 − ⟹ 𝐴𝐴1�𝐶𝐶1𝐵𝐵1 𝐴𝐴�1𝑃𝑃𝑃𝑃1 ′ ~ ′ = ′ ′ ⟹ 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 𝑃𝑃 ⟹ 𝐴𝐴�1𝑃𝑃𝑃𝑃1 𝐴𝐴�𝑃𝑃𝐵𝐵 � ⟹ = ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ⟹ 𝐴𝐴𝐴𝐴𝐴𝐴� 𝐴𝐴1�𝐶𝐶1𝐵𝐵1 inscribable′ ′ = ⟹ 𝐴𝐴�𝑃𝑃𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴� ⟹ = 90 ′ ′ =′90 ′ � � ′ 0 ⟹ 𝐴𝐴 𝑃𝑃𝐵𝐵 ′𝐶𝐶 − ⟹′ 𝑃𝑃𝐵𝐵=𝐶𝐶 0 𝑃𝑃(𝐴𝐴 𝐵𝐵) � � � ⟹ 𝑚𝑚�𝑃𝑃𝐵𝐵 𝐶𝐶� ⟹ 𝑃𝑃𝐴𝐴 ⊥ 𝐵𝐵𝐵𝐵 ⟹ 𝑚𝑚�𝑃𝑃𝐴𝐴 𝐵𝐵� ′ ⟹ 𝐵𝐵 𝑝𝑝𝑟𝑟𝐴𝐴𝐴𝐴 𝐵𝐵1

Figure 80

152 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Theorem 27 (Lemoine) The geometric place of the points in the plane of a triangle , whose podal triangles in relation to it are triorthological 𝑀𝑀 with , is Lemoine line of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 Proof𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 We consider the triangle such that in Cartesian frame of reference we have (0, ), ( , 0), ( , 0). 𝐴𝐴𝐴𝐴𝐴𝐴 𝑥𝑥𝑥𝑥𝑥𝑥 It is known that the equation of a circle determined by three noncollinear 𝐴𝐴 𝑎𝑎 𝐵𝐵 𝑏𝑏 𝐶𝐶 𝑐𝑐 points, , , , ( , ), {1, 2, 3}, is given by: + 1 1 2 3 𝑖𝑖 𝑖𝑖 𝑖𝑖 2 𝐴𝐴+ 𝐴𝐴2 𝐴𝐴 𝐴𝐴 𝑥𝑥 1𝑦𝑦 𝑖𝑖 ∈ 𝑥𝑥 𝑦𝑦 = 0. 𝑥𝑥2 𝑦𝑦2 + 1 1 1 𝑥𝑥1 𝑦𝑦1 𝑥𝑥 𝑦𝑦 � 2 + 2 1� �𝑥𝑥2 𝑦𝑦2 𝑥𝑥2 𝑦𝑦2 � For2 the 2points3 ,3 , with the above coordinates, this equation is the 𝑥𝑥3 𝑦𝑦3 𝑥𝑥 𝑦𝑦 following, after development of the determinant: 𝐴𝐴 𝐵𝐵 𝐶𝐶 ( + ) ( + ) ( + ) + = 0. We 2write2 the Lemoine equation2 of the triangle . The equation is 𝑎𝑎 𝑥𝑥 𝑦𝑦 − 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑥𝑥 − 𝑎𝑎 𝑏𝑏𝑏𝑏 𝑦𝑦 𝑎𝑎𝑎𝑎𝑎𝑎 determined by the intersection of tangents to the circle circumscribed with the opposite sides of the triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 The equation of the tangent in (0, ) to the circumscribed circle is: ( + ) ( ) + ( ) = 0. 2 𝐴𝐴2 𝑎𝑎 Intersecting this tangent with 0 , we find the point , with ; 0 . 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑥𝑥 − 𝑦𝑦 𝑎𝑎 − 𝑏𝑏𝑏𝑏 𝑎𝑎 𝑎𝑎 − 𝑏𝑏𝑏𝑏 2 𝑏𝑏𝑏𝑏−𝑎𝑎 The equation of the tangent in ( , 0) to the circumscribed circle of the 𝑥𝑥 𝐷𝐷 𝐷𝐷 � 𝑏𝑏+𝑐𝑐 � triangle is: 𝐵𝐵 𝑏𝑏 ( ) ( + ) ( ) = 0. 𝐴𝐴𝐴𝐴𝐴𝐴 We intersect this2 tangent with the line : 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 𝑥𝑥 − 𝑎𝑎 𝑏𝑏𝑏𝑏 𝑦𝑦 − 𝑎𝑎𝑎𝑎 𝑏𝑏 − 𝑐𝑐 + = 0. 𝐴𝐴𝐴𝐴 We obtain: 𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐 − 𝑎𝑎𝑎𝑎 ( ) ; . 2 2 2 𝑐𝑐�𝑎𝑎 +𝑏𝑏 � −𝑎𝑎 𝑏𝑏−𝑐𝑐 The 2slope2 of Lemoine2 2 line of the triangle is: 𝐸𝐸 �𝑎𝑎 −𝑐𝑐 +2𝑏𝑏𝑏𝑏 𝑎𝑎 −𝑐𝑐 +2𝑏𝑏𝑏𝑏� ( )( ) = . 2 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 𝑏𝑏+𝑐𝑐 𝑏𝑏−𝑐𝑐 The𝐷𝐷𝐷𝐷 equation4 of2 Lemoine3 3 line2 of2 the triangle is: 𝑚𝑚 −𝑎𝑎 −2𝑎𝑎 𝑏𝑏𝑏𝑏−𝑏𝑏𝑐𝑐 −𝑏𝑏 𝑐𝑐−𝑏𝑏 𝑐𝑐 ( + )( ) + ( + 2 + + ) ( )( ) = 0 . 2 4 2 3 3 2 2 𝐴𝐴𝐴𝐴𝐴𝐴 2 2 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑏𝑏 − 𝑐𝑐 𝑥𝑥 𝑎𝑎 𝑎𝑎 𝑏𝑏𝑏𝑏 𝑏𝑏𝑐𝑐 𝑏𝑏 𝑐𝑐 − 𝑏𝑏 𝑐𝑐 𝑦𝑦 − 𝑎𝑎 𝑏𝑏𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑐𝑐

Ion Pătrașcu, Florentin Smarandache 153 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOLOGICAL TRIANGLES WITH THE SAME ORTHOLOGY CENTER

Let us consider now a point ( , ) that has the property from statement, namely its podal triangle, which we denote , is triorthological with the 𝑀𝑀 𝑥𝑥0 𝑦𝑦0 triangle . We have that ( , 0). The equation of the perpendicular taken 𝐴𝐴1𝐵𝐵1𝐶𝐶1 from to is: = (1 0 ). 𝐴𝐴𝐴𝐴𝐴𝐴 𝑐𝑐𝐴𝐴 𝑥𝑥 Its intersection with 0 : + 0 = 0 is: 𝑀𝑀 𝐴𝐴𝐴𝐴 𝑦𝑦 − 𝑦𝑦 𝑎𝑎 𝑥𝑥 − 𝑥𝑥 ; . 2 𝐴𝐴𝐴𝐴2 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 − 𝑎𝑎𝑎𝑎 𝑏𝑏�𝑎𝑎 +𝑏𝑏𝑥𝑥0−𝑎𝑎𝑦𝑦0� 𝑎𝑎�𝑏𝑏 −𝑏𝑏𝑥𝑥0+𝑎𝑎𝑦𝑦0� The1 perpendicular2 2 taken2 from2 to has the equation: 𝐶𝐶 � 𝑎𝑎 +𝑏𝑏 𝑎𝑎 +𝑏𝑏 � + = 0. 𝑀𝑀 𝐴𝐴𝐴𝐴 Intersecting this line with : + = 0, we obtain: 𝑐𝑐𝑐𝑐 − 𝑎𝑎𝑎𝑎 − 𝑐𝑐𝑥𝑥0 𝑎𝑎𝑦𝑦0 ; . 2 2 𝐴𝐴𝐴𝐴 𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐 − 𝑎𝑎𝑎𝑎 𝑐𝑐�𝑎𝑎 +𝑐𝑐𝑥𝑥0−𝑎𝑎𝑦𝑦0� 𝑎𝑎�𝑐𝑐 −𝑐𝑐𝑥𝑥0+𝑎𝑎𝑦𝑦0� The1 equation2 2 of the perpendicular2 2 taken from to is: 𝐵𝐵 � 𝑎𝑎 +𝑐𝑐 𝑎𝑎 +𝑐𝑐 � = 0. 𝐴𝐴1 𝐴𝐴𝐴𝐴 The equation of0 the perpendicular taken from to is: 𝑐𝑐𝑐𝑐 −(𝑎𝑎𝑎𝑎+− 𝑐𝑐𝑥𝑥) ( + ) ( + ) + ( + ) 𝐵𝐵1 𝐴𝐴𝐴𝐴 2 2 2 ( 2 ) = 0.2 2 2 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑥𝑥 − 𝑎𝑎 𝑎𝑎 𝑐𝑐 𝑦𝑦 − 𝑥𝑥0 𝑏𝑏𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑦𝑦0 ∙ 𝑎𝑎 𝑎𝑎 𝑏𝑏𝑏𝑏 The equation of the perpendicular2 2 taken from to is: − 𝑎𝑎 𝑐𝑐 − 𝑏𝑏𝑏𝑏 ( + ) ( + ) = 0. 𝐶𝐶1 𝐵𝐵𝐵𝐵 These2 perpendicular2 2 s are concurrent if and only if: 𝑎𝑎 𝑐𝑐 𝑥𝑥 − 𝑏𝑏 𝑎𝑎 𝑐𝑐𝑥𝑥0 − 𝑎𝑎𝑦𝑦0 ( + ) ( + ) ( + ) + ( + + ) 0 = 0. 2+𝑐𝑐 2 −20𝑎𝑎 2 2 2 ( +−𝑐𝑐3𝑥𝑥 ) 2 2 2 0 0 �𝑏𝑏 𝑎𝑎2 𝑐𝑐2 −𝑎𝑎 𝑎𝑎 𝑐𝑐 −𝑥𝑥 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎 𝑦𝑦2 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎 𝑐𝑐 − 𝑎𝑎 𝑏𝑏𝑏𝑏 � Developing this determinant, making reductions,0 we0 obtain: 𝑎𝑎 𝑏𝑏 −𝑏𝑏 𝑎𝑎 𝑏𝑏𝑥𝑥 − 𝑎𝑎𝑦𝑦 ( + )( ) + ( + 2 + + ) 2 ( 4 )(2 ) =3 0 3 2 2 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑏𝑏 − 𝑐𝑐 𝑥𝑥0 𝑎𝑎 𝑎𝑎 𝑏𝑏𝑏𝑏 𝑏𝑏𝑐𝑐 𝑏𝑏 𝑐𝑐 − 𝑏𝑏 𝑐𝑐 𝑦𝑦0 which shows that the point belongs to2 the Lemoine2 line of the triangle . − 𝑎𝑎 𝑏𝑏𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑐𝑐

𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴

154 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

In 1922, J. Neuberg proposed the name of bilogical triangles for the triangles that are simultaneously homological and orthological.

6.1 Sondat’s theorem. Proofs

Theorem 28 (P. Sondat, 1894) If two triangles and are bilogical with the homology center and the orthology centers and , then , and are 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 on the same line, which is perpendicular to the axis of homology. 𝑃𝑃 𝑄𝑄1 𝑄𝑄 𝑃𝑃 𝑄𝑄 𝑄𝑄1 Proof 1 (V. Thébault, 1952) We denote by the axis of homology of the given triangle. On this line, there exist the points: { } = , { } = , { } = 𝑑𝑑 1 1 1 1 1 1. ′ ′ ′ 𝐴𝐴 𝐵𝐵𝐵𝐵 ∩ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐴𝐴𝐴𝐴 ∩ 𝐴𝐴 𝐶𝐶 𝐶𝐶 𝐴𝐴𝐴𝐴 ∩ The orthology center is the intersection of the perpendiculars taken from 𝐴𝐴 𝐵𝐵 1 , , respectively on ; and . The idea of proving the 𝑄𝑄 collinearity of points , , is to show that and that . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵11𝐶𝐶1 𝐴𝐴1𝐶𝐶1 𝐴𝐴1𝐵𝐵1 To prove that , it is necessary and sufficient to prove the relation: 𝑃𝑃 𝑄𝑄 𝑄𝑄 𝑃𝑃𝑃𝑃 ⊥ 𝑑𝑑 𝑃𝑃𝑄𝑄1 ⊥ 𝑑𝑑 = . (1) 𝑃𝑃𝑃𝑃 ⊥ 𝑑𝑑 We′ employ2 ′ Stewart’s2 ′ t2heorem′ in2 the triangle ; ; we have: 𝐵𝐵 𝑃𝑃 − 𝐵𝐵 𝑄𝑄 𝐴𝐴 𝑃𝑃 − 𝐴𝐴 𝑄𝑄 + = ′ . (2) 𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵 ∈ 𝐴𝐴𝐴𝐴 The′ 2point being2 the homology′ 2 center′ , we have′ : ′ 𝐵𝐵 𝑃𝑃 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶𝑃𝑃 ∙ 𝐵𝐵 𝐴𝐴 − 𝑃𝑃𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 = , = , = , (3) 𝑃𝑃 , , real numbers (in the case of Figure 81, , , are strictly positive). 𝑃𝑃��𝑃𝑃��1⃗ 𝛼𝛼 ∙ 𝐴𝐴𝐴𝐴��1⃗ 𝑃𝑃��𝑃𝑃��1⃗ 𝛽𝛽 ∙��𝐵𝐵𝐵𝐵��1⃗ 𝑃𝑃��𝑃𝑃��1⃗ 𝛾𝛾 ∙ 𝐶𝐶𝐶𝐶��1⃗ Menelaus’s theorem applied in the triangle for the transverse 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 implies: ′ 𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵 − 1 1 = 1. (4) 𝐶𝐶 −′𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐴𝐴 𝐶𝐶1𝑃𝑃 ′ 𝐵𝐵 𝐴𝐴 ∙ 𝐴𝐴1𝑃𝑃 ∙ 𝐶𝐶1𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 155 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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C1 Q Q1

B A' C

Figure 81

156 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Taking into account (3), we get from (4) that: = . (5) ′ 𝐵𝐵 𝐶𝐶 𝛼𝛼 ′ 𝐵𝐵Stewart’s𝐴𝐴 𝛾𝛾 theorem applied in the triangle , , leads to: + = ′ . (6) 𝑄𝑄𝑄𝑄𝑄𝑄 𝐵𝐵 ∈ 𝐴𝐴𝐴𝐴 We′ equal2 the relations2 ′ (2) and2 (6);′ we get′: ′ 𝐵𝐵 𝑄𝑄 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶𝑄𝑄 ∙ 𝐵𝐵 𝐴𝐴 − 𝑄𝑄𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 + = + or: ′ 2 2 ′ 2 ′ ′ 2 2 ′ 2 ′ 𝐵𝐵 𝑃𝑃 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶𝑃𝑃 ∙ 𝐵𝐵 𝐴𝐴 − 𝑃𝑃𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝑄𝑄 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶𝑄𝑄 ∙ 𝐵𝐵 𝐴𝐴 − 𝑄𝑄𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 ( ) = ( ) + ( ) . (7) We′ denote2 :′ 2 2 2 ′ 2 2 ′ 𝐵𝐵 𝑃𝑃 − 𝐵𝐵 𝑄𝑄 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶𝑄𝑄 − 𝐶𝐶𝑃𝑃 ∙ 𝐵𝐵 𝐴𝐴 𝑃𝑃𝐴𝐴 − 𝑄𝑄𝐴𝐴 ∙ 𝐵𝐵 𝐶𝐶 = , = , = . (8) We2 find: 2 2 2 2 2 𝑃𝑃𝐴𝐴 − 𝑄𝑄𝐴𝐴 𝑢𝑢 𝑃𝑃𝐵𝐵 − 𝑄𝑄𝐵𝐵 𝑣𝑣 𝑃𝑃𝐶𝐶 − 𝑄𝑄𝐶𝐶 𝑡𝑡 = . (9) ′ 2 ′ 2 𝛼𝛼𝛼𝛼−𝛾𝛾𝛾𝛾 𝐵𝐵We𝑃𝑃 apply− 𝐵𝐵 Stewart’s𝑄𝑄 𝛼𝛼− 𝛾𝛾theorem in the triangles and , : + = , ′ (10) 𝑃𝑃𝑃𝑃𝑃𝑃 𝑄𝑄𝑄𝑄𝑄𝑄 𝐴𝐴 ∈ 𝐵𝐵𝐵𝐵 ′ 2 2 ′ + 2 ′ =′ ′ . (11) 𝐴𝐴 𝑃𝑃 ∙ 𝐵𝐵𝐵𝐵 − 𝑃𝑃𝐵𝐵 ∙ 𝐴𝐴 𝐶𝐶 𝑃𝑃𝐶𝐶 ∙ 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 ∙ 𝐵𝐵𝐵𝐵 ∙ 𝐴𝐴 𝐶𝐶 Menelaus’s′ 2 theorem2 in′ the triangle2 ′ for′ the transverse′ , 𝐴𝐴 𝑄𝑄 ∙ 𝐵𝐵𝐵𝐵 − 𝑄𝑄𝐵𝐵 ∙ 𝐴𝐴 𝐶𝐶 𝑄𝑄𝐶𝐶 ∙ 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 ∙ 𝐵𝐵𝐵𝐵 ∙ 𝐴𝐴 𝐶𝐶 leads to: ′ 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴 − 𝐶𝐶1 − 𝐵𝐵1 = . (12) ′ 𝐴𝐴 𝐵𝐵 𝛾𝛾 ′ We𝐴𝐴 𝐶𝐶 equal𝛽𝛽 the relations (10) and (11), taking into account (8) and (12): = . (13) ′ 2 ′ 2 𝑣𝑣𝑣𝑣−𝑡𝑡𝑡𝑡 𝐴𝐴The𝑃𝑃 relation− 𝐴𝐴 𝑄𝑄 (1) is𝛽𝛽 −equivalent𝛾𝛾 to: ( ) + ( ) + ( ) = 0. (14) To prove (14), we apply Stewart’s theorem in the triangles and , 𝛼𝛼𝛼𝛼 𝑢𝑢 − 𝑣𝑣 𝛽𝛽𝛽𝛽 𝑣𝑣 − 𝑡𝑡 𝛾𝛾𝛾𝛾 𝑡𝑡 − 𝑢𝑢 1 ; we get: 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 + = , (15) 𝐴𝐴 ∈ 𝐴𝐴𝐴𝐴 2 2 + 2 = . (16) 𝐶𝐶𝐴𝐴 ∙ 𝑃𝑃𝐴𝐴1 − 𝐶𝐶𝑃𝑃 ∙ 𝐴𝐴𝐴𝐴1 𝐶𝐶𝐴𝐴1 ∙ 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 ∙ 𝑃𝑃𝐴𝐴1 ∙ 𝐴𝐴𝐴𝐴1 We2 equal these rela2 tions and we2 get: 𝐵𝐵𝐴𝐴 ∙ 𝑃𝑃𝐴𝐴1 − 𝐵𝐵𝑃𝑃 ∙ 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐴𝐴1 ∙ 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 ∙ 𝑃𝑃𝐴𝐴1 ∙ 𝐴𝐴𝐴𝐴1 ( ) + ( ) + ( ) = 0. (17) 2 2 2 2 2= 2 Because 1 , we have that: 1 1 1 . 𝐵𝐵𝐴𝐴 − 𝐶𝐶𝐴𝐴 𝑃𝑃𝐴𝐴 𝑃𝑃𝐶𝐶 − 𝑃𝑃𝐵𝐵 𝐴𝐴𝐴𝐴2 𝐵𝐵𝐴𝐴2 − 𝐶𝐶𝐴𝐴2 𝑃𝑃𝑃𝑃 2 Substituting1 this relation into (17); taking1 into1 account that = ; and the 𝐴𝐴 𝑄𝑄 ⊥ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐴𝐴 − 𝐶𝐶𝐴𝐴 𝑄𝑄𝐵𝐵 − 𝑄𝑄𝑃𝑃𝐶𝐶𝐴𝐴1 relations (8), – we obtain: 𝐴𝐴𝐴𝐴1 𝛼𝛼 + = . (18) 𝑣𝑣−𝑡𝑡 Similarly2 , 2we get the2 relations2 : 𝐵𝐵𝐴𝐴 − 𝐶𝐶𝐴𝐴 𝑄𝑄𝐶𝐶 − 𝑄𝑄𝐵𝐵 𝛼𝛼

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+ = , (19) 2 2 2 2 𝑡𝑡−𝑢𝑢 𝐶𝐶𝐵𝐵 − 𝐴𝐴𝐵𝐵 + 𝑄𝑄𝐴𝐴 − 𝑄𝑄𝐶𝐶 = 𝛽𝛽 . (20) 2 2 2 2 𝑢𝑢−𝑣𝑣 𝐴𝐴Adding𝐶𝐶 − 𝐵𝐵 the𝐶𝐶 last𝑄𝑄 thr𝐵𝐵 ee− member𝑄𝑄𝐴𝐴 -to𝛾𝛾-member relations, we get: + + = 0. (21) 𝑣𝑣−𝑡𝑡 𝑡𝑡−𝑢𝑢 𝑢𝑢−𝑣𝑣 The𝛼𝛼 relations𝛽𝛽 𝛾𝛾(14) and (21) are equivalent because . Similarly, it is proved that , which ends the proof of P. Sondat’s theorem. 𝑃𝑃𝑃𝑃 ⊥ 𝑑𝑑 1 Proof 2 𝑃𝑃(adapted𝑄𝑄 ⊥ 𝑑𝑑 after the proof given by Jean-Louis Aymé) Let be the intersection of the perpendicular taken from to with ; – the intersection of the parallel taken through to with ; 𝐴𝐴2 𝑄𝑄1 𝐵𝐵𝐵𝐵 and – the intersection of the parallel taken through to with (see 𝐴𝐴𝐴𝐴1 𝐵𝐵2 𝐴𝐴2 𝐴𝐴1𝐵𝐵1 𝐵𝐵𝐵𝐵1 Figure2 82). 2 1 1 1 𝐶𝐶 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶𝐶𝐶

P A A1 A 2

C" C1

C2 B A" C

B1 B"

Figure 82

158 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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The triangles and are homothetic by a homothety of center (they have respectively parallel sides). Because and 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 , it follows that ; similarly, and , 𝑃𝑃 𝐴𝐴𝑄𝑄1 ⊥ 𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 ⊥ therefore is common orthology center for the triangles and . 𝐵𝐵2𝐶𝐶2 𝐴𝐴𝑄𝑄1 ⊥ 𝐵𝐵2𝐶𝐶2 𝐵𝐵𝑄𝑄1 ⊥ 𝐴𝐴2𝐶𝐶2 𝐶𝐶𝑄𝑄1 ⊥ 𝐴𝐴2𝐵𝐵2 The triangle being homothetic with , and being 𝑄𝑄1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 homological with (of center ), we denote by the homology axis 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 of triangles and , we have that ′′ ′′ ( ′′ is the homology 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 axis of triangles and ). Applying′′ now′′ Proposition′ ′ ′ ′ 67, it follows 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 ∥ 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 that , therefore: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ′′ . ′′ (1) 𝑃𝑃𝑄𝑄1 ⊥ 𝐴𝐴 𝐵𝐵 We denote′ by′ the intersection of the perpendiculars taken from to 𝑃𝑃𝑄𝑄1 ⊥ 𝐴𝐴 𝐵𝐵 with ; let – the intersection of the parallel taken from to with 𝐴𝐴3 𝑄𝑄 𝐵𝐵1𝐶𝐶1 ; and let – the intersection of the parallel taken from to with . 𝐴𝐴𝐴𝐴1 𝐵𝐵3 𝐴𝐴3 𝐴𝐴𝐵𝐵1 The triangles and are homothetic of center (they have 𝐵𝐵𝐵𝐵1 𝐶𝐶3 𝐵𝐵3 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶1 respectively parallel sides). Having and , it follows that 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝑃𝑃 ; from and , we obtain that . In 𝐵𝐵1𝑄𝑄 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴3𝐶𝐶3 ∥ 𝐴𝐴𝐴𝐴 conclusion, the triangles and are orthological, with the 𝐵𝐵1𝑄𝑄 ⊥ 𝐴𝐴3𝐶𝐶3 𝐴𝐴1𝑄𝑄 ⊥ 𝐵𝐵𝐵𝐵 𝐵𝐵3𝐶𝐶3 ∥ 𝐵𝐵𝐵𝐵 𝐴𝐴1𝑄𝑄 ⊥ 𝐵𝐵3𝐶𝐶3 common orthology center the point , and homological, of center . Applying 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴3𝐵𝐵3𝐶𝐶3 the Proposition 67, it follows that (where is the homology 𝑄𝑄 𝑃𝑃 axis of triangles and . Since′′′ ′′′ is h′′′omot′′′ hetic with , 𝑃𝑃𝑃𝑃 ∥ 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 it follows that is parallel with ; consequently: 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴𝐴𝐴𝐴𝐴 . ′′′ ′′′ ′ ′ (2) 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 The relations′ ′ (1) and (2) lead to the conclusion. 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴 𝐵𝐵 6.2 Remarkable bilogical triangles

6.2.1 A triangle and its first Brocard triangle

Definition 37 We call the first Brocard triangle of a given triangle – the triangle determined by the projections of the symmedian center on the mediators of the given triangle.

Observation 56 In Figure 83, is the intersection of symmedians of triangles ; , , are its mediators; and is the first Brocard triangle. ′ ′ ′ 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝐴𝐴 𝑂𝑂𝐵𝐵 𝑂𝑂𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1

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B' C' C 1 O

K A1

B A' C

Figure 83

Proposition 69 The first Brocard triangle is similar with the given triangle.

Proof We observe that = 90 , therefore belongs to the circle of diameter ; similarly , belong to0 this circle. We have: 𝑚𝑚�𝐾𝐾�𝐴𝐴1𝑂𝑂� 𝐴𝐴1 (subtending the same arc in the circle circumscribed to the first Brocard 𝑂𝑂𝑂𝑂 𝐵𝐵1 𝐶𝐶1 𝐵𝐵1�𝐴𝐴1𝐶𝐶1 ≡ 𝐵𝐵�1𝑂𝑂𝐶𝐶1 triangle). On the other hand, (they have sides respectively perpendicular). We obtain that . Similarly, it follows that ∢𝐵𝐵1𝑂𝑂𝐶𝐶1 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 , and, consequently, the given triangle and its first ∢𝐵𝐵1𝐴𝐴1𝐶𝐶1 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 Brocard triangle are similar. ∢𝐴𝐴1𝐵𝐵1𝐶𝐶1 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Observation 𝐴𝐴57𝐵𝐵 𝐶𝐶 1. The circle circumscribed to the first Brocard triangle is called Brocard circle. 2. The previous Proposition is proved in the same way in the case of the obtuse (or right) triangle.

160 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Theorem 29 The triangle and its first Brocard triangle, , are bilogical triangles. 1 1 1 We will prove 𝐴𝐴𝐴𝐴𝐴𝐴this theorem in two steps: 𝐴𝐴 𝐵𝐵 𝐶𝐶

I. We prove that the triangles and are orthological. Indeed, the perpendiculars taken from , , to , , are 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 mediators of the triangle and, consequently, – the center of the circle 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 circumscribed to the triangle , is the orthology center of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 in relation to the triangle . According to the theorem of 𝐴𝐴𝐴𝐴𝐴𝐴 orthological triangles, the perpendiculars taken from , , respectively 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 to the sides of the first Brocard triangle are concurrent. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 Because this point is important in the geometry of the triangle, we will 𝐴𝐴1𝐵𝐵1𝐶𝐶1 define it and prove the concurrency of previous lines.

A' 1 A

C' C' B' B'1 1 B1 O

A1 K

B A' C

Figure 84

Ion Pătrașcu, Florentin Smarandache 161 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Definition 38 It is called a Tarry point of the triangle the orthology center of the triangle in relation to the first Brocard triangle. 𝑨𝑨𝑨𝑨𝑨𝑨 Proposition 70 𝑨𝑨𝑨𝑨𝑨𝑨 The orthology center of the triangle in relation to – its first Brocard triangle, is the Tarry point, , and this point belongs to the circle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 circumscribed to the triangle . 𝑇𝑇 Proof 𝐴𝐴𝐴𝐴𝐴𝐴

We denote: { 1} = 1 1, { 1} = 1 1 (see Figure 84). We ′ ′ = 180 have: 𝐵𝐵 ; it𝐵𝐵𝐵𝐵 follows∩ 𝐴𝐴 𝐶𝐶 that 𝐶𝐶 𝐶𝐶𝐶𝐶 ∩ 𝐴𝐴 𝐵𝐵 , consequently ′ ′ 0 ∢𝐶𝐶1𝐴𝐴1𝐵𝐵, 1which≡ ∢𝐴𝐴 shows that belongs𝑚𝑚 �𝐵𝐵 1�to𝐴𝐴1 the𝐶𝐶1� circle circumscribed− 𝐴𝐴 to the triangle . ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴 𝑇𝑇

𝐴𝐴𝐴𝐴𝐴𝐴 II. To prove that the triangles and are homological, we need some helpful results: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Definition 39 The points and from the interior of triangle , with the properties: ′ 𝜴𝜴 𝜴𝜴 𝑨𝑨𝑨𝑨𝑨𝑨 ( ) = ( ) = ( ) = , ( ) = ( ) = ( ) = , 𝒎𝒎 ∢𝜴𝜴𝜴𝜴𝜴𝜴 𝒎𝒎 ∢𝜴𝜴𝜴𝜴𝜴𝜴 𝒎𝒎 ∢𝜴𝜴𝜴𝜴𝜴𝜴 𝝎𝝎 are called′ Brocard points′ , and is called′ Brocard angle. 𝒎𝒎 ∢𝜴𝜴 𝑩𝑩𝑩𝑩 𝒎𝒎 ∢𝜴𝜴 𝑨𝑨𝑨𝑨 𝒎𝒎 ∢𝜴𝜴 𝑪𝑪𝑪𝑪 𝝎𝝎 Lemma 7 𝝎𝝎 In the triangle , where is the first Brocard point and = { }, we have = . ′′𝐴𝐴𝐴𝐴𝐴𝐴2 Ω 𝐴𝐴Ω ∩ 𝐵𝐵𝐵𝐵 ′′ 𝐵𝐵𝐴𝐴 𝑐𝑐 ′′ 2 𝐴𝐴 𝐶𝐶𝐴𝐴 𝑎𝑎 Proof Area = , (1) ′′ 1 ′′ Area∆𝐴𝐴𝐴𝐴𝐴𝐴 = 2 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝑠𝑠 ). (2) 1 From (1) and′′ (2), it follows′′ that: ∆𝐴𝐴𝐴𝐴𝐴𝐴 2 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴 − 𝜔𝜔

162 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

= . (3) ′′ ( ) Area∆𝐴𝐴𝐴𝐴𝐴𝐴 𝑐𝑐∙𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ′′ OnArea the∆𝐴𝐴𝐴𝐴 𝐴𝐴other 𝑏𝑏hand∙𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴,− 𝜔𝜔the mentioned triangles have the same altitude taken from , hence: = . (4) 𝐴𝐴 ′′ ′′ Area∆𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐴𝐴 ′′ ′′ TheArea ∆relations𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶(3)𝐴𝐴 and (4) lead to: = . (5) ′′ ( ) 𝐵𝐵𝐴𝐴 𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ′′ 𝐶𝐶𝐴𝐴 𝑎𝑎 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴−𝜔𝜔

Figure 85

Applying the sinus theorem in the triangles and , we get: = ( ) ( ), (6) 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 𝑠𝑠𝑠𝑠𝑠𝑠 𝐴𝐴−=𝜔𝜔 sin 𝐴𝐴𝐴𝐴𝐴𝐴 ( ). (7) 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝑠𝑠Because𝑠𝑠𝑠𝑠𝑠𝑠 sin 𝐵𝐵𝐵𝐵𝐵𝐵 = 180 and = 180 , from relations (6) and (7) – it follows that: 0 0 𝑚𝑚�𝐴𝐴𝐴𝐴𝐴𝐴�� − 𝐴𝐴 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� − 𝐶𝐶 = ( ) . (8) 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑏𝑏 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠The𝑠𝑠𝑠𝑠 𝐴𝐴 sinus−𝜔𝜔 theorem𝑎𝑎 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 in the triangle provides: = . (9) 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 𝑠𝑠The𝑠𝑠𝑠𝑠𝑠𝑠 relations𝑎𝑎 (5), (8) and (9) lead to: = . ′′ 2 𝐵𝐵𝐴𝐴 𝑐𝑐 ′′ 2 𝐶𝐶𝐴𝐴 𝑎𝑎

Ion Pătrașcu, Florentin Smarandache 163 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 58 1. Denoting { } = and { } = , we obtain similarly the relations: ′′ ′′ 𝐵𝐵 𝐵𝐵𝐵𝐵 ∩ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐶𝐶𝐶𝐶 ∩ 𝐴𝐴𝐴𝐴 = = ′′ 2, ′′ 2. 𝐶𝐶𝐵𝐵 𝑎𝑎 𝐴𝐴𝐶𝐶 𝑏𝑏 ′′ 2 {′′ } 2 { } { } 2. 𝐵𝐵Denoting𝐴𝐴 𝑏𝑏 𝐶𝐶 𝐵𝐵 =𝑐𝑐 , = , = , and proceeding similarly′′′ , we′ find: ′′′ ′ ′′′ ′ 𝐴𝐴 𝐴𝐴𝛺𝛺 ∩ 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐵𝐵𝛺𝛺 ∩ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐶𝐶𝛺𝛺 ∩ 𝐴𝐴𝐴𝐴 = = = ′′′ 2; ′′′ 2; ′′′ 2. 𝐵𝐵𝐴𝐴 𝑎𝑎 𝐶𝐶𝐵𝐵 𝑏𝑏 𝐴𝐴𝐶𝐶 𝑐𝑐 ′′′ 2 ′′′ 2 ′′′ 2 𝐶𝐶𝐴𝐴 𝑏𝑏 𝐴𝐴𝐵𝐵 𝑐𝑐 𝐵𝐵𝐶𝐶 𝑎𝑎 Lemma 8 In a triangle , the following relation is true: cot = cot + cot + cot . (10) 𝐴𝐴𝐴𝐴𝐴𝐴 Proof𝜔𝜔 𝐴𝐴 𝐵𝐵 𝐶𝐶 From relation (8), it follows that: sin( ) = sin . (11) 𝑎𝑎 sin 𝐴𝐴 But 𝐴𝐴 − 𝜔𝜔= ; replacing𝑏𝑏 ∙ sin 𝐶𝐶 ∙ in (11),𝜔𝜔 we find that: sin 𝐴𝐴 𝑎𝑎 sin(sin 𝐵𝐵 )𝑏𝑏= . 2 sin 𝐴𝐴∙sin 𝜔𝜔 We develop: sin( ) = sin sin sin cos . 𝐴𝐴 − 𝜔𝜔 sin 𝐵𝐵∙sin 𝐶𝐶 We have: sin cos sin cos = . (12) 𝐴𝐴 − 𝜔𝜔 𝐴𝐴 ∙ 𝜔𝜔 − 2 𝜔𝜔 ∙ 𝐴𝐴 sin 𝐴𝐴∙sin 𝜔𝜔 We divide the relation (12) by sin sin , and considering that sin = 𝐴𝐴 ∙ 𝜔𝜔 − 𝜔𝜔 ∙ 𝐴𝐴 sin 𝐵𝐵∙sin 𝐶𝐶 sin( + ), and sin( + ) = sin cos + sin cos , we get the relation 𝐴𝐴 ∙ 𝜔𝜔 𝐴𝐴 (10). 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐵𝐵 ∙ 𝐶𝐶 𝐶𝐶 ∙ 𝐵𝐵 Observation 59 From (10), we obtain that: tan = (13) 4𝑆𝑆 2 2 2 𝜔𝜔 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 Lemma 9 If, in the triangle , is symmedian center, and is the projection of on , then: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐾𝐾 𝐾𝐾1 = tan . 𝐾𝐾 𝐵𝐵𝐵𝐵1 𝐾𝐾𝐾𝐾1 2 𝑎𝑎 ∙ 𝜔𝜔

164 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Proof If and are symmedians, applying Menelaus’s theorem in the triangle 2 and taking2 into account that = and = , we get that 𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 2 2 𝐵𝐵𝐴𝐴2 𝑐𝑐 𝐶𝐶2𝐴𝐴 𝑏𝑏 2 2 = 𝐴𝐴𝐴𝐴2, 𝐵𝐵and from here = . 𝐶𝐶𝐴𝐴2 𝑏𝑏 𝐶𝐶2𝐵𝐵 𝑎𝑎 2 2 2 2 2 𝐴𝐴𝐴𝐴 𝑏𝑏 +𝑐𝑐 𝐴𝐴𝐴𝐴2 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 2 2 𝐾𝐾𝐴𝐴2But 𝑎𝑎 = ( is the𝐾𝐾 altitude𝐴𝐴2 from𝑎𝑎 of the triangle ). 𝐴𝐴𝐴𝐴2 ℎ𝑎𝑎 From𝐾𝐾 𝐴𝐴2 =𝐾𝐾𝐾𝐾1 ℎ𝑎𝑎 , since = , it𝐴𝐴 follows that 𝐴𝐴𝐴𝐴𝐴𝐴= tan . 2 2 2 ℎ𝑎𝑎 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 2𝑆𝑆 1 2 𝐾𝐾𝐾𝐾1 𝑎𝑎 ℎ𝑎𝑎 𝑎𝑎 𝐾𝐾𝐾𝐾1 2 𝑎𝑎 ∙ 𝜔𝜔 Observation 60 If , are the projections of on and , then the following relation takes place: 𝐾𝐾2 𝐾𝐾3 𝐾𝐾 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 = = = tan . 𝐾𝐾𝐾𝐾1 𝐾𝐾𝐾𝐾2 𝐾𝐾𝐾𝐾3 1 𝑎𝑎 𝑏𝑏 𝑐𝑐 2 𝜔𝜔 Lemma 10 In the triangle the cevians of Brocard and intersect in the point (vertex of the first Brocard triangle). ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝛺𝛺 1 Proof𝐴𝐴

Figure 86

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Let { } = (see Figure 86). Because = , it follows that′ the triangle ′ is isosceles if ′ ( is the ′midpoint of 𝐴𝐴1 𝐵𝐵𝐵𝐵 ∩ 𝐶𝐶𝛺𝛺 ∢𝐴𝐴1𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴1𝐶𝐶𝐶𝐶 𝜔𝜔 ). Moreover, having: ′ = tan , therefore′ ′ ′ = , it follows 𝐵𝐵𝐵𝐵1𝐶𝐶 𝐴𝐴1𝐴𝐴 ⊥ 𝐵𝐵𝐵𝐵 𝐴𝐴 ′ ′ 1 ′ ′ that = . 1 𝐵𝐵𝐵𝐵 𝐴𝐴1𝐴𝐴 2 ∙ 𝑎𝑎 ∙ 𝜔𝜔 𝐴𝐴1𝐴𝐴 𝐾𝐾𝐾𝐾 ′ 1 1 Lemma𝐴𝐴 𝐴𝐴 11 Let – a triangle, and – its first Brocard triangle. The lines , , are the isotomics of the symmedians , , . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 1 1 1 2 2 2 𝐴𝐴𝐴𝐴Proof𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 We denote by = { }, = { } and = { }. Since , and are concurrent′′ ′ , applying ′′′Ceva’s theorem, we′ have 𝐵𝐵𝐵𝐵 ∩ 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶𝛺𝛺 ∩ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴1 𝐴𝐴2 that: ′ 𝐵𝐵𝐵𝐵 𝐶𝐶𝛺𝛺 𝐴𝐴𝐴𝐴1 = 1, but = and = , we obtain that: = . ′′ ′′′ ′ ′′ 2 ′′′ 2 ′ 2 𝐶𝐶𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐴𝐴2𝐵𝐵 𝐶𝐶𝐵𝐵 𝑎𝑎 𝐶𝐶 𝐴𝐴 𝑐𝑐 𝐴𝐴2𝐵𝐵 𝑏𝑏 ′′ ′′′ ′ ′′ 2 ′′′ 2 ′ 2 Because𝐵𝐵 𝐴𝐴 ∙ 𝐶𝐶 𝐵𝐵 ∙is𝐴𝐴 2symmedian𝐶𝐶 𝐵𝐵and𝐴𝐴 𝑏𝑏 = , 𝐶𝐶it follows𝐵𝐵 𝑎𝑎 that the points 𝐴𝐴2 𝐶𝐶and 𝑐𝑐 2 𝐴𝐴2𝐵𝐵 𝑐𝑐 ′ 2 are symmetric𝐴𝐴𝐴𝐴2 with respect to midpoint𝐴𝐴2𝐶𝐶 𝑏𝑏 of , therefore is the𝐴𝐴 2isotomic𝐴𝐴2 of symmedian ; similarly, we have that′ and are′ the isotomic of 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴2 symmedian ; respectively of symmedian ′ . ′ 𝐴𝐴𝐴𝐴2 𝐵𝐵𝐵𝐵2 𝐶𝐶𝐶𝐶2 2 2 We are now𝐵𝐵𝐵𝐵 completing the second step of𝐶𝐶𝐶𝐶 the proof. It is known that the isotomics of concurrent cevians in a triangle are concurrent cevians and, since , , are isotomics of symmedian, it follows that they are concurrent and, consequently, the triangle and its first Brocard triangle are 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 homological triangles. The center of these homologies is denoted, by some 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 authors, , and it is called the third Brocard point. ′′ Remark𝛺𝛺 20 Sondat’s theorem implies the collinearity of points: – the center of the circle circumscribed to the triangle ; – Tary point; and – third Brocard point of 𝑂𝑂 the triangle . ′′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇 Ω 𝐴𝐴𝐴𝐴𝐴𝐴

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6.2.2 A triangle and its Neuberg triangle

Definition 40 Two triangles that have the same Brocard angle are called equibrocardian triangles.

Observation 61 a) Two similar triangles are equibrocardian triangles. b) A given triangle and its first Brocard triangle are equibrocardian triangles.

Theorem 30 The geometric place of the points in plane located on the same section of the side of a given triangle , that has the 𝑀𝑀 property that the triangle is equibrocardian with , is a 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 circle of center located on the mediator of the side , such 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 that ( ) 𝑎𝑎= 2 , and of radius = cot 3 𝑁𝑁 1 𝐵𝐵𝐵𝐵 (Neuberg circle – 1882). 2 𝑚𝑚 ∢𝐵𝐵𝑁𝑁𝑎𝑎𝐶𝐶 𝜔𝜔 𝑛𝑛𝑎𝑎 2 ∙ 𝑎𝑎 ∙ √ 𝜔𝜔 − Proof Let be a given triangle (see Figure 87). We start the proof by building some points of the geometric place in the idea of "identifying" the shape of the place. It𝐴𝐴𝐴𝐴𝐴𝐴 is clear that the Brocard point of a triangle that is equibrocardian with can be considered on the semi-line ( . We choose – the intersection between ( and the mediator of the side . ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝛺𝛺 We build the geometric place of the points in the boundary plane , 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 containing the point , of which the segment "is seen" from an angle of 𝑀𝑀 𝐵𝐵𝐵𝐵 measure . This geometric place is the circle of cent′ er – the intersection of 𝐴𝐴 𝐵𝐵𝛺𝛺 mediator of the segment with the perpendicular in ′to , having as 𝜔𝜔 𝑂𝑂 radius. ′ ′ 𝐵𝐵𝛺𝛺 𝐵𝐵 𝐵𝐵𝐵𝐵 𝑂𝑂 𝐵𝐵 We build now a secant to this circle, , such that = ; let be the second intersection point of this secant with the circle′ ( ; ). 𝐶𝐶𝑀𝑀1 𝑚𝑚�𝛺𝛺�𝐶𝐶𝑀𝑀1� 𝜔𝜔 The triangles and have the same Brocard point and the same 𝑀𝑀2 𝒞𝒞 𝑂𝑂′ 𝑂𝑂′𝐵𝐵 angle , therefore they are equibrocardian with ; therefore ′the points , 𝑀𝑀1𝐵𝐵𝐵𝐵 𝑀𝑀2𝐵𝐵𝐵𝐵 𝛺𝛺 belong to the sought geometric place. We have now three points of the 𝜔𝜔 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀1 sought geometric place, , , . 𝑀𝑀2 𝐴𝐴 𝑀𝑀1 𝑀𝑀2

Ion Pătrașcu, Florentin Smarandache 167 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Obviously, we can build the symmetric of this figure with respect to mediator of the segment , and we get other three points, , , . It is possible that the sought geometric place to be a circle and its′ center′ ′be, for 𝐵𝐵𝐵𝐵 𝐴𝐴 𝑀𝑀 1 𝑀𝑀 2 reasons of symmetry, located on the mediator of the segment . We denote by the intersection of this mediator with the circle ( ; ). We prove 𝐵𝐵𝐵𝐵 that is the center of the circle – geometric place – the Neuberg′ ′ circle. We 𝑁𝑁𝑎𝑎 𝒞𝒞 𝑂𝑂 𝑂𝑂 𝐵𝐵 have = , also = ( ). The triangle is 𝑁𝑁𝑎𝑎 isosceles, hence′ = ′ . ′ ′ ′ ∢𝐵𝐵𝑁𝑁𝑎𝑎𝛺𝛺 𝜔𝜔 ∢𝑂𝑂 𝐵𝐵𝑁𝑁𝑎𝑎 𝜔𝜔 𝑂𝑂 𝐵𝐵 ∥ 𝑁𝑁𝑎𝑎𝛺𝛺 𝑂𝑂 𝐵𝐵𝑁𝑁𝑎𝑎 For reasons of symmetry′ , we have that = , consequently is a ∢𝑂𝑂 𝑁𝑁𝑎𝑎𝐵𝐵 𝜔𝜔 fixed point located on the mediator of , because′ = 2 . We ∢𝐶𝐶𝑁𝑁𝑎𝑎𝛺𝛺 𝜔𝜔 𝑁𝑁𝑎𝑎 show that = . ′ 𝑁𝑁𝛺𝛺 𝐵𝐵𝐵𝐵 𝑚𝑚�𝐵𝐵�𝑁𝑁𝑎𝑎𝐶𝐶� 𝜔𝜔

𝑁𝑁𝑎𝑎𝑀𝑀1 𝑁𝑁𝑎𝑎𝑀𝑀2

Figure 87

We denote by I the intersection between and , and by the intersection between and ; because = 2 ′ , , 𝐶𝐶𝑀𝑀1 𝛺𝛺 𝑁𝑁𝑎𝑎 𝐽𝐽 and = 2 , it follows′ that ; consequently: = 𝐶𝐶𝑀𝑀1 𝑂𝑂 𝑁𝑁𝑎𝑎 𝑚𝑚�𝑁𝑁�𝑁𝑁𝑁𝑁� 𝜔𝜔 ∢𝑁𝑁𝑁𝑁𝑁𝑁 ≡ ∢𝑁𝑁𝑎𝑎𝐼𝐼𝐼𝐼 . ′ 𝑚𝑚�𝐽𝐽�𝑁𝑁𝑎𝑎𝑁𝑁� 𝜔𝜔 𝑁𝑁𝑎𝑎𝑂𝑂 ⊥ 𝑀𝑀1𝑀𝑀2 𝑁𝑁𝑎𝑎𝑀𝑀1 𝑁𝑁𝑎𝑎𝑀𝑀2 168 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

We prove now that = cot 3. We denote by the projection 1 2 of on , and we have𝑎𝑎 : 𝑁𝑁 𝐴𝐴 2 𝑎𝑎 ∙ √ 𝜔𝜔 − 𝑃𝑃 = 2 + 2 (generalized Pythagorean theorem). 𝐴𝐴 𝑁𝑁𝑎𝑎𝑁𝑁 From2 the median theorem2 in the triangle , we have: 𝑁𝑁𝑎𝑎𝐴𝐴 𝐴𝐴𝑁𝑁 𝑁𝑁𝑁𝑁𝑎𝑎 − 𝑁𝑁𝑁𝑁𝑎𝑎 ∙ 𝑃𝑃𝑃𝑃 4 = 2 2 + 2 2, apoi = cot . 𝐴𝐴𝐴𝐴𝐴𝐴 2 1 We established that: cot = 𝑎𝑎 . 2 𝐴𝐴𝑁𝑁 �𝑏𝑏 𝑐𝑐 � − 𝑎𝑎 2 𝑁𝑁𝑁𝑁2 2 ∙ 𝑎𝑎 ∙ 𝜔𝜔 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 Area = 𝜔𝜔 cos 4𝑆𝑆 . 1 From cosine theorem applied in the triangle , we have: ∆𝑀𝑀𝑀𝑀𝑀𝑀 2 ∙ 𝑎𝑎 ∙ 𝑀𝑀𝑀𝑀 ∙ �𝑀𝑀�𝑀𝑀𝑁𝑁𝑎𝑎� 2 2 ( ) = + 2 cos . 𝑎𝑎 2 𝑀𝑀𝑀𝑀𝑁𝑁 We𝑎𝑎 established that𝑎𝑎 : = 𝑎𝑎cot . 𝑎𝑎 𝑀𝑀 𝑀𝑀𝑁𝑁 𝑁𝑁 𝑁𝑁 − 𝑀𝑀𝑀𝑀𝑀𝑀1 𝑁𝑁 ∙ 𝑀𝑀�𝑀𝑀𝑁𝑁 By replacing in the preceding formula, we have: 𝑁𝑁𝑎𝑎𝑁𝑁 2 𝑎𝑎 ∙ 𝜔𝜔

= + cot cot 2 3 2. 2𝑀𝑀𝑀𝑀 + 𝑎𝑎 2 2 1 2 2 2 𝑛𝑛The𝑎𝑎 radius𝑁𝑁𝑀𝑀 of Neuberg4 𝑎𝑎 ∙ circle𝜔𝜔 −: 𝑎𝑎 ∙ = 𝜔𝜔 ∙ 2𝑎𝑎∙cot 𝜔𝜔 3. 1 We have: 2 𝑛𝑛𝑎𝑎 2 ∙ 𝑎𝑎 ∙ √ 𝜔𝜔 − cot = + cot . 1 2 2 3 2 2 1 2 2 3 2 cot 𝜔𝜔 cot 𝜔𝜔 2 It follows that: ′ ′ 4 𝑎𝑎 ∙ 𝜔𝜔 − 4 𝑎𝑎 𝑀𝑀𝑁𝑁 4 𝑎𝑎 ∙ 𝜔𝜔 − 4 𝑎𝑎 ∙ cot 𝜔𝜔 − cot 𝜔𝜔 ∙ 𝑀𝑀𝑁𝑁 + = 0. ′ cot 𝜔𝜔 −cot 𝜔𝜔 3 From this relation2 , we get2 cot = cot , which implies = . cot 𝜔𝜔 �𝑀𝑀𝑁𝑁 4 𝑎𝑎 � ′ ′ Remark 21 𝜔𝜔 𝜔𝜔 𝜔𝜔 𝜔𝜔 Let us reformulate the statement of Theorem 30 as follows: Find the geometric place of the points in the plane of the triangle with the property that the triangles with a vertex in and the other two to be vertices of 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 the given triangle and have the same Brocard angle as ; the answer will be 𝑀𝑀 given in the same way as before, but there are six Neuberg circles (two symmetrical 𝐴𝐴𝐴𝐴𝐴𝐴 with respect to each side of the given triangle).

We observe that 2 = , therefore denoting = we have: cot = . 𝑎𝑎 2 2 𝑆𝑆 𝑎𝑎 ∙ 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴 𝑚𝑚 3𝑎𝑎 +4𝑚𝑚𝑎𝑎 =4𝑎𝑎∙𝑃𝑃𝑃𝑃 + cot cot . 𝜔𝜔 2 2 2 2 1 2 2 3𝑎𝑎 +4𝑚𝑚𝑎𝑎 𝑁𝑁We𝑎𝑎𝐴𝐴 get: 𝑚𝑚𝑎𝑎 =4 𝑎𝑎 ∙ cot𝜔𝜔 − 𝑎𝑎 ∙3. 𝜔𝜔 ∙ 4𝑎𝑎∙cot 𝜔𝜔 1 2 𝑁𝑁𝑎𝑎𝐴𝐴 2 𝑎𝑎 ∙ √∙ 𝜔𝜔 −

Ion Pătrașcu, Florentin Smarandache 169 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

We proved that any vertex of a triangle that is equibrocardian with and has the common side with belongs to the Neuberg circle ( ; ). 𝐴𝐴𝐴𝐴𝐴𝐴 We prove the reciprocal, namely that any point that belongs to the circle 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝒞𝒞 𝑁𝑁𝑎𝑎 𝑛𝑛𝑎𝑎 ( ; ) is the vertex of the triangle that is equibrocardian with the 𝑀𝑀 given triangle . 𝒞𝒞 𝑁𝑁𝑎𝑎 𝑛𝑛𝑎𝑎 𝑀𝑀𝑀𝑀𝑀𝑀 We denote by – Brocard angle of the triangle (see Figure 88), 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ( ; ), then cot = . 𝜔𝜔 2 2 2 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀 ∈ 𝑀𝑀𝐵𝐵 +𝑀𝑀𝐶𝐶 +𝐵𝐵𝐶𝐶 From the median theorem′ applied in the triangle MBC, we note that: 𝒞𝒞 𝑁𝑁𝑎𝑎 𝑛𝑛𝑎𝑎 𝜔𝜔 4∙Area ∆𝑀𝑀𝑀𝑀𝑀𝑀 + = 2 + . 2 2 2 2 𝑎𝑎 𝑀𝑀𝐵𝐵 𝑀𝑀𝐶𝐶 𝑀𝑀𝑁𝑁 4

Figure 88

Definition 41 We call Neuberg triangle of a given triangle – the triangle formed by the centers of Neuberg circles (located in 𝑨𝑨𝑨𝑨𝑨𝑨 semi-planes determined by a side and by a vertex of the triangle 𝑵𝑵𝒂𝒂𝑵𝑵𝒃𝒃𝑵𝑵𝒄𝒄 ).

𝑨𝑨𝑨𝑨𝑨𝑨 170 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Theorem 31 The triangle and its Neuberg triangle are bilogical triangles. The homology center is the Tarry point of the triangle , 𝑎𝑎 𝑏𝑏 𝑐𝑐 and an orthology𝐴𝐴𝐴𝐴𝐴𝐴 center is – the center of the𝑁𝑁 circle𝑁𝑁 𝑁𝑁 circumscribed 𝐴𝐴𝐴𝐴𝐴𝐴 to the triangle . 𝑂𝑂 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 The perpendiculars taken from , , to the sides of the triangle are its mediators, hence – the center of the circle circumscribed to the triangle 𝑁𝑁𝑎𝑎 𝑁𝑁𝑏𝑏 𝑁𝑁𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 , is orthology center of Neuberg triangle in relation to the triangle . 𝑂𝑂 In Proposition 70, we proved that the perpendiculars taken from , , to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 the sides of the first Brocard triangle are concurrent in the Tarry point, , of the 𝐴𝐴 𝐵𝐵 𝐶𝐶 triangle . To prove that is the homology center of triangles and 𝑇𝑇 , it is sufficient to prove that the points , , ; , , ; , , are 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 collinear. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁𝑎𝑎 𝐴𝐴 𝑇𝑇 𝑁𝑁𝑏𝑏 𝐵𝐵 𝑇𝑇 𝑁𝑁𝑐𝑐 𝐶𝐶 𝑇𝑇 The condition , , to be collinear is equivalent to 1 1, namely to 1 1 = 0𝑁𝑁;𝑎𝑎 𝐴𝐴 𝑇𝑇= + ; = +𝑁𝑁𝑎𝑎𝐴𝐴 ⊥+𝐵𝐵 𝐶𝐶 ; = ′ ′ ′ ′ ′ ′ 𝐴𝐴��𝐴𝐴��+𝑎𝑎⃗ ∙ 𝐵𝐵��𝐶𝐶��;�⃑ =𝐴𝐴��𝐴𝐴��𝑎𝑎⃗ �𝐴𝐴𝐴𝐴��;� �⃗′ 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗1𝐵𝐵��1�1�𝐶𝐶��=1⃑ 𝐵𝐵��1��𝐵𝐵��+⃗ 𝐵𝐵��𝐶𝐶��⃗ 𝐶𝐶�1��𝐶𝐶��1⃗+ �𝐴𝐴𝐴𝐴��⃗ + 1 1 ′ ′ ′ ′ ′ 1 ′ ′ 1 1 1 1 2 �𝐴𝐴𝐴𝐴��⃗ =𝐴𝐴𝐴𝐴��⃗� 𝐵𝐵��𝐶𝐶��⃗ + − 2 𝐵𝐵𝐵𝐵��⃗ 𝐴𝐴��𝐴𝐴+��𝑎𝑎⃗ ∙ 𝐵𝐵��𝐶𝐶��⃑ +𝐴𝐴𝐴𝐴��⃗ 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗ +𝐵𝐵��𝐵𝐵��⃗ 𝐵𝐵��𝐶𝐶��⃗ + 1 2 1 2 2 1� 2 1 � � 2 ′ ′ ′ ′ ′ ′ ′ ′ 1 𝐶𝐶��𝐶𝐶��⃗ 𝐴𝐴𝐴𝐴��+��⃗ ∙ 𝐵𝐵��𝐵𝐵��⃗ 𝐴𝐴𝐴𝐴��+��⃗ ∙ 𝐵𝐵��𝐶𝐶��⃗ 𝐴𝐴𝐴𝐴��+�⃗ ∙ 𝐶𝐶��𝐶𝐶��⃗ 𝐴𝐴𝐴𝐴��.�⃗ ∙ 𝐵𝐵��𝐵𝐵��⃗ 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐵𝐵��𝐶𝐶��⃗ 2 � 1 1 1 ′ ′ ′ ′ ′ ′ ′ ′ 𝐴𝐴𝐴𝐴��⃗But∙ 𝐶𝐶�� 𝐶𝐶��⃗ 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗=∙ 𝐵𝐵�0��;�� 𝐵𝐵��⃗ 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗ =∙ 𝐵𝐵��0��𝐶𝐶�;� ⃗ 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗ ∙ 𝐶𝐶��𝐶𝐶=��⃗0. ′ = tan ′ sin , ′ ′ ′ 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐶𝐶��𝐶𝐶��1⃗ 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐵𝐵��1��𝐵𝐵��⃗ �𝐴𝐴��𝑁𝑁��𝑎𝑎⃗ ∙ 𝐵𝐵��𝐶𝐶��⃗ 1 ′ 1 2 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐵𝐵��1��𝐵𝐵��⃗= − 4 𝑏𝑏𝑏𝑏tan∙ 𝜔𝜔sin∙ , 𝐴𝐴 1 ′ 1 2 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐶𝐶��𝐶𝐶��1⃗ = 4 𝑏𝑏𝑏𝑏 ∙ cos𝜔𝜔,∙ 𝐴𝐴 1 ′ ′ 1 2 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐵𝐵��𝐶𝐶��⃗ = 4 𝑎𝑎𝑎𝑎 ∙ cos𝐵𝐵 , 1 1 ′ ′ 1 2 ��⃗ ��⃗ = 4 cos , 𝐴𝐴𝐴𝐴 ∙ 𝐵𝐵 1𝐶𝐶 −4 𝑎𝑎𝑎𝑎 ∙ 𝐶𝐶 ′ ′ 1 𝐴𝐴��𝑁𝑁��𝑎𝑎⃗ ∙ 𝐵𝐵��𝐵𝐵��⃗ = 𝑎𝑎𝑎𝑎 ∙cos 𝐶𝐶. 1 4 ′ ′ 𝐴𝐴We��𝑁𝑁�� 𝑎𝑎⃗consequently∙ 𝐶𝐶��𝐶𝐶��⃗ 𝑎𝑎 𝑎𝑎obtain∙ 𝐵𝐵that = 0. Similarly, we show that the points , , ; , , are collinear. �𝐴𝐴��𝐴𝐴��𝑎𝑎⃗ ∙ 𝐵𝐵��1��𝐶𝐶��1⃗

𝑁𝑁𝑏𝑏 𝐵𝐵 𝑇𝑇 𝑁𝑁𝑐𝑐 𝐶𝐶 𝑇𝑇

Ion Pătrașcu, Florentin Smarandache 171 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

C' B' K A1

B1 O B A' C

Figure 89

Remark 22 The Neuberg triangle and the first Brocard triangle of a given triangle are bilogical triangles. The homology center is the center of the circle circumscribed to the given triangle, and one of the orthology centers is the Tarry point of the triangle.

Theorem 32 If is a given triangle, is its Neuberg triangle, and is its first Brocard triangle, then: these triangles are 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 bilogical two by two, have the same axis of orthology and the 1 1 1 same𝐴𝐴 𝐵𝐵 𝐶𝐶axis of homology.

172 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Proof We noticed that the points , , are collinear; applying Theorem 19 from [24], it follows that the triangles ′′ , , have two by two the 𝑂𝑂 𝑇𝑇 Ω same axis of homology. If we denote by the orthology center of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 in relation to and by – the orthology center of the triangle 𝑈𝑈 in relation to , then, according to Sondat’s theorem, it follows 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 𝑉𝑉 that the orthology axis of the triangles and , namely , is 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 perpendicular to their homology axis. Because is orthology center of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑂𝑂𝑂𝑂 triangles and , it means that their axis of orthology is the 𝑂𝑂 perpendicular from to their axis of homology, hence it coincides with , 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 and the orthology axis of the triangles and , passing through 𝑂𝑂 𝑂𝑂𝑂𝑂 and being perpendicular to the axis of homology, is . Sondat’s theorem 𝑁𝑁𝑎𝑎𝑁𝑁𝑏𝑏𝑁𝑁𝑐𝑐 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑇𝑇 implies the collinearity of points , , , and . ′′ 𝑂𝑂𝑂𝑂 6.2.3 A triangle and the triangle𝑇𝑇 that𝑂𝑂 Ωdetermines𝑈𝑈 𝑉𝑉 on its sides three congruent antiparallels

Theorem 33 (R. Tucker) Three congruent antiparallels in relation to the sides of a triangle determines on these sides six concyclic points.

Proof Let ( ), ( ), ( ) be the three antiparallels respectively to the congruent sides , , (see Figure 90). 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 We denote by , , the intersection of the pairs of antiparallels 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 ( ; ), ( ; ) and ( ; ). The triangles ; 𝐴𝐴3 𝐵𝐵3 𝐶𝐶3 ; are isosceles. Indeed, and , 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 𝐶𝐶1𝐶𝐶2 𝐴𝐴1𝐴𝐴2 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐴𝐴3𝐵𝐵1𝐶𝐶2 therefore . 𝐵𝐵3𝐶𝐶1𝐴𝐴2 𝐶𝐶3𝐵𝐵2𝐴𝐴1 ∢𝐵𝐵𝐵𝐵1𝐵𝐵2 ≡ ∢𝐴𝐴 ∢𝐶𝐶1𝐶𝐶2𝐶𝐶 ≡ ∢𝐴𝐴 These angles being opposite at vertex with and , we obtain ∢𝐶𝐶1𝐶𝐶2𝐶𝐶 ≡ ∢𝐵𝐵𝐵𝐵1𝐵𝐵2 that the triangle is isosceles. Similarly, it is shown that the triangles 𝐴𝐴3�𝐶𝐶2𝐵𝐵1 𝐴𝐴3�𝐵𝐵1𝐶𝐶2 and are isosceles. We obtain that the bisectors of triangle 𝐴𝐴3𝐵𝐵1𝐶𝐶2 are mediators of the segments ( ); ( ); ( ). 𝐵𝐵3𝐶𝐶1𝐴𝐴2 𝐶𝐶3𝐵𝐵2𝐴𝐴1 Let be the intersection of these bisectors (the center of the circle inscribed 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐵𝐵1𝐶𝐶2 𝐶𝐶1𝐴𝐴2 𝐴𝐴1𝐵𝐵2 in the triangle ), we have the relations = ; = ; = 𝑇𝑇 . The triangles and are congruent (S.S.S.), it follows that 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝑇𝑇𝐵𝐵1 𝑇𝑇𝐶𝐶2 𝑇𝑇𝐶𝐶1 𝑇𝑇𝐴𝐴2 𝑇𝑇𝐵𝐵2 , with the consequence: . 𝑇𝑇𝐴𝐴1 𝑇𝑇𝐵𝐵1𝐴𝐴3 𝑇𝑇𝐶𝐶2𝐴𝐴3 ∢𝑇𝑇𝐵𝐵1𝐴𝐴3 ≡ ∢𝑇𝑇𝐶𝐶2𝐴𝐴3 ∢𝑇𝑇𝐵𝐵1𝐵𝐵2 ≡ ∢𝑇𝑇𝐶𝐶2𝐶𝐶1

Ion Pătrașcu, Florentin Smarandache 173 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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This relation (together with = and = ) leads to , hence = . 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 𝑇𝑇𝐵𝐵1 𝑇𝑇𝐶𝐶2 Similarly, it follows that , with the consequence: ∆𝑇𝑇𝐵𝐵1𝐵𝐵2 ≡ ∆𝑇𝑇𝐶𝐶2𝐶𝐶1 𝑇𝑇𝐵𝐵2 𝑇𝑇𝐶𝐶1 . ∆𝑇𝑇𝐴𝐴2𝐴𝐴1 ≡ ∆𝑇𝑇𝐶𝐶1𝐶𝐶2 𝑇𝑇𝐴𝐴1 ≡ Thus, we obtain that: = = = = = , which 𝑇𝑇𝐶𝐶2 shows that the points , , , , , are on a circle with the center . 𝑇𝑇𝐴𝐴1 𝑇𝑇𝐴𝐴2 𝑇𝑇𝐵𝐵1 𝑇𝑇𝐵𝐵2 𝑇𝑇𝐶𝐶1 𝑇𝑇𝐶𝐶2 This circle is called Tucker circle. 𝐴𝐴1 𝐴𝐴2 𝐵𝐵1 𝐵𝐵2 𝐶𝐶1 𝐶𝐶2 𝑇𝑇 B3

O K T

B2 C1

B B1 C2 C

Figure 90

174 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Theorem 34 The triangles and are bilogical. The homology center is the symmedian center of the triangle , and the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴3𝐵𝐵3𝐶𝐶3 orthology centers are the center of the Tucker circle, , and the 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 center of the circle circumscribed to the triangle . 𝑇𝑇 Proof 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 We build the tangents in and to the circumscribed circle of the triangle , and we denote by their intersection. It is known that is symmedian 𝐵𝐵 𝐶𝐶 in the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑆𝑆 𝐴𝐴𝐴𝐴 We prove that the points , , are collinear (see Figure 91). 𝐴𝐴𝐴𝐴𝐴𝐴 B3 𝐴𝐴 𝐴𝐴3 𝑆𝑆

A2 A1

M C4 B4 T

B2 C1

A4 B B1 C2 C

Figure 91

Indeed, because is isosceles trapezoid, we have that ; on the other hand, is antiparallel with , therefore ; similarly, 𝐵𝐵1𝐶𝐶2𝐶𝐶1𝐵𝐵2 𝐵𝐵2𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 . The triangles and have respectively parallel sides, 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 ∥ 𝐵𝐵2𝐴𝐴3 hence they are homothetic, because { } = , it follows that the 𝐶𝐶𝐶𝐶 ∥ 𝐶𝐶1𝐴𝐴3 𝐴𝐴3𝐶𝐶1𝐵𝐵2 𝑆𝑆𝑆𝑆𝑆𝑆 homothety center is , consequently the points , , are collinear, so is 𝐴𝐴 𝐵𝐵𝐵𝐵2 ∩ 𝐶𝐶𝐶𝐶1 symmedian in the triangle ; similarly, it follows that and are the 𝐴𝐴 𝐴𝐴 𝐴𝐴3 𝑆𝑆 𝐴𝐴𝐴𝐴3 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵3 𝐶𝐶𝐶𝐶3 Ion Pătrașcu, Florentin Smarandache 175 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES other symmedians, therefore they are concurrent in – the symmedian center. This point is the homology center of the triangles and . We noticed 𝐾𝐾 that the perpendiculars from , , to , , are bisectors of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴3𝐵𝐵3𝐶𝐶3 triangle , therefore they are concurrent in – the center of Tucker circle. 𝐴𝐴3 𝐵𝐵3 𝐶𝐶3 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 This point is orthology center in the triangle in relation to . The 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝑇𝑇 perpendiculars from , , to the antiparallels , , are also 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴𝐴𝐴𝐴𝐴 concurrent. Because these antiparallels are parallel with the tangents taken from 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵3𝐶𝐶3 𝐶𝐶3𝐴𝐴3 𝐴𝐴3𝐵𝐵3 , respectively to the circle circumscribed, it means that the perpendiculars taken in , , to tangents pass through – the center of the circumscribed 𝐴𝐴 𝐵𝐵 𝐶𝐶 circle, and this point is consequently the orthology center of the triangle 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 in relation to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 3 3 3 Remark 23 𝐴𝐴 𝐵𝐵 𝐶𝐶

1. The homology of the triangles and 3 3 3 can be proved with the help of Pascal’s theorem relative to an inscribed hexagon (see [24]). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Indeed, applying this theorem in the inscribed hexagon , we obtain that its opposite sides, videlicet and ; and ; 𝐴𝐴1𝐴𝐴2𝐶𝐶1𝐶𝐶2𝐵𝐵1𝐵𝐵2 and , intersect respectively in the points , , , and these are collinear 𝐴𝐴1𝐴𝐴2 𝐵𝐵𝐵𝐵 𝐵𝐵1𝐵𝐵2 𝐴𝐴𝐴𝐴 𝐶𝐶1𝐶𝐶2 points. They determine the homology axis of triangles and ; 𝐴𝐴𝐴𝐴 𝑀𝑀 𝑁𝑁 𝑃𝑃 3 3 3 according to Desargues’s theorem, the lines , , are concurrent. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 2. From Sondat’s theorem, we obtain that the points , , are collinear. 𝐴𝐴𝐴𝐴3 𝐵𝐵𝐵𝐵3 𝐶𝐶𝐶𝐶3 Also from this theorem, we obtain that is perpendicular to the 𝐾𝐾 𝑂𝑂 𝑇𝑇 homology axis of the triangles and . 3 𝑂𝑂3 𝑂𝑂3 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Proposition 71 The triangles and , formed by the intersections to the diagonals of the trapezoids ; ; , are 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴4𝐵𝐵4𝐶𝐶4 homological. Their homology center is , the center of Tucker circle, and 𝐵𝐵1𝐶𝐶2𝐶𝐶1𝐵𝐵2 𝐶𝐶2𝐶𝐶1𝐴𝐴2𝐴𝐴1 𝐴𝐴1𝐵𝐵2𝐵𝐵1𝐴𝐴2 the line determined by and by the center of the circle circumscribed to the 𝑇𝑇 triangle – is perpendicular to the homology axis of the triangles. 𝑇𝑇 4 4 4 Proof𝐴𝐴 𝐵𝐵 𝐶𝐶 The quadrilateral is isosceles trapezoid; the triangle is isosceles; it follows that is mediator of the segment , hence it passes 𝐴𝐴1𝐴𝐴2𝐵𝐵1𝐵𝐵2 𝐶𝐶3𝐴𝐴1𝐴𝐴2 through , the center of Tucker circle of the triangle (see Figure 92). 𝐶𝐶3𝐶𝐶4 𝐴𝐴1𝐵𝐵2 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴

176 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Similarly, and pass through , hence the center of Tucker circle of the triangle is homology center of triangles 3 3 3 and . 𝐴𝐴3𝐴𝐴4 𝐵𝐵3𝐵𝐵4 𝑇𝑇 We denote by: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴4𝐵𝐵4𝐶𝐶4 { } = , { } = , 𝐿𝐿 𝐴𝐴3𝐵𝐵3 𝐴𝐴4𝐵𝐵4 { } = ∩ . 𝑀𝑀 𝐵𝐵3𝐶𝐶3 𝐵𝐵4𝐶𝐶4 According to∩ Desargues’s theorem, the points , , are collinear and 𝑁𝑁 𝐴𝐴3𝐶𝐶3 𝐴𝐴4𝐶𝐶4 belong to the homolog∩ y axis of the previously indicated triangles. 𝑀𝑀 𝑁𝑁 𝑃𝑃

A2 A1

M C4 B4 T

B2 C1

A4 B B1 C2 C

Figure 92

Because (subtending congruent arcs in Tucker circle), it follows that the quadrilateral is inscribable, we have = 𝐶𝐶4�𝐴𝐴1𝐵𝐵4 ≡ 𝐶𝐶4�𝐴𝐴2𝐵𝐵4 , therefore the point has equal powers over Tucker circle and over 𝐴𝐴1𝐴𝐴2𝐵𝐵4𝐶𝐶4 𝑀𝑀𝑀𝑀1 ∙ 𝑀𝑀𝑀𝑀2 the circumscribed circle of the triangle , so it belongs to the radical axis 𝑀𝑀𝑀𝑀4 ∙ 𝑀𝑀𝑀𝑀4 𝑀𝑀 of these circles. 𝐴𝐴4𝐵𝐵4𝐶𝐶4 Similarly, it is shown that the points and belong to the same radical axis. 𝐿𝐿 𝑁𝑁

Ion Pătrașcu, Florentin Smarandache 177 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

6.2.4 A triangle and the triangle of the projections of the center of the circle inscribed on its mediators

Theorem 35 Let be a scalene triangle and let be the triangle determined by the projections of the center ′of ′the′ inscribed circle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 in the triangle on its mediators; then, the triangles and are bilogical. 𝐼𝐼 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 Proof𝐴𝐴 𝐵𝐵 𝐶𝐶 We prove by vector that the homology center of the triangles and is the Nagel point, . We have: 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ ′ = + = + + . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑁𝑁 ′ ′ 𝑏𝑏 𝑐𝑐 𝑏𝑏−𝑐𝑐 𝑏𝑏−𝑐𝑐 𝐴𝐴�We��𝐴𝐴��⃗ considered�𝐴𝐴��𝐴𝐴⃗ 𝐼𝐼��𝐴𝐴�� ⃗the triangle2𝑝𝑝 ∙ 𝐴𝐴𝐴𝐴��⃗ 2𝑝𝑝 ,∙ with𝐵𝐵𝐵𝐵��⃗ − 2𝑎𝑎 ∙ 𝐴𝐴𝐴𝐴��⃗, and2𝑎𝑎 took∙ 𝐴𝐴𝐴𝐴�� into⃗ account that: = + , = , = , and is the projection 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ≤ 𝐴𝐴𝐴𝐴 𝑏𝑏 𝑐𝑐 ′ of𝐴𝐴𝐴𝐴 on2 𝑝𝑝 ∙ 𝐴𝐴𝐴𝐴��.� �⃗ 2𝑝𝑝 ∙ 𝐴𝐴𝐴𝐴��⃗ �𝐼𝐼��𝐴𝐴��⃗ 𝛼𝛼𝐵𝐵𝐵𝐵��⃗ �𝐵𝐵𝐵𝐵��⃗ 𝐴𝐴𝐴𝐴��⃗ − 𝐴𝐴𝐴𝐴��⃗ 𝐶𝐶𝑎𝑎 the midpoint of ( ), having = and = , we find 𝐼𝐼 𝐵𝐵𝐵𝐵 𝑎𝑎 𝑀𝑀𝑎𝑎= , therefore =𝐵𝐵𝐵𝐵 . 𝐵𝐵𝐶𝐶𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝐵𝐵𝑀𝑀𝑎𝑎 2 𝑏𝑏−𝑐𝑐 𝑏𝑏−𝑐𝑐 𝐶𝐶𝑎𝑎𝑀𝑀𝑎𝑎 =2𝑎𝑎 𝛼𝛼 + 2𝑎𝑎 + . 𝑏𝑏 𝑏𝑏−𝑐𝑐 𝑐𝑐 𝑏𝑏−𝑐𝑐 ′{ } = 𝐴𝐴�Let��𝐴𝐴��⃗ �2𝑝𝑝 − 2𝑎𝑎 � 𝐴𝐴𝐴𝐴��⃗, we�2 have𝑝𝑝 :2 𝑎𝑎 � 𝐴𝐴𝐴𝐴��⃗ ( ′ ) ( ) =𝑎𝑎 + . 𝐷𝐷 𝑎𝑎𝑎𝑎−𝐴𝐴𝑝𝑝𝐴𝐴𝑏𝑏−∩𝑐𝑐 𝐵𝐵𝐵𝐵 𝑎𝑎𝑎𝑎+𝑝𝑝 𝑏𝑏−𝑐𝑐 𝐴𝐴On��𝐷𝐷�� 𝑎𝑎⃗the other𝜆𝜆 2 hand𝑎𝑎𝑎𝑎 , 𝐴𝐴𝐴𝐴��⃗ =𝜆𝜆 2𝑎𝑎𝑎𝑎 𝐴𝐴𝐴𝐴��⃗. The scalars and are such that: 𝐷𝐷��𝑎𝑎��𝐶𝐶�⃗ 𝜇𝜇�𝐴𝐴𝐴𝐴��⃗ − 𝐴𝐴𝐴𝐴��⃗� = , therefore: 𝜆𝜆 𝜇𝜇 ( ) ( ) ��𝑎𝑎��⃗ ��⃗ =��𝑎𝑎⃗ . 𝐷𝐷 𝐶𝐶 𝐴𝐴𝐴𝐴 − 𝐴𝐴𝐷𝐷 𝑎𝑎𝑎𝑎−𝑝𝑝 𝑏𝑏−𝑐𝑐 𝑎𝑎𝑎𝑎+𝑝𝑝 𝑏𝑏−𝑐𝑐 𝜇𝜇It� derives�𝐴𝐴𝐴𝐴��⃗ − 𝐴𝐴𝐴𝐴�� that��⃗� : 𝐴𝐴𝐴𝐴��⃗ − 𝜆𝜆 2𝑎𝑎𝑎𝑎 𝐴𝐴𝐴𝐴��⃗ − 𝜆𝜆 2𝑎𝑎𝑎𝑎 𝐴𝐴𝐴𝐴��⃗ ( ) ( ) 1 + + = 0. 𝑎𝑎𝑎𝑎+𝑝𝑝 𝑏𝑏−𝑐𝑐 𝑎𝑎𝑎𝑎−𝑝𝑝 𝑏𝑏−𝑐𝑐 �The𝜇𝜇 − vectors𝜆𝜆 2 𝑎𝑎and𝑎𝑎 � 𝐴𝐴𝐴𝐴�� are�⃗ non�𝜆𝜆-collinear2𝑎𝑎𝑎𝑎 ; −it 𝜇𝜇follows� 𝐴𝐴𝐴𝐴��⃗ that: ( ) 1 + ��⃗ =��0�⃗, 𝑎𝑎𝑎𝑎+𝐴𝐴𝐴𝐴𝑝𝑝 𝑏𝑏−𝑐𝑐 𝐴𝐴𝐴𝐴 ( ) 𝜇𝜇 −+ 𝜆𝜆 2𝑎𝑎𝑎𝑎 = 0. 𝑎𝑎𝑎𝑎−𝑝𝑝 𝑏𝑏−𝑐𝑐 −𝜇𝜇 𝜆𝜆 2𝑎𝑎𝑎𝑎

178 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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We find: = and = , consequently = , therefore 𝑝𝑝 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑏𝑏 = , and the point is the isotomic of 𝑎𝑎(contact of the inscribed 𝜆𝜆 𝑏𝑏+𝑐𝑐 𝜇𝜇 𝑎𝑎 𝐷𝐷��𝐶𝐶�⃗ 𝑎𝑎 ∙ 𝐵𝐵𝐵𝐵��⃗ circle with ), hence pass through the Nagel point, . 𝐷𝐷𝑎𝑎𝐶𝐶 𝑝𝑝 − 𝑏𝑏 𝐷𝐷𝑎𝑎 𝐶𝐶𝑎𝑎 Similarly, it is shown that and pass through . Obviously, the 𝐵𝐵𝐵𝐵 𝐴𝐴𝐷𝐷𝑎𝑎 𝑁𝑁 triangles and are orthological′ , and′ the orthology center is – the �𝐵𝐵��𝐵𝐵��⃗ 𝐶𝐶��𝐶𝐶��⃗ 𝑁𝑁 center of the′ circle′ ′ circumscribed to the triangle . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 According to the the orthological triangles theorem, it follows that the 𝐴𝐴𝐴𝐴𝐴𝐴 perpendiculars taken from , , respectively to , and are concurrent. ′ ′ ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 We denote this orthology center by and we show that belongs to the circle circumscribed to the triangle . 𝛷𝛷 𝛷𝛷 A 𝐴𝐴𝐴𝐴𝐴𝐴

Mc I A' Mb C'

B B' T a M a D a C O

Figure 93

We have = 180 . On the other hand, the points , , belong to the circle of 0diameter′ ′ ′, therefore: ′; the′ ∢𝐵𝐵𝛷𝛷𝐴𝐴 − ∢𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐴𝐴 𝐵𝐵 latter′ has the sides perpendicular to the mediators of ′the′ sides′ ′and′ ; 𝐶𝐶 𝑂𝑂𝑂𝑂 ∢𝐴𝐴 𝐶𝐶 𝐵𝐵 ≡ −∢𝐴𝐴 𝐼𝐼𝐼𝐼 hence, it is congruent with . 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 = 180 Having ∢𝐵𝐵𝐵𝐵𝐵𝐵 , it means that the quadrilateral is inscribable, therefore belongs0 to the circle circumscribed to the triangle 𝑚𝑚 �𝐵𝐵�𝛷𝛷𝐴𝐴 − 𝑚𝑚�𝐶𝐶̂�� 𝐴𝐴𝛷𝛷𝛷𝛷𝛷𝛷 . 𝛷𝛷 Sondat’s theorem implies the collinearity of points , , . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁 𝑂𝑂 𝛷𝛷

Ion Pătrașcu, Florentin Smarandache 179 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Remark 24 In the anticomplementary triangle of the triangle , the circle circumscribed to the triangle is the circle of the nine points, and – Nagel point, is the center 𝐴𝐴𝐴𝐴𝐴𝐴 of the circle inscribed in the anticomplementary triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁 From Feuerbach's theorem (see [15]), these circles are tangent, the point of tangency being the point (the orthology center of the triangle in relation to the triangle ) – called Feuerbach point. Φ 𝐴𝐴𝐴𝐴𝐴𝐴 The collinearity′ ′ ′ of points , , indicated above, and the fact that the 𝐴𝐴 𝐵𝐵 𝐶𝐶 mentioned circles are interior tangents lead to = 2 . 𝑁𝑁 𝑂𝑂 Φ 6.2.5 A triangle and the triangle of the projections𝑂𝑂𝑂𝑂 𝑅𝑅 − of 𝑟𝑟the centers of ex-inscribed circles on its mediators

Tc I

B Ta T'a C

T'b

T'c

Figure 94

180 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Proposition 72 Let be a given triangle and the triangle whose vertices are the projections of the centers , ′ ,′ ′ of the ex-inscribed circles 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 respectively on the sides mediators ( ), ( ) and ( ). 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 𝐼𝐼𝑐𝑐 Then the triangles and are bilogical. The orthology center 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 of the triangle in relation to′ ′ ′ belongs to the line ( is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Gergonne point of the triangle and′ ′ –′ the center of its circumscribed 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Γ𝑂𝑂 Γ circle). 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 Proof Let and the projections of centers and on (see Figure 94). These points are isotomic. The inscribed circle and the -ex-inscribed circle 𝐶𝐶𝑎𝑎 𝐷𝐷𝑎𝑎 𝐼𝐼 𝐼𝐼𝑎𝑎 𝐵𝐵𝐵𝐵 are homothetic by homothety of center and by ratio . We denote by the 𝐴𝐴 𝐼𝐼𝐶𝐶𝑎𝑎 ′′ diameter of in the circle -ex-inscribed𝐴𝐴 ; we have𝐼𝐼𝐷𝐷𝑎𝑎 that and 𝐷𝐷 are homotetical points, therefore , , are collinear. The point ′′, the 𝐷𝐷𝑎𝑎 𝐴𝐴 𝐶𝐶𝑎𝑎 𝐷𝐷 projection of on the mediator of the side′′ , is the midpoint of the segment′ 𝐴𝐴 𝐶𝐶𝑎𝑎 𝐷𝐷 𝐴𝐴 because contains the midline of the rectangle , hence 𝐼𝐼𝑎𝑎 𝐵𝐵𝐵𝐵 belongs′′ to Gergonne′ cevian , similarly and belong to′′ Gergonne′ 𝐶𝐶𝑎𝑎𝐷𝐷 𝑂𝑂𝑂𝑂 𝐶𝐶𝑎𝑎𝐷𝐷𝑎𝑎𝐷𝐷 𝐴𝐴 cevian , . As these cevians are concurrent′ in ′ (Gergonne point), it 𝐴𝐴𝐴𝐴𝑎𝑎 𝐵𝐵 𝐶𝐶 follows that this point is homology center of the triangles and . 𝐵𝐵𝐵𝐵𝑎𝑎 𝐶𝐶𝐶𝐶𝑎𝑎 𝛤𝛤 Obviously, the triangle is orthological in relation to ′ and′ ′ the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthology center is . ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 From Sondat’s theorem, it follows that the orthology center of the triangle 𝑂𝑂 in relation to belongs to the line . ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴6.2.6 A triangle and𝐴𝐴 𝐵𝐵its𝐶𝐶 Napoleon triangle 𝛤𝛤𝛤𝛤

Definition 42 If is a triangle and we build in its exterior the equilateral triangles , , , we say about the triangle 𝑨𝑨𝑨𝑨𝑨𝑨 which has the′ centers′ of ′ the circles circumscribed to the 𝑩𝑩𝑩𝑩𝑨𝑨𝟏𝟏 𝑪𝑪𝑪𝑪𝑩𝑩𝟏𝟏 𝑨𝑨𝑨𝑨𝑪𝑪𝟏𝟏 𝑶𝑶𝟏𝟏𝑶𝑶𝟐𝟐𝑶𝑶𝟑𝟑 equilateral triangles that we just built – that it is a Napoleon exterior triangle corresponding to the triangle .

𝑨𝑨𝑨𝑨𝑨𝑨

Ion Pătrașcu, Florentin Smarandache 181 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 62 In Figure 95, the Napoleon exterior triangle is .

𝑂𝑂1𝑂𝑂2𝑂𝑂3 B1

A C 1 O2

O3 T

B C

Figure 95

Definition 43 The circles circumscribed to the equilateral triangles , , – are called Toricelli circles. 𝑩𝑩𝑩𝑩𝑨𝑨𝟏𝟏 𝟏𝟏 𝟏𝟏 𝑪𝑪𝑪𝑪Definition𝑩𝑩 𝑨𝑨𝑨𝑨 44𝑪𝑪 The triangle determined by the centers of the circles circumscribed ′to ′the′ equilateral triangles , , 𝑶𝑶𝟏𝟏𝑶𝑶𝟐𝟐𝑶𝑶𝟑𝟑 built on the sides of the triangle to its interior′ – is′ called a′ 𝑩𝑩𝑩𝑩𝑨𝑨𝟏𝟏 𝑪𝑪𝑪𝑪𝑩𝑩𝟏𝟏 𝑨𝑨𝑨𝑨𝑪𝑪𝟏𝟏 Napoleon interior triangle. 𝑨𝑨𝑨𝑨𝑨𝑨

182 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 36 On the sides of the triangle , we build in its exterior the equilateral triangles , , ; then: 𝐴𝐴𝐴𝐴𝐴𝐴 i) The Toricelli circles intersect in a point ; 𝐵𝐵𝐵𝐵𝐴𝐴1 𝐶𝐶𝐶𝐶𝐵𝐵1 𝐴𝐴𝐴𝐴𝐶𝐶1 ii) The Napoleon exterior triangle is equilateral; 𝑇𝑇 iii) 1 = 1 = 1; iv) The triangles and 1 1 1 are bilogical. 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 i) We denote by the second point of intersection of Toricelli circles circumscribed to the triangles and . We have = 𝑇𝑇 = 120 ; assuming that is located in the interior of triangle , it 𝐴𝐴𝐴𝐴𝐶𝐶1 𝐴𝐴𝐴𝐴𝐵𝐵1 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� follows that 0 = 120 , consequently the quadrilateral is 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴 inscribable, and, consequently,0 belongs to the circles circumscribed to the 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝐵𝐵𝐵𝐵𝐵𝐵𝐴𝐴1 equilateral triangle . 𝑇𝑇 1 Observation 63𝐵𝐵𝐵𝐵 𝐴𝐴 a) If > 120 , then is in the exterior of the triangle and = 0 = 60 ; it follows that = 120 , 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑇𝑇 therefore T belongs to Toricelli0 circle circumscribed to . 0 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝐶𝐶𝐶𝐶𝐶𝐶�� 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� b) In case that = 120 , the point coincides with the vertex 𝐵𝐵𝐵𝐵𝐴𝐴1 . 0 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑇𝑇 c) The point is called Toricelli-Fermat point of the triangle . 𝐴𝐴

ii) If the measures𝑇𝑇 of the angles of the triangle are smaller than𝐴𝐴𝐴𝐴𝐴𝐴 120 , then: = 120 + , = 120 + , = 120 +0 𝐴𝐴𝐴𝐴𝐴𝐴 . 0 0 0 𝑚𝑚�𝐵𝐵�1𝐴𝐴𝐶𝐶1� 𝐴𝐴 𝑚𝑚�𝐴𝐴�1𝐵𝐵𝐶𝐶1� 𝐵𝐵 𝑚𝑚�𝐵𝐵�1𝐶𝐶𝐴𝐴1� Also: = 60 + , = 60 + and = 𝐶𝐶 60 + . 0 0 𝑚𝑚�𝑂𝑂�2𝐴𝐴𝑂𝑂3� 𝐴𝐴 𝑚𝑚�𝑂𝑂�1𝐵𝐵𝑂𝑂3� 𝐵𝐵 𝑚𝑚�𝑂𝑂�1𝐶𝐶𝑂𝑂2� 0We calculate the sides of Napoleon triangle with the help of the cosine theorem𝐶𝐶. We have: = + 2 cos(60 + ). 2 2 2 0 𝑂𝑂2𝑂𝑂3 𝑂𝑂3𝐴𝐴 𝑂𝑂2𝐴𝐴 − 𝑂𝑂3𝐴𝐴 ∙ 𝑂𝑂2𝐴𝐴 ∙ 𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 183 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Because = , = and cos(60 + ) = cos √3 √3 0 1 sin , we have3 =3 +2 cos3 + 3 sin . 2 𝑂𝑂 𝐴𝐴 𝑐𝑐 ∙ 2 𝑂𝑂 𝐴𝐴2 𝑏𝑏 ∙ 𝐴𝐴 𝐴𝐴 − √3 2 𝑏𝑏 𝑐𝑐 𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 But 2 cos =2 3 + and sin = 2 . 2 𝐴𝐴 𝑂𝑂 𝑂𝑂 3 3 − 3 𝐴𝐴 3 ∙ √ 𝐴𝐴 2 2 2 We obtain: = . 𝑏𝑏𝑏𝑏 𝐴𝐴 𝑏𝑏 2𝑐𝑐 −2 𝑎𝑎2 𝑏𝑏𝑏𝑏 ∙ 𝐴𝐴 𝑆𝑆 2 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 +4𝑆𝑆√3 Similarly, 𝑂𝑂2𝑂𝑂3 and 6 are given by the same expression, hence the triangle is equilateral2 . 2 𝑂𝑂3𝑂𝑂2 𝑂𝑂2𝑂𝑂1

𝑂𝑂1𝑂𝑂2𝑂𝑂3 iii) We consider in the interior of the triangle , therefore no angle of the triangle has a measure greater than or equal to 120 . We have 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴 = = = 120 (this property makes that0 the point 𝐴𝐴𝐴𝐴𝐴𝐴 in this case to be called isogon center of the0 triangle ABC). On the other hand, 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝐶𝐶𝐶𝐶𝐶𝐶�� 𝑇𝑇 = 60 , hence = 120 + 60 = 180 , therefore the points , , 0are collinear (similarly , 0, and0 , , 0 are collinear). 𝑚𝑚�𝐵𝐵𝐵𝐵�𝐴𝐴1� 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐵𝐵1� From Van Schooten relation, we have that + = , since , , are 𝐴𝐴 𝑇𝑇 𝐴𝐴1 𝐵𝐵 𝑇𝑇 𝐴𝐴1 𝐶𝐶 𝑇𝑇 𝐴𝐴1 collinear, it follows that = + + , similarly = = + 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 𝑇𝑇𝐴𝐴1 𝐴𝐴 𝑇𝑇 𝐴𝐴1 + . 𝐴𝐴𝐴𝐴1 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 Observation𝑇𝑇𝑇𝑇 64 It can be shown that = = by direct calculation with cosine theorem; it is found that =1 1 1 . 𝐴𝐴𝐴𝐴 2𝐵𝐵𝐵𝐵2 2 𝐶𝐶𝐶𝐶 2 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 +4𝑆𝑆√3 𝐴𝐴𝐴𝐴1 2 iv) We have shown before that , , ; , , ; , , are collinear; hence, the triangles and are homological, and the homology center 𝐴𝐴 𝑇𝑇 𝐴𝐴1 𝐵𝐵 𝑇𝑇 𝐵𝐵1 𝐶𝐶 𝑇𝑇 𝐶𝐶1 is the Toricelli-Fermat point, . The perpendiculars taken from , , to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 , respectively are the mediators of these sides, consequently – the 𝑇𝑇 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 center of the circle circumscribed to the triangle is the orthology center of 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑂𝑂 the triangle in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 Theorem𝐴𝐴 3𝐵𝐵7 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 The triangle where no angle is greater than 1200 and its Napoleon exterior triangle – are bilogical triangles. i) The 𝐴𝐴𝐴𝐴𝐴𝐴orthology centers are the center of the circle circumscribed to the triangle and the isogonal center of 𝑂𝑂 the triangle ; 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴 184 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

ii) The homology center belongs to the axis of orthology and the axis of orthology is perpendicular to the axis of homology.

Proof

i) The perpendiculars taken from 1, 2, 3 to , respectively are the mediators of these sides; hence, is the orthology center of the 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 Napoleon triangle 1 2 3 in relation to . The segments , , 𝑂𝑂 are common chords in the Toricelli circles and, consequently, 2 3 is 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 mediator of [ ], 3 1 is mediator of [ ] and 1 2 is mediator of [ ]. 𝑂𝑂 𝑂𝑂 It follows that is the second orthology center of the triangles indicated 𝑇𝑇𝑇𝑇 𝑂𝑂 𝑂𝑂 𝑇𝑇𝑇𝑇 𝑂𝑂 𝑂𝑂 𝑇𝑇𝑇𝑇 in the statement. 𝑇𝑇 B1

C A 1 O2

O3 T P

O B C

Figure 96

Ion Pătrașcu, Florentin Smarandache 185 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

ii) Let , , be the intersections of the cevians 1, , 3 ′ ′ ′ respectively with , , . We have: 2 Area𝐴𝐴 𝐵𝐵 𝐶𝐶 Area Area ( ) 𝐴𝐴𝐴𝐴 𝐵𝐵𝑂𝑂 𝐶𝐶𝑂𝑂 = = 1 = 1 . ′ Area 𝐵𝐵𝐵𝐵′ 𝐶𝐶𝐶𝐶Area𝐴𝐴𝐴𝐴 ′ Area ( ) 𝐵𝐵𝐴𝐴 ∆�𝐴𝐴𝐴𝐴𝐴𝐴 � ∆�𝐵𝐵𝐵𝐵1𝐴𝐴 � ∆ 𝐴𝐴𝐴𝐴𝐴𝐴1 ′ ′ ′ We𝐶𝐶𝐴𝐴 obtain∆:� 𝐴𝐴𝐴𝐴𝐴𝐴 � ∆�𝐶𝐶𝐶𝐶 𝐴𝐴 � ∆ 𝐴𝐴𝐴𝐴𝐴𝐴 ( ) sin +300 = 1 1 = . ′ ( ) sin +300 𝐵𝐵𝐴𝐴 𝐴𝐴𝐴𝐴∙𝐵𝐵𝐵𝐵1∙sin 𝐴𝐴𝐴𝐴𝑂𝑂1 𝑐𝑐 �𝐵𝐵 � ′ Similarly𝐶𝐶𝐴𝐴 𝐴𝐴𝐴𝐴∙𝐶𝐶𝐶𝐶: ∙sin 𝐴𝐴𝐴𝐴𝑂𝑂 𝑏𝑏 ∙ �𝐶𝐶 � sin +300 = , ′ sin +300 𝐶𝐶𝐵𝐵 𝑐𝑐 �𝐶𝐶 � ′ sin +300 𝐵𝐵 𝐴𝐴 = 𝑎𝑎 ∙ �𝐴𝐴 �. ′ sin +300 𝐶𝐶 𝐴𝐴 𝑏𝑏 �𝐴𝐴 � ′ 𝐶𝐶 𝐵𝐵 𝑎𝑎 ∙ �𝐵𝐵 � = 1 From ′ ′ ′ and Ceva’s reciprocal theorem, we obtain that 1, 𝐵𝐵𝐴𝐴 𝐶𝐶𝐵𝐵 𝐴𝐴𝐶𝐶 , are′ concurrent′ ′ , hence the triangle and the Napoleon exterior 𝐶𝐶𝐴𝐴 ∙ 𝐴𝐴𝐵𝐵 ∙ 𝐵𝐵𝐶𝐶 𝐴𝐴𝐴𝐴 triangle, 1 2 3, are homological. We denote by the center of this 𝐵𝐵𝑂𝑂2 𝐶𝐶𝑂𝑂3 𝐴𝐴𝐴𝐴𝐴𝐴 homology. 𝑂𝑂 𝑂𝑂 𝑂𝑂 𝑃𝑃 From Sondat’s theorem, it follows that the points , , are collinear and that the axis of orthology is perpendicular to the axis of homology of 𝑇𝑇 𝑂𝑂 𝑃𝑃 bilogical triangles and 1 2 3. 𝑂𝑂𝑂𝑂 Theorem 38 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 𝑂𝑂 On the sides of the given triangle we build the equilateral triangles , , (whose interiors intersect the 𝐴𝐴𝐴𝐴𝐴𝐴 interior of the triangle ). Then: 𝐵𝐵𝐵𝐵𝐴𝐴2 𝐶𝐶𝐶𝐶𝐵𝐵2 𝐴𝐴𝐴𝐴𝐶𝐶2 i) The circumscribed circles of these equilateral triangles 𝐴𝐴𝐴𝐴𝐴𝐴 have a common point . ii) The Napoleon interior triangle′ , , is equilateral. 𝑇𝑇 iii) = = . ′ ′ ′ 𝑂𝑂1𝑂𝑂2𝑂𝑂3 iv) The triangles and are bilogical triangles. 𝐴𝐴𝐴𝐴2 𝐵𝐵𝐵𝐵2 𝐶𝐶𝐶𝐶2 2 2 2 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 i) Let be the second intersection point of the circles circumscribed to the triangles ′ and (see Figure 97). 𝑇𝑇 = 60 = = 60 We have𝐴𝐴𝐴𝐴:𝐵𝐵 2 𝐴𝐴𝐴𝐴𝐶𝐶2 and . ′ 0 ′ ′ 0 From the last𝑚𝑚 � relation𝐵𝐵�𝑇𝑇 𝐶𝐶� , it follows 𝑚𝑚the� 𝐴𝐴collinearity�𝑇𝑇 𝐶𝐶� 𝑚𝑚 of�𝐴𝐴 �the𝑇𝑇 𝐶𝐶 points2� , , . ′ 𝑇𝑇 𝐶𝐶2 𝐶𝐶 186 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Because = = 60 , we obtain that belongs to the ′ 0 ′ circle circumscribed𝑚𝑚 �𝐵𝐵�𝑇𝑇 𝐶𝐶to� the𝑚𝑚 equilateral�𝐵𝐵�𝐴𝐴2𝐶𝐶� triangle . 𝑇𝑇 2 ii) We calculate the length of sides using 𝐵𝐵𝐵𝐵the𝐴𝐴 cosine theorem, bearing in mind that, in general, the angles , , have measures equal to: 60 , 60 or 60′ (or′ 60 ′ ′, 60 ′ ,′ 60 ) in the ∢𝑂𝑂1𝐴𝐴𝑂𝑂3 ∢𝑂𝑂2𝐵𝐵𝑂𝑂3 ∢𝑂𝑂2𝐶𝐶𝑂𝑂1 case that <0 30 or 0 < 30 0 or 0 < 60 .0 0 𝐴𝐴 − 𝐵𝐵 − 𝐶𝐶 − − 𝐴𝐴 − 𝐵𝐵 − 𝐶𝐶 It is obtained that:0 0 0 𝑚𝑚�𝐴𝐴̂� 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐶𝐶̂� = = = . 2 2 2 2 2 2 𝑎𝑎 +𝑏𝑏 +𝑐𝑐 −4𝑆𝑆√3 ′ ′ ′ ′ ′ ′ 𝑂𝑂1𝑂𝑂3 𝑂𝑂2𝑂𝑂3 𝑂𝑂1𝑂𝑂2 6 A2

O'1 O'3

C2 O'2 B C

Figure 97

Ion Pătrașcu, Florentin Smarandache 187 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

Remark 25 From the preceding expression, it is obtained that, in a triangle , the inequality 2 + 2 + 2 4 3 is true. The equality takes place if and only if 𝐴𝐴𝐴𝐴𝐴𝐴 is equilateral. 𝑎𝑎 𝑏𝑏 𝑐𝑐 ≥ 𝑆𝑆 𝐴𝐴𝐴𝐴𝐴𝐴 √ iii) (SAS), = , = and = = 60 (in the case of Figure 97), it follows that = . ∆𝐵𝐵𝐵𝐵𝐵𝐵2 ≡ ∆𝐶𝐶2𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶2𝐴𝐴 𝐴𝐴𝐵𝐵2 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐵𝐵𝐵𝐵�𝐵𝐵2� Similarly, 0 , it follows that = . 𝑚𝑚�𝐶𝐶�2𝐴𝐴𝐴𝐴� 𝐴𝐴 − 𝐵𝐵𝐵𝐵2 𝐶𝐶𝐶𝐶2

∆𝐴𝐴𝐴𝐴𝐴𝐴2 ≡ ∆𝐶𝐶2𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴2 𝐶𝐶𝐶𝐶2 iv). Similarly, since the colinearity , , was proved, the colinearity of the points , , and , , is also′ proved. This collinearity implies that 𝑇𝑇 𝐴𝐴 𝐴𝐴2 and ′ are homological′ . The homology center is the point , which 𝑇𝑇 𝐵𝐵 𝐵𝐵2 𝑇𝑇 𝐶𝐶 𝐶𝐶2 is called the second Torricelli-Fermat point. The perpendiculars taken from′ , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑇𝑇 , respectively to , and are the mediators of these sides; 𝐴𝐴2 consequently, the center of the circle circumscribed to the triangle , , is 𝐵𝐵2 𝐶𝐶2 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 the orthology center of these triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 Theorem 39 The given triangle and its Napoleon interior triangle are bilogical. ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2𝑂𝑂3 Proof The perpendiculars taken from , , to , respectively are mediators of these three sides, hence′ they′ are′ concurrent in – the center of 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 the circle circumscribed to the triangle , point that is the orthology center 𝑂𝑂 of the triangles and . 𝐴𝐴𝐴𝐴𝐴𝐴 Because is′ a′ common′ chord in the Toricelli circle of centers and , 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝐴𝐴𝐴𝐴𝐴𝐴 it follows that ′ is mediator of the segment , therefore the perpendicular′ ′ 𝑇𝑇 𝐴𝐴 𝑂𝑂3 𝑂𝑂2 from to ′pass′ es through ; similarly, it follows′ that the perpendiculars 𝑂𝑂3𝑂𝑂2 𝑇𝑇 𝐴𝐴 taken from ′ to′ and from ′ to pass through the second Toricelli- 𝐴𝐴 𝑂𝑂3𝑂𝑂2 𝑇𝑇 Fermat point, - ′point′ that is the orthology′ ′ center of the triangles and 𝐵𝐵 𝑂𝑂1𝑂𝑂3 𝐶𝐶 𝑂𝑂1𝑂𝑂2 . Let ′ , , be the intersections of cevians , , 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴 respectively′ ′ ′ with′ ′, ′ and . ′ ′ ′ 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵2 𝐶𝐶𝑂𝑂3 We have: 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 = = = . ′ ′ ′ ′ ′ 𝐵𝐵𝐴𝐴 Arie∆𝐴𝐴𝐴𝐴𝐴𝐴 Arie∆𝐵𝐵𝑂𝑂1𝐴𝐴 Arie∆𝐴𝐴𝐴𝐴𝑂𝑂1 ′ ′ ′ ′ ′ 𝐶𝐶𝐴𝐴 Arie∆𝐴𝐴𝐴𝐴𝐴𝐴 Arie∆𝐶𝐶𝑂𝑂1𝐴𝐴 Arie∆𝐴𝐴𝐴𝐴𝑂𝑂1

188 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

BILOGICAL TRIANGLES

We obtain: = = . ′ ′ ′ ( 0) 𝐵𝐵𝐴𝐴 𝐴𝐴𝐴𝐴∙𝐵𝐵𝑂𝑂1∙sin 𝐴𝐴�𝐴𝐴𝑂𝑂1 𝑐𝑐∙sin�𝐵𝐵−30 � ′ ′ ′ 0 Similarly𝐶𝐶𝐴𝐴 𝐴𝐴𝐴𝐴∙𝐶𝐶, 𝑂𝑂it1 ∙followssin 𝐴𝐴�𝐴𝐴𝑂𝑂1 that𝑏𝑏:∙sin 𝐶𝐶−30 = , ′ ( 0) 𝐶𝐶𝐵𝐵 𝑎𝑎 sin�𝐶𝐶−30 � ′ 0 𝐵𝐵 𝐴𝐴 = 𝑐𝑐 ∙ sin 𝐴𝐴−30 . ′ ( 0) 𝐶𝐶 𝐴𝐴 𝑏𝑏 sin�𝐴𝐴−30 � ′ 0 Because𝐶𝐶 𝐵𝐵 𝑎𝑎 ∙ sin 𝐵𝐵−30 = 1, the Ceva’s reciprocal theorem implies the ′ ′ ′ 𝐵𝐵𝐴𝐴 𝐶𝐶𝐵𝐵 𝐴𝐴𝐶𝐶 concurrency of′ lines′ ′ , , and, consequently, the homology of 𝐶𝐶𝐴𝐴 ∙ 𝐴𝐴𝐵𝐵 ∙ 𝐵𝐵𝐶𝐶 triangles and ′. We ′denote′ by the homology center. 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵2 𝐶𝐶𝑂𝑂3 Sondat’s theorem′ shows′ ′ that the points ′ , , are collinear and is 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝑃𝑃 perpendicular to the homology axis of bilogical′ triangles′ and .′ 𝑇𝑇 𝑂𝑂 𝑃𝑃 𝑂𝑂𝑇𝑇 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2𝑂𝑂3

Ion Pătrașcu, Florentin Smarandache 189 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

7.1. Orthogonal triangles

Definition 42 Two triangles and are called orthogonal if they have the sides respectively perpendicular. 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 A

C1 B C

Figure 98

In Figure 98, the triangles and are orthogonal. We have: , and . 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 𝟏𝟏 Observation𝑨𝑨𝑨𝑨 ⊥ 𝑨𝑨 62𝑩𝑩 𝑩𝑩𝑩𝑩 ⊥ 𝑩𝑩 𝑪𝑪 𝑪𝑪𝑪𝑪 ⊥ 𝑪𝑪 𝑨𝑨 If the triangles and are orthogonal, and the vertices of the triangle are respectively on the sides of the triangle , we say that the triangles 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 are orthogonal, and is inscribed in . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴

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Problem 9 Being given a triangle , build a triangle such that and to be orthogonal triangles and to be inscribed in the triangle 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴Solution If we build the perpendicular in to , the perpendicular in to and the perpendicular in to , then, denoting { } = , 𝑑𝑑1 𝐴𝐴1 𝐴𝐴1𝐶𝐶1 𝑑𝑑2 𝐵𝐵1 { } = and { } = , the triangle is orthogonal with , 𝐵𝐵1𝐴𝐴1 𝑑𝑑3 𝐶𝐶1 𝐶𝐶1𝐵𝐵1 𝐴𝐴 𝑑𝑑2 ∩ 𝑑𝑑3 and the latter is inscribed in (see Figure 99). 𝐵𝐵 𝑑𝑑1 ∩ 𝑑𝑑3 𝐶𝐶 𝑑𝑑1 ∩ 𝑑𝑑2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 B

d1 A1

B1 C1

d2 A

Figure 99

If we build the perpendicular in to , the perpendicular in to and the perpendicular in to , and we denote by { } = 𝑑𝑑1 𝐴𝐴1 𝐴𝐴1𝐵𝐵1 𝑑𝑑2 𝐵𝐵1 , { } = and { } = , the triangle is also a solution for 𝐵𝐵1𝐶𝐶1 𝑑𝑑3 𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴 𝑑𝑑1 ∩ the proposed problem. 𝑑𝑑3 𝐵𝐵 𝑑𝑑2 ∩ 𝑑𝑑1 𝐶𝐶 𝑑𝑑2 ∩ 𝑑𝑑3 𝐴𝐴𝐴𝐴𝐴𝐴

192 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Problem 10 Build the triangle inscribed in the given triangle such as to be orthogonal with it. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Solution We suppose the problem is solved and we ratiocinate about the configuration in Figure 100, where the triangle 1 1 1 is inscribed in the given triangle , and it is orthogonal with it. 𝐴𝐴 𝐵𝐵 𝐶𝐶 We build the perpendiculars in , , respectively to , and ; 𝐴𝐴𝐴𝐴𝐴𝐴 consequently, we obtain the triangle orthogonal with . Next, we 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 build the perpendiculars in , , respectively to , and , 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 obtaining at their intersections the triangle orthogonal with . 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐴𝐴2𝐶𝐶2 𝐵𝐵2𝐴𝐴2 𝐶𝐶2𝐵𝐵2 3 3 3 2 2 2 C3 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶

A2 C A3

B2 B1

A1 A C 1 B

Figure 100

Ion Pătrașcu, Florentin Smarandache 193 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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We note that , , ; hence, the triangles and are homothetic. The center of homothety is { } = . We 𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴3𝐶𝐶3 𝐵𝐵𝐵𝐵 ∥ 𝐵𝐵3𝐶𝐶3 𝐴𝐴𝐴𝐴 ∥ 𝐴𝐴3𝐵𝐵3 𝐴𝐴𝐴𝐴𝐴𝐴 also observe that , and , therefore the 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝑂𝑂 𝐴𝐴𝐴𝐴3 ∩ 𝐵𝐵𝐵𝐵3 triangles and are homothetic as well. 𝐴𝐴1𝐵𝐵1 ∥ 𝐴𝐴2𝐵𝐵2 𝐴𝐴1𝐶𝐶1 ∥ 𝐴𝐴2𝐶𝐶2 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵2𝐶𝐶2 Because the homothetic of the side is the side by homothety of 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 center and ratio , since , if we take and we denote by 𝐴𝐴3𝐶𝐶3 𝐴𝐴𝐴𝐴 𝑂𝑂𝐴𝐴3 ′ the intersection with , we have2 : 3 3 2 1 𝑂𝑂 𝑂𝑂𝑂𝑂 𝐴𝐴 ∈ 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝑂𝑂 𝐴𝐴 = . 𝑂𝑂𝐴𝐴2 𝑂𝑂𝐴𝐴3 𝐴𝐴𝐴𝐴 ′ 𝑂𝑂Similarly𝐴𝐴1 𝑂𝑂𝑂𝑂 we find that: = and = , hence the triangle is the homothetic of 𝑂𝑂𝐵𝐵2 𝑂𝑂𝐵𝐵3 𝑂𝑂𝐶𝐶2 𝑂𝑂𝐶𝐶3 ′ ′ ′ ′ ′ the 𝑂𝑂triange𝐵𝐵1 𝑂𝑂𝑂𝑂 𝑂𝑂𝐶𝐶 1by homothety𝑂𝑂𝑂𝑂 of center 𝐴𝐴 1and𝐵𝐵1 ratio𝐶𝐶1 . Because the 𝑂𝑂𝐴𝐴3 homothety is a transformation that preserves the measures of angles and 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑂𝑂 𝑂𝑂𝑂𝑂 transforms the lines into lines, and the triangle is orthogonal with the triangle , it means that the triangle is also orthogonal with 𝐴𝐴2𝐵𝐵2𝐶𝐶2 , hence = , = , = . ′ ′ ′ 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐴𝐴1 𝐵𝐵1 𝐵𝐵1 𝐶𝐶1 𝐶𝐶1 We can build the triangle in the following way: 1. We build the triangle orthogonal with the given triangle , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and inscribed in (we build effectively the perpendiculars 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 to , , , respectively to , and ). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2𝐵𝐵2𝐶𝐶2 2. We build the triangle orthogonal with , such that 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 to be inscribed in . 𝐴𝐴3𝐵𝐵3𝐶𝐶3 𝐴𝐴2𝐵𝐵2𝐶𝐶2 3. We unite with , with and we denote { } = . 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴3𝐵𝐵3𝐶𝐶3 4. We unite with , with , with . At the intersection of these 𝐴𝐴 𝐴𝐴3 𝐵𝐵 𝐵𝐵3 𝑂𝑂 𝐴𝐴𝐴𝐴3 ∩ 𝐵𝐵𝐵𝐵3 lines with , , we find the points , , – the vertices of 𝐴𝐴2 𝑂𝑂 𝐵𝐵2 𝑂𝑂 𝐶𝐶2 𝑂𝑂 the requested triangle. 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 Observation 63 Because the triangle can be build in two ways, it follows that we can obtain at least two solutions for the proposed problem. 𝐴𝐴2𝐵𝐵2𝐶𝐶2 Problem 11 Being given a triangle , build a triangle such that the triangles to be orthogonal. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

194 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Solution

We consider a point 1 in the plane of the triangle (see Figure 101). We take the orthogonal projections of on , , , denoted , , . 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 We consider ( ); we take from the perpendicular to ′ and′ we′ 𝐶𝐶1 1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 denote by its intersec′ tion with ( ). The triangles and are 1 𝐵𝐵1 ∈ 𝐴𝐴 𝐶𝐶1 𝐵𝐵 𝐴𝐴𝐴𝐴 orthogonal. ′ 𝐴𝐴 𝐶𝐶1𝐵𝐵 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 A

C A' B C' B1

Figure 101

7.2. Simultaneously orthogonal and orthological triangles

Proposition 73 (Ion Pătrașcu) If the orthogonal triangles and are given, and they are orthological as well, then the orthology is in the sense that is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 orthological with , the orthology centers are the vertices and , 𝐴𝐴𝐴𝐴𝐴𝐴 and the triangles and are similar. 𝐵𝐵1𝐴𝐴1𝐶𝐶1 𝐶𝐶 𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Ion Pătrașcu, Florentin Smarandache 195 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Proof Because and are orthogonal, we have , , (see Figure 102). 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴 ⊥ 𝐴𝐴1𝐵𝐵1 𝐵𝐵𝐵𝐵 ⊥ 𝐵𝐵1𝐶𝐶1 Let us consider that and are orthological in the sense that the 𝐶𝐶𝐶𝐶 ⊥ 𝐶𝐶1𝐴𝐴1 perpendicular taken from to , the perpendicular taken from to , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and the perpendicular taken from to are concurrent in a point . Then, 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵 𝐴𝐴1𝐶𝐶1 the perpendicular from B to will be parallel with , the perpendicular 𝐶𝐶 𝐴𝐴1𝐵𝐵1 𝑂𝑂 from to will be parallel with ; which leads to the conclusion that the 𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 point is such that the quadrilateral is parallelogram. On the other hand, 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 the perpendicular taken from to must be parallel with and it must 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 pass through , which is absurd, because it is not possible that in the 𝐶𝐶 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 parallelogram the diagonal to be parallel with the diagonal . 𝑂𝑂 Let us consider that and are orthological in the sense that the 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 perpendicular from to , the perpendicular taken from to , the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 perpendicular taken from to are concurrent in a point . Then, the 𝐴𝐴 𝐴𝐴1𝐵𝐵1 𝐵𝐵 𝐵𝐵1𝐶𝐶1 perpendicular from to is , the perpendicular from to is ; 𝐶𝐶 𝐶𝐶1𝐴𝐴1 𝑂𝑂 in this moment, the point coincides with ; the perpendicular from C to , 𝐴𝐴 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 ie. , should pass though , which is impossible. 𝑂𝑂 𝐵𝐵 𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 𝐵𝐵 A1

B 1 C1

Figure 102

196 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Finally, let us consider that and are orthological triangles in the sense that the perpendicular taken from to , the perpendicular taken 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 from to and the perpendicular taken from to are concurrent in a 𝐴𝐴 𝐶𝐶1𝐴𝐴1 point . Because the perpendicular from to must be parallel with , it 𝐵𝐵 𝐵𝐵1𝐶𝐶1 𝐶𝐶 𝐴𝐴1𝐵𝐵1 follows that this perpendicular is actually . The perpendicular from to 𝑂𝑂 𝐴𝐴 𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 is actually , hence the point coincides with . The perpendicular 𝐴𝐴𝐴𝐴 𝐵𝐵 from to will be parallel with ; which is possible, and hence, the 𝐶𝐶1𝐵𝐵1 𝐵𝐵𝐵𝐵 𝑂𝑂 𝐶𝐶 vertex is the orthology center of the triangle in relation to the triangle 𝐶𝐶 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 . According to the theorem of orthological triangles, the triangle 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 is orthological as well in relation to . We find that the orthology center is 𝐵𝐵1𝐴𝐴1𝐶𝐶1 𝐵𝐵1𝐴𝐴1𝐶𝐶1 the vertex . 𝐴𝐴𝐴𝐴𝐴𝐴 In Figure 102, it is observed that the angles and have the sides 𝐶𝐶1 respectively perpendicular, hence they are congruent. So does the angle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐶𝐶1𝐵𝐵1 and the angle (they also have the sides respectively perpendicular), 𝐵𝐵𝐵𝐵𝐵𝐵 therefore they are congruent. It consequently follows that: ~ . 𝐵𝐵1𝐴𝐴1𝐶𝐶1 1 1 1 Proposition 74 (Ion Pătrașcu) ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴 𝐵𝐵 𝐶𝐶

If is a right triangle in and 1 1 1 is an orthogonal triangle with it, then: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 i) The triangle is a right triangle in ; ii) The triangles and are triorthological. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1 1 1 1 Proof 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 i) From , and = 90 , it follows that = 90 (see Figure 103). 0 𝐴𝐴1𝐶𝐶1 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1 ⊥ 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐴𝐴̂� 𝑚𝑚�𝐴𝐴�1� 0ii) The perpendicular taken from to is , and the perpendicular taken from to is . These are concurrent in the point , through which passes, 𝐴𝐴 𝐴𝐴1𝐶𝐶1 𝐴𝐴𝐴𝐴 obviously, the perpendicular taken from to . Consequently, the triangles 𝐵𝐵 𝐶𝐶1𝐵𝐵1 𝐶𝐶𝐶𝐶 𝐶𝐶 and are orthological, and the orthology center is the vertex . The 𝐶𝐶 𝐴𝐴1𝐵𝐵1 perpendicular taken from to is , and the perpendicular taken from to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1𝐴𝐴1𝐶𝐶1 𝐶𝐶 is . These perpendiculars intersect in the point ; the perpendicular taken 𝐴𝐴 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 𝐶𝐶 from to passes, obviously, through , hence the point is the orthology 𝐵𝐵1𝐶𝐶1 𝐶𝐶𝐶𝐶 𝐵𝐵 center of the triangle in relation to the triangle . We can affirm that the 𝐵𝐵 𝐴𝐴1𝐶𝐶1 𝐵𝐵 𝐵𝐵 triangle is biorthological with the triangle and, applying Pantazi’s 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶1𝐵𝐵1𝐴𝐴1 theorem, we have that and are triorthological. The fact that the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangles and are orthological can be proved as above. The orthology 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 center is . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐶𝐶1𝐵𝐵1 𝐴𝐴 Ion Pătrașcu, Florentin Smarandache 197 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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A B

A1 C1

Figure 103

Observation 64 Obviously, the triangle is also triorthological in relation to the triangle ; the orthology centers are the vertices of the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 7.3. Orthohomological triangles

Definition 43 Two triangles that are simultaneously orthological and homological are called orthohomological triangles (J. Neuberg).

Problem 12

The triangle being given, build the triangle 1 1 1, such that and 1 1 1 to be orthohomological triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

To𝐴𝐴 solve𝐵𝐵 𝐶𝐶 this problem, we prove:

198 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Lemma 12 Let ( , ) and ( , ) be two secant circles with the common points and . We take through the secants , , and , , . The angle 𝒞𝒞 𝑂𝑂 𝑟𝑟 𝒞𝒞 𝑂𝑂1 𝑟𝑟1 of the chords and is congruent with the angle of the given circles. 𝑀𝑀 𝑁𝑁 𝑀𝑀 𝐴𝐴 𝑀𝑀 𝐴𝐴1 𝐵𝐵 𝑀𝑀 𝐵𝐵1 1 1 Definition 𝐴𝐴𝐴𝐴44 𝐴𝐴 𝐵𝐵 The angle of two secant circles is the angle created by the tangents taken to the circles in one of the common points.

Proof We denote: { } = , and , – the tangents taken in M to the two circles (see Figure 104). We have: , 𝑃𝑃 𝐴𝐴𝐴𝐴 ∩ 𝐴𝐴1𝐵𝐵1 𝑀𝑀𝑀𝑀 𝑀𝑀𝑇𝑇1 . ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝑃𝑃𝐴𝐴1𝑀𝑀 ≡ 1 1 ∢𝐵𝐵 𝑀𝑀𝑇𝑇 P

B T1

A1 M

A O O 1 B1

Figure 104

Adding these relations, and considering that supplements of the measures found in this addition are equal, we obtain: .

∢𝐴𝐴𝐴𝐴𝐴𝐴1 ≡ ∢𝑇𝑇𝑇𝑇𝑇𝑇1 Ion Pătrașcu, Florentin Smarandache 199 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Definition 45 Two circles are called orthogonal if the angle they create is a right angle.

Observation 65 If we consider two orthogonal circles, ( , ) and ( , ), and two chords, and , in these circles, such that , , and , , to be collinear ( 𝒞𝒞 𝑂𝑂 𝑟𝑟 𝒞𝒞 𝑂𝑂1 𝑟𝑟1 is a common point for the given circles), then, according to Lemma 12, it follows 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1 𝐴𝐴 𝑀𝑀 𝐴𝐴1 𝐵𝐵 𝑀𝑀 𝐵𝐵1 𝑀𝑀 that .

1 1 Solution𝐴𝐴𝐴𝐴 ⊥ 𝐴𝐴 𝐵𝐵of Problem 12 We build the circumscribed circle of the triangle and then we build an orthogonal circle of this circle. We denote by one of their common points 𝐴𝐴𝐴𝐴𝐴𝐴 (see Figure 105). 𝑀𝑀 We take the lines , , and we denote by , , their second point of intersection with the orthogonal circle to the circle circumscribed to the 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 A

M B1 B

O A1 O 1 C1

Figure 105

200 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

According to Lemma 12, it follows that the triangles and have their sides respectively perpendiculars, therefore they are orthogonal triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 On the other hand, , , are concurrent in , therefore the triangles are homological as well. 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑀𝑀 Remark 24 In order to build two orthogonal circles ( , ) and ( , ), we take into account: 𝒞𝒞 𝑂𝑂 𝑟𝑟 𝒞𝒞 𝑂𝑂1 𝑟𝑟1 Theorem 40 Two circles ( , ) and ( , ) are orthogonal if and only if + = . 1 1 2 2 𝒞𝒞 2𝑂𝑂 𝑟𝑟 𝒞𝒞 𝑂𝑂 𝑟𝑟 1 1 Proof𝑟𝑟 𝑟𝑟 𝑂𝑂𝑂𝑂 If the circles are orthogonal and is one of their common points, then and are tangent to the circles, the triangle is a right triangle and, 𝑀𝑀 𝑀𝑀𝑀𝑀 consequently, + = . Reciprocally, if the circles are such that + 𝑀𝑀𝑂𝑂1 𝑂𝑂𝑂𝑂𝑂𝑂1 = , it follows2 2that the2 angle is a right angle and, also, the angle2 𝑟𝑟 𝑟𝑟1 𝑂𝑂𝑂𝑂1 𝑟𝑟 2 , created2 by the tangents in to the circles, is also a right angle, hence, 𝑟𝑟1 𝑂𝑂𝑂𝑂1 𝑂𝑂𝑂𝑂𝑂𝑂1 the circles are orthogonal. 𝑇𝑇𝑇𝑇𝑀𝑀1 𝑀𝑀

T1 T

r1 O O1

Figure 106

Ion Pătrașcu, Florentin Smarandache 201 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Theorem 41 If and are two orthohomological triangles, then: i. Their circumscribed circles are secant, and one of their 1 1 1 𝐴𝐴𝐴𝐴𝐴𝐴common𝐴𝐴 𝐵𝐵 points𝐶𝐶 is the center of homology; ii. The circles circumscribed to the given triangles are orthogonal; iii. The other common point of the circles circumscribed to the given triangles is the similarity center of these triangles; iv. The Simson lines of the homology center in relation to the given triangles are parallel to their homology axis; v. The Simson lines of the similarity center of the triangles in relation to the given triangles are orthogonal in a point belonging to their homology axis.

Proof i) We denote by the homology center of the given triangles, therefore { } = ; also, we denote by , , the homology axis (see 𝑀𝑀 Figure 107): 𝑀𝑀 𝐴𝐴𝐴𝐴1 ∩ 𝐵𝐵𝐵𝐵1 ∩ 𝐶𝐶𝐶𝐶1 𝑃𝑃 𝑄𝑄 𝑅𝑅 { } = , { } = , 𝑃𝑃 𝐴𝐴1𝐵𝐵1 ∩ 𝐴𝐴𝐴𝐴 { } = . 𝑄𝑄 𝐵𝐵1𝐶𝐶1 ∩ 𝐵𝐵𝐵𝐵 , , are collinear and due to the orthogonality of the given triangles, we 𝑅𝑅 𝐶𝐶1𝐴𝐴1 ∩ 𝐴𝐴𝐴𝐴 have that the angles from , and are right angles. 𝑃𝑃 𝑄𝑄 𝑅𝑅 ii) Because the chords and in the two circles are perpendicular, 𝑃𝑃 𝑄𝑄 𝑅𝑅 taking into account Lemma 12, it follows that the circles circumscribed to the 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1 triangles and are orthogonal.

1 1 1 iii) It 𝐴𝐴𝐴𝐴𝐴𝐴follows from𝐴𝐴 𝐵𝐵 T𝐶𝐶he Theory of Similar Figures, see Annex no. 2.

iv) (Mihai Miculița; see Figure 108) inscribable 1 𝑀𝑀𝑅𝑅 ⊥ 𝐴𝐴𝐴𝐴, 1 1 1 1 1 1 � ⟹ 𝑀𝑀𝑀𝑀𝑅𝑅 𝑄𝑄 − ⟹ 𝑄𝑄�𝑅𝑅 𝐴𝐴 ≡ 𝑄𝑄�𝑀𝑀𝑀𝑀⎫ 𝑀𝑀𝑄𝑄 ⊥ 𝐵𝐵𝐵𝐵 ⎪ 𝑀𝑀𝑄𝑄1 𝐶𝐶1𝑄𝑄 ⊥ 𝐵𝐵𝐵𝐵 ⟹ 𝑀𝑀𝑄𝑄inscribable1 ∥ 𝐶𝐶1𝑄𝑄 ⟹ 𝑄𝑄�1𝑀𝑀𝑀𝑀 ≡ 𝑄𝑄�𝐶𝐶1𝐶𝐶 ⟹ 𝐶𝐶1𝑄𝑄 ⊥ 𝐵𝐵𝐵𝐵 ⎬ � ⟹ 𝐶𝐶1𝐶𝐶𝐶𝐶𝐶𝐶 − ⟹ 𝑄𝑄�𝐶𝐶1𝐶𝐶.≡ 𝑄𝑄�𝑄𝑄𝑄𝑄 ⎪ 𝐶𝐶1𝑅𝑅 ⊥ 𝐴𝐴𝐴𝐴 ⎭ ⟹ 𝑄𝑄�1𝑅𝑅1𝐴𝐴 ≡ 𝑄𝑄�𝑄𝑄𝑄𝑄 ⟹ 𝑄𝑄1𝑅𝑅1 ∥ 𝑄𝑄𝑄𝑄 202 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

C Q B

C1 P M N

Figure 107

v) We denote by the second point of intersection of the circles circumscribed to the given triangles, the quadrilateral is inscribable 𝑁𝑁 (from , is seen from a right angle, and from , also, is seen from a 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴1 right angle, being its own homologous point). From the same considerations, 𝑅𝑅 𝐴𝐴𝐴𝐴1 𝑁𝑁 𝐴𝐴𝐴𝐴1 the quadrilateral is inscribable, we obtain that the points , , , , 𝑁𝑁 are on the circle of diameter . Considering the triangles and , 𝐴𝐴𝐴𝐴𝐴𝐴1𝑁𝑁 𝐴𝐴 𝑃𝑃 𝐴𝐴1 𝑁𝑁 𝑅𝑅 and applying Proposition 53, we obtain that the Simson lines of the point in 𝐴𝐴𝐴𝐴1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝑃𝑃𝑃𝑃 relation to the triangles and are perpendicular and intersect in a 𝑁𝑁 point situated on the line . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑃𝑃𝑃𝑃 Ion Pătrașcu, Florentin Smarandache 203 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

R B Q 1 Q C P1 C1 M N

Figure 108

Theorem 42 (P. Sondat) The homology axis of two orthohomological triangles and passes through the midpoint of the segment 𝐴𝐴𝐴𝐴𝐴𝐴 determined by the ortocenters of these triangles. 1 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻𝐻𝐻

204 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

M1 R1

B Q1 Q C P Z 1 C Y 1 N M Q2 M2 X P R2

A1 H1

Figure 109

Proof Let , the Simson lines of homology center of the triangles and , and their homology axis (see Figure 109). We denote 𝑃𝑃1𝑄𝑄1𝑅𝑅11 𝑃𝑃12𝑄𝑄2𝑅𝑅2 𝑀𝑀 by respectively the symmetrics of towards respectively ; 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃𝑃𝑃𝑃𝑃 because the Simson line passes through the middle of the segment 𝑀𝑀1 𝑀𝑀2 𝑀𝑀 𝑄𝑄1 𝑄𝑄2 𝑃𝑃1𝑄𝑄1𝑅𝑅1 𝑀𝑀𝑀𝑀

Ion Pătrașcu, Florentin Smarandache 205 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

(Theorem 17), we have that is parallel with , therefore with , similarly is parallel with . The quadrilateral is rectangle, if 𝑀𝑀1𝐻𝐻 𝑃𝑃1𝑄𝑄1 𝑃𝑃𝑃𝑃 we denote by its center; we obviously have – – collinear and 𝑀𝑀2𝐻𝐻1 𝑃𝑃𝑃𝑃 𝑀𝑀𝑄𝑄1𝑄𝑄2𝑄𝑄3 – – collinear. The line is the homothety of the line by 𝑋𝑋 𝑄𝑄1 𝑋𝑋 𝑄𝑄2 homothety of center and ratio 2, consequently the point is the midpoint of 𝑀𝑀 𝑋𝑋 𝑄𝑄 𝑀𝑀1𝑀𝑀2 𝑄𝑄1𝑄𝑄2 the segment . The quadrilateral is trapeze (its bases are parallel 𝑀𝑀 𝑄𝑄 to the axis of homology ), because is the midpoint of and the 𝑀𝑀1𝑀𝑀2 𝑀𝑀1𝐻𝐻𝑀𝑀2𝐻𝐻1 parallel taken through to is the axis of homology , according to a 𝑃𝑃𝑃𝑃 𝑄𝑄 𝑀𝑀1𝑀𝑀2 theorem in trapeze, will also contain the midpoint of the diagonal . 𝑄𝑄 𝑀𝑀1𝐻𝐻 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 A 𝐻𝐻𝐻𝐻1

B Q Q C 1 Z Y C1 P1

M Q2 N

R X 2

Figure 110

206 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Proposition 75 (Ion Pătrașcu) Let and be two orthohomological triangles. The Simson lines of the homology center of triangles to the circumscribed circles are , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 , , respectively , , . Then the midpoints of the segments , 𝑃𝑃1 , are collinear. 𝑄𝑄1 𝑅𝑅1 𝑃𝑃2 𝑄𝑄2 𝑅𝑅2 𝑃𝑃1𝑃𝑃2 1 2 1 2 𝑄𝑄 𝑄𝑄Proof𝑅𝑅 𝑅𝑅 Let be the homology center and the homology axis of the given triangles (see Figure 110). 𝑀𝑀 𝑃𝑃 − 𝑄𝑄 − 𝑅𝑅 We denote by and the Simson lines of in relation to the triangles respectively . 𝑃𝑃1 − 𝑄𝑄1 − 𝑅𝑅1 𝑃𝑃2 − 𝑄𝑄2 − 𝑅𝑅2 𝑀𝑀 The quadrilateral is rectangle, therefore the midpoint of is the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 center of this rectangle; we denote it by ; similarly, let the midpoint of 𝑀𝑀𝑃𝑃1𝑃𝑃𝑃𝑃2 𝑃𝑃1𝑃𝑃2 (therefore the center of this rectangle ) and the midpoint of . 𝑋𝑋 𝑌𝑌 𝑄𝑄1𝑄𝑄2 Because , , are the midpoints of the segments , , respectively 𝑀𝑀𝑄𝑄1𝑄𝑄𝑄𝑄2 𝑍𝑍 𝑅𝑅1𝑅𝑅2 and , , are collinear points, it follows that , , are collinear as well, 𝑋𝑋 𝑌𝑌 𝑍𝑍 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 they belong to the homothetic of the line by the homothety of center and 𝑀𝑀𝑀𝑀 𝑃𝑃 𝑄𝑄 𝑅𝑅 𝑋𝑋 𝑌𝑌 𝑍𝑍 ratio . 1 𝑃𝑃𝑃𝑃 𝑀𝑀 2 Remark 25 We can prove in the same way that the midpoints of the segments determined by the feet of altitudes of the orthohomological triangles and are collinear points. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 Theorem 43

Let and 1 1 1 two orthohomological triangles having as its axis of homology the line . Then: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 i) The complete quadrilaterals ( , ) and ( 1 1 1, ) 𝑑𝑑 have the same Miquel’s point and the same Miquel’s 𝑑𝑑 circle; 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴 𝐵𝐵 𝐶𝐶 ii) The centers of the circles circumscribed to the triangles and 1 1 1 are the extremities of a diameter of the Miquel’s circle, and the homology center of these triangles 𝐴𝐴𝐴𝐴𝐴𝐴belongs to𝐴𝐴 this𝐵𝐵 Miq𝐶𝐶 uel’s circle.

Ion Pătrașcu, Florentin Smarandache 207 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ORTHOHOMOLOGICAL TRIANGLES

Proof i) Let , , be the points in which the axis of homology intersects the sides , , respectively (see Figure 111). The triangles and 𝑃𝑃 𝑄𝑄 𝑅𝑅 𝑑𝑑 being orthogonal, we have that = 90 , therefore the 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangles and have the same circumscribed circle with0 center . ∢𝐴𝐴𝐴𝐴𝐴𝐴1 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝑃𝑃𝑃𝑃 𝑂𝑂2 A

O C P

Q O2 B O4 B1

N M O 3 d R

Figure 111

The quadrilaterals and are inscribable, consequently: , (1) 𝑃𝑃𝑃𝑃𝑃𝑃𝐵𝐵1 𝑄𝑄𝑄𝑄𝑄𝑄𝐶𝐶1 ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵1𝐴𝐴1 208 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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. (2) We also have: 1 ∢𝐶𝐶𝐶𝐶𝐶𝐶 ≡ ∢𝑅𝑅𝐶𝐶 𝐶𝐶 (opposite vertex), (3) (opposite vertex). (4) ∢From𝐵𝐵𝐵𝐵𝐵𝐵 these≡ ∢ relations𝐶𝐶𝐶𝐶𝐶𝐶 , we obtain that: 1 1 1 ∢𝑀𝑀𝐶𝐶 𝐴𝐴 ≡ ∢𝑅𝑅𝐶𝐶 𝐶𝐶 . (5)

This relations1 1 show1 that1 the circumscribed circle of the triangle contains∢𝑀𝑀𝐵𝐵 the𝐴𝐴 point≡ ∢ 𝑀𝑀𝐶𝐶. The𝐴𝐴 condition of orthogonality of the triangles and 1 1 1 implies their direct similarity (the angles of the triangles have the𝐴𝐴 𝐵𝐵sides𝐶𝐶 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 respectively1 1 1 perpendicular). From the conciclicity of the points , , , , it𝐴𝐴 derives𝐵𝐵 𝐶𝐶 that: 𝑀𝑀 1 1 1 . 𝐴𝐴 (6)𝐵𝐵 𝐶𝐶

But1: 1 1 1 1 . (7) On∢𝐴𝐴 the𝑀𝑀 𝐵𝐵other≡ ∢hand𝐴𝐴 𝐶𝐶: 𝐵𝐵 1 1 1 ∢𝐴𝐴 𝐶𝐶 𝐵𝐵 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 (opuse la vârf). (8) We get in this way that , condition showing that the point ∢𝐴𝐴1𝑀𝑀𝐵𝐵1 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 , the homology center of triangles to the circle circumscribed to the triangle 𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 . ∢𝐵𝐵 ≡ ∢ 𝑀𝑀 Similarly, the triangles and have the same circumscribed circle 𝐴𝐴𝐴𝐴𝐴𝐴 of center , and the triangles and have the same circumscribed 𝐶𝐶𝐶𝐶𝐶𝐶 𝐶𝐶1𝑄𝑄𝑄𝑄 circle of center – the midpoint of the segment . If we denote by and 𝑂𝑂3 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵1𝑃𝑃𝑃𝑃 the points of intersection of the circles circumscribed to the triangles 𝑂𝑂4 𝐵𝐵𝐵𝐵1 𝑀𝑀 and (of centers , respectively ), then, due to the fact that the circles 𝑁𝑁 𝐴𝐴𝐴𝐴𝐴𝐴 circumscribed to the triangles and coincide with the circles 1 1 1 𝑂𝑂 𝑂𝑂1 circumscribed𝐴𝐴 𝐵𝐵 𝐶𝐶 to the triangles and , it means that their second point 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶𝐶𝐶 of intersection is , the common point of the circles of the complete 𝐴𝐴1𝑃𝑃𝑃𝑃 𝐶𝐶1𝑄𝑄𝑄𝑄 quadrilaterals ( , ), ( 1 1 1, ), hence is the Miquel’s point of these 𝑁𝑁 quadrilaterals (see Annex no. 3). Also, the Miquel’s circle of the complete 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 quadrilateral ( , ), ie.𝐴𝐴 the𝐵𝐵 circle𝐶𝐶 𝑑𝑑 that contains𝑁𝑁 the points , , , and the point coincides with Miquel’s circle of the quadrilateral ( 1 1 1, ), 𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝑂𝑂4 which contains the points , , , , . 𝑁𝑁 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑 𝑁𝑁 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝑂𝑂4 ii) The circles circumscribed to the triangles and 1 1 1 are orthogonal; because is one of their common points, we have that 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 = 90 , and because , and are on the Miquel’s circle, it means 𝑁𝑁 that and are0 diametrically opposed in this circle. 𝑚𝑚�𝑂𝑂�𝑂𝑂𝑂𝑂1� 𝑂𝑂 𝑁𝑁 𝑂𝑂1 𝑂𝑂 𝑂𝑂1

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If we denote by the homology center of the triangles and 1 1 1, then will be the second point of intersection of these circles; because is 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 situated on the Miquel’s circle and is its symmetric towards the diameter 𝑀𝑀 𝑁𝑁 , it follows that also belongs to the common Miquel’s circle – complete 𝑀𝑀 quadrilaterals ( ; ), ( ; 1). 𝑂𝑂𝑂𝑂1 𝑀𝑀 1 1 1 Definition 46𝐴𝐴𝐴𝐴𝐴𝐴 𝑑𝑑 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑 If is a triangle and a transverse ( , , ), and the perpendiculars raised in , , 𝑨𝑨𝑨𝑨𝑨𝑨 𝑷𝑷 − 𝑸𝑸 − 𝑹𝑹 𝑷𝑷 ∈ 𝑨𝑨𝑨𝑨 𝑸𝑸 ∈ respectively to , and determine a triangle , 𝑩𝑩𝑩𝑩 𝑹𝑹 ∈ 𝑨𝑨𝑨𝑨 𝑷𝑷 𝑸𝑸 𝑹𝑹 which is called paralogical triangle of . 𝑨𝑨𝑨𝑨 𝑨𝑨𝑨𝑨 𝑪𝑪𝑪𝑪 𝑨𝑨𝟏𝟏𝑩𝑩𝟏𝟏𝑪𝑪𝟏𝟏 Observation 66 𝑨𝑨𝑨𝑨𝑨𝑨

In Figure 112, 1 1 1 is the paralogical triangle of .

𝐴𝐴 𝐵𝐵 𝐶𝐶 A 𝐴𝐴𝐴𝐴𝐴𝐴

B Q C

Figure 112

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7.4. Metaparallel triangles or parallelologic triangles

Definition 47 Two triangles and , with the property that the parallels taken through , , respectively to , , 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨′𝑩𝑩′𝑪𝑪′ are concurrent in a point , are called metaparallel triangles or 𝑨𝑨 𝑩𝑩 𝑪𝑪 𝑩𝑩′𝑪𝑪′ 𝑪𝑪′𝑨𝑨′ 𝑨𝑨′𝑩𝑩′ parallelologic triangles. The point is called center of 𝑷𝑷 parallelogy. 𝑷𝑷 Theorem 44 If the triangles ABC and are parallelologic, then the triangle is also parallelologic in relation to (parallels 𝐴𝐴′𝐵𝐵′𝐶𝐶′ taken through the vertices , , respectively to , , 𝐴𝐴′𝐵𝐵′𝐶𝐶′ 𝐴𝐴𝐴𝐴𝐴𝐴 are concurrent in a point ’ - center of parallelology of the triangle 𝐴𝐴′ 𝐵𝐵′ 𝐶𝐶′ 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 in relation to the triangle ). 𝑃𝑃 Proof𝐴𝐴′𝐵𝐵′𝐶𝐶 ′ 𝐴𝐴𝐴𝐴𝐴𝐴 Let and be two parallelologic triangles and the parallelology center of the triangle′ ′ ′ in relation to (see Figure 113). We denote by 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 , , the intersections of the parallel′ s ′taken′ from , , to , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 , respectively with , , . Therefore = ′{ ′}. ′ ′ 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 ′ According′ to Ceva’s theorem, we have that: 𝐴𝐴 𝐵𝐵 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴1 ∩ 𝐵𝐵𝐵𝐵1 ∩ 𝐶𝐶𝐶𝐶1 𝑃𝑃 = 1. (1) 𝐴𝐴1𝐵𝐵 𝐵𝐵1𝐶𝐶 𝐶𝐶1𝐴𝐴 We𝐴𝐴1𝐶𝐶 ∙denote𝐵𝐵1𝐴𝐴 ∙ 𝐶𝐶 1by𝐵𝐵 , , the intersections of the parallels taken through , , , respectively′ to′ ′, , with the sides , and ; we 𝐴𝐴 1 𝐵𝐵 1 𝐶𝐶 1 note′ that′ ′ ~ (they have parallel sides respectively′ ′ ′ ′ ); it follows′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 that: ′ ′ ′ ∆𝐴𝐴 1𝐴𝐴 𝐵𝐵 ∆𝐴𝐴1𝐶𝐶𝐶𝐶 = . (2) ′ ′ ′ ′ 𝐴𝐴 1𝐴𝐴 𝐴𝐴 1𝐵𝐵 Also𝐴𝐴1𝐶𝐶 , we𝐴𝐴 have1𝑃𝑃 ~ , from where: ′ ′ ′ = . 1 1 (3) ′ ′ ′ ′ ∆𝐴𝐴 𝐴𝐴 𝐶𝐶 ∆𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴 1𝐴𝐴 𝐴𝐴 1𝐶𝐶 From𝐴𝐴1𝐵𝐵 relations𝐴𝐴1𝑃𝑃 (2) and (3), we get: = . (4) ′ ′ 𝐴𝐴 1𝐵𝐵 𝐴𝐴1𝐵𝐵 ′ ′ Similarly𝐴𝐴 1𝐶𝐶 𝐴𝐴1, 𝐶𝐶the following relations are obtained:

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= , (5) ′ ′ 𝐵𝐵 1𝐶𝐶 𝐵𝐵1𝐶𝐶 ′ ′ 𝐵𝐵 1𝐴𝐴 = 𝐵𝐵1𝐴𝐴. (6) ′ ′ 𝐶𝐶 1𝐴𝐴 𝐶𝐶1𝐴𝐴 ′ ′ 𝐶𝐶The1𝐵𝐵 relations𝐶𝐶1𝐵𝐵 (4), (5), (6) and (1) show, together with Ceva’s theorem, that , , are concurrent in the parallelology center of the triangle ′ ′ ′ ′ ′ ′ 1 in relation1 1 to the triangle . 𝐴𝐴′𝐴𝐴′ ′𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 A 𝐴𝐴𝐴𝐴𝐴𝐴

C1 B1 P C'

B A1 C B'1

A' P' A'1

C'1

Figure 113

Observation 67 Two orthogonal triangles are parallelologic. Their parallelology centers are the orthocenters.

Remark 26 If and are parallelologic triangles, then, obviously, they are orthogonal triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

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From the reciprocal of Desargues’s theorem (see [24]), it follows that and are also homological triangles; hence, two parallelologic triangles are 𝐴𝐴𝐴𝐴𝐴𝐴 orthohomological triangles. 𝐴𝐴1𝐵𝐵1𝐶𝐶1 The theorem 40 can be formulated for the parallelologic triangles. In the same way, it can be demonstrated:

Theorem 45 If and are two triangles in the same plane, and through the vertices′ ′ ′ , , some lines that create with , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 and angles of measure are taken, and these lines are 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵′𝐶𝐶′ 𝐶𝐶′𝐴𝐴′ concurrent, then the lines that pass through , , and create 𝐴𝐴′𝐵𝐵′ 𝜑𝜑 with , , angles of measure 1800 ′ are′ concurrent′ . 𝐴𝐴 𝐵𝐵 𝐶𝐶 (The triangles and are called isologic). 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 − 𝜑𝜑 ′ ′ ′ This theorem generalizes𝐴𝐴𝐴𝐴𝐴𝐴 the𝐴𝐴 theorem𝐵𝐵 𝐶𝐶 of the orthological triangles.

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8.1 Annex 1: Barycentric Coordinates

8.1.1 Barycentric coordinates of a point in a plane Let us consider a scalene triangle which we will call the reference triangle, and an arbitrary point in the plane of the triangle. We denote by 𝐴𝐴𝐴𝐴𝐴𝐴 the intersections of the lines , , with the sides , , 𝑀𝑀 (see Figure 113). 𝐴𝐴′𝐵𝐵′𝐶𝐶′ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 III. A

I. IV. C' II. B' M

B A' C

V. VI. VII. Figure 113 The point determines with two points of the triangle, generally three triangles , , . The areas of these triangles are considered 𝑀𝑀 positive or negative according to the following rule: 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀

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If a triangle has a common side with the reference triangle and its other vertex is on the same section of the common side with the “remaining” vertex of the reference triangle, then its area is positive, and if the common side separates its vertex with the other vertex of the reference triangle, the area is negative. If the point is situated on a side of the reference triangle, then the area of the "degenerate" triangle determined by it with the vertices of the 𝑀𝑀 reference triangle determining the respective side is zero. Denoting by , , the areas of the three triangle , , , it is observed that 𝑆𝑆𝑎𝑎 the signs of these areas correspond to those in the following table: 𝑆𝑆𝑏𝑏 𝑆𝑆𝑐𝑐 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀

Region Sa Sb Sc I + + + II + - + III + - - IV + + - V - + - VI - + + VII - - +

Definition 48 Three real numbers , , proportional to the three algebraic considered areas , , are called barycentric coordinates of 𝜶𝜶 𝜷𝜷 𝜸𝜸 the point in relation to the triangle . 𝑺𝑺𝒂𝒂 𝑺𝑺𝒃𝒃 𝑺𝑺𝒄𝒄

𝑴𝑴 𝑨𝑨𝑨𝑨𝑨𝑨 We denote ( , , ). If , , are such that + + = 1, then , , are barycentric absolute coordinates of the point . 𝑀𝑀 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 If we denote by S the area of the triangle , then the barycentric 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝑀𝑀 absolute coordinates of the point are , , . 𝑆𝑆𝑎𝑎 𝑆𝑆𝑏𝑏 𝑆𝑆𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 For example, if is the center of gravity of the triangle , then the 𝑀𝑀 𝑆𝑆 𝑆𝑆 𝑆𝑆 , , absolute barycentric𝐺𝐺 coordinates of are . 𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 𝐺𝐺 �3 3 3�

216 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 46 If is a given triangle and is a point in its plane, then there exists and it is unique the ordered triplet ( , , ) , + + 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 = 1, such that + + = 0. 3 𝛼𝛼 𝛽𝛽 𝛾𝛾 ∈ ℝ 𝛼𝛼 𝛽𝛽 𝛾𝛾Reciprocally 𝛼𝛼𝑀𝑀��𝑀𝑀�⃗ 𝛽𝛽𝑀𝑀��𝑀𝑀�⃗ 𝛾𝛾�𝑀𝑀��𝑀𝑀�⃗ �⃗ For any triplet ( , , ) , + + = 1, there exists and it is unique a point in the plane of the3 triangle , such that + + 𝛼𝛼 𝛽𝛽 𝛾𝛾 ∈ ℝ 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 0. 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼𝑀𝑀��𝑀𝑀�⃗ 𝛽𝛽𝑀𝑀��𝑀𝑀�⃗ 𝛾𝛾�𝑀𝑀��Observation𝑀𝑀�⃗ �⃗ 68 From the previous theorem, it follows that the point satisfies the condition ’ ’ = or = . ’ 𝑀𝑀 𝑀𝑀𝑀𝑀��⃗ 𝛼𝛼 𝑀𝑀𝑀𝑀��⃗ 𝐴𝐴��𝐴𝐴��⃗ We𝛽𝛽 +note𝛾𝛾 that𝐴𝐴𝐴𝐴��⃗ is𝛼𝛼 negative when separates the points and , and negative if and are on the same section of . 𝛼𝛼 ’ 𝐵𝐵𝐵𝐵 𝐴𝐴 𝑀𝑀 On the other hand, = , in the sign convention for ; in conclusion, the 𝐴𝐴 𝑀𝑀 ’ 𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀��⃗ 𝑆𝑆𝑎𝑎 triplet ( , , ) from the theorem constitutes the barycentric absolute𝑎𝑎 coordinates of 𝐴𝐴𝐴𝐴��⃗ 𝑆𝑆 𝑆𝑆 the point . 𝛼𝛼 𝛽𝛽 𝛾𝛾 Theorem𝑀𝑀 47 (The position vector of a point) Let be a certain point in the plane of the triangle and ( , , ), + + = 1. Then: = + + . 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀Observation𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 69 𝛽𝛽 𝛾𝛾 𝑂𝑂��𝑂𝑂��⃗ 𝛼𝛼𝑂𝑂��𝑂𝑂�⃗ 𝛽𝛽�𝑂𝑂��𝑂𝑂�⃗ 𝛾𝛾�𝑂𝑂��𝑂𝑂�⃗ 1. We can denote = + + . 2. From the previous theorem, it follows that = + , = 𝑟𝑟��𝑀𝑀�⃗ 𝛼𝛼𝑟𝑟��𝐴𝐴⃗ 𝛽𝛽�𝑟𝑟��𝐵𝐵⃗ 𝛾𝛾𝑟𝑟��𝐶𝐶⃗ + , = + . 𝐴𝐴��𝐴𝐴��⃗ 𝛽𝛽𝐴𝐴𝐴𝐴��⃗ 𝛾𝛾𝐴𝐴𝐴𝐴��⃗ 𝐵𝐵𝐵𝐵��⃗ ��⃗ ��⃗ ��⃗ ��⃗ ��⃗ Theorem𝛼𝛼 4𝐵𝐵𝐵𝐵8 𝛾𝛾𝐵𝐵𝐵𝐵 𝐶𝐶𝑀𝑀 𝛼𝛼𝐶𝐶𝐶𝐶 𝛽𝛽𝐶𝐶𝐶𝐶 If ( , , ), ( , , ), with + + = 1, = 1, 2 two given points in the plane of the triangle , then = 𝑄𝑄1 𝛼𝛼1 𝛽𝛽1 𝛾𝛾1 𝑄𝑄2 𝛼𝛼2 𝛽𝛽2 𝛾𝛾2 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 �� ( ) + ( ) + ( ) . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄��1��𝑄𝑄��2⃗ 𝛼𝛼2 − 𝛼𝛼1 𝑟𝑟��𝐴𝐴⃗ 𝛽𝛽2 − 𝛽𝛽1 𝑟𝑟��𝐵𝐵⃗ 𝛾𝛾2 − 𝛾𝛾1 𝑟𝑟��𝐶𝐶⃗

Ion Pătrașcu, Florentin Smarandache 217 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 49 (Barycentric Coordinates of a vector) Let be a given triangle and a point in its plane considered to be the origin of the plane. We denote by , , the position 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 vectors of the points , , and with a vector in the plane. 𝑟𝑟��𝐴𝐴⃗ 𝑟𝑟��𝐵𝐵⃗ 𝑟𝑟��𝐶𝐶⃗ Then there exist and they are unique three real numbers , , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑢𝑢�⃗ with + + = 0, such that = + + . 𝛼𝛼 𝛽𝛽 𝛾𝛾 Reciproc𝛼𝛼 ally𝛽𝛽 𝛾𝛾 𝑢𝑢�⃗ 𝛼𝛼𝑟𝑟��𝐴𝐴⃗ 𝛽𝛽𝑟𝑟��𝐵𝐵⃗ 𝛾𝛾𝑟𝑟��𝑐𝑐⃗ For any triplet ( , , ) with + + 0, there exists and it is unique a vector that verifies the3 relation = + + . 𝛼𝛼 𝛽𝛽 𝛾𝛾 ∈ ℝ 𝛼𝛼 𝛽𝛽 𝛾𝛾 − 𝐴𝐴 𝐵𝐵 𝑐𝑐 Definition 49𝑢𝑢�⃗ 𝑢𝑢�⃗ 𝛼𝛼𝑟𝑟��⃗ 𝛽𝛽𝑟𝑟��⃗ 𝛾𝛾𝑟𝑟��⃗ The triplet ( , , ) , + + = , with the property that + + =𝟑𝟑 constitutes the barycentric 𝜶𝜶 𝜷𝜷 𝜸𝜸 ∈ ℝ 𝜶𝜶 𝜷𝜷 𝜸𝜸 𝟎𝟎 coordinates of the vector . We denote it by ( , , ). 𝜶𝜶 ∙ 𝒓𝒓��𝑨𝑨⃗ 𝜷𝜷 ∙ 𝒓𝒓��𝑩𝑩⃗ 𝜸𝜸 ∙ �𝒓𝒓��𝒄𝒄⃗ �𝒖𝒖�⃗ Observation 70 𝒖𝒖��⃗ 𝒖𝒖��⃗ 𝜶𝜶 𝜷𝜷 𝜸𝜸 The barycentric coordinates of vector do not depend on choice of origin .

Consequence 𝑢𝑢�⃗ 𝑂𝑂 If ( , , ), then: = + ; 𝑢𝑢�⃗ 𝛼𝛼 𝛽𝛽 𝛾𝛾 = + ; 𝑢𝑢�⃗ 𝛽𝛽�𝐴𝐴𝐴𝐴��⃗ 𝛾𝛾𝐴𝐴𝐴𝐴��⃗ = + . 𝑢𝑢�⃗ 𝛼𝛼𝐵𝐵𝐵𝐵��⃗ 𝛾𝛾�𝐵𝐵𝐵𝐵��⃗ ��⃗ ��⃗ Theorem𝑢𝑢�⃗ 𝛼𝛼𝐶𝐶𝐶𝐶 50𝛽𝛽 𝐶𝐶𝐶𝐶(The position vector of a point that divides a segment into a given ratio) Let ( , , ), ( , , ), + + = 1, = 1,2 and

the point𝑄𝑄1 𝛼𝛼 1 𝛽𝛽that1 𝛾𝛾 1divides𝑄𝑄2 𝛼𝛼 the2 𝛽𝛽 segment2 𝛾𝛾2 𝛼𝛼 𝑖𝑖 𝛽𝛽𝑖𝑖 thus𝛾𝛾𝑖𝑖: =𝑖𝑖 . 𝑃𝑃��𝑄𝑄��1⃗ Then 𝑃𝑃 , , . 𝑄𝑄1𝑄𝑄2 𝑃𝑃��𝑄𝑄��2⃗ 𝑘𝑘 𝛼𝛼1−𝑘𝑘𝛼𝛼2 𝛽𝛽1−𝑘𝑘𝛽𝛽2 𝛾𝛾1−𝑘𝑘𝛾𝛾2 𝑃𝑃 � 1−𝑘𝑘 1−𝑘𝑘 1−𝑘𝑘 �

218 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Consequences 1. The barycentric coordinates of the midpoint of the segment [ , ], ( , , ) with + + = 1, = 1,2 are given by 𝑄𝑄1 𝑄𝑄2 , , 𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝛼𝛼𝑖𝑖. 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 �� 𝛼𝛼1+𝛼𝛼2 𝛽𝛽1+𝛽𝛽2 𝛾𝛾1+𝛾𝛾2 2. If ( , , ), + + = 1, = 1,3 are the vertices of a 𝑀𝑀 � 2 2 2 � triangle, then the center of gravity of the triangle has the 𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 �� barycentric coordinates: + + + + 𝐺𝐺 + + , , . 3 3 3 𝛼𝛼1 𝛼𝛼2 𝛼𝛼3 𝛽𝛽1 𝛽𝛽2 𝛽𝛽3 𝛾𝛾1 𝛾𝛾2 𝛾𝛾3 𝐺𝐺 � � Theorem 51 (The collinearity condition of two vectors) Let us have the vectors ( , , ), ( , , ), + + = 0, + + = 0. 𝑢𝑢��1⃗ α1 β1 γ1 𝑢𝑢��2⃗ α2 β2 γ2 α1 β1 The1 vectors2 , 2 are collinear2 if and only if = = . γ α β γ α1 β1 γ1 �𝑢𝑢��1⃗ 𝑢𝑢��2⃗ α2 β2 γ2 Theorem 52 (The condition of perpendicularity of two vectors) Let be a given triangle, = , = , = and ( , , ), ( , , ), with , , = 0, = 1, 2. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑎𝑎 𝐴𝐴𝐴𝐴 𝑏𝑏 𝐴𝐴𝐴𝐴 𝑐𝑐 Then: ( + ) + ( + ) + 𝑢𝑢��1⃗ α1 β1 γ1 𝑢𝑢��2⃗ α2 β2 γ2 α𝑖𝑖 β𝑖𝑖 γ𝑖𝑖 𝑖𝑖 ( + ) = 0 2 2 1 2 . 1 2 2 1 1 2 2 1 𝑢𝑢��⃗ ⊥2𝑢𝑢��⃗ ⟺ β γ β γ 𝑎𝑎 𝛼𝛼 𝛾𝛾 𝛼𝛼 𝛾𝛾 𝑏𝑏 Conse𝛼𝛼1𝛽𝛽2quence𝛼𝛼2𝛽𝛽 1 𝑐𝑐 If ( , , ), with + + = 1, = 1,2 and ( , , ), with + + = 0, then: 𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 𝑄𝑄0 𝛼𝛼0 𝛽𝛽0 𝛾𝛾0 = 90 [( )( ) + ( )( 𝛼𝛼0 𝛽𝛽0 𝛾𝛾0 )] + [( )(0 ) + ( )( )] + [( 𝑚𝑚�𝑄𝑄1�𝑄𝑄0𝑄𝑄2� ⇔ 𝛽𝛽1 − 𝛽𝛽0 𝛾𝛾1 − 𝛾𝛾0 𝛽𝛽2 − 𝛽𝛽0 𝛾𝛾1 − )( 2 ) + ( )( )] = 0 2 0 1 0 2 0 2 0 . 1 0 1 𝛾𝛾 𝑎𝑎 𝛼𝛼 − 𝛼𝛼 𝛾𝛾 − 𝛾𝛾 𝛼𝛼 −2𝛼𝛼 𝛾𝛾 − 𝛾𝛾 𝑏𝑏 𝛼𝛼 − 0 2 0 2 0 1 0 𝛼𝛼 Theorem𝛽𝛽 − 𝛽𝛽 53 𝛼𝛼 − 𝛼𝛼 𝛽𝛽 − 𝛽𝛽 𝑐𝑐 If ( , , ), + + = 0, then | | = ( + + ). 2 2 2 𝑢𝑢�⃗2 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝑢𝑢�⃗ − 𝛽𝛽𝛽𝛽𝑎𝑎 𝛾𝛾𝛾𝛾𝑏𝑏 Conse𝛼𝛼𝛼𝛼𝑐𝑐 quence Let ( , , ), with + + = 1, = 1,2.

𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 Ion Pătrașcu, Florentin Smarandache 219 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ANNEXES

The distance between and is given by = [( )( )] + [( )( )] + [( )( 2 )] . 𝑄𝑄1 𝑄𝑄2 𝑄𝑄1𝑄𝑄2 −� 𝛽𝛽2 − 2 2 2 1 2 1 2 1 2 1 2 1 2 1 𝛽𝛽 Theorem𝛾𝛾 − 𝛾𝛾 5𝑎𝑎4 𝛾𝛾 − 𝛾𝛾 𝛼𝛼 − 𝛼𝛼 𝑏𝑏 𝛼𝛼 − 𝛼𝛼 𝛽𝛽 − 𝛽𝛽 𝑐𝑐 � Let ( , , ) with + + = 0, = 1,2; then:

𝑖𝑖 𝑖𝑖 𝑖𝑖 [(𝑖𝑖 + 𝑖𝑖 )𝑖𝑖 +𝑖𝑖( + ) + ( + 𝑢𝑢�⃗ 𝛼𝛼 𝛽𝛽1 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝑖𝑖 ) ]. 2 2 𝑢𝑢�⃗1 ∙ 𝑢𝑢�⃗2 − − 2 𝛽𝛽1𝛾𝛾2 𝛽𝛽2𝛾𝛾1 𝑎𝑎 𝛾𝛾1𝛼𝛼2 𝛼𝛼2𝛾𝛾2 𝑏𝑏 𝛼𝛼1𝛽𝛽2 2 2 1 𝛼𝛼Theorem𝛽𝛽 𝑐𝑐 55 The points , , with + + = 1, = 1, 3 are collinear if and only if: 𝑄𝑄𝑖𝑖�𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖� 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 = = . 𝛼𝛼2−𝛼𝛼1 𝛽𝛽2−𝛽𝛽1 𝛾𝛾2−𝛾𝛾1 0 𝛼𝛼If3 a− 𝛼𝛼denominator1 𝛽𝛽3−𝛽𝛽1 is 𝛾𝛾3,− then𝛾𝛾1 it is agreed that the appropriate numerator to be zero.

Theorem 56 (The three-point collinearity condition)

The points 1, 2, 3 with ( , , ), + + = 1, = 1,3 are collinear𝑄𝑄 𝑄𝑄, if 𝑄𝑄and only 𝑄𝑄if:𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 = 0. 𝛼𝛼1 𝛽𝛽1 𝛾𝛾1 �𝛼𝛼2 𝛽𝛽2 𝛾𝛾2� Conse𝛼𝛼3 𝛽𝛽quence3 𝛾𝛾3 The point ( , , ), + + = 1 is located on the line , ( , , ) with + + = 1, = 1, 2, if and only if 𝑃𝑃 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑄𝑄1𝑄𝑄2 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑄𝑄 𝛼𝛼 𝛽𝛽 𝛾𝛾 =𝛼𝛼 0. 𝛽𝛽 𝛾𝛾 𝑖𝑖 �� 𝑥𝑥 𝑦𝑦 𝑧𝑧 1 1 1 �𝛼𝛼 𝛽𝛽 𝛾𝛾 � 2 2 2 Observation𝛼𝛼 𝛽𝛽 𝛾𝛾 71 From those previously established, it follows that the equation of a line in barycentric coordinates is + + = 0, , , . The director vector of the line : + + = 0 is given by = ( 𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛 𝑝𝑝𝑝𝑝 𝑚𝑚 𝑛𝑛 𝑝𝑝 ∈ ℝ ) + ( ) + ( ) . 𝑑𝑑 𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛 𝑝𝑝 𝑢𝑢��𝑑𝑑⃗ 𝑛𝑛 − 𝑝𝑝 𝑟𝑟��𝐴𝐴⃗ 𝑝𝑝 − 𝑚𝑚 𝑟𝑟��𝐵𝐵⃗ 𝑚𝑚 − 𝑛𝑛 𝑟𝑟��𝐶𝐶⃗

220 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 72 The barycentric coordinates of the director vector of the line : + + = 0 are: ( , , ). 𝑑𝑑 𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛 𝑝𝑝𝑝𝑝 Theorem𝑛𝑛 5−7 𝑝𝑝(The𝑝𝑝 − condition𝑚𝑚 𝑚𝑚 − 𝑛𝑛 of parallelism of two lines) The lines : + + = 0 , : + + = 0 are parallels if and only if: 𝑑𝑑1 𝑚𝑚1𝑥𝑥 𝑛𝑛1𝑦𝑦 𝑝𝑝1𝑧𝑧 𝑑𝑑2 𝑚𝑚2𝑥𝑥 𝑛𝑛2𝑦𝑦 𝑝𝑝2𝑧𝑧 = = , 𝑚𝑚1−𝑛𝑛1 𝑛𝑛1−𝑝𝑝1 𝑝𝑝1−𝑚𝑚1 𝑚𝑚or2:− 𝑛𝑛2 𝑛𝑛2−𝑝𝑝2 𝑝𝑝2−𝑚𝑚2 1 1 1 = 0.

𝑑𝑑1 ∥ 𝑑𝑑2 ⇔ �𝑚𝑚1 𝑛𝑛1 𝑝𝑝1� 2 2 2 Theorem 58 𝑚𝑚(The condition𝑛𝑛 𝑝𝑝 of perpendicularity of two lines) The lines : + + = 0 and : + + = 0 are perpendicular1 1 1 if and1 only if: 2 2 2 [( )( 𝑑𝑑 𝑚𝑚 𝑥𝑥) + (𝑛𝑛 𝑦𝑦 𝑝𝑝 )𝑧𝑧( )] 𝑑𝑑+ [𝑚𝑚( 𝑥𝑥 𝑛𝑛)𝑦𝑦( 2 𝑝𝑝 𝑧𝑧 ) + ( )( ] 2 + [( ) ( 𝑝𝑝1 − 𝑚𝑚1 𝑚𝑚2 − 𝑛𝑛2 𝑚𝑚1 − 𝑛𝑛1 𝑝𝑝2 − 𝑚𝑚2 𝑎𝑎 𝑛𝑛1 − 𝑝𝑝1 𝑚𝑚2 ) + ( ) ( )] 2 = 0. 2 1 1 2 2 1 1 2 − 𝑛𝑛 𝑚𝑚 − 𝑛𝑛 𝑚𝑚 − 𝑝𝑝 𝑏𝑏 2 𝑛𝑛 − 𝑝𝑝 𝑝𝑝 2 1 1 2 2 Theorem 59 − 𝑚𝑚 𝑝𝑝 − 𝑚𝑚 𝑛𝑛 − 𝑝𝑝 𝑐𝑐 The equation of the line determined by the point ( , , ), + + = 1 and by the director vector ( , , ), + + 𝑃𝑃 𝑥𝑥0 𝑦𝑦0 𝑧𝑧0 = 0 is: 𝑥𝑥0 𝑦𝑦0 𝑧𝑧0 �𝑢𝑢�⃗ 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 = 0. 𝑥𝑥 𝑦𝑦 𝑧𝑧 0 0 0 �𝑥𝑥 𝑦𝑦 𝑧𝑧 � Conse𝛼𝛼 𝛽𝛽quence𝛾𝛾 The equation of the line that passes through ( , , ), + + = 1 and is parallel with the line ÷ + + = 0 is: 𝑃𝑃 𝑥𝑥0 𝑦𝑦0 𝑧𝑧0 𝑥𝑥0 𝑦𝑦0 0 𝑧𝑧 = 0. 𝑑𝑑 𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛 𝑝𝑝𝑝𝑝 𝑥𝑥 𝑦𝑦 𝑧𝑧 0 0 0 � 𝑥𝑥 𝑦𝑦 𝑧𝑧 � 𝑛𝑛 − 𝑝𝑝 𝑝𝑝 − 𝑚𝑚 𝑚𝑚 − 𝑛𝑛

Ion Pătrașcu, Florentin Smarandache 221 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 60 (Barycentric Coordinates of a vector perpendicular to a given vector)

If ( , , ), + + = 0 is a given vector, and 1 is the vector perpendicular to , then: ��⃗ 𝑢𝑢 𝛼𝛼( 𝛽𝛽 𝛾𝛾 ) 𝛼𝛼 𝛽𝛽 𝛾𝛾+ , + ( )𝑢𝑢 , �𝑢𝑢�⃗ 2 + (2 )2 . 2 2 2 2 2 𝑢𝑢⊥� γ − β 𝑎𝑎 − 𝛼𝛼𝑏𝑏 𝛼𝛼𝑐𝑐 𝛽𝛽𝑎𝑎 − 𝛽𝛽𝑐𝑐 α − γ 𝑏𝑏 𝛾𝛾𝑏𝑏 − 𝛾𝛾𝑎𝑎 2 Theorem 61 𝛽𝛽 − 𝑑𝑑 𝑐𝑐 � In the plane of the triangle , we consider the point ( , , ), + + = 1. We denote by { } = , { } = , { } = 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 . The barycentric coordinates of the points , , are: 𝛽𝛽 𝛾𝛾 𝑀𝑀 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 𝑁𝑁 𝐵𝐵𝐵𝐵 ∩ 𝐶𝐶𝐶𝐶 𝑃𝑃 𝐶𝐶𝐶𝐶 ∩ 0, , ; ,0, ; , , 0 . 𝐴𝐴𝐴𝐴 𝑀𝑀 𝑁𝑁 𝑃𝑃 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝑀𝑀 � 𝛽𝛽+𝛾𝛾 𝛽𝛽+𝛾𝛾� 𝑁𝑁 �𝛼𝛼+𝛾𝛾 𝛼𝛼+𝛾𝛾� 𝑃𝑃 �𝛼𝛼+𝛽𝛽 𝛼𝛼+𝛽𝛽 � Remark If the coordinates of the point ( , , ) are not absolute, then the barycentric (non-absolute) coordinates of the points , , are (0, , ), ( , 0, ), 𝑄𝑄 𝛼𝛼 𝛽𝛽 𝛾𝛾 ( , , 0). 𝑀𝑀 𝑁𝑁 𝑃𝑃 𝑀𝑀 𝛽𝛽 𝛾𝛾 𝑁𝑁 𝛼𝛼 𝛾𝛾 𝑃𝑃 𝛼𝛼Theorem𝛽𝛽 62 (The condition of concurrency of three lines) Let the line of equations: + + = 0, = 1, 3. 𝑑𝑑𝑖𝑖 The lines , , are concurrent, if and only if: 𝑚𝑚𝑖𝑖𝑥𝑥 𝑛𝑛𝑖𝑖𝑥𝑥 𝑝𝑝𝑖𝑖𝑧𝑧 𝑖𝑖 𝑑𝑑1 𝑑𝑑2=𝑑𝑑03. 𝑚𝑚1 𝑛𝑛1 𝑝𝑝1 2 2 2 �𝑚𝑚 𝑛𝑛 𝑝𝑝 � Conse𝑚𝑚3 quence𝑛𝑛3 𝑝𝑝3 (The condition of concurrency of three cevians) If the points , , situated in the plane of the triangle have the coordinates ( , , ), = 1, 3, then the lines , , are 𝑄𝑄1 𝑄𝑄2 𝑄𝑄3 𝐴𝐴𝐴𝐴𝐴𝐴 concurrent if and only if: = . 𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 𝐴𝐴𝑄𝑄1 𝐵𝐵𝑄𝑄2 𝐶𝐶𝑄𝑄3 3 1 2 2 3 1 Theorem 63 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛼𝛼 𝛽𝛽 𝛾𝛾 Let , in the plane of the triangle such that: + + = 0, 𝑄𝑄1 𝑄𝑄2 𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼1𝑄𝑄��1��𝐴𝐴�⃗ 𝛽𝛽1𝑄𝑄��1��𝐵𝐵�⃗ 𝛾𝛾1𝑄𝑄��1��𝐶𝐶�⃗

222 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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+ + = 0; then the lines , intersect the sides of the triangle in the 𝛼𝛼2𝑄𝑄��2��𝐴𝐴�⃗ 𝛽𝛽2𝑄𝑄��2��𝐵𝐵�⃗ 𝛾𝛾2𝑄𝑄��2��𝐶𝐶�⃗ points , , that verifies the relations: 𝑄𝑄1 𝑄𝑄2 𝐴𝐴𝐴𝐴𝐴𝐴

= 𝑀𝑀𝛾𝛾1∈ 𝐵𝐵𝐵𝐵𝛼𝛼1 , 𝑁𝑁 ∈= 𝐶𝐶𝐶𝐶𝛼𝛼1𝑃𝑃 ∈𝛽𝛽1 𝐵𝐵𝐵𝐵, = 𝛽𝛽1 𝛾𝛾1 . � � � � 𝑀𝑀𝑀𝑀��⃗ �𝛾𝛾2 𝛼𝛼2� 𝑁𝑁𝑁𝑁��⃗ 𝛼𝛼2 𝛽𝛽2 𝑃𝑃��𝑃𝑃�⃗ 𝛽𝛽2 𝛾𝛾2 1 1 ��⃗ 𝛽𝛽1 𝛼𝛼1 ��⃗ 𝛾𝛾1 𝛽𝛽1 ��⃗ 𝛼𝛼 𝛾𝛾 𝑀𝑀𝑀𝑀 − � � 𝑁𝑁𝑁𝑁 − � � 𝑃𝑃𝑃𝑃 − � � 𝛽𝛽2 𝛼𝛼2 𝛾𝛾2 𝛽𝛽2 𝛼𝛼2 𝛾𝛾2 Theorem 64 Let a triangle and a point in its plane such that + + = 0, where , , 0 and + + 0. We 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄 𝛼𝛼�𝑄𝑄��𝑄𝑄�⃗ denote = { }, = { }, = { }. 𝛽𝛽𝑄𝑄��𝑄𝑄�⃗ 𝛾𝛾�𝑄𝑄��𝑄𝑄�⃗ �⃗ 𝛼𝛼 𝛽𝛽 𝛾𝛾 ≠ 𝛼𝛼 𝛽𝛽 𝛾𝛾 ≠ Then: 𝐴𝐴𝐴𝐴=∩ 𝐵𝐵𝐵𝐵, =𝑀𝑀 𝐵𝐵𝐵𝐵, =∩ 𝐴𝐴𝐴𝐴. 𝑁𝑁 𝐶𝐶𝐶𝐶 ∩ 𝐴𝐴𝐴𝐴 𝑃𝑃 𝑀𝑀𝑀𝑀��⃗ 𝛾𝛾 𝑁𝑁𝑁𝑁��⃗ 𝛼𝛼 𝑃𝑃��𝑃𝑃�⃗ 𝛽𝛽 𝑀𝑀𝑀𝑀��⃗ − 𝛽𝛽 𝑁𝑁𝑁𝑁��⃗ − 𝛾𝛾 𝑃𝑃��𝑃𝑃�⃗ − 𝛼𝛼 Theorem 65 Let the triangle and , , , such that:

= , = 𝐴𝐴𝐴𝐴𝐴𝐴, = 𝑀𝑀. ∈Then𝐵𝐵𝐵𝐵 𝑁𝑁the∈ lines𝐴𝐴𝐴𝐴 𝑃𝑃 ∈ 𝐴𝐴𝐴𝐴, , are �𝑀𝑀𝑀𝑀��⃗ −𝛾𝛾 𝑁𝑁𝑁𝑁��⃗ −𝛼𝛼 𝑃𝑃��𝑃𝑃�⃗ −𝛽𝛽 concurrent𝑀𝑀𝑀𝑀��⃗ 𝛽𝛽 𝑁𝑁𝑁𝑁�� in⃗ the𝛾𝛾 point𝑃𝑃��𝑃𝑃�⃗ 𝛼𝛼 whose barycentric𝐴𝐴 𝐴𝐴coordinates𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 are ( , , ). 𝑄𝑄 Conse𝑄𝑄 𝛼𝛼 𝛽𝛽quence𝛾𝛾 = If , , are three concurrent cevians in the point and ′ ′ ′ ′ 𝐴𝐴 𝐵𝐵 ′ , = , = , then , , . 𝐴𝐴 𝐶𝐶 ′ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵′ 𝐶𝐶𝐶𝐶 𝑋𝑋 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 1 1 1 ′ ′ 𝛼𝛼 𝐵𝐵 𝐴𝐴 𝛽𝛽 𝐶𝐶 𝐵𝐵 𝛾𝛾 𝑋𝑋 �1−𝛾𝛾+𝛾𝛾𝛾𝛾 1−𝛼𝛼+𝛼𝛼𝛼𝛼 1− 𝛽𝛽+𝛽𝛽𝛽𝛽� Theorem 66 Let ( ), ( ) be two isotomic points in the triangle ; then = = . 𝑃𝑃 𝛼𝛼𝛼𝛼𝛼𝛼 𝑃𝑃′ 𝛼𝛼′𝛽𝛽′𝛾𝛾′ Theorem𝐴𝐴𝐴𝐴𝐴𝐴 67𝛼𝛼𝛼𝛼 ′ 𝛽𝛽𝛽𝛽′ 𝛾𝛾𝛾𝛾′ Let ( , , ), ( ) be two isogonal points in the triangle , with = , = , = ; then: = = . 𝑃𝑃 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝑃𝑃′ 𝛼𝛼′𝛽𝛽′𝛾𝛾′ ′ ′ ′ αα ββ γγ 2 2 2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑎𝑎 𝐶𝐶𝐶𝐶 𝑏𝑏 𝐴𝐴𝐴𝐴 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐

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8.1.2 The Barycentric Coordinates of some important points in the triangle geometry – THE CENTER OF GRAVITY 1 1 1 : , , 3 3 3 – THE CENTER OF THE INSCRIBED𝐺𝐺 � � CIRCLE , , 2 2 2 𝑎𝑎 𝑏𝑏 𝑐𝑐 – THE ORTHOCENTER 𝐼𝐼 � � 𝑝𝑝 𝑝𝑝 𝑝𝑝 (cot cot , cot cot , cot cot ) – THE CENTER OF THE CIRCUMSCRIBED CIRCLE 𝐻𝐻 𝐵𝐵sin 2𝐶𝐶 𝐴𝐴sin 2 𝐶𝐶 sin𝐴𝐴 2 𝐵𝐵 , , 2 2 2 2 2 2 𝑅𝑅 𝐴𝐴 𝑅𝑅 𝐵𝐵 𝑅𝑅 𝐶𝐶 – THE CENTER𝑂𝑂 OF� THE A-EX-INSCRIBED CIRCLE� 𝑆𝑆 𝑆𝑆 𝑆𝑆 , , 2( ) 2( ) 2( ) −𝑎𝑎 𝑏𝑏 𝑐𝑐 – THE NAGEL’S𝐼𝐼𝑎𝑎 POINT� � 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑎𝑎 , , 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 – THE GERGONNE’S𝑁𝑁 POINT� � ( )( ) (𝑝𝑝 )𝑝𝑝( 𝑝𝑝) ( )( ) , , (4 + ) (4 + ) (4 + ) 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑐𝑐 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 – THE 𝛤𝛤LEMOINE� ’S POINT � 𝑟𝑟 𝑅𝑅 𝑟𝑟 𝑟𝑟 𝑅𝑅 𝑟𝑟 𝑟𝑟 𝑅𝑅 𝑟𝑟 , , + 2 + + 2 + + 2 + 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐾𝐾 � 2 2 2 2 2 2 2 2 2� Observation 7𝑎𝑎3 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 The above barycentric coordinates are absolute, and the relative barycentric coordinates are: (1, 1, 1), ( , , ), (sin 2 , sin 2 , sin 2 ), ( , , ), ( , , ) , , ( , , ) 𝐺𝐺 , 𝐼𝐼 𝑎𝑎 𝑏𝑏 𝑐𝑐 ,𝑂𝑂 𝐴𝐴 . 𝐵𝐵 𝐶𝐶 𝐼𝐼𝑎𝑎 −𝑎𝑎 𝑏𝑏 𝑐𝑐 1 1 1 2 2 2 𝑁𝑁 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 𝛤𝛤 �𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐� 𝐾𝐾 𝑎𝑎 𝑏𝑏 𝑐𝑐 8.1.3 Other Barycentric Coordinates and useful equations – The coordinates of the vertices of the reference triangle are: (1, 0, 0), (0, 1, 0), (0, 0, 1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶

224 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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– The coordinates of the midpoints , , of the sides of the reference triangle are: 1 1 1𝑀𝑀 𝑁𝑁1 𝑃𝑃 1 1 0, , , , 0, , , , 0 𝐴𝐴𝐴𝐴𝐴𝐴 2 2 2 2 2 2 – The coordinates of𝑀𝑀 the� point� 𝑁𝑁 � , � =𝑃𝑃 � are: � �� 1 𝑀𝑀𝑀𝑀 𝑀𝑀0, ∈ 𝐵𝐵𝐵𝐵 , 𝑢𝑢𝑢𝑢�� 𝑘𝑘 1 1 −𝑘𝑘 – The coordinates of an arbitrary𝑀𝑀 � point � are (0, , ) – The coordinates of an arbitrary point− 𝑘𝑘 − 𝑘𝑘 are ( , 0, ) 𝑀𝑀 ∈ 𝐵𝐵𝐵𝐵 𝑀𝑀 𝑏𝑏 𝑐𝑐 – The coordinates of an arbitrary point are ( , , 0) 𝑁𝑁 ∈ 𝐶𝐶𝐶𝐶 𝑁𝑁 𝑎𝑎 𝑐𝑐 – The equation of the line is = 0 𝑃𝑃 ∈ 𝐴𝐴𝐴𝐴 𝑃𝑃 𝑎𝑎 𝑏𝑏 – The equation of the line is = 0 𝐵𝐵𝐵𝐵 𝑥𝑥 – The equation of the line is = 0 𝐴𝐴𝐴𝐴 𝑦𝑦 – The equation of a line that passes through is + = 0 𝐴𝐴𝐴𝐴 𝑧𝑧 – The equation of a line that passes through is + = 0 𝐴𝐴 𝑛𝑛𝑛𝑛 𝑝𝑝𝑝𝑝 – The equation of a line that passes through is + = 0 𝐵𝐵 𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝 – The coordinates of a direction vector are: 𝐶𝐶 𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛 (0, 1, 1) 𝐵𝐵𝐵𝐵 – The coordinates of a direction vector are: 𝑢𝑢�⃗𝐵𝐵��𝐵𝐵�⃗ − (1,0, 1) 𝐶𝐶𝐶𝐶 – The coordinates of a direction vector are: 𝑢𝑢�⃗𝐶𝐶��𝐶𝐶�⃗ − ( 1,1, 0) 𝐴𝐴𝐴𝐴 – The coordinates of a vector perpendicular to are: 𝑢𝑢�⃗𝐴𝐴��𝐴𝐴�⃗ − (2 , + , + ) 𝐵𝐵𝐵𝐵 – The coordinates of a vector2 perpendicular2 2 2 to 2 are2 : 2 𝑢𝑢�⃗⊥𝐵𝐵𝐵𝐵 𝑎𝑎 −𝑎𝑎 − 𝑏𝑏 𝑐𝑐 −𝑎𝑎 𝑏𝑏 − 𝑐𝑐 ( + , 2 , ) 𝐶𝐶𝐶𝐶 – The coordinates of a vector2 perpendicular2 2 2 to2 are2 : 2 𝑢𝑢�⃗⊥𝐶𝐶𝐶𝐶 −𝑎𝑎 − 𝑏𝑏 𝑐𝑐 𝑏𝑏 𝑎𝑎 − 𝑏𝑏 − 𝑐𝑐 ( + , , 2 ) 𝐴𝐴𝐵𝐵 – The equation of the mediator2 of2 the2 side2 2is: 2 2 𝑢𝑢�⃗⊥𝐴𝐴𝐴𝐴 −𝑎𝑎 𝑏𝑏 − 𝑐𝑐 𝑎𝑎 − 𝑏𝑏 − 𝑐𝑐 𝑐𝑐 ( ) + = 0 𝐵𝐵𝐵𝐵 – The equation of the mediator2 2 of the side2 2 is: 𝑏𝑏 − 𝑐𝑐 𝑥𝑥 𝑎𝑎 𝑦𝑦 − 𝑎𝑎 𝑧𝑧 + ( ) + = 0 𝐶𝐶𝐶𝐶 – The equation of the mediator2 of2 the side2 2is: −𝑏𝑏 𝑥𝑥 𝑐𝑐 − 𝑎𝑎 𝑦𝑦 𝑏𝑏 𝑧𝑧 + ( ) = 0 2 2 2 𝐴𝐴𝐴𝐴2 𝑐𝑐 𝑥𝑥 − 𝑐𝑐 𝑦𝑦 𝑎𝑎 − 𝑏𝑏 𝑧𝑧

Ion Pătrașcu, Florentin Smarandache 225 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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8.1.4 Aplications 1. In a triangle , – the center of gravity, – the center of the inscribed circle and the Nagel’s point, , are collinear points. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐺𝐺 𝐼𝐼 Solution 𝑁𝑁 1 1 1 1 1 1 1 1 1 = = 0. p a p b p c � 𝑎𝑎 𝑏𝑏 𝑐𝑐 � �𝑎𝑎 𝑏𝑏 𝑐𝑐� − �𝑎𝑎 𝑏𝑏 𝑐𝑐� − − − 𝑝𝑝 𝑝𝑝 𝑝𝑝 2. Prove that in a triangle the Gergonne’s𝑎𝑎 𝑏𝑏 𝑐𝑐 point, , the Nagel’s point, , and the point – the isotomic of the orthocenter , are collinear 𝐴𝐴𝐴𝐴𝐴𝐴 𝛤𝛤 points. 𝑁𝑁 𝑅𝑅 𝐻𝐻 Solution The barycentric coordinates of are: (cot cot , cot cot , cot cot ). 𝐻𝐻 The barycentric coordinates of its isotomic are: 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 , , ′ , 𝐻𝐻 1 1 1 therefore (tan tan , tan tan , tan ). �cot 𝐵𝐵 cot 𝐶𝐶 cot 𝐶𝐶 cot 𝐴𝐴 cot 𝐴𝐴 cot 𝐵𝐵� We have ′ , , and ( , , ). 𝐻𝐻 1 𝐵𝐵1 𝐶𝐶1 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 The collinearity𝛤𝛤 �𝑝𝑝−𝑎𝑎 of𝑝𝑝 −points𝑏𝑏 𝑝𝑝−𝑐𝑐 � , 𝑁𝑁and𝑝𝑝 − is𝑎𝑎 equivalent𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − to𝑐𝑐: tan tan tan tan ′tan tan 𝐻𝐻 𝛤𝛤 𝑁𝑁 = 0. 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 � 𝑝𝑝 −1 𝑎𝑎 𝑝𝑝 −1 𝑏𝑏 𝑝𝑝 −1 𝑐𝑐 � The preceding𝑝𝑝−𝑎𝑎 condition𝑝𝑝−𝑏𝑏 is equivalent𝑝𝑝−𝑐𝑐 to: cot cot cot = = 0. ( 𝐴𝐴) ( 𝐵𝐵) ( 𝐶𝐶) 𝐷𝐷 � 𝑝𝑝 − 𝑎𝑎−1 𝑝𝑝 − 𝑏𝑏−1 𝑝𝑝 −( 𝑐𝑐−1) �( )( ) We know that cot = = and the analogs. 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏2 2 2 𝑝𝑝 − 𝑐𝑐 𝑏𝑏 +𝑐𝑐 −𝑎𝑎 𝑃𝑃 𝑝𝑝−𝑎𝑎 − 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 = Area . 𝐴𝐴 2𝑆𝑆 2𝑆𝑆

𝑆𝑆 ∆𝐴𝐴𝐴𝐴𝐴𝐴= 2 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 𝑃𝑃 ( ) ( ) ( ) 𝐷𝐷 � 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 � − 𝑆𝑆 (−1 ) (−1 ) (−1 ) ( )( )( 𝑝𝑝 − )𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 −1 −1 −1 = 0. 2 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 ( ) ( ) ( ) − � 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 � 𝑆𝑆 −1 −1 −1 𝑝𝑝 − 𝑎𝑎 𝑝𝑝 − 𝑏𝑏 𝑝𝑝 − 𝑐𝑐 226 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 74 The point , the isotomic of the orthocenter of a triangle, is also called the retrocenter of a triangle. 𝑅𝑅

3. If , are points in the plane of the triangle , ( , , ) with + + = 1, = 1, 2, deduce the formula: 𝑄𝑄1 𝑄𝑄2 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄𝑖𝑖 𝛼𝛼𝑖𝑖 𝛽𝛽𝑖𝑖 𝛾𝛾𝑖𝑖 𝑖𝑖 𝑖𝑖 𝑖𝑖 = + + . 𝛼𝛼 𝛽𝛽 𝛾𝛾 1 𝑖𝑖 2 1 2 1 2 1 2 2 2 2 2 2 2 2 Solution𝑄𝑄 𝑄𝑄 𝛼𝛼 𝑄𝑄 𝐴𝐴 𝛽𝛽 𝑄𝑄 𝐵𝐵 𝛾𝛾 𝑄𝑄 𝐶𝐶 − � 𝑎𝑎 𝛽𝛽 𝛾𝛾 Let (1, 0, 0), (0, 1, 0), (0, 0, 1) be the barycentric coordinates of vertices. We have: 𝐴𝐴 𝐵𝐵 2 𝐶𝐶 2 1 = 2 1 2 1 2 1 ( 2 1) 2 2( ) . 𝑄𝑄 𝑄𝑄2 −𝑎𝑎 �𝛽𝛽 − 𝛽𝛽 ��𝛾𝛾2 − 𝛾𝛾1 � −2𝑏𝑏 �𝛾𝛾1 − 𝛾𝛾 � 𝛼𝛼 − 𝛼𝛼 By doing the calculations, we get: − 𝑐𝑐 𝛼𝛼 − 𝛼𝛼 �𝛽𝛽 − 𝛽𝛽 � 1 = + + (1) 2 2 2 2 2 We calculate2 : 2 2 1 1 1 2 2 1 𝑄𝑄 𝑄𝑄 − ∑ 𝑎𝑎2 𝛽𝛽 𝛾𝛾 − ∑2 𝑎𝑎 𝛽𝛽 𝛾𝛾 ∑ 𝑎𝑎 𝛽𝛽2 𝛾𝛾 ∑ 𝑎𝑎 𝛽𝛽 𝛾𝛾 1 = + (1 ) + (1 ) 2 = 2 2 2 + 2 + 2 . 𝑄𝑄 𝐴𝐴 −𝑎𝑎 𝛽𝛽1𝛾𝛾1 𝑏𝑏 𝛾𝛾1 − 𝛼𝛼1 𝑐𝑐 𝛽𝛽1 − 𝛼𝛼1 Therefore: 𝑎𝑎 1 1 𝑏𝑏 1 1 𝑐𝑐 1 1 𝑏𝑏 1 𝑐𝑐 1 = + 𝛽𝛽 +𝛾𝛾 − . 𝛾𝛾 𝛼𝛼 − 𝛼𝛼 𝛽𝛽 𝛾𝛾 𝛽𝛽 Similarly2 : 2 2 2 𝑄𝑄1𝐴𝐴 − ∑ 𝑎𝑎 𝛽𝛽1𝛾𝛾1 𝑏𝑏 𝛾𝛾1 𝑐𝑐 𝛽𝛽1 = + + . 2 = 2 + 2 + 2 . 𝑄𝑄1𝐵𝐵 − ∑ 𝑎𝑎 𝛽𝛽1𝛾𝛾1 𝑐𝑐 𝛼𝛼1 𝑎𝑎 𝛾𝛾1 We e2valuate: 2 2 2 𝑄𝑄1𝐶𝐶 − ∑ 𝑎𝑎 𝛽𝛽1𝛾𝛾1 𝑎𝑎 𝛽𝛽1 𝑏𝑏 𝛾𝛾1 + + . 2 2 2 We have: 2 1 2 1 2 1 + 𝛼𝛼+𝑄𝑄 𝐴𝐴 𝛽𝛽 𝑄𝑄 𝐵𝐵 𝛾𝛾 𝑄𝑄 𝐶𝐶 2 2 2 2 1 2 1 = 2 1 + 𝛼𝛼 𝑄𝑄 𝐴𝐴 𝛽𝛽 𝑄𝑄 𝐵𝐵 𝛾𝛾 𝑄𝑄 𝐶𝐶 1 1 1 + 1 2 2 2 2 + 𝑎𝑎 �− � 𝑎𝑎 𝛽𝛽1𝛾𝛾1 + 𝑏𝑏 𝛾𝛾1 + 𝑐𝑐 𝛽𝛽1� 2 2 2 2 + 𝛽𝛽 �− � 𝑎𝑎 𝛽𝛽1 𝛾𝛾1 + 𝑐𝑐 𝛼𝛼1 + 𝑎𝑎 𝛾𝛾1 � 2 2 2 2 = 𝛾𝛾 �− � 𝑎𝑎1 𝛽𝛽1 +𝛾𝛾 𝑎𝑎 𝛽𝛽1 2 +𝑏𝑏 𝛾𝛾 � 2 1 . (2) 2 2 2 Comparing the relations− (1)� and𝑎𝑎 𝛽𝛽 (2),𝛾𝛾 we� find𝑎𝑎 that𝛽𝛽 𝛾𝛾: � 𝑎𝑎 𝛽𝛽 𝛾𝛾 2 2 2 2 1 = 2 1 + 2 1 + 2 1 2 2 . (3) 2 𝑄𝑄 𝑄𝑄2 𝛼𝛼 𝑄𝑄 𝐴𝐴 𝛽𝛽 𝑄𝑄 𝐵𝐵 𝛾𝛾 𝑄𝑄 𝐶𝐶 − ∑ 𝑎𝑎 𝛽𝛽 𝛾𝛾

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4. If is a given scalene triangle, and are respectively the centers of its circumscribed and inscribed circles; prove the relation: = 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝐼𝐼 2 . 2 2 𝑂𝑂𝐼𝐼 𝑅𝑅 − 𝑅𝑅𝑅𝑅Solution We use the formula (3) from the previous aplication, where = and

= . The barycentric coordinates of are , , . 1 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑄𝑄 𝑄𝑄 𝑄𝑄2 We𝐼𝐼 have = = 𝐼𝐼 2𝑝𝑝 2𝑝𝑝 =2𝑝𝑝 . 2 2 2 𝑏𝑏𝑏𝑏 2 𝑎𝑎𝑎𝑎𝑎𝑎 2 𝑎𝑎𝑎𝑎𝑎𝑎 2 2 = = 4 Taking into𝑂𝑂𝐼𝐼 account𝑅𝑅 − the∑ known𝑎𝑎 4𝑝𝑝 formulas𝑅𝑅 − 4 𝑝𝑝 ∑ 𝑎𝑎 and𝑅𝑅 − 2𝑝𝑝 , we get: = = 2 . 𝑆𝑆 𝑝𝑝𝑝𝑝 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅𝑅𝑅 2 2 4𝑅𝑅𝑅𝑅 2 𝑂𝑂𝐼𝐼 𝑅𝑅 − 2𝑝𝑝 𝑅𝑅 − 𝑅𝑅𝑅𝑅 Observation 75 The resulting relation is called Euler’s relation. From here, it follows that, in a triangle, 2 (Euler’s inequality).

𝑅𝑅 ≥ 𝑟𝑟 5. Let be a scalene triangle, – its circumscribed center, and – the Nagel’s point. Show that = 2 ( – radius of the circumscribed 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑁𝑁 circle, – radius of the circle inscribed in the triangle ). 𝑂𝑂𝑂𝑂 𝑅𝑅 − 𝑟𝑟 𝑅𝑅 Solution𝑟𝑟 𝐴𝐴𝐴𝐴𝐴𝐴 We employ the formula (3) from Application 3, where = and = .

The barycentric coordinates of the Nagel’s point are 1 , , 2 . We 𝑄𝑄𝑝𝑝−𝑎𝑎 𝑄𝑄𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑄𝑄𝑐𝑐 𝑁𝑁 have: 𝑁𝑁 � 𝑝𝑝 𝑝𝑝 𝑝𝑝 � ( )( ) = . 2 2 2 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 1 2 1 𝑂𝑂𝑁𝑁= 𝑅𝑅 − ∑ 𝑎𝑎 ( 𝑝𝑝 + )( + ) = [ ( ) ] 4 4 2 2 2 2 2 2 2 2 1 2 𝑂𝑂𝑁𝑁 𝑅𝑅 − �=𝑎𝑎 𝑎𝑎 − 𝑏𝑏 𝑐𝑐 𝑎𝑎( 𝑏𝑏 − 𝑐𝑐 𝑅𝑅 +−2 ) �= 𝑎𝑎 𝑎𝑎 − 𝑏𝑏 − 𝑐𝑐 𝑝𝑝 4 𝑝𝑝 2 2 2 2 2 2 1 2 + 𝑅𝑅 −2 � 𝑎𝑎 𝑎𝑎 − 𝑏𝑏 −2𝑐𝑐 ( 𝑏𝑏+𝑏𝑏 +𝑅𝑅 ) 4 𝑝𝑝 2 2 4 2 1 1 = +� �(𝑏𝑏 6𝑐𝑐 − �4 𝑎𝑎 −) =𝑎𝑎𝑎𝑎𝑎𝑎 +𝑎𝑎 𝑏𝑏 (16𝑐𝑐 � 16 ) 𝑝𝑝 4 4 2 2 2 2 2 2 = 𝑅𝑅 + 4 2 𝑎𝑎4𝑆𝑆 − (𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝2 ) 𝑅𝑅 2 𝑝𝑝 𝑟𝑟 − 𝑝𝑝 𝑅𝑅 ∙ 𝑟𝑟 𝑝𝑝 𝑝𝑝 The following formulas2 were2 used: 2 𝑅𝑅 𝑟𝑟 − 𝑝𝑝𝑝𝑝 − 𝑅𝑅 − 𝑟𝑟 16 = 2( + + ) , 2= 4 2 and2 =2 2 . 2 2 4 4 4 𝑆𝑆 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑐𝑐 𝑎𝑎 − 𝑎𝑎 − 𝑏𝑏 − 𝑐𝑐 𝑎𝑎𝑎𝑎𝑎𝑎 𝑅𝑅 ∙ 𝑆𝑆 𝑆𝑆 𝑝𝑝 ∙ 𝑟𝑟 228 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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8.2 Annex 2: The similarity of two figures

The study of the geometrical properties of the plane figures often calls for plane transformations that increase (or decrease) the distance between points, but retain the shape of the figures.

8.2.1 Properties of the similarity on a plane

Definition 50 An application : , where , is called a similarity ratio if ( ) ( ) = , ( ) , ∗ . 𝒂𝒂𝒌𝒌 𝓟𝓟 → 𝓟𝓟 𝒌𝒌 ∈ ℝ+ We denoted ( ) = , ( ) = and we say about and 𝒌𝒌 𝒂𝒂𝒌𝒌 𝑨𝑨 𝒂𝒂𝒌𝒌 𝑩𝑩 𝒌𝒌 ∙ 𝑨𝑨𝑨𝑨 ∀ 𝑨𝑨 𝑩𝑩 ∈ 𝓟𝓟 that are the homologues′ or the similar′ s of the points , ′ . ′ 𝒂𝒂𝒌𝒌 𝑨𝑨 𝑨𝑨 𝒂𝒂𝒌𝒌 𝑩𝑩 𝑩𝑩 𝑨𝑨 𝑩𝑩 The constant ratio is called a similarity or a similitude ratio. 𝑨𝑨 𝑩𝑩 From definition, it follows that, if is a given triangle and 𝒌𝒌 is the triangle obtained from , applying to it the 𝑨𝑨𝑨𝑨𝑨𝑨 similarity′ ′ ′ , having = = = , the triangle is 𝑨𝑨 𝑩𝑩 𝑪𝑪 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨𝑨𝑨 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 ′ ′ ′ similar to the triangle ′ ′ . ′ ′ ′ ′ 𝒂𝒂𝒌𝒌 𝑨𝑨 𝑩𝑩 𝑨𝑨 𝑪𝑪 𝑪𝑪 𝑪𝑪 𝒌𝒌 𝑨𝑨 𝑩𝑩 𝑪𝑪 We denote ~ . 𝑨𝑨𝑨𝑨𝑨𝑨 Furthermore, ′if ′we′ consider the triangle , oriented such as ∆𝑨𝑨 𝑩𝑩 𝑪𝑪 ∆𝑨𝑨𝑨𝑨𝑨𝑨 the vertices , , are read in the trigonometric sense, and if , 𝑨𝑨𝑨𝑨𝑨𝑨 , have the same orientation, we say that the similarity is′ 𝑨𝑨 𝑩𝑩 𝑪𝑪 𝑨𝑨 direct′ ′. 𝑩𝑩 𝑪𝑪 If the similarity is inverse, and are inversely- orientated. ′ ′ ′ 𝑨𝑨𝑨𝑨𝑨𝑨 𝑨𝑨 𝑩𝑩 𝑪𝑪 During this presentation, we will say that two figures are similar instead of directly-similar and we will make the special mention in the case of the inversely-similar figures.

Proposition 76 A similarity transforms three collinear points in three other collinear points keeping the points in order.

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Proof Let , , be three collinear points and let = ( ), = ( ) and = ( ) be their similars. ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝑎𝑎𝑘𝑘 𝐴𝐴 𝐵𝐵 𝑎𝑎𝑘𝑘 𝐵𝐵 ′ The points , , are in the order in which they were written, therefore 𝐶𝐶 𝑎𝑎𝑘𝑘 𝐶𝐶 + = , having = ( ) ( ) = , = ( ) ( ) = 𝐴𝐴= 𝐵𝐵 (𝐶𝐶 ) ′( ′) = ′ ′ + = and 𝑘𝑘 ; it𝑘𝑘 follows that 𝑘𝑘 𝑘𝑘 if 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 ′𝐴𝐴𝐴𝐴′ 𝐴𝐴 𝐵𝐵 𝑎𝑎 𝐴𝐴 𝑎𝑎 𝐵𝐵 𝐾𝐾𝐾𝐾𝐾𝐾 𝐵𝐵 ′𝐶𝐶 ′ 𝑎𝑎 ′ 𝐵𝐵′ 𝑎𝑎 ′𝐶𝐶 ′ , , are collinear𝑘𝑘 in the𝑘𝑘 order - - . 𝐾𝐾′𝐾𝐾𝐾𝐾 ′ ′ 𝐴𝐴 𝐶𝐶 𝑎𝑎 𝐴𝐴 𝑎𝑎 𝐶𝐶 𝐾𝐾𝐾𝐾𝐾𝐾′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵Remark𝐶𝐶 27 𝐴𝐴 𝐵𝐵 𝐶𝐶 From the previous property, it follows that a similarity transforms a segment into another segment, a semi-line into another semi-line, a line into another line (keeping the points in order).

Property 77 A similarity transforms two parallel lines into two other parallel lines.

d1 d2 B A C D

A' B' d'1

D' C' d'2

Figure 114

Proof Let and = ( ), = ( ) (see Figure 114). If , , , = ( ) = 1 such2 that 1 𝑘𝑘is parallelogram1 2 𝑘𝑘 , 2then we have , 1 ( ) 𝑑𝑑 ∥=𝑑𝑑 ( )𝑑𝑑 =𝑎𝑎 𝑑𝑑( )𝑑𝑑 𝑎𝑎 𝑑𝑑 = ′ = 𝐴𝐴 𝐵𝐵 ∈ ′𝑑𝑑 , 2 , . Since , 𝑘𝑘 and 𝐶𝐶 𝐷𝐷=∈ 𝑑𝑑 ′ 𝐴𝐴𝐴𝐴𝐴𝐴′ 𝐴𝐴 = ′ ′ 𝐴𝐴′ ′ 𝑎𝑎 𝐴𝐴 𝐵𝐵 𝑘𝑘 , it follows𝑘𝑘 that 𝑘𝑘 (1). 𝑎𝑎 𝐵𝐵 𝐶𝐶 𝑎𝑎 𝐶𝐶 𝐷𝐷 ′𝑎𝑎 ′ 𝐷𝐷 ′ ′ 𝐴𝐴 𝐵𝐵 𝑘𝑘 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐷𝐷 𝑘𝑘 ∙ 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷 230 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Also, = , = and = , it follows that = (2). The′ ′ relations (1)′ and′ (2) show that is a parallelogram′ ′ , 𝐴𝐴 𝐷𝐷 𝑘𝑘 ∙ 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑘𝑘 ∙ 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐷𝐷 therefore′ ′ . ′ ′ ′ ′ 𝐵𝐵 𝐶𝐶 ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷 1 2 Remark𝑑𝑑 28∥ 𝑑𝑑 The image of an angle by a similarity is an angle and . ′ ′ ′ � 𝑘𝑘 � � ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 𝐴𝐴 𝑂𝑂 𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 ≡ � 𝐴𝐴 𝑂𝑂Definition𝐵𝐵 51 Two figures and of the plane are called similar figures with the similarity ratio′ if there exists a similarity : 𝓕𝓕 𝓕𝓕 𝓟𝓟 such that ( ) = . 𝒌𝒌 𝒂𝒂𝒌𝒌 𝓟𝓟 → 𝓟𝓟 ~ ′ We denote 𝒌𝒌 and read is similar to . 𝒂𝒂 𝓕𝓕 ′ 𝓕𝓕 ′ Observation𝓕𝓕 76 𝓕𝓕 𝓕𝓕 𝓕𝓕 If ~ , then any triangle with the vertices belonging to the figure are as image a similar′ triangle with the vertices belonging to the figure . ℱ ℱ 𝐴𝐴𝐴𝐴𝐴𝐴 ℱ In two similar figures, ′ the′ ′homologous segments are proportional to′ the 𝐴𝐴 𝐵𝐵 𝐶𝐶 ℱ similarity ratio, and the homologous angles are congruent.

Definition 52 It is called center of similarity (or double point) of two similar figures the point of a figure that coincides with its homologue (similar) in the other figure.

Remark 29 From previous Definition, it follows that, if two figures and are similar and if is their similarity center, and are two homologous segment′ s from ℱ ℱ the figures respectively , then the triangles′ ′ and are similar. 𝑂𝑂 ′ 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 ′ ′ Propositionℱ 78 ℱ 𝑂𝑂𝑂𝑂𝑂𝑂 𝑂𝑂𝐴𝐴 𝐵𝐵 If and two homologous segments from the similar figures and , such that′ ′the lines and intersect in a point , then the 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 ℱ intersection′ of the circles circumscribed′ ′ to triangles and ℱ 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝑀𝑀 contain the center of similarity. ′ ′ 𝐴𝐴𝐴𝐴 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑀𝑀

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Proof We denote by the second common point of the circles circumscribed to the triangles and (see Figure 115). 𝑂𝑂 The quadrilateral′ ′ is inscribed, therefore (1). 𝐴𝐴𝐴𝐴 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑀𝑀 Also, we have that ′ (1). The quadrilateral ′ is 𝑀𝑀𝑀𝑀𝐵𝐵 𝑂𝑂 ∢𝑀𝑀𝑀𝑀𝑀𝑀 ≡ ∢𝑀𝑀𝐵𝐵 𝑂𝑂 inscribed, hence ′ (2). From′ relations (2) and (3), we note′ that ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀𝐴𝐴 𝑂𝑂 (4). ′ ′ ′ ∢′ 𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 M

Figure 115

Because = + and = + , we obtain that ′ (5). ′ ′ ′ ′ ′ 𝐴𝐴�𝐴𝐴𝐴𝐴 𝐴𝐴�𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵�𝐴𝐴 𝐵𝐵𝐵𝐵�𝐵𝐵 𝐵𝐵𝐵𝐵�𝐴𝐴 𝐴𝐴�𝑂𝑂𝐵𝐵 The relations (1)′ and′ (5) show that ~ and, consequently, is ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴 𝑂𝑂𝐵𝐵 the double point of similarity. ′ ′ ∆𝑂𝑂𝑂𝑂𝑂𝑂 ∆𝑂𝑂𝐴𝐴 𝐵𝐵 𝑂𝑂 Remark 30 i) The homothety is a particular case of similarity. ii) In the case of homothety, the similarity center is the homothety center.

Teorema68 Two similar figures can become homothetic by a rotation around their similarity center.

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Proof If the figures and are similar and have their similarity center , their similarity ratio being , we′ rotate the figure around the point with an angle ℱ ℱ 𝑂𝑂 = , where and are homologous′ points belonging to the figures 𝑘𝑘 ℱ 𝑂𝑂 and . Then′ the point occupies′ the position on the radius , also 𝜑𝜑 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴 � 𝐴𝐴 𝐴𝐴 because ′ ~ , the′ point occupies the position′′ on the radius ℱ ℱ ′ ′ 𝐴𝐴 ′ 𝐴𝐴 ′′ 𝑂𝑂𝑂𝑂 , and we have = . The reasoning applied to is valid for any point ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴 𝑂𝑂′′ 𝐵𝐵 𝐵𝐵 𝐵𝐵 𝑂𝑂𝐵𝐵 ; this will pass after rotation in on the homologous′ radius and 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 𝑘𝑘 𝐵𝐵 ′ ′ ′′ we have = . The figure occupies a new position after rotation, , and 𝑋𝑋 ∈ ℱ ′′ 𝑋𝑋 𝑂𝑂𝑂𝑂 𝑂𝑂𝑋𝑋 is homothetic to the figure ′ by homothety of center and ratio . ′′ 𝑂𝑂𝑂𝑂 𝑘𝑘 ℱ ℱ ′′ ℱ Remark 31 ℱ 𝑂𝑂 𝑘𝑘 i) Two homologous lines and form between them an angle of measure equal to the measure′ ′of the angle of rotation which 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 transforms the similar figures into homothetic figures. 𝜑𝜑 ii) The ratio of the distances of the similarity center of two homologous lines and is constant and equal with the similarity ratio. ′ ′ Theorem 69 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 The geometric place of the similarity centers of two non- concentric and non-congruent given circles is the circle with the diameter determined by the center of homothety of the two circles.

Proof Let ( , ) and ( , ), < , be the given circles (see Figure 116). We denote by and their center of direct and inverse homothety. If is 𝒞𝒞 𝑂𝑂1 𝑟𝑟1 𝒞𝒞 𝑂𝑂2 𝑟𝑟2 𝑟𝑟1 𝑟𝑟2 a similarity center of the two circles, then = (obviously the points and 𝐴𝐴 𝐵𝐵 𝑀𝑀𝑂𝑂1 𝑟𝑟1 𝑀𝑀 belong to the geometric place because they𝑀𝑀𝑂𝑂2 are𝑟𝑟 2similarity centers). 𝐴𝐴 From = = , it follows that is an internal bisector in the 𝐵𝐵 𝐵𝐵𝑂𝑂1 𝑟𝑟1 𝑀𝑀𝑂𝑂1 triangle 𝐵𝐵𝑂𝑂2 , 𝑟𝑟and2 from𝑀𝑀𝑂𝑂2 = , we obtain𝑀𝑀𝑀𝑀 that is an external bisector 𝐴𝐴𝑂𝑂1 𝑀𝑀𝑂𝑂1 in the triangle𝑀𝑀𝑂𝑂1𝑂𝑂 2 . 𝐴𝐴𝑂𝑂2 𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 𝑀𝑀𝑂𝑂1𝑂𝑂2

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Because the internal and external bisectors corresponding to the same angle of a triangle are perpendicular, we have that = 90 and consequently the point belongs to the circle of diameter [ ]. 0 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� 𝑀𝑀 𝐴𝐴𝐴𝐴 M T2

T1 T'1

A O1 B O2

T'2

Figure 116

Remark 32 i) If the circles ( , ) and ( , ) are secants in the points and , then obviously these points are similarity centers for the given 𝒞𝒞 𝑂𝑂1 𝑟𝑟1 𝒞𝒞 𝑂𝑂2 𝑟𝑟2 𝑀𝑀 circles. 𝑁𝑁 ii) If is a given scalene triangle, the geometric place of the points in the plane of the triangle for which = is a circle, called 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴 𝑀𝑀 the -Apollonius circle of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 8.2.2 Applications 1. Let and two squares in plane with the same orientation. Prove that , , are concurrent. 𝑂𝑂𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑂𝑂𝐴𝐴2𝐵𝐵2𝐶𝐶2 Solution 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 ~ . Because [ ] and [ ] are analogous segments in the directly-similar given squares and is their similarity point, it follows that ∆𝑂𝑂𝐴𝐴1𝐵𝐵1 ∆𝑂𝑂𝐴𝐴2𝐵𝐵2 𝐴𝐴1𝐵𝐵1 𝐴𝐴2𝐵𝐵2 and intersect in the second point of intersection of the circles 𝑂𝑂 circumscribed to the squares, point that was denoted by in Figure 117. 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝑀𝑀

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The same reasoning applies also to the homologous segments and , therefore intersects with in as well. 𝐵𝐵1𝐶𝐶1 𝐵𝐵2𝐶𝐶2 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 𝑀𝑀 C2

O C1

A1 B1

Figure 117 2. Let be a triangle and - - a transverse ( , , ). Prove that the circles circumscribed to the triangles , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ , and have a common point (the Miquel’s circles). 𝐴𝐴𝐴𝐴 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐶𝐶Solution𝐵𝐵𝐴𝐴1𝐶𝐶 1 𝐴𝐴1𝐶𝐶𝐵𝐵1 Let be the second common point of the circles circumscribed to the triangles and . This point is the similarity center of the homologous 𝑂𝑂 segments ( ) and ( ) (see Figure 118). 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 Therefore there exists a similarity such that ( ) = , ( ) = , 𝐶𝐶1𝐵𝐵 𝐵𝐵1𝐶𝐶 ( ) = ; there exists also a similarity such that ( ) = , ( ) = , 𝑎𝑎 𝑎𝑎 𝑂𝑂 𝑂𝑂 𝑎𝑎 𝐶𝐶1 𝐵𝐵1 ( ) = ; then the circles circumscribed′ to the triangles′ ′and 𝑎𝑎 𝐵𝐵 𝐶𝐶 𝑎𝑎 𝑎𝑎 𝑂𝑂 𝑂𝑂 𝑎𝑎 𝐶𝐶1 𝐵𝐵1 pass′ through the point . 𝑎𝑎 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐴𝐴1𝐶𝐶 𝐶𝐶1𝐴𝐴1𝐵𝐵 𝑂𝑂

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A O

B C A1

Figure 118

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8.3 Annex 3: The point, the triangle and the Miquel’s circles

8.3.1 Definitions and theorems

Theorem 70 (J. Steiner, 1827) The four circles circumscribed to the triangles formed by four lines intersecting two by two pass through the same point (Miquel’s point).

Proof Let , , and , , be the intersection points of the four lines in the statement (the quadrilateral is a complete quadrilateral, see Figure 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 119). We denote by the second point of intersection of the circles 𝐵𝐵𝐵𝐵𝐵𝐵1𝐶𝐶1𝐴𝐴1𝐴𝐴 circumscribed to the triangles and . 𝑀𝑀 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝐶𝐶𝐵𝐵1𝐴𝐴1

A M

B C A1

Figure 119

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The quadrilaterals and being inscribable, we obtain the relation: 𝑀𝑀𝑀𝑀𝐶𝐶1𝐵𝐵1 𝑀𝑀𝐵𝐵1𝐶𝐶𝐴𝐴1 = . (1) From (1), we note that: ∢𝑀𝑀𝐶𝐶1𝐴𝐴 ≡ ∢𝑀𝑀𝐵𝐵1𝐴𝐴 ≡ ∢𝑀𝑀𝐴𝐴1𝐶𝐶 𝜑𝜑 , (2) a relation showing that the quadrilateral is inscribable, hence the circle ∢𝑀𝑀𝐶𝐶1𝐴𝐴 ≡ ∢𝑀𝑀𝐴𝐴1𝐵𝐵 circumscribed to passes through the point . Also from the 𝑀𝑀𝐶𝐶1𝐵𝐵𝐴𝐴1 inscribability of the previous quadrilaterals, we obtain that: 𝐵𝐵𝐴𝐴1𝐶𝐶1 𝑀𝑀 . (3) Noting from this relation that: , we conclude that the ∢𝑀𝑀𝑀𝑀𝐶𝐶1 ≡ ∢𝑀𝑀𝐵𝐵1𝐴𝐴1 ≡ ∢𝑀𝑀𝑀𝑀𝐴𝐴1 quadrilateral is inscribable, consequently the circumscribed circle of the ∢𝑀𝑀𝑀𝑀𝐶𝐶1 ≡ ∢𝑀𝑀𝑀𝑀𝐴𝐴1 triangle passes through ; the theorem is proved. 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 Observation𝐴𝐴𝐴𝐴𝐴𝐴 77 𝑀𝑀 a) The concurrency point of the four circles is called the Miquel’s point of the complete quadrilateral . 𝑀𝑀 b) The preceding theorem can be reformulated; thus: The circles 𝐵𝐵𝐵𝐵𝐵𝐵1𝐶𝐶1𝐴𝐴1𝐴𝐴 circumscribed to the four triangles of a complete quadrilateral have a common point.

Theorem 71 (J. Steiner, 1827) The centers of the circles circumscribed to the four triangles of a complete quadrilateral and the Miquel’s point of this quadrilateral have five concyclic points (Miquel’s circle).

Proof Let , , , respectively be the centers of the circles circumscribed to the triangles , , , (see Figure 119). We denoted 𝑂𝑂 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 = , then = 2 . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐶𝐶1𝐴𝐴1 𝐶𝐶𝐵𝐵1𝐴𝐴1 Because , it follows that = 180 . On the other 𝑚𝑚�𝑀𝑀�𝐶𝐶1𝐴𝐴� 𝜑𝜑 𝑚𝑚�𝑀𝑀�𝑂𝑂1𝐴𝐴� 𝜑𝜑 hand, = and = 2 ; also 0 , leading to 𝑂𝑂𝑂𝑂1 ⊥ 𝐴𝐴𝐴𝐴 𝑚𝑚�𝑀𝑀�𝑂𝑂1𝑂𝑂� − 𝜑𝜑 = . From = 180 and = , it appears 𝑚𝑚�𝑀𝑀�𝐴𝐴1𝐵𝐵� 𝜑𝜑 𝑚𝑚�𝑀𝑀�𝑂𝑂2𝐵𝐵� 𝜑𝜑 𝑂𝑂2𝑂𝑂 ⊥ 𝑀𝑀𝑀𝑀 that the quadrilateral is inscribable0 if the points , , , are 𝑚𝑚�𝑀𝑀�𝑂𝑂2𝑂𝑂� 𝜑𝜑 𝑚𝑚�𝑀𝑀�𝑂𝑂1𝑂𝑂� − 𝜑𝜑 𝑚𝑚�𝑀𝑀�𝑂𝑂2𝑂𝑂� 𝜑𝜑 concyclic. 𝑀𝑀𝑂𝑂1𝑂𝑂𝑂𝑂2 𝑀𝑀 𝑂𝑂1 𝑂𝑂 𝑂𝑂2

238 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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A similar reasoning leads to = , and since = 180 , we have the points , , , . We note that the points , , , 𝑚𝑚�𝑀𝑀�𝑂𝑂3𝑂𝑂� 𝜑𝜑 𝑚𝑚�𝑀𝑀�𝑂𝑂1𝑂𝑂� , 0 are concyclic. − 𝜑𝜑 𝑀𝑀 𝑂𝑂1 𝑂𝑂 𝑂𝑂3 𝑀𝑀 𝑂𝑂 𝑂𝑂1 Their circle is called Miquel’s circle. 𝑂𝑂2 𝑂𝑂3 Remark 33 a) From Theorem 70, we can observe that the collinear points - - are the projections of the point (that belongs to the circle circumscribed to 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 the triangle ), under the angle of measurement and in the same 𝑀𝑀 direction (measured) on the sides of the triangle . Thus occurs the 𝐴𝐴𝐴𝐴𝐴𝐴 𝜑𝜑 generalized Simson’s theorem (of angle ). 𝐴𝐴𝐴𝐴𝐴𝐴

𝜑𝜑 The generalized Simson’s theorem has been proved by L. Carnot and it is thus stated: The projections of a point of the circle circumscribed under the same angle and direction on the sides of the triangle are collinear. In the case of Figure 119, we have: ( , ) = ( , ) = ( , ) = .

It can be1 shown that the1 Theorem 70 and1 the generalized Simson’s theorem are equivalent𝑚𝑚 𝑀𝑀𝐴𝐴. 𝐵𝐵𝐵𝐵 𝑚𝑚 𝑀𝑀𝐴𝐴 𝐶𝐶𝐶𝐶 𝑚𝑚 𝑀𝑀𝐶𝐶 𝐴𝐴𝐴𝐴 𝜑𝜑

b) The equivalence of the previous theorems leads to:

Theorem 72 The circles described on the chords , , of the circle circumscribed to the given triangle capable of the same 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 angle intersect two by two in the collinear points , , 𝐴𝐴𝐴𝐴𝐴𝐴 situated on the sides of the triangle . 1 1 1 𝜑𝜑 𝐴𝐴 𝐵𝐵 𝐶𝐶 This theorem for the right angle case is𝐴𝐴𝐴𝐴𝐴𝐴 owed to the Irish mathematician G. Salmon (1819-1904). In Theorem 70, the points - - that belong to the sides of the triangle are collinear; in the following, we show that we can define the Miquel’s point 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 even if , , are not collinear. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1

Ion Pătrașcu, Florentin Smarandache 239 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Theorem 73

If the points 1, 1, 1 repectively belong to the sides , and of the triangle , then the circles circumscribed to the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 triangles , and pass through the same point 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 (Miquel’s point). 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐶𝐶1𝐴𝐴1 𝐶𝐶𝐴𝐴1𝐵𝐵1 Proof We consider , , on the sides ( ), ( ), ( ) (see Figure 120). We denote by the second point of intersection of the circles circumscribed to 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 the triangles and . The quadrilateral and are 𝑀𝑀 inscribable, hence: 𝐴𝐴𝐶𝐶1𝐵𝐵1 𝐵𝐵𝐶𝐶1𝐴𝐴1 𝐴𝐴𝐶𝐶1𝑀𝑀𝐵𝐵1 𝐵𝐵𝐶𝐶1𝑀𝑀𝐴𝐴1 , (1) . (2) ∢𝑀𝑀𝐵𝐵1𝐴𝐴 ≡ ∢𝑀𝑀𝐶𝐶1𝐵𝐵 The preceding relations imply that: ∢𝑀𝑀𝐶𝐶1𝐵𝐵 ≡ ∢𝑀𝑀𝐴𝐴1𝐶𝐶 , (3) and this relations shows that the points , , , are situated on a circle. ∢𝑀𝑀𝐵𝐵1𝐴𝐴 ≡ ∢𝑀𝑀𝐴𝐴1𝐶𝐶 A 𝑀𝑀 𝐴𝐴1 𝐶𝐶 𝐵𝐵1

C1 B1 M

B A1 C

Figure 120

240 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Observation 78 a) The theorem can be proved in the same way even if one of the points , , are situated on a side, and the other two points are situated on the extensions of the other two sides. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 b) The triangle was called the Miquel’s triangle, corresponding to the Miquel’s point M, and the circles circumscribed to the triangles 𝐴𝐴1𝐵𝐵1𝐶𝐶1 , and were called the Miquel’s circles.

1 1 1 1 1 1 Proposition𝐴𝐴𝐶𝐶 𝐵𝐵79 𝐵𝐵𝐶𝐶 𝐴𝐴 𝐶𝐶𝐴𝐴 𝐵𝐵 If is a fixed point in the plane of the triangle , then we can build any Miquel’s triangles we want, corresponding to the point . 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 Indeed, we can build from the lines , , that create with 𝑀𝑀 , , the same angle measured in the same direction, or we can 𝑀𝑀 𝑀𝑀𝐴𝐴1 𝑀𝑀𝐵𝐵1 𝑀𝑀𝐶𝐶1 proceed in this way: 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 We take through and a certain circle that cuts and in and . We build the circumscribed circle of the triangle ; this will 𝑀𝑀 𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐶𝐶1 intersect the second time in the point . The triangle is a 𝐵𝐵1 𝐵𝐵𝐵𝐵𝐶𝐶1 Miquel’s triangle corresponding to the point . 𝐵𝐵𝐵𝐵 𝐴𝐴1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Theorem 71 𝑀𝑀 All the Miquel’s triangles corresponding to a given point in the plane of the triangle are directly-similar, and the point is 𝑀𝑀 the double point (their similarity center): 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 Proof Let be a Miquel’s point for the triangle and the Miquel’s triangle corresponding to this point M, supposedly fixed. Then we know the 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 angular coordinates of , therefore the angles , , (see Figure 121). It is not difficult to establish that: 𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵� 𝐶𝐶𝐶𝐶𝐶𝐶� 𝐴𝐴�𝐴𝐴𝐴𝐴 = ; = ; ∢𝐵𝐵1𝐴𝐴1𝐶𝐶1 ∢𝐵𝐵𝐵𝐵𝐵𝐵 − ∢𝐵𝐵𝐵𝐵𝐵𝐵 = . ∢𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∢𝐴𝐴𝐴𝐴𝐴𝐴 − ∢𝐴𝐴𝐴𝐴𝐴𝐴 As it can be seen, the measures of these angles are well determined when ∢𝐵𝐵1𝐶𝐶1𝐴𝐴1 ∢𝐴𝐴𝐴𝐴𝐴𝐴 − ∢𝐴𝐴𝐴𝐴𝐵𝐵 the point is fixed. If we will consider the triangle – Miquel’s triangle corresponding to the point , the angles , and will 𝑀𝑀 𝐴𝐴2𝐵𝐵2𝐶𝐶2 be given by the same formulas, hence ~ . 𝑀𝑀 ∢𝐵𝐵1𝐴𝐴2𝐶𝐶2 ∢𝐴𝐴2𝐵𝐵2𝐶𝐶2 ∢𝐵𝐵2𝐶𝐶2𝐴𝐴2 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 Ion Pătrașcu, Florentin Smarandache 241 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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Then, = 180 , , , consequently 0~ , which shows that the ∢𝐴𝐴1𝑀𝑀𝐵𝐵1 ≡ ∢𝐴𝐴2𝑀𝑀𝐵𝐵2 − 𝑚𝑚�𝐶𝐶̂� ∢𝑀𝑀𝐴𝐴1𝐵𝐵1 ≡ ∢𝑀𝑀𝑀𝑀𝑀𝑀 point is the similarity center of the two Miquel’s triangles. ∢𝑀𝑀𝐴𝐴2𝐵𝐵2 ≡ ∢𝑀𝑀𝑀𝑀𝑀𝑀 ∆𝑀𝑀𝐴𝐴1𝐵𝐵1 ∆𝑀𝑀𝐴𝐴2𝐵𝐵2 𝑀𝑀 A

C1 B1

C 2 M

B A1 A 2 C

Figure 121

8.3.2 Applications 1. Let be a Miquel’s point in the interior of the triangle and let be its corresponding Miquel’s triangle. We denote by , , 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 the intersections of semi-lines ( ,( , ( with the circumscribed 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 circle of the triangle . Prove that ~ . 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴2𝐵𝐵2𝐶𝐶2 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 242 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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B1 C2

C1 M

B A1 C

Figure 122 Solution = + (see Figure 122). But = and , and . We ∢𝐵𝐵2𝐴𝐴2𝐶𝐶2 ∢𝐵𝐵2𝐴𝐴2𝐴𝐴 ∢𝐴𝐴𝐴𝐴2𝐶𝐶2 ∢𝐵𝐵2𝐴𝐴2𝐴𝐴 ∢𝐵𝐵2𝐵𝐵𝐵𝐵 obtain that . Similarly, we show that ∢𝐵𝐵2𝐵𝐵𝐵𝐵 ≡ ∢𝑀𝑀𝑀𝑀1𝐶𝐶1 ∢𝐴𝐴𝐴𝐴2𝐶𝐶2 ≡ ∢𝐴𝐴𝐴𝐴𝐶𝐶2 ∢𝐴𝐴𝐴𝐴𝐶𝐶2 ≡ ∢𝑀𝑀𝑀𝑀1𝐵𝐵1 . ∢𝐵𝐵2𝐴𝐴2𝐶𝐶2 ≡ ∢𝐵𝐵1𝐴𝐴1𝐶𝐶1 ∢𝐴𝐴2𝐵𝐵2𝐶𝐶2 ≡

∢𝐴𝐴1𝐵𝐵1𝐶𝐶1 2. Let be the Miquel’s point corresponding to the Miquel’s triangle , with the vertices on the sides of the triangle . Three cevians, 𝑀𝑀 concurrent in the point from the interior of the triangle , intersect a 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 second time the Miquel’s circles in the points , , . Prove that the 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 points , , , are concyclic. 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝑀𝑀

Ion Pătrașcu, Florentin Smarandache 243 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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B A 2 2 B1 M P

B A1 C

Figure 123 Solution In Figure 123, let the intersection of cevian with the Miquel’s circle ( ). We denote = = = . We have = 𝐴𝐴2 𝐴𝐴𝐴𝐴 180 , = 180 = 180 . It follows that = 𝐴𝐴𝐵𝐵1𝐶𝐶1 𝑀𝑀�𝐴𝐴1𝐶𝐶 𝑀𝑀�𝐵𝐵1𝐴𝐴 𝑀𝑀�𝐶𝐶1𝐵𝐵 𝜑𝜑 ∢𝑀𝑀𝐴𝐴2𝑃𝑃 , therefore0 the points , 0 , , , are concyclic0 . − 𝜑𝜑 ∢𝑀𝑀𝐵𝐵2𝑃𝑃 − ∢𝐵𝐵𝐵𝐵2𝑀𝑀 − 𝜑𝜑 ∢𝑀𝑀𝐶𝐶2𝑃𝑃

𝜑𝜑 𝑀𝑀 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝑃𝑃 3. Three circles passing respectively through the vertices , , of the triangle intersect in a point in the interior of the triangle and a second 𝐴𝐴 𝐵𝐵 𝐶𝐶 time in the points , , belonging to the sides ( ), ( ), ( ). We take 𝐴𝐴𝐴𝐴𝐴𝐴 𝑆𝑆 through , , three parallels with a given direction that intersect the 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 second time each circle respectively in the points , and . Prove that the 𝐴𝐴 𝐵𝐵 𝐶𝐶 points , , are collinear. Van Khea𝑃𝑃 𝑄𝑄 – Peru,𝑅𝑅 Geometrico 2017 𝑃𝑃 𝑄𝑄 𝑅𝑅

244 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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E F Q

B D C

Figure 124 Solution The quadrilateral is inscribed, therefore: . (1) 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 From the parallelism of lines and , we note that: ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐹𝐹𝐹𝐹𝐹𝐹 . (2) 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 The quadrilateral is inscribed, hence: . (3) ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐹𝐹𝐹𝐹𝐹𝐹 The relations (1) – (3) lead to . (4) 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 ∢𝐹𝐹𝐹𝐹𝐹𝐹 ≡ ∢𝐹𝐹𝐹𝐹𝐹𝐹 This relation shows that the points , , are collinear. Because the ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝐹𝐹𝐹𝐹𝐹𝐹 quadrilateral is inscribed, we have that . (5) 𝑃𝑃 𝑄𝑄 𝑆𝑆 We denoted by the intersection of the line with the circle′ that passes 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝑄𝑄𝑄𝑄𝑄𝑄 ≡ ∢𝐷𝐷𝐷𝐷𝑅𝑅 through and (see ′Figure 124). 𝑅𝑅 𝑄𝑄𝑄𝑄 The quadrilateral is inscribed, hence . (6) 𝐶𝐶 𝑆𝑆 The relations (5) and (6)′ lead to . ′ ′ (7) 𝐷𝐷𝐷𝐷𝐷𝐷𝑅𝑅 ∢𝐷𝐷𝐷𝐷′𝑅𝑅 ≡ ∢𝐷𝐷𝐷𝐷𝑅𝑅 ∢𝑄𝑄𝑄𝑄𝑄𝑄 ≡ ∢𝐷𝐷𝐷𝐷𝑅𝑅 Ion Pătrașcu, Florentin Smarandache 245 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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On the other hand, is parallel with , therefore: . (8) 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 The relations (7) and (8) show that = . Having , , collinear and ∢𝑄𝑄𝑄𝑄𝑄𝑄 ≡ ∢𝐷𝐷𝐷𝐷𝐷𝐷 , , also collinear, it follows that , , ′ are collinear as well. 𝑅𝑅 𝑅𝑅 𝑃𝑃 𝑄𝑄 𝑆𝑆 𝑄𝑄 𝑆𝑆 𝑅𝑅 A 𝑃𝑃 𝑄𝑄 𝑅𝑅

E F B'

B D C

Figure 125 4. Let be a scalene triangle and the points ( ), ( ), ( ). The circles circumscribed to the triangles , and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷 ∈ 𝐵𝐵𝐵𝐵 𝐸𝐸 ∈ 𝐴𝐴𝐴𝐴 intersect in a point . An arbitrary line which passes through the point 𝐹𝐹 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶 intersects the second time the circles circumscribed to the triangles , 𝑃𝑃 𝑃𝑃 and respectively in the points , and . Show that the lines 𝐴𝐴𝐴𝐴𝐴𝐴 , , are parallel. ′ ′ ′ 𝐵𝐵𝐵𝐵𝐵𝐵′ ′ 𝐶𝐶𝐶𝐶𝐶𝐶′ Mihai Miculița – A𝐴𝐴 reciprocal𝐵𝐵 𝐶𝐶of a problem by Van Khea 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶

246 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

ANNEXES

Solution (Mihai Miculița) The quadrilateral is inscribed, then . (1) The quadrilateral ′ is inscribed, therefore: ′ ′ 𝐴𝐴𝐴𝐴 𝐹𝐹𝐹𝐹 ∢𝐹𝐹𝐹𝐹𝐴𝐴 ≡ ∢𝐹𝐹𝐹𝐹𝐴𝐴 . ′ (2) 𝐹𝐹 From relations′ (1)′𝐵𝐵𝐵𝐵 and𝐵𝐵 (2), we obtain that . This implies ∢𝐹𝐹𝐹𝐹𝐴𝐴 ≡ ∢𝐹𝐹𝐹𝐹𝐵𝐵 consequently that (see Figure 125). ′ ′ (3) ∢𝐹𝐹𝐹𝐹𝐴𝐴 ≡ ∢𝐹𝐹𝐹𝐹𝐵𝐵 The quadrilateral ′ ′ is inscribed, therefore: 𝐴𝐴𝐴𝐴 ∥ 𝐵𝐵𝐵𝐵 . ′ (4) 𝐵𝐵𝐵𝐵 The′ inscribed quadrilateral𝑃𝑃𝐵𝐵 leads to . (5) ∢𝐵𝐵 𝐵𝐵𝐵𝐵 ≡ ∢𝐷𝐷𝐷𝐷𝐷𝐷 The relations (4) and (5) imply that′ . ′ 𝐶𝐶 ∢𝐷𝐷𝐷𝐷𝐷𝐷 ≡ ∢𝐷𝐷𝐷𝐷𝐶𝐶 The consequence is that 𝑃𝑃𝑃𝑃𝐶𝐶 . ′ ′ (6) ∢𝐵𝐵 𝐵𝐵𝐵𝐵 ≡ ∢𝐷𝐷𝐷𝐷𝐶𝐶 Finally, the relations (3) and′ (6) lead′ to the requested conclusion: 𝐵𝐵𝐵𝐵 ∥ 𝐶𝐶𝐶𝐶 . ′ ′ ′ 𝐴𝐴𝐴𝐴 ∥ 𝐵𝐵𝐵𝐵 ∥ 𝐶𝐶𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 247 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

9.1 Proposed Problems

1. The triangles and are symmetrical to the line . Prove that and are orthological triangles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑑𝑑

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 2. In the triangle , denote by and the contacts of the inscribed circle (of center ) with , respectively . Let , , be the midpoints 𝐴𝐴𝐴𝐴𝐴𝐴 𝐸𝐸 𝐹𝐹 of segments , and respectively . Prove that the perpendiculars 𝐼𝐼 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝑀𝑀 𝑁𝑁 𝑃𝑃 taken from the points , and respectively to , and are 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 concurrent. 𝐼𝐼 𝐵𝐵 𝐶𝐶 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 Ion𝑀𝑀 Pătrașcu𝑀𝑀

3. Let , , be respectively the centers of the circles circumscribed to the triangles , , , where is a certain point in the interior 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 of the triangle . Prove that the triangles and are 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀 orthological, and specify the orthology centers. 1 2 3 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 𝑂𝑂 4. Let be a non-right triangle, – its orthocenter and – a point located on . The perpendiculars taken from to and to intersect 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝑃𝑃 and in , respectively . Prove that the lines and are 𝐴𝐴𝐴𝐴 𝐻𝐻 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 parallel. 1 1 1 1 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 Ion Pătrașcu𝐵𝐵𝐵𝐵

5. If: and ’ are points in the interior of the triangle ; ’ ’ ’ and ’’ ’’ ’’ are their pedal triangles; the set of the orthology centers of the 𝑃𝑃 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 triangles , ’ ’ ’ and with ’’ ’’ ’’ is formed only by the points 𝐴𝐴 𝐵𝐵 𝐶𝐶 , ’; – then show that the points and ’ are isogonal conjugate points. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 𝑃𝑃 𝑃𝑃 𝑃𝑃 Ion Pătrașcu, Florentin Smarandache 249 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

6. Let be a right triangle in , and its altitude, ( ). Denote by the midpoint of , and by the projection of on the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴𝐴𝐴 𝐷𝐷 ∈ 𝐵𝐵𝐵𝐵 mediator of the side . Let be the intersection of the semi-line ( with 𝐾𝐾 𝐴𝐴𝐴𝐴 𝑃𝑃 𝐾𝐾 the circumscribed circle of the triangle (whose center is the point ), 𝐵𝐵𝐵𝐵 𝑄𝑄 𝐴𝐴𝐴𝐴 and the symmetric of to . Prove that the Simson line of the point 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 is parallel with . 𝑆𝑆 𝑄𝑄 𝐵𝐵𝐵𝐵 Ion Pătrașcu𝑆𝑆 𝑂𝑂𝑂𝑂 7. Let be an equilateral triangle. Find the positions of the points , , on the sides , respectively , such that the lines , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 to be concurrent, and the perpendiculars raised on the sides respectively 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 in the points , , also to be concurrent. 𝐶𝐶𝐶𝐶1

𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 8. In the triangle of orthocenter , let be the diametral of in the circumscribed circle, – the projection of on , and , – 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝐴𝐴1 𝐴𝐴 analogous points. Let , be the intersections of the parallels taken 𝐴𝐴2 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵2 𝐶𝐶2 through and to with , respectively , and , , , – 𝐴𝐴𝑏𝑏 𝐴𝐴𝑐𝑐 analogous points. Show that the perpendiculars from , , respectively 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝑐𝑐 𝐵𝐵𝑎𝑎 𝐶𝐶𝑎𝑎 𝐶𝐶𝑏𝑏 to , , are concurrent in . 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 Nicolae Mihăileanu – The Correlative of a Proposition – 𝑏𝑏 𝑐𝑐 𝐶𝐶 𝑄𝑄 𝑎𝑎 𝑏𝑏 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐻𝐻 Victor Thébault, G.M. vol. 41, 1936

9. Let , , be the concurrent cevians in the point in the interior of the triangle and let be the orthology center of the triangle 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑃𝑃 in relation to the triangle . Denote { } = , { } = 𝐴𝐴𝐴𝐴𝐴𝐴 𝑄𝑄 , { } = ; denote by the′ orthology center of′ the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵1𝐶𝐶1 ∩ 𝐴𝐴𝐴𝐴1 𝐵𝐵 triangle in′ relation to the triangle . Prove that: 𝐶𝐶1𝐴𝐴1 ∩ 𝐵𝐵𝐵𝐵1 𝐶𝐶 𝐴𝐴1𝐵𝐵1 ∩ 𝐶𝐶𝐶𝐶1 𝑅𝑅 i) ′ The′ ′ points , , are collinear if and only if is the gravity 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 center of the triangle ; 𝑅𝑅 𝑃𝑃 𝑄𝑄 𝑃𝑃 ii) If is the gravity center of the triangle , then the points 𝐴𝐴1𝐵𝐵1𝐶𝐶1 , , , are collinear ( is the orthology center of the triangle 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 in relation to triangle ). 𝑅𝑅 𝑃𝑃 𝑄𝑄 𝑆𝑆 𝑆𝑆 ′ ′ ′ Vincențiu Pașol – Ion Pătrașcu 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 10. Let be a scalene triangle, – its orthocenter and – an arbitrary point on . Denote by and the midpoints of the sides 𝐴𝐴𝐴𝐴𝐴𝐴 ′ 𝐻𝐻 ′ 𝑃𝑃 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 250 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES and , and by the point of intersection of the perpendicular taken from to the line with the perpendicular taken from to the line . Show 𝐴𝐴𝐴𝐴 𝑄𝑄 that′ the point is found on the mediator of the side ′ . 𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶Cities Tour – Ru𝐵𝐵𝐵𝐵ssia, 2010 𝑄𝑄 𝐵𝐵𝐵𝐵 11. Let be a right triangle. Build the rectangle on the hypotenuse , in triangle’s exterior. Denote by the intersection of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 perpendicular taken from to with the perpendicular taken from to 𝐵𝐵𝐵𝐵 𝐼𝐼 . Prove that the triangles and are orthological. 𝐷𝐷 𝐴𝐴𝐴𝐴 Ion Pătrașcu𝐸𝐸 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼𝐼𝐼 12. Let be an acute triangle, – the center of its circumscribed circle and , , – the symmetrics of with respect to the sides , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 respectively . 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑂𝑂 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 a) Prove that the triangle and are bilogical; 𝐴𝐴𝐴𝐴 b) Prove that the homology center of the triangles and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 belongs to the Euler line of the triangle ; 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 c) If is the homology center of triangles and , calculate 𝐴𝐴𝐴𝐴𝐴𝐴 1, where is the orthocenter of the triangle 1. 1 1 𝑂𝑂1𝐻𝐻𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂1𝑂𝑂 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 13. Let be a right triangle in and – the center of its circumscribed circle. Build the points ’ respectively ’, such that ’ = 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝑂𝑂 ’ = , on the semi-lines ( and ( . On the semi-line ( , where 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 is the foot of the altitude from , build ’ such that ’ = . Prove 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴1 that the perpendiculars taken from , , , respectively ’ ’, ’ ’ and ’ ’ 𝐴𝐴1 𝐴𝐴 𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 are concurrent. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 14. In a certain triangle , let be parallel with , with ( ), ( ), and – the orthogonal projection of on . Prove 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ that the perpendicular taken from to , the perpendicular taken from 𝐴𝐴𝐴𝐴 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐴𝐴 𝐵𝐵𝐵𝐵 to , and are concurrent. 1 1 𝐵𝐵 𝐴𝐴 𝐶𝐶 Ion Pătrașcu 1 1 1 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐴𝐴𝐴𝐴 15. Let be an equilateral triangle and – a point in its plane. Denote by ’, ’, ’ the symmetrics of with respect to , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑀𝑀 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 251 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES respectively , and note that ’ = ’ = ’. Prove that the triangles , ’ ’ ’ are orthological and have a common orthology center. 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 16. The triangles and are orthological, and the orthology centers are respectively . Let be the translation of a triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 by vector translation . Prove that the triangles and 𝑂𝑂 𝑂𝑂1 𝐴𝐴′1𝐵𝐵′1𝐶𝐶′1 are reciprocal polar to a circle of center . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑂𝑂��1��𝑂𝑂�⃗ 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴′1𝐵𝐵′1𝐶𝐶′1 𝑂𝑂 17. Let be an acute triangle, – its orthocenter and ’ ’ ’ – its orthic triangle. Denote: { } = , { } = , { } = 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝐴𝐴 𝐵𝐵 𝐶𝐶 , { } = , { } = ′ ′ , { } = ′ ′ . Prove that′ the′ 𝑃𝑃 𝐵𝐵 𝐶𝐶 ∩ 𝐵𝐵𝐵𝐵 𝑄𝑄 𝐴𝐴 𝐵𝐵 ∩ 𝐴𝐴𝐴𝐴 𝑅𝑅 𝐴𝐴 𝐶𝐶 ∩ median triangle of the triangle and the triangle are 𝐴𝐴𝐴𝐴 𝑈𝑈 𝐴𝐴𝐴𝐴 ∩ 𝐶𝐶𝐶𝐶 𝑉𝑉 𝐵𝐵𝐵𝐵 ∩ 𝐶𝐶𝐶𝐶 𝑊𝑊 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 orthological. 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑀𝑀 𝑀𝑀 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 Ion𝑈𝑈𝑈𝑈𝑈𝑈 Pătrașcu

18. Let be an acute triangle and let be its orthic triangle. Denote by , respectively the projections of vertices , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 respectively on , and . Prove that the triangles and 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐴𝐴 𝐵𝐵 𝐶𝐶 are bilogical triangles. 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴2𝐵𝐵2𝐶𝐶2 19. Let be a scalene triangle. Denote by and the intersections of a circle taken through the points and with the sides ( ) respectively 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵1 𝐶𝐶1 ( ). Denote by , , the midpoints of the segments , 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 respectively . Prove that the triangles and are 𝐴𝐴𝐴𝐴 𝑀𝑀𝑎𝑎 𝑀𝑀𝑏𝑏 𝑀𝑀𝑐𝑐 𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶 orthological. 1 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐵𝐵𝐶𝐶 𝑀𝑀 𝑀𝑀 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 20. Let and be two homological triangles located in different planes, and let be their center of homology. Denote by , , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 the projections of the points , , on the plane ( ). Prove 𝑂𝑂 𝐴𝐴0 𝐵𝐵0 that, if the two triangles and are orthological, then the 𝐶𝐶0 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 triangles and 1 are1 1 bilogical2 triangles2 2 . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 Ion Pătrașcu 0 0 0 2 2 2 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 21. Let be a triangle and ( ), ( ), ( ). Prove that: ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵 ∈ 𝐴𝐴𝐴𝐴 𝐶𝐶 ∈ 𝐴𝐴𝐴𝐴

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PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

a) The perpendiculars from , , respectively to , , are concurrent if and only if: ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 + + = ( + + ). 1 b) In the case′ of a), the′ following inequality′ 2takes place2 : 2 𝐵𝐵𝐵𝐵 ∙ 𝐵𝐵𝐴𝐴 𝐶𝐶𝐶𝐶 ∙ 𝐶𝐶𝐵𝐵 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐶𝐶 2 𝐴𝐴𝐵𝐵 𝐵𝐵𝐶𝐶 𝐶𝐶𝐴𝐴 + + ( + + ). 2 2 2 1 a) If the′ points′ , , ′ are mobile2 and 2the hypothesis2 from a) is true, 𝐵𝐵𝐴𝐴 𝐶𝐶𝐵𝐵 𝐴𝐴𝐶𝐶 ≥ 4 𝐴𝐴𝐵𝐵 𝐵𝐵𝐶𝐶 𝐶𝐶𝐴𝐴 the sum ′ +′ ′ + is minimal if and only if the 𝐴𝐴 𝐵𝐵 𝐶𝐶 concurrency point′2 of ′the2 three′2 perpendiculars is the center of the 𝐵𝐵𝐴𝐴 𝐶𝐶𝐵𝐵 𝐴𝐴𝐶𝐶 circle circumscribed to the triangle . Ovidiu Pop, professor, Satu Mare (The annual𝐴𝐴𝐴𝐴𝐴𝐴 contest of G.M. resolvers, 1990)

22. Let the triangle and the points , ( ), , ( ), , ( ) be such that: = ; = ; = . Prove 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐴𝐴2 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵1 𝐵𝐵2 ∈ 𝐶𝐶𝐶𝐶 that the orthology center of the triangle in relation to the triangle 𝐶𝐶1 𝐶𝐶2 ∈ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐴𝐴1 𝐶𝐶𝐴𝐴2 𝐶𝐶𝐵𝐵1 𝐴𝐴𝐵𝐵2 𝐴𝐴𝐶𝐶1 𝐵𝐵𝐶𝐶2 determined by the intersections of the lines of the centers of the circles 𝐴𝐴𝐴𝐴𝐴𝐴 circumscribed to the triangles: and ; and ; and – is the gravity center of the1 triangle 2 . 1 2 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵Ion𝐵𝐵 Pătrașcu𝐶𝐶𝐶𝐶𝐶𝐶 2 𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 23. Let be a convex quadrilateral and , , the orthocenters of the triangles ; and . Prove that the perpendiculars taken 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 from , and respectively to , and are concurrent. 𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 24. Show that, if and are two orthological equilateral triangles, and the triangle is inscribed in the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ( ), ( ), ( ) , then is the median triangle of 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 the triangle . �𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ 𝐶𝐶𝐶𝐶 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 � 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Ion Pătrașcu 𝐴𝐴𝐴𝐴𝐴𝐴 25. Let be the pedal triangle of the center of the inscribed circle in the triangle′ ′ ′ . Prove that the triangles and are 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthological if and only if the triangle is isosceles. ′ ′ ′ 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴Reformulation of the problem O: 695, 𝐴𝐴𝐴𝐴𝐴𝐴G.M. nr. 710𝐴𝐴-11𝐵𝐵-12/1992.𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 Autor: M. Bârsan, Iași

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26. Let be an equilateral triangle of side , and – a point in its interior. Denote by , , the projections of on the sides. Show that: 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 𝑀𝑀 a) + +1 1 =1 . 𝐴𝐴 𝐵𝐵 𝐶𝐶 3 𝑀𝑀 b) The lines , , are concurrent if and only if is located 𝐴𝐴1𝐵𝐵 𝐵𝐵1𝐶𝐶 𝐶𝐶1𝐴𝐴 2 𝑎𝑎 on one of the1 altitude1 s of1 the triangles. 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 Laurențiu Panaitopol, County Olympics𝑀𝑀 1990.

27. In the triangle , denote by , , the feet of symmedians from , , . Prove that the perpendiculars in the points , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 respectively to the lines , , are concurrent if and only if the triangle 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 is isosceles. F. Enescu,𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 student𝐴𝐴𝐴𝐴 , Bucharest – Problem C. 1125, G.M. 5/1991. The annual contest of the resolvers, grades 7-8

28. Let be an equilateral triangle and – a triangle inscribed in , such that + + = 0. Show that: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 a) The triangles and are orthological. 𝐴𝐴𝐴𝐴𝐴𝐴 �𝐴𝐴��𝐴𝐴��1⃗ ∙ �𝐵𝐵𝐵𝐵��⃗ �𝐵𝐵��𝐵𝐵��1⃗ ∙ �𝐶𝐶𝐶𝐶��⃗ 𝐶𝐶��𝐶𝐶��1⃗ ∙ 𝐴𝐴𝐴𝐴��⃗ b) If is the orthology center of the triangle in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and is the center of the circle circumscribed to , then: + 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 + = . ��1⃗ 𝑂𝑂 3 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝐴𝐴

𝑃𝑃��𝐵𝐵��1⃗ 𝑃𝑃��𝐶𝐶��1⃗ 2 𝑃𝑃��𝑃𝑃�⃗ 29. Let be a given equilateral triangle and – a point in its interior. Show that there is an infinity of equilateral triangles – orthological 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 with , and having the point as orthology center.′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 Ion Pătrașcu 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 30. Two triangles , are homological (the lines , , meet in a point ). The perpendiculars′ ′ ′ in to , meet ′ ,′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 respectively in , . Similarly, the perpendiculars taken in ′and′ ′on′ 𝐼𝐼 𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐶𝐶 the lines ( , ), ( , ) determine on ( , ), ( , ) the 𝐴𝐴𝑐𝑐 𝐴𝐴𝑏𝑏 𝐵𝐵 𝐶𝐶 points , , , . Show that the perpendiculars′ ′ descending′ ′ ′ from′ ′ ′, , 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐴𝐴 on the lines𝑐𝑐 𝑎𝑎 𝑎𝑎 , 𝑏𝑏 , are concurrent. 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 Gh.𝐴𝐴 Țițeica𝐵𝐵 𝐶𝐶 𝑏𝑏 𝑐𝑐 𝑐𝑐 𝑎𝑎 𝑎𝑎 𝑏𝑏 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶

254 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

31. Let be a right triangle in and the foot of the perpendicular from . Denote by the midpoint of , and – { } = 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐴𝐴1 𝐷𝐷 , { } = . Also, { } = and , – 𝐴𝐴1 𝐾𝐾 𝐴𝐴1𝐷𝐷 𝑁𝑁 the projections of on respectively . Prove that and 𝐴𝐴2𝐾𝐾 ∩ 𝐴𝐴1𝐴𝐴3 𝑀𝑀 𝐴𝐴3𝐾𝐾 ∩ 𝐴𝐴1𝐴𝐴2 𝐵𝐵1 𝑀𝑀𝑀𝑀 ∩ 𝐴𝐴2𝐴𝐴3 𝐵𝐵2 𝐵𝐵3 are triorthological1 .1 3 1 2 1 2 3 𝐵𝐵 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴Ion𝐴𝐴 Pătrașcu𝐴𝐴 1 2 3 𝐵𝐵 𝐵𝐵 𝐵𝐵 32. Denote by , , the orthogonal projections of the point from the interior of triangle′ ′ ′ on the sides , respectively . The 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 circumscribed circle of the triangle intersect the second time the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 sides , , in the points , ′ ′, respectively′ . Prove that the 𝐴𝐴 𝐵𝐵 𝐶𝐶 perpendiculars taken from , and respectively to , and 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 are concurrent. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1

33. Take the triangle ; the equilateral triangles , , are built on its sides, in the exterior. Let , , be the midpoints of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐶𝐶1 𝐵𝐵𝐵𝐵𝐴𝐴1 𝐶𝐶𝐶𝐶𝐵𝐵1 segments ; ; . Show that the perpendiculars taken from , , 𝛼𝛼 𝛽𝛽 𝛾𝛾 on the sides1 1 1, 1 respectively1 1 are concurrent. 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 Dan Voiculescu𝛼𝛼 𝛽𝛽 𝛾𝛾 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 34. Let be a scalene triangle inscribed in a circle of center . The parallels taken through , , with , respectively intersect the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 circle the second time in the points , , . Prove that the perpendiculars 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 taken from , , to the sides ′, ′, ′ are concurrent. 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 35. Let be a scalene triangle and , , be the feet of its altitudes. Denote by , , the feet of the altitudes of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 . Show that the circles circumscribed to the triangles , , 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 1 1 1still have a common point. 1 2 1 2 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴Ion 𝐴𝐴Pătrașcu𝐵𝐵 𝐵𝐵 1 2 𝐴𝐴𝐶𝐶 𝐶𝐶 36. Let be a triangle and ( ), ( ), ( ) such that , and . Show that, if the Lemoine point of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 ∈ 𝐴𝐴𝐴𝐴 𝑁𝑁 ∈ 𝐴𝐴𝐴𝐴 𝑃𝑃 ∈ 𝐵𝐵𝐵𝐵 triangle coincides with the gravity center of the triangle , then 𝑀𝑀𝑀𝑀 ⊥ 𝐴𝐴𝐴𝐴 𝑁𝑁𝑁𝑁 ⊥ 𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 ⊥ 𝐵𝐵𝐵𝐵 the triangle is equilateral. 𝐴𝐴𝐴𝐴𝐴𝐴 Ciprian Manolescu, Problem O: 830, G.M.𝑀𝑀 no𝑀𝑀𝑀𝑀. 10/1996. 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 255 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

37. Let be a rectangle of center . Denote by and the intersections of mediator of the diagonal with and . Let , , 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝐸𝐸 𝐹𝐹 be respectively the midpoints of the sides , , and let – the 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑀𝑀 𝑁𝑁 𝑃𝑃 intersection with of the perpendicular taken from to . Prove that 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐷𝐷𝐷𝐷 𝐿𝐿 the triangles and are orthological. 𝐴𝐴𝐴𝐴 𝐷𝐷 𝑃𝑃𝑃𝑃 Ion Pătrașcu 𝐷𝐷𝐷𝐷𝐷𝐷 𝑁𝑁𝑁𝑁𝑁𝑁 38. Let be a quadrilateral inscribed in the circle of diameter . It is known that there exist the point on ( ) and the point on ( ) 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 such that the line is perpendicular to and the line is perpendicular 𝐸𝐸 𝐶𝐶𝐶𝐶 𝐹𝐹 𝐵𝐵𝐵𝐵 to . Show that = . 𝐴𝐴𝐴𝐴Problem no. 4, National𝐷𝐷 Mathematical𝐷𝐷 Olympiad𝐴𝐴𝐴𝐴 , grade 9, 2014 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 39. Let be a right isosceles triangle, = . Denote by the midpoint of . The point is defined by 4 = , , such that 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝑀𝑀 is collinear with , and is the midpoint of . Prove that the 𝐴𝐴𝐴𝐴 𝑄𝑄 ∙ 𝐴𝐴��𝐴𝐴�⃗ �𝐴𝐴𝐴𝐴��⃗ 𝑅𝑅 ∈ 𝐵𝐵𝐵𝐵 triangles and are orthological and specify the orthology centers. ��⃗ ��⃗ 𝑄𝑄𝑄𝑄 𝐴𝐴𝐴𝐴 𝑃𝑃 𝐶𝐶𝐶𝐶 Ion Pătrașcu 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 40. Let , , be concurrent cevians in triangle , ( ), ( ), ( ), such that is median and the triangles 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 ∈ and are orthological. Prove that the triangle is isosceles 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ 𝐴𝐴𝐴𝐴 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴1 or that 1 and1 1 are medians. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Florentin𝐴𝐴𝐴𝐴𝐴𝐴 Smarandache 1 1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 41. Let be an acute triangle; prove that there exists a triangle inscribed in , with ( ), ( ), ( ), such 𝐴𝐴𝐴𝐴𝐴𝐴 that , , . Denote by , , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐵𝐵1 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 respectively the midpoints of segments , , . Prove that the 𝐴𝐴1𝐵𝐵1 ⊥ 𝐵𝐵𝐵𝐵 𝐶𝐶1𝐵𝐵1 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐶𝐶1 ⊥ 𝐴𝐴𝐴𝐴 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 triangles and are orthological1. 1 1 𝐵𝐵𝐴𝐴 𝐶𝐶𝐵𝐵 𝐴𝐴𝐶𝐶 Ion Pătrașcu 1 1 1 1 2 3 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝑂𝑂 𝑂𝑂 𝑂𝑂 42. Let ( ), ( ), ( ) be three circles having their centers in noncollinear points, exterior two by two. Denote by the point located on 𝒞𝒞 𝑂𝑂1 𝒞𝒞 𝑂𝑂2 𝒞𝒞 𝑂𝑂3 which has equal powers over the circles ( ) and ( ); by the 𝐴𝐴 point located on which has equal powers over the circles ( ) and 𝑂𝑂2𝑂𝑂3 𝒞𝒞 𝑂𝑂2 𝒞𝒞 𝑂𝑂3 𝐵𝐵 𝑂𝑂3𝑂𝑂1 𝒞𝒞 𝑂𝑂3 256 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

( ); and by the point belonging to the line which has equal powers over the circles ( ) and ( ). Show that the triangles and 𝒞𝒞 𝑂𝑂1 𝐶𝐶 𝑂𝑂1𝑂𝑂2 are orthological. Specify the orthology centers. 𝒞𝒞 𝑂𝑂1 𝒞𝒞 𝑂𝑂2 𝐴𝐴𝐴𝐴𝐴𝐴

𝑂𝑂1𝑂𝑂2𝑂𝑂3 43. Let be a scalene triangle and – its contact triangle. Denote by , , respectively the orthocenters of triangles , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 , . Prove that the triangles and are 𝐻𝐻𝑎𝑎 𝐻𝐻𝑏𝑏 𝐻𝐻𝑐𝑐 𝐴𝐴𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 orthological𝑎𝑎 𝑐𝑐 .𝑎𝑎 Specify𝑏𝑏 their orthology centers. 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐵𝐵𝐶𝐶 𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶 𝐻𝐻 𝐻𝐻 𝐻𝐻 𝐶𝐶 Ion𝐶𝐶 𝐶𝐶Pătrașcu

44. Show that the triangle determined by the points of tangent with the sides of the triangle of its -ex-inscribed circle, and the triangle are orthological, and they have the same orthology center. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 45. The triangles and are orthological; is inscribed in , ( ), ( ), ( ), , . Prove 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 that is the complementary (median) triangle of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵1 ∈ 𝐶𝐶𝐶𝐶 𝐶𝐶1 ∈ 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝐵𝐵1𝐴𝐴1 ∥ 𝐴𝐴𝐴𝐴

𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 46. Let be a non-right triangle, and the center of its circumscribed circle. The mediators of segments , and determine the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 ( , belong to the mediator of the segment ). Prove that the 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 triangles and are orthological, and have a common orthology 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴𝐴𝐴 center. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

47. Let , , be three cevians, concurrent in the point inside the triangle ′ . B′uild the′ mediators of segments , , , and denote 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑃𝑃 the triangle determined by them ( and belong to mediator of 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 the segment ). Show that is orthological in relation to the triangle 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1 𝐶𝐶1 , and specify their orthology centers. 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 48. The triangles and are orthological of center ( is the orthology center of the triangle in relation to ). Let 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑀𝑀 𝑀𝑀 be the contact triangle of the triangle . Denote by , ′ , ′ ′ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 respectively the intersection points with the circle inscribed′ of′ the′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 perpendiculars taken from , , to , respectively . Denote by the intersection between the tangent1 1 taken1 1 in to the inscribed1 1 circle 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 ′ 𝐴𝐴 𝐵𝐵 𝑋𝑋 𝐴𝐴1 Ion Pătrașcu, Florentin Smarandache 257 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES with the parallel taken through (the center of the inscribed circle) with ; let be the intersection between the tangent taken through to the 𝐼𝐼 inscribed circle with the parallel taken through with ; and let ′ be the 𝐵𝐵1𝐶𝐶1 𝑌𝑌 𝐵𝐵1 intersection of the tangent taken through to the inscribed circle with the 𝐼𝐼 𝐶𝐶1𝐴𝐴1 𝑍𝑍 parallel taken through with . Prove′ that the points , , are 𝐶𝐶1 collinear. 1 1 𝐼𝐼 𝐴𝐴 𝐵𝐵 𝑋𝑋 Ion𝑌𝑌 Pătrașcu𝑍𝑍

49. Let be a scalene triangle, and – the midpoint of the side . The perpendiculars taken from to and intersect respectively in 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝐵𝐵𝐵𝐵 and the perpendiculars raised in and to . The points and 𝑀𝑀 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝑃𝑃 are symmetric with respect to , and′ are located′ on . Denote:′ { } =′ 𝑄𝑄 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐶𝐶 and { } = . Prove that the triangles and are 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑅𝑅 orthological. 𝐴𝐴𝐴𝐴 ∩ 𝑃𝑃𝑃𝑃 𝑆𝑆 𝐴𝐴𝐴𝐴 ∩ 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 50. Let be a right trapeze, = = 90 . Consider the point on the side and let – the midpoint of the side 0 . Build , 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴̂ 𝐵𝐵� 𝐸𝐸 and , . Denote by and the intersections of the line 𝐶𝐶𝐶𝐶 𝑀𝑀 𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 ⊥ 𝐴𝐴𝐴𝐴 𝑃𝑃 ∈ with respectively . Prove that the triangle and are 𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 ⊥ 𝐵𝐵𝐵𝐵 𝑄𝑄 ∈ 𝐵𝐵𝐵𝐵 𝑅𝑅 𝑆𝑆 orthological. 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐸𝐸𝐸𝐸𝐸𝐸 Ion𝐸𝐸 𝐸𝐸𝐸𝐸Pătrașcu

51. Let be a triangle, – a point in its interior, and – the isogonal conjugate of . If the -circumpedal triangle of the triangle and the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑈𝑈 𝑉𝑉 triangle are orthological, then the -circumpedal triangle of the 𝑈𝑈 𝑈𝑈 𝐴𝐴𝐴𝐴𝐴𝐴 triangle and the triangle are also orthological. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑉𝑉

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 52. Let be a scalene triangle. The following squares are built in the triangle’s exterior, on its sides: , and . Denote: { } = 𝐴𝐴𝐴𝐴𝐴𝐴 , { } = and { } = . Prove that the triangles 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 and are orthological. What important point in the triangle 𝑀𝑀𝑀𝑀 ∩ 𝑁𝑁𝑁𝑁 𝐵𝐵1 𝑀𝑀𝑀𝑀 ∩ 𝑆𝑆𝑆𝑆 𝐶𝐶1 𝑁𝑁𝑁𝑁 ∩ 𝑆𝑆𝑆𝑆 is its orthology center in relation to the triangle ? 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴1𝐵𝐵1𝐶𝐶1 53. If the segments , , that join the vertices of the orthological triangles and′ ′ are′ divided by the points , , 𝐴𝐴𝐴𝐴 𝐵𝐵′ 𝐵𝐵′ ′ 𝐶𝐶𝐶𝐶 ′′ ′′ ′′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶

258 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES and , , in proportional segments, then the triangles and ′′′ ′′′ are also′′′ orthological. ′′ ′′ ′′ ′′′ 𝐴𝐴′′′ ′′′𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵J. Neuberg𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 54. Let be the orthocenter of an acute triangle . Consider the points , , , with , and , on the semi- 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 lines ( ′ , (′ ,′ ( . Prove′ that: ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐻𝐻𝐻𝐻 ≡ 𝐵𝐵𝐵𝐵 𝐻𝐻𝐵𝐵 ≡ 𝐶𝐶𝐶𝐶 𝐻𝐻𝐶𝐶 ≡ 𝐴𝐴𝐴𝐴 a) The orthology center of the triangle in relation to the triangle 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 𝐻𝐻𝐻𝐻 is the gravity center of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 b) If′ { ′ }′ = , { } = , then . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 ′′ ′ ′ ′′ ′ ′ ′′ ′′ 𝐴𝐴 𝐵𝐵𝐵𝐵 ∩ 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶𝐶𝐶 ∩ 𝐶𝐶 𝐴𝐴 𝐻𝐻𝐻𝐻 ⊥ 𝐴𝐴 𝐵𝐵 55. Build the squares , and in the exterior of the triangle . Let – the intersection of lines and , – the 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 intersection of lines and , and – the intersection of lines and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶1 𝑆𝑆𝑆𝑆 𝑅𝑅𝑅𝑅 𝐵𝐵1 . Prove that the perpendiculars from , and to the lines , 𝑆𝑆𝑆𝑆 𝑀𝑀𝑀𝑀 𝐴𝐴1 𝑀𝑀𝑀𝑀 respectively are concurrent. 1 1 1 𝑅𝑅𝑅𝑅 Petru Braica, Satu𝐵𝐵 Mare,𝐴𝐴 Problem𝐶𝐶 27089, G.M. nr.𝐴𝐴𝐴𝐴 1/2015𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 56. Let be a scalene triangle. Build with , , as centers three congruent circles that cut the sides and in , ; the sides and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 in , ; the sides , in , . Denote:′ ′′ = { }, 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐵𝐵𝐵𝐵 ′ ′′ = { } and ′ ′′= { }. Prove′ ′′that the′ ′′ triangles 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐶𝐶 𝐶𝐶 𝐵𝐵 𝐵𝐵 ∩ 𝐶𝐶 𝐶𝐶 𝐴𝐴1 ′ ′′ and ′ ′′ are orthological′ ′′ , and′ s′′pecify their orthology centers. 𝐴𝐴 𝐴𝐴 ∩ 𝐶𝐶 𝐶𝐶 𝐵𝐵1 𝐴𝐴 𝐴𝐴 ∩ 𝐵𝐵 𝐵𝐵 𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 57. Let be a convex hexagon with: = , = , = . Show that the perpendiculars taken from , and respectively 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐷𝐷𝐷𝐷 to the lines , and are concurrent. 𝐸𝐸𝐸𝐸 𝐹𝐹𝐹𝐹 𝐴𝐴 𝐶𝐶 𝐸𝐸

𝐹𝐹𝐹𝐹 𝐵𝐵𝐵𝐵 𝐷𝐷𝐷𝐷 58. Let , and be the feet of altitudes of a triangle . Consider the points ′ ′ , ′ , and denote = { }, 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 = { }. If the points′ and′ are fixe′ d, and mobile, it is required to: 𝑀𝑀 ∈ 𝐴𝐴𝐴𝐴 𝑁𝑁 ∈ 𝐵𝐵𝐵𝐵 𝑃𝑃 ∈ 𝐶𝐶𝐶𝐶 𝑀𝑀𝑀𝑀 ∩ 𝐴𝐴𝐴𝐴 𝐾𝐾 𝑀𝑀𝑀𝑀 ∩ a) Prove that rotates around a fixed point. 𝐴𝐴𝐴𝐴 𝐿𝐿 𝑁𝑁 𝑃𝑃 𝑀𝑀 b) Find the geometric place of the intersections of the perpendiculars 𝑀𝑀𝑀𝑀 descending from and respectively to and . V. Sergiescu, student, Bucharest, Problem 8794, G.M. nr. 1/1969 𝐵𝐵 𝐶𝐶 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀

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59. Let be a scalene triangle, and – its cevian. Denote by , and the projections of vertex on and the projections of and ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 on′ . Prove′ that the triangle is orthological in relation to the triangle 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐶𝐶 . ′ 𝐴𝐴′𝐴𝐴′ 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu 𝐴𝐴 𝐵𝐵 𝐶𝐶 60. Let be an acute triangle, – its orthocenter and – a point on . The perpendiculars taken from to and to intersect and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝑃𝑃 in , respectively . Prove that the lines and are parallel. 𝐴𝐴𝐴𝐴 𝐻𝐻 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴Ion Pătrașcu𝐴𝐴𝐴𝐴 1 1 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 61. Let be a given triangle and – the triangle formed by the intersections of parallels taken through , , to the interior bisectors of 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the triangle (the parallel taken through to the bisector intersects 𝐴𝐴 𝐵𝐵 𝐶𝐶 with the parallel taken from to the bisector in , …). Prove′ that the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵𝐵𝐵 triangle is orthological in relation to Fuhrmann′ triangle of the 𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶1 triangle 1 1. 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 Ion Pătrașcu 𝐴𝐴𝐴𝐴𝐴𝐴 62. Let be the midpoint of the side of the triangle , and ( ), ( ) – two points such that . Show that the 𝐾𝐾 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐿𝐿 ∈ perpendiculars raised in the points , and respectively to , and 𝐴𝐴𝐴𝐴 𝑀𝑀 ∈ 𝐵𝐵𝐵𝐵 ∢𝐶𝐶𝐶𝐶𝐶𝐶 ≡ ∢𝐶𝐶𝐶𝐶𝐶𝐶 are concurrent in a point . Middle𝐾𝐾 𝐿𝐿 European𝑀𝑀 MO: 2012 Team𝐴𝐴𝐴𝐴 Competition𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑃𝑃 63. Let be a given triangle and – the podal triangle of center of the -ex-inscribed circle to the triangle. Denote by the intersections 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 of cevians , , , and – { } = , { } = . 𝐼𝐼𝑎𝑎 𝐴𝐴 𝛤𝛤𝑎𝑎 Prove that . 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑋𝑋 𝐵𝐵1𝐶𝐶1 ∩ 𝐵𝐵𝐵𝐵 𝑌𝑌 𝐴𝐴1𝐶𝐶1 ∩ 𝐴𝐴𝐴𝐴

𝐼𝐼𝑎𝑎𝛤𝛤𝑎𝑎 ⊥ 𝑋𝑋𝑋𝑋 64. Let be an acute triangle inscribed in the circle of center . Denote by , , respectively the midpoints of the circle’s high arcs, 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 supported by the chords , and . Prove that the triangles and 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 are orthological. 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴1𝐵𝐵1𝐶𝐶1

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65. In a certain triangle , let , , be the projections of centers of ex-inscribed circles, , , respectively on mediators of the sides , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 and . Prove that the triangles and are bilogical triangles. 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 𝐼𝐼𝑐𝑐 𝐵𝐵𝐵𝐵

𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 66. Let be a scalene triangle and – the midpoint of the side . Build congruent circles which pass through and , and through and , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐾𝐾 𝐴𝐴𝐴𝐴 and which have the centers on the same section of the line as the point 𝐴𝐴 𝐾𝐾 𝐵𝐵 𝐾𝐾 . These circles cut the second time the sides and in respectively 𝐴𝐴𝐴𝐴 . Prove that the triangles and are orthological. Determine the 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐿𝐿 geometric place of the orthology center of these triangles. 𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu, Mihai Miculița 𝑃𝑃 67. Let be a rhombus of center . Denote by the projection of on and by the symmetric of with respect to the midpoint of the 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝐸𝐸 segment . The perpendicular taken from to intersects the 𝑂𝑂 𝐴𝐴𝐴𝐴 𝐹𝐹 𝑂𝑂 perpendicular taken from to in . Prove that . 𝐴𝐴𝐴𝐴 Ion Pătrașcu, Problem S:𝐹𝐹 L.17.299𝐴𝐴𝐴𝐴 – G.M. nr. 11/2017 𝐷𝐷 𝐸𝐸𝐸𝐸 𝐻𝐻 𝐴𝐴𝐴𝐴 ⊥ 𝐶𝐶𝐶𝐶 68. Let be a rectangle with = 2 and = 3. Denote by the midpoint of the side , by – the midpoint of the segment , and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑀𝑀 by – the symmetric of with respect to . Denote the intersection√ of the 𝐴𝐴𝐴𝐴 𝑃𝑃 𝐷𝐷𝐷𝐷 lines and by . be the intersection of perpendicular from to 𝑆𝑆 𝑃𝑃 𝐴𝐴𝐴𝐴 with the parallel taken through with . Let be the intersection of 𝐴𝐴𝐴𝐴 𝐷𝐷𝐷𝐷 𝑇𝑇 𝑉𝑉 𝐷𝐷 𝐷𝐷𝐷𝐷 bisector of the angle with the perpendicular taken from to . Prove 𝑃𝑃 𝐴𝐴𝐴𝐴 𝑄𝑄 that the points , , are collinear. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷 I𝐶𝐶𝐶𝐶on Pătrașcu 𝑃𝑃 𝑄𝑄 𝑇𝑇 69. Let and be two orthological triangles of center . Denote by , and the symmetrics of with respect to the midpoints 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 of the sides of the triangle . Prove that the triangle is 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝑃𝑃 orthological with the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 2 2 2 70. The following are required𝐴𝐴 𝐵𝐵: 𝐶𝐶 a) Find the condition which an acute triangle must satisfy in order that the orthic triangle of its orthic triangle does not exist;

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b) Find a triangle such that it does not exist an orthic triangle of the orthic triangle of the orthic triangle of the given triangle; c) Let be a triangle, – its othic triangle, and – the orthic triangle of the triangle′ ′ ′ . What can you say′ about′ ′ the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 relation of orthology relative to′′ the′′ triangles′′ and ? 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′′ ′′ ′′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 71. Let be an acute triangle. Denote by the projection of on , and by – the projection of on , and by , , respectively the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷 𝐵𝐵 midpoints of the segments , and . Prove that: 𝐴𝐴𝐴𝐴 𝐸𝐸 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐾𝐾 𝐿𝐿 𝑀𝑀 a) The triangles and are orthological; 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐷𝐷𝐷𝐷 b) The axis of orthology is perpendicular to . 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴

𝐾𝐾𝐾𝐾 72. Let be a scalene triangle and – its -circumpedal triangle ( – the center of the circle inscribed in the′ triangle′ ′ ). Prove that: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐼𝐼 a) The triangle is orthological in relation to the triangle ; 𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 b) The circles ( ; ), ( ; ), ( ; ) intersect′ in′ the′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthology center ′of the′ triangle′ from′ the point′ a)′ ; 𝒞𝒞 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝒞𝒞 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝒞𝒞 𝐶𝐶 𝐶𝐶 𝐴𝐴 c) Let – { } = , { } = , and – the center of the circle circumscribed′ ′ to the triangle′ ′ABC; then: . 𝑋𝑋 𝐵𝐵 𝐶𝐶 ∩ 𝐵𝐵𝐵𝐵 𝑌𝑌 𝐴𝐴 𝐵𝐵 ∩ 𝐴𝐴𝐴𝐴 𝑂𝑂

𝑂𝑂𝑂𝑂 ⊥ 𝑋𝑋𝑋𝑋 73. Let be an acute triangle with < , and of orthocenter . The mediator of the side intersects the sides , and respectively 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐻𝐻 on the points , and . Denote by the midpoint of the segment . 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 Prove that the triangles and are orthogonal. 𝑀𝑀 𝑄𝑄 𝑃𝑃 𝑁𝑁 𝑃𝑃𝑃𝑃

𝐵𝐵𝐵𝐵𝐵𝐵 𝑄𝑄𝑄𝑄𝑄𝑄 74. The circle intersect the sides ( ), ( ) and ( ) of the triangle in ; , , . Prove that, if the triangles and are 𝜔𝜔 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 orthological1 ,2 then1 the2 triangle1 2 s and are also1 orthological1 1 . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴Reformulation𝐵𝐵 𝐵𝐵 𝐶𝐶 of𝐶𝐶 a problem proposed at the Hungarian𝐴𝐴 𝐵𝐵 Competition𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴, 1914. 2 2 2 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 75. Let and be two triangles located in distinct planes such that the perpendiculars taken from , , respectively to , and 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 are concurrent in a point . Prove that the triangle (the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 projection of on the plane ) and the triangle ′ ′ ′ are 𝐴𝐴1𝐵𝐵1 𝐻𝐻 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthological. 1 1 1 1 1 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴Ion𝐵𝐵 Pătrașcu𝐶𝐶

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PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

76. Let be an isosceles triangle, = ; is its orthocenter, and is its orthic triangle. Show that the triangle is orthological in 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐻𝐻 relation to . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐻𝐻𝐻𝐻𝐻𝐻

𝐴𝐴1𝐵𝐵1𝐶𝐶1 77. Let be the center of the circle circumscribed to a non-isosceles triangle . The circumscribed circle of the triangle intersects the 𝑂𝑂 second time the lines and in the points respectively ; the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂𝑂𝑂 circumscribed circle of the triangle intersects the second time the lines 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝑐𝑐 𝐴𝐴𝑏𝑏 and in the points respectively ; and the circumscribed circle of 𝑂𝑂𝑂𝑂𝑂𝑂 the triangle intersects the second time the lines and in the 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵𝑐𝑐 𝐵𝐵𝑎𝑎 points respectively . Show that , and are three 𝑂𝑂𝑂𝑂𝑂𝑂 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 concurent𝑎𝑎 lines. 𝑏𝑏 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑐𝑐 𝑏𝑏 𝐶𝐶 𝐶𝐶 National𝐴𝐴 Mathematics𝐵𝐵 𝐴𝐴 𝐶𝐶 Olympiad,𝐵𝐵 𝐶𝐶 Brazil, 2009

78. Let , , be the feet of altitudes of the acute triangle , and ( ), ( ), ( ), such that: 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 1= 1 , 1=1 and1 1 = . 𝑋𝑋 ∈𝐶𝐶1𝑋𝑋𝐵𝐵 𝐶𝐶𝑏𝑏 cos𝑌𝑌𝐶𝐶∈𝐴𝐴1𝐶𝐶𝑌𝑌 𝐴𝐴 𝑐𝑐 cos𝑍𝑍 𝐴𝐴∈ 𝐴𝐴 𝐵𝐵1𝑍𝑍 𝑎𝑎 cos 𝐵𝐵 𝑋𝑋Show𝐵𝐵1 that𝑐𝑐 cos the𝐵𝐵 𝑌𝑌 lines𝐶𝐶1 𝑎𝑎 cos, 𝐶𝐶 and𝑍𝑍 𝐴𝐴1 are𝑏𝑏 cos concurrent𝐴𝐴 . Petru Braica, Problem 27309, G.M. nr. 12/2016 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 79. Let be a given triangle, and – a triangle inscribed in and orthologic with it. Denote by the orthology center of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 in relation to . Consider the point on the perpendicular raised 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 in in the plane ; and the points , , on the segments , , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 . Prove that the triangles and are orthological with a single 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝑃𝑃𝐴𝐴1 𝑃𝑃𝐵𝐵1 orthology1 center. 2 2 2 𝑃𝑃𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu

80. Let and be the feet of altitudes from the vertices and of the acute triangle , and – the midpoint of the side . Denote: { } = 𝐸𝐸 𝐹𝐹 ( ) ( ) ( ) 𝐵𝐵 𝐶𝐶 , = , = , = . Show that is the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑁𝑁 𝑁𝑁 𝑃𝑃 𝑃𝑃 orthocenter of the triangle𝐵𝐵𝐵𝐵 . 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ∩ 𝐸𝐸𝐸𝐸 𝑃𝑃 𝑃𝑃𝑃𝑃 𝑅𝑅 𝑃𝑃𝑃𝑃 𝑆𝑆 𝑃𝑃𝑃𝑃 Nguy𝑁𝑁ễn Minh Hả 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 263 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

81. On the sides of the triangle , consider the points ( ), ( ) and ( ), such that: 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 ∈ 𝐵𝐵𝐵𝐵 𝑁𝑁 ∈ = = = . 𝐶𝐶𝐶𝐶𝑀𝑀𝑀𝑀 𝑁𝑁𝑁𝑁𝑃𝑃 ∈ 𝑃𝑃𝐴𝐴𝐴𝐴𝑃𝑃 Let be the perpendicular from to . Define similarly the lines 𝑀𝑀𝑀𝑀 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 𝑘𝑘 and . Then: , , are concurrent if and only if = 1. M.𝑎𝑎 Monea, Problem 4, The National𝑀𝑀 Mathematical𝐵𝐵𝐵𝐵 Olympiad, local stage, 2003𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑘𝑘 82. Let ( ), ( ), ( ) be tangents in the vertices , , of the triangle to the circumscribed circle of the triangle. Prove that the 𝑇𝑇𝑎𝑎 𝑇𝑇𝑏𝑏 𝑇𝑇𝑐𝑐 𝐴𝐴 𝐵𝐵 𝐶𝐶 perpendiculars taken from the midpoints of the sides opposed to ( ), ( ), 𝐴𝐴𝐴𝐴𝐴𝐴 ( ) are concurrent, and determine their concurrency point. 𝑇𝑇𝑎𝑎 𝑇𝑇𝑏𝑏

𝑇𝑇𝑐𝑐 83. An equilateral triangle is given, and – an arbitrary point in its plane. Denote by , and the centers of the circles inscribed in the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐷𝐷 triangles , and . Prove that the perpendiculars taken from the 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 vertices , , respectively on the sides , and are 𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 concurrent. 1 1 1 1 1 1 𝐴𝐴 𝐵𝐵 𝐶𝐶 I. Shariguin, Collection𝐵𝐵 𝐶𝐶 of 𝐶𝐶problem𝐴𝐴 s, Problem𝐴𝐴 𝐵𝐵 II.17

84. Let be a given line and , , – three lines perpendicular to . Consider , , – points on , such that: 𝑑𝑑 𝑑𝑑1 𝑑𝑑2 𝑑𝑑3 𝑑𝑑 ( , ) = , ( , ) = , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑 ( , ) = , ( , ) = , 𝑑𝑑 𝐴𝐴 𝑑𝑑2 𝑎𝑎1 𝑑𝑑 𝐴𝐴 𝑑𝑑3 𝑎𝑎2 ( , ) = , ( , ) = . 𝑑𝑑 𝐵𝐵 𝑑𝑑3 𝑏𝑏1 𝑑𝑑 𝐵𝐵 𝑑𝑑1 𝑏𝑏2 Find the condition that needs to be met by , , , , , such that 𝑑𝑑 𝐶𝐶 𝑑𝑑1 𝑐𝑐1 𝑑𝑑 𝐶𝐶 𝑑𝑑2 𝑐𝑐2 whatever the points , , on , respectively , the perpendiculars 𝑎𝑎1 𝑎𝑎2 𝑏𝑏1 𝑏𝑏2 𝑐𝑐1 𝑐𝑐2 in , , to , , to be concurrent. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑑𝑑1 𝑑𝑑2 𝑑𝑑3

𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 85. Let be the median triangle of the triangle . If is a point in the plane of the triangle , and is the triangle formed by 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 the orthogonal projections of the point on the sides , respectively 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 , then show that the triangles and are orthological 𝑀𝑀 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 triangles. 𝑎𝑎 𝑏𝑏 𝑐𝑐 1 1 1 𝐴𝐴𝐴𝐴 𝑀𝑀 𝑀𝑀 𝑀𝑀 𝐴𝐴 𝐵𝐵 𝐶𝐶 86. Let and be orthological triangles, and a certain point in their plane. Denote 1by1 1 the symmetrical triangle with respect to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶′ ′ ′ 𝑂𝑂 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 264 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES of the triangle . Show that the triangles and are ′ ′ ′ orthological. 1 1 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 87. Let be an orthodiagonal trapezoid of bases and ; denote by the intersection of diagonals, by – the projection of on , and 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 by – the symmetric of with respect to the midpoint of the segment . 𝑂𝑂 𝐸𝐸 𝑂𝑂 𝐴𝐴𝐴𝐴 The perpendicular from to intersects with the perpendicular taken 𝐹𝐹 𝑂𝑂 𝐴𝐴𝐴𝐴 from to in . Prove that . 𝐹𝐹 𝐴𝐴𝐴𝐴Ion Pătrașcu, Problem 27447, G.M. no. 11/2017 𝐷𝐷 𝐸𝐸𝐸𝐸 𝐻𝐻 𝐴𝐴𝐴𝐴 ⊥ 𝐶𝐶𝐶𝐶 88. Show that the orthology center of a triangle in relation to the podal triangle of symmedian center is the gravity center of the triangle , 𝐴𝐴𝐴𝐴𝐴𝐴 and the orthology center of the podal triangle of the symmedian center in 𝐴𝐴𝐴𝐴𝐴𝐴 relation to the triangle is the gravity center of the podal triangle of the symmedian center. 𝐴𝐴𝐴𝐴𝐴𝐴

89. Show that: a) Two equilateral triangles and ’ ’ ’, inversely oriented, are three times parallelogic, namely with the orders: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 ; ; . ’ ’ ’ ’ ’ ’ ’ ’ ’ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 b) Denoting� � by� , � , � the �points of parallelogy corresponding to the𝐴𝐴 𝐵𝐵terns𝐶𝐶 above𝐵𝐵 𝐶𝐶 𝐴𝐴, then𝐶𝐶 the𝐴𝐴 𝐵𝐵triangle is equilateral, with the 𝑃𝑃1 𝑃𝑃2 𝑃𝑃3 vertices on the circumscribed circle of1 the2 3 triangle ABC. 𝑃𝑃 𝑃𝑃 𝑃𝑃 Constantin Cocea

90. Draw the triangle , and – its orthic triangle. , , are the projections of vertices of the triangle respectively on , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 and . Prove that the triangles and are orthological1 1. I. Sharig𝐴𝐴𝐴𝐴𝐴𝐴uin, Collection of problems𝐵𝐵 𝐶𝐶 1 1 1 1 2 2 2 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 91. On the sides of the acute triangle , the equilateral triangles , , are being built on the exterior. Show that the median triangle of 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 the triangle and the triangle are orthological. 𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

𝐾𝐾𝐾𝐾𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 92. Let be a scalene triangle, and let be the podal triangle of the symmedian center of the triangle . Denote by , , the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 Ion Pătrașcu, Florentin Smarandache 265 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES symmetrics of the points , , in relation to . Prove that the triangles and are orthological. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐾𝐾

𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 93. Let be a convex quadrilateral inscribed in the circle of center , where is not parallel with . Denote by and the midpoints of 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 the sides and , and by the intersection of the lines and . 𝑂𝑂 𝐵𝐵𝐵𝐵 𝐷𝐷𝐷𝐷 𝑄𝑄 𝑅𝑅 Prove that the perpendiculars taken from , , respectively to , 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐷𝐷𝐷𝐷 and are concurrent. 𝐴𝐴 𝐵𝐵 𝐶𝐶 Ion Pătrașcu𝑅𝑅𝑅𝑅 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 94. Let be a regular hexagon. Show that the triangles and are triorthological. Specify the orthology centers. 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu 𝐵𝐵𝐵𝐵𝐵𝐵 𝐸𝐸𝐸𝐸𝐸𝐸 95. Let be a right triangle in with = 60 and = 7. 0 Draw parallel𝐴𝐴𝐴𝐴𝐴𝐴s with , and 𝐵𝐵located𝑚𝑚 at�𝐴𝐴 ̂distances� 𝐵𝐵𝐵𝐵, √and √21 2√7 respectively which 𝐵𝐵𝐵𝐵intersects𝐴𝐴𝐴𝐴 the 𝐶𝐶𝐶𝐶interior of the triangle. 7 7 √7 Prove that7: a) The parallels are concurrent in a point ; b) The podal triangle of the point , denoted by , is equilateral; and 𝑀𝑀 c) The triangle are not bilogical1. 1 1 𝑀𝑀 𝐴𝐴 𝐵𝐵 𝐶𝐶 Ion Pătrașcu 1 1 1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 96. Let be an acute triangle, – the center of its circumscribed circle, and , – the intersections of the semi-line ( with 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 respectively with the circumscribed circle. The tangent in to the 𝑀𝑀 𝐷𝐷 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 circumscribed circle intersects in and in . The circles 𝐷𝐷 circumscribed to triangles and intersect the second time 𝐴𝐴𝐴𝐴 𝐾𝐾 𝐴𝐴𝐴𝐴 𝐿𝐿 and respectively in and . Prove that the triangle is 𝐷𝐷𝐷𝐷𝐷𝐷 𝐷𝐷𝐷𝐷𝐷𝐷 orthological with the triangle , and the orthology center is the 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐹𝐹 𝐸𝐸 𝐷𝐷𝐷𝐷𝐷𝐷 symmetric of the point to . 𝐴𝐴𝐴𝐴𝐴𝐴

𝐷𝐷 𝑀𝑀 97. Let be an acute triangle, – its orthic triangle, and – its median triangle. Denote by , , the midpoints of medians 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2

266 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

( ), ( ), ( ). Prove that the triangles and are orthological. 1 1 1 2 2 2 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴Ion𝐵𝐵 Pătrașcu𝐶𝐶

98. Let and be two orthological triangles. Denote by – the orthology center of the triangle in relation to the triangle ; 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑃𝑃 and by – the orthology center of the triangle in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 Then, the barycentric coordinates of in relation to are equal with the 𝑃𝑃1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 barycentric coordinates of in relation to . 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴

𝑃𝑃1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 99. Let a triangle inscribed in the circle of center . The bisectors , , are concurrent in . The perpendiculars taken from to , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 and intersect , and respectively in , , . Show that 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐼𝐼 𝐼𝐼 𝐵𝐵𝐵𝐵 , , are concurrent in a point situated on . 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸 𝐷𝐷𝐷𝐷 𝑀𝑀 𝑁𝑁Nguy𝑃𝑃 ễn Minh Hả 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑂𝑂𝑂𝑂 100. Let be a triangle and be a point in its interior. Denote by , , the feet of perpendiculars taken from to , , respectively . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 𝐷𝐷 Suppose that: 𝐸𝐸 𝐹𝐹 𝑃𝑃 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 + = + = + . Denote2 by ,2 , the2 centers2 of the2 ex-in2scribed circles to the triangle 𝐴𝐴𝑃𝑃 𝑃𝑃𝐷𝐷 𝐵𝐵𝑃𝑃 𝑃𝑃𝐸𝐸 𝐶𝐶𝑃𝑃 𝑃𝑃𝐸𝐸 . Show that is the center of the circle circumscribed to the triangle 𝐼𝐼𝑎𝑎 𝐼𝐼𝑏𝑏 𝐼𝐼𝑐𝑐 . 𝐴𝐴𝐴𝐴𝐴𝐴 Problem𝑃𝑃 G3 – Short-listed – 44th International Mathematical Olimpiad, 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐼𝐼 𝐼𝐼 𝐼𝐼 Tokyo, Japan, 2003

9.2 Open Problems

1. The pedal triangle of the isogonal of Gergonne point in the triangle is orthological with the triangle if and only if the triangle is an isosceles triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 2. The pedal triangle of the isogonal of Nagel point in the triangle is orthological with the triangle if and only if is an isosceles 𝐴𝐴𝐴𝐴𝐴𝐴 triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 267 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

PROBLEMS CONCERNING ORTHOLOGICAL TRIANGLES

3. If ’ ’ ’ is the -pedal triangle of the point from the interior of triangle , and the triangles and ’ ’ ’ are orthological, and 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑀𝑀 𝑀𝑀 ’’ ’’ ’’ is the ’-pedal triangle of the isogonal ’ of the point in relation 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 to the triangle , the triangles and ’’ ’’ ’’ are orthological. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑀𝑀 𝑀𝑀 𝑀𝑀

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 4. Let be the -circumpedal triangle of the triangle ( – the gravity center in the triangle ). Is it true that and are 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 𝐺𝐺 orthological triangles if and only if is an equilateral triangle? 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 5. Let be an isosceles triangle, with = , and – its gravity center. If and are the orthology centers of the orthological and - 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐺𝐺 circumpedal triangles , and these points are symmetrical with respect 𝑈𝑈 𝑉𝑉 𝐺𝐺 to , find the measures of angles of the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴

𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 6. The podal triangle of the orthocenter of the triangle is orthological with the -circumpedal triangle. What conditions must the 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 point from the interior of triangle meet in order that its podal and 𝐻𝐻 its -circumpedal triangles to be orthological triangles? 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴

𝑀𝑀 7. Let and be two equilateral inversely similar triangles. Denote by , , the orthology centers of triangle in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the triangles , and . Let , , be the orthology 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝐴𝐴𝐴𝐴𝐴𝐴 centers of the triangle in relation to the′ triangles′ ′ , and 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1𝐴𝐴1 𝐶𝐶1𝐴𝐴1𝐵𝐵1 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 . It is known that and are equilateral, inversely similar 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 and triorthological triangles. If one′ ′ continue′ s for these triangles the 𝐶𝐶𝐶𝐶𝐶𝐶 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝑂𝑂1𝑂𝑂2𝑂𝑂3 construction made for and , and then for those determined by their orthology centers, etc., will the process continue endlessly or stop? 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

8. Let be the pedal triangle of the center of the circumscribed circle of the′ ′acute′ triangle . Prove that the triangles and 𝐴𝐴 𝐵𝐵 𝐶𝐶 are orthological if and only if is an isosceles triangle. ′ ′ ′ 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 Ion Pătrașcu,𝐴𝐴𝐴𝐴𝐴𝐴 Mihai𝐴𝐴 𝐵𝐵 Dinu𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

268 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

SOLUTIONS, INDICATIONS, ANSWERS TO THE PROPOSED ORTHOLOGY PROBLEMS

1. Solution 1. The triangles and are orthological if and only if + + = + + . Being symmetrical with 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 respect2 to the line2 , we2 have that2 2 = 2, = and = , 𝐴𝐴𝐵𝐵1 𝐵𝐵𝐶𝐶1 𝐶𝐶𝐴𝐴1 𝐴𝐴𝐶𝐶1 𝐵𝐵𝐴𝐴1 𝐶𝐶𝐵𝐵1 hence the above relation is verified. 𝑑𝑑 𝐴𝐴𝐵𝐵1 𝐵𝐵𝐴𝐴1 𝐵𝐵𝐶𝐶1 𝐶𝐶𝐵𝐵1 𝐶𝐶𝐴𝐴1 𝐴𝐴𝐶𝐶1 Solution 2. The triangles and are similar and inversely oriented. We apply now Theorem 26. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

2. We prove that the triangles and are orthological. The perpendicular from to is mediator of . The perpendicular from 𝐵𝐵𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀𝑀𝑀 to is the radical axis of the inscribed circle and the null circle , and the 𝑀𝑀 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝑁𝑁 perpendicular from to is the radical axis of the inscribed circle and the 𝐶𝐶𝐶𝐶 𝒞𝒞 null circle . The radical axis of the null circles , is mediator of . The 𝑃𝑃 𝐵𝐵𝐵𝐵 radical axes of three circles are concurrent in their radical center . Because ℬ ℬ 𝒞𝒞 𝐵𝐵𝐵𝐵 perpendiculars from , , to , and are concurrent, it means that Ω the triangles and are orthological, therefore the perpendiculars 𝑀𝑀 𝑁𝑁 𝑃𝑃 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 taken from , , respectively to , and will be concurrent. 𝑀𝑀𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵𝐵𝐵 Observation𝐼𝐼 𝐵𝐵 𝐶𝐶 𝑁𝑁𝑁𝑁 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 The problem remains valid even if instead of the inscribed circle ( ) we consider the -ex-inscribed circle ( ). 𝐼𝐼

𝐴𝐴 𝐼𝐼𝑎𝑎 3. The perpendiculars taken from , , respectively to , , are the mediators of these sides, therefore they are concurrent in the center 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 of the circle circumscribed to the triangle , point that is the orthology center of the triangles and . 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2𝑂𝑂3 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 269 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

The perpendiculars taken from , , to , respectively are the chords , , , therefore is the second orthology center. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂2𝑂𝑂3 𝑂𝑂3𝑂𝑂1 𝑂𝑂1𝑂𝑂2

𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑀𝑀 4. The triangles and are orthological, because the perpendicular taken from to , the perpendicular from to and the 𝐵𝐵1𝐶𝐶1𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 perpendicular taken from to are concurrent in . According to the 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵1 𝐵𝐵𝐵𝐵 theorem of orthological triangles, the property is also true vice versa: the 𝐶𝐶1 𝐶𝐶𝐶𝐶 𝐻𝐻 perpendicular taken from to and the perpendicular taken from to are concurrent. Because the first two perpendiculars are altitudes in 𝐶𝐶 𝐶𝐶1𝐴𝐴 𝑃𝑃 the triangle , they are concurrent in , and it follows that the third 𝐵𝐵1𝐶𝐶1 perpendicular passes through , therefore is perpendicular in . On 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 the other hand, is perpendicular to , consequently and are 𝐻𝐻 𝑃𝑃𝑃𝑃 𝐵𝐵1𝐶𝐶1 parallel. 𝑃𝑃𝑃𝑃 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐵𝐵1𝐶𝐶1

5. The triangles and are orthological triangles. One of the orthology centers is , and let′ ′ ′be the second orthology center. It is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 known that is the isogonal conjugate′′ of the point . If is the second 𝑃𝑃 𝑃𝑃 orthology center′′ of the triangle (the first is ), then ′′′ is the isogonal 𝑃𝑃 𝑃𝑃 𝑃𝑃 conjugate of the point . Because the set of orthology′ centers′′′ consists only 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 𝑃𝑃 of and , it means that′ and must coincide with , respectively 𝑃𝑃 , then and′ are isogonal′′ conjugate′′′ points. ′ 𝑃𝑃 𝑃𝑃 𝑃𝑃 𝑃𝑃 𝑃𝑃 ′ 𝑃𝑃 𝑃𝑃 𝑃𝑃 6. If , , is the Simson line of the point , we have (the′ quadrilateral′ ′ is inscribable). Because , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑆𝑆 ∢𝑆𝑆𝑆𝑆𝑆𝑆 ≡ it follows′ ′ that ′ ′, therefore . The quadrilateral ∢𝑆𝑆𝐴𝐴 𝐶𝐶 𝑆𝑆𝐴𝐴 𝐵𝐵𝐶𝐶 ∢𝑆𝑆𝑆𝑆𝑆𝑆 ≡ ∢𝑆𝑆𝑆𝑆𝑆𝑆 is parallelogram′ ′ because and′ ′ ; it follows that ∢𝑆𝑆𝐴𝐴 𝐶𝐶 ≡ ∢𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴 𝐶𝐶 ∥ 𝐴𝐴𝐴𝐴 . and lead to . 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ∥ 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴 ≡ 𝑃𝑃𝑃𝑃 ′ ′ ′ ′ 𝑂𝑂𝑂𝑂7.∥ 𝐴𝐴We𝐴𝐴 consider𝐴𝐴 𝐶𝐶 ∥ 𝐴𝐴 𝐴𝐴the side𝐴𝐴𝐴𝐴 of∥ 𝑂𝑂an𝑂𝑂 equilateral𝐴𝐴 𝐶𝐶triangle∥ 𝑂𝑂𝑂𝑂 of length 1 and let = , = , = . From Ceva’s theorem and Carnot’s theorem, it follows that: 𝐴𝐴𝐶𝐶1 𝑥𝑥 𝐵𝐵𝐴𝐴1 𝑦𝑦 𝐶𝐶𝐵𝐵1 𝑧𝑧 = (1 )(1 )(1 ) and + + = (1 ) + (1 ) + (1 ) . 𝑥𝑥𝑥𝑥𝑥𝑥 − 𝑥𝑥 − 𝑦𝑦 − 𝑧𝑧 From2 the2 second2 , we note2 that: +2 + = 2, therefore (1 ) + 𝑥𝑥 𝑦𝑦 𝑧𝑧 − 𝑥𝑥 − 𝑦𝑦 − 𝑧𝑧 3 (1 ) + (1 ) = . 𝑥𝑥 𝑦𝑦 𝑧𝑧 2 − 𝑥𝑥 3 − 𝑦𝑦 − 𝑧𝑧 2 270 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

We also have: + + = ( + + ) . 1 9 2 2 2 𝑥𝑥Analogously𝑥𝑥 𝑦𝑦𝑦𝑦 𝑧𝑧𝑧𝑧: 2 �4 − 𝑥𝑥 𝑦𝑦 𝑧𝑧 � (1 )(1 ) + (1 )(1 ) + (1 )(1 ) =

= (− 𝑥𝑥+ −+𝑦𝑦 ) . − 𝑦𝑦 − 𝑧𝑧 − 𝑧𝑧 − 𝑥𝑥 1 9 2 2 2 { } If 2 �=4 − 𝑥𝑥 , then𝑦𝑦 we𝑧𝑧 have� the solutions , , 1 , 1 , from where it follows that = 1 , and in any situation we obtain again that = = 𝑥𝑥 𝑦𝑦 ≠ 𝑧𝑧 𝑥𝑥 𝑧𝑧 − 𝑥𝑥 − 𝑧𝑧 = . 1 𝑥𝑥 − 𝑧𝑧 𝑥𝑥 𝑦𝑦 If , , are all different, then { , , } = {1 , 1 , 1 }. If = 𝑧𝑧 2 1 , then = , and if = 1 , then = 1 and = 1 lead to 𝑥𝑥 𝑦𝑦 𝑧𝑧 1 𝑥𝑥 𝑦𝑦 𝑧𝑧 − 𝑥𝑥 − 𝑦𝑦 − 𝑧𝑧 𝑥𝑥 =− 𝑥𝑥. Hence𝑥𝑥, the2 solutions𝑥𝑥 of the− equation𝑦𝑦 𝑦𝑦 are −, 1𝑥𝑥 , 𝑧𝑧 . − 𝑧𝑧 1 1 In conclusion, all the positions of the points , , are given by the 𝑧𝑧 2 �𝑥𝑥 − 𝑥𝑥 2� , 1 , (0, 1) triplets and their permutations, with𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 , hence one of 1 the points is the midpoint of one side, and the other is equally spaced from �𝑥𝑥 − 𝑥𝑥 2� 𝑥𝑥 ∈ the vertices of the respective side.

8. Let { } = ; we suppose that the triangle is acute and > . We have = tan , = tan , tan = tan 𝑋𝑋 𝐴𝐴𝑏𝑏𝐴𝐴𝑐𝑐 ∩ 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 tan . It is observed that is the isotomic of – the feet of the altitude 𝐵𝐵� 𝐶𝐶̂ 𝐴𝐴𝑐𝑐𝐶𝐶 𝑎𝑎 𝐵𝐵 𝐴𝐴𝑏𝑏𝐵𝐵 𝑎𝑎 𝐶𝐶 �𝐴𝐴�𝑏𝑏𝑋𝑋𝑋𝑋� 𝐵𝐵 − from A; since = cos , it follows that: ′ 𝐶𝐶 𝐴𝐴2 𝐴𝐴 =′ 2 cos = 2 sin 4 sin cos 𝐵𝐵𝐵𝐵 𝑐𝑐 ∙ 𝐵𝐵 ′ = 2 (sin 2 sin cos ). 𝐴𝐴 𝐴𝐴2 𝑎𝑎 − 𝑐𝑐 ∙ 𝐵𝐵 𝑅𝑅 𝐴𝐴 − 𝑅𝑅 𝐶𝐶 𝐵𝐵 = cot = = 2 cos cos . 𝑅𝑅 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 ′ ′ 𝐶𝐶∙cos 𝐵𝐵∙coscos𝐶𝐶 cos 𝐻𝐻tan𝐴𝐴 𝐶𝐶 ∙ =𝐵𝐵𝐴𝐴 = sin 𝐶𝐶 𝑅𝑅 𝐵𝐵 𝐶𝐶 ′ sin 2 sin cos ′ 𝐻𝐻𝐴𝐴 𝐵𝐵 ∙ 𝐶𝐶 �2 ′ cos cos �𝐻𝐻𝐴𝐴 𝐴𝐴 � = 2 𝐴𝐴 𝐴𝐴sin( +𝐴𝐴 −) 2 sin𝐶𝐶 ∙ cos𝐵𝐵 𝐵𝐵 ∙ cos𝐶𝐶 cos = sin 𝐵𝐵 cos𝐶𝐶 −+ sin 𝐶𝐶 ∙cos 𝐵𝐵 2 sin cos cos cos 𝐵𝐵 ∙ 𝐶𝐶 1 = = . sin 𝐵𝐵 ∙ cos 𝐶𝐶 sin 𝐶𝐶 ∙ cos 𝐵𝐵 − tan 𝐶𝐶 ∙ tan 𝐵𝐵 𝐵𝐵 ∙ 𝐶𝐶 𝐵𝐵 ∙ 𝐶𝐶 − 𝐶𝐶 ∙ 𝐵𝐵 𝐵𝐵 − 𝐶𝐶

Ion Pătrașcu, Florentin Smarandache 271 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

Because tan = cot , it follows that . Similarly, we prove that and′ that . �𝐴𝐴�𝑏𝑏𝑋𝑋𝑋𝑋� �𝐻𝐻�𝐴𝐴2𝐴𝐴 � 𝐴𝐴2𝐻𝐻 ⊥ 𝐴𝐴𝑏𝑏𝐴𝐴𝑐𝑐 2 𝑎𝑎 𝑐𝑐 2 𝑏𝑏 𝑎𝑎 Observation 𝐵𝐵 𝐻𝐻 ⊥ 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐻𝐻 ⊥ 𝐶𝐶 𝐶𝐶 The problem states that the triangles and the one with the sides determined by the lines , , – are orthological, and is their center 𝐴𝐴2𝐵𝐵2𝐶𝐶2 of orthology. 𝐴𝐴𝑏𝑏𝐴𝐴𝑐𝑐 𝐵𝐵𝑎𝑎𝐵𝐵𝑐𝑐 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏 𝐻𝐻

9. i) We suppose that , , are collinear; let = , 𝑅𝑅𝑅𝑅 ~ = = 𝑃𝑃𝑃𝑃 𝑅𝑅′ 𝑃𝑃 𝑄𝑄′ 𝜆𝜆 (1) ′ 𝐴𝐴 𝑅𝑅 𝐴𝐴 𝑃𝑃 ΔAnalogously𝐴𝐴 𝑅𝑅𝑅𝑅 ∆𝐴𝐴1,𝑄𝑄 𝑄𝑄 ⟹ 𝐴𝐴1𝑄𝑄 𝐴𝐴1𝑃𝑃 𝜆𝜆 ~ = = ′ ′ , (2) ′ 𝐵𝐵 𝑅𝑅 𝐵𝐵 𝑃𝑃 ~ 1 𝐵𝐵1𝑄𝑄 = 𝐵𝐵1𝑃𝑃 = Δ𝐵𝐵 𝑅𝑅𝑅𝑅 ∆𝐵𝐵 𝑄𝑄𝑄𝑄 ⟹ ′ ′ 𝜆𝜆. (3) ′ 𝐶𝐶 𝑅𝑅 𝐶𝐶 𝑃𝑃 ΔThe𝐶𝐶 𝑅𝑅 relations𝑅𝑅 ∆𝐶𝐶1𝑄𝑄 (1),𝑄𝑄 ⟹ (2)𝐶𝐶 1and𝑄𝑄 (3)𝐶𝐶1 𝑃𝑃 lead𝜆𝜆 to , , . The triangles and have′ ′ respectively′ ′parallel sides′ ;′ it 𝐴𝐴 𝐵𝐵 ∥ 𝐴𝐴1𝐵𝐵1 𝐵𝐵 𝐶𝐶 ∥ 𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐶𝐶 ∥ follows that the quadrilaterals ′ ′ ′ and are parallelograms, 𝐴𝐴1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 therefore = and ′ =′ ′ , consequently′ ′ ′ is the midpoint 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶1 𝐴𝐴 𝐶𝐶 𝐴𝐴1𝐵𝐵 of the side ′ ′ , analogously′ ′ ′ is the midpoint′ of the side′ , and is 𝐴𝐴 𝐶𝐶 𝐶𝐶1𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴1𝐵𝐵 𝐵𝐵 the midpoint of the side ′; therefore is the gravity center of ′the 𝐴𝐴1𝐶𝐶1 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐶𝐶 triangle . 𝐴𝐴1𝐵𝐵1 𝑃𝑃 Let now be the gravity center of the triangle and – the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 orthology center of the triangle in relation to . If is the 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑄𝑄 orthology center of the triangle in relation to , then, because the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑅𝑅 triangles and are ′ homological′ ′ of center , according to 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 Sondat’s theorem, we obtain that the points , and are collinear ( is 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 the orthology center of the triangle in relation to ). On the other 𝑄𝑄 𝑃𝑃 𝑆𝑆 𝑆𝑆 hand, the triangle is the median triangle of the triangle , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 therefore the perpendiculars′ ′ ′ taken from , , to the sides of are 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1 perpendicular to the sides of the triangle as well, consequently is 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1 the orthology center of the triangle in ′relation′ ′ to . According to 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑆𝑆 Sondat’s theorem, taking into account that and ′ ′ ′are homothetic 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 with the homothety center, we have that the′ ′points′ , , are collinear. 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 From , , and , , – collinear, it follows that , , , are collinear. 𝑃𝑃 𝑆𝑆 𝑅𝑅 𝑃𝑃 𝑄𝑄 𝑃𝑃 𝑆𝑆 𝑆𝑆 𝑅𝑅 𝑃𝑃 𝑄𝑄 𝑅𝑅 𝑆𝑆 𝑃𝑃 272 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

ii) It follows from i).

10. The triangle is orthological with the triangle (we denoted by the midpoint of ). The orthology centers are ′and′ ′ . 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐻𝐻 𝑄𝑄 11. The triangles and are congruent (A.S.A.). From and = , it follows that the quadrilateral is parallelogram, hence 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼𝐼𝐼 𝐴𝐴𝐴𝐴 ∥ 𝐼𝐼𝐼𝐼 . Because , it follows that , therefore is the 𝐴𝐴𝐴𝐴 𝐼𝐼𝐼𝐼 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 altitude from of the triangle . We obtained that the perpendiculars 𝐴𝐴𝐴𝐴 ∥ 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 ⊥ 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 taken from , , respectively to , and are concurrent, hence the 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 triangles and are orthological. 𝐼𝐼 𝐷𝐷 𝐸𝐸 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴

𝐼𝐼𝐼𝐼𝐼𝐼 𝐴𝐴𝐴𝐴𝐴𝐴 12. a) Obviously, the triangles and are orthological, the orthology center being . If we denote by , , the midpoints of the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 sides , , , we observe that is parallel with ; and since 𝑂𝑂 𝑀𝑀 𝑁𝑁 𝑃𝑃 is parallel with , it follows that the perpendicular taken from to 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑁𝑁𝑁𝑁 𝐵𝐵1𝐶𝐶1 𝑁𝑁𝑁𝑁 is the altitude . Reasoning analogously, we obtain that the second 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐵𝐵1𝐶𝐶1 orthology center of′ the triangles and is the orthocenter of 𝐴𝐴𝐴𝐴 the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐻𝐻 b) The quadrilateral is parallelogram because = 2 and 𝐴𝐴𝐴𝐴𝐴𝐴 ; it follows that passes through the midpoint of the 𝐴𝐴𝐴𝐴𝐴𝐴1𝑂𝑂 𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂 segment . Analogously, we obtain that and pass through , 𝐴𝐴𝐴𝐴 ∥ 𝑂𝑂𝐴𝐴1 𝐴𝐴𝐴𝐴1 𝑂𝑂1 therefore the homology center is , and it belongs to Euler line of the 𝑂𝑂𝑂𝑂 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝑂𝑂1 triangle . 𝑂𝑂1 𝑂𝑂𝑂𝑂 c) From b), we have that = 1. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝐻𝐻 𝑂𝑂1𝑂𝑂 13. The triangle is orthological in relation to , and the orthology center is . Then′ ′ ′ is also orthological in relation to . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ ′ 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 14. The triangles and are orthological. Indeed, the perpendicular taken from to , the perpendicular taken from to 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 (the altitude of the triangle ), and the perpendicular taken from to 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵1 𝐴𝐴𝐴𝐴 (the altitude of the triangle ) are concurrent. The orthology center 𝐴𝐴𝐴𝐴1𝐶𝐶1 𝐶𝐶1 is the the orthocenter of the triangle . 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴1𝐶𝐶1

𝐻𝐻1 𝐴𝐴𝐴𝐴1𝐶𝐶1

Ion Pătrașcu, Florentin Smarandache 273 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

15. It is clear that must be in the interior of the triangle . We have: (S.S.S.); it follows that , therefore 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 also ′ ; hence′ is bisector, therefore′ an altitude′ in . ∆𝐴𝐴𝐴𝐴𝐵𝐵 ≡ ∆𝐴𝐴𝐴𝐴𝐶𝐶 ∢𝐵𝐵 𝐴𝐴𝐴𝐴 ≡ ∢𝐶𝐶 𝐴𝐴𝐴𝐴 Similarly, we show that is an altitude in the triangle , hence 𝐵𝐵𝐵𝐵𝐵𝐵� ≡ 𝐶𝐶𝐶𝐶𝐶𝐶� 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 must be the orthocenter of this triangle. Moreover, , , are 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 altitudes in . The common orthology center is . ′ ′ ′ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶

𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 16. We denote by , , the orthogonal projections of the point on the sides of the triangle , and by , , – the projections of on the 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑂𝑂 sides of the triangle , , . The quadrilaterals , , , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎1 𝑏𝑏1 𝑐𝑐1 𝑂𝑂 , , ′ ′ are′ inscribable, having two′ opposite right′ angles′ . 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝑎𝑎𝐴𝐴1𝑏𝑏1𝐵𝐵 𝑏𝑏1𝐵𝐵𝐵𝐵𝐶𝐶1 𝑐𝑐𝐶𝐶1𝑎𝑎1𝐴𝐴 It follows′ ′ that: ′ 𝑎𝑎1𝐴𝐴𝐴𝐴𝐵𝐵1 𝑏𝑏𝐵𝐵1𝑐𝑐1𝐶𝐶 𝑐𝑐1𝐶𝐶𝐶𝐶𝐴𝐴1 = = = ′ = ′ = 𝑂𝑂��𝑎𝑎⃗ ∙ �𝑂𝑂��𝑂𝑂��1⃗ �𝑂𝑂��𝑏𝑏⃗ ∙ �𝑂𝑂��𝑂𝑂��1⃗ ′ = = 1 = . 𝑂𝑂��𝑐𝑐⃗ ∙ �𝑂𝑂��𝑂𝑂��1⃗ �𝑂𝑂��𝑎𝑎��⃗ ∙ �𝑂𝑂��𝑂𝑂�⃗ 2 This shows that , , 1 – the poles1 of the sides , respectively �𝑂𝑂��𝑏𝑏��⃗ ∙ �𝑂𝑂��𝑂𝑂�⃗ �𝑂𝑂��𝑐𝑐�⃗ ∙ �𝑂𝑂��𝑂𝑂�⃗ 𝑘𝑘 with respect to the′ circle′ ′ ( ; ), videlicet the triangles and 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 are reciprocal polar to this circle. (G.M. XXII) ′ ′ ′ 𝐴𝐴𝐴𝐴 𝒞𝒞 𝑂𝑂 𝑘𝑘 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 17. We consider the circle circumscribed to the quadrilateral ; let be the center of this circle. The polar of with respect to this circle′ ′ 𝐶𝐶𝐵𝐵 𝐶𝐶 𝐵𝐵 is , and the polar of with respect to this circle is ; it follows that 𝑀𝑀𝑎𝑎 𝑃𝑃 is the pole of the line and, hence, or . 𝐴𝐴𝐴𝐴 𝐴𝐴 𝑃𝑃𝑃𝑃 𝐻𝐻 Analogously, it is shown that and that . 𝐴𝐴𝐴𝐴 𝑀𝑀𝑎𝑎𝐻𝐻 ⊥ 𝐴𝐴𝐴𝐴 𝑀𝑀𝑎𝑎𝐻𝐻 ⊥ 𝑈𝑈𝑈𝑈 Consequently, the triangles and are orthological, with the 𝑀𝑀𝑏𝑏𝐻𝐻 ⊥ 𝑉𝑉𝑉𝑉 𝑀𝑀𝑐𝑐𝐻𝐻 ⊥ 𝑈𝑈𝑈𝑈 orthology center being . 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝑈𝑈𝑈𝑈𝑈𝑈

𝐻𝐻 18. It is obvious that the triangles and are homological with the help of Ceva’s theorem; =1 1 1 . The 2ort2hology2 derives from 𝐴𝐴2𝐶𝐶1 𝐴𝐴 𝐵𝐵𝑏𝑏∙cos𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 the fact that – being perpendicular𝐴𝐴2𝐵𝐵1 to𝑐𝑐∙cos 𝐵𝐵 (which is antiparallel with ) – passes through the center of the circle circumscribed to the triangle 𝐴𝐴𝐴𝐴2 𝐵𝐵1𝐶𝐶1 . 𝐵𝐵𝐵𝐵 𝑂𝑂

𝐴𝐴𝐴𝐴𝐴𝐴

274 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

19. We denote by the midpoint of the side . The quadrilateral is parallelogram. The perpendiculars raised in the points , 𝑀𝑀𝑑𝑑 𝐵𝐵𝐵𝐵 , , to , , respectively are concurrent in the center 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑑𝑑𝑀𝑀𝑐𝑐 𝑀𝑀𝑎𝑎 of𝑏𝑏 the circle𝑑𝑑 𝑐𝑐. 1 1 1 𝑀𝑀 𝑀𝑀 𝑀𝑀 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 20. Let , , be the projections of points , , on ; respectively ′ ′ . We′ denote by { } = . From the 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵2𝐶𝐶2 𝐶𝐶2𝐴𝐴2 reciprocal of the three perpendicular theorem, it′ follows′ that ′ , 𝐴𝐴2𝐵𝐵2 𝑄𝑄 𝐴𝐴1𝐴𝐴1 ∩ 𝐵𝐵1𝐵𝐵1 ∩ 𝐶𝐶1𝐶𝐶1 and . On the other hand, the planes (′ ), 𝐴𝐴0𝐴𝐴1 ⊥ 𝐵𝐵2𝐶𝐶2 ( ′ ), ( ) have′ in common the point and contain respectively′ 𝐵𝐵0𝐵𝐵1 ⊥ 𝐴𝐴2𝐶𝐶2 𝐶𝐶0𝐶𝐶1 ⊥ 𝐴𝐴2𝐵𝐵2 𝐴𝐴0𝐴𝐴1𝐴𝐴1 the paral′ lel lines ′ , , ; it means that they have a line in 𝐵𝐵0𝐵𝐵1𝐵𝐵1 𝐶𝐶0𝐶𝐶1𝐶𝐶1 𝑄𝑄 common with them, which passes through and intersects the plane 𝐴𝐴1𝐴𝐴0 𝐵𝐵1𝐵𝐵0 𝐶𝐶1𝐶𝐶0 in . This point is on each of the lines , , , hence 𝑄𝑄 these lines are′ concurrent in′ , and this point is the orthology′ ′ center′ of the 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑄𝑄 𝑄𝑄 𝐴𝐴0𝐴𝐴1 𝐵𝐵0𝐵𝐵1 𝐶𝐶0𝐶𝐶1 triangle with respect′ to . We denote by , , the 𝑄𝑄 homology axis of triangles and (the line of intersection of 𝐴𝐴0𝐵𝐵0𝐶𝐶0 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑀𝑀 𝑁𝑁 𝑃𝑃 their planes). Because the projection of the line on the plane 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 ( ) is and ( ) = { }, it follows that , 𝐵𝐵1𝐶𝐶1 therefore { } = , analogously { } = and { } = 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐵𝐵0𝐶𝐶0 𝐵𝐵1𝐶𝐶1 ∩ 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝑀𝑀 𝑀𝑀 ∈ 𝐵𝐵0𝐶𝐶0 . 𝑀𝑀 𝐵𝐵0𝐶𝐶0 ∩ 𝐵𝐵2𝐶𝐶2 𝑁𝑁 𝐴𝐴0𝐶𝐶0 ∩ 𝐴𝐴2𝐶𝐶2 𝑃𝑃

𝐴𝐴0𝐵𝐵0 ∩ 𝐴𝐴2𝐵𝐵2 21. a) The perpendiculars in , , respectively to , , are concurrent if and only if: ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 + + = 0. This′ 2 relation′ 2 is equivalent′ 2 ′to2: ′ 2 ′ 2 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 ( ) + ( ) + ( ) = 0. But′ : =′ ; ′ = ′ ; ′ = ′ . 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 ∙ 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 ∙ 𝐴𝐴𝐴𝐴 Replacing′ these relation′ s ′in the preceding′ relation′ , we get′ the required 𝐴𝐴 𝐶𝐶 𝐵𝐵𝐵𝐵 − 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐴𝐴 𝐴𝐴𝐴𝐴 − 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐵𝐵 𝐴𝐴𝐴𝐴 − 𝐶𝐶 𝐴𝐴 relation. b) It is known that, if , , and , , , the Cauchy- Buniakovski-Schwarz inequality takes place: 𝑎𝑎 𝑏𝑏 𝑐𝑐 ∈ ℝ 𝑥𝑥 𝑦𝑦 𝑧𝑧 ∈ ℝ ( + + )( + + ) ( + + ) . Taking2 2 =2 , 2 = 2, 2= and = , 2 = , = , the 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑥𝑥 𝑦𝑦 𝑧𝑧 ≥ 𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐 relation b) is obtained taking into account′ the Cauchy′ -Buniakovski′ - Schwarz inequality𝐵𝐵𝐵𝐵 𝑎𝑎 and𝐶𝐶𝐶𝐶 the𝑏𝑏 relation𝐴𝐴𝐴𝐴 𝑐𝑐from a)𝐴𝐴. 𝐵𝐵 𝑥𝑥 𝐵𝐵 𝐶𝐶 𝑦𝑦 𝐶𝐶 𝐴𝐴 𝑧𝑧

Ion Pătrașcu, Florentin Smarandache 275 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

c) + + is minimal if and only if the equality holds in the inequality′2 from′2 b), ie. ′if2 and only if = = . We denote: 𝐵𝐵𝐵𝐵 𝐶𝐶𝐵𝐵 𝐴𝐴𝐴𝐴 𝑥𝑥 𝑦𝑦 𝑧𝑧 = = = . 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑦𝑦 𝑦𝑦 𝑧𝑧 From a), it derives that: 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑘𝑘 + + = ( + + ); 1 2 2 2 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 𝑐𝑐+𝑐𝑐 2 =𝑎𝑎 ( 𝑏𝑏+ 𝑐𝑐 + ), therefore = , which shows 1 1 that 2 , ,2 are2 the midpoints2 2 of the2 sides , , ; hence, the 𝑎𝑎 𝑘𝑘 𝑏𝑏 𝑘𝑘 𝑐𝑐 𝑘𝑘 2 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑘𝑘 2 intersection′ ′ point′ of the perpendiculars in , , to , , is the 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 center of the circle circumscribed to the triangle′ ′ ′ . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴 22. It is sufficient to prove that the medians of the triangle are perpendicular to the lines of the centers of the circles circumscribed to the 𝐴𝐴𝐴𝐴𝐴𝐴 respective triangles. We prove that (the median din ) is perpendicular to ( is the center of the circle circumscribed to the triangle , 𝐴𝐴𝐴𝐴 𝐴𝐴 and is the center of the circle circumscribed to the triangle ). 𝑂𝑂1𝑂𝑂2 𝑂𝑂1 𝐴𝐴𝐴𝐴𝐴𝐴1 We denote: = r , = r , ( , ) = , ( , ) = , 𝑂𝑂2 𝐴𝐴𝐴𝐴𝐴𝐴2 = and 1= 1. 2 2 1 1 2 2 1 𝐴𝐴𝑂𝑂 1 𝐴𝐴𝑂𝑂 𝑑𝑑 𝑂𝑂 𝐵𝐵𝐵𝐵 𝑑𝑑 𝑑𝑑 𝑂𝑂 𝐵𝐵𝐵𝐵 𝑑𝑑 1 = . 2 𝐵𝐵𝐴𝐴 𝑥𝑥 𝑀𝑀𝑀𝑀 2 𝑎𝑎 2 2 2 2 = 1 2+ 1 ; 2 1 2 𝐴𝐴𝐴𝐴 ⊥ 𝑂𝑂 𝑂𝑂 ⟺ 𝐴𝐴𝑂𝑂 −2 𝐴𝐴𝑂𝑂 𝑀𝑀𝑂𝑂 − 𝑀𝑀𝑂𝑂 2 2 𝑎𝑎 𝑀𝑀𝑂𝑂1 = 𝑑𝑑1 + �2 − 𝑥𝑥� . 2 2 2 𝑎𝑎 On the2 other2 hand, = r , = r . 𝑀𝑀𝑂𝑂 𝑑𝑑 �2 − 𝑥𝑥� It follows that 2 2 = 2 2 =2 2 . 𝑑𝑑1 1 − 𝑥𝑥 𝑑𝑑2 2 − 𝑥𝑥 2 2 2 2 2 2 𝑀𝑀𝑂𝑂1 − 𝑀𝑀𝑂𝑂2 𝑟𝑟1 − 𝑟𝑟2 𝑂𝑂1𝐴𝐴 − 𝑂𝑂2𝐴𝐴 23. The perpendicular from to , the perpendicular from to and the perpendicular from to intersect in ; in other words, the 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵1 𝐶𝐶𝐶𝐶 triangle is orthological with the triangle , and the orthology 𝐶𝐶1 𝐴𝐴𝐴𝐴 𝐷𝐷 center is the point . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴

𝐷𝐷 24. Solution 1. We denote: = and = , = , = . From = , we get: 𝐵𝐵𝐵𝐵 𝑎𝑎 𝐵𝐵𝐴𝐴1 𝑥𝑥 𝐶𝐶𝐵𝐵1 𝑦𝑦 𝐴𝐴𝐶𝐶1 𝑧𝑧 ( ) + ( ) = ( ) + ( ) , 𝐴𝐴1𝐵𝐵1 𝐵𝐵1𝐶𝐶1 equivalent to2 : 2 2 2 𝑎𝑎 − 𝑥𝑥 𝑦𝑦 − 𝑎𝑎 − 𝑥𝑥 𝑦𝑦 𝑎𝑎 − 𝑦𝑦 𝑧𝑧 − 𝑎𝑎 − 𝑦𝑦 ∙ 𝑧𝑧

276 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

( )( + + ) = ( ). (1) Analogously, from = , it follows that: 𝑥𝑥 − 𝑧𝑧 𝑥𝑥 𝑦𝑦 𝑧𝑧 − 𝑎𝑎 𝑎𝑎 𝑥𝑥 − 𝑦𝑦 ( )( + + ) = ( ). (2) 𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐶𝐶1 The triangles and being orthological, we have that: 𝑦𝑦 − 𝑥𝑥 𝑥𝑥 𝑦𝑦 𝑧𝑧 − 𝑎𝑎 𝑎𝑎 𝑦𝑦 − 𝑧𝑧 ( ) + ( ) + ( ) = 0 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 2 22 +2 2 + 2 2 = 32 2 𝑥𝑥 − 𝑎𝑎 − 𝑥𝑥 𝑦𝑦 − 𝑎𝑎 − 𝑦𝑦 𝑧𝑧 − 𝑎𝑎 − 𝑧𝑧 ⟺ + + = . 2 (3) 3 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎 ⟺ Substituting + + from (3) in (1) and (2), we obtain that = = 𝑥𝑥 𝑦𝑦 𝑧𝑧 2 𝑎𝑎 = , which shows that is the median triangle of the triangle . 𝑎𝑎 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑥𝑥 𝑦𝑦 Solution 2. Keeping the previous notations, we prove that = = , 𝑧𝑧 2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 showing that . 𝑥𝑥 𝑦𝑦 𝑧𝑧 Indeed, if we denote = , then = 120 , and ∆𝐴𝐴𝐵𝐵1𝐶𝐶1 ≡ ∆𝐵𝐵𝐶𝐶1𝐴𝐴1 ≡ ∆𝐶𝐶𝐴𝐴1𝐵𝐵1 since = 60 , we obtain that: = , therefore 0 = 𝑚𝑚𝐴𝐴�𝐵𝐵1𝐶𝐶1 𝛼𝛼 𝑚𝑚𝐴𝐴�𝐶𝐶1𝐵𝐵1 − 𝛼𝛼 120 . Analogously0 , we get = and = 120 . 𝑚𝑚𝐴𝐴1�𝐶𝐶1𝐵𝐵1 𝑚𝑚𝐵𝐵�𝐶𝐶1𝐴𝐴1 𝛼𝛼 𝑚𝑚𝐵𝐵�𝐴𝐴1𝐶𝐶1 The0 congruence of the triangles derives now from A.S.A. Then0 the − 𝛼𝛼 𝑚𝑚𝐵𝐵�1𝐴𝐴1𝐶𝐶 𝛼𝛼 𝑚𝑚𝐴𝐴�1𝐵𝐵1𝐶𝐶 − 𝛼𝛼 continuation is as in Solution 1.

25. If the triangle is isosceles, for example = , we prove that and are orthological, showing that the relation: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ′ ′ ′ + + = 0 (1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 is true′ . 2 ′ 2 ′ 2 ′ 2 ′ 2 ′ 2 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 Because = , it remains to show that + = 0. ′ and ′ are bisectors and = ′ ; 2we obtain′ 2 without′ 2 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − difficulty′ 2 that ′ = ′ and = , therefore (1) is verified. 𝐶𝐶 𝐵𝐵 𝐵𝐵𝐵𝐵 ′ 𝐶𝐶𝐶𝐶 ′ ′ ′ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 Observation𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 In this hypothesis, we can also prove the concurrency of perpendiculars taken in , , to , , in this way: We′ ′denote′ by the intersection of the perpendicular in to with . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶1 𝐴𝐴𝐴𝐴 From the congruence of triangles and , it follows that′ = 90 ′, 𝑂𝑂 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 therefore also the perpendicular in ′to pass′es through . ′ 0 𝐴𝐴𝐵𝐵 𝑂𝑂 𝐴𝐴𝐴𝐴 𝑂𝑂 ∢𝐴𝐴𝐶𝐶 𝑂𝑂1 Reciprocally, if the triangles ′ and are orthological, the relation (1) 𝐶𝐶 𝐴𝐴𝐴𝐴 𝑂𝑂1 takes place, and let us prove that the triangle ′ ′ ′ is isosceles. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 277 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

Using the bisector theorem, we find: = , = , = , = ′ 𝑎𝑎𝑎𝑎 ′ 𝑎𝑎𝑎𝑎 ′ 𝑎𝑎𝑎𝑎 ′ , = , = . 𝐴𝐴 𝐵𝐵 𝑏𝑏+𝑐𝑐 𝐴𝐴 𝐶𝐶 𝑏𝑏+𝑐𝑐 𝐵𝐵 𝐶𝐶 𝑎𝑎+𝑐𝑐 𝐵𝐵 𝐴𝐴 𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎 We′ get: ′ 𝑎𝑎+𝑐𝑐 𝐶𝐶 𝐴𝐴 𝑎𝑎+𝑏𝑏 𝐶𝐶 𝐵𝐵 𝑎𝑎+𝑏𝑏 ( ) ( ) ( ) + + = 0 2 2 2 . 𝑎𝑎 𝑐𝑐−𝑏𝑏 𝑏𝑏 𝑎𝑎−𝑐𝑐 𝑐𝑐 𝑏𝑏−𝑎𝑎 By𝑏𝑏+ doing𝑐𝑐 the𝑎𝑎+ calculations,𝑐𝑐 𝑏𝑏+𝑎𝑎 we obtain: ( )( )( )( + + ) = 0, from where it follows that = sau 2= sau = , therefore the triangle 𝑎𝑎 − 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 is isosceles. 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑐𝑐 𝑎𝑎 𝐴𝐴𝐴𝐴𝐴𝐴

26. a) The condition of concurrency of lines , , is equivalent to: 𝑀𝑀𝐴𝐴1 𝑀𝑀𝐵𝐵1 𝑀𝑀𝐶𝐶1 + + = 0, or: 2( 2 )( 2 + 2 ) + ( 2 2 )( + ) 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 + ( )( + ) = 0. 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐴𝐴1𝐵𝐵 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐵𝐵1𝐶𝐶 𝐵𝐵1𝐴𝐴 We get: 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐶𝐶1𝐴𝐴 𝐶𝐶1𝐵𝐵 + + = + + = . 3𝑎𝑎 b) We denote: = , = , = . The condition of 𝐴𝐴1𝐵𝐵 𝐵𝐵1𝐶𝐶 𝐶𝐶1𝐴𝐴 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐴𝐴 𝐶𝐶1𝐵𝐵 2 concurrency of lines 1 , , 1 is equivalent1 to = 1 or: 𝐴𝐴 𝐵𝐵 𝑥𝑥 𝐵𝐵 𝐶𝐶 𝑦𝑦 𝐶𝐶 𝐴𝐴 𝑧𝑧𝑎𝑎−𝑥𝑥 𝑎𝑎−𝑦𝑦 𝑎𝑎−𝑧𝑧 ( + + ) 1+ ( 1 + 1 + ) 2 = 0 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑥𝑥 ∙. 𝑦𝑦 ∙ 𝑧𝑧 Because3 : + + 3= , it follows that: 𝑎𝑎 − 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑎𝑎 3𝑎𝑎𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 𝑧𝑧𝑧𝑧 𝑎𝑎 − 𝑥𝑥𝑥𝑥𝑥𝑥 + 2( + + ) 4 = 0. (1) 𝑥𝑥 𝑦𝑦 𝑧𝑧 2 On 3the other hand: (−𝑎𝑎 2 )(𝑥𝑥𝑥𝑥 2𝑦𝑦𝑦𝑦)( 𝑧𝑧𝑧𝑧2𝑎𝑎 )−= 𝑥𝑥𝑥𝑥𝑥𝑥 2( + + ) + 4( + + ) 8 = 3 + 4( +3 + ) 8 2 = 2 + 𝑎𝑎 − 𝑥𝑥 𝑎𝑎 − 𝑦𝑦 𝑎𝑎 − 𝑧𝑧 𝑎𝑎 − 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝑎𝑎 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 4( + + ) 3 8 3 . 3 (2) 𝑧𝑧𝑧𝑧 𝑎𝑎 − 𝑥𝑥𝑥𝑥𝑥𝑥 𝑎𝑎 − 𝑎𝑎 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 𝑧𝑧𝑧𝑧 𝑎𝑎 − 𝑥𝑥𝑥𝑥𝑥𝑥 − 𝑎𝑎 From relations (1) and (2) we have that: 𝑥𝑥𝑥𝑥 𝑦𝑦𝑦𝑦 𝑧𝑧𝑧𝑧 𝑎𝑎 − 𝑥𝑥𝑥𝑥𝑥𝑥 ( 2 )( 2 )( 2 ) = 0, from where = or = or = , therefore is on one of the altitudes. 𝑎𝑎 − 𝑥𝑥 𝑎𝑎 −𝑎𝑎 𝑦𝑦 𝑎𝑎 −𝑎𝑎 𝑧𝑧 𝑎𝑎

𝑥𝑥 2 𝑦𝑦 2 𝑧𝑧 2 𝑀𝑀 27. Let be the Lemoine point in the triangle ; it is known that: = ; = ; = . ′ 2 ′𝐾𝐾 2 ′ 2 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝑐𝑐 𝐵𝐵 𝐶𝐶 𝑎𝑎 𝐶𝐶 𝐴𝐴 𝑏𝑏 ′ 2 ′ 2 ′ 2 𝐴𝐴 𝐶𝐶 We𝑏𝑏 obtain𝐵𝐵 𝐴𝐴 : 𝑐𝑐 𝐶𝐶 𝐵𝐵 𝑎𝑎 = = = = 2 ; 2 ; 2 ; 2 ; ′ 𝑎𝑎𝑐𝑐 ′ 𝑎𝑎𝑏𝑏 ′ 𝑏𝑏𝑎𝑎 ′ 𝑏𝑏𝑐𝑐 2 2 2 2 2 2 2 2 𝐴𝐴 𝐵𝐵 𝑏𝑏 +𝑐𝑐 𝐴𝐴 𝐶𝐶 𝑏𝑏 +𝑐𝑐 𝐵𝐵 𝐶𝐶 𝑎𝑎 +𝑐𝑐 𝐵𝐵 𝐴𝐴 𝑎𝑎 +𝑐𝑐 278 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= = 2 ; 2 . ′ 𝑐𝑐𝑏𝑏 ′ 𝑐𝑐𝑎𝑎 The triangles2 2 and2 2 are orthological if and only if: 𝐶𝐶 𝐴𝐴 𝑎𝑎 +𝑏𝑏 𝐶𝐶 𝐵𝐵 𝑎𝑎 +𝑏𝑏 + ′ ′ ′+ = 0. (1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 If′ we2 suppose′ 2 that′ the2 triangle′ 2 ′ 2is isosceles′ 2 , = , then is 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 median, therefore = ; also, we find that = and ′ = 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 , and the relation′ (1) is satisfied′ . ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ If the relation (1) takes place, then we obtain that: 𝐶𝐶 𝐵𝐵 ( ) ( ) ( ) + + = 0 2 2 2 2 2 2 2 2 2 . 𝑎𝑎 𝑐𝑐 −𝑏𝑏 𝑏𝑏 𝑎𝑎 −𝑐𝑐 𝑐𝑐 𝑏𝑏 −𝑎𝑎 2 2 2 2 2 2 This𝑏𝑏 + relation𝑐𝑐 is𝑎𝑎 equivalent+𝑐𝑐 𝑎𝑎 to+:𝑏𝑏 ( )( + )( + ) + ( )( + )( + ) 2 2 2 2 + 2 ( 2 2)( +2 2)( 2+ 2) = 02. 2 2 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑎𝑎 − 𝑐𝑐 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 ( )( +2 2)( 2+ 2) 2 2 2 𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝑎𝑎 𝑐𝑐 𝑏𝑏 𝑐𝑐 2 2 2 2 + (2 +2 )[2 ( + ) 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 + (2 2 2 +4 2 2 2)]2= 0. 2 2 𝑏𝑏 𝑐𝑐 𝑏𝑏 𝑎𝑎 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝑐𝑐 ( )( + 2 )(2 2+ )4 2 2 2 2 𝑐𝑐 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 𝑐𝑐 − 𝑎𝑎 𝑐𝑐 2 2 2 2 + (2 +2 )[2 ( ) ( ) 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 + 2 ( 2 )4] =2 0 2 2 4 4 𝑏𝑏 𝑐𝑐 −𝑎𝑎 𝑐𝑐 − 𝑏𝑏 − 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 ( ){ ( + )2( 2 +2 ) 2 𝑏𝑏 𝑐𝑐 𝑐𝑐 − 𝑏𝑏 2 2 2 2+ ( 2 + 2 )[ 2 + ]} = 0 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 ( )[ + 2 +2 4 2 2 2 2 2 2 + 𝑏𝑏 𝑐𝑐 −𝑎𝑎 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑐𝑐 𝑏𝑏 𝑐𝑐 2 2 6 4 2 4 2 2 2 2 +4 2 ] =2 04 2 2 2 4 2 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 − 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑏𝑏 𝑐𝑐 We obtain that: 4 2 2 2 2 2 4 2 4 − 𝑎𝑎 𝑐𝑐 − 𝑎𝑎 𝑏𝑏 𝑐𝑐 − 𝑎𝑎 𝑐𝑐 𝑏𝑏 𝑐𝑐 ( )( )( )( + + ) = 0, therefore: ( 2 )(2 2 )( 2 )2( +2 )( 2 + )2( +2 )( + + ) = 0. 𝑎𝑎 − 𝑏𝑏 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 − 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 Hence, = sau = or = , consequently2 the 2triangle2 is 𝑎𝑎 − 𝑏𝑏 𝑐𝑐 − 𝑏𝑏 𝑎𝑎 − 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 isosceles. 𝑎𝑎 𝑏𝑏 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 28. a) The given condition is equivalent to the orthology of triangles. b) We take through the point the parallels , , with , respectively . Obviously, the triangles , and are 𝑃𝑃 𝑀𝑀𝑀𝑀 𝑅𝑅𝑅𝑅 𝑈𝑈𝑈𝑈 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 equilateral. We have: 𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 = + ; 1 𝑃𝑃��𝐴𝐴��1⃗ = 2 �𝑃𝑃��𝑃𝑃�⃗ + 𝑃𝑃��𝑃𝑃�⃗�; 1 𝑃𝑃��𝐵𝐵��1⃗ 2 �𝑃𝑃��𝑃𝑃��⃗ 𝑃𝑃��𝑃𝑃�⃗� Ion Pătrașcu, Florentin Smarandache 279 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= + . 1 𝑃𝑃��𝐶𝐶��1⃗ + 2 �𝑃𝑃��𝑃𝑃⃗+ 𝑃𝑃��𝑃𝑃��⃗=� + + . 1 We took into account that the quadrilaterals ; ; are 𝑃𝑃��𝐴𝐴��1⃗ 𝑃𝑃��𝐵𝐵��1⃗ 𝑃𝑃��𝐶𝐶��1⃗ 2 �𝑃𝑃��𝑃𝑃�⃗ 𝑃𝑃��𝑃𝑃�⃗ �𝑃𝑃��𝑃𝑃�⃗� parallelograms. For any point from the plane of the triangle the relation 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 + + = 3 (where is the gravity center) takes place. In 𝑃𝑃 our case��⃗, �=��⃗0, because��⃗ ��⃗ is equilateral, hence + + = . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 𝐺𝐺 3

𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃��𝑃𝑃�⃗ 𝑃𝑃��𝑃𝑃�⃗ �𝑃𝑃��𝑃𝑃�⃗ 2 𝑃𝑃��𝑃𝑃�⃗ 29. Let be the podal triangle of . We have ( ) = ( ) = ( ) = 120 . If, on the semi-lines ( , ( , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑀𝑀 𝑚𝑚 ∢𝐴𝐴1𝑀𝑀𝐵𝐵1 ( we consider the points , , 0 , such as = = , then 𝑚𝑚 ∢𝐵𝐵1𝑀𝑀𝐶𝐶1 𝑚𝑚 ∢𝐶𝐶1𝑀𝑀𝐴𝐴1 𝑀𝑀𝐴𝐴1 𝑀𝑀𝐵𝐵1 and are orthological′ , and′ the′ orthology center′ is′ the point′ . 𝑀𝑀𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑀𝑀𝐴𝐴 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 ′ ′ ′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 30. The perpendiculars taken from to , , meet respectively , , in , , . The triangles and are 𝐼𝐼 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 orthological′ ′ ′ ′ ( ′is ′one of their orthology centers). The lines , , 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝑇𝑇1 𝑇𝑇2 𝑇𝑇3 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇1𝑇𝑇2𝑇𝑇3 are parallel with the sides of the triangle , therefore the 𝐼𝐼 𝐴𝐴𝑏𝑏𝐴𝐴𝑐𝑐 𝐵𝐵𝑐𝑐𝐵𝐵𝑎𝑎 perpendiculars taken to them from , , respectively are concurrent in 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏 𝑇𝑇1𝑇𝑇2𝑇𝑇3 the second orthology center of the triangles and . 𝐴𝐴 𝐵𝐵 𝐶𝐶

𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇1𝑇𝑇2𝑇𝑇3 31. We observe that the triangle is the podal triangle of the point in relation to the triangle , consequently the triangles 𝐵𝐵1𝐵𝐵2𝐵𝐵3 and are orthological, their orthology center being . Also, the 𝐵𝐵1 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐵𝐵1𝐵𝐵2𝐵𝐵3 triangle is orthological with the triangle , their orthology 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐵𝐵1 center being the point . Let us prove now that the perpendicular from 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐵𝐵1𝐵𝐵2𝐵𝐵3 to , the perpendicular from to , and the perpendicular from 𝐴𝐴1 𝐴𝐴1 to are concurrent. We observe that the first two perpendiculars are 𝐵𝐵3𝐵𝐵1 𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐴𝐴3 concurrent in ; we prove that the perpendicular from to also 𝐵𝐵2𝐵𝐵3 passes through ; basically, we prove that . 𝐴𝐴2 𝐴𝐴3 𝐵𝐵2𝐵𝐵3 The quadrilateral is a rectangle; it follows that 𝐴𝐴2 𝐵𝐵2𝐵𝐵3 ⊥ 𝐴𝐴2𝐴𝐴3 . From , we obtain that 𝐵𝐵1𝐵𝐵2𝐴𝐴1𝐵𝐵3 ∢𝐵𝐵1𝐵𝐵2𝐵𝐵3 ≡ . ∢𝐵𝐵1𝐴𝐴1𝐵𝐵3 ∢𝐴𝐴1𝐴𝐴3𝐴𝐴2 ≡ ∢𝐴𝐴2𝐴𝐴1𝐷𝐷 ∢𝐵𝐵1𝐵𝐵2𝐵𝐵3 ≡ These angles have ; it follows that ; since ∢𝐴𝐴2𝐴𝐴1𝐷𝐷 , it follows that . 𝐵𝐵1𝐵𝐵2 ∥ 𝐴𝐴2𝐴𝐴1 𝐵𝐵2𝐵𝐵3 ∥ 𝐴𝐴1𝐷𝐷 𝐴𝐴1𝐷𝐷 ⊥ 𝐴𝐴2𝐴𝐴3 𝐵𝐵2𝐵𝐵3 ⊥ 𝐴𝐴2𝐴𝐴3

280 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

We obtained that the second orthology center of the triangles and is . It follows that the triangle is two times 𝐴𝐴1𝐴𝐴2𝐴𝐴3 orthological with the triangle . We show that the second orthology 𝐵𝐵1𝐵𝐵2𝐵𝐵3 𝐴𝐴2 𝐵𝐵1𝐵𝐵2𝐵𝐵3 center is . Indeed, the perpendicular from to is , therefore 𝐴𝐴1𝐴𝐴2𝐴𝐴3 it passes through . The perpendicular from to passes through , 𝐵𝐵2 𝐵𝐵1 𝐴𝐴3𝐴𝐴1 𝐵𝐵1𝐵𝐵2 and we have previously shown that the perpendicular from to is 𝐵𝐵2 𝐵𝐵2 𝐴𝐴1𝐴𝐴2 𝐵𝐵2 . It is not difficult to show that the perpendicular from to , the 𝐵𝐵3 𝐴𝐴2𝐴𝐴3 perpendicular from to and the perpendicular from to are 𝐵𝐵2𝐵𝐵3 𝐴𝐴1 𝐵𝐵1𝐵𝐵2 concurrent in , therefore is the third orthology center of the triangle 𝐴𝐴2 𝐵𝐵2𝐵𝐵3 𝐴𝐴3 𝐵𝐵3𝐵𝐵1 in relation to . The third orthology center of the triangle 𝐴𝐴3 𝐴𝐴3 in relation to is . 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐵𝐵1𝐵𝐵2𝐵𝐵3

𝐵𝐵1𝐵𝐵2𝐵𝐵3 𝐴𝐴1𝐴𝐴2𝐴𝐴3 𝐵𝐵3 32. We show that the triangle is orthological with . We denote by the center of the circle circumscribed to the triangle ; 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 obviously the′ mediators of the segments , , pass through′ ′ ′. 𝑂𝑂 𝐴𝐴 𝐵𝐵 𝐶𝐶 The perpendicular in to passes through′ ′, the symmetric′ of with′ 𝐴𝐴 𝐴𝐴1 𝐵𝐵 𝐵𝐵1 𝐶𝐶 𝐶𝐶1 𝑂𝑂 respect to . (Indeed, the quadrilateral ′ is rectangular trapeze and 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝑃𝑃 𝑃𝑃 the mediator′ of contains the midline of′ this′ trapeze.) Analogously, it 𝑂𝑂 𝐴𝐴1𝐴𝐴 𝑃𝑃𝑃𝑃 follows that belongs′ also to the perpendiculars in to and in to 𝐴𝐴1𝐴𝐴 . Consequently′ , the triangles and are orthological. 𝑃𝑃 𝐵𝐵1 𝐴𝐴𝐴𝐴 𝐶𝐶1

𝐴𝐴𝐴𝐴1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 33. We show that: + + = 0. We2 calculate2 and2 2as medians2 in 2the triangles and . 𝛼𝛼𝐵𝐵 − 𝛼𝛼𝐶𝐶 𝛽𝛽𝐶𝐶 − 𝛽𝛽𝐴𝐴 𝛾𝛾𝐴𝐴 − 𝛾𝛾𝐵𝐵 It is easy to note the congruence of triangles and , since 𝛼𝛼𝛼𝛼 𝛼𝛼𝛼𝛼 𝐵𝐵𝐵𝐵1𝐶𝐶1 𝐶𝐶𝐶𝐶1𝐵𝐵1 = . We have: 𝐵𝐵𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶𝐶𝐶1 = [2( + ) ], 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 2 1 2 2 2 𝛼𝛼𝐵𝐵 = 4 [2(𝑐𝑐 + 𝐵𝐵𝐵𝐵1 ) − 𝐵𝐵1𝐶𝐶1 ]. 2 1 2 2 2 𝛼𝛼Therefore𝐶𝐶 4 𝑏𝑏 𝐶𝐶𝐶𝐶1 =− 𝐵𝐵(1𝐶𝐶1 ). 1 Analogously, 2we find2: 2 2 𝛼𝛼𝐵𝐵 − 𝛼𝛼𝐶𝐶 2 𝑐𝑐 − 𝑏𝑏 = ( ), 2 2 1 2 2 𝛽𝛽𝐶𝐶 − 𝛽𝛽𝐴𝐴 = 2( 𝑎𝑎 − 𝑐𝑐 ). 1 2 2 2 2 𝛾𝛾𝐴𝐴 − 𝛾𝛾𝐵𝐵 2 𝑏𝑏 − 𝑎𝑎

Ion Pătrașcu, Florentin Smarandache 281 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

34. The quadrilaterals ; ; are isosceles trapezoids. We have = ; = ′ etc. We′ show′ that is orthological 𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 with ; we′ calculate′: ′ ′ ′ 𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 + + ; 𝐴𝐴𝐴𝐴𝐴𝐴 it is ′ obtained2 ′ 2zero, ′ therefore2 ′ 2 ′ 2 is orthological′ 2 with . The 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐶𝐶 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 − 𝐶𝐶 𝐵𝐵 orthology center of the triangle ′ ′in ′relation to is . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 35. Let be the orthocenter of the triangle and – the intersection of semi-line ( with the circumscribed circle of the triangle 𝐻𝐻1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑋𝑋1 . We have: = (1) (the power of the point 𝐴𝐴𝐻𝐻1 over the circle ). We denote by the intersection of semi-line ( 𝐴𝐴𝐴𝐴1𝐴𝐴2 𝐴𝐴𝐻𝐻1 ∙ 𝐻𝐻1𝑋𝑋1 𝐴𝐴1𝐻𝐻1 ∙ 𝐻𝐻1𝐴𝐴2 𝐻𝐻1 with the circle . We have: = (2). 𝐴𝐴𝐴𝐴1𝐴𝐴2 𝑋𝑋2 𝐴𝐴𝐻𝐻1 Analogously, if is the intersection of semi-line ( with the circle 𝐴𝐴𝐴𝐴1𝐵𝐵2 𝐴𝐴𝐻𝐻1 ∙ 𝐻𝐻1𝑋𝑋2 𝐵𝐵1𝐻𝐻1 ∙ 𝐻𝐻1𝐵𝐵2 , we have: = (3). From relations (1), (2), (3), 𝑋𝑋3 𝐴𝐴𝐻𝐻1 because of the fact that = = (4), it 𝐴𝐴𝐴𝐴1𝐶𝐶2 𝐴𝐴𝐻𝐻1 ∙ 𝐻𝐻1𝑋𝑋3 𝐶𝐶1𝐻𝐻1 ∙ 𝐻𝐻1𝐶𝐶2 follows that = = , consequently the circles still have a common 𝐴𝐴1𝐻𝐻1 ∙ 𝐻𝐻1𝐴𝐴2 𝐵𝐵1𝐻𝐻1 ∙ 𝐻𝐻1𝐵𝐵2 𝐶𝐶1𝐻𝐻1 ∙ 𝐻𝐻1𝐶𝐶2 point. The relation1 (4)2 derives3 from the property of the symmetrics of an orthocenter of𝑋𝑋 a triangle𝑋𝑋 𝑋𝑋 to be on its circumscribed circle, and from the power of orthocenter over the circumscribed circle of the triangle. The reasoning is the same in the case of the obtuse triangle.

36. Let be the Lemoine point of the triangle and , , – the projections of on , respectively . We use the following result: 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 “the Lemoine point of the triangle is the gravity center of its podal 𝐾𝐾 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 triangle and reciprocally”. 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 We know about that it is the gravity center of the triangle and of the triangle ; we have: 𝐾𝐾 𝐴𝐴1𝐵𝐵1𝐶𝐶1 + + = 0, 𝑀𝑀𝑀𝑀𝑀𝑀 + + = 0. 𝐾𝐾��𝐴𝐴��1⃗ 𝐾𝐾��𝐵𝐵��1⃗ 𝐾𝐾��𝐶𝐶��1⃗ �⃗ Because = + ; = + and = + , 𝐾𝐾��𝐾𝐾�⃗ �𝐾𝐾��𝐾𝐾��⃗ 𝐾𝐾��𝐾𝐾�⃗ �⃗ we obtain that: + + = 0. �𝐾𝐾��𝐾𝐾�⃗ 𝐾𝐾��𝐴𝐴��1⃗ 𝐴𝐴��1��𝑃𝑃�⃗ 𝐾𝐾��𝐾𝐾��⃗ �𝐾𝐾��𝐵𝐵��1⃗ 𝐵𝐵��1��𝑀𝑀��⃗ 𝐾𝐾��𝐾𝐾�⃗ 𝐾𝐾��𝐶𝐶��1⃗ 𝐶𝐶��1��𝑁𝑁�⃗ If we denote by , , the projections of on the sides of the �𝐴𝐴��1��𝑃𝑃�⃗ 𝐵𝐵��1��𝑀𝑀��⃗ 𝐶𝐶��1��𝑁𝑁�⃗ triangle , it follows′ that′ the′ relation + + = 0, which 𝑀𝑀 𝑁𝑁 𝑃𝑃 𝐾𝐾 expresses that the point is the gravity center ′of its podal′ triangle′ ��⃗ ��⃗ ��⃗ 𝑀𝑀𝑀𝑀𝑀𝑀 𝐾𝐾𝑃𝑃 𝐾𝐾𝑀𝑀 𝐾𝐾𝑁𝑁 ′ ′ ′ 𝐾𝐾 𝑀𝑀 𝑁𝑁 𝑃𝑃

282 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS in relation to the triangle , hence is the symmedian center of the triangle . 𝑀𝑀𝑀𝑀𝑀𝑀 𝐾𝐾 Because in the triangle the symmedian center coincides with the 𝑀𝑀𝑀𝑀𝑀𝑀 gravity center, we obtain that this triangle is equilateral. On the other hand, 𝑀𝑀𝑀𝑀𝑀𝑀 the triangle built as in the statement is similar with the triangle . The triangle being equilateral, it follows that is also equilateral. 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴

𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 37. Let { } = ; since is middle line in , it follows that , also . Since , it follows that is the 𝐻𝐻 𝐸𝐸𝐸𝐸 ∩ 𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 orthocenter of the triangle , hence . The quadrilateral 𝑁𝑁𝑁𝑁 ∥ 𝐵𝐵𝐵𝐵 𝐸𝐸𝐸𝐸 ⊥ 𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁 ⊥ 𝐸𝐸𝐸𝐸 𝐻𝐻 is parallelogram of center , hence ; and since , 𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁 ⊥ 𝐸𝐸𝐸𝐸 it follows that . The perpendicular taken from to is , the 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑂𝑂 𝑃𝑃𝑃𝑃 ∥ 𝐻𝐻𝐻𝐻 𝐷𝐷𝐷𝐷 ⊥ 𝑁𝑁𝑁𝑁 perpendicular taken from to is , and the perpendicular taken from 𝑀𝑀𝑀𝑀 ⊥ 𝐷𝐷𝐷𝐷 𝑁𝑁 𝐿𝐿𝐿𝐿 𝑁𝑁𝑁𝑁 to is ; these perpendiculars are altitudes in the triangle , and 𝐸𝐸 𝐷𝐷𝐷𝐷 𝐸𝐸𝐸𝐸 are concurrent in . The orthology center of the triangle in relation 𝑀𝑀 𝐿𝐿𝐿𝐿 𝑀𝑀𝑀𝑀 𝑁𝑁𝑁𝑁𝑁𝑁 to is – the orthocenter of the triangle , and the orthology center 𝐻𝐻 𝑁𝑁𝑁𝑁𝑁𝑁 of the triangle in relation to the triangle is the orthocenter of the 𝐷𝐷𝐷𝐷𝐷𝐷 𝐻𝐻 𝑁𝑁𝑁𝑁𝑁𝑁 triangle . 𝐷𝐷𝐷𝐷𝐷𝐷 𝑁𝑁𝑁𝑁𝑁𝑁

𝐷𝐷𝐷𝐷𝐷𝐷 38. We observe that the triangles and are orthological. Indeed, the perpendicular from to , the perpendicular from to , and the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 perpendicular from to are concurrent in . The point is the 𝐴𝐴 𝐹𝐹𝐹𝐹 𝐶𝐶 𝐴𝐴𝐴𝐴 orthology center of the triangle in relation to . According to the 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐸𝐸 𝐸𝐸 theorem of orthological triangles, we have that the triangle is also 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 orthological in relation to , hence the perpendicular from to , the 𝐴𝐴𝐴𝐴𝐴𝐴 perpendicular from to , the perpendicular from to are concurrent. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐶𝐶𝐶𝐶 Because the perpendicular from to and the perpendicular from to 𝐹𝐹 𝐴𝐴𝐴𝐴 𝐷𝐷 𝐴𝐴𝐴𝐴 are concurrent in , it follows that the perpendicular from to must 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐹𝐹 also pass through ; and since is diameter in the circle, it follows that 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐷𝐷 𝐴𝐴𝐴𝐴 must be the symmetric of with respect to , hence = . 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐵𝐵

𝐷𝐷 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 39. From , – the midpoint of , and { } = , we obtain that is the midpoint of . Also, = = , 𝑄𝑄𝑄𝑄 ∥ 𝐴𝐴𝐴𝐴 𝑀𝑀 𝐴𝐴𝐴𝐴 1 𝑁𝑁 𝑄𝑄𝑄𝑄 ∩ 𝐶𝐶𝐶𝐶1 therefore 𝑁𝑁 = , videlicet 𝑄𝑄 𝑄𝑄is the gravity𝐴𝐴𝐴𝐴 center4 ∙ 𝐴𝐴𝐴𝐴 of⟹ the𝑀𝑀 triangle𝑀𝑀 4 ∙ 𝐶𝐶𝐶𝐶 1 (isosceles rectangle), consequently . The perpendicular from to 𝑃𝑃𝑃𝑃 2 ∙ 𝐶𝐶𝐶𝐶 𝑃𝑃 𝐶𝐶𝐶𝐶𝐶𝐶 𝑄𝑄𝑄𝑄 ⊥ 𝐶𝐶𝐶𝐶 𝑅𝑅 Ion Pătrașcu, Florentin Smarandache 283 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

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is , and the perpendicular from to is . From the above, it follows that the triangle is orthological with the triangle , the 𝐴𝐴𝐴𝐴 𝑅𝑅𝑅𝑅 𝑄𝑄 𝐴𝐴𝐴𝐴 𝑄𝑄𝑄𝑄 orthology center being the point . The theorem of orthological triangles 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 shows that the triangle is also orthological in relation to , the 𝑄𝑄 orthology center being the vertex . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃𝑃𝑃

𝐶𝐶 40. We denote = = , = , = , = , 𝑎𝑎 = . From Ceva’s theorem, we find: 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 2 𝐵𝐵1𝐶𝐶 𝑦𝑦 𝐵𝐵1𝐴𝐴 𝑏𝑏 − 𝑦𝑦 𝐶𝐶1𝐴𝐴 𝑧𝑧 1 = ( ). (1) 𝐶𝐶 𝐵𝐵 𝑐𝑐𝑏𝑏− 𝑧𝑧 The triangle being orthological in relation to , we have: 𝑦𝑦 𝑐𝑐 𝑐𝑐 − 𝑧𝑧 + + = 0. 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 We obtain2 : 2 2 2 2 2 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 2 + 2 = 0. (2) Replacing2 in (2), it2 follows that: 𝑏𝑏𝑏𝑏 − 𝑏𝑏 𝑐𝑐𝑐𝑐 − 𝑐𝑐 2 ( ) ( ) = 0 ( )( + )(2 ) = 0. 𝑦𝑦 If =2 , then2 is2 median2 , and as well. If = , then the triangle 𝑧𝑧 𝑐𝑐 −𝑐𝑐 𝑏𝑏 − 𝑐𝑐 𝑐𝑐 − 𝑏𝑏 ⟺ 𝑐𝑐 − 𝑏𝑏 𝑐𝑐 𝑏𝑏 𝑧𝑧 − 𝑐𝑐 is isosceles. 𝑧𝑧 2 𝐶𝐶𝐶𝐶1 𝐵𝐵𝐵𝐵1 𝑏𝑏 𝑐𝑐

𝐴𝐴𝐴𝐴𝐴𝐴41. a) Assuming the problem solved, given conditions, we find that , , hence ~ . We firstly build a triangle with ( ), ( ), ( ) and ~ . We ∢𝐴𝐴1 ≡ ∢𝐵𝐵 ∢𝐵𝐵1 ≡ ∢𝐶𝐶 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∆𝐵𝐵𝐵𝐵𝐵𝐵 fix′ ′ ′ ( ) and′ we build′ ( ), ′such that ′ ′ . We′ then build 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵 ∈ 𝐶𝐶𝐶𝐶 𝐶𝐶 ∈ 𝐴𝐴𝐴𝐴 ∆𝐴𝐴 𝐵𝐵 𝐶𝐶 ∆𝐵𝐵𝐵𝐵𝐵𝐵 the perpendicula′ r in to ′. We build on this perpendicular′ ′ the point , 𝐴𝐴 ∈ 𝐵𝐵𝐵𝐵 𝐵𝐵 ∈ 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 ⊥ 𝐵𝐵𝐵𝐵 such that ′ . Now we draw the line and we denote by ′ 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐶𝐶 its intersection′ ′with′ ( ). We build with′ ( ); we build ∢𝐶𝐶 𝐴𝐴 𝐵𝐵 ≡ ∢𝐵𝐵 𝐶𝐶𝐶𝐶 𝐶𝐶1 with ( ). The triangle is the required triangle. 𝐴𝐴𝐴𝐴 𝐶𝐶1𝐵𝐵1 ⊥ 𝐴𝐴𝐴𝐴 𝐵𝐵1 ∈ 𝐴𝐴𝐴𝐴 The proof of the construction results from the fact that is similar 𝐵𝐵1𝐴𝐴1 ⊥ 𝐵𝐵𝐵𝐵 𝐴𝐴1 ∈ 𝐵𝐵𝐵𝐵 𝐴𝐴1𝐵𝐵1𝐶𝐶1 with ( = and ). The triangle ′and′ the′ triangle 𝐴𝐴 𝐵𝐵 𝐶𝐶 are homot′ hetic, therefore′ ~ . ′Also′ ′, it follows that 𝐵𝐵𝐵𝐵𝐵𝐵 ∢𝐵𝐵 ∢𝐶𝐶 ∢𝐴𝐴 ≡ ∢𝐵𝐵 𝐴𝐴 𝐵𝐵 𝐶𝐶 . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∆𝐴𝐴1𝐵𝐵1𝐶𝐶1 ∆𝐵𝐵𝐵𝐵𝐵𝐵 b) The perpendiculars taken from , , respectively to , 𝐴𝐴1𝐶𝐶1 ⊥ 𝐴𝐴𝐴𝐴 and are mediators of the triangle . 𝑂𝑂1 𝑂𝑂2 𝑂𝑂3 𝐴𝐴1𝐶𝐶1 𝐴𝐴1𝐵𝐵1

𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1

284 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

42. The perpendiculars raised in , , respectively to , and are the radical axes of the pairs of circles: ( ), ( ) and 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂2𝑂𝑂3 𝑂𝑂3𝑂𝑂1 ( ), ( ) respectively ( ), ( ) . As it is known, these radical 𝑂𝑂1𝑂𝑂2 �𝒞𝒞 𝑂𝑂2 𝒞𝒞 𝑂𝑂3 � axes are concurrent in the radical center of the given circles. �𝒞𝒞 𝑂𝑂3 𝒞𝒞 𝑂𝑂1 � �𝒞𝒞 𝑂𝑂1 𝒞𝒞 𝑂𝑂2 � Consequently, the triangle and the triangle are orthological, Ω and the orthology center is . The orthology center of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2𝑂𝑂3 in relation to will be – the isogonal conjugate of (considering that Ω 𝑂𝑂1𝑂𝑂2𝑂𝑂3 is the podal triangle of′ in relation to ). 𝐴𝐴𝐴𝐴𝐴𝐴 Ω Ω

𝐴𝐴𝐴𝐴𝐴𝐴 Ω 𝑂𝑂1𝑂𝑂2𝑂𝑂3 43. The triangle is isosceles. Its altitude from is the bisector from of the triangle , hence the perpendicular from to passes 𝐴𝐴𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 𝐴𝐴 through – the center of the inscribed circle. Analogously, the 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 perpendiculars taken from respectively to pass through , 𝐼𝐼 consequently the triangles and are orthological, and the 𝐻𝐻𝑏𝑏 𝐻𝐻𝑐𝑐 𝐶𝐶𝑎𝑎𝐶𝐶𝑐𝑐 𝐼𝐼 orthology center is . The quadrilaterals , and 𝐻𝐻𝑎𝑎𝐻𝐻𝑏𝑏𝐻𝐻𝑐𝑐 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑐𝑐 are rhombuses. In the triangle , the line , where is the midpoint 𝐼𝐼 𝐻𝐻𝑎𝑎𝐶𝐶𝑐𝑐𝐼𝐼𝐶𝐶𝑏𝑏 𝐻𝐻𝑏𝑏𝐶𝐶𝑎𝑎𝐼𝐼𝐶𝐶𝑐𝑐 𝐶𝐶𝑎𝑎𝐻𝐻𝑐𝑐𝐶𝐶𝑏𝑏𝐼𝐼 of and is the midpoint of , is midline, therefore . 𝐼𝐼𝐼𝐼𝑏𝑏𝐻𝐻𝑐𝑐 𝑀𝑀𝑀𝑀 𝑀𝑀 Then, the perpendicular taken from to is perpendicular to ; 𝐶𝐶𝑎𝑎𝐶𝐶𝑐𝑐 𝑁𝑁 𝐶𝐶𝑎𝑎𝐶𝐶𝑏𝑏 𝑀𝑀𝑀𝑀 ∥ 𝐻𝐻𝑏𝑏𝐻𝐻𝑐𝑐 hence it is the altitude from of the contact triangle. The orthology center 𝐶𝐶𝑎𝑎 𝐻𝐻𝑏𝑏𝐻𝐻𝑐𝑐 𝑀𝑀𝑀𝑀 of the triangle in relation to is the orthocenter of the contact 𝐶𝐶𝑎𝑎 triangle. 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐻𝐻 𝐻𝐻 𝐻𝐻 𝐶𝐶 𝐶𝐶 𝐶𝐶 44. Let , , the contacts with , respectively of the -ex- inscribed circle. The triangle is orthological in relation to because 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 the perpendiculars taken in , , to , and are concurrent in the 𝐷𝐷𝐷𝐷𝐷𝐷 𝐴𝐴𝐴𝐴𝐴𝐴 center of the -ex-inscribed circle. Because = (tangents taken 𝐷𝐷 𝐸𝐸 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 from a point exterior to the circle), it means that the perpendicular from 𝐼𝐼𝑎𝑎 𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 to is a bisector in the triangle ; hence it contains the point ; 𝐴𝐴 analogously, the perpendiculars taken from and from to 𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎 respectively pass through . 𝐵𝐵 𝐶𝐶 𝐸𝐸𝐸𝐸

𝐷𝐷𝐷𝐷 𝐼𝐼𝑎𝑎 45. From it follows that = (1). From , it 𝐴𝐴𝐶𝐶1 𝐴𝐴𝐵𝐵1 follows that 𝐶𝐶1𝐵𝐵1=∥ 𝐵𝐵𝐵𝐵 (2). The triangle𝐶𝐶1𝐵𝐵 𝐵𝐵1𝐶𝐶 being orth𝐵𝐵ologic1𝐴𝐴1 ∥al𝐴𝐴𝐴𝐴 with 𝐵𝐵1𝐶𝐶 𝐴𝐴1𝐶𝐶 and 𝐵𝐵1𝐴𝐴 𝐴𝐴1𝐵𝐵, , it means𝐴𝐴𝐴𝐴𝐴𝐴 that the orthology center is , 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝐴𝐴1𝐵𝐵1 ∥ 𝐴𝐴𝐴𝐴 𝐻𝐻 Ion Pătrașcu, Florentin Smarandache 285 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS the orthocenter of ; this would mean that must be parallel with

. From (1) and (2) we get: = (3). Because1 1 , we have: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐶𝐶1 𝐴𝐴1𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴 = (4). The relations (3)𝐶𝐶 1and𝐵𝐵 (4)𝐴𝐴1𝐶𝐶 lead to = 𝐴𝐴1𝐶𝐶,1 therefore∥ 𝐴𝐴𝐴𝐴 = 𝐴𝐴𝐶𝐶1 𝐶𝐶𝐴𝐴1 𝐴𝐴1𝐵𝐵 𝐴𝐴1𝐶𝐶 𝐶𝐶1𝐵𝐵 , hence𝐵𝐵𝐵𝐵1 is the midpoint of . Having 𝐴𝐴1𝐶𝐶 𝐴𝐴1𝐵𝐵, we obtain 𝐴𝐴that1𝐵𝐵 is the midpoint of , and leads to the conclusion that is the 𝐴𝐴1𝐶𝐶 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐴𝐴1𝐵𝐵1 ∥ 𝐴𝐴𝐴𝐴 𝐵𝐵1 midpoint of . Consequently, is the complementary (median) 𝐴𝐴𝐴𝐴 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵𝐵𝐵 𝐶𝐶1 triangle of the triangle . 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 46. Obviously, is the orthology center of the triangle in relation to the triangle . Because is the mediator of , we have that 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 = ; since is the mediator of , we have that = . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴 From the above equalities, we note that = , therefore is on the 𝐵𝐵1𝐴𝐴 𝐵𝐵𝑂𝑂1 𝐵𝐵1𝐴𝐴1 𝐶𝐶𝐶𝐶 𝐵𝐵1𝑂𝑂 𝐵𝐵1𝐶𝐶 mediator of the side of the triangle ; it follows analogously that the 𝐵𝐵1𝐴𝐴 𝐵𝐵1𝐶𝐶 𝐵𝐵1 perpendicular from to is the mediator of , and the perpendicular 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 from to is the mediator of . Consequently, – the center of the 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 circumscribed circle, is also the orthology center of the triangle in 𝐶𝐶1 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝑂𝑂 relation to . 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 47. The triangles and are obviously orthological, the orthology center being . From the theorem of the orthological triangles, 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 we have that is also orthological in relation to . Because 𝑃𝑃 belongs to the mediators of the segments and , it follows that is 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 the center of the circle circumscribed to the triangle , therefore the 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴1 perpendicular taken from to is the mediator of , consequently the 𝐵𝐵𝐵𝐵𝐵𝐵 orthology center of the triangle in relation to the triangle is 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 – the center of the circle circumscribed to the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂

𝐴𝐴𝐴𝐴𝐴𝐴 48. The triangle being orthological in relation to , it means that the perpendiculars taken from , , to , respectively 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 are concurrent in the orthology center . The pole of the perpendicular 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 taken from to is the point , the pole of the perpendicular taken from 𝑁𝑁 to is the point , and the pole of the perpendicular taken from to 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝑋𝑋 is the point . The previous perpendiculars being concurrent in , it 𝐵𝐵 𝐶𝐶1𝐴𝐴1 𝑌𝑌 𝐶𝐶 𝐴𝐴1𝐵𝐵1 𝑍𝑍 𝑁𝑁

286 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS means that their poles in relation to the duality relative to the inscribed circle are collinear points. In conclusion, , , are collinear points.

𝑋𝑋 𝑌𝑌 𝑍𝑍 49. Because the perpendicular from to and the perpendicular from to are concurrent in , the triangles and , in order to be 𝑄𝑄 𝐴𝐴𝐴𝐴 orthological, it is necessary that the perpendicular taken from to to 𝑃𝑃 𝐴𝐴𝐴𝐴 𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 pass as well through . We build the circumscribed circle of the triangle 𝐴𝐴 𝑅𝑅𝑅𝑅 and the circumscribed circle of the triangle . The point has 𝑀𝑀 equal′ powers over these circles because: ′ 𝑃𝑃𝐵𝐵 𝐵𝐵 𝑄𝑄𝐶𝐶 𝐶𝐶 𝑀𝑀 = . (1) We build′ the circumscribed′ circle of the triangle and the 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝐵𝐵 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝐶𝐶 circumscribed circle of the triangle , where { } = ′ , { } = 𝐴𝐴𝐴𝐴𝑃𝑃 ; let and be their centers′, and let ′– their second point′ of 𝐴𝐴𝐴𝐴𝑄𝑄 𝑃𝑃 𝑀𝑀𝑀𝑀 ∩ 𝐴𝐴𝐴𝐴 𝑄𝑄 intersection. From = and = (2) 𝑀𝑀𝑀𝑀 ∩ 𝐴𝐴𝐴𝐴 𝑂𝑂1 𝑂𝑂2 𝑇𝑇 and from the relation (1), it follows′ that the′ point has equal′ powers ′over 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝑃𝑃 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝐵𝐵 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝑄𝑄 𝑀𝑀𝑀𝑀 ∙ 𝑀𝑀𝐶𝐶 these circles, therefore it belongs to the radical axis . Since 𝑀𝑀 and (midline in the triangle ), it follows that . 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ⊥ 𝑂𝑂1𝑂𝑂2

𝑂𝑂1𝑂𝑂2 ∥ 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ⊥ 𝑅𝑅𝑅𝑅 50. Because the perpendicular taken from to and the perpendicular taken from to are concurrent in , the triangles and must 𝑃𝑃 𝐸𝐸𝐸𝐸 have the perpendicular taken from to passing through in order to 𝑄𝑄 𝐸𝐸 𝑀𝑀 𝐸𝐸𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸𝐸𝐸 be orthological. Therefore, we need to prove that . We build the 𝐸𝐸 𝑅𝑅𝑅𝑅 𝑀𝑀 circles circumscribed to triangles and ; we denote by their 𝑀𝑀𝑀𝑀 ⊥ 𝑃𝑃𝑃𝑃 second point of intersection, and we denote by respectively by their 𝑀𝑀𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀𝑀𝑀 𝑇𝑇 second point of intersection with respectively′ . From ′ = 𝐴𝐴 𝐵𝐵 = 90 , it follows that belongs to the line . The point is 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑚𝑚�𝑀𝑀�𝑀𝑀𝑀𝑀� the symmetric of0 with respect to , and is the symmetric of with′ 𝑚𝑚�𝑀𝑀�𝑀𝑀𝑀𝑀� 𝑇𝑇 𝑃𝑃𝑃𝑃 𝐴𝐴 respect to . The quadrilateral is′ inscribable ( = = 𝐴𝐴 𝑀𝑀𝑀𝑀 𝐵𝐵 𝐵𝐵 = ), then = ′, hence′ has equal powers over ′the 𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴 𝐵𝐵 𝐵𝐵 𝑀𝑀𝑀𝑀 𝑀𝑀𝐴𝐴 circles′ and ′ , consequently′ belongs to the radical axis of these 𝑀𝑀𝐵𝐵 𝑀𝑀𝑀𝑀 𝐸𝐸𝐴𝐴 ∙ 𝐸𝐸𝐸𝐸 𝐸𝐸𝐵𝐵 ∙ 𝐸𝐸𝐸𝐸 𝐸𝐸 circles, namely to the common chord . This is perpendicular to the line 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵 𝐸𝐸 of the centers of the circles that is parallel to , consequently . 𝑀𝑀𝑀𝑀

𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀 ⊥ 𝑃𝑃𝑃𝑃

Ion Pătrașcu, Florentin Smarandache 287 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

51. We denote: – the -circumpedal triangle of and = , = , = , = , = , = . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑈𝑈 𝐴𝐴𝐴𝐴𝐴𝐴 𝛼𝛼 The triangles and are orthological, hence: ∢𝐵𝐵𝐵𝐵𝐴𝐴1 𝛽𝛽 ∢𝐶𝐶𝐶𝐶𝐴𝐴1 𝛿𝛿 ∢𝐵𝐵1𝐵𝐵𝐵𝐵 𝛾𝛾 ∢𝐵𝐵1𝐵𝐵𝐵𝐵 𝜑𝜑 ∢𝐶𝐶1𝐶𝐶𝐶𝐶 𝜀𝜀 ∢𝐶𝐶1𝐶𝐶𝐶𝐶 + + = 0. (1) 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 From2 the sinus2 theorem,2 it follows2 that2 =2 2 sin , = 2 sin 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 and analogs. is the radius in the circle circumscribed to the triangle . 𝐵𝐵𝐴𝐴1 𝑅𝑅 𝛼𝛼 𝐶𝐶𝐴𝐴1 𝑅𝑅 𝛽𝛽 With these substitutions, the relation (1) becomes: 𝑅𝑅 𝐴𝐴𝐴𝐴𝐴𝐴 sin sin + sin sin + sin sin = 0. (2) Because2 is2 the isogonal2 conjugate2 of2 , we have2 (denoting by 𝛼𝛼 − 𝛽𝛽 𝛾𝛾 − 𝛿𝛿 𝜀𝜀 − 𝜑𝜑 the circumpedal of ): = , = , = , = 𝑉𝑉 𝑈𝑈 𝐴𝐴2𝐵𝐵2𝐶𝐶2 , = , = . 𝑉𝑉 ∢𝐴𝐴2𝐴𝐴𝐴𝐴 𝛽𝛽 ∢𝐴𝐴2𝐴𝐴𝐴𝐴 𝛼𝛼 ∢𝐵𝐵2𝐵𝐵𝐵𝐵 𝛾𝛾 ∢𝐵𝐵2𝐵𝐵𝐵𝐵 The relation (2) can be written in this way: 𝛿𝛿 ∢𝐶𝐶2𝐶𝐶𝐶𝐶 𝜑𝜑 ∢𝐶𝐶2𝐶𝐶𝐶𝐶 𝜀𝜀 sin sin + sin sin + sin sin = 0. This2 relation2 is equivalent2 to: 2 2 2 𝛽𝛽 − 𝛼𝛼 𝛿𝛿 − 𝛾𝛾 𝜑𝜑 − 𝜀𝜀 + + = 0, which expresses2 2 the orthology2 of2 the -circumpedal2 2 triangle with the 𝐴𝐴2𝐵𝐵 − 𝐴𝐴2𝐶𝐶 𝐵𝐵2𝐶𝐶 − 𝐵𝐵2𝐴𝐴 𝐶𝐶2𝐴𝐴 − 𝐶𝐶2𝐵𝐵 triangle . 𝑉𝑉 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴 52. We show that the perpendicular taken from to is also a median in the triangle . Let be the fourth vertex of the parallelogram . 𝐴𝐴 𝑆𝑆𝑆𝑆 It is noticed that the triangle is obtained from the triangle if to the 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 latter a vector translation is applied, then a rotation of center and of 𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝐴𝐴𝐴𝐴 right angle. By these transformations, the median of the triangle 𝐴𝐴��𝑆𝑆⃗ 𝑆𝑆 ({ } = ) is the transformation of the median from of the triangle 𝑆𝑆𝑆𝑆 𝐴𝐴𝐴𝐴𝐴𝐴 . These lines are perpendicular, hence the median from of triangle 𝑂𝑂 𝑆𝑆𝑆𝑆 ∩ 𝐴𝐴𝐴𝐴 𝐴𝐴 is perpendicular to . Analogously, the median from is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 perpendicular to , and that from is perpendicular to . The 𝐴𝐴𝐴𝐴𝐴𝐴 𝑆𝑆𝑆𝑆 𝐵𝐵 orthology center is – the gravity center in the triangle . 𝐴𝐴1𝐶𝐶1 𝐶𝐶 𝐴𝐴1𝐵𝐵1

𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 53. We have that ( ), ( ), ( ), ( ), ( ), ( ); and are orthological triangles; we know′ that′ : ′ ′ ′ ′ 𝐴𝐴 𝑎𝑎 𝐵𝐵 𝑏𝑏 𝐶𝐶 𝑐𝑐 𝐴𝐴 𝑎𝑎 𝐵𝐵 𝑏𝑏 𝐶𝐶 𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′( ′ ) + ( ) + ( ) = 0. (1) 𝐴𝐴 𝐵𝐵 𝐶𝐶 ′ ′ ′ ′ ′ ′ Because = = = , we have that: 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 ′′ 𝑏𝑏 𝑎𝑎 −′′ 𝑐𝑐 ′′𝑐𝑐 𝑏𝑏 − 𝑎𝑎 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 ′′ ′ ′′ ′ ′′ ′ 𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐵𝐵 𝐶𝐶 𝐶𝐶 ′ , ′ , 𝜆𝜆 ′ . ′′ 𝑎𝑎+𝜆𝜆𝑎𝑎 ′′ 𝑏𝑏+𝜆𝜆𝑏𝑏 ′′ 𝑐𝑐+𝜆𝜆𝑐𝑐 𝐴𝐴 � 1+𝜆𝜆 � 𝐵𝐵 � 1+𝜆𝜆 � 𝐶𝐶 � 1+𝜆𝜆 � 288 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

( ) ′ ′ ; 𝑏𝑏−𝑎𝑎+𝜆𝜆 𝑏𝑏 −𝑎𝑎 ′′ ′′ ( ) ��⃗ 1+𝜆𝜆 𝐴𝐴 𝐵𝐵 � ′ ′ �; 𝑐𝑐−𝑏𝑏+𝜆𝜆 𝑐𝑐 −𝑏𝑏 ′′ ′′ ( ) ��⃗ 1+𝜆𝜆 𝐴𝐴 𝐵𝐵 � ′ ′ �. 𝑎𝑎−𝑐𝑐+𝜆𝜆 𝑐𝑐 −𝑎𝑎 We′′ evaluate′′ : 𝐶𝐶��𝐴𝐴��⃗ � 1+𝜆𝜆 � [( ) + ( )] [( ) + ( )] + 1 + ′ ′ 1 + ′ ′ 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝜆𝜆 𝑐𝑐 − 𝑏𝑏[( 𝑏𝑏) +𝑎𝑎 −(𝑐𝑐 𝜆𝜆 )𝑎𝑎] − 𝑐𝑐 + . 𝜆𝜆 1 + ′ 𝜆𝜆′ Taking into account (1) and𝑐𝑐 the𝑏𝑏 fact− 𝑎𝑎 that 𝜆𝜆 (𝑏𝑏 − 𝑎𝑎) + ( ) + ( ) = 0, we obtain that the previous sum is zero;𝜆𝜆 hence and 𝑎𝑎 𝑐𝑐 − 𝑏𝑏 𝑏𝑏 𝑎𝑎 − 𝑐𝑐 𝑐𝑐 𝑏𝑏 − are orthological triangles. ′′ ′′ ′′ 𝑎𝑎 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 Analogously, it is proved that and are orthological triangles. Now is taking over′ the′ ′role of ′′ ′′ , ′′ is taking 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 over the role of ′ ′ ′ , and is taking over the ′′role′′ of′′ ; 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 from previous proof′ ′, we′ find that′′′ the′′′ triangles′′′ and ′′ ′′ are′′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 orthological. ′′ ′′ ′′ ′′′ ′′′ ′′′ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 54. a) Let be the midpoint of the side . We show that . We have: ′ ′ 𝑀𝑀1 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵 𝐶𝐶 = + 2 ′ ′ 1 ′1 ′ 1 1 𝐴𝐴��𝐴𝐴��⃗ ∙ 𝐵𝐵��𝐶𝐶��⃗ �𝐴𝐴𝐴𝐴��=��⃗ 𝐴𝐴𝐴𝐴��⃗� ∙ ��𝐻𝐻��𝐶𝐶��⃗ − 𝐻𝐻��𝐵𝐵��⃗� + . 2 2 2 2 ′ ′ ′ ′ Because 𝐴𝐴𝐴𝐴�=��⃗ ∙ 𝐻𝐻��𝐶𝐶��⃗ − =𝐴𝐴𝐴𝐴��0��⃗, we∙ 𝐻𝐻�� 𝐵𝐵�have��⃗ : 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐻𝐻��𝐶𝐶��⃗ − 𝐴𝐴𝐴𝐴��⃗ ∙ 𝐻𝐻��𝐵𝐵��⃗ = ′ cos ′ cos = 𝐴𝐴𝐴𝐴��⃗ ∙ �𝐻𝐻��𝐶𝐶��⃗ �𝐴𝐴𝐴𝐴��⃗ ∙ 𝐻𝐻��𝐵𝐵��⃗ ′ ′ 1 ′ ′ 𝐴𝐴��𝐴𝐴�cos�⃗ ∙ 𝐵𝐵��𝐶𝐶��+�⃗ 2 �𝑏𝑏 ∙ 𝑐𝑐 ∙ +�𝐶𝐶�𝐶𝐶𝐶𝐶=� 0−, 𝑐𝑐 ∙ 𝑏𝑏 ∙ �𝐵𝐵�𝐵𝐵𝐵𝐵�� 1 𝜋𝜋 𝜋𝜋 hence , . 2 𝑏𝑏𝑏𝑏 � �2 𝐴𝐴� − 𝑐𝑐𝑐𝑐𝑐𝑐 �2 𝐴𝐴�� Analogously′ , ′it follows ′that′ and , in conclusion 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵 𝐶𝐶 the triangle is orthological in relation′ ′ to and′ ′ the orthology 𝐵𝐵𝐵𝐵 ⊥ 𝐶𝐶 𝐴𝐴 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴 𝐵𝐵 center is the gravity center of the triangle . ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 b) The orthology center of the triangle in relation to the triangle 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 is obviously . The triangles and′ ′ ′ are homological, of 𝐴𝐴 𝐵𝐵 𝐶𝐶 center . From Sondat’s theorem, it follows that′ the′ ′orthology axis is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 perpendicular to the homology axis . 𝐻𝐻 ′′ ′′ 𝐻𝐻𝐻𝐻 𝐴𝐴 𝐵𝐵 Ion Pătrașcu, Florentin Smarandache 289 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

55. We prove that the triangle is orthological in relation to , and respectively that the orthology center is the gravity center of the triangle 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 . We denote by and the intersections of the perpendicular taken from to with ′ respectively′′ with . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐴𝐴 Then: 𝐴𝐴 𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 = = ′′ and ′′ . 𝐵𝐵𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴 ′′ ′′ ′′ ′′ sinIt follows�𝐵𝐵�𝐵𝐵𝐴𝐴 � thatsin: � 𝐵𝐵�𝐴𝐴 𝐴𝐴� sin�𝐶𝐶�𝐶𝐶𝐴𝐴 � sin�𝐶𝐶�𝐴𝐴 𝐴𝐴� = = = = = 1 ′′ ′′ ′ . 𝐵𝐵𝐴𝐴 𝐴𝐴𝐴𝐴 sin�𝐵𝐵�𝐴𝐴𝐴𝐴 � 𝐴𝐴𝐴𝐴 cos�𝑆𝑆�𝐴𝐴𝐴𝐴 � 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ′′ ′′ ′ Consequently𝐶𝐶𝐴𝐴 𝐴𝐴𝐴𝐴 ∙, sin�𝐶𝐶�𝐶𝐶 is𝐴𝐴 a� median𝐴𝐴𝐴𝐴 ∙ cos in �the𝑄𝑄�𝑄𝑄 𝐴𝐴triangle� 𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 . Analogously𝐴𝐴𝐴𝐴 ∙ 𝐴𝐴𝐴𝐴 , it follows that the perpendiculars′′ from and are also medians. 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴

𝐵𝐵 𝐶𝐶 56. The perpendiculars taken from , , to , and are bisectors in triangle , therefore – the center of the inscribed circle, is 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 orthology center. The orthology center of the triangle in relation to 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 the triangle is – the center of the center circumscribed to the 𝐴𝐴1𝐵𝐵1𝐶𝐶1 triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂1

𝐴𝐴1𝐵𝐵1𝐶𝐶1 57. The triangles and are orthological. The orthology center is the center of the circle circumscribed to the triangle . 𝐷𝐷𝐷𝐷𝐷𝐷 𝐴𝐴𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴 58. a) The triangles and are obviously homological. We denote { } = ; since and are fixed, it follows that is fixed. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀 The homology axis of triangles and is – – , therefore 𝐼𝐼 𝑁𝑁𝑁𝑁 ∩ 𝐵𝐵𝐵𝐵 𝑁𝑁𝑁𝑁 𝐵𝐵𝐵𝐵 𝐼𝐼 passes through . 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀 𝐼𝐼 𝐾𝐾 𝐿𝐿 𝐾𝐾𝐾𝐾 b) The triangles and are orthological, the orthology center is 𝐼𝐼 H – the orthocenter of . Also, the perpendiculars taken from to , 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀𝑀𝑀 from to and from to are concurrent. The points of the geometric 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝑀𝑀𝑀𝑀 place belong to the perpendicular taken from to (fixed line). 𝐶𝐶 𝑀𝑀𝑀𝑀 𝐴𝐴 𝑁𝑁𝑁𝑁

𝐴𝐴 𝑁𝑁𝑁𝑁 59. The perpendicular from to is the altitude from of the triangle ; the perpendicular from to′ and the perpendicular from to 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴 intersect in a point on the altitude′ ; this point is the orthopole′ of the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐴𝐴𝐴𝐴 line in relation to the triangle ′and is the orthology center of the 𝑃𝑃 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 290 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS triangle in relation to . The theorem of the orthological triangle implies the′ conclusion′ ′ that is orthological with . 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 60. The triangle and the triangle are orthological: indeed, the perpendicular taken from to , the perpendicular taken from to 𝐵𝐵1𝐶𝐶1𝐴𝐴 𝐶𝐶𝐶𝐶𝐶𝐶 and the perpendicular taken from to are concurrent in . From 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵1 the theorem of orthological triangles, it follows that the triangle is also 𝐵𝐵𝐵𝐵 𝐶𝐶1 𝐶𝐶𝐶𝐶 𝐻𝐻 orthological in relation to . Therefore the perpendicular taken from 𝐶𝐶𝐶𝐶𝐶𝐶 to , from to and from to are concurrent; since the 𝐵𝐵1𝐶𝐶1𝐴𝐴 𝐶𝐶 perpendiculars taken from to and from to are concurrent in ; 𝐶𝐶1𝐴𝐴 𝐵𝐵 𝐵𝐵1𝐴𝐴 𝑃𝑃 𝐵𝐵1𝐶𝐶1 it follows that the perpendicular from to passes through , so 𝐶𝐶 𝐶𝐶1𝐴𝐴 𝐵𝐵 𝐵𝐵1𝐴𝐴 𝐻𝐻 ; but , therefore . 𝑃𝑃 𝐵𝐵1𝐶𝐶1 𝐻𝐻 𝑃𝑃𝑃𝑃 ⊥ 1 1 1 1 𝐵𝐵 𝐶𝐶61. We𝑃𝑃 use𝑃𝑃 ⊥ the𝐵𝐵𝐵𝐵 property: the𝐵𝐵 perpendiculars𝐶𝐶 ∥ 𝐵𝐵𝐵𝐵 taken from the vertices of Fuhrmann triangle on the interior bisectors are concurrent in the orthocenter of the triangle. The orthology center of the Fuhrmann triangle in relation to is – the orthocenter of the triangle . 𝐹𝐹𝑎𝑎𝐹𝐹𝑏𝑏𝐹𝐹𝑐𝑐

𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 62. Solution 1 (Mihai Miculița).We denote by the intersection point of the perpendicular raised in to with the perpendicular raised in to 𝑃𝑃 the side , and by the projection of on the side . To solve the 𝐿𝐿 𝐴𝐴𝐴𝐴 𝑀𝑀 problem, it must be shown that the point is the midpoint of the side ( ). 𝐵𝐵𝐵𝐵 𝐾𝐾 𝑃𝑃 𝐴𝐴𝐴𝐴 From the quadrilateral𝐾𝐾 is inscribable, therefore𝐴𝐴𝐴𝐴: 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 �. ⟹ 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 (1) 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 From∢𝐾𝐾𝐾𝐾𝐾𝐾 ≡ ∢𝐾𝐾𝐾𝐾𝐾𝐾 the quadrilateral is inscribable, therefore: 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 �.⟹ 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 (2) From 𝑃𝑃the𝑃𝑃 hypothesis⊥ 𝐵𝐵𝐵𝐵 , , and from here it follows that ∢𝐾𝐾𝐾𝐾𝐾𝐾 ≡ ∢𝐾𝐾𝐾𝐾𝐾𝐾 their supplements are congruent, therefore: ∢𝐾𝐾𝐾𝐾𝐾𝐾 ≡ ∢𝐾𝐾𝐾𝐾𝐾𝐾 . (3) The relations (1), (2) and (3) lead to . This relation, ∢𝐾𝐾𝐾𝐾𝐾𝐾 ≡ ∢𝐾𝐾𝐾𝐾𝐾𝐾 (together with ) implies that [ ] [ ]. ∢𝐾𝐾𝐾𝐾𝐾𝐾 ≡ ∢𝐾𝐾𝐾𝐾𝐾𝐾 Solution 2 (Ion Pătrașcu). The relations from the hypothesis, [ ] 𝐾𝐾𝐾𝐾 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 ≡ 𝐵𝐵𝐵𝐵 [ ] and , and the sinus theorem applied in the triangle 𝐴𝐴𝐴𝐴 ≡ and lead to the conclusion that the circles circumscribed to these 𝐵𝐵𝐵𝐵 ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵𝐵𝐵

Ion Pătrașcu, Florentin Smarandache 291 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS triangles are congruent. Let respectively be the centers of these circles; they are located on the same side of the line as the vertex , and 𝑂𝑂1 𝑂𝑂2 we have that . We denote by the second intersection point of 𝐴𝐴𝐴𝐴 𝐶𝐶 previous circles. We proved that is the sought point. The point being 𝑂𝑂1𝑂𝑂2 ∥ 𝐴𝐴𝐴𝐴 𝑃𝑃 the symmetric of with respect to , and , we have that 𝑃𝑃 𝑃𝑃 . From , it follows that ( ) is diameter in the circle of 𝐾𝐾 𝑂𝑂1𝑂𝑂2 𝑂𝑂1𝑂𝑂2 ∥ 𝐴𝐴𝐴𝐴 center . Joining with , we have that (the angle is 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃 ⊥ 𝐾𝐾𝐾𝐾 𝐴𝐴𝐴𝐴 inscribed in semicircle). Reasoning analogously in the circle of center , 𝑂𝑂1 𝑃𝑃 𝐿𝐿 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 we have that ( ) is diameter, and . 𝑂𝑂2 Notes and comments𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 The problem in discussion required basically, in the respective hypotheses, to prove that the triangle is orthological in relation to the triangle , and that the orthology center of in relation to is the point . 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 It is known that, if is an orthological triangle in relation to , and the 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃 orthology center is , then is orthological in relation to , ′the′ orthology′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 center being , the isogonal′ conjugate′ ′ of (see [2]). Thus, it follows that the 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵𝐵𝐵 perpendiculars ′taken from to , from to and from to are concurrent 𝑃𝑃 𝑃𝑃 in a point – the isogonal conjugate of in the triangle . Indeed, if we take 𝐴𝐴 𝐿𝐿𝐿𝐿 𝐵𝐵 𝐾𝐾𝐾𝐾 𝐶𝐶 𝑀𝑀𝑀𝑀 the perpendicular′ from to and we denote by its foot, we have that 𝑃𝑃 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴 , because their complements, respectively′ , are congruent′ , 𝐴𝐴 𝐾𝐾𝐾𝐾 𝐴𝐴 ∢𝐾𝐾𝐾𝐾𝐴𝐴 ≡ consequently is the isogonal of . ′ ∢𝑃𝑃𝑃𝑃𝑃𝑃 ∢𝐴𝐴𝐴𝐴𝐴𝐴 ∢𝐴𝐴𝐴𝐴𝐴𝐴 In this regard,′ we propose to the reader to solve this problem, proving that the 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 perpendicular from to , the perpendicular from to , and the perpendicular from to are concurrent. 𝐴𝐴 𝐾𝐾𝐾𝐾 𝐵𝐵 𝐾𝐾𝐾𝐾

𝐶𝐶 𝑀𝑀𝑀𝑀 63. is the orthology center of bilogical triangles and . The homology center of these triangles is , and is axis of homology. 𝐼𝐼𝑎𝑎 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 According to Sondat’s theorem, we have . 𝛤𝛤𝑎𝑎 𝑋𝑋𝑋𝑋

𝐼𝐼𝑎𝑎𝛤𝛤𝑎𝑎 ⊥ 𝑋𝑋𝑋𝑋 64. The perpendicular from to is the mediator of , therefore – the center of the circle circumscribed to the triangle is the orthology 𝐴𝐴1 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝑂𝑂 center of the triangle in relation to the triangle . From the 𝐴𝐴𝐴𝐴𝐴𝐴 theorem of orthological triangles, it follows that the triangle is also 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 orthological in relation to the triangle . 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴

292 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

65. We first prove that , , are concurrent. Let be the contact of the -ex-inscribed circle with and let be the diametral of 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝐷𝐷𝑎𝑎 in the -exinscribed circle. We denote by and ′ the contact of the 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐷𝐷𝑎𝑎 inscribed circle with and its diametral in the inscribed′ circle. 𝐷𝐷𝑎𝑎 𝐴𝐴 𝐶𝐶𝑎𝑎 𝐶𝐶𝑎𝑎 The inscribed and -ex-inscribed circles are homothetic by homothety 𝐵𝐵𝐵𝐵 of center and ratio ; by the same homothety, the point corresponds to 𝑟𝑟𝑎𝑎𝐴𝐴 the point , the point corresponds to the point , and the point 𝐴𝐴 𝑟𝑟 𝐼𝐼𝑎𝑎 corresponds to the point . The projection of on the ′mediator of the side′ 𝐼𝐼 𝐷𝐷𝑎𝑎 𝐶𝐶𝑎𝑎 𝐷𝐷𝑎𝑎 , denoted by , is such that contains the Nagel point of the triangle 𝐶𝐶𝑎𝑎 𝐼𝐼 . Because ′ and are isotomic′ points, it follows that belongs to 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐴𝐴𝐴𝐴 , therefore passes through (the Gergonne point). Analogously, it 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝑎𝑎 𝐷𝐷𝑎𝑎 𝐴𝐴1 is shown that and contain . 𝐴𝐴𝐴𝐴𝑎𝑎 𝐴𝐴𝐴𝐴1 Γ The triangle is obviously orthological in relation to 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 Γ because the perpendiculars taken from , , to , and are 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 mediators in the triangle , hence is the orthology center. 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 66. We denote by and the centers of the circles; the circles being congruent, it follows that (the chords and being equal, 𝑂𝑂1 𝑂𝑂2 they are equally spaced from the centers). We denote by the symmetric 𝑂𝑂1𝑂𝑂2 ∥ 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 of with respect to ; obviously, is the second intersection point of 𝑃𝑃 circles of centers and . Since and , we obtain 𝐾𝐾 𝑂𝑂1𝑂𝑂2 𝑃𝑃 that , , are collinear and also , , are collinear. Because is 𝑂𝑂1 𝑂𝑂2 𝑂𝑂1𝑂𝑂2 ∥ 𝐴𝐴𝐴𝐴 𝑂𝑂1𝑂𝑂2 ⊥ 𝑃𝑃𝑃𝑃 diameter, we have that = 90 ; also is diameter, so = 90 . 𝐴𝐴 𝑂𝑂1 𝑃𝑃 𝐵𝐵 𝑂𝑂2 𝑃𝑃 𝐴𝐴𝐴𝐴 Having , and 0 , it means that is the orthology0 ∢𝑃𝑃𝑃𝑃𝑃𝑃 𝐵𝐵𝐵𝐵 ∢𝑃𝑃𝑃𝑃𝑃𝑃 center of the triangle in relation to . The geometric place of the 𝑃𝑃𝑃𝑃 ⊥ 𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 𝑃𝑃𝑃𝑃 ⊥ 𝐴𝐴𝐴𝐴 𝑃𝑃 point (the orthology center of the triangle in relation to ) is the 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 semi-line of origin, perpendicular to , and located in the boundary 𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 semi-plane containing the point . 𝐾𝐾 𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴 𝐶𝐶 67. The triangle is orthological in relation to the triangle . Indeed, the perpendicular from to is , the perpendicular from to 𝐸𝐸𝐸𝐸𝐸𝐸 𝐹𝐹𝐹𝐹𝐹𝐹 is (because is rectangle), and the perpendicular from to 𝐸𝐸 𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 𝐵𝐵 is , hence is the orthology center. The orthology relation being 𝐹𝐹𝐹𝐹 𝐵𝐵𝐵𝐵 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝐶𝐶 symmetrical, it means that the triangle is also orthological in relation 𝐹𝐹𝐹𝐹 𝐶𝐶𝐶𝐶 𝑂𝑂 to the triangle , therefore the perpendicular from to is (it is 𝐹𝐹𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸𝐸𝐸 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐹𝐹𝐹𝐹 Ion Pătrașcu, Florentin Smarandache 293 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS perpendicular to ), the perpendicular from to and the perpendicular from to are concurrent. Because the perpendicular from to and 𝐴𝐴𝐴𝐴 𝐷𝐷 𝐸𝐸𝐸𝐸 the perpendicular from to intersect in , it follows that the 𝐴𝐴 𝐸𝐸𝐸𝐸 𝐹𝐹 𝐴𝐴𝐴𝐴 perpendicular from to passes through . 𝐷𝐷 𝐸𝐸𝐸𝐸 𝐻𝐻

𝐴𝐴 𝐸𝐸𝐸𝐸 𝐻𝐻 68. The triangles and are homological, and is the homology center ( is the Toricelli point of the triangle ). The triangle is 𝐷𝐷𝐷𝐷𝐷𝐷 𝑆𝑆𝑆𝑆𝑆𝑆 𝑇𝑇 orthological in relation to , and the orthology center is the point 𝑇𝑇 𝐷𝐷𝐷𝐷𝐷𝐷 𝑆𝑆𝑆𝑆𝑆𝑆 (indeed, the perpendicular from to , the perpendicular from to 𝐷𝐷𝐷𝐷𝐷𝐷 𝑃𝑃 and the perpendicular from to are concurrent in – the midpoint of 𝑆𝑆 𝐴𝐴𝐴𝐴 𝐶𝐶 𝐷𝐷𝐷𝐷 ). From the theorem of orthological triangles, it follows that the triangle 𝑉𝑉 𝐴𝐴𝐴𝐴 𝑃𝑃 is orthological in relation to , and the orthology center is . 𝐷𝐷𝐷𝐷 Sondat’s theorem implies the collinearity of points , , . 𝐷𝐷𝐷𝐷𝐷𝐷 𝑆𝑆𝑆𝑆𝑆𝑆 𝑄𝑄

𝑇𝑇 𝑃𝑃 𝑄𝑄 69. Let be the median triangle of . Then the triangle is the′ homot′ ′ hetic of the triangle by homothety of center 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and by ratio 2. The triangle is orthological′ ′ ′ in relation to ; the 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑃𝑃 orthology center is the orthocenter of . The triangle ′ ′having′ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 the sides parallel to that of the triangle , it means that the orthocenter 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 of is orthology center of the triangle′ ′ ′ in relation to . 𝐴𝐴 𝐵𝐵 𝐶𝐶

𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 70. i) If is an acute triangle and is its orthic triangle, then is acute and = 180 2 , ′ ′ ′ = 180 2 , = 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 180′ ′ ′ 2 . If is′ the orthic0 triangle of′ the triangle0 , then′ : 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑚𝑚�𝐴𝐴� � − 𝐴𝐴̂ 𝑚𝑚�𝐵𝐵�� − 𝐵𝐵� 𝑚𝑚�𝐶𝐶�� 0 = 4 180′′ ′′ , ′′ = 4 180 , = 4 ′ 180′ ′ . We − 𝐶𝐶̂ 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 observed′′ that a right0 triangle′′ does not have 0an orthic′′ triangle, therefore0 it 𝑚𝑚�𝐴𝐴�� 𝐴𝐴̂ − 𝑚𝑚�𝐵𝐵� � 𝐵𝐵� − 𝑚𝑚�𝐶𝐶� � 𝐶𝐶̂ − would be necessary for the triangle to have an angle of measure 45 . ii) If the triangle has, for example, = 112 30 , then 0its 𝐴𝐴𝐴𝐴𝐴𝐴 orthic triangle has = 45 , = 2 0 , ′ = 𝐴𝐴𝐴𝐴𝐴𝐴 𝑚𝑚�𝐴𝐴̂� 2 . The triangle′ ′ ′ – the′ orthic triangle0 of′ , has ′ = 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑚𝑚�𝐴𝐴� � 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐶𝐶�� 90 , = 180 ′′ 4′′ ′′ , = 180 4′ ′ ′ . The triangle′′ 𝑚𝑚�𝐶𝐶̂� 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑚𝑚�𝐴𝐴�� 0 being′′ rectangular0 , it does not ′′have an orthic0 triangle. 𝑚𝑚�𝐵𝐵� � − 𝑚𝑚�𝐵𝐵�� 𝑚𝑚�𝐶𝐶� � − 𝑚𝑚�𝐶𝐶̂� ′′ iii)′′ If′′ is an equilateral triangle, then and are 𝐴𝐴 𝐵𝐵 𝐶𝐶 equilateral triangles, and is orthological in ′relation′ ′ to ′′ ′′ ′′ . In 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 general, and are not orthological. ′′ ′′ ′′ ′′ ′′ ′′𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 294 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

71. a) The perpendicular taken from to is because, being midline in the triangle , we have , therefore . 𝐾𝐾 𝐴𝐴𝐴𝐴 𝐾𝐾𝐾𝐾 Analogously, the perpendicular from to is ; obviously, the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐾𝐾𝐾𝐾 ∥ 𝐵𝐵𝐵𝐵 𝐾𝐾𝐾𝐾 ⊥ 𝐴𝐴𝐴𝐴 perpendicular from to passes through , hence is the orthology 𝐿𝐿 𝐴𝐴𝐴𝐴 𝐿𝐿𝐿𝐿 center of the triangle in relation to . 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑀𝑀 𝑀𝑀 b) The other orthology center is ; indeed, the perpendicular from to 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 is , the perpendicular from to is and the perpendicular 𝐴𝐴 𝐵𝐵 from to passes obviously through . Since in the triangle the 𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵 𝐶𝐶 𝐾𝐾𝐾𝐾 𝐶𝐶𝐶𝐶 point is orthocenter ( and are altitudes), it follows that is an 𝐴𝐴 𝐾𝐾𝐾𝐾 𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 altitude, hence . 𝑀𝑀 𝐾𝐾𝐾𝐾 𝐿𝐿𝐿𝐿 𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴 ⊥ 𝐾𝐾𝐾𝐾 72. a) Because is bisector, we have the arcs , therefore = , and the perpendicular′ from to is the′ mediator′ of . 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵� ≡ 𝐶𝐶𝐶𝐶� Thus′ , we obtain′ that the -circumpedal triangle′ is orthological in relation to 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 , and the orthology center is – the center of the circumscribed circle. 𝐼𝐼 It can be shown directly that is orthological in relation to , and 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 the orthology center is . ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 b) We noticed that = ; the triangle is isosceles ( = 𝐼𝐼 ) because ′ , ′ therefore the circle′ ( ; ) pass′ es 𝐴𝐴 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴 𝐵𝐵 through′ ; analogously′ , it follows′ that the other circles pass′ through′ as 𝐴𝐴 𝐼𝐼 ∢𝐴𝐴 𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴 𝐼𝐼𝐼𝐼 𝒞𝒞 𝐴𝐴 𝐴𝐴 𝐵𝐵 well. 𝐼𝐼 𝐼𝐼 c) The triangles and are homological; the axis of homology is ; the axis of orthology is ′ ;′ according′ to Sondat’s theorem, . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶

𝑋𝑋𝑋𝑋 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ⊥ 𝑋𝑋𝑋𝑋 73. Obviously, , . We prove that we also have . We observe that = = 90 = . Also, 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 ⊥ 𝑄𝑄𝑄𝑄 𝑀𝑀𝑀𝑀 ⊥ = = 180 , therefore: 0 ~ . It also 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� 𝑚𝑚�𝑀𝑀�𝑀𝑀𝑀𝑀� − 𝐶𝐶̂ 𝐻𝐻�𝐻𝐻𝐻𝐻 follows from here that 0 ~ . We note that ; 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵� � 𝑚𝑚�𝑃𝑃�𝑃𝑃𝑃𝑃� − 𝐴𝐴̂ ∆𝐵𝐵𝐵𝐵𝐵𝐵 ∆𝑄𝑄𝑄𝑄𝑄𝑄 because , we have that + = 90 , ∆𝐵𝐵𝐵𝐵𝐵𝐵 ∆𝑄𝑄𝑄𝑄𝑄𝑄 ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐻𝐻𝐻𝐻𝐻𝐻 consequently . 0 𝑁𝑁𝑁𝑁 ⊥ 𝐵𝐵𝐵𝐵 𝑚𝑚�𝐻𝐻�𝐻𝐻𝐻𝐻� 𝑚𝑚�𝐴𝐴�𝐴𝐴𝐴𝐴� Observation𝑀𝑀𝑀𝑀 ⊥ 𝑁𝑁𝑁𝑁 It can be seen that the point { } = belongs to the circle circumscribed to the triangle . 𝑆𝑆 𝑀𝑀𝑀𝑀 ∩ 𝑁𝑁𝑁𝑁

𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 295 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

74. If is the orthology center of the triangle in relation to the triangle , and is the center of the circle , ie. the intersection of 𝑃𝑃 𝐴𝐴1𝐵𝐵1𝐶𝐶1 mediators of the segment ( ), ( ), ( ), and the perpendicular in 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 𝜔𝜔 on intersects in , the point will be the symmetric of towards 𝐴𝐴1𝐴𝐴2 𝐵𝐵1𝐵𝐵2 𝐶𝐶1𝐶𝐶2 . Similarly, the perpendicular from on passes through the 𝐴𝐴2 𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃 𝑄𝑄 𝑄𝑄 𝑃𝑃 symmetric of with respect to , therefore through , and in the same way 𝑂𝑂 𝐵𝐵2 𝐴𝐴𝐴𝐴 belongs to the perpendicular in to . Consequently: the point is 𝑃𝑃 𝑂𝑂 𝑄𝑄 the orthology center of the triangle in relation to the triangle . 𝑄𝑄 𝐶𝐶2 𝐴𝐴𝐴𝐴 𝑄𝑄

𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 75. We denote by the projection of on the plane and by , , the projections′ of points , , respectively on , 𝐻𝐻 𝐻𝐻 𝐴𝐴1𝐵𝐵1𝐶𝐶1 and′′ ′′ . ′′From ( ) and , we obtain that 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 (the reciprocal ′theorem of the three perpendicular′′ ). On the other′ hand′′ , 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 ⊥ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴 ⊥ 𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐴𝐴 ⊥ being perpendicular to ( ), the plane ( ) contains , and 𝐵𝐵1𝐶𝐶1 even more: . Similarly, we show that ′ ′′ and ′ . 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴 𝐴𝐴 𝐻𝐻𝐻𝐻 The point ′is the′ orthology′′ center of the triangle′ ′ ′′ in relation′ ′ ′′to 𝐻𝐻 ∈ 𝐴𝐴 𝐴𝐴 𝐻𝐻 ∈ 𝐵𝐵 𝐵𝐵 𝐻𝐻 ∈ 𝐶𝐶 𝐶𝐶 . ′ ′ ′ ′ 𝐻𝐻 𝐴𝐴 𝐵𝐵 𝐶𝐶

𝐴𝐴1𝐵𝐵1𝐶𝐶1 76. The perpendicular from to is the mediator of , the perpendicular from to passes through – the center of the circle 𝐻𝐻 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 circumscribed to the triangle , and also, the perpendicular from to 𝐵𝐵 𝐴𝐴1𝐶𝐶1 𝑂𝑂 , which is antiparallel with , passes through . Therefore – the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶 center of the circle circumscribed to the triangle , is the orthology 𝐴𝐴1𝐵𝐵1 𝐴𝐴𝐴𝐴 𝑂𝑂 𝑂𝑂 center of the triangle in relation to . 𝐴𝐴𝐴𝐴𝐴𝐴

𝐻𝐻𝐻𝐻𝐻𝐻 𝐴𝐴1𝐵𝐵1𝐶𝐶1 77. Solution (Mihai Miculița). We prove that the line is mediator = of the side . We have: . 𝑏𝑏 𝑎𝑎 inscribable 𝐴𝐴 𝐵𝐵

𝐴𝐴𝐴𝐴 = 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ⟹ ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐵𝐵𝑎𝑎𝐶𝐶 − ⟹ ∢𝑂𝑂𝑂𝑂𝐵𝐵𝑎𝑎 ≡ ∢𝑂𝑂𝑂𝑂𝑂𝑂 = + = � ⟹ ∢𝑂𝑂+𝑂𝑂𝐵𝐵𝑎𝑎 ≡ ∢𝑂𝑂𝑂𝑂𝑂𝑂= 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ⟹ ∢𝑂𝑂𝑂𝑂𝑂𝑂 ≡ ∢𝑂𝑂𝑂𝑂𝑂𝑂 = . (1) ⟹ 𝑚𝑚�𝐵𝐵𝐵𝐵�𝐵𝐵𝑎𝑎� 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝑂𝑂�𝑂𝑂𝐵𝐵𝑎𝑎� 𝑚𝑚�𝐴𝐴𝐴𝐴�𝐴𝐴� 𝑚𝑚�𝑂𝑂�𝑂𝑂𝑂𝑂� On the other hand, from: 𝑚𝑚�𝐴𝐴𝐴𝐴�𝐵𝐵𝑎𝑎� ⟹ 𝐵𝐵𝑎𝑎𝐴𝐴 𝐵𝐵𝑎𝑎𝐵𝐵 = , = 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ⟹inscribable∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂𝑂𝑂 𝑂𝑂𝑂𝑂 ⟹ ∢𝑂𝑂𝑂𝑂𝑂𝑂 ≡ ∢𝑂𝑂𝑂𝑂𝑂𝑂 � ⟹ ∢𝑂𝑂𝑂𝑂𝑂𝑂 ≡ ∢𝑂𝑂𝑂𝑂𝐴𝐴𝑏𝑏 𝐶𝐶𝐶𝐶𝐵𝐵𝐵𝐵𝑏𝑏 − ⟹ ∢𝑂𝑂𝑂𝑂𝑂𝑂 ≡ ∢𝑂𝑂𝑂𝑂𝐴𝐴𝑏𝑏

296 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= + = + = = . (2) ⟹ 𝑚𝑚�𝐵𝐵𝐵𝐵�𝐴𝐴𝑏𝑏� 𝑚𝑚�𝐵𝐵𝐵𝐵𝐵𝐵�� 𝑚𝑚�𝑂𝑂�𝑂𝑂𝑂𝑂� 𝑚𝑚�𝐴𝐴𝐴𝐴�𝐴𝐴� 𝑚𝑚�𝑂𝑂�𝑂𝑂𝐴𝐴𝑏𝑏� The relations (1) and (2) show that the line is mediator of the side 𝑚𝑚�𝐴𝐴𝐴𝐴�𝐴𝐴𝑏𝑏� ⟹ 𝐵𝐵�𝐴𝐴𝐴𝐴𝑏𝑏 ≡ 𝐴𝐴𝐴𝐴�𝐴𝐴𝑏𝑏 ⟹ 𝐴𝐴𝑏𝑏𝐴𝐴 𝐴𝐴𝑏𝑏𝐵𝐵 , so we have: = { }. 𝐴𝐴𝑏𝑏𝐵𝐵𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 𝑐𝑐 𝑏𝑏 𝐴𝐴𝐴𝐴Observation 𝐴𝐴 𝐵𝐵 ∩ 𝐴𝐴 𝐶𝐶 ∩ 𝐵𝐵 𝐶𝐶 𝑂𝑂 The triangles and are orthological; the orthology center is . Also, the triangles and are orthological of center . 𝐴𝐴𝑏𝑏𝐴𝐴𝑐𝑐𝐵𝐵𝑐𝑐 𝐶𝐶𝐶𝐶𝐶𝐶 𝑂𝑂

𝐵𝐵𝑎𝑎𝐶𝐶𝑏𝑏𝐶𝐶𝑎𝑎 𝐶𝐶𝐶𝐶𝐶𝐶 𝑂𝑂 78. We have = = . We take , ( ). 𝐴𝐴1𝐶𝐶 𝑏𝑏 cos 𝐶𝐶 𝐶𝐶1𝑋𝑋 ′ ′ = cos = cos 1 1 1 1 𝐵𝐵1𝐶𝐶 𝑐𝑐, cos 𝐵𝐵 𝑋𝑋𝐵𝐵1 . 𝐴𝐴𝑋𝑋 ⊥ 𝐵𝐵 𝐶𝐶 𝑋𝑋 ∈ 𝐵𝐵 𝐶𝐶 ′= cos , = ′ cos . 𝐵𝐵1𝑋𝑋 𝐴𝐴𝐵𝐵1 ∙ 𝐵𝐵 𝐶𝐶1𝑋𝑋 𝐴𝐴𝐶𝐶1 ∙ 𝐶𝐶 1 = = ,1 it follows that = . 𝐴𝐴𝐵𝐵′ 𝑐𝑐 ∙ 𝐴𝐴 𝐴𝐴𝐶𝐶 𝑏𝑏 ∙ 𝐴𝐴 𝐶𝐶1𝑋𝑋 𝑏𝑏 cos 𝐶𝐶 𝐶𝐶1𝑋𝑋 ′ ′ 𝐵𝐵The1𝑋𝑋 triangles𝑐𝑐 cos 𝐵𝐵 𝑋𝑋𝐵𝐵1 and its orthic𝑋𝑋 triangle𝑋𝑋 are orthological, therefore , and are concurrent. The concurrency point is the 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 orthology center of the triangle in relation to , therefore – 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 the center of the circle circumscribed to the triangle . 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝑂𝑂

𝐴𝐴𝐴𝐴𝐴𝐴 79. Obviously, the triangle is orthological in relation to , because and ( ); from the theorem of the three 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 perpendiculars, it follows that ; similarly and 𝑃𝑃𝐴𝐴1 ⊥ 𝐵𝐵𝐵𝐵 𝐴𝐴2 ∈ 𝑃𝑃𝐴𝐴1 , and = { }, hence is orthology center. 𝐴𝐴1𝐴𝐴2 ⊥ 𝐵𝐵𝐵𝐵 𝐵𝐵1𝐵𝐵2 ⊥ 𝐴𝐴𝐴𝐴 Let and let be the orthology center of the triangle in 𝐶𝐶2𝐶𝐶1 ⊥ 𝐴𝐴𝐴𝐴 𝐴𝐴1𝐴𝐴2 ∩ 𝐵𝐵1𝐵𝐵2 ∩ 𝐶𝐶1𝐶𝐶2 𝑃𝑃 𝑃𝑃 ′ ( ) = relation to 1 1 , 1 . From . 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴 𝐶𝐶 ′ 𝑂𝑂 ′ 2 ′ 2 2 2𝐴𝐴𝐴𝐴𝐴𝐴 1 1 1 1 1 1 1 1 1 80. Solution𝐴𝐴 𝐵𝐵 𝐶𝐶(Mihai𝐵𝐵 ∈Miculița).𝐴𝐴 𝐶𝐶 𝐵𝐵 𝐴𝐴 − 𝐵𝐵 𝐶𝐶 𝑃𝑃𝐴𝐴 − 𝑃𝑃𝐶𝐶 We denote the midpoint of the segment by ; we show that [ ]. 𝐸𝐸𝐸𝐸 𝑄𝑄 𝑄𝑄 ∈ = 𝐴𝐴𝐴𝐴 = = 1 . (1) 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴𝐴𝐴 = � ⇒ 𝑀𝑀𝑀𝑀 = 2 𝐵𝐵𝐵𝐵 = 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 1 � ⟹ � ⟹ 𝑀𝑀𝑀𝑀 ⊥ 𝐸𝐸𝐸𝐸 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴 𝑄𝑄𝑄𝑄 𝑄𝑄𝑄𝑄 On the other� ⇒hand𝑀𝑀𝑀𝑀, whereas2 𝐵𝐵𝐵𝐵 : 𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀 inscribable . (2, 3) 𝐵𝐵𝐵𝐵 ⊥ 𝐴𝐴𝐴𝐴 ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 � ⟹ 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 − ⟹ � 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴 ∢𝑄𝑄𝑄𝑄𝑄𝑄 ≡ ∢𝑀𝑀𝑀𝑀𝑀𝑀 Ion Pătrașcu, Florentin Smarandache 297 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

Taking the relations (1) and (3) into consideration, we obtain that: inscribable ( = ( . (4) 𝑀𝑀𝑀𝑀 ⊥ 𝐸𝐸𝐸𝐸 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 From relations� ⟹ (3) and (4), it follows� ⇒ 𝑄𝑄 that∈ 𝐴𝐴 𝐴𝐴the⇒ points𝐴𝐴𝐴𝐴 𝐴𝐴and𝐴𝐴 are ∢𝐴𝐴𝐴𝐴𝐴𝐴 ≡ ∢𝐴𝐴𝐴𝐴𝐴𝐴 homolo𝑁𝑁𝑁𝑁 g⊥ical𝐵𝐵𝐵𝐵 points of the similar triangles ~ . Hence we have: 𝑁𝑁 𝑃𝑃 = . (5) 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 ∆𝐴𝐴𝐴𝐴𝐴𝐴 ∆𝐴𝐴𝐴𝐴𝐴𝐴 From𝑁𝑁𝑁𝑁 𝑃𝑃relations𝑃𝑃 (4) and (5), it follows now that: = . (6) 𝑁𝑁𝑁𝑁 𝑅𝑅𝑅𝑅 𝑅𝑅𝑅𝑅 ∥ 𝐶𝐶𝐶𝐶 S𝑁𝑁𝑁𝑁imilarly𝑅𝑅𝑅𝑅 ⟹, it is shown �that⟹ 𝑅𝑅𝑅𝑅 ⊥ 𝐴𝐴𝐴𝐴. (7) Finally, the 𝐶𝐶𝐶𝐶relations⊥ 𝐴𝐴𝐴𝐴 (6) and (7) shows that the lines and are 𝑆𝑆𝑆𝑆 ⊥ 𝐴𝐴𝐴𝐴 altitudes in the triangle , therefore is the orthocenter of this triangle. 𝐴𝐴𝐴𝐴 𝑆𝑆𝑆𝑆 Observation 𝐴𝐴𝐴𝐴𝐴𝐴 𝑁𝑁 During the solution, we solved the following problem: Let and – be the feet of the altitudes taken from the vertices and of an acute triangle , and – the midpoint of the side . We denote by { } = 𝐸𝐸 𝐹𝐹 𝐵𝐵 𝐶𝐶 and by the projection of on . Show that the semi-line ( is a 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀 𝐵𝐵𝐵𝐵 𝑁𝑁 symmedian of the triangle . 𝐴𝐴𝐴𝐴 ∩ 𝐸𝐸𝐸𝐸 𝑃𝑃 𝑁𝑁 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴

𝐴𝐴𝐴𝐴𝐴𝐴 81. Obviously, if = 1, the lines , , are concurrent being the mediators of the triangle . Reciprocally, let = { }. Then 𝑘𝑘 𝑎𝑎 𝑏𝑏 𝑐𝑐 = 0, ie. ( )( ) = 0. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑎𝑎 ∩ 𝑏𝑏 ∩ 𝑐𝑐 𝑆𝑆 𝑀𝑀��𝑀𝑀⃗ ∙ ( ) = ( ) �𝐵𝐵𝐵𝐵��⃗From here𝑟𝑟��,�𝑆𝑆⃗ − �𝑟𝑟��𝑀𝑀�⃗ 𝑟𝑟��𝐶𝐶⃗ − 𝑟𝑟��𝐵𝐵⃗ . 𝑟𝑟��𝐵𝐵�⃗+𝑘𝑘�𝑟𝑟��𝐶𝐶�⃗ We write the analogous relations, and we add them. We obtain: 𝑟𝑟��𝑆𝑆⃗ ∙ �𝑟𝑟��𝐶𝐶⃗ − 𝑟𝑟��𝐵𝐵⃗ 1+𝑘𝑘 𝑟𝑟��𝐶𝐶⃗ − 𝑟𝑟��𝐵𝐵⃗ ( + ) ( ) = 0. (1) We consider a system of axes with the origin in the center of the circle ∑ �𝑟𝑟��𝐵𝐵⃗ 𝑘𝑘�𝑟𝑟��𝐶𝐶⃗ 𝑟𝑟��𝐶𝐶⃗ − 𝑟𝑟��𝐵𝐵⃗ circumscribed to the triangle . Suppose this circle has the radius 1. Then: = 1. 𝐴𝐴𝐴𝐴𝐴𝐴 The equality (1) becomes: �𝑟𝑟��𝐵𝐵⃗ ∙ 𝑟𝑟��𝐵𝐵⃗ ( + 1 ) = 0, namely: ( 1)(3 ) = 0. ∑ �𝑟𝑟��𝐵𝐵⃗ ∙ �𝑟𝑟��𝐶𝐶⃗ 𝑘𝑘 − − 𝑘𝑘 ∙ 𝑟𝑟��𝐵𝐵⃗ ∙ �𝑟𝑟��𝐶𝐶⃗ But: 𝑘𝑘 − − 𝑟𝑟��𝐴𝐴⃗ ∙ �𝑟𝑟��𝐵𝐵⃗ − 𝑟𝑟��𝐵𝐵⃗ ∙ �𝑟𝑟��𝐶𝐶⃗ − �𝑟𝑟��𝐶𝐶⃗ ∙ �𝑟𝑟��𝐴𝐴⃗ | + + | < | || | + | || | + | || | = 3, consequently = 1. 𝑟𝑟��𝐴𝐴⃗ ∙ �𝑟𝑟��𝐵𝐵⃗ 𝑟𝑟��𝐵𝐵⃗ ∙ �𝑟𝑟��𝐶𝐶⃗ �𝑟𝑟��𝐶𝐶⃗ ∙ �𝑟𝑟��𝐴𝐴⃗ 𝑟𝑟��𝐴𝐴⃗ 𝑟𝑟��𝐵𝐵⃗ �𝑟𝑟��𝐵𝐵⃗ 𝑟𝑟��𝐶𝐶⃗ �𝑟𝑟��𝐶𝐶⃗ 𝑟𝑟��𝐴𝐴⃗ 𝑘𝑘 298 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

Observation The problem expresses that only the triangle – inscribed in the triangle , with the property = = , and orthological with , is the median 𝑀𝑀𝑀𝑀 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀 triangle. 𝐴𝐴𝐴𝐴𝐴𝐴 𝑀𝑀𝑀𝑀 𝑁𝑁𝑁𝑁 𝑃𝑃𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴

82. We denote: the median triangle of the triangle , and – the tangential triangle of the triangle . The triangles 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 and are orthological, and is their orthology center. Indeed, the 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 perpendiculars taken from , , to , , are bisectors in 𝐴𝐴𝐴𝐴𝐴𝐴 𝑂𝑂 and consequently pass through , which is the center of the circle inscribed 𝑇𝑇𝑎𝑎 𝑇𝑇𝑏𝑏 𝑇𝑇𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 in the triangle . Moreover, is mediator of ( ), and it 𝑂𝑂 consequently passes through ; being mediator, it is perpendicular 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑇𝑇𝑎𝑎𝑂𝑂 𝐵𝐵𝐵𝐵 to , but also to , which is midline. From the theorem of 𝑀𝑀𝑎𝑎 𝑇𝑇𝑎𝑎𝑀𝑀𝑎𝑎 orthological triangles, it follows that the perpendiculars taken from , , 𝐵𝐵𝐵𝐵 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 to , respectively are concurrent as well. The point of 𝑀𝑀𝑎𝑎 𝑀𝑀𝑏𝑏 concurrency is – the center of the Euler circle of the triangle . We 𝑀𝑀𝑐𝑐 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑇𝑇𝑐𝑐𝑇𝑇𝑎𝑎 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏 prove this fact. We take ; we denote { } = , 𝑂𝑂9 𝐴𝐴𝐴𝐴𝐴𝐴 where is the orthocenter of the triangle . It is known that = 𝑀𝑀𝑎𝑎𝑀𝑀1 ⊥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝐻𝐻1 𝑀𝑀𝑎𝑎𝑀𝑀1 ∩ 𝐴𝐴𝐴𝐴 2 . We join with ; we have that ; since and 𝐻𝐻 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 , it follows that the quadrilateral is parallelogram. 𝑂𝑂𝑀𝑀𝑎𝑎 𝑂𝑂 𝐴𝐴 𝑂𝑂𝑂𝑂 ⊥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 𝑀𝑀𝑎𝑎𝑀𝑀1 ⊥ 𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 From = and = 2 , we obtain that is the midpoint of 𝐴𝐴𝐴𝐴 ∥ 𝑂𝑂𝑀𝑀𝑎𝑎 𝑂𝑂𝑀𝑀𝑎𝑎𝐻𝐻1𝐴𝐴 ( ), therefore is on the circle of the nine points of the triangle . 𝐴𝐴𝐴𝐴1 𝑂𝑂𝑀𝑀𝑎𝑎 𝐴𝐴𝐴𝐴 𝑂𝑂𝑀𝑀𝑎𝑎 𝐻𝐻1 On this circle, we find also the points (the feet of the altitude from ) and 𝐴𝐴𝐴𝐴 𝐻𝐻1 𝐴𝐴𝐴𝐴𝐴𝐴 . Since = 90 , it follows that′ is diameter in Euler circle, 𝐴𝐴 𝐴𝐴 therefore the midpoint′ of0 is the center of Euler circle, which we 𝑀𝑀𝑎𝑎 ∢𝐴𝐴𝐴𝐴 𝑀𝑀𝑎𝑎 𝑀𝑀𝑎𝑎𝐻𝐻1 denote by . We observe that the quadrilateral is parallelogram 𝑀𝑀𝑎𝑎𝐻𝐻1 as well, and it follows that is the midpoint of . 𝑂𝑂9 𝐻𝐻1𝐻𝐻𝑀𝑀𝑎𝑎𝑂𝑂 9 Observation 𝑂𝑂 𝑂𝑂𝑂𝑂 The triangles and are bilogical.

𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝑇𝑇𝑎𝑎𝑇𝑇𝑏𝑏𝑇𝑇𝑐𝑐 83. We denote: = , = , = and = = = ; let , , be the contacts of the circles inscribed in the triangles , 𝐴𝐴𝐴𝐴 𝑥𝑥 𝐵𝐵𝐵𝐵 𝑦𝑦 𝐶𝐶𝐶𝐶 𝑧𝑧 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑎𝑎 and with , respectively . 𝐴𝐴2 𝐵𝐵2 𝐶𝐶2 𝐵𝐵𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶𝐶𝐶 It is shown without difficulty that: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 299 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= ; = ; = ; 𝑦𝑦+𝑎𝑎−𝑧𝑧 𝑎𝑎+𝑧𝑧−𝑦𝑦 𝑧𝑧+𝑎𝑎−𝑥𝑥 𝐵𝐵𝐴𝐴2 = 2 ; 𝐶𝐶𝐴𝐴2 = 2 ; 𝐶𝐶𝐵𝐵2 = 2 . 𝑎𝑎+𝑥𝑥−𝑧𝑧 𝑥𝑥+𝑎𝑎−𝑦𝑦 𝑦𝑦+𝑎𝑎−𝑥𝑥 We show that the triangle is orthological in relation to 𝐴𝐴𝐵𝐵2 2 𝐴𝐴𝐶𝐶2 2 𝐵𝐵𝐶𝐶2 2 proving the equality: 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 + + = + + . Consequently2 ,2 it also follows2 2that 2 is orthological2 in relation to . 𝐵𝐵𝐴𝐴2 𝐶𝐶𝐵𝐵2 𝐴𝐴𝐶𝐶2 𝐶𝐶𝐴𝐴1 𝐴𝐴𝐵𝐵2 𝐵𝐵𝐶𝐶1

𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 84. We denote by , , the distances of the points , , from the line . In order that the perpendiculars taken from , , respectively to 𝑥𝑥 𝑦𝑦 𝑧𝑧 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 , , to be concurrent, the following condition must be 𝑑𝑑 𝐴𝐴 𝐵𝐵 𝐶𝐶 satisfied: 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 + + = 0. 2 2 2 2 2 2 This1 condition1 is equivalent1 1 to: 1 1 𝐴𝐴𝐵𝐵( −+𝐴𝐴𝐶𝐶 ) 𝐵𝐵(𝐶𝐶 −+𝐵𝐵𝐴𝐴) + (𝐶𝐶𝐴𝐴 +− 𝐶𝐶)𝐵𝐵 ( + ) + ( + ) 2 2 2 ( 2 + )2 = 02. 2 2 2 2 𝑎𝑎1 𝑦𝑦 − 𝑎𝑎2 𝑧𝑧 𝑏𝑏1 𝑧𝑧 − 𝑏𝑏2 𝑥𝑥 𝑐𝑐1 𝑥𝑥 From where we find: 2 2 − 𝑐𝑐2 𝑦𝑦 + + = 0 (it does not depend of , , ). 2 2 2 2 2 2 1 2 1 2 1 2 𝑎𝑎Remar− 𝑎𝑎k 𝑏𝑏 − 𝑏𝑏 𝑐𝑐 − 𝑐𝑐 𝑥𝑥 𝑦𝑦 𝑧𝑧 If the points , , belong to the lines , , , the preceding condition is obviously fulfilled, and the concurrency point of the perpendiculars taken from , 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑑𝑑1 𝑑𝑑2 𝑑𝑑3 , to , respectively is the orthopole of the line in relation to the 𝐴𝐴 triangle . 𝐵𝐵 𝐶𝐶 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 𝐴𝐴1𝐵𝐵1 𝑑𝑑

𝐴𝐴1𝐵𝐵1𝐶𝐶1 85. The perpendicular from to is , the perpendicular from to is , and the perpendicular from to is , hence 𝐴𝐴1 𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 𝐴𝐴1𝑀𝑀 is the orthology center of the triangle in relation to the median 𝐵𝐵1 𝑀𝑀𝑐𝑐𝑀𝑀𝑎𝑎 𝐵𝐵1𝑀𝑀 𝐶𝐶1 𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏 𝐶𝐶1𝑀𝑀 triangle . 𝑀𝑀 𝐴𝐴1𝐵𝐵1𝐶𝐶1

𝑀𝑀𝑎𝑎𝑀𝑀𝑏𝑏𝑀𝑀𝑐𝑐 86. The triangle is orthological in relation to , therefore the perpendiculars taken from , , to , , are concurrent in a 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴𝐴𝐴𝐴𝐴 point . The triangle , the symmetric with respect to of the triangle 𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 , has the sides respectively′ ′ ′ parallel with its sides. The perpendicular 𝑃𝑃 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑂𝑂 from to will be perpendicular to as well. The orthology center 𝐴𝐴𝐴𝐴𝐴𝐴 ′ ′ of the triangle1 in relation to will be the point P. 𝐴𝐴 𝐵𝐵𝐵𝐵 ′ 𝐵𝐵′ 𝐶𝐶′ 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴 𝐵𝐵 𝐶𝐶 300 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

87. The triangle is orthological in relation to the triangle , the orthology center being . Indeed, the perpendicular taken from to , 𝐸𝐸𝐸𝐸𝐸𝐸 𝐹𝐹𝐹𝐹𝐹𝐹 the perpendicular taken from to , and the perpendicular taken from 𝑂𝑂 𝐸𝐸 𝐴𝐴𝐴𝐴 to intersect in . The relation of orthology being symmetrical, it follows 𝐵𝐵 𝐹𝐹𝐹𝐹 𝐶𝐶 that the triangle is orthological in relation to the triangle as well. 𝐹𝐹𝐹𝐹 𝑂𝑂 Then the perpendicular from to , the perpendicular from to and 𝐹𝐹𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸𝐸𝐸 the perpendicular from to are concurrent. Because the perpendicular 𝐹𝐹 𝐵𝐵𝐵𝐵 𝐷𝐷 𝐸𝐸𝐸𝐸 from to and the perpendicular from to intersect in , it follows 𝐴𝐴 𝐸𝐸𝐸𝐸 that the perpendicular from to passes through as well, therefore 𝐷𝐷 𝐸𝐸𝐸𝐸 𝐹𝐹 𝐴𝐴𝐴𝐴 𝐻𝐻 . 𝐴𝐴 𝐸𝐸𝐸𝐸 𝐻𝐻

𝐴𝐴𝐴𝐴 ⊥ 𝐸𝐸𝐸𝐸 88. We denote by the podal triangle of the symmedian center of the triangle ; also′ , ′we′ denote by the intersection of perpendicular 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐾𝐾 from to with . We have that (1) (they have 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1 the same complement)′ ′ ′, ′ (2)′ (the quadrilateral′ ′ is 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐵𝐵 𝐶𝐶 ∢𝐶𝐶 𝐴𝐴𝐴𝐴1 ≡ ∢𝐾𝐾𝐶𝐶 𝐵𝐵 inscribable). From relations (1)′ ′and (2), it ′follows that ′ ′ , ∢𝐾𝐾𝐶𝐶 𝐵𝐵 ≡ ∢𝐾𝐾𝐾𝐾𝐵𝐵 𝐾𝐾𝐶𝐶 𝐴𝐴𝐵𝐵 therefore is the isogonal of symmedian , hence ′ passes through′ ∢𝐶𝐶 𝐴𝐴𝐴𝐴1 ≡ ∢𝐾𝐾𝐾𝐾𝐵𝐵 – the gravity center of the triangle . It is obvious that the symmedian 𝐴𝐴𝐴𝐴1 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴1 center is orthology center. To prove that is the gravity center of the 𝐺𝐺 𝐴𝐴𝐴𝐴𝐴𝐴 triangle , it is shown that: 𝐾𝐾 𝐾𝐾 Area( ′ ′ ′ ) = Area( ) = Area( ). 𝐴𝐴 𝐵𝐵 𝐶𝐶 We use the′ relation′ : ′ ′ ′ ′ ∆𝐾𝐾𝐵𝐵 𝐶𝐶 ∆𝐾𝐾𝐶𝐶 𝐴𝐴 ∆𝐾𝐾𝐴𝐴 𝐵𝐵 = = ′ ′ ′. 𝐾𝐾𝐴𝐴 𝐾𝐾𝐵𝐵 𝐾𝐾𝐶𝐶 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴 89. a) We denote by the intersection of the parallel taken through with , with the parallel taken through with . We have that 𝑃𝑃1 𝐴𝐴 ′ ′ , therefore ( ) = 60 , which′ ′ shows that 𝐵𝐵 𝐶𝐶 𝐶𝐶 𝐴𝐴 𝐵𝐵 belongs to the′ ′circle′ circumscribed to the triangle0 . From angle ∢𝐴𝐴𝑃𝑃1𝐶𝐶 ≡ ∢𝐵𝐵 𝐶𝐶 𝐴𝐴 𝑚𝑚 𝐴𝐴𝑃𝑃1𝐶𝐶 𝑃𝑃1 measurement, is 60 ; and from the reciprocal theorem of 𝐴𝐴𝐴𝐴𝐴𝐴 the angle with parallel 0sides; we have′ ′that , hence is 𝐵𝐵𝑃𝑃1𝐶𝐶 𝐶𝐶𝑃𝑃1 ∥ 𝐴𝐴 𝐵𝐵 parallelogy center of the triangle in relation to the′ triangle′ . 𝐵𝐵𝑃𝑃1 ∥ 𝐴𝐴 𝐶𝐶 𝑃𝑃1 Similarly reasoning, we find that and , parallelogy centers of′ ′the′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴 𝐵𝐵 𝐶𝐶 triangle in relation to and , are located on the 𝑃𝑃2 𝑃𝑃3 circumscribed circle of the triangle′ ′ ′ . ′ ′ ′ 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐴𝐴 𝐶𝐶 𝐴𝐴 𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴

Ion Pătrașcu, Florentin Smarandache 301 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

b) is parallel with , and is also parallel with ; it follows that . The trapeze′ ′ , being inscribed, is isosceles′ ′ , therefore 𝐴𝐴𝑃𝑃2 𝐴𝐴 𝐶𝐶 𝐵𝐵𝑃𝑃1 𝐴𝐴 𝐶𝐶 the diagonals and are congruent. Similarly, it is shown that = 𝐴𝐴𝑃𝑃2 ∥ 𝐵𝐵𝑃𝑃1 𝐴𝐴𝐴𝐴2𝐵𝐵𝑃𝑃1 and = , consequently the triangle is equilateral. 𝐴𝐴𝐴𝐴 𝑃𝑃1𝑃𝑃2 𝑃𝑃2𝑃𝑃3

𝐵𝐵𝐵𝐵 𝑃𝑃1𝑃𝑃3 𝐴𝐴𝐴𝐴 𝑃𝑃1𝑃𝑃2𝑃𝑃3 90. To prove that the triangle is orthological in relation to , the following relation must be verified: 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐴𝐴𝐴𝐴𝐴𝐴 + + = 0. (1) Because2 2, , 2 are concurrent2 2, we have2 that: 𝐴𝐴2𝐵𝐵 − 𝐴𝐴2𝐶𝐶 𝐵𝐵2𝐶𝐶 − 𝐵𝐵2𝐴𝐴 𝐶𝐶2𝐴𝐴 − 𝐶𝐶2𝐵𝐵 + + = 0. (2) 𝐴𝐴𝐴𝐴1 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 The 2perpendicular2 2 to 2 – which2 is antiparal2 lel to – passes 𝐶𝐶1𝐴𝐴 − 𝐶𝐶1𝐵𝐵 𝐴𝐴1𝐵𝐵 − 𝐴𝐴1𝐶𝐶 𝐵𝐵1𝐶𝐶 − 𝐵𝐵1𝐴𝐴 through – the center of the circle circumscribed to the triangle . Also, 𝐴𝐴𝐴𝐴2 𝐵𝐵1𝐶𝐶1 𝐵𝐵𝐵𝐵 and pass through . The cevians , and being 𝑂𝑂 𝐴𝐴𝐴𝐴𝐴𝐴 concurrent, we have that: 𝐵𝐵𝐵𝐵2 𝐶𝐶𝐶𝐶2 𝑂𝑂 𝐴𝐴𝐴𝐴2 𝐵𝐵𝐵𝐵2 𝐶𝐶𝐶𝐶2 + + = 0. (3) We 2calculate2 2 . We2 will2 apply the2 cosine theorem in the 𝐴𝐴2𝐶𝐶1 − 𝐴𝐴2𝐵𝐵1 𝐵𝐵2𝐶𝐶1 − 𝐵𝐵2𝐴𝐴1 𝐶𝐶2𝐴𝐴1 − 𝐶𝐶2𝐵𝐵1 triangles respectively2 2 . The lines and being 𝐴𝐴2𝐵𝐵 − 𝐵𝐵2𝐴𝐴 antiparallels with respectively , it follows that . 𝐴𝐴2𝐵𝐵𝐶𝐶1 𝐵𝐵2𝐴𝐴𝐶𝐶1 𝐵𝐵1𝐶𝐶1 𝐶𝐶1𝐴𝐴1 The triangles and are similar; we obtain that: 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 ∢𝐴𝐴𝐶𝐶1𝐵𝐵1 ≡ ∢𝐵𝐵𝐶𝐶1𝐴𝐴1 = . (4) 𝐴𝐴𝐶𝐶1𝐴𝐴2 𝐵𝐵𝐵𝐵1𝐵𝐵2 = + 2 cos , 𝐴𝐴𝐶𝐶1 ∙ 𝐶𝐶1𝐵𝐵2 𝐵𝐵𝐶𝐶1 ∙ 𝐶𝐶1𝐴𝐴2 2 = 2 + 2 2 cos 2 1 1 2 1 1 2 1 2. 𝐴𝐴 𝐵𝐵2 𝐶𝐶 𝐵𝐵2 𝐶𝐶 𝐴𝐴2 − 𝐶𝐶 𝐵𝐵 ∙ 𝐶𝐶 𝐴𝐴 ∙ ∢𝐵𝐵𝐶𝐶 𝐴𝐴 Since2 1 1 2 , taking1 1 (4)2 into consideration1 2 , we obtain that: 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐴𝐴 =𝐶𝐶 𝐵𝐵 − 𝐶𝐶 𝐴𝐴 ∙+𝐶𝐶 𝐵𝐵 ∙ ∢𝐴𝐴𝐶𝐶 𝐵𝐵 1 2 1 2 . (5) 2 ∢𝐵𝐵𝐶𝐶 𝐴𝐴2 ≡ ∢𝐴𝐴𝐶𝐶2 𝐵𝐵 2 2 2 Similarly2 , 2 we calculate1 :1 1 1 and2 and, 𝐴𝐴 𝐵𝐵 − 𝐵𝐵 𝐴𝐴 𝐶𝐶 𝐵𝐵 − 𝐶𝐶 𝐴𝐴 𝐶𝐶2 𝐴𝐴 − 𝐶𝐶2 𝐵𝐵 2 2 considering the relations (5), (2) and2 (3), we2 obtain the relation2 (1).2 𝐵𝐵 𝐶𝐶 − 𝐶𝐶 𝐵𝐵 𝐶𝐶 𝐴𝐴 − 𝐴𝐴 𝐶𝐶 91. Let the midpoint of , the midpoint of , and the midpoint of . It is shown that: 𝑀𝑀1 𝐿𝐿𝐿𝐿 𝑀𝑀2 𝑀𝑀𝑀𝑀 𝑀𝑀3 + + = 0. (1) 2 𝐾𝐾𝐾𝐾 2 2 2 2 2 We1 calculate1 2 with the2 median3 theorem3 applied in the triangle 𝑀𝑀 𝐵𝐵 − 𝑀𝑀 𝐶𝐶 𝑀𝑀2 𝐶𝐶 − 𝑀𝑀 𝐴𝐴 𝑀𝑀 𝐴𝐴 − 𝑀𝑀 𝐵𝐵 , and we have: 1 ( 𝑀𝑀 𝐵𝐵) = , 𝑀𝑀𝑀𝑀𝑀𝑀 2 2 2 2 𝐴𝐴𝐴𝐴 +𝐵𝐵𝐵𝐵 −𝑀𝑀𝑀𝑀 2 ( ) 1 = 4 𝑀𝑀 𝐵𝐵 2 2 2. 2 2 𝐴𝐴𝐴𝐴 +𝐶𝐶𝐶𝐶 −𝑀𝑀𝑀𝑀 𝑀𝑀1𝐶𝐶 4 302 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

Because (S.A.S), it follows that = and 2 = . 1 2∆𝐴𝐴𝐴𝐴2𝐴𝐴 ≡ ∆𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀 𝑀𝑀 𝐵𝐵 − 𝐴𝐴𝐵𝐵 −𝐴𝐴𝐶𝐶 Similarly2 , we calculate: and . 𝑀𝑀1𝐶𝐶 2 2 2 2 2 Replacing in (1), this is verified2 . 2 3 3 𝑀𝑀 𝐶𝐶 − 𝑀𝑀 𝐴𝐴 𝑀𝑀 𝐴𝐴 − 𝑀𝑀 𝐵𝐵 92. The symmedian center of the triangle is the gravity center of the triangle . The triangle is the symmetric, in relation to , 𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 of the triangle ; we have ; and 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2 𝐾𝐾 . The triangles and are orthological, and their common 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 ∥ 𝐵𝐵2𝐶𝐶2 𝐴𝐴1𝐵𝐵1 ∥ 𝐴𝐴2𝐵𝐵2 𝐴𝐴1𝐶𝐶1 ∥ orthology center is the orthocenter of the triangle . 𝐴𝐴2𝐶𝐶2 𝐴𝐴1𝐵𝐵1𝐶𝐶1 𝐴𝐴2𝐵𝐵2𝐶𝐶2

𝐴𝐴1𝐵𝐵1𝐶𝐶1 93. We denote by the midpoint of . The perpendiculars taken from , , to , and are concurrent in the point – the symmetric of 𝑃𝑃 𝐷𝐷𝐷𝐷 with respect to the center of the parallelogram ( is the midpoint 𝑃𝑃 𝑄𝑄 𝑅𝑅 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑀𝑀 of ). The point is called the Mathot point of the inscribable 𝑂𝑂 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑆𝑆 quadrilateral . In other words, the triangle is orthological in 𝐵𝐵𝐵𝐵 𝑀𝑀 relation to . It follows that is orthological as well in relation to 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑃𝑃𝑃𝑃𝑃𝑃 , therefore the perpendicular taken from to , the perpendicular 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴 taken from to , and the perpendicular taken from to are 𝑃𝑃𝑃𝑃𝑃𝑃 𝐴𝐴 𝑅𝑅𝑅𝑅 concurrent. Because and , we get to the required 𝐵𝐵 𝑃𝑃𝑃𝑃 𝐶𝐶 𝑃𝑃𝑃𝑃 conclusion. 𝑃𝑃𝑃𝑃 ∥ 𝐶𝐶𝐶𝐶 𝑅𝑅𝑅𝑅 ∥ 𝐵𝐵𝐵𝐵

94. The perpendiculars taken from , , respectively to , and are , and , and they intersect in – the center of the circle 𝐵𝐵 𝐹𝐹 𝐷𝐷 𝐶𝐶𝐶𝐶 𝐸𝐸𝐸𝐸 circumscribed to the hexagon, the point which is the common center of 𝐸𝐸𝐸𝐸 𝐵𝐵𝐵𝐵 𝐹𝐹𝐹𝐹 𝐷𝐷𝐷𝐷 𝑂𝑂 orthology of triangles and . The triangle is orthological in relation to the triangle , and the orthology center is the point . The 𝐵𝐵𝐵𝐵𝐵𝐵 𝐸𝐸𝐸𝐸𝐸𝐸 𝐵𝐵𝐵𝐵𝐵𝐵 orthology center of the triangle in relation to is the point . The 𝐶𝐶𝐶𝐶𝐶𝐶 𝐴𝐴 triangle is orthological in relation to the triangle , and the 𝐶𝐶𝐶𝐶𝐶𝐶 𝐵𝐵𝐵𝐵𝐵𝐵 𝐷𝐷 orthology center is the point . The orthology center of the triangle in 𝐵𝐵𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴𝐴𝐴 relation to is the point . 𝐶𝐶 𝐴𝐴𝐴𝐴𝐴𝐴

𝐵𝐵𝐵𝐵𝐵𝐵 𝐹𝐹 95. a) Let be the intersection of the parallel with , with the parallel to . The quadrilateral𝑀𝑀 is a rectangle; 𝐵𝐵𝐵𝐵= , = , √21 2√7 𝐴𝐴𝐴𝐴 𝑀𝑀𝐶𝐶1𝐵𝐵𝐴𝐴1 𝑀𝑀𝐴𝐴1 7 𝑀𝑀𝐶𝐶1 7

Ion Pătrașcu, Florentin Smarandache 303 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= . The quadrilateral is inscribable; from sinus theorem, 5√7 𝐴𝐴we1𝐶𝐶 have 7that: = sin𝑀𝑀30𝐴𝐴1𝐶𝐶, 𝐵𝐵1 = . 1 From the triangle , it follows0 that = 2; therefore = 1. 𝐴𝐴1𝐵𝐵1 𝑀𝑀𝑀𝑀 ∙ 𝐴𝐴1𝐵𝐵1 2 𝑀𝑀𝑀𝑀 = + 1 = 1, therefore = 1. 1 1 2 𝑀𝑀𝐴𝐴 𝐶𝐶 2 𝑀𝑀𝑀𝑀 𝐴𝐴 𝐵𝐵 2 √21 2√7 We1 1denote = ; applying cosine theorem1 1 in triangle , the 𝐴𝐴 𝐶𝐶 � 7 � � 7 � 𝐴𝐴 𝐶𝐶 equation 7 + 3 7 4 = 0 is obtained, with the appropriate solution 𝑀𝑀𝐵𝐵1 𝑥𝑥 𝐴𝐴1𝑀𝑀𝐵𝐵1 = 2 , which𝑥𝑥 shows√ 𝑥𝑥 that− also belongs to the parallel with . √7 b) The quadrilateral is inscribable; = 120 ; we 𝑥𝑥 7 𝑀𝑀 𝐴𝐴𝐴𝐴 calculate with cosine theorem; it follows that = 1, therefore0 the 𝑀𝑀𝐵𝐵1𝐴𝐴𝐴𝐴1 ∢𝐵𝐵1𝑀𝑀𝐶𝐶1 triangle is equilateral. 𝐵𝐵1𝐶𝐶1 𝐵𝐵1𝐶𝐶1 c) In the triangles and , the cevians are obviously 𝐴𝐴1𝐵𝐵1𝐶𝐶1 orthological. They are not bilogical because they are not homological; , 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴1𝐵𝐵1𝐶𝐶1 , are not concurrent. Indeed, it is verified that: 11. 𝐴𝐴1𝐵𝐵 𝐵𝐵1𝐶𝐶 𝐶𝐶1𝐴𝐴 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵1 𝐶𝐶𝐶𝐶1 𝐴𝐴1𝐶𝐶 ∙ 𝐵𝐵1𝐴𝐴 ∙ 𝐶𝐶1𝐵𝐵 ≠ 96. Obviously, (the radius is perpendicular to the tangent). Since = 90 , it follows that = 90 as well (1). Because 𝐴𝐴𝐴𝐴 ⊥ 𝐾𝐾𝐾𝐾 = 90 ( 0diameter), it follows that 0 = 90 as well (2). 𝑚𝑚�𝐷𝐷�𝐷𝐷𝐷𝐷� 𝑚𝑚�𝐷𝐷�𝐷𝐷𝐷𝐷� The relations (1)0 and (2) lead to , , collinear. The triangles0 and 𝑚𝑚�𝐷𝐷�𝐷𝐷𝐷𝐷� 𝐴𝐴𝐴𝐴 𝑚𝑚�𝐸𝐸�𝐸𝐸𝐸𝐸� have ; it is obvious that the orthocenter of the triangle is 𝐸𝐸 𝑀𝑀 𝐹𝐹 𝐷𝐷𝐷𝐷𝐷𝐷 their orthology center; we denote it by . The quadrilateral is 𝐴𝐴𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 ∥ 𝐾𝐾𝐾𝐾 𝐴𝐴𝐴𝐴𝐴𝐴 inscribable (3), (4) (sides respectively perpendicular). 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 From (3) and (4), it follows that , therefore ∢𝐵𝐵𝐵𝐵𝐵𝐵 ≡ ∢𝑃𝑃𝑃𝑃𝑃𝑃 (5). In the triangle , is an altitude and a bisector, therefore ∢𝑃𝑃𝑃𝑃𝑃𝑃 ≡ ∢𝑀𝑀𝑀𝑀𝑀𝑀 ∢𝑀𝑀𝑀𝑀𝑀𝑀 ≡ is isosceles, hence = . ∢𝑀𝑀𝑀𝑀𝑀𝑀 𝑃𝑃𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀

𝑃𝑃𝑃𝑃𝑃𝑃 𝑀𝑀𝑀𝑀 𝑃𝑃𝑃𝑃 97. It is obvious that the triangle is orthological in relation to , and the orthology center is . We prove that , , are concurrent 𝑀𝑀𝑀𝑀𝑀𝑀 𝐴𝐴𝐴𝐴𝐴𝐴 using Ceva’s theorem – trigonometric𝑎𝑎 variant. We take ; 𝐼𝐼 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶′ ′ ; , we have = ; = + , = cos ; ′ ′ 𝐹𝐹𝐹𝐹 ⊥ 𝐵𝐵𝐵𝐵 𝑀𝑀𝑀𝑀 ⊥ ′ 𝐹𝐹𝐹𝐹 𝐹𝐹 𝑀𝑀 ′ ′ ′ ′ ′ ′ ′ 𝐵𝐵𝐵𝐵from𝐸𝐸 the𝐸𝐸 t⊥heorem𝐵𝐵𝐵𝐵 of the exterior𝑀𝑀𝑀𝑀 𝐸𝐸 bisector𝑀𝑀 𝐹𝐹 𝑀𝑀, it follows𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵that 𝐵𝐵𝐹𝐹= ; we𝐵𝐵𝐵𝐵 fi∙nd that𝐵𝐵 ( ) 𝐹𝐹𝐹𝐹 𝑎𝑎 = ; = . Then: = . 𝐹𝐹𝐹𝐹 𝑏𝑏 𝑎𝑎𝑎𝑎 ′ ′ ′ 𝑐𝑐 𝑝𝑝−𝑎𝑎 𝐹𝐹𝐹𝐹 𝑏𝑏−𝑎𝑎 𝐵𝐵𝐵𝐵 𝑝𝑝 − 𝑐𝑐 𝐹𝐹 𝑀𝑀 𝑏𝑏−𝑎𝑎 304 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

= + = cos + . ( ) ′ ′ = ′ ′ = = = 𝐸𝐸But𝑀𝑀 𝐶𝐶𝐶𝐶 ,𝐶𝐶𝐶𝐶 it follows𝐶𝐶𝐶𝐶 that 𝐶𝐶 𝑝𝑝 − 𝑏𝑏 , , . 𝑎𝑎 ′ ′ 𝑏𝑏 𝑝𝑝−𝑐𝑐 𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 We get: 𝑐𝑐−𝑎𝑎 𝑐𝑐−𝑎𝑎 𝑏𝑏−𝑎𝑎 𝑐𝑐−𝑎𝑎 𝐶𝐶𝐶𝐶 ( ) 𝐸𝐸 𝑀𝑀 𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸 = ( ). (2) sin 𝐵𝐵𝐵𝐵𝐵𝐵 𝑐𝑐 𝑝𝑝−𝑏𝑏 Calculatingsin 𝑀𝑀𝑀𝑀𝑀𝑀 𝑏𝑏 𝑝𝑝in− 𝑐𝑐a similar way, we find: = ( ), (3) sin 𝐶𝐶𝐶𝐶𝐶𝐶 𝑄𝑄𝑄𝑄 ( ) sin 𝑁𝑁𝑁𝑁𝑁𝑁 = 𝑐𝑐 𝑝𝑝−𝑏𝑏 . (4) sin 𝐸𝐸𝐸𝐸𝐸𝐸 𝑏𝑏 𝑝𝑝−𝑐𝑐 sinReplacing𝑃𝑃𝑃𝑃𝑃𝑃 𝑎𝑎(2)𝑎𝑎 , (3), (4) in relation (1), the equality implying the concurrency of lines , , is verified.

𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 98. The barycentric coordinates of a point in relation to the triangle are three numbers , , proportional to the areas of the triangles 𝑃𝑃 , and . Because , and , we 𝐴𝐴𝐴𝐴𝐴𝐴 𝑥𝑥 𝑦𝑦 𝑧𝑧 have that sin = sin , sin = sin and sin = 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃1𝐴𝐴1 ⊥ 𝐵𝐵𝐵𝐵 𝑃𝑃1𝐵𝐵1 ⊥ 𝐶𝐶𝐶𝐶 𝑃𝑃1𝐶𝐶1 ⊥ 𝐴𝐴𝐴𝐴 sin . 𝐵𝐵1𝑃𝑃1𝐶𝐶1 𝐴𝐴 𝑃𝑃1𝐵𝐵1𝐶𝐶1 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃1𝐶𝐶1𝐵𝐵1 We have: 𝑃𝑃𝑃𝑃𝑃𝑃 = = = . Area𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃∙𝑏𝑏∙sin 𝑃𝑃𝑃𝑃𝑃𝑃 𝑏𝑏∙sin 𝑃𝑃1𝐵𝐵1𝐶𝐶1 𝑏𝑏 𝑃𝑃1𝐶𝐶1 AreaSimilarly𝑃𝑃𝑃𝑃𝑃𝑃 , we𝑃𝑃𝑃𝑃∙ 𝑐𝑐get∙sin: 𝑃𝑃𝑃𝑃𝑃𝑃 𝑐𝑐∙sin 𝑃𝑃1𝐶𝐶1𝐵𝐵1 𝑐𝑐 ∙ 𝑃𝑃1𝐵𝐵1 = ÷ . Area𝑃𝑃𝑃𝑃𝑃𝑃 𝑎𝑎 𝑃𝑃1𝐶𝐶1 OnArea the𝑃𝑃𝑃𝑃𝑃𝑃 other𝑐𝑐 hand𝑃𝑃1𝐴𝐴,1 we find: ( ) = = . ( ) Area 𝑃𝑃1𝐴𝐴1𝐶𝐶1 𝑃𝑃1𝐴𝐴1∙𝑃𝑃1𝐶𝐶1∙sin 𝐵𝐵 𝑏𝑏 𝑃𝑃1𝐶𝐶1 AreaTherefore𝑃𝑃1𝐴𝐴1𝐵𝐵:1 𝑃𝑃1𝐴𝐴1∙𝑃𝑃1𝐵𝐵1∙sin 𝐶𝐶 𝑐𝑐 ∙ 𝑃𝑃1𝐵𝐵1 ( ) ( ) = . ( ) ( ) Area 𝑃𝑃𝑃𝑃𝑃𝑃 Area 𝑃𝑃1𝐴𝐴1𝐶𝐶1 AreaSimilarly𝑃𝑃𝑃𝑃𝑃𝑃 : Area 𝑃𝑃1𝐴𝐴1𝐵𝐵1 ( ) ( ) = , ( ) ( ) Area 𝑃𝑃𝑃𝑃𝑃𝑃 Area 𝑃𝑃1𝐵𝐵1𝐶𝐶1 1 Area1 1 ( ) Area( ) Area 𝑃𝑃𝑃𝑃𝑃𝑃 Area 𝑃𝑃 𝐶𝐶 𝐴𝐴 = Area( ) Area( ) 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃1𝐴𝐴1𝐵𝐵1 1 1 1 ( ) 𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃 𝐵𝐵 𝐶𝐶 = = 99. ( ) ( ). 𝐹𝐹𝐹𝐹 Area 𝐹𝐹𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹 sin 𝛼𝛼 = We 𝑀𝑀𝑀𝑀denotedArea 𝐸𝐸𝐸𝐸𝐸𝐸 𝐸𝐸𝐸𝐸. ⋅ sin 𝐴𝐴−𝛼𝛼 𝛼𝛼 ∢𝐵𝐵𝐵𝐵𝐵𝐵

Ion Pătrașcu, Florentin Smarandache 305 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

We take , and perpendicular to ; we have = ′ ′ ′ ′ , = , = cos . From bisector’s theorem, we find: ′ ′𝐹𝐹𝐹𝐹 𝐸𝐸𝐸𝐸 𝑀𝑀𝑀𝑀 𝐵𝐵𝐵𝐵 𝐵𝐵𝐵𝐵 𝑝𝑝 − 𝐹𝐹𝐹𝐹 𝐹𝐹 𝑀𝑀 ′ ′ ′ 𝑀𝑀𝑀𝑀 =𝐸𝐸 𝑀𝑀 cos = 𝑏𝑏 ;𝐵𝐵𝐵𝐵 𝐹𝐹𝐹𝐹 ∙ 2𝐵𝐵( 2 ) 2; 𝑎𝑎𝑎𝑎 𝑎𝑎 +𝑐𝑐 −𝑏𝑏 ( ) =𝑎𝑎+𝑏𝑏 2 𝑎𝑎+𝑏𝑏= 𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹 ∙ 𝐵𝐵 2 ( 2 ) 2 ; 𝑎𝑎+𝑏𝑏+𝑐𝑐 𝑎𝑎 +𝑐𝑐 −𝑏𝑏 𝑐𝑐 𝑝𝑝−𝑐𝑐 ′ ′ = = cos 𝐹𝐹 𝑀𝑀 2 − 𝑏𝑏 − 2 𝑎𝑎+𝑏𝑏 𝑎𝑎+𝑏𝑏 ; ( ) ′ =′ ′ ′ = 𝐸𝐸 𝑀𝑀 𝐶𝐶;𝑀𝑀 we− find𝐶𝐶𝐸𝐸 that 𝑝𝑝 − 𝑐𝑐 − 𝐸𝐸𝐸𝐸 ∙ ; 𝐶𝐶 𝑎𝑎𝑎𝑎 𝑏𝑏 𝑝𝑝−𝑏𝑏 ( )( ) ′ ′ = 𝑎𝑎+𝑐𝑐 = 𝑎𝑎+𝑐𝑐 𝐸𝐸𝐸𝐸 ′ ′ ( )( 𝐸𝐸); 𝑀𝑀 𝐹𝐹𝐹𝐹 𝐹𝐹 𝑀𝑀 𝑐𝑐 𝑝𝑝−𝑐𝑐 𝑎𝑎+𝑐𝑐 ′ ′ ( ) 𝑀𝑀𝑀𝑀 𝐸𝐸 𝑀𝑀= 𝑏𝑏 𝑝𝑝−𝑏𝑏 =𝑎𝑎+𝑏𝑏 ( ) ′ ′ ( ). (1) sin 𝛼𝛼 𝐹𝐹 𝑀𝑀 𝐸𝐸𝐸𝐸 𝑐𝑐 𝑝𝑝−𝑐𝑐 ′ ′ ( ) sin 𝐴𝐴−𝛼𝛼 𝐸𝐸 𝑀𝑀 𝐹𝐹𝐹𝐹 𝑏𝑏 𝑝𝑝−𝑏𝑏 = Similarly, we find∙ that ( ) ( ), (2) sin 𝛽𝛽 𝑎𝑎 𝑝𝑝−𝑎𝑎 ( ) = sin 𝐵𝐵−𝛽𝛽 𝑐𝑐 𝑝𝑝−𝑐𝑐 ( ) ( ). (3) sin 𝛾𝛾 𝑏𝑏 𝑝𝑝−𝑏𝑏 sinWe𝐶𝐶 denoted−𝛾𝛾 𝑎𝑎 𝑝𝑝−=𝑎𝑎 and = . The relations (1), (2), (3) and Ceva’s theorem – trigonometric variant, lead to the concurrency of the 𝛽𝛽 ∢𝑁𝑁𝑁𝑁𝑁𝑁 𝛾𝛾 ∢𝑃𝑃𝑃𝑃𝑃𝑃 cevians , , . If we denote { } = , { } = , { } = and by – the concurrency point of cevians , , , 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑋𝑋 𝐴𝐴𝐴𝐴 ∩ 𝐵𝐵𝐵𝐵 𝑌𝑌 𝐵𝐵𝐵𝐵 ∩ 𝐴𝐴𝐴𝐴 we find: 𝑍𝑍 𝐶𝐶𝐶𝐶 ∩( 𝐴𝐴𝐴𝐴) 𝐿𝐿 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 = 2( ). (4) 𝑋𝑋𝑋𝑋 𝑐𝑐 𝑝𝑝−𝑐𝑐 2 The𝑋𝑋𝑋𝑋 relation𝑏𝑏 𝑝𝑝−𝑏𝑏 (4) and the Steiner relation relative to the isogonal cevians show that the cevian is the isogonal of Nagel cevian relative to . Thus, we find that the point is the Nagel isogonal point of the triangle 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 . The barycentric coordinates of the points , , , where is the Nagel 𝐿𝐿 point, are: 𝐴𝐴𝐴𝐴𝐴𝐴 𝐼𝐼 𝐽𝐽 𝑂𝑂 𝐽𝐽 , , ; 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐼𝐼 �2𝑝𝑝 2,𝑝𝑝 2𝑝𝑝,� ; 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 (sin 2 , sin 2 , sin 2 ) 𝐽𝐽 � 𝑝𝑝 𝑝𝑝 𝑝𝑝 � . The coordinates of – the isogonal of – are: 𝑂𝑂 𝐴𝐴 𝐵𝐵 𝐶𝐶 , , . 2 2 2 𝐿𝐿 𝐽𝐽 𝑎𝑎 𝑝𝑝 𝑏𝑏 𝑝𝑝 𝑐𝑐 𝑝𝑝 𝐿𝐿To� 𝑝𝑝show−𝑎𝑎 𝑝𝑝− that𝑏𝑏 𝑝𝑝 −𝑐𝑐,� , are collinear, the following condition must be proved: 𝐼𝐼 𝐿𝐿 𝑂𝑂

306 Ion Pătrașcu, Florentin Smarandache THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES

SOLUTIONS, INDICATIONS, ANSWERS

𝑎𝑎 𝑏𝑏 𝑐𝑐 = 0 2𝑝𝑝 2𝑝𝑝 2𝑝𝑝 . (5) 2 2 2 𝑎𝑎 𝑝𝑝 𝑏𝑏 𝑝𝑝 𝑐𝑐 𝑝𝑝 �sin 2 sin 2 sin 2 � � 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 � Because sin 2 = 2 sin cos and sin = , condition (5) is 𝐴𝐴 𝐵𝐵 𝐶𝐶 𝑎𝑎 equivalent to: 𝐴𝐴 𝐴𝐴 𝐴𝐴 𝐴𝐴 2𝑅𝑅 1 1 1 = 0. (6) 𝑎𝑎 𝑏𝑏 𝑐𝑐 cos cos cos � 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 𝑝𝑝−𝑐𝑐 � Condition (6) implies that: 1 𝐴𝐴 𝐵𝐵 0 𝐶𝐶 0 = 0. (7) 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 cos cos cos cos cos � 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑏𝑏 − 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑐𝑐 − 𝑝𝑝−𝑎𝑎 � Therefore: 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 − 𝐴𝐴 = 0 𝑏𝑏 𝑎𝑎 𝑐𝑐 𝑎𝑎 . (8) cos cos cos cos 𝑝𝑝−𝑏𝑏 − 𝑝𝑝−𝑎𝑎 𝑝𝑝−𝑐𝑐 − 𝑝𝑝−𝑎𝑎 � cos = � cos = cos = Considering that 2 2 2, 2 2 2, 𝐵𝐵 − 𝐴𝐴 𝐶𝐶 − 𝐴𝐴𝑏𝑏 +𝑐𝑐 −𝑎𝑎 𝑎𝑎 +𝑐𝑐 −𝑏𝑏 2𝑏𝑏𝑏𝑏 2𝑎𝑎𝑎𝑎 2 2 2, it is found that the relation𝐴𝐴 (8) is verified𝐵𝐵. 𝐶𝐶 𝑎𝑎 +𝑏𝑏 −𝑐𝑐 2 𝑎𝑎𝑎𝑎 100. From the given relation, it follows that = , therefore = ; we denote: = = , and2 we find2 similarly2 that2 𝐴𝐴𝑃𝑃 − 𝑃𝑃𝐸𝐸 𝐵𝐵𝑃𝑃 − 𝑃𝑃𝐷𝐷 = = and = = . We show that all the points , , 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑧𝑧 belong to the sides of the triangle . Indeed, if, for example, , and 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝑦𝑦 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝑥𝑥 𝐷𝐷 𝐸𝐸 𝐹𝐹 are in this order, then we have: + = ( + ) + ( ) = + = 𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵 𝐶𝐶 𝐷𝐷 – contradiction. Denoting by , , the lengths of the sides , , 𝐴𝐴𝐴𝐴 𝐵𝐵𝐵𝐵 𝑥𝑥 𝑦𝑦 𝑧𝑧 − 𝑦𝑦 𝑥𝑥 𝑧𝑧 and = , we find that = , = , = . These 𝐴𝐴𝐴𝐴 𝑎𝑎+𝑏𝑏+𝑐𝑐 𝑎𝑎 𝑏𝑏 𝑐𝑐 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 relations show that the points , , are contacts of the ex-inscribed circles 𝑝𝑝 2 𝑥𝑥 𝑝𝑝 − 𝑎𝑎 𝑦𝑦 𝑝𝑝 − 𝑏𝑏 𝑧𝑧 𝑝𝑝 − 𝑐𝑐 with , , , hence , , are collinear, and also , , and , , 𝐷𝐷 𝐸𝐸 𝐹𝐹 . Thus, we find that is orthology center of the antisupplementary triangle 𝐵𝐵𝐵𝐵 𝐶𝐶𝐶𝐶 𝐴𝐴𝐴𝐴 𝐼𝐼𝑎𝑎 𝐷𝐷 𝑃𝑃 𝐼𝐼𝑏𝑏 𝐸𝐸 𝑃𝑃 𝐼𝐼𝐶𝐶 𝐹𝐹 in relation to the given triangle . We noticed that this triangle 𝑃𝑃 𝑃𝑃 and have as orthology center the center of the circle circumscribed to 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝐴𝐴𝐴𝐴𝐴𝐴 the triangle , therefore the point and the center of the circle inscribed 𝐴𝐴𝐴𝐴𝐴𝐴 in the triangle . 𝐼𝐼𝑎𝑎𝐼𝐼𝑏𝑏𝐼𝐼𝑐𝑐 𝑃𝑃 𝐴𝐴𝐴𝐴𝐴𝐴

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BIBLIOGRAPHY

[1] Barbu, C. Teoreme din geometria triunghiului (Theorems of triangle geometry). Editura Unique, Bacău, 2008. [2] Botez, M. Șt. Probleme de geometrie (Geometry problems). Editura Tehnică, București, 1976. [3] Buricea B., Pasol V. Teoreme și probleme de geometrie clasică (Theorems and problems of classical geometry). Editura Lotus, Craiova, 1995. [4] Brânzei, D. Bazele raționamentului geometric (The basis of geometric reasoning). Editura Academiei, București, 1983. [5] Brânzei, D. Geometrie circumstanțială (Circumstantial geometry). Editura Junimea, Iași, 1984. [6] Boju V., Funar L. The Math problems Notebook, Birichäuser, Boston Basel Berlin, 2007. [7] Brânzei, D., Negrescu, A. Probleme de pivotare (Pivot’s problems). Editura “Recreații Matematice”, Iași, 2011. [8] Cocea, C. Noi probleme de geometrie (New geometry problems). Editura Spiru Haret, Iași, 1997. [9] Coșniță, C. Teoreme și probleme de matematici (Mathematical theorems and problems). Editura Didactică and Pedagogică, București, 1958. [10] Coșniță, C. Coordonées Baricentiques. București, 1941. [11] Coandă, C. Geometrie analitică in coordonate baricentrice (Analytical geometry in barycentric coordinates). Editura Reprograph, Craiova, 2005. [12] Efremov, D. Noua geometrie a triunghiului (The new geometry of the triangle). Odesa, 1902. Traducere de Mihai Miculița, Editura Cril, Zalău, 2010. [13] Hadamand, J. Lecții de geometrie plană (Lessons in plane geometry). Editura Tehnică, București, 1960. [14] Lalescu, T. Geometria triunghiului (The geometry of the triangle). Editura Tineretului, București, 1958. [15] Mihalescu, C. Geometria elementelor remarcabile (The geometry of the remarkable elements). Editura Tehnică, București, 1957. [16] Mihăileanu, N.N. Lecții complementare de geometrie (Complementary geometry lessons). Editura Didactică și Pedagogică, București, 1976. [17] Nicolaescu, L., Boskof V. Probleme practice de geometrie (Practical problems of geometry). Editura Tehnică, București, 1990.

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[18] Pop, O., Minculete, N., Bencze, M. An Introduction to Quadrilateral Geometry. Editura Didactică și Pedagogică, București, 2013. [19] Johnson, A. R. Advanced Euclidean Geometry. Dover Publications, Inc., Mineola, New York, 2007. [20] Pătrașcu, Ion. Probleme de geometrie plană (Problems of plane geometry). Editura Cardinal, Craiova, 1996. [21] Pătrașcu, I., Smarandache, F. Variance on topics of plane geometry. Educational Publishing, Columbus, Ohio, 2013. [22] Pătrașcu, I., Smarandache, F. Complements to Classic Topics of Circles Geometry. Pons Editions, Brussels, 2016. [23] Smaranda, D., Soare, N. Transformări geometrice (Geometric transformations). Editura Academiei R.S.R., București, 1988. [24] Smarandache, F., Pătrașcu, I. The Geometry of Homological Triangles. The Educational Publishing, Columbus, Ohio, 2012. [25] Smarandache, F. Multispace & Multistructure Neutrosophic Transdisciplinarity (100 Collected Papers of Science), vol. IV. North- European Scientific Publishers, Honco, Finland, 2010. [26] Țițeica, Gh. Culegere de probleme de geometrie (Collection of geometry problems). Editura Tehnică, București, 1965.

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BIBLIOGRAPHY

The idea of this book came up when writing our previous book, The Geometry of Homological Triangles (2012). As there, we try to graft on the central theme, of the orthological triangles, many results from the elementary geometry. In particular, we approach the connection between the orthological and homological triangles; also, we review the "S" triangles, highlighted for the first time by the great Romanian mathematician Traian Lalescu. The book is addressed to both those who have studied and love geometry, as well as to those who discover it now, through study and training, in order to obtain special results in school competitions. In this regard, we have sought to prove some properties and theorems in several ways: synthetic, vectorial, analytical. Prof. ION PĂTRAŞCU Prof. FLORENTIN SMARANDACHE

Ion Pătrașcu, Florentin Smarandache 311 THE GEOMETRY OF THE ORTHOLOGICAL TRIANGLES