Complements to Classic Topics of Circles Geometry

Home , Central angle, Concyclic points, Harmonic quadrilateral, Homothetic center, Inscribed angle, Simson line

Ion Patrascu | Florentin Smarandache

Pons Editions Brussels | 2016

Complements to Classic Topics of Circles Geometry

Ion Patrascu | Florentin Smarandache

Complements to Classic Topics of Circles Geometry

Ion Patrascu, Florentin Smarandache

In the memory of the first author's father Mihail Patrascu and the second author's mother Maria (Marioara) Smarandache, recently passed to eternity...

Complements to Classic Topics of Circles Geometry

Ion Patrascu | Florentin Smarandache

Complements to Classic Topics

of Circles Geometry

Pons Editions Brussels | 2016

Ion Patrascu, Florentin Smarandache

All rights reserved. This book is protected by copyright. No part of this book may be reproduced in any form or by any means, including photocopying or using any information storage and retrieval system without written permission from the copyright owners.

ISBN 978-1-59973-465-1

Complements to Classic Topics of Circles Geometry

Contents

Introduction ...... 15 Lemoine’s Circles ...... 17 1st Theorem...... 17 Proof...... 17 2nd Theorem...... 19 Proof...... 19 Remark...... 21 References...... 22 Lemoine’s Circles Radius Calculus ...... 23 1st Theorem ...... 23 Proof...... 23 Consequences...... 25 1st Proposition...... 25 Proof...... 25 2nd Proposition...... 26 Proof...... 27 Remarks...... 28 3rd Proposition...... 28 Proof...... 28 Remark...... 29 References...... 29 Radical Axis of Lemoine’s Circles ...... 31 1st Theorem...... 31 2nd Theorem...... 31 1st Remark...... 31 1st Proposition...... 32 Proof...... 32 5

Ion Patrascu, Florentin Smarandache

Comment...... 34 2nd Remark...... 34 2nd Proposition...... 34 References...... 34 Generating Lemoine’s circles ...... 35 1st Definition...... 35 1st Proposition...... 35 2nd Definition...... 35 1st Theorem...... 36 3rd Definition...... 36 1st Lemma...... 36 Proof...... 36 Remark...... 38 2nd Theorem...... 38 Proof...... 38 Further Remarks...... 40 References...... 40 The Radical Circle of Ex-Inscribed Circles of a Triangle ...... 41 1st Theorem...... 41 Proof...... 41 2nd Theorem...... 43 Proof...... 43 Remark...... 44 3rd Theorem...... 44 Proof...... 45 References...... 48 The Polars of a Radical Center ...... 49 1st Theorem...... 49 Proof...... 50

Complements to Classic Topics of Circles Geometry

2nd Theorem...... 51 Proof...... 52 Remarks...... 52 References...... 53 Regarding the First Droz-Farny’s Circle ...... 55 1st Theorem...... 55 Proof...... 55 Remarks...... 57 2nd Theorem...... 57 Remark...... 58 Definition...... 58 Remark...... 58 3rd Theorem...... 58 Proof...... 59 Remark...... 60 4th Theorem...... 60 Proof...... 60 Reciprocally...... 61 Remark...... 62 References...... 62 Regarding the Second Droz-Farny’s Circle...... 63 1st Theorem...... 63 Proof...... 63 1st Proposition...... 65 Proof...... 65 Remarks...... 66 2nd Theorem...... 66 Proof...... 67 Reciprocally...... 67 Remarks...... 68

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2nd Proposition...... 68 Proof...... 69 References...... 70 Neuberg’s Orthogonal Circles ...... 71 1st Definition...... 71 2nd Definition...... 72 3rd Definition...... 72 1st Proposition...... 72 Proof...... 73 Consequence...... 74 2nd Proposition...... 74 Proof...... 74 4th Definition...... 75 3rd Proposition...... 75 Proof...... 75 Reciprocally...... 76 4th Proposition...... 76 Proof...... 77 References...... 78 Lucas’s Inner Circles ...... 79 1. Definition of the Lucas’s Inner Circles ...... 79 Definition...... 80 2. Calculation of the Radius of the A-Lucas Inner Circle ...... 80 Note...... 81 1st Remark...... 81 3. Properties of the Lucas’s Inner Circles ...... 81 1st Theorem...... 81 Proof...... 81 2nd Definition...... 83

Complements to Classic Topics of Circles Geometry

Remark...... 83 2nd Theorem...... 84 3rd Definition...... 84 3rd Theorem...... 84 Proof...... 85 Remarks...... 86 4th Definition...... 87 1st Property...... 87 Proof...... 87 2nd Property...... 88 Proof...... 88 References...... 88 Theorems with Parallels Taken through a Triangle’s Vertices and Constructions Performed only with the Ruler ...... 89 1st Problem...... 89 2nd Problem...... 89 3rd Problem...... 90 1st Lemma...... 90 Proof...... 91 Remark...... 92 2nd Lemma...... 92 Proof...... 92 3rd Lemma...... 93 Solution...... 93 Construction’s Proof...... 93 Remark...... 94 1st Theorem...... 95 Proof...... 96 2nd Theorem...... 97

Ion Patrascu, Florentin Smarandache

Proof...... 98 Remark...... 99 3rd Theorem...... 99 Proof...... 99 Reciprocally...... 100 Solution to the 1st problem...... 101 Solution to the 2nd problem...... 101 Solution to the 3rd problem...... 102 References...... 102 Apollonius’s Circles of kth Rank ...... 103 1st Definition...... 103 2nd Definition...... 103 1st Theorem...... 103 Proof...... 104 3rd Definition...... 105 2nd Theorem...... 105 Proof...... 106 3rd Theorem...... 107 Proof...... 107 4th Theorem...... 108 Proof...... 108 1st Remark...... 108 5th Theorem...... 108 Proof...... 109 6th Theorem...... 109 Proof...... 109 4th Definition...... 110 1st Property...... 110 2nd Remark...... 110 7th Theorem...... 111

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Proof...... 111 References...... 112 Apollonius’s Circle of Second Rank ...... 113 1st Definition...... 113 1st Theorem...... 113 1st Proposition...... 114 Proof...... 114 Remarks...... 115 2nd Definition...... 115 3rd Definition...... 116 2nd Proposition...... 116 Proof...... 116 Remarks...... 118 Open Problem...... 119 References...... 120 A Sufficient Condition for the Circle of the 6 Points to Become Euler’s Circle ...... 121 1st Definition...... 121 1st Remark...... 122 2nd Definition...... 122 2nd Remark...... 122 1st Proposition...... 122 Proof...... 122 3rd Remark...... 123 3rd Definition...... 123 4th Remark...... 124 1st Theorem...... 124 Proof...... 124 4th Definition...... 126 2nd Proposition...... 126

Ion Patrascu, Florentin Smarandache

1st Proof...... 126 2nd Proof...... 127 5th Remark...... 128 References...... 128 An Extension of a Problem of Fixed Point ...... 129 Proof...... 130 1st Remark...... 132 2nd Remark...... 132 3rd Remark...... 133 References...... 133 Some Properties of the Harmonic Quadrilateral 135 1st Definition...... 135 2nd Definition...... 135 1st Proposition...... 135 2nd Proposition...... 136 Proof...... 136 Remark 1...... 137 3rd Proposition...... 137 Remark 2...... 138 Proposition 4...... 138 Proof...... 139 3rd Remark...... 140 5th Proposition...... 140 6th Proposition...... 140 Proof...... 140 3rd Definition...... 142 4th Remark...... 142 7th Proposition...... 143 Proof...... 143 5th Remark...... 144

Complements to Classic Topics of Circles Geometry

8th Proposition...... 145 Proof...... 145 4th Definition...... 146 6th Remark...... 146 9th Proposition...... 146 Proof...... 146 10th Proposition...... 147 Proof...... 147 7th Remark...... 148 11th Proposition...... 148 Proof...... 149 References...... 149 Triangulation of a Triangle with Triangles having Equal Inscribed Circles ...... 151 Solution...... 151 Building the point D...... 153 Proof...... 154 Discussion...... 155 Remark...... 156 Open problem...... 156 An Application of a Theorem of Orthology ...... 157 Proposition...... 157 Proof...... 158 Remark...... 160 Note...... 160 References...... 162 The Dual of a Theorem Relative to the Orthocenter of a Triangle ...... 163 1st Theorem...... 163 Proof...... 164

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Note...... 165 2nd Theorem...... 165 Proof...... 166 References...... 168 The Dual Theorem Concerning Aubert’s Line .... 169 1st Theorem...... 169 Proof...... 169 Note...... 171 1st Definition...... 171 Note...... 171 2nd Definition...... 172 2nd Theorem...... 172 Note...... 172 3rd Theorem...... 172 Proof...... 173 Notes...... 174 4th Theorem...... 174 Proof...... 174 References...... 176 Bibliography ...... 177

Complements to Classic Topics of Circles Geometry

Introductory Note

We approach several themes of classical geometry of the circle and complete them with some original results, showing that not everything in traditional math is revealed, and that it still has an open character. The topics were chosen according to authors’ aspiration and attraction, as a poet writes lyrics about spring according to his emotions.

Ion Patrascu, Florentin Smarandache

Complements to Classic Topics of Circles Geometry

Lemoine’s Circles

In this article, we get to Lemoine's circles in a different manner than the known one.

1st Theorem.

Let 퐴퐵퐶 a triangle and 퐾 its simedian center. We take through K the parallel 퐴1퐴2 to 퐵퐶, 퐴1 ∈ (퐴퐵), 퐴2 ∈ (퐴퐶); through 퐴2 we take the antiparallels 퐴2퐵1 to 퐴퐵 in relation to 퐶퐴 and 퐶퐵 , 퐵1 ∈ (퐵퐶) ; through 퐵1 we take the parallel 퐵1퐵2 to 퐴퐶, 퐵2 ∈ 퐴퐵; through 퐵2 we take the antiparallels 퐵1퐶1 to 퐵퐶 , 퐶1 ∈ (퐴퐶) , and through 퐶1 we take the parallel 퐶1퐶2 to 퐴퐵, 퐶1 ∈ (퐵퐶). Then:

i. 퐶2퐴1 is an antiparallel of 퐴퐶; ii. 퐵1퐵2 ∩ 퐶1퐶2 = {퐾}; iii. The points 퐴1, 퐴2 , 퐵1 , 퐵2, 퐶1, 퐶2 are concyclical (the first Lemoine’s circle).

Proof.

i. The quadrilateral 퐵퐶2퐾퐴 is a parallelogram, and its center, i.e. the

middle of the segment (퐶2퐴1), belongs to the simedian 퐵퐾; it follows that 퐶2퐴2 is an antiparallel to 퐴퐶(see Figure 1).

Ion Patrascu, Florentin Smarandache

ii. Let {퐾′} = 퐴1퐴2 ∩ 퐵1퐵2 , because the quadrilateral 퐾′퐵1퐶퐴2 is a parallelogram; it follows that 퐶퐾′ is a simedian; also,

퐶퐾 is a simedian, and since 퐾, 퐾′ ∈ 퐴1퐴2, it follows that we have 퐾′ = 퐾.

iii. 퐵2퐶1 being an antiparallel to 퐵퐶 and 퐴1퐴2 ∥ 퐵퐶 , it means that 퐵2퐶1 is an antiparallel to 퐴1퐴2 , so the points 퐵2, 퐶1, 퐴2, 퐴1 are concyclical. From 퐵1퐵2 ∥ 퐴퐶 , ∢퐵2퐶1퐴 ≡ ∢퐴퐵퐶 , ∢퐵1퐴2퐶 ≡ ∢퐴퐵퐶 we get that the quadrilateral 퐵2퐶1퐴2퐵1 is an isosceles trapezoid, so the points

퐵2, 퐶1, 퐴2, 퐵1 are concyclical. Analogously, it can be shown that the quadrilateral

퐶2퐵1퐴2퐴1 is an isosceles trapezoid, therefore the points 퐶2, 퐵1, 퐴2, 퐴1 are concyclical.

Figure 1

Complements to Classic Topics of Circles Geometry

From the previous three quartets of concyclical points, it results the concyclicity of the points belonging to the first Lemoine’s circle.

2nd Theorem.

In the scalene triangle 퐴퐵퐶, let K be the simedian center. We take from 퐾 the antiparallel 퐴1퐴2 to 퐵퐶; 퐴1 ∈ 퐴퐵, 퐴2 ∈ 퐴퐶; through 퐴2 we build 퐴2퐵1 ∥ 퐴퐵; 퐵1 ∈ (퐵퐶), then through 퐵1 we build 퐵1퐵2 the antiparallel to 퐴퐶, 퐵2 ∈ (퐴퐵), and through 퐵2 we build 퐵2퐶1 ∥ 퐵퐶, 퐶1 ∈ 퐴퐶 , and, finally, through 퐶1 we take the antiparallel 퐶1퐶2 to 퐴퐵, 퐶2 ∈ (퐵퐶). Then:

i. 퐶2퐴1 ∥ 퐴퐶; ii. 퐵1퐵2 ∩ 퐶1퐶2 = {퐾}; iii. The points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are concyclical (the second Lemoine’s circle).

Proof.

i. Let {퐾′} = 퐴1퐴2 ∩ 퐵1퐵2 , having ∢퐴퐴1퐴2 = ∢퐴퐶퐵 and ∢퐵퐵1퐵2 ≡ ∢퐵퐴퐶 because 퐴1퐴2 și 퐵1퐵2 are antiparallels to 퐵퐶 , 퐴퐶 , respectively, it follows that ∢퐾′퐴1퐵2 ≡ ∢퐾′퐵2퐴1 , so 퐾′퐴1 = 퐾′퐵2 ; having 퐴1퐵2 ∥ 퐵1퐴2 as well, it follows that also 퐾′퐴2 = 퐾′퐵1 , so 퐴1퐴2 = 퐵1퐵2 . Because 퐶1퐶2 and 퐵1퐵2 are antiparallels to 퐴퐵 and 퐴퐶 , we

Ion Patrascu, Florentin Smarandache

have 퐾"퐶2 = 퐾"퐵1; we noted {퐾"} = 퐵1퐵2 ∩ 퐶1퐶2; since 퐶1퐵2 ∥ 퐵1퐶2, we have that the triangle 퐾"퐶1퐵2 is also isosceles, therefore 퐾"퐶1 = 퐶1퐵2, and we get that 퐵1퐵2 = 퐶1퐶2. Let {퐾′′′} = 퐴1퐴2 ∩ 퐶1퐶2 ; since 퐴1퐴2 and 퐶1퐶2 are antiparallels to 퐵퐶 and 퐴퐵 , we get that the triangle 퐾′′′퐴2퐶1 is isosceles, so 퐾′′′퐴2 = 퐾′′′퐶1, but 퐴1퐴2 = 퐶1퐶2 implies that 퐾′′′퐶2 = 퐾′′′퐴1 , then ∢퐾′′′퐴1퐶2 ≡ ∢퐾′′′퐴2퐶1 and, accordingly, 퐶2퐴1 ∥ 퐴퐶.

Figure 2

ii. We noted {퐾′} = 퐴1퐴2 ∩ 퐵1퐵2 ; let {푋} = 퐵2퐶1 ∩ 퐵1퐴2 ; obviously, 퐵퐵1푋퐵2 is a parallelogram; if 퐾0 is the middle of (퐵1퐵2), then 퐵퐾0 is a simedian, since 퐵1퐵2 is an antiparallel to 퐴퐶, and the middle of the antiparallels of 퐴퐶 are situated on the

Complements to Classic Topics of Circles Geometry

simedian 퐵퐾 . If 퐾0 ≠ K, then 퐾0퐾 ∥ 퐴1퐵2 (because 퐴1퐴2 = 퐵1퐵2 and 퐵1퐴2 ∥ 퐴1퐵2 ), on the other hand, 퐵, 퐾0, K are collinear (they belong to the simedian 퐵퐾 ),

therefore 퐾0퐾 intersects 퐴퐵 in 퐵, which is absurd, so 퐾0 =K, and, accordingly, 퐵1퐵2 ∩ 퐴1퐴2 = {퐾} . Analogously, we prove that 퐶1퐶2 ∩ 퐴1퐴2 = {퐾}, so 퐵1퐵2 ∩ 퐶1퐶2 = {퐾}. iii. K is the middle of the congruent

antiparalells 퐴1퐴2 , 퐵1퐵2 , 퐶1퐶2 , so 퐾퐴1 = 퐾퐴2 = 퐾퐵1 = 퐾퐵2 = 퐾퐶1 = 퐾퐶2 . The simedian center 퐾 is the center of the second Lemoine’s circle.

Remark.

The center of the first Lemoine’s circle is the middle of the segment [푂퐾], where 푂 is the center of the circle circumscribed to the triangle 퐴퐵퐶. Indeed, the perpendiculars taken from 퐴, 퐵, 퐶 on the antiparallels 퐵2퐶1 , 퐴1퐶2 , 퐵1퐴2 respectively pass through O, the center of the circumscribed circle (the antiparallels have the directions of the tangents taken to the circumscribed circle in 퐴, 퐵, 퐶). The mediatrix of the segment 퐵2퐶1 pass though the middle of 퐵2퐶1 , which coincides with the middle of 퐴퐾 , so is the middle line in the triangle 퐴퐾푂 passing through the middle of (푂퐾) . Analogously, it follows that the mediatrix of 퐴1퐶2 pass through the middle 퐿1 of [푂퐾]. 21

Ion Patrascu, Florentin Smarandache

References.

[1] D. Efremov: Noua geometrie a triunghiului [The New Geometry of the Triangle], translation from Russian into Romanian by Mihai Miculița, Cril Publishing House, Zalau, 2010. [2] Gh. Mihalescu: Geometria elementelor remarcabile [The Geometry of the Outstanding Elements], Tehnica Publishing House, Bucharest, 1957. [3] Ion Patrascu, Gh. Margineanu: Cercul lui Tucker [Tucker’s Circle], in Alpha journal, year XV, no. 2/2009.

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Lemoine’s Circles Radius Calculus

For the calculus of the first Lemoine’s circle, we will first prove:

1st Theorem (E. Lemoine – 1873)

The first Lemoine’s circle divides the sides of a triangle in segments proportional to the squares of the triangle’s sides. Each extreme segment is proportional to the corresponding adjacent side, and the chord-segment in the Lemoine’s circle is proportional to the square of the side that contains it.

Proof.

퐵퐶 퐶 퐵 퐵 퐶 We will prove that 2 = 2 1 = 1 . 푐2 푎2 푏2 In figure 1, 퐾 represents the symmedian center of the triangle 퐴퐵퐶 , and 퐴1퐴2 ; 퐵1퐵2 ; 퐶1퐶2 represent Lemoine parallels.

The triangles 퐵퐶2퐴1 ; 퐶퐵1퐴2 and 퐾퐶2퐴1 have heights relative to the sides 퐵퐶2; 퐵1퐶 and 퐶2퐵1 equal (퐴1퐴2 ∥ 퐵퐶).

Ion Patrascu, Florentin Smarandache

Hence: 퐴푟푒푎 퐵퐴 퐶 퐴푟푒푎 퐾퐶 퐴 퐴푟푒푎 퐶퐵 퐴 ∆ 1 2 = ∆ 2 1 = ∆ 1 2 . (1) 퐵퐶2 퐶2퐵1 퐵1퐶

Figure 1

On the other hand: 퐴1퐶2 and 퐵1퐴2 being antiparallels with respect to 퐴퐶 and 퐴퐵, it follows that

Δ퐵퐶2퐴1~훥퐵퐴퐶 and Δ퐶퐵1퐴2~훥퐶퐴퐵 , likewise 퐾퐶2 ∥ 퐴퐶 implies: Δ퐾퐶2퐵1~훥퐴퐵퐶. We obtain: 2 퐴푟푒푎∆ 퐵퐶2퐴1 퐵퐶2 = 2 ; 퐴푟푒푎∆ 퐴퐵퐶 푐 2 퐴푟푒푎∆ 퐾퐶2퐵1 퐶2퐵1 = 2 ; 퐴푟푒푎∆ 퐴퐵퐶 푎 2 퐴푟푒푎∆ 퐶퐵1퐴2 퐶퐵1 = 2 . (2) 퐴푟푒푎∆ 퐴퐵퐶 푏 If we denote 퐴푟푒푎∆ 퐴퐵퐶 = 푆, we obtain from the relations (1) and (2) that: 퐵퐶 퐶 퐵 퐵 퐶 2 = 2 1 = 1 . 푐2 푎2 푏2

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Consequences.

1. According to the 1st Theorem, we find that: 푎푐2 푎푏2 푎3 퐵퐶 = ; 퐵 퐶 = ; 퐵 퐶 = . 2 푎2+푏2+푐2 1 푎2+푏2+푐2 1 2 푎2+푏2+푐2 2. We also find that: 퐵 퐶 퐴 퐶 퐴 퐵 1 2 = 2 1 = 1 2 , 푎3 푏3 푐3 meaning that: “The chords determined by the first Lemoine’s circle on the triangle’s sides are proportional to the cubes of the sides.” Due to this property, the first Lemoine’s circle is known in England by the name of triplicate ratio circle.

1st Proposition.

The radius of the first Lemoine’s circle, 푅퐿1 is given by the formula: 1 푅2(푎2+푏2+푐2)+ 푎2푏2푐2 푅2 = ∙ , (3) 퐿1 4 (푎2+푏2+푐2)2 where 푅 represents the radius of the circle inscribed in the triangle 퐴퐵퐶.

Proof.

Let 퐿 be the center of the first Lemoine’s circle that is known to represent the middle of the segment (푂퐾) – 푂 being the center of the circle inscribed in the triangle 퐴퐵퐶.

Ion Patrascu, Florentin Smarandache

푎 (푐2+푎2) Considering C1, we obtain 퐵퐵 = . 1 푎2+푏2+푐2 Taking into account the power of point 퐵 in relation to the first Lemoine’s circle, we have: 2 2 퐵퐶2 ∙ 퐵퐵1 = 퐵푇 − 퐿푇 , (퐵푇 is the tangent traced from 퐵 to the first Lemoine’s circle, see Figure 1). 2 2 Hence: 푅퐿1 = 퐵퐿 − 퐵퐶2 ∙ 퐵퐵1. (4) The median theorem in triangle 퐵푂퐾 implies that: 2∙(퐵퐾2+퐵푂2)−푂퐾2 퐵퐿2 = . 4 (푎2+푐2)∙푆 2푎푐∙푚 It is known that 퐾 = 푏 ; 푆 = 푏 , 푎2+푏2+푐2 푏 푎2+푐2 where 푆푏 and 푚푏 are the lengths of the symmedian 3푎2푏2푐2 and the median from 퐵, and 푂퐾2 = 푅2 − , see (푎2+푏2+푐2) (3). 2푎2푐2(푎2+푐2)−푎2푏2푐2 Consequently: 퐵퐾2 = , and (푎2+푏2+푐2)2 4푎2푐2(푎2+푐2)+푎2푏2푐2 4퐵퐿2 = 푅2 + . (푎2+푏2+푐2)2 푎2푐2(푎2+푐2) As: 퐵퐶 ∙ 퐵퐵 = , by replacing in (4), 2 1 (푎2+푏2+푐2)2 we obtain formula (3).

2nd Proposition.

The radius of the second Lemoine’s circle, 푅퐿2, is given by the formula: 푎푏푐 푅 = . (5) 퐿2 푎2+푏2+푐2

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Proof.

Figure 2

In Figure 2 above, 퐴1퐴2; 퐵1퐵2; 퐶1퐶2 are Lemoine antiparallels traced through symmedian center 퐾 that is the center of the second Lemoine’s circle, thence:

푅퐿2 = 퐾퐴1 = 퐾퐴2. If we note with 푆 and 푀 the feet of the symmedian and the median from 퐴, it is known that: 퐴퐾 푏2+푐2 = . 퐾푆 푎2 From the similarity of triangles 퐴퐴2퐴1 and 퐴퐵퐶, 퐴 퐴 퐴퐾 we have: 1 2= . 퐵퐶 퐴푀 퐴퐾 푏2+푐2 2푏푐 But: = and 퐴푆 = ∙ 푚 . 퐴푆 푎2+푏2+푐2 푏2+푐2 푎

퐴1퐴2 = 2푅퐿2, 퐵퐶 = 푎, therefore: 퐴퐾∙푎 푅퐿2 = , 2푚푎 2푏푐 ∙ 푚 and as 퐴퐾 = 푎 , formula (5) is a consequence. 푎2+푏2+푐2

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Remarks.

4푆 1. If we use 푡푔휔 = , 휔 being the Brocard’s 푎2+푏2+푐2 angle (see [2]), we obtain: 푅퐿2 = 푅 ∙ 푡푔휔. 2. If, in Figure 1, we denote with 푀1 the middle of the antiparallel 퐵2퐶1, which is equal to 푅퐿2 (due to their similarity), we thus find from the rectangular triangle 퐿푀1퐶1 that: 1 퐿퐶2 = 퐿푀2 + 푀 퐶2 , but 퐿푀2 = 푎2 and 푀 퐶 = 1 1 1 1 1 4 1 2 1 푅 ; it follows that: 2 퐿2 1 푅2 푅2 = (푅2 + 푅2 ) = (1 + 푡푔2휔). 퐿1 4 퐿2 4 We obtain: 푅 푅 = ∙ √1 + 푡푔2휔 . 퐿1 2

3rd Proposition.

The chords determined by the sides of the triangle in the second Lemoine’s circle are respectively proportional to the opposing angles cosines.

Proof.

퐾퐶2퐵1 is an isosceles triangle, ∢퐾퐶2퐵1 = 퐶 퐵 ∢퐾퐵 퐶 = ∢퐴; as 퐾퐶 = 푅 we have that cos 퐴 = 2 1 , 1 2 2 퐿2 2푅 퐿2 퐶 퐵 퐴 퐶 퐵 퐴 deci 2 1 = 2푅 , similary: 2 1 = 2 1 = 2푅 cos 퐴 퐿2 cos 퐵 cos 퐶 퐿2.

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Remark.

Due to this property of the Lemoine’s second circle, in England this circle is known as the cosine circle.

References.

[1] D. Efremov, Noua geometrie a triunghiului [The New Geometry of the Triangle], translation from Russian into Romanian by Mihai Miculița, Cril Publishing House, Zalau, 2010. [2] F. Smarandache and I. Patrascu, The Geometry of Homological Triangles, The Education Publisher, Ohio, USA, 2012. [3] I. Patrascu and F. Smarandache, Variance on Topics of Plane Geometry, Educational Publisher, Ohio, USA, 2013.

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Radical Axis of Lemoine’s Circles

In this article, we emphasize the radical axis of the Lemoine’s circles. For the start, let us remind:

1st Theorem.

The parallels taken through the simmedian center 퐾 of a triangle to the sides of the triangle determine on them six concyclic points (the first Lemoine’s circle).

2nd Theorem.

The antiparallels taken through the triangle’s simmedian center to the sides of a triangle determine six concyclic points (the second Lemoine’s circle).

1st Remark.

If 퐴퐵퐶 is a scalene triangle and 퐾 is its simmedian center, then 퐿 , the center of the first Lemoine’s circle, is the middle of the segment [푂퐾], where 푂 is the center of the circumscribed circle, and

Ion Patrascu, Florentin Smarandache the center of the second Lemoine’s circle is 퐾 . It follows that the radical axis of Lemoine’s circles is perpendicular on the line of the centers 퐿퐾, therefore on the line 푂퐾.

1st Proposition.

The radical axis of Lemoine’s circles is perpendicular on the line 푂퐾 raised in the simmedian center 퐾.

Proof.

Let 퐴1퐴2 be the antiparallel to 퐵퐶 taken through

퐾, then 퐾퐴1 is the radius 푅퐿2 of the second Lemoine’s circle; we have: 푎푏푐 푅 = . 퐿2 푎2+푏2+푐2

Figure 1

Complements to Classic Topics of Circles Geometry

′ ′ Let 퐴1퐴2 be the Lemoine’s parallel taken to 퐵퐶; we evaluate the power of 퐾 towards the first Lemoine’s circle. We have: ⃗⃗⃗⃗⃗⃗⃗⃗′ ⃗⃗⃗⃗⃗⃗⃗⃗′ 2 2 퐾퐴1 ∙ 퐾퐴2 = 퐿퐾 − 푅퐿1. (1) Let 푆 be the simmedian leg from 퐴; it follows that: 퐾퐴′ 퐴퐾 퐾퐴′ 1 = − 2 . 퐵푆 퐴푆 푆퐶 We obtain: 퐴퐾 퐴퐾 퐾퐴′ = 퐵푆 ∙ and 퐾퐴′ = 푆퐶 ∙ , 1 퐴푆 2 퐴푆 퐵푆 푐2 퐴퐾 푏2+푐2 but = and = . 푆퐶 푏2 퐴푆 푎2+푏2+푐2 Therefore: 퐴퐾 2 −푎2푏2푐2 퐾퐴⃗⃗⃗⃗⃗⃗⃗⃗′ ∙ 퐾퐴⃗⃗⃗⃗⃗⃗⃗⃗′ = −퐵푆 ∙ 푆퐶 ∙ ( ) = ; 1 2 퐴푆 (푏2 + 푐2)2 2 (푏2+푐2) = −푅2 . (2) (푎2+푏2+푐2)2 퐿2 We draw the perpendicular in 퐾 on the line 퐿퐾 and denote by 푃 and 푄 its intersection to the first Lemoine’s circle. ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 2 We have 퐾푃 ∙ 퐾푄 = −푅퐿2 ; by the other hand, 퐾푃 = 퐾푄 (푃푄 is a chord which is perpendicular to the diameter passing through 퐾).

It follows that 퐾푃 = 퐾푄 = 푅퐿2 , so 푃 and 푄 are situated on the second Lemoine’s circle. Because 푃푄 is a chord which is common to the Lemoine’s circles, it obviously follows that 푃푄 is the radical axis.

Ion Patrascu, Florentin Smarandache

Comment.

Equalizing (1) and (2), or by the Pythagorean theorem in the triangle 푃퐾퐿, we can calculate 푅퐿1. 3푎2푏2푐2 It is known that: 푂퐾2 = 푅2 − , and (푎2+푏2+푐2)2 1 since 퐿퐾 = 푂퐾, we find that: 2 1 푎2푏2푐2 푅2 = ∙ [푅2 + ]. 퐿1 4 (푎2+푏2+푐2)2

2nd Remark.

The 1st Proposition, ref. the radical axis of the Lemoine’s circles, is a particular case of the following Proposition, which we leave to the reader to prove.

2nd Proposition.

If 풞(푂1, 푅1) şi 풞(푂2, 푅2) are two circles such as 2 the power of center 푂1 towards 풞(푂2, 푅2) is −푅1 , then the radical axis of the circles is the perpendicular in

푂1 on the line of centers 푂1푂2.

References.

[1] F. Smarandache, Ion Patrascu: The Geometry of Homological Triangles, Education Publisher, Ohio, USA, 2012. [2] Ion Patrascu, F. Smarandache: Variance on Topics of Plane Geometry, Education Publisher, Ohio, USA, 2013.

Complements to Classic Topics of Circles Geometry

Generating Lemoine’s circles

In this paper, we generalize the theorem relative to the first Lemoine’s circle and thereby highlight a method to build Lemoine’s circles. Firstly, we review some notions and results.

1st Definition.

It is called a simedian of a triangle the symmetric of a median of the triangle with respect to the internal bisector of the triangle that has in common with the median the peak of the triangle.

1st Proposition.

In the triangle 퐴퐵퐶, the cevian 퐴푆, 푆 ∈ (퐵퐶), is a 푆퐵 퐴퐵 2 simedian if and only if = ( ) . For Proof, see [2]. 푆퐶 퐴퐶

2nd Definition.

It is called a simedian center of a triangle (or Lemoine’s point) the intersection of triangle’s simedians.

Ion Patrascu, Florentin Smarandache

1st Theorem.

The parallels to the sides of a triangle taken through the simedian center intersect the triangle’s sides in six concyclic points (the first Lemoine’s circle - 1873). A Proof of this theorem can be found in [2].

3rd Definition.

We assert that in a scalene triangle 퐴퐵퐶 the line 푀푁, where 푀 ∈ 퐴퐵 and 푁 ∈ 퐴퐶, is an antiparallel to 퐵퐶 if ∢푀푁퐴 ≡ ∢퐴퐵퐶.

1st Lemma.

In the triangle 퐴퐵퐶 , let 퐴푆 be a simedian, 푆 ∈ (퐵퐶). If 푃 is the middle of the segment (푀푁), having 푀 ∈ (퐴퐵) and 푁 ∈ (퐴퐶), belonging to the simedian 퐴푆, then 푀푁 and 퐵퐶 are antiparallels.

Proof.

We draw through 푀 and 푁, 푀푇 ∥ 퐴퐶 and 푁푅 ∥ 퐴퐵, 푅, 푇 ∈ (퐵퐶) , see Figure 1. Let {푄} = 푀푇 ∩ 푁푅 ; since 푀푃 = 푃푁 and 퐴푀푄푁 is a parallelogram, it follows that 푄 ∈ 퐴푆.

Complements to Classic Topics of Circles Geometry

Figure 1.

Thales's Theorem provides the relations: 퐴푁 퐵푅 = ; (1) 퐴퐶 퐵퐶 퐴퐵 퐵퐶 = . (2) 퐴푀 퐶푇 From (1) and (2), by multiplication, we obtain: 퐴푁 퐴퐵 퐵푅 ∙ = . (3) 퐴푀 퐴퐶 푇퐶 Using again Thales's Theorem, we obtain: 퐵푅 퐴푄 = , (4) 퐵푆 퐴푆 푇퐶 퐴푄 = . (5) 푆퐶 퐴푆 From these relations, we get 퐵푅 푇퐶 = , (6) 퐵푆 푆퐶 or 퐵푆 퐵푅 = . (7) 푆퐶 푇퐶 In view of Proposition 1, the relations (7) and (3) 퐴푁 퐴퐵 lead to = , which shows that ∆퐴푀푁~∆퐴퐶퐵 , so 퐴퐵 퐴퐶 ∢퐴푀푁 ≡ ∢퐴퐵퐶.

Ion Patrascu, Florentin Smarandache

Therefore, 푀푁 and 퐵퐶 are antiparallels in relation to 퐴퐵 and 퐴퐶.

Remark.

1. The reciprocal of Lemma 1 is also valid, meaning that if 푃 is the middle of the antiparallel 푀푁 to 퐵퐶, then 푃 belongs to the simedian from 퐴.

2nd Theorem. (Generalization of the 1st Theorem) Let 퐴퐵퐶 be a scalene triangle and 퐾 its simedian center. We take 푀 ∈ 퐴퐾 and draw 푀푁 ∥ 퐴퐵, 푀푃 ∥ 퐴퐶, where 푁 ∈ 퐵퐾, 푃 ∈ 퐶퐾. Then: i. 푁푃 ∥ 퐵퐶; ii. 푀푁, 푁푃 and 푀푃 intersect the sides of triangle 퐴퐵퐶 in six concyclic points.

Proof.

In triangle 퐴퐵퐶, let 퐴퐴1, 퐵퐵1, 퐶퐶1 the simedians concurrent in 퐾 (see Figure 2). We have from Thales' Theorem that: 퐴푀 퐵푁 = ; (1) 푀퐾 푁퐾 퐴푀 퐶푃 = . (2) 푀퐾 푃퐾 From relations (1) and (2), it follows that 퐵푁 퐶푃 = , (3) 푁퐾 푃퐾 which shows that 푁푃 ∥ 퐵퐶.

Complements to Classic Topics of Circles Geometry

Let 푅, 푆, 푉, 푊, 푈, 푇 be the intersection points of the parallels 푀푁, 푀푃, 푁푃 of the sides of the triangles to the other sides. Obviously, by construction, the quadrilaterals 퐴푆푀푊; 퐶푈푃푉; 퐵푅푁푇 are parallelograms. The middle of the diagonal 푊푆 falls on 퐴푀, so on the simedian 퐴퐾, and from 1st Lemma we get that 푊푆 is an antiparallel to 퐵퐶. Since 푇푈 ∥ 퐵퐶 , it follows that 푊푆 and 푇푈 are antiparallels, therefore the points 푊, 푆, 푈, 푇 are concyclic (4).

Figure 2.

Analogously, we show that the points 푈, 푉, 푅, 푆 are concyclic (5). From 푊푆 and 퐵퐶 antiparallels, 푈푉 and 퐴퐵 antiparallels, we have that ∢푊푆퐴 ≡ ∢퐴퐵퐶 and ∢푉푈퐶 ≡ ∢퐴퐵퐶 , therefore: ∢푊푆퐴 ≡ ∢푉푈퐶 , and since 39

Ion Patrascu, Florentin Smarandache

푉푊 ∥ 퐴퐶, it follows that the trapeze 푊푆푈푉 is isosceles, therefore the points 푊, 푆, 푈, 푉 are concyclic (6). The relations (4), (5), (6) drive to the concyclicality of the points 푅, 푈, 푉, 푆, 푊, 푇, and the theorem is proved.

Further Remarks.

2. For any point 푀 found on the simedian 퐴퐴1, by performing the constructions from hypothesis, we get a circumscribed circle of the 6 points of intersection of the parallels taken to the sides of triangle. 3. The 2nd Theorem generalizes the 1st Theorem because we get the second in the case the parallels are taken to the sides through the simedian center 푘. 4. We get a circle built as in 2nd Theorem from the first Lemoine’s circle by homothety of pole 푘 and of ratio 휆 ∈ ℝ. 5. The centers of Lemoine’s circles built as above belong to the line 푂퐾, where 푂 is the center of the circle circumscribed to the triangle 퐴퐵퐶.

References.

[1] Exercices de Géométrie, par F.G.M., Huitième édition, Paris VIe, Librairie Générale, 77, Rue Le Vaugirard. [2] Ion Patrascu, Florentin Smarandache: Variance on topics of Plane Geometry, Educational Publishing, Columbus, Ohio, 2013. 40

Complements to Classic Topics of Circles Geometry

The Radical Circle of Ex- Inscribed Circles of a Triangle

In this article, we prove several theorems about the radical center and the radical circle of ex-inscribed circles of a triangle and calculate the radius of the circle from vectorial considerations.

1st Theorem.

The radical center of the ex-inscribed circles of the triangle 퐴퐵퐶 is the Spiecker’s point of the triangle (the center of the circle inscribed in the median triangle of the triangle 퐴퐵퐶).

Proof.

We refer in the following to the notation in

Figure 1. Let 퐼푎, 퐼푏, 퐼푐 be the centers of the ex-inscribed circles of a triangle (the intersections of two external bisectors with the internal bisector of the other angle). Using tangents property taken from a point to a circle to be congruent, we calculate and find that:

퐴퐹푎 = 퐴퐸푎 = 퐵퐷푏 = 퐵퐹푏 = 퐶퐷푐 = 퐶퐸푐 = 푝, 퐵퐷푐 = 퐵퐹푐 = 퐶퐷푏 = 퐶퐸푏 = 푝 − 푎, 41

Ion Patrascu, Florentin Smarandache

퐶퐸푎 = 퐶퐷푎 = 퐴퐹푐 = 퐴퐸푐 = 푝 − 푏, 퐴퐹푏 = 퐴퐸푏 = 퐵퐹푐 = 퐵퐷푐 = 푝 − 푐. If 퐴1 is the middle of segment 퐷푐퐷푏 , it follows that 퐴1 has equal powers to the ex-inscribed circles (퐼푏) and (퐼푐). Of the previously set, we obtain that 퐴1 is the middle of the side 퐵퐶.

Figure 1.

Also, the middles of the segments 퐸푏퐸푐 and 퐹푏퐹푐, which we denote 푈 and 푉, have equal powers to the circles (퐼푏) and (퐼푐). The radical axis of the circles ( 퐼푏 ), ( 퐼푐 ) will include the points 퐴1, 푈, 푉. Because 퐴퐸푏 = 퐴퐹푏 and 퐴퐸푐 = 퐴퐹푐, it follows that 1 퐴푈 = 퐴푌 and we find that ∢퐴푈푉 = ∢퐴, therefore the 2 42

Complements to Classic Topics of Circles Geometry

radical axis of the ex-inscribed circles (퐹푏) and (퐼푐) is the parallel taken through the middle 퐴1 of the side 퐵퐶 to the bisector of the angle 퐵퐴퐶.

Denoting 퐵1 and 퐶1 the middles of the sides 퐴퐶, 퐴퐵, respectively, we find that the radical center of the ex-inscribed circles is the center of the circle inscribed in the median triangle 퐴1퐵1퐶1 of the triangle 퐴퐵퐶. This point, denoted 푆, is the Spiecker’s point of the triangle ABC.

2nd Theorem.

The radical center of the inscribed circle (퐼) and of the 퐵 −ex-inscribed and 퐶 −ex-inscribed circles of the triangle 퐴퐵퐶 is the center of the 퐴1 − ex-inscribed circle of the median triangle 퐴1퐵1퐶1, corresponding to the triangle 퐴퐵퐶).

Proof.

If 퐸 is the contact of the inscribed circle with 퐴퐶 and 퐸푏 is the contact of the 퐵 −ex-inscribed circle with 퐴퐶 , it is known that these points are isotomic, therefore the middle of the segment 퐸퐸푏 is the middle of the side 퐴퐶, which is 퐵1. This point has equal powers to the inscribed circle (퐼) and to the 퐵 −ex-inscribed circle (퐼푏), so it belongs to their radical axis.

Ion Patrascu, Florentin Smarandache

Analogously, 퐶1 is on the radical axis of the circles (퐼) and (퐼푐). The radical axis of the circles (퐼), (퐼푏 ) is the perpendicular taken from 퐵1 to the bisector 퐼퐼푏. This bisector is parallel with the internal bisector of the angle 퐴1퐵1퐶1 , therefore the perpendicular in 퐵1 on 퐼퐼푏 is the external bisector of the angle 퐴1퐵1퐶1 from the median triangle. Analogously, it follows that the radical axis of the circles (퐼), (퐼푐) is the external bisector of the angle 퐴1퐶1퐵1 from the median triangle. Because the bisectors intersect in the center of the circle 퐴1 -ex-inscribed to the median triangle 퐴1퐵1퐶1, this point 푆푎 is the center of the radical center of the circles (퐼), (퐼푏), (퐼푐).

Remark.

The theorem for the circles (퐼), (퐼푎), (퐼푏) and (퐼), (퐼푎 ), (퐼푐 ) can be proved analogously, obtaining the points 푆푐 and 푆푏.

3rd Theorem.

The radical circle’s radius of the circles ex- inscribed to the triangle 퐴퐵퐶 is given by the formula: 1 √푟2 + 푝2, where 푟 is the radius of the inscribed circle. 2

Complements to Classic Topics of Circles Geometry

Proof.

The position vector of the circle 퐼 of the inscribed circle in the triangle ABC is: 1 푃퐼⃗⃗⃗⃗ = (푎푃퐴⃗⃗⃗⃗⃗ + 푏푃퐵⃗⃗⃗⃗⃗ + 푐푃퐶⃗⃗⃗⃗⃗ ). 2푝 Spiecker’s point 푆 is the center of radical circle of ex-inscribed circle and is the center of the inscribed circle in the median triangle 퐴1퐵1퐶1, therefore: 1 1 1 1 푃푆⃗⃗⃗⃗ = ( 푎푃⃗⃗⃗⃗퐴⃗⃗⃗ + 푏푃⃗⃗⃗⃗퐵⃗⃗⃗ + 푐푃⃗⃗⃗⃗퐶⃗⃗⃗ ). 푝 2 1 2 1 2 1

Figure 2.

We denote by 푇 the contact point with the 퐴-ex- inscribed circle of the tangent taken from 푆 to this circle (see Figure 2).

Ion Patrascu, Florentin Smarandache

The radical circle’s radius is given by: 2 2 푆푇 = √푆퐼푎 − 퐼푎 1 퐼⃗⃗⃗⃗⃗⃗푆 = (푎퐼⃗⃗⃗⃗⃗퐴⃗⃗⃗ + 푏퐼⃗⃗⃗⃗⃗퐵⃗⃗⃗ + 푐퐼⃗⃗⃗⃗⃗퐶⃗⃗⃗ ). 푎 2푝 푎 1 푎 1 푎 1

We evaluate the product of the scales 퐼⃗⃗푎⃗⃗⃗⃗푆 ∙ 퐼⃗⃗푎⃗⃗⃗⃗푆 ; we have: 1 퐼 푆2 = (푎2퐼 퐴2 + 푏2퐼 퐵2 + 푐2퐼 퐶2 + 2푎푏퐼⃗⃗⃗⃗⃗퐴⃗⃗⃗ ∙ 푎 4푝2 푎 1 푎 1 푎 1 푎 1

퐼⃗⃗푎⃗⃗⃗퐵⃗⃗⃗1 + 2푏푐퐼⃗⃗푎⃗⃗⃗퐵⃗⃗⃗1 ∙ 퐼⃗⃗푎⃗⃗⃗퐶⃗⃗⃗1 + 2푎푐퐼⃗⃗푎⃗⃗⃗퐴⃗⃗⃗1 ∙ 퐼⃗⃗푎⃗⃗⃗퐶⃗⃗⃗1 ). From the law of cosines applied in the triangle

퐼푎퐴1퐵1, we find that: 1 2퐼⃗⃗⃗⃗⃗퐴⃗⃗⃗ ∙ 퐼⃗⃗⃗⃗⃗퐵⃗⃗⃗ = 퐼 퐴2 + 퐼 퐵2 − 푐2, therefore: 푎 1 푎 1 푎 1 푎 1 4 1 2푎푏퐼⃗⃗⃗⃗⃗퐴⃗⃗⃗ ∙ 퐼⃗⃗⃗⃗⃗퐵⃗⃗⃗ = 푎푏(퐼 퐴2 + 퐼 퐵2 − 푎푏푐2. 푎 1 푎 1 푎 1 푎 1 4 Analogously, we obtain: 1 2푏푐퐼⃗⃗⃗⃗⃗퐵⃗⃗⃗ ∙ 퐼⃗⃗⃗⃗⃗퐶⃗⃗⃗ = 푏푐(퐼 퐵2 + 퐼 퐶2 − 푎2푏푐, 푎 1 푎 1 푎 1 푎 1 4 1 2푎푐퐼⃗⃗⃗⃗⃗퐴⃗⃗⃗ ∙ 퐼⃗⃗⃗⃗⃗퐶⃗⃗⃗ = 푎푐(퐼 퐴2 + 퐼 퐶2 − 푎푏2푐. 푎 1 푎 1 푎 1 푎 1 4 1 퐼 푆2 = [(푎2 + 푎푏 + 푎푐)퐼 퐴2 + (푏2 + 푎푏 + 푎 4푝2 푎 1 푎푏푐 푏푐)퐼 퐵2 + (푐2 + 푏푐 + 푎푐)퐼 퐶2 − (푎 + 푏 + 푐)], 푎 1 푎 1 4 1 퐼 푆2 = [2푝(푎퐼 퐴2 + 푏퐼 퐵2 + 푐퐼 퐶2) − 2푅푆 ], 푎 4푝2 푎 1 푎 1 푎 1 푝 1 1 퐼 푆2 = (푎퐼 퐴2 + 푏퐼 퐵2 + 푐퐼 퐶2) − 푅푟. 푎 2푝 푎 1 푎 1 푎 1 2

From the right triangle 퐼푎퐷푎퐴1, we have that: 푎 2 퐼 퐴2 = 푟2 + 퐴 퐷2 = 푟2 + [ − (푝 − 푐)] = 푎 1 푎 1 푎 푎 2 (푐−푏)2 = 푟2 + . 푎 4 From the right triangles 퐼푎퐸푎퐵1 și 퐼푎퐹푎퐶1 , we find:

Complements to Classic Topics of Circles Geometry

푏 2 퐼 퐵2 = 푟2 + 퐵 퐸2 = 푟2 + [ − (푝 − 푏)] = 푎 1 푎 1 푎 푎 2 1 = 푟2 + (푎 + 푐)2, 푎 4 1 퐼 퐶2 = 푟2 + (푎 + 푏)2. 푎 1 푎 4 2 2 2 Evaluating 푎퐼푎퐴1 + 푏퐼푎퐵1 + 푐퐼푎퐶1 , we obtain: 2 2 2 푎퐼푎퐴1 + 푏퐼푎퐵1 + 푐퐼푎퐶1 = 1 1 = 2푝푟2 + 푝(푎푏 + 푎푐 + 푏푐) − 푎푏푐. 푎 2 4 But: 푎푏 + 푎푐 + 푏푐 = 푟2 + 푝2 + 4푅푟. It follows that: 1 1 1 [푎퐼 퐴2 + 푏퐼 퐵2 + 푐퐼 퐶2] = 푟2 + (푟2 + 푝2) + 푅푟 2푝 푎 1 푎 1 푎 1 푎 4 2 and 1 퐼 푆2 = 푟2 + (푟2 + 푝2). 푎 푎 4 Then, we obtain: 1 푆푇 = √푟2 + 푝2. 2

Ion Patrascu, Florentin Smarandache

References.

[1] C. Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental Theorems of Triangle Geometry]. Bacau, Romania: Editura Unique, 2008. [2] I. Patrascu, F. Smarandache: Variance on topics of plane geometry. Columbus: The Educational Publisher, Ohio, USA, 2013.

Complements to Classic Topics of Circles Geometry

The Polars of a Radical Center

In [1] , the late mathematician Cezar Cosnita, using the barycenter coordinates, proves two theorems which are the subject of this article. In a remark made after proving the first theorem, C. Cosnita suggests an elementary proof by employing the concept of polar. In the following, we prove the theorems based on the indicated path, and state that the second theorem is a particular case of the former. Also, we highlight other particular cases of these theorems.

1st Theorem.

Let 퐴퐵퐶 be a given triangle; through the pairs of points (퐵, 퐶) , (퐶, 퐴) and (퐴, 퐵) we take three circles such that their radical center is on the outside. The polar lines of the radical center of these circles in relation to each of them cut the sides 퐵퐶, 퐶퐴 and 퐴퐵 respectively in three collinear points.

Ion Patrascu, Florentin Smarandache

Proof.

We denote by 퐷, 푃, 퐹 the second point of intersection of the pairs of circles passing through (퐴, 퐵) and (퐵, 퐶) ; (퐵, 퐶) and (퐴, 퐶) , (퐵, 퐶) and (퐴, 퐵) respectively (see Figure 1).

Figure 1

Let 푅 be the radical center of those circles. In fact, {푅} = 퐴퐹 ∩ 퐵퐷 ∩ 퐶퐸.

We take from 푅 the tangents 푅퐷1 = 푅퐷2 to the circle (퐵, 퐶), 푅퐸1 = 푅퐸2 to the circle (퐴, 퐶) and 푅퐹1 = 푅퐹2 to the circle passing through (퐴, 퐵). Actually, we build the radical circle 풞(푅, 푅퐷1) of the given circles. The polar lines of 푅 to these circles are the lines

퐷1퐷2, 퐸1퐸2, 퐹1퐹2. These three lines cut 퐵퐶, 퐴퐶 and 퐴퐵 in the points 푋, 푌 and 푍 , and these lines are respectively the polar lines of 푅 in respect to the 50

Complements to Classic Topics of Circles Geometry circles passing through (퐵, 퐶), (퐶, 퐴) and (퐴, 퐵) . The polar lines are the radical axis of the radical circle with each of the circles passing through (퐵, 퐶), (퐶, 퐴), (퐴, 퐵), respectively. The points belong to the radical axis having equal powers to those circles, thereby 푋퐷1 ∙ 푋퐷2 = 푋퐶 ∙ 푋퐵. This relationship shows that the point 푋 has equal powers relative to the radical circle and to the circle circumscribed to the triangle 퐴퐵퐶; analogically, the point 푌 has equal powers relative to the radical circle and to the circle circumscribed to the triangle 퐴퐵퐶 ; and, likewise, the point 푍 has equal powers relative to the radical circle and to the circle circumscribed to the triangle 퐴퐵퐶. Because the locus of the points having equal powers to two circles is generally a line, i.e. their radical axis, we get that the points 푋, 푌 and 푍 are collinear, belonging to the radical axis of the radical circle and to the circle circumscribed to the given triangle.

2nd Theorem.

If 푀 is a point in the plane of the triangle 퐴퐵퐶 and the tangents in this point to the circles circumscribed to triangles 퐶, 푀퐴퐶, 푀퐴퐵, respectively, cut 퐵퐶, 퐶퐴 and 퐴퐵, respectively, in the points 푋, 푌, 푍, then these points are collinear.

Ion Patrascu, Florentin Smarandache

Proof.

The point 푀 is the radical center for the circles (푀퐵퐶), (푀퐴퐶), and (푀퐴퐵), and the tangents in 푀 to these circles are the polar lines to 푀 in relation to these circles. If 푋, 푌, 푍 are the intersections of these tangents (polar lines) with 퐵퐶, 퐶퐴, 퐴퐵, then they belong to the radical axis of the circumscribed circle to the triangle 퐴퐵퐶 and to the circle “reduced” to the point 푀 (푋푀2 = 푋퐵 ∙ 푋퐶, etc.). Being located on the radical axis of the two circles, the points 푋, 푌, 푍 are collinear.

Remarks.

1. Another elementary proof of this theorem is to be found in [3]. 2. If the circles from the 1st theorem are adjoint circles of the triangle 퐴퐵퐶 , then they intersect in Ω (the Brocard’s point). Therefore, we get that the tangents taken in Ω to the adjoin circles cut the sides 퐵퐶, 퐶퐴 and 퐴퐵 in collinear points.

Complements to Classic Topics of Circles Geometry

References.

[1] C. Coșniță: Coordonées baricentrique [Barycentric Coordinates], Bucarest – Paris: Librairie Vuibert, 1941. [3] I. Pătrașcu: Axe și centre radicale ale cercului adjuncte unui triunghi [Axis and radical centers of the adjoin circle of a triangle], in “Recreații matematice”, year XII, no. 1, 2010. [3] C. Mihalescu: Geometria elementelor remarcabile [The Geometry of Outstanding Elements], Bucharest: Editura Tehnica, 1957. [4] F. Smarandache, I. Patrascu: Geometry of Homological Triangle, Columbus: The Educational Publisher Inc., 2012.

Ion Patrascu, Florentin Smarandache

Complements to Classic Topics of Circles Geometry

Regarding the First Droz- Farny’s Circle

In this article, we define the first Droz- Farny’s circle, we establish a connection between it and a concyclicity theorem, then we generalize this theorem, leading to the generalization of Droz-Farny’s circle. The first theorem of this article was enunciated by J. Steiner and it was proven by Droz-Farny (Mathésis, 1901).

1st Theorem.

Let 퐴퐵퐶 be a triangle, 퐻 its orthocenter and

퐴1, 퐵1, 퐶1 the means of sides (퐵퐶), (퐶퐴), (퐴퐵). If a circle, having its center 퐻, intersects 퐵1퐶1 in 푃1, 푄1 ; 퐶1퐴1 in 푃2, 푄2 and 퐴1퐵1 in 푃3, 푄3 , then 퐴푃1 = 퐴푄1 = 퐵푃2 = 퐵푄2 = 퐶푃3 = 퐶푄3.

Proof.

Naturally, 퐻푃1 = 퐻푄1, 퐵1퐶1 ∥ 퐵퐶, 퐴퐻 ⊥ 퐵퐶. It follows that 퐴퐻 ⊥ 퐵1퐶1.

Ion Patrascu, Florentin Smarandache

Figure 1.

Therefore, 퐴퐻 is the mediator of segment 푃1푄1; similarly, 퐵퐻 and 퐶퐻 are the mediators of segments

푃2푄2 and 푃3푄3. Let 푇1 be the intersection of lines 퐴퐻 and 퐵1퐶1 2 2 2 2 (see Figure 1); we have 푄1퐴 − 푄1퐻 = 푇1퐴 − 푇1퐻 . We 2 2 denote 푅퐻 = 퐻푃1. It follows that 푄1퐴 = 푅퐻 + (푇1퐴 + 2 푇1퐻)(푇1퐴 − 푇1퐻) = 푅퐻 + 퐴퐻 ∙ (푇1퐴 − 푇1퐻). However, 푇1퐴 = 푇1퐻1, where 퐻1 is the projection 2 2 of 퐴 on 퐵퐶; we find that 푄1퐴 = 푅퐻 + 퐴퐻 ∙ 퐻퐻1. It is known that the symmetric of orthocenter 퐻 towards 퐵퐶 belongs to the circle of circumscribed triangle 퐴퐵퐶. ′ ′ Denoting this point by 퐻1 , we have 퐴퐻 ∙ 퐻퐻1 = 푅2 − 푂퐻2 (the power of point 퐻 towards the circumscribed circle).

Complements to Classic Topics of Circles Geometry

1 We obtain that 퐴퐻 ∙ 퐻퐻 = ∙ (푅2 − 푂퐻2) , and 1 2 1 therefore 퐴푄2 = 푅2 + ∙ (푅2 − 푂퐻2) , where 푂 is the 1 퐻 2 center of the circumscribed triangle 퐴퐵퐶. 1 Similarly, we find 퐵푄2 = 퐶푄2 = 푅2 + (푅2 − 2 3 퐻 2 2 푂퐻 ), therefore 퐴푄1 = 퐵푄2 = 퐶푄3.

Remarks.

a. The proof adjusts analogously for the obtuse triangle. b. 1st Theorem can be equivalently formulated in this way:

2nd Theorem.

If we draw three congruent circles centered in a given triangle’s vertices, the intersection points of these circles with the sides of the median triangle of given triangle (middle lines) are six points situated on a circle having its center in triangle’s orthocenter. If we denote by 휌 the radius of three congruent circles having 퐴, 퐵, 퐶 as their centers, we get: 1 푅2 = 휌2 + (푂퐻2 − 푅2). 퐻 2 However, in a triangle, 푂퐻2 = 9푅2 − (푎2 + 푏2 + 푐2), 푅 being the radius of the circumscribed circle; it follows that: 1 푅2 = 휌2 + 4푅2 − (푎2 + 푏2 + 푐2). 퐻 2

Ion Patrascu, Florentin Smarandache

Remark.

A special case is the one in which 휌 = 푅, where 1 1 we find that 푅2 = 푅2 = 5푅2 − (푎2 + 푏2 + 푐2) = (푅2 + 퐻 1 2 2 푂퐻2).

Definition.

The circle 풞(퐻, 푅1), where: 1 푅 = √5푅2 − (푎2 + 푏2 + 푐2), 1 2 is called the first Droz-Farny’s circle of the triangle 퐴퐵퐶.

Remark.

Another way to build the first Droz-Farny’s circle is offered by the following theorem, which, according to [1], was the subject of a problem proposed in 1910 by V. Thébault in the Journal de Mathématiques Elementaire.

3rd Theorem.

The circles centered on the feet of a triangle’s altitudes passing through the center of the circle circumscribed to the triangle cut the triangle’s sides in six concyclical points.

Complements to Classic Topics of Circles Geometry

Proof.

We consider 퐴퐵퐶 an acute triangle and 퐻1, 퐻2, 퐻3 the altitudes’ feet. We denote by 퐴1, 퐴2; 퐵1, 퐵2; 퐶1, 퐶2 the intersection points of circles having their centers

퐻1, 퐻2, 퐻3 to 퐵퐶, 퐶퐴, 퐴퐵, respectively. We calculate 퐻퐴2 of the right angled triangle 2 2 2 퐻퐻1퐴2 (see Figure 2). We have 퐻퐴2 = 퐻퐻1 + 퐻1퐴2. 2 Because 퐻1퐴2 = 퐻1푂 , it follows that 퐻퐴2 = 2 2 퐻퐻1 + 퐻1푂 . We denote by 푂9 the mean of segment 푂퐻 ; the median theorem in triangle 퐻1퐻푂 leads to 1 퐻 퐻2 + 퐻 푂2 = 2퐻 푂2 + 푂퐻2. 1 1 1 9 2 It is known that 푂9퐻1 is the nine-points circle’s 1 1 radius, so 퐻 푂 = 푅 ; we get: 퐻퐴2 = (푅2 + 푂퐻2) ; 1 9 2 1 2 1 similarly, we find that 퐻퐵2 = 퐻퐶2 = (푅2 + 푂퐻2) , 1 1 2 which shows that the points 퐴1, 퐴2; 퐵1, 퐵2; 퐶1, 퐶2 belong to the first Droz-Farny’s circle.

Figure 2.

Ion Patrascu, Florentin Smarandache

Remark.

The 2nd and the 3rd theorems show that the first Droz-Farny’s circle pass through 12 points, situated two on each side of the triangle and two on each side of the median triangle of the given triangle. The following theorem generates the 3rd Theorem.

4th Theorem.

The circles centered on the feet of altitudes of a given triangle intersect the sides in six concyclical points if and only if their radical center is the center of the circle circumscribed to the given triangle.

Proof.

Let 퐴1, 퐴2; 퐵1, 퐵2; 퐶1, 퐶2 be the points of intersection with the sides of triangle 퐴퐵퐶 of circles having their centers in altitudes’ feet 퐻1, 퐻2, 퐻3. Suppose the points are concyclical; it follows that their circle’s radical center and of circles centered in 퐻2 and 퐻3 is the point 퐴 (sides 퐴퐵 and 퐴퐶 are radical axes), therefore the perpendicular from 퐴 on

퐻2퐻3 is radical axis of centers having their centers 퐻2 and 퐻3. Since 퐻2퐻3 is antiparallel to 퐵퐶, it is parallel to tangent taken in 퐴 to the circle circumscribed to triangle 퐴퐵퐶.

Complements to Classic Topics of Circles Geometry

Consequently, the radical axis is the perpendicular taken in 퐴 on the tangent to the circumscribed circle, therefore it is 퐴푂. Similarly, the other radical axis of circles centered in 퐻1, 퐻2 and of circles centered in 퐻1, 퐻3 pass through 푂, therefore 푂 is the radical center of these circles.

Reciprocally.

Let 푂 be the radical center of the circles having their centers in the feet of altitudes. Since 퐴푂 is perpendicular on 퐻2퐻3 , it follows that 퐴푂 is the radical axis of circles having their centers in 퐻2, 퐻3, therefore 퐴퐵1 ∙ 퐴퐵2 = 퐴퐶1 ∙ 퐴퐶2. From this relationship, it follows that the points

퐵1, 퐵2; 퐶1, 퐶2 are concyclic; the circle on which these points are located has its center in the orthocenter 퐻 of triangle 퐴퐵퐶.

Indeed, the mediators’ chords 퐵1퐵2 and 퐶1퐶2 in the two circles are the altitudes from 퐶 and 퐵 of triangle 퐴퐵퐶, therefore 퐻퐵1 = 퐻퐵2 = 퐻퐶1 = 퐻퐶2. This reasoning leads to the conclusion that 퐵푂 is the radical axis of circles having their centers 퐻1 and 퐻3 , and from here the concyclicality of the points 퐴1, 퐴2; 퐶1, 퐶2 on a circle having its center in 퐻 , therefore 퐻퐴1 = 퐻퐴2 = 퐻퐶1 = 퐻퐶2 . We obtained that 퐻퐴1 = 퐻퐴2 = 퐻퐵1 = 퐻퐵2 = 퐻퐶1 = 퐻퐶2 , which shows the concyclicality of points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2. 61

Ion Patrascu, Florentin Smarandache

Remark.

The circles from the 3rd Theorem, passing through 푂 and having noncollinear centers, admit 푂 as radical center, and therefore the 3rd Theorem is a particular case of the 4th Theorem.

References.

[1] N. Blaha: Asupra unor cercuri care au ca centre două puncte inverse [On some circles that have as centers two inverse points], in “Gazeta Matematica”, vol. XXXIII, 1927. [2] R. Johnson: Advanced Euclidean Geometry. New York: Dover Publication, Inc. Mineola, 2004 [3] Eric W. Weisstein: First Droz-Farny’s circle. From Math World – A Wolfram WEB Resurse, http://mathworld.wolfram.com/. [4] F. Smarandache, I. Patrascu: Geometry of Homological Triangle. Columbus: The Education Publisher Inc., Ohio, SUA, 2012.

Complements to Classic Topics of Circles Geometry

Regarding the Second Droz- Farny’s Circle

In this article, we prove the theorem relative to the second Droz-Farny’s circle, and a sentence that generalizes it. The paper [1] informs that the following Theorem is attributed to J. Neuberg (Mathesis, 1911).

1st Theorem.

The circles with its centers in the middles of triangle 퐴퐵퐶 passing through its orthocenter 퐻 intersect the sides 퐵퐶, 퐶퐴 and 퐴퐵 respectively in the points 퐴1, 퐴2, 퐵1, 퐵2 and 퐶1, 퐶2, situated on a concentric circle with the circle circumscribed to the triangle 퐴퐵퐶 (the second Droz-Farny’s circle).

Proof.

We denote by 푀1, 푀2, 푀3 the middles of 퐴퐵퐶 triangle’s sides, see Figure 1. Because 퐴퐻 ⊥ 푀2푀3 and 퐻 belongs to the circles with centers in 푀2 and 푀3, it follows that 퐴퐻 is the radical axis of these circles,

Ion Patrascu, Florentin Smarandache

therefore we have 퐴퐶1 ∙ 퐴퐶2 = 퐴퐵2 ∙ 퐴퐵1. This relation shows that 퐵1, 퐵2, 퐶1, 퐶2 are concyclic points, because the center of the circle on which they are situated is 푂, the center of the circle circumscribed to the triangle 퐴퐵퐶, hence we have that:

푂퐵1 = 푂퐶1 = 푂퐶2 = 푂퐵2. (1)

Figure 1.

Analogously, 푂 is the center of the circle on which the points 퐴1, 퐴2, 퐶1, 퐶2 are situated, hence: 푂퐴1 = 푂퐶1 = 푂퐶2 = 푂퐴2. (2) Also, 푂 is the center of the circle on which the points 퐴1, 퐴2, 퐵1, 퐵2 are situated, and therefore: 푂퐴1 = 푂퐵1 = 푂퐵2 = 푂퐴2. (3) The relations (1), (2), (3) show that the points

퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are situated on a circle having the center in 푂, called the second Droz-Farny’s circle.

Complements to Classic Topics of Circles Geometry

1st Proposition.

The radius of the second Droz-Farny’s circle is given by: 1 푅2 = 5푒2 − (푎2 + 푏2 + 푐2). 2 2

Proof.

From the right triangle 푂푀1퐴1, using Pitagora’s theorem, it follows that: 2 2 2 2 푂퐴1 = 푂푀1 + 퐴1푀1 = 푂푀1 + 푀1푀2. From the triangle 퐵퐻퐶 , using the median theorem, we have: 1 퐻푀2 = [2(퐵퐻2 + 퐶퐻2) − 퐵퐶2]. 1 4 But in a triangle,

퐴퐻 = 2푂푀1, 퐵퐻 = 2푂푀2, 퐶퐻 = 2푂푀3, hence: 푎2 퐻푀2 = 2푂푀2 + 2푂푀2 = . 1 2 3 4 But: 푎2 푂푀2 = 푅2 − ; 1 4 푏2 푂푀2 = 푅2 − ; 2 4 푐2 푂푀2 = 푅2 − , 3 4 where R is the radius of the circle circumscribed to the triangle 퐴퐵퐶. 1 We find that 푂퐴2 = 푅2 = 5푅2 − (푎2 + 푏2 + 푐2). 1 2 2

Ion Patrascu, Florentin Smarandache

Remarks.

2 a. We can compute 푂푀1 + 푀1푀2 using the median theorem in the triangle 푂푀1퐻 for the median 푀1푂9 (푂9 is the center of the nine points circle, i.e. the middle of (푂퐻)). 1 Because 푂 푀 = 푅 , we obtain: 푅2 = 9 1 2 2 1 (푂푀2 + 푅2). In this way, we can prove 2 2 2 the Theorem computing 푂퐵1 and 푂퐶1 . b. The statement of the 1st Theorem was the subject no. 1 of the 49th International Olympiad in Mathematics, held at Madrid in 2008. c. The 1st Theorem can be proved in the same way for an obtuse triangle; it is obvious that for a right triangle, the second Droz- Farny’s circle coincides with the circle circumscribed to the triangle 퐴퐵퐶. d. The 1st Theorem appears as proposed problem in [2].

2nd Theorem.

The three pairs of points determined by the intersections of each circle with the center in the middle of triangle’s side with the respective side are on a circle if and only these circles have as radical center the triangle’s orthocenter.

Complements to Classic Topics of Circles Geometry

Proof.

Let 푀1, 푀2, 푀3 the middles of the sides of triangle 퐴퐵퐶 and let 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 the intersections with 퐵퐶, 퐶퐴, 퐴퐵 respectively of the circles with centers in

푀1, 푀2, 푀3. Let us suppose that 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are concyclic points. The circle on which they are situated has evidently the center in 푂, the center of the circle circumscribed to the triangle 퐴퐵퐶.

The radical axis of the circles with centers 푀2, 푀3 will be perpendicular on the line of centers 푀2푀3, and because A has equal powers in relation to these circles, since 퐴퐵1 ∙ 퐴퐵2 = 퐴퐶1 ∙ 퐴퐶2, it follows that the radical axis will be the perpendicular taken from A on 푀2푀3, i.e. the height from 퐴 of triangle 퐴퐵퐶. Furthermore, it ensues that the radical axis of the circles with centers in 푀1 and 푀2 is the height from 퐵 of triangle 퐴퐵퐶 and consequently the intersection of the heights, hence the orthocenter 퐻 of the triangle 퐴퐵퐶 is the radical center of the three circles.

Reciprocally.

If the circles having the centers in 푀1, 푀2, 푀3 have the orthocenter with the radical center, it follows that the point 퐴, being situated on the height from A which is the radical axis of the circles of centers 푀2, 푀3

Ion Patrascu, Florentin Smarandache will have equal powers in relation to these circles and, consequently, 퐴퐵1 ∙ 퐴퐵2 = 퐴퐶1 ∙ 퐴퐶2 , a relation that implies that 퐵1, 퐵2, 퐶1, 퐶2 are concyclic points, and the circle on which these points are situated has 푂 as its center.

Similarly, 퐵퐴1 ∙ 퐵퐴2 = 퐵퐶1 ∙ 퐵퐶2 , therefore 퐴1, 퐴2, 퐶1, 퐶2 are concyclic points on a circle of center 푂. Having 푂퐵1 = 푂퐵2 = 푂퐶1 = 푂퐶2 and 푂퐴1 ∙ 푂퐴2 = 푂퐶1 ∙ 푂퐶2 , we get that the points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are situated on a circle of center 푂.

Remarks.

1. The 1st Theorem is a particular case of the 2nd Theorem, because the three circles of

centers 푀1, 푀2, 푀3 pass through 퐻, which means that 퐻 is their radical center. 2. The Problem 525 from [3] leads us to the following Proposition providing a way to

construct the circles of centers 푀1, 푀2, 푀3 intersecting the sides in points that belong to a Droz-Farny’s circle of type 2.

2nd Proposition.

1 1 The circles ∁ (푀 , √푘 + 푎2), ∁ (푀 , √푘 + 푏2), 1 2 2 2 1 ∁ (푀 , √푘 + 푐2) intersect the sides 퐵퐶 , 퐶퐴 , 퐴퐵 3 2 respectively in six concyclic points; 푘 is a conveniently

Complements to Classic Topics of Circles Geometry chosen constant, and 푎, 푏, 푐 are the lengths of the sides of triangle 퐴퐵퐶.

Proof.

According to the 2nd Theorem, it is necessary to prove that the orthocenter 퐻 of triangle 퐴퐵퐶 is the radical center for the circles from hypothesis.

Figure 2.

The power of 퐻 in relation with 1 1 ∁ (푀 , √푘 + 푎2) is equal to 퐻푀2 − (푘 + 푎2) . We 1 2 1 4 푏2 푐2 푎2 observed that 푀2 = 4푅2 − − − , therefore 1 2 2 4 1 푎2+푏2+푐2 1 퐻푀2 − (푘 + 푎2) = 4푅2 − − 푘. We use the 1 4 4 4 same expression for the power of H in relation to the circles of centers 푀2, 푀3, hence H is the radical center of these three circles. 69

Ion Patrascu, Florentin Smarandache

References.

[1] C. Mihalescu: Geometria elementelor remarcabile [The Geometry of Outstanding Elements]. Bucharest: Editura Tehnică, 1957. [2] I. Pătraşcu: Probleme de geometrie plană [Some Problems of Plane Geometry], Craiova: Editura Cardinal, 1996. [3] C. Coşniţă: Teoreme şi probleme alese de matematică [Theorems and Problems], Bucureşti: Editura de Stat Didactică şi Pedagogică, 1958. [4] I. Pătrașcu, F. Smarandache: Variance on Topics of Plane Geometry, Educational Publishing, Columbus, Ohio, SUA, 2013.

Complements to Classic Topics of Circles Geometry

Neuberg’s Orthogonal Circles

In this article, we highlight some metric properties in connection with Neuberg's circles and triangle. We recall some results that are necessary.

1st Definition.

It's called Brocard’s point of the triangle 퐴퐵퐶 the point Ω with the property: ∢Ω퐴퐵 = ∢Ω퐵퐶 = ∢Ω퐶퐴 . The measure of the angle Ω퐴퐵 is denoted by 휔 and it is called Brocard's angle. It occurs the relationship: ctg휔 = ctg퐴 + ctg퐵 + ctg퐶 (see [1]).

Figure 1.

Ion Patrascu, Florentin Smarandache

2nd Definition.

Two triangles are called equibrocardian if they have the same Brocard’s angle.

3rd Definition.

The locus of points 푀 from the plane of the triangle located on the same side of the line 퐵퐶 as 퐴 and forming with 퐵퐶 an equibrocardian triangle with 퐴퐵퐶, containing the vertex 퐴 of the triangle, it's called A-Neuberg’ circle of the triangle 퐴퐵퐶.

We denote by 푁푎 the center of A-Neuberg’ circle by radius 푛푎 (analogously, we define B-Neuberg’ and C-Neuberg’ circles). 푎 We get that 푚(퐵푁 퐶) = 2휔 and 푛 = √ctg2휔 − 3 푎 푎 2 (see [1]).

The triangle 푁푎푁푏푁푐 formed by the centers of Neuberg’s circles is called Neuberg’s triangle.

1st Proposition.

The distances from the center circumscribed to the triangle 퐴퐵퐶 to the vertex of Neuberg’s triangle are proportional to the cubes of 퐴퐵퐶 triangle’s sides lengths.

Complements to Classic Topics of Circles Geometry

Proof.

Let 푂 be the center of the circle circumscribed to the triangle 퐴퐵퐶 (see Figure 2).

Figure 2.

The law of cosines applied in the triangle 푂푁푎퐵 provides: 푂푁 푅 푎 = . sin(푁푎퐵푂) sin휔 But 푚(∢푁푎퐵푂) = 푚(∢푁푎퐵퐶) − 푚(∢푂퐵퐶) = 퐴 − 휔. 푂푁 푅 We have that 푎 = . sin(퐴−휔) sin휔 But 푎 ∙2푅sin휔 3 3 sin(퐴−휔) 퐶Ω 푐 푎 푎 = = 푏 = = , sin휔 AΩ ∙2푅sin휔 푎푏푐 4푅푆 푎 푆 being the area of the triangle 퐴퐵퐶.

Ion Patrascu, Florentin Smarandache

푎3 푂푁 It follows that 푂푁 = , and we get that 푎 = 푎 4푆 푎3 푂푁 푂푁 푏 = 푐 . 푏3 푐3

Consequence.

In a triangle ABC, we have: 3 1) 푂푁푎 ∙ 푂푁푏 ∙ 푂푁푐 = 푅 ; 푂푁 푂푁 푂푁 2) ctg휔 = 푎 + 푏 + 푐 . 푎 푏 푐

2nd Proposition.

If 푁푎푁푏푁푐 is the Neuberg’s triangle of the triangle 퐴퐵퐶, then: (푎2 + 푏2)(푎4 + 푏4) − 푎2푏2푐2 푁 푁 2 = . 푎 푏 2푎2푏2 + 2푏2푐2 + 2푐2푎2 − 푎4 − 푏4 − 푐4 (The formulas for 푁푏푁푐 and 푁푐푁푎 are obtained from the previous one, by circular permutations.)

Proof.

We apply the law of cosines in the triangle

푁푎푂푁푐: 푎3 푏3 푂푁 = , 푂푁 = , 푚(∢푁 푂푁 ) = 1800 − 푐̂. 푎 4푆 푏 4푆 푎 푏 푎6 + 푏6 − 2푎3푏3cos (1800 − 푐) 푁 푁 2 = 푎 푏 16푆2 푎6 + 푏6 + 2푎3푏3cos퐶 = . 16푆2

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But the law of cosines in the triangle 퐴퐵퐶 provides 푎2 + 푏2 − 푐2 2cos퐶 = 2푎푏 and, from din Heron’s formula, we find that 16푆2 = 2푎2푏2 + 2푏2푐2 + 2푐2푎2 − 푎4 − 푏4 − 푐4. Substituting the results above, we obtain, after a few calculations, the stated formula.

4th Definition.

Two circles are called orthogonal if they are secant and their tangents in the common points are perpendicular.

3rd Proposition. (Gaultier – 18B)

Two circles 풞(푂1, 푟1), 풞(푂2, 푟2) are orthogonal if and only if 2 2 2 푟1 + 푟2 = 푂1푂2 .

Proof.

Let 풞(푂1, 푟1), 풞(푂2, 푟2) be orthogonal (see Figure 3); then, if 퐴 is one of the common points, the triangle

푂1퐴푂2 is a right triangle and the Pythagorean Theorem 2 2 2 applied to it, leads to 푟1 + 푟2 = 푂1푂2 .

Ion Patrascu, Florentin Smarandache

Reciprocally.

If the metric relationship from the statement occurs, it means that the triangle 푂1퐴푂2 is a right triangle, therefore 퐴 is their common point (the 2 2 2 2 2 relationship 푟1 + 푟2 = 푂1푂2 implies 푟1 + 푟2 > 2 푂1푂2 ), then 푂1퐴 ⊥ 푂2퐴, so 푂1퐴 is tangent to the circle 풞(푂2, 푟2) because it is perpendicular in 퐴 on radius 푂2퐴, and as well 푂2퐴 is tangent to the circle 풞(푂1, 푟1) , therefore the circles are orthogonal.

Figure 3.

4th Proposition.

B-Neuberg’s and C-Neuberg’s circles associated to the right triangle 퐴퐵퐶 (in 퐴) are orthogonal.

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Proof.

푏6+푐6 If 푚(퐴̂) = 900, then 푁 푁2 = . 푏 푐 16푆 푏 푐 푛 = √ctg2휔 − 3 ; 푛 = √ctg2휔 − 3. 푏 2 푐 2 푎2+푏2+푐2 푏2+푐2 푎2 But ctg휔 − = = . 4푆 2푆 푏푐 It was taken into account that 푎2 = 푏2 + 푐2 and 2푆 = 푏푐. 푎4 (푏2 + 푐2)2 − 3푏2푐2 ctg2휔 − 3 = − 3 = 푏2푐2 푏2푐2 푏4 + 푐4 − 푏2푐2 ctg2휔 − 3 = 푏2푐2 푏4 + 푐4 − 푏2푐2 푏2 + 푐2 푛2 + 푛2 = ( ) 푏 푐 푏2푐2 4 (푏2 + 푐2)(푏4 + 푐4 − 푏2푐2) 푏6 + 푐6 = = . 4푏2푐2 16푆2 2 2 2 By 푁푏 + 푁푐 = 푁푏푁푐 , it follows that B-Neuberg’s and C-Neuberg’s circles are orthogonal.

Ion Patrascu, Florentin Smarandache

References.

[1] F. Smarandache, I. Patrascu: The Geometry of Homological Triangles. Columbus: The Educational Publisher, Ohio, USA, 2012. [2] T. Lalescu: Geometria triunghiului [The Geometry of the Triangle]. Craiova: Editura Apollo, 1993. [3] R. A. Johnson: Advanced Euclidian Geometry. New York: Dover Publications, 2007.

Complements to Classic Topics of Circles Geometry

Lucas’s Inner Circles

In this article, we define the Lucas’s inner circles and we highlight some of their properties.

1. Definition of the Lucas’s Inner Circles

Let 퐴퐵퐶 be a random triangle; we aim to construct the square inscribed in the triangle 퐴퐵퐶 , having one side on 퐵퐶.

Figure 1.

In order to do this, we construct a square 퐴,퐵,퐶,퐷, with 퐴, ∈ (퐴퐵), 퐵,, 퐶, ∈ (퐵퐶) (see Figure 1). , We trace the line 퐵퐷 and we note with 퐷푎 its intersection with (퐴퐶) ; through 퐷푎 we trace the

Ion Patrascu, Florentin Smarandache

parallel 퐷푎퐴푎 to 퐵퐶 with 퐴푎 ∈ (퐴퐵) and we project onto 퐵퐶 the points 퐴푎, 퐷푎 in 퐵푎 respectively 퐶푎. We affirm that the quadrilateral 퐴푎퐵푎퐶푎퐷푎 is the required square. 퐷 퐶 Indeed, 퐴 퐵 퐶 퐷 is a square, because 푎 푎 = 푎 푎 푎 푎 퐷,퐶, 퐵퐷 퐴 퐷 푎 = 푎 푎 and, as 퐷,퐶, = 퐴,퐷, , it follows that 퐴 퐷 = 퐵퐷, 퐴,퐷, 푎 푎 퐷푎퐶푎.

Definition.

It is called A-Lucas’s inner circle of the triangle 퐴퐵퐶 the circle circumscribed to the triangle AAaDa.

We will note with 퐿푎 the center of the A-Lucas’s inner circle and with 푙푎 its radius. Analogously, we define the B-Lucas’s inner circle and the C-Lucas’s inner circle of the triangle 퐴퐵퐶.

2. Calculation of the Radius of the A-Lucas Inner Circle

We note 퐴푎퐷푎 = 푥, 퐵퐶 = 푎; let ℎ푎 be the height from 퐴 of the triangle 퐴퐵퐶.

The similarity of the triangles 퐴퐴푎퐷푎 and 퐴퐵퐶 푥 ℎ −푥 푎ℎ leads to: = 푎 , therefore 푥 = 푎 . 푎 ℎ푎 푎+ℎ푎

푙푎 푥 푅.ℎ푎 From = we obtain 푙푎 = . (1) 푅 푎 푎+ℎ푎

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Note.

Relation (1) and the analogues have been deduced by Eduard Lucas (1842-1891) in 1879 and they constitute the “birth certificate of the Lucas’s circles”.

1st Remark.

2푆 If in (1) we replace ℎ = and we also keep into 푎 푎 consideration the formula 푎푏푐 = 4푅푆, where 푅 is the radius of the circumscribed circle of the triangle 퐴퐵퐶 and 푆 represents its area, we obtain: 푅 푙푎 = 2푎푅 [see Ref. 2]. 1+ 푏푐

3. Properties of the Lucas’s Inner Circles

1st Theorem.

The Lucas’s inner circles of a triangle are inner tangents of the circle circumscribed to the triangle and they are exteriorly tangent pairwise.

Proof.

The triangles 퐴퐴푎퐷푎 and 퐴퐵퐶 are homothetic ℎ through the homothetic center 퐴 and the rapport: 푎 . 푎+ℎ푎

Ion Patrascu, Florentin Smarandache

푙 ℎ Because 푎 = 푎 , it means that the A-Lucas’s inner 푅 푎+ℎ푎 circle and the circle circumscribed to the triangle 퐴퐵퐶 are inner tangents in 퐴.

Analogously, it follows that the B-Lucas’s and C- Lucas’s inner circles are inner tangents of the circle circumscribed to 퐴퐵퐶.

Figure 2.

We will prove that the A-Lucas’s and C-Lucas’s circles are exterior tangents by verifying

퐿푎퐿푐 = 푙푎 + 푙푐. (2) We have:

푂퐿푎 = 푅 − 푙푎; 푂퐿푐 = 푅 − 푙푐 and 푚(퐴̂0퐶) = 2퐵 (if 푚(퐵̂) > 90° then 푚(퐴̂0퐶) = 360° − 2퐵). 82

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The theorem of the cosine applied to the triangle

푂퐿푎퐿푐 implies, keeping into consideration (2), that: 2 2 (푅 − 푙푎) + (푅 − 푙푎) − 2(푅 − 푙푎)(푅 − 푙푐)푐표푠2퐵 = 2 = (푙푎 + 푙푐) . Because 푐표푠2퐵 = 1 − 2푠𝑖푛2퐵, it is found that (2) is equivalent to: 푙 푙 푠𝑖푛2퐵 = 푎 푐 . (3) (푅−푙푎)(푅−푙푐) 푅2푎푏2푐 But we have: 푙 푙 = , 푎 푐 (2푎푅+푏푐)(2푐푅+푎푏) 푐 푎 푙 + 푙 = 푅푏( + ). 푎 푐 2푎푅+푏푐 2푐푅+푎푏 푎푏2푐 By replacing in (3), we find that 푠𝑖푛 2퐵 = = 4푎푐푅2 푏2 푏 ⟺ sin 퐵 = is true according to the sines theorem. 4푎2 2푅 So, the exterior tangent of the A-Lucas’s and C-Lucas’s circles is proven. Analogously, we prove the other tangents.

2nd Definition.

It is called an A-Apollonius’s circle of the random triangle 퐴퐵퐶 the circle constructed on the segment determined by the feet of the bisectors of angle 퐴 as diameter.

Remark.

Analogously, the B-Apollonius’s and C- Apollonius’s circles are defined. If 퐴퐵퐶 is an isosceles triangle with 퐴퐵 = 퐴퐶 then the A-Apollonius’s circle

Ion Patrascu, Florentin Smarandache isn’t defined for 퐴퐵퐶 , and if 퐴퐵퐶 is an equilateral triangle, its Apollonius’s circle isn’t defined.

2nd Theorem.

The A-Apollonius’s circle of the random triangle is the geometrical point of the points 푀 from the plane 푀퐵 푐 of the triangle with the property: = . 푀퐶 푏

3rd Definition.

We call a fascicle of circles the bunch of circles that do not have the same radical axis. a. If the radical axis of the circles’ fascicle is exterior to them, we say that the fascicle is of the first type. b. If the radical axis of the circles’ fascicle is secant to the circles, we say that the fascicle is of the second type. c. If the radical axis of the circles’ fascicle is tangent to the circles, we say that the fascicle is of the third type.

3rd Theorem.

The A-Apollonius’s circle and the B-Lucas’s and C-Lucas’s inner circles of the random triangle 퐴퐵퐶 form a fascicle of the third type.

Complements to Classic Topics of Circles Geometry

Proof.

Let {푂퐴} = 퐿푏퐿푐 ∩ 퐵퐶 (see Figure 3). Menelaus’s theorem applied to the triangle 푂퐵퐶 implies that: 푂 퐵 퐿 퐵 퐿 푂 퐴 . 푏 . 푐 = 1, 푂퐴퐶 퐿푏푂 퐿푐퐶 so: 푂 퐵 푙 푅−푙 퐴 . 푏 . 푐 = 1 푂퐴퐶 푅−푙푏 푙푐 and by replacing 푙푏 and 푙푐, we find that: 2 푂퐴퐵 푏 = 2. 푂퐴퐶 푐 This relation shows that the point 푂퐴 is the foot of the exterior symmedian from 퐴 of the triangle 퐴퐵퐶 (so the tangent in 퐴 to the circumscribed circle), namely the center of the A-Apollonius’s circle.

Let 푁1 be the contact point of the B-Lucas’s and C-Lucas’s circles. The radical center of the B-Lucas’s, C-Lucas’s circles and the circle circumscribed to the triangle 퐴퐵퐶 is the intersection 푇퐴 of the tangents traced in 퐵 and in 퐶 to the circle circumscribed to the triangle 퐴퐵퐶.

It follows that 퐵푇퐴 = 퐶푇퐴 = 푁1푇퐴, so 푁1 belongs to the circle 풞퐴 that has the center in 푇퐴 and orthogonally cuts the circle circumscribed in 퐵 and 퐶. The radical axis of the B-Lucas’s and C-Lucas’s circles is 푇퐴푁1, and 푂퐴푁1 is tangent in 푁1 to the circle 풞퐴. Considering the power of the point 푂퐴 in relation to 풞퐴, we have: 2 푂퐴푁1 = 푂퐴퐵. 푂퐴퐶.

Ion Patrascu, Florentin Smarandache

Figure 3.

2 Also, 푂퐴푂 = 푂퐴퐵 ∙ 푂퐴퐶 ; it thus follows that 푂퐴퐴 = 푂퐴푁1, which proves that 푁1 belongs to the A- Apollonius’s circle and is the radical center of the A- Apollonius’s, B-Lucas’s and C-Lucas’s circles.

Remarks.

1. If the triangle 퐴퐵퐶 is right in 퐴 then

퐿푏퐿푐||퐵퐶, the radius of the A-Apollonius’s 푎푏푐 circle is equal to: . The point 푁 is |푏2−푐2| 1 the foot of the bisector from 퐴. We find 푎푏푐 that 푂 푁 = , so the theorem stands 퐴 1 |푏2−푐2| true.

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2. The A-Apollonius’s and A-Lucas’s circles are orthogonal. Indeed, the radius of the A-Apollonius’s circle is perpendicular to the radius of the circumscribed circle, 푂퐴, so, to the radius of the A-Lucas’s circle also.

4th Definition.

The triangle 푇퐴푇퐵푇퐶 determined by the tangents traced in 퐴, 퐵, 퐶 to the circle circumscribed to the triangle 퐴퐵퐶 is called the tangential triangle of the triangle 퐴퐵퐶.

1st Property.

The triangle 퐴퐵퐶 and the Lucas’s triangle 퐿푎퐿푏퐿푐 are homological.

Proof.

Obviously, 퐴퐿푎, 퐵퐿푏, 퐶퐿푐 are concurrent in 푂 , therefore 푂, the center of the circle circumscribed to the triangle 퐴퐵퐶, is the homology center.

We have seen that {푂퐴} = 퐿푏퐿푐 ∩ 퐵퐶 and 푂퐴 is the center of the A-Apollonius’s circle, therefore the homology axis is the Apollonius’s line 푂퐴푂퐵푂퐶 (the line determined by the centers of the Apollonius’s circle).

Ion Patrascu, Florentin Smarandache

2nd Property.

The tangential triangle and the Lucas’s triangle of the triangle 퐴퐵퐶 are orthogonal triangles.

Proof.

The line 푇퐴푁1 is the radical axis of the B-Lucas’s inner circle and the C-Lucas’s inner circle, therefore it is perpendicular on the line of the centers 퐿푏퐿푐 . Analogously, 푇퐵푁2 is perpendicular on 퐿푐퐿푎 , because the radical axes of the Lucas’s circles are concurrent in 퐿, which is the radical center of the Lucas’s circles; it follows that 푇퐴푇퐵푇퐶 and 퐿푎퐿푏퐿푐 are orthological and 퐿 is the center of orthology. The other center of orthology is 푂 the center of the circle circumscribed to 퐴퐵퐶.

References.

[1] D. Brânzei, M. Miculița. Lucas circles and spheres. In “The Didactics of Mathematics”, volume 9/1994, Cluj-Napoca (Romania), p. 73-80. [2] P. Yiu, A. P. Hatzipolakis. The Lucas Circles of a Triangle. In “Amer. Math. Monthly”, no. 108/2001, pp. 444-446. http://www.math.fau. edu/yiu/monthly437-448.pdf. [3] F. Smarandache, I. Patrascu. The Geometry of Homological Triangles. Educational Publisher, Columbus, Ohio, U.S.A., 2013.

Complements to Classic Topics of Circles Geometry

Theorems with Parallels Taken through a Triangle’s Vertices and Constructions Performed only with the Ruler

In this article, we solve problems of geometric constructions only with the ruler, using known theorems.

1st Problem.

Being given a triangle 퐴퐵퐶 , its circumscribed circle (its center known) and a point 푀 fixed on the circle, construct, using only the ruler, a transversal line 퐴1, 퐵1, 퐶1, with 퐴1 ∈ 퐵퐶, 퐵1 ∈ 퐶퐴, 퐶1 ∈ 퐴퐵, such that ∢푀퐴1퐶 ≡ ∢푀퐵1퐶 ≡ ∢푀퐶1퐴 (the lines taken though 푀 to generate congruent angles with the sides 퐵퐶, 퐶퐴 and 퐴퐵, respectively).

2nd Problem.

Being given a triangle 퐴퐵퐶 , its circumscribed circle (its center known) and 퐴1, 퐵1, 퐶1, such that 퐴1 ∈

Ion Patrascu, Florentin Smarandache

퐵퐶, 퐵1 ∈ 퐶퐴, 퐶1 ∈ 퐴퐵 and 퐴1, 퐵1, 퐶1 collinear, construct, using only the ruler, a point 푀 on the circle circumscribing the triangle, such that the lines

푀퐴1, 푀퐵1, 푀퐶1 to generate congruent angles with 퐵퐶, 퐶퐴 and 퐴퐵, respectively.

3rd Problem.

Being given a triangle 퐴퐵퐶 inscribed in a circle of given center and 퐴퐴′ a given cevian, 퐴′ a point on the circle, construct, using only the ruler, the isogonal ′ cevian 퐴퐴1 to the cevian 퐴퐴 .

To solve these problems and to prove the theorems for problems solving, we need the following Lemma:

1st Lemma. (Generalized Simpson's Line) If 푀 is a point on the circle circumscribed to the triangle 퐴퐵퐶 and we take the lines 푀퐴1, 푀퐵1, 푀퐶1 which generate congruent angles ( 퐴1 ∈ 퐵퐶, 퐵1 ∈ 퐶퐴, 퐶1 ∈ 퐴퐵) with 퐵퐶, 퐶퐴 and 퐴퐵 respectively, then the points 퐴1, 퐵1, 퐶1 are collinear.

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Proof.

Let 푀 on the circle circumscribed to the triangle 퐴퐵퐶 (see Figure 1), such that:

∢푀퐴1퐶 ≡ ∢푀퐵1퐶 ≡ ∢푀퐶1퐴 = 휑. (1)

Figure 1.

From the relation (1), we obtain that the quadrilateral 푀퐵1퐴1퐶 is inscriptible and, therefore: ∢퐴1퐵퐶 ≡ ∢퐴1푀퐶. (2). Also from (1), we have that 푀퐵1퐴퐶1 is inscriptible, and so

∢퐴퐵1퐶1 ≡ ∢퐴푀퐶1. (3) 91

Ion Patrascu, Florentin Smarandache

The quadrilateral MABC is inscribed, hence:

∢푀퐴퐶1 ≡ ∢퐵퐶푀. (4) On the other hand, 0 ∢퐴1푀퐶 = 180 − (퐵퐶푀̂ + 휑), 0 ∢퐴푀퐶1 = 180 − (푀퐴̂퐶1 + 휑). The relation (4) drives us, together with the above relations, to:

∢퐴1푀퐶 ≡ ∢퐴푀퐶1. (5) Finally, using the relations (5), (2) and (3), we conclude that: ∢퐴1퐵1퐶 ≡ 퐴퐵1퐶1 , which justifies the collinearity of the points 퐴1, 퐵1, 퐶1.

Remark.

The Simson’s Line is obtained in the case when 휑 = 900.

2nd Lemma.

If 푀 is a point on the circle circumscribed to the triangle 퐴퐵퐶 and 퐴1, 퐵1, 퐶1 are points on 퐵퐶 , 퐶퐴 and 퐴퐵 , respectively, such that ∢푀퐴1퐶 = ∢푀퐵1퐶 = ∢푀퐶1퐴 = 휑, and 푀퐴1 intersects the circle a second time in 퐴’, then 퐴퐴′ ∥ 퐴1퐵1.

Proof.

The quadrilateral 푀퐵1퐴1퐶 is inscriptible (see Figure 1); it follows that:

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′ ∢퐶푀퐴 ≡ ∢퐴1퐵1퐶. (6) On the other hand, the quadrilateral 푀퐴퐴′퐶 is also inscriptible, hence: ∢퐶푀퐴′ ≡ ∢퐴′퐴퐶. (7) The relations (6) and (7) imply: ∢퐴′푀퐶 ≡ ∢퐴′퐴퐶, ′ which gives 퐴퐴 ∥ 퐴1퐵1.

3rd Lemma. (The construction of a parallel with a given diameter using a ruler)

In a circle of given center, construct, using only the ruler, a parallel taken through a point of the circle at a given diameter.

Solution.

In the given circle 풞(푂, 푅) , let be a diameter (퐴퐵)] and let 푀 ∈ 풞(푂, 푅). We construct the line 퐵푀 (see Figure 2). We consider on this line the point 퐷 (푀 between 퐷 and 퐵). We join 퐷 with 푂, 퐴 with 푀 and denote 퐷푂 ∩ 퐴푀 = {푃}. We take 퐵푃 and let {푁} = 퐷퐴 ∩ 퐵푃. The line 푀푁 is parallel to 퐴퐵.

Construction’s Proof.

In the triangle 퐷퐴퐵, the cevians 퐷푂, 퐴푀 and 퐵푁 are concurrent. Ceva’s Theorem provides: 93

Ion Patrascu, Florentin Smarandache

푂퐴 푀퐵 푁퐷 ∙ ∙ = 1. (8) 푂퐵 푀퐷 푁퐴 But 퐷푂 is a median, 퐷푂 = 퐵푂 = 푅. 푀퐵 푁퐴 From (8), we get = , which, by Thales 푀퐷 푁퐷 reciprocal, gives 푀푁 ∥ 퐴퐵.

Figure 2.

Remark.

If we have a circle with given center and a certain line 푑, we can construct though a given point 푀 a parallel to that line in such way: we take two diameters [푅푆] and [푈푉] through the center of the given circle (see Figure 3).

Complements to Classic Topics of Circles Geometry

Figure 3.

We denote 푅푆 ∩ 푑 = {푃}; because [푅푂] ≡ [푆푂], we can construct, applying the 3rd Lemma, the parallels through 푈 and 푉 to 푅푆 which intersect 푑 in 퐾 and 퐿, respectively. Since we have on the line 푑 the points 퐾, 푃, 퐿 , such that [퐾푃] ≡ [푃퐿] , we can construct the parallel through 푀 to 푑 based on the construction from 3rd Lemma.

1st Theorem. (P. Aubert – 1899)

If, through the vertices of the triangle 퐴퐵퐶, we take three lines parallel to each other, which intersect the circumscribed circle in 퐴′ , 퐵′ and 퐶′ , and 푀 is a

Ion Patrascu, Florentin Smarandache point on the circumscribed circle, as well 푀퐴′ ∩ 퐵퐶 = ′ ′ {퐴1}, 푀퐵 ∩ 퐶퐴 = {퐵1}, 푀퐶 ∩ 퐴퐵 = {퐶1}, then 퐴1, 퐵1, 퐶1 are collinear and their line is parallel to 퐴퐴′.

Proof.

The point of the proof is to show that 푀퐴1, 푀퐵1, 푀퐶1 generate congruent angles with 퐵퐶, 퐶퐴 and 퐴퐵, respectively. 1 푚(푀̂퐴 퐶) = [푚(푀퐶̆ ) + 푚(퐵̆퐴′)] (9) 1 2 1 푚(푀̂퐵 퐶) = [푚(푀퐶̆ ) + 푚(퐴̆퐵′)] (10) 1 2 But 퐴퐴′ ∥ 퐵퐵′ implies 푚(퐵퐴̆′) = 푚(퐴̆퐵′) , hence, from (9) and (10), it follows that:

∢푀퐴1퐶 ≡ ∢푀퐵1퐶, (11) 1 푚(푀̂퐶 퐴) = [푚(퐵푀̆ ) − 푚(퐴̆퐶′)]. (12) 1 2 But 퐴퐴′ ∥ 퐶퐶′ implies that 푚(퐴̆퐶′) = 푚(퐴̆′퐶); by returning to (12), we have that: 1 푚(푀̂퐶 퐴) = [푚(퐵푀̆ ) − 푚(퐴̆퐶′)] = 1 2 1 = [푚(퐵퐴̆′) + 푚(푀퐶̆ )]. (13) 2 The relations (9) and (13) show that:

∢푀퐴1퐶 ≡ ∢푀퐶1퐴. (14) From (11) and (14), we obtain: ∢푀퐴1퐶 ≡ st ∢푀퐵1퐶 ≡ ∢푀퐶1퐴 , which, by 1 Lemma, verifies the nd collinearity of points 퐴1, 퐵1, 퐶1. Now, applying the 2 Lemma, we obtain the parallelism of lines 퐴퐴′ and

퐴1퐵1.

Complements to Classic Topics of Circles Geometry

Figure 4.

2nd Theorem. (M’Kensie – 1887)

If 퐴1퐵1퐶1 is a transversal line in the triangle 퐴퐵퐶 (퐴1 ∈ 퐵퐶, 퐵1 ∈ 퐶퐴, 퐶1 ∈ 퐴퐵), and through the triangle’s vertices we take the chords 퐴퐴′ , 퐵퐵′, 퐶퐶′ of a circle circumscribed to the triangle, parallels with the transversal line, then the lines 퐴퐴′ , 퐵퐵′, 퐶퐶′ are concurrent on the circumscribed circle.

Ion Patrascu, Florentin Smarandache

Proof.

′ We denote by 푀 the intersection of the line 퐴1퐴 with the circumscribed circle (see Figure 5) and with ′ ′ ′ 퐵1 , respectively 퐶1 the intersection of the line 푀퐵 with 퐴퐶 and of the line 푀퐶′ with 퐴퐵.

Figure 5.

According to the P. Aubert’s theorem, we have ′ ′ that the points 퐴1, 퐵1, 퐶1 are collinear and that the line ′ ′ 퐴1퐵1 is parallel to 퐴퐴 . ′ From hypothesis, we have that 퐴1퐵1 ∥ 퐴퐴 ; from ′ the uniqueness of the parallel taken through 퐴1 to 퐴퐴 , ′ ′ it follows that 퐴1퐵1 ≡ 퐴1퐵1 , therefore 퐵1 = 퐵1 , and ′ analogously 퐶1 = 퐶1.

Complements to Classic Topics of Circles Geometry

Remark.

We have that: 푀퐴1, 푀퐵1, 푀퐶1 generate congruent angles with 퐵퐶, 퐶퐴 and 퐴퐵, respectively.

3rd Theorem. (Beltrami – 1862)

If three parallels are taken through the three vertices of a given triangle, then their isogonals intersect each other on the circle circumscribed to the triangle, and vice versa.

Proof.

Let 퐴퐴′, 퐵퐵′, 퐶퐶′ the three parallel lines with a certain direction (see Figure 6).

Ion Patrascu, Florentin Smarandache

Figure 6.

To construct the isogonal of the cevian 퐴퐴′, we take 퐴′푀 ∥ 퐵퐶, 푀 belonging to the circle circumscribed to the triangle, having 퐵̆퐴′ ≡ 퐶푀̆ , it follows that 퐴푀 will be the isogonal of the cevian 퐴퐴′. (Indeed, from 퐵̆퐴′ ≡ 퐶푀̆ it follows that ∢퐵퐴퐴′ ≡ ∢퐶퐴푀.) On the other hand, 퐵퐵′ ∥ A 퐴′ implies 퐵퐴̆′ ≡ 퐴퐵̆ ′, and since 퐵̆퐴′ ≡ 퐶푀̆ we have that 퐴퐵̆′ ≡ 퐶푀̆ , which shows that the isogonal of the parallel 퐵퐵′ is 퐵푀. From 퐶퐶′ ∥ 퐴퐴′, it follows that 퐴′퐶 ≡ 퐴퐶′, having ∢퐵′퐶푀 ≡ ∢퐴퐶퐶′, therefore the isogonal of the parallel 퐶퐶′ is 퐶푀′.

Reciprocally.

If 퐴푀, 퐵푀, 퐶푀 are concurrent cevians in 푀, the point on the circle circumscribed to the triangle 퐴퐵퐶, let us prove that their isogonals are parallel lines. To construct an isogonal of 퐴푀 , we take 푀퐴′ ∥ 퐵퐶 , 퐴′ belonging to the circumscribed circle. We have 푀퐶̆ ≡ 퐵̆퐴′. Constructing the isogonal 퐵퐵′ of 퐵푀, with 퐵′ on the circumscribed circle, we will have 퐶푀̆ ≡ 퐴퐵̆′ , it follows that 퐵̆퐴′ ≡ 퐴퐵̆′ and, consequently, ∢퐴퐵퐵′ ≡ ∢퐵퐴퐴′, which shows that 퐴퐴′ ∥ 퐵퐵′. Analogously, we show that 퐶퐶′ ∥ 퐴퐴′. We are now able to solve the proposed problems.

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Solution to the 1st problem.

Using the 3rd Lemma, we construct the parallels 퐴퐴′, 퐵퐵′, 퐶퐶′ with a certain directions of a diameter of the circle circumscribed to the given triangle. We join 푀 with 퐴′ , 퐵′, 퐶′ and denote the ′ ′ intersection between 푀퐴 and 퐵퐶, 퐴1; 푀퐵 ∩ 퐶퐴 = {퐵1} ′ and 푀퐴 ∩ 퐴푉 = {퐶1}. According to the Aubert’s Theorem, the points ′ ′ ′ 퐴1, 퐵1, 퐶1 will be collinear, and 푀퐴 , 푀퐵 , 푀퐶 generate congruent angles with 퐵퐶, 퐶퐴 and 퐴퐵, respectively.

Solution to the 2nd problem.

Using the 3rd Lemma and the remark that follows it, we construct through 퐴, 퐵, 퐶 the parallels to 퐴1퐵1; we denote by 퐴′, 퐵′, 퐶′ their intersections with the circle circumscribed to the triangle 퐴퐵퐶. (It is enough to build a single parallel to the transversal line 퐴1퐵1퐶1, for example 퐴퐴′). ′ We join 퐴 with 퐴1 and denote by 푀 the intersection with the circle. The point 푀 will be the point we searched for. The construction’s proof follows from the M’Kensie Theorem.

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Solution to the 3rd problem.

We suppose that 퐴′ belongs to the little arc determined by the chord 퐵퐶̆ in the circle circumscribed to the triangle 퐴퐵퐶.

In this case, in order to find the isogonal 퐴퐴1, we construct (by help of the 3rd Lemma and of the remark ′ that follows it) the parallel 퐴 퐴1 to 퐵퐶, 퐴1 being on the ′ circumscribed circle, it is obvious that 퐴퐴 and 퐴퐴1 will be isogonal cevians. We suppose that 퐴′ belongs to the high arc determined by the chord 퐵퐶̆ ; we consider 퐴′ ∈ 퐴퐵̆ (the arc 퐴퐵̆ does not contain the point 퐶). In this situation, we firstly construct the parallel 퐵푃 to 퐴퐴′, 푃 belongs to the circumscribed circle, and then through 푃 we construct the parallel 푃퐴1 to 퐴퐶 , 퐴1 belongs to the circumscribed circle. The isogonal of the line 퐴퐴′ will rd be 퐴퐴1 . The construction’s proof follows from 3 Lemma and from the proof of Beltrami’s Theorem.

References.

[1] F. G. M.: Exercices de Géometrie. VIII-e ed., Paris, VI-e Librairie Vuibert, Rue de Vaugirard,77. [2] T. Lalesco: La Géométrie du Triangle. 13-e ed., Bucarest, 1937; Paris, Librairie Vuibert, Bd. Saint Germain, 63. [3] C. Mihalescu: Geometria elementelor remarcabile [The Geometry of remarkable elements]. Bucharest: Editura Tehnică, 1957.

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Apollonius’s Circles of kth Rank

The purpose of this article is to introduce the notion of Apollonius’s circle of kth rank.

1st Definition.

It is called an internal cevian of kth rank the line 퐵퐴 퐴퐵 푘 퐴퐴푘 where 퐴푘 ∈ (퐵퐶), such that = ( ) (푘 ∈ ℝ). 퐴푘퐶 퐴퐶 ′ If 퐴푘 is the harmonic conjugate of the point 퐴푘 in ′ relation to 퐵 and 퐶, we call the line 퐴퐴푘 an external cevian of kth rank.

2nd Definition.

We call Apollonius’s circle of kth rank with respect to the side 퐵퐶 of 퐴퐵퐶 triangle the circle which ′ has as diameter the segment line 퐴푘퐴푘.

1st Theorem.

Apollonius’s circle of kth rank is the locus of points 푀 from 퐴퐵퐶 triangle's plan, satisfying the 푀퐵 퐴퐵 푘 relation: = ( ) . 푀퐶 퐴퐶 103

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Proof.

th Let 푂퐴푘 the center of the Apollonius’s circle of k rank relative to the side 퐵퐶 of 퐴퐵퐶 triangle (see Figure 1) and 푈, 푉 the points of intersection of this circle with the circle circumscribed to the triangle 퐴퐵퐶. We denote by 퐷 the middle of arc 퐵퐶, and we extend ′ 퐷퐴푘 to intersect the circle circumscribed in 푈 . In 퐵푈′퐶 triangle, 푈′퐷 is bisector; it follows that ′ 푘 퐵퐴푘 푈 퐵 퐴퐵 ′ = ′ = ( ) , so 푈 belongs to the locus. 퐴푘퐶 푈 퐶 퐴퐶 ′ ′ The perpendicular in 푈 on 푈 퐴푘 intersects BC on ′′ 퐴푘 , which is the foot of the 퐵푈퐶 triangle's outer bisector, so the harmonic conjugate of 퐴푘 in relation ′′ ′ to 퐵 and 퐶, thus 퐴푘 = 퐴푘. Therefore, 푈′ is on the Apollonius’s circle of rank 푘 relative to the side 퐵퐶, hence 푈′ = 푈.

Figure 3

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Let 푀 a point that satisfies the relation from the 푀퐵 퐵퐴 statement; thus = 푘 ; it follows – by using the 푀퐶 퐴푘퐶 reciprocal of bisector's theorem – that 푀퐴푘 is the internal bisector of angle 퐵푀퐶. Now let us proceed as before, taking the external bisector; it follows that 푀 belongs to the Apollonius’s circle of center 푂퐴푘 . We consider now a point 푀 on this circle, and we ′ ′ construct 퐶 such that ∢퐵푁퐴푘 ≡ ∢퐴푘푁퐶 (thus (푁퐴푘 is the internal bisector of the angle 퐵푁̂퐶′ ). Because ′ ′ 퐴푘푁 ⊥ 푁퐴푘, it follows that 퐴푘 and 퐴푘 are harmonically conjugated with respect to 퐵 and 퐶′ . On the other hand, the same points are harmonically conjugated with respect to 퐵 and 퐶; from here, it follows that 퐶′ = 푁퐵 퐵퐴 퐴퐵 푘 퐶, and we have = 푘 = ( ) . 푁퐶 퐴푘퐶 퐴퐶

3rd Definition.

It is called a complete quadrilateral the geometric figure obtained from a convex quadrilateral by extending the opposite sides until they intersect. A complete quadrilateral has 6 vertices, 4 sides and 3 diagonals.

2nd Theorem.

In a complete quadrilateral, the three diagonals' middles are collinear (Gauss - 1810).

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Proof.

Let 퐴퐵퐶퐷퐸퐹 a given complete quadrilateral (see

Figure 2). We denote by 퐻1, 퐻2, 퐻3, 퐻4 respectively the orthocenters of 퐴퐵퐹, 퐴퐷퐸, 퐶퐵퐸, 퐶퐷퐹 triangles, and let 퐴1, 퐵1, 퐹1 the feet of the heights of 퐴퐵퐹 triangle.

Figure 4

As previously shown, the following relations occur: 퐻1퐴. 퐻1퐴1 − 퐻1퐵. 퐻1퐵1 = 퐻1퐹. 퐻1퐹1; they express that the point 퐻1 has equal powers to the circles of diameters 퐴퐶, 퐵퐷, 퐸퐹 , because those circles contain respectively the points 퐴1, 퐵1, 퐹1, and 퐻1 is an internal point.

It is shown analogously that the points 퐻2, 퐻3, 퐻4 have equal powers to the same circles, so those points are situated on the radical axis (common to the circles), therefore the circles are part of a fascicle, as

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3rd Theorem.

The Apollonius’s circle of kth rank of a triangle are part of a fascicle.

Proof.

th Let 퐴퐴푘 , 퐵퐵푘, 퐶퐶푘 be concurrent cevians of k ′ ′ ′ th rank and 퐴퐴푘, 퐵퐵푘, 퐶퐶푘 be the external cevians of k ′ ′ ′ rank (see Figure 3). The figure 퐵푘퐶푘퐵푘퐶푘퐴푘퐴푘 is a complete quadrilateral and 2nd theorem is applied.

Figure 5

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4th Theorem.

The Apollonius’s circle of kth rank of a triangle are the orthogonals of the circle circumscribed to the triangle.

Proof.

We unite 푂 to 퐷 and 푈 (see Figure 1), 푂퐷 ⊥ 퐵퐶 ̂′ 0 ̂′ ̂ and 푚(퐴푘푈퐴푘) = 90 , it follows that 푈퐴푘퐴푘 = 푂퐷퐴푘 =

푂푈̂퐴푘. ̂′ ̂ The congruence 푈퐴푘퐴푘 ≡ 푂푈퐴푘 shows that 푂푈 is tangent to the Apollonius’s circle of center 푂퐴푘. Analogously, it can be demonstrated for the other Apollonius’s Circle.

1st Remark.

The previous theorem indicates that the radical axis of Apollonius’s circle of kth rank is the perpendicular taken from 푂 to the line 푂퐴푘푂퐵푘.

5th Theorem.

The centers of Apollonius’s Circle of kth rank of a triangle are situated on the trilinear polar associated to the intersection point of the cevians of 2kth rank.

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Proof.

From the previous theorem, it results that 푂푈 ⊥

푈푂퐴푘, so 푈푂퐴푘 is an external cevian of rank 2 for 퐵퐶푈 triangle, thus an external symmedian. Henceforth,

푂퐴 퐵 퐵푈 2 퐴퐵 2푘 푘 = ( ) = ( ) (the last equality occurs because 푂 퐶 퐶푈 퐴퐶 퐴푘 푈 belong to the Apollonius’s circle of rank 푘 associated to the vertex 퐴).

6th Theorem.

The Apollonius’s circle of kth rank of a triangle intersects the circle circumscribed to the triangle in two points that belong to the internal and external cevians of k+1th rank.

Proof.

Let 푈 and 푉 points of intersection of the

Apollonius’s circle of center 푂퐴푘 with the circle circumscribed to the 퐴퐵퐶 (see Figure 1). We take from

푈 and 푉 the perpendiculars 푈푈1 , 푈푈2 and 푉푉1, 푉푉2 on 퐴퐵 and 퐴퐶 respectively. The quadrilaterals 퐴퐵푉퐶 , 퐴퐵퐶푈 are inscribed, it follows the similarity of triangles 퐵푉푉1, 퐶푉푉2 and 퐵푈푈1, 퐶푈푈2, from where we get the relations: 퐵푉 푉푉 푈퐵 푈푈 = 1 , = 1. 퐶푉 푉푉2 푈퐶 푈푈2

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퐵푉 퐴퐵 푘 푈퐵 퐴퐵 푘 푉푉 퐴퐵 푘 푈푈 But = ( ) , = ( ) , 1 = ( ) and 1 = 퐶푉 퐴퐶 푈퐶 퐴퐶 푉푉2 퐴퐶 푈푈2 퐴퐵 푘 ( ) , relations that show that 푉 and 푈 belong 퐴퐶 respectively to the internal cevian and the external cevian of rank 푘 + 1.

4th Definition.

If the Apollonius’s circle of kth rank associated with a triangle has two common points, then we call these points isodynamic points of kth rank (and we ′ denote them 푊푘, 푊푘).

1st Property.

′ th If 푊푘, 푊푘 are isodynamic centers of k rank, then: 푘 푘 푘 푊푘퐴. 퐵퐶 = 푊푘퐵. 퐴퐶 = 푊푘퐶. 퐴퐵 ; ′ 푘 ′ 푘 ′ 푘 푊푘퐴. 퐵퐶 = 푊푘퐵. 퐴퐶 = 푊푘퐶. 퐴퐵 . The proof of this property follows immediately from 1st Theorem.

2nd Remark.

The Apollonius’s circle of 1st rank is the investigated Apollonius’s circle (the bisectors are cevians of 1st rank). If 푘 = 2, the internal cevians of 2nd rank are the symmedians, and the external cevians of 2nd rank are the external symmedians, i.e. the tangents

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Complements to Classic Topics of Circles Geometry in triangle’s vertices to the circumscribed circle. In this case, for the Apollonius’s circle of 2nd rank, the 3rd Theorem becomes:

7th Theorem.

The Apollonius’s circle of 2nd rank intersects the circumscribed circle to the triangle in two points belonging respectively to the antibisector's isogonal and to the cevian outside of it.

Proof.

It follows from the proof of the 6th theorem. We mention that the antibisector is isotomic to the bisector, and a cevian of 3rd rank is isogonic to the antibisector.

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References.

[1] N. N. Mihăileanu: Lecții complementare de geometrie [Complementary Lessons of Geometry], Editura Didactică și Pedagogică, București, 1976. [2] C. Mihalescu: Geometria elementelor remarcabile [The Geometry of Outstanding Elements], Editura Tehnică, București, 1957. [3] V. Gh. Vodă: Triunghiul – ringul cu trei colțuri [The Triangle-The Ring with Three Corners], Editura Albatros, București, 1979. [4] F. Smarandache, I. Pătrașcu: Geometry of Homological Triangle, The Education Publisher Inc., Columbus, Ohio, SUA, 2012.

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Apollonius’s Circle of Second Rank

This article highlights some properties of Apollonius’s circle of second rank in connection with the adjoint circles and the second Brocard’s triangle.

1st Definition.

It is called Apollonius’s circle of second rank relative to the vertex 퐴 of the triangle 퐴퐵퐶 the circle constructed on the segment determined on the simedians’ feet from 퐴 on 퐵퐶 as diameter.

1st Theorem.

The Apollonius’s circle of second rank relative to the vertex 퐴 of the triangle 퐴퐵퐶 intersect the circumscribed circle of the triangle 퐴퐵퐶 in two points belonging respectively to the cevian of third rank (antibisector’s isogonal) and to its external cevian. The theorem’s proof follows from the theorem relative to the Apollonius’s circle of kth rank (see [1]).

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1st Proposition.

The Apollonius’s circle of second rank relative to the vertex 퐴 of the triangle 퐴퐵퐶 intersects the circumscribed circle in two points 푄 and 푃 (푄 on the same side of 퐵퐶 as 퐴). Then, (푄푆 is a bisector in the triangle 푄퐵퐶, S is the simedian’s foot from 퐴 of the triangle 퐴퐵퐶.

Proof.

푄 belongs to the Apollonius’s circle of second rank, therefore: 푄퐵 퐴퐵 2 = ( ) . (1) 푄퐶 퐴퐶

Figure 1.

On the other hand, 푆 being the simedian’s foot, we have:

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푆퐵 퐴퐵 2 = ( ) . (2) 푆퐶 퐴퐶 From relations (1) and (2), we note that 푄퐵 푆퐵 = , 푄퐶 푆퐶 a relation showing that 푄푆 is bisector in the triangle 푄퐵퐶.

Remarks.

1. The Apollonius’s circle of second relative to the vertex 퐴 of the triangle 퐴퐵퐶 (see Figure 1) is an Apollonius’s circle for the triangle 푄퐵퐶. Indeed, we proved that 푄푆 is an internal bisector in the triangle 푄퐵퐶, and since 푆′, the external simedian’s foot of the triangle 퐴퐵퐶 , belongs to the Apollonius’s Circle of second rank, we have 푚(∢푆′푄푆) = 900 , therefore 푄푆’ is an external bisector in the triangle 푄퐵퐶. 2. 푄푃 is a simedian in 푄퐵퐶 . Indeed, the Apollonius’s circle of second rank, being an Apollonius’s circle for 푄퐵퐶, intersects the circle circum-scribed to 푄퐵퐶 after 푄푃, which is simedian in this triangle.

2nd Definition.

It is called adjoint circle of a triangle the circle that passes through two vertices of the triangle and in

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3rd Definition.

It is called the second Brocard’s triangle the triangle 퐴2퐵2퐶2 whose vertices are the projections of the center of the circle circumscribed to the triangle 퐴퐵퐶 on triangle’s simedians.

2nd Proposition.

The Apollonius’s circle of second rank relative to the vertex 퐴 of triangle 퐴퐵퐶 and the adjoint circles relative to the same vertex 퐴 intersect in vertex 퐴2 of the second Brocard’s triangle.

Proof.

It is known that the adjoint circles (퐵퐴̅) and (퐶퐴̅) intersect in a point belonging to the simedian 퐴푆; we denote this point 퐴2 (see [3]). We have:

∢퐵퐴2푆 = ∢퐴2퐵퐴 + ∢퐴2퐴퐵, but:

∢퐴2퐵퐴 ≡ ∢퐵퐴2푆 = ∢퐴2퐴퐵 + 퐴2퐴퐶 = ∢퐴.

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Analogously, ∢퐶퐴2푆 = ∢퐴, therefore (퐴2푆 is the bisector of the angle 퐵퐴2퐶. The bisector’s theorem in this triangle leads to: 푆퐵 퐵퐴 = 2 , 푆퐶 퐶퐴2 but: 푆퐵 퐴퐵 2 = ( ) , 푆퐶 퐴퐶 consequently: 퐵퐴 퐴퐵 2 2 = ( ) , 퐶퐴2 퐴퐶 so 퐴2 is a point that belongs to the Apollonius’s circle of second rank.

Figure 2.

We prove that 퐴2 is a vertex in the second Brocard’s triangle, i.e. 푂퐴2 ⊥ 퐴푆, O the center of the circle circumscribed to the triangle 퐴퐵퐶.

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We pointed (see Figure 2) that 푚(퐵̂퐴2퐶) = 2퐴, if ∢퐴 is an acute angle, then also 푚(퐵푂퐶̂) = 2퐴 , therefore the quadrilateral 푂퐶퐴2퐵 is inscriptible. Because 푚(푂퐶퐵̂) = 900 − 푚(퐴̂) , it follows that 0 푚(퐵̂퐴2푂) = 90 + 푚(퐴̂). 0 On the other hand, 푚(퐴̂퐴2퐵) = 180 − 푚(퐴̂), so 0 푚(퐵̂퐴2푂) + 푚(퐴̂퐴2퐵) = 270 and, consequently, 푂퐴2 ⊥ 퐴푆.

Remarks.

1. If 푚(퐴̂) < 900 , then four remarkable

circles pass through 퐴2 : the two circles adjoint to the vertex 퐴 of the triangle 퐴퐵퐶, the circle circumscribed to the triangle 퐵푂퐶 (where 푂 is the center of the circumscribed circle) and the Apollonius’s circle of second rank corresponding to the vertex 퐴.

2. The vertex 퐴2 of the second Brocard’s triangle is the middle of the chord of the circle circumscribed to the triangle 퐴퐵퐶 containing the simedian 퐴푆.

3. The points 푂 , 퐴2 and 푆’ (the foot of the external simedian to 퐴퐵퐶) are collinear.

Indeed, we proved that 푂퐴2 ⊥ 퐴푆; on the other hand, we proved that (퐴2푆 is an internal bisector in the triangle 퐵퐴2퐶, and

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since 푆′퐴2 ⊥ 퐴푆, the outlined collinearity follows from the uniqueness of the

perpendicular in 퐴2 on 퐴푆.

Open Problem.

The Apollonius’s circle of second rank relative to the vertex 퐴 of the triangle 퐴퐵퐶 intersects the circle circumscribed to the triangle 퐴퐵퐶 in two points 푃 and 푄 (푃 and 퐴 apart of 퐵퐶). We denote by 푋 the second point of intersection between the line 퐴푃 and the Apollonius’s circle of second rank. What can we say about 푋? Is 푋 a remarkable point in triangle’s geometry?

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References.

[1] I. Patrascu, F. Smarandache: Cercurile Apollonius de rangul k [The Apollonius’s Circle of kth rank]. In: “Recreații matematice”, Anul XVIII, nr. 1/2016, Iași, România, p. 20-23. [2] I. Patrascu: Axe și centre radicale ale cercurilor adjuncte ale unui triunghi [Axes and Radical Centers of Adjoint Circles of a Triangle]. In: “Recreații matematice”, Anul XVII, nr. 1/2010, Iași, România. [3] R. Johnson: Advanced Euclidean Geometry. New York: Dover Publications, Inc. Mineola, 2007. [4] I. Patrascu, F. Smarandache: Variance on topics of plane geometry. Columbus: The Educational Publisher, Ohio, USA, 2013.

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A Sufficient Condition for the Circle of the 6 Points to Become Euler’s Circle

In this article, we prove the theorem relative to the circle of the 6 points and, requiring on this circle to have three other remarkable triangle’s points, we obtain the circle of 9 points (the Euler’s Circle).

1st Definition.

It is called cevian of a triangle the line that passes through the vertex of a triangle and an opposite side’s point, other than the two left vertices.

Figure 1.

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1st Remark.

The name cevian was given in honor of the italian geometrician Giovanni Ceva (1647 - 1734).

2nd Definition.

Two cevians of the same triangle’s vertex are called isogonal cevians if they create congruent angles with triangle’s bisector taken through the same vertex.

2nd Remark.

In the Figure 1 we represented the bisector 퐴퐷 and the isogonal cevians 퐴퐴1 and 퐴퐴2. The following relations take place:

퐴̂1퐴퐷 ≡ 퐴̂2퐴퐷; 퐵퐴̂퐴1 ≡ 퐶퐴퐴̂2.

1st Proposition.

In a triangle, the height and the radius of the circumscribed circle corresponding to a vertex are isogonal cevians.

Proof.

Let 퐴퐵퐶 an acute-angled triangle with the height 퐴퐴′ and the radius 퐴푂 (see Figure 2).

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Figure 2.

The angle 퐴푂퐶 is a central angle, and the angle 퐴퐵퐶 is an inscribed angle, so 퐴푂퐶̂ = 2퐴퐵퐶̂. It follows that 퐴푂퐶̂ = 900 − 퐵̂. On the other hand, 퐵퐴퐴̂′ = 900 − 퐵̂ , so 퐴퐴′ and 퐴푂 are isogonal cevians. The theorem can be analogously proved for the obtuse triangle.

3rd Remark.

One can easily prove that in a triangle, if 퐴푂 is circumscribed circle’s radius, its isogonal cevian is the triangle’s height from vertex 퐴.

3rd Definition.

Two points 푃1, 푃2 in the plane of triangle 퐴퐵퐶 are called isogonals (isogonal conjugated) if the cevians’ pairs (퐴푃1, 퐴푃2), (퐵푃1, 퐵푃2), (퐶푃1, 퐶푃2), are composed of isogonal cevians.

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4th Remark.

In a triangle, the orthocenter and circumscribed circle’s center are isogonal points.

1st Theorem. (The 6 points circle)

If 푃1 and 푃2 are two isogonal points in the triangle 퐴퐵퐶 , and 퐴1, 퐵1, 퐶1 respectively 퐴2, 퐵2, 퐶2 are their projections on the sides 퐵퐶 , 퐶퐴 and 퐴퐵 of the triangle, then the points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are concyclical.

Proof.

The mediator of segment [퐴1퐴2] passes through the middle 푃 of segment [푃1, 푃2] because the trapezoid 푃1퐴1퐴2푃2 is rectangular and the mediator of [퐴1퐴2 ] contains its middle line, therefore (see Figure 3), we have: 푃퐴1 = 푃퐴2 (1). Analogously, it follows that the mediators of segments [퐵1퐵2] and [퐶1퐶2] pass through 푃, so 푃퐵1 = 푃퐵2 (2) and 푃퐶1 = 푃퐶2 (3). We denote by 퐴3 and 퐴4 respectively the middles of segments [퐴푃1] and [ 퐴푃2 ]. We prove that the triangles 푃퐴3퐶1 and 1 퐵 퐴 푃 are congruent. Indeed, 푃퐴 = 퐴푃 (middle 2 4 3 2 2 1 line), and 퐵 퐴 = 퐴푃 , because it is a median in the 2 4 2 2 rectangular triangle 푃2퐵2퐴 , so 푃퐴3 = 퐵2퐴4 ; analogously, we obtain that 퐴4푃 = 퐴3퐶1 . 124

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Figure 3.

We have that:

푃̂퐴3퐶1 = 푃̂퐴3푃1 + 푃1̂퐴3퐶1 = 푃̂1퐴푃2 + 2푃̂1퐴퐶1 = = 퐴̂ + 푃̂1퐴퐵; 퐵̂2퐴4푃 = 퐵2̂퐴4푃2 + 푃̂퐴4푃2 = 푃̂1퐴푃2 + 2푃̂2퐴퐵2 = = 퐴̂ + 푃̂2퐴퐶. But 푃̂1퐴퐵 = 푃̂2퐴퐶 , because the cevians 퐴푃1 and 퐴푃2 are isogonal and therefore 푃̂퐴3퐶1 = 퐵̂2퐴4푃. Since ∆푃퐴3퐶1 = ∆퐵2퐴4푃, it follows that 푃퐵2 = 푃퐶1 (4). Repeating the previous reasoning for triangles

푃퐵3퐶1 and 퐴2퐵4푃 , where 퐵3 and 퐵4 are respectively the middles of segments (퐵푃1) and (퐵푃2), we find that they are congruent and it follows that 푃퐶1 = 푃퐴2 (5). The relations (1) - (5) lead to 푃퐴1 = 푃퐴2 = 푃퐵1 = 푃퐵2 = 푃퐶1 = 푃퐶2 , which shows that the points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are located on a circle centered in 푃, 125

Ion Patrascu, Florentin Smarandache the middle of the segment determined by isogonal points given by 푃1 and 푃2.

4th Definition.

It is called the 9 points circle or Euler’s circle of the triangle 퐴퐵퐶 the circle that contains the middles of triangle’s sides, the triangle heights’ feet and the middles of the segments determined by the orthocenter with triangle’s vertex.

2nd Proposition.

If 푃1, 푃2 are isogonal points in the triangle 퐴퐵퐶 and if on the circle of their corresponding 6 points there also are the middles of the segments (퐴푃1), (퐵푃1), ( 퐶푃1 ), then the 6 points circle coincides with the Euler’s circle of the triangle 퐴퐵퐶.

1st Proof.

We keep notations from Figure 3; we proved that the points 퐴1, 퐴2, 퐵1, 퐵2, 퐶1, 퐶2 are on the 6 points circle of the triangle 퐴퐵퐶, having its center in 푃, the middle of segment [푃1푃2]. If on this circle are also situated the middles

퐴3, 퐵3, 퐶3 of segments (퐴푃1), (퐵푃1), (퐶푃1), then we have 푃퐴3 = 푃퐵3 = 푃퐶3.

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We determined that 푃퐴3 is middle line in the 1 triangle 푃 푃 퐴 , therefore 푃퐴 = 퐴푃 , analogously 1 2 3 2 2 1 1 푃퐵 = 퐵푃 and 푃퐶 = 퐶푃 , and we obtain that 푃 퐴 = 3 2 2 3 2 2 2 푃2퐵 = 푃2퐶, consequently 푃 is the center of the circle circumscribed to the triangle 퐴퐵퐶, so 푃2 = 푂. Because 푃1 is the isogonal of 푂, it follows that 푃1 = 퐻, therefore the circle of 6 points of the isogonal points 푂 and 퐻 is the circle of 9 points.

2nd Proof.

Because 퐴3퐵3 is middle line in the triangle 푃1퐴퐵, it follows that

∢푃1퐴퐵 ≡ ∢푃1퐴2퐵3. (1) Also, 퐴3퐶3 is middle line in the triangle 푃1퐴퐶, and 퐴3퐶3 is middle line in the triangle 푃1퐴푃2, therefore we get

∢푃퐴3퐶3 ≡ ∢푃2퐴퐶. (2) The relations (1), (2) and the fact that 퐴푃1 and 퐴푃2 are isogonal cevians lead to: ∢푃1퐴2퐵3 ≡ 푃퐴3퐶3. (3) The point 푃 is the center of the circle circumscribed to 퐴3퐵3퐶3 ; then, from (3) and from isogonal cevians’ properties, one of which is circumscribed circle radius, it follows that in the triangle 퐴3퐵3퐶3 the line 푃1퐴3 is a height, as 퐵3퐶3 ∥ 퐵퐶, we get that 푃1퐴 is a height in the triangle ABC and, therefore, 푃1 will be the orthocenter of the triangle

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퐴퐵퐶 , and 푃2 will be the center of the circle circumscribed to the triangle 퐴퐵퐶.

5th Remark.

Of those shown, we get that the center of the circle of 9 points is the middle of the line determined by triangle’s orthocenter and by the circumscribed circle’s center, and the fact that Euler’s circle radius is half the radius of the circumscribed circle.

References.

[1] Roger A. Johnson: Advanced Euclidean Geometry. New York: Dover Publications, 2007. [2] Cătălin Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental theorems of triangle’s geometry]. Bacău: Editura Unique, 2008.

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An Extension of a Problem of Fixed Point

In this article, we extend the requirement of the Problem 9.2 proposed at Varna 2015 Spring Competition, both in terms of membership of the measure 훾, and the case for the problem for the ex-inscribed circle 퐶. We also try to guide the student in the search and identification of the fixed point, for succeeding in solving any problem of this type.

The statement of the problem is as follows: “We fix an angle 훾 ∈ (0, 900) and the line 퐴퐵 which divides the plane in two half-planes 훾 and 훾̅. The point 퐶 in the half-plane 훾 is situated such that 푚(퐴퐶퐵̂) = 훾. The circle inscribed in the triangle 퐴퐵퐶 with the center 퐼 is tangent to the sides 퐴퐶 and 퐵퐶 in the points 퐹 and 퐸 , respectively. The point 푃 is located on the segment line (퐼퐸 , the point 퐸 between 퐼 and 푃 such that 푃퐸 ⊥ 퐵퐶 and 푃퐸 = 퐴퐹. The point 푄 is situated on the segment line (퐼퐹, such that 퐹 is between 퐼 and 푄 ; 푄퐹 ⊥ 퐴퐶 and 푄퐹 = 퐵퐸 . Prove 129

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that the mediator of segment 푃푄 passes through a fixed point.” (Stanislav Chobanov)

Proof.

Firstly, it is useful to note that the point 퐶 varies in the half-plane 휓 on the arc capable of angle 훾; we 훾 know as well that 푚(퐴퐼퐵̂ ) = 900 + , so 퐼 varies on the 2 훾 arc capable of angle of measure 900 + situated in the 2 half-plane 휓. Another useful remark is about the segments 퐴퐹 and 퐵퐸, which in a triangle have the lengths 푝 − 푎, respectively 푝 − 푏 , where 푝 is the half-perimeter of the triangle 퐴퐵퐶 with 퐴퐵 = 푐 – constant; therefore, we have ∆푃퐸퐵 ≡ ∆퐴퐹푄 with the consequence 푃퐵 = 푄퐴. Considering the vertex 퐶 of the triangle 퐴퐵퐶 the middle of the arc capable of angle 훾 built on 퐴퐵, we observe that 푃푄 is parallel to 퐴퐵 ; more than that, 퐴퐵푃푄 is an isosceles trapezoid, and segment 푃푄 mediator will be a symmetry axis of the trapezoid, so it will coincide with the mediator of 퐴퐵, which is a fixed line, so we're looking for the fixed point on mediator of 퐴퐵. Let 퐷 be the intersection of the mediators of segments 푃푄 and 퐴퐵 , see Figure 1, where we considered 푚(퐴̂) < 푚(퐵̂) . The point 퐷 is on the mediator of 퐴퐵, so we have 퐷퐴 = 퐷퐵; the point 퐷 is also on the mediator of 푃푄, so we have 퐷푃 = 퐷푄; it

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Figure 1

If we denote 푚(푄퐴퐹̂) = 푥 and 푚(퐷퐴퐵̂) = 푦, we have 푄퐴퐷̂ = 푥 + 퐴 + 푦, 푃퐵퐷̂ = 3600 − 퐵 − 푦 − (900 − 푥). From 푥 + 퐴 + 푦 = 3600 − 퐵 − 푦 − 900 + 푥, we find that 퐴 + 퐵 + 2푦 = 2700, and since 퐴 + 퐵 = 1800 − 훾, we find that 2푦 = 900 − 훾, therefore the requested fixed point 퐷 is the vertex of triangle 퐷퐴퐵, situated in 휓̅ such that 푚(퐴퐷퐵̂) = 900 − 훾.

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1st Remark.

If 훾 = 900, we propose to the reader to prove that the quadrilateral 퐴퐵푃푄 is a parallelogram; in this case, the requested fixed point does not exist (or is the point at infinity of the perpendicular lines to 퐴퐵).

2nd Remark.

If 훾 ∈ (900, 1800), the problem applies, and we find that the fixed point 퐷 is located in the half-plane 휓 , such that the triangle 퐷퐴퐵 is isosceles, having 푚(퐴푂퐵̂) = 훾 − 900. We suggest to the reader to solve the following problem: We fix an angle 훾 ∈ (00, 1800) and the line AB which divides the plane in two half-planes, 휓 and 휓̅. The point 퐶 in the half-plane 휓 is located such that 푚(퐴퐶퐵̂) = 훾 . The circle 퐶 – ex-

inscribed to the triangle 퐴퐵퐶 with center 퐼푐 is tangent to the sides 퐴퐶 and 퐵퐶 in the points 퐹 and 퐸, respectively. The point 푃 is located on the

line segment (퐼푐퐸 , 퐸 is between 퐼푐 and 푃 such that 푃퐸 ⊥ 퐵퐶 and 푃퐸 = 퐴퐹 . The point 푄 is

located on the line segment (퐼푐퐹 such that 퐹 is between 퐼 and 푄 , 푄퐹 ⊥ 퐴퐶 and 푄퐹 = 퐵퐸 . Prove that the mediator of the segment 푃푄 passes through a fixed point.

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3rd Remark.

As seen, this problem is also true in the case 훾 = 900, more than that, in this case, the fixed point is the middle of 퐴퐵. Prove!

References.

[1] Ion Patrascu: Probleme de geometrie plană [Planar Geometry Problems]. Craiova: Editura Cardinal, 1996.

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Some Properties of the Harmonic Quadrilateral

In this article, we review some properties of the harmonic quadrilateral related to triangle simedians and to Apollonius’s Circle.

1st Definition.

A convex circumscribable quadrilateral 퐴퐵퐶퐷 having the property 퐴퐵 ∙ 퐶퐷 = 퐵퐶 ∙ 퐴퐷 is called harmonic quadrilateral.

2nd Definition.

A triangle simedian is the isogonal cevian of a triangle median.

1st Proposition.

In the triangle 퐴퐵퐶, the cevian 퐴퐴1, 퐴1 ∈ (퐵퐶) is 퐵퐴 퐴퐵 2 a simedian if and only if 1 = ( ) . For Proof of this 퐴1퐶 퐴퐶 property, see infra.

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Figura 1.

2nd Proposition.

In an harmonic quadrilateral, the diagonals are simedians of the triangles determined by two consecutive sides of a quadrilateral with its diagonal.

Proof.

Let 퐴퐵퐶퐷 be an harmonic quadrilateral and {퐾} = 퐴퐶 ∩ 퐵퐷 (see Figure 1). We prove that 퐵퐾 is simedian in the triangle 퐴퐵퐶. From the similarity of the triangles 퐴퐵퐾 and 퐷퐶퐾, we find that: 퐴퐵 퐴퐾 퐵퐾 = = . (1) 퐷퐶 퐷퐾 퐶퐾 From the similarity of the triangles 퐵퐶퐾 şi 퐴퐷퐾, we conclude that:

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퐵퐶 퐶퐾 퐵퐾 = = . (2) 퐴퐷 퐷퐾 퐴퐾 From the relations (1) and (2), by division, it follows that: 퐴퐵 퐴퐷 퐴퐾 . = . (3) 퐵퐶 퐷퐶 퐶퐾 But 퐴퐵퐶퐷 is an harmonic quadrilateral; consequently, 퐴퐵 퐴퐷 = ; 퐵퐶 퐷퐶 substituting this relation in (3), it follows that: 퐴퐵 2 퐴퐾 ( ) = ; 퐵퐶 퐶퐾 As shown by Proposition 1, 퐵퐾 is a simedian in the triangle 퐴퐵퐶. Similarly, it can be shown that 퐴퐾 is a simedian in the triangle 퐴퐵퐷, that 퐶퐾 is a simedian in the triangle 퐵퐶퐷, and that 퐷퐾 is a simedian in the triangle 퐴퐷퐶.

Remark 1.

The converse of the 2nd Proposition is proved similarly, i.e.:

3rd Proposition.

If in a convex circumscribable quadrilateral, a diagonal is a simedian in the triangle formed by the other diagonal with two consecutive sides of the quadrilateral, then the quadrilateral is an harmonic quadrilateral. 137

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Remark 2.

From 2nd and 3rd Propositions above, it results a simple way to build an harmonic quadrilateral. In a circle, let a triangle 퐴퐵퐶 be considered; we construct the simedian of A, be it 퐴퐾, and we denote by D the intersection of the simedian 퐴퐾 with the circle. The quadrilateral 퐴퐵퐶퐷 is an harmonic quadrilateral.

Proposition 4.

In a triangle 퐴퐵퐶, the points of the simedian of A are situated at proportional lengths to the sides 퐴퐵 and 퐴퐶.

Figura 2.

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Proof.

We have the simedian 퐴퐴1 in the triangle 퐴퐵퐶 (see Figure 2). We denote by 퐷 and 퐸 the projections of 퐴1 on 퐴퐵, and 퐴퐶 respectively. We get: 퐵퐴 퐴푟푒푎 (퐴퐵퐴 ) 퐴퐵 ∙ 퐴 퐷 1 = ∆ 1 = 1 . 퐶퐴1 퐴푟푒푎∆(퐴퐶퐴1) 퐴퐶 ∙ 퐴1퐸 Moreover, from 1st Proposition, we know that 퐵퐴 퐴퐵 2 1 = ( ) . 퐴1퐶 퐴퐶 Substituting in the previous relation, we obtain that: 퐴 퐷 퐴퐵 1 = . 퐴1퐸 퐴퐶 On the other hand, 퐷퐴1 = 퐴퐴1. From 퐵퐴퐴1 and 퐴1퐸 = 퐴퐴1 ∙ 푠𝑖푛퐶퐴̂퐴1, hence: 퐴 퐷 푠𝑖푛퐵퐴̂퐴 퐴퐵 1 = 1 = . (4) 퐴1퐸 푠𝑖푛퐶퐴̂퐴1 퐴퐶 If M is a point on the simedian and 푀푀1 and 푀푀2 are its projections on 퐴퐵 , and 퐴퐶 respectively, we have:

푀푀1 = 퐴푀 ∙ 푠𝑖푛퐵퐴̂퐴1, 푀푀2 = 퐴푀 ∙ 푠𝑖푛퐶퐴̂퐴1, hence: 푀푀 푠𝑖푛퐵퐴̂퐴 1 = 1 . 푀푀2 푠𝑖푛퐶퐴̂퐴1 Taking (4) into account, we obtain that: 푀푀 퐴퐵 1 = . 푀푀2 퐴퐶

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3rd Remark.

The converse of the property in the statement above is valid, meaning that, if 푀 is a point inside a triangle, its distances to two sides are proportional to the lengths of these sides. The point belongs to the simedian of the triangle having the vertex joint to the two sides.

5th Proposition.

In an harmonic quadrilateral, the point of intersection of the diagonals is located towards the sides of the quadrilateral to proportional distances to the length of these sides. The Proof of this Proposition relies on 2nd and 4th Propositions.

6th Proposition. (R. Tucker)

The point of intersection of the diagonals of an harmonic quadrilateral minimizes the sum of squares of distances from a point inside the quadrilateral to the quadrilateral sides.

Proof.

Let 퐴퐵퐶퐷 be an harmonic quadrilateral and 푀 any point within.

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We denote by 푥, 푦, 푧, 푢 the distances of 푀 to the 퐴퐵 , 퐵퐶 , 퐶퐷 , 퐷퐴 sides of lenghts 푎, 푏, 푐, and 푑 (see Figure 3).

Figure 3.

Let 푆 be the 퐴퐵퐶퐷 quadrilateral area. We have: 푎푥 + 푏푦 + 푐푧 + 푑푢 = 2푆. This is true for 푥, 푦, 푧, 푢 and 푎, 푏, 푐, 푑 real numbers. Following Cauchy-Buniakowski-Schwarz Ine- quality, we get: (푎2 + 푏2 + 푐2 + 푑2)(푥2 + 푦2 + 푧2 + 푢2) ≥ (푎푥 + 푏푦 + 푐푧 + 푑푢)2 , and it is obvious that: 4푆2 푥2 + 푦2 + 푧2 + 푢2 ≥ . 푎2 + 푏2 + 푐2 + 푑2 We note that the minimum sum of squared distances is:

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4푆2 = 푐표푛푠푡. 푎2 + 푏2 + 푐2 + 푑2 In Cauchy-Buniakowski-Schwarz Inequality, the equality occurs if: 푥 푦 푧 푢 = = = . 푎 푏 푐 푑 Since {퐾} = 퐴퐶 ∩ 퐵퐷 is the only point with this property, it ensues that 푀 = 퐾, so 퐾 has the property of the minimum in the statement.

3rd Definition.

We call external simedian of 퐴퐵퐶 triangle a cevian 퐴퐴1’ corresponding to the vertex 퐴, where 퐴1’ is the harmonic conjugate of the point 퐴1 – simedian’s foot from 퐴 relative to points 퐵 and 퐶.

4th Remark.

In Figure 4, the cevian 퐴퐴1 is an internal simedian, and 퐴퐴1’ is an external simedian. We have: 퐴 퐵 퐴 ′퐵 1 = 1 . 퐴1퐶 퐴1′퐶 In view of 1st Proposition, we get that: 퐴 ′퐵 퐴퐵 2 1 = ( ) . 퐴1′퐶 퐴퐶

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7th Proposition.

The tangents taken to the extremes of a diagonal of a circle circumscribed to the harmonic quadrilateral intersect on the other diagonal.

Proof.

Let 푃 be the intersection of a tangent taken in 퐷 to the circle circumscribed to the harmonic quadrilateral 퐴퐵퐶퐷 with 퐴퐶 (see Figure 4).

Figure 4.

Since triangles PDC and PAD are alike, we conclude that: 푃퐷 푃퐶 퐷퐶 = = . (5) 푃퐴 푃퐷 퐴퐷 From relations (5), we find that: 푃퐴 퐴퐷 2 = ( ) . (6) 푃퐶 퐷퐶

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This relationship indicates that P is the harmonic conjugate of K with respect to A and C, so 퐷푃 is an external simedian from D of the triangle 퐴퐷퐶. Similarly, if we denote by 푃’ the intersection of the tangent taken in B to the circle circumscribed with 퐴퐶, we get: 푃′퐴 퐵퐴 2 = ( ) . (7) 푃′퐶 퐵퐶 From (6) and (7), as well as from the properties of the harmonic quadrilateral, we know that: 퐴퐵 퐴퐷 = , 퐵퐶 퐷퐶 which means that: 푃퐴 푃′퐴 = , 푃퐶 푃′퐶 hence 푃 = 푃’. Similarly, it is shown that the tangents taken to A and C intersect at point Q located on the diagonal 퐵퐷.

5th Remark.

a. The points P and Q are the diagonal poles of 퐵퐷 and 퐴퐶 in relation to the circle circumscribed to the quadrilateral. b. From the previous Proposition, it follows that in a triangle the internal simedian of an angle is consecutive to the external simedians of the other two angles.

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Figure 5.

8th Proposition.

Let 퐴퐵퐶퐷 be an harmonic quadrilateral inscribed in the circle of center O and let P and Q be the intersections of the tangents taken in B and D, respectively in A and C to the circle circumscribed to the quadrilateral. If {퐾} = 퐴퐶 ∩ 퐵퐷, then the orthocenter of triangle 푃퐾푄 is O.

Proof.

From the properties of tangents taken from a point to a circle, we conclude that 푃푂 ⊥ 퐵퐷 and 푄푂 ⊥ 퐴퐶. These relations show that in the triangle 푃퐾푄, 푃푂 and 푄푂 are heights, so 푂 is the orthocenter of this triangle. 145

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4th Definition.

The Apollonius’s circle related to the vertex A of the triangle 퐴퐵퐶 is the circle built on the segment [퐷퐸] in diameter, where D and E are the feet of the internal, respectively external, bisectors taken from A to the triangle 퐴퐵퐶.

6th Remark.

If the triangle 퐴퐵퐶 is isosceles with 퐴퐵 = 퐴퐶, the Apollonius’s circle corresponding to vertex A is not defined.

9th Proposition.

The Apollonius’s circle relative to the vertex A of the triangle 퐴퐵퐶 has as center the feet of the external simedian taken from A.

Proof.

Let Oa be the intersection of the external simedian of the triangle 퐴퐵퐶 with 퐵퐶 (see Figure 6). Assuming that 푚(퐵̂) > 푚(퐶̂), we find that: 1 푚(퐸퐴퐵̂) = [푚(퐵̂) + 푚(퐶̂)]. 2 Oa being a tangent, we find that 푚(푂̂푎퐴퐵) = 푚(퐶̂). Withal, 1 푚(퐸퐴푂 ) = [푚(퐵̂) − 푚(퐶̂)] 푎 2 146

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and 1 푚(퐴퐸푂 ) = [푚(퐵̂) − 푚(퐶̂)]. 푎 2 It results that 푂푎퐸 = 푂푎퐴; onward, 퐸퐴퐷 being a right angled triangle, we obtain: 푂푎퐴 = 푂푎퐷; hence Oa is the center of Apollonius’s circle corresponding to the vertex 퐴.

Figura 6.

10th Proposition.

Apollonius’s circle relative to the vertex A of triangle 퐴퐵퐶 cuts the circle circumscribed to the triangle following the internal simedian taken from A.

Proof.

Let S be the second point of intersection of Apollonius’s Circle relative to vertex A and the circle circumscribing the triangle 퐴퐵퐶.

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Because 푂푎퐴 is tangent to the circle circumscribed in A, it results, for reasons of symmetry, that

푂푎푆 will be tangent in S to the circumscribed circle. For triangle 퐴퐶푆 , 푂푎퐴 and 푂푎푆 are external simedians; it results that 퐶푂푎 is internal simedian in the triangle 퐴퐶푆 , furthermore, it results that the quadrilateral 퐴퐵푆퐶 is an harmonic quadrilateral. Consequently, 퐴푆 is the internal simedian of the triangle 퐴퐵퐶 and the property is proven.

7th Remark.

From this, in view of Figure 5, it results that the circle of center Q passing through A and C is an Apollonius’s circle relative to the vertex A for the triangle 퐴퐵퐷. This circle (of center Q and radius QC) is also an Apollonius’s circle relative to the vertex C of the triangle 퐵퐶퐷. Similarly, the Apollonius’s circle corresponding to vertexes B and D and to the triangles ABC, and ADC respectively, coincide. We can now formulate the following:

11th Proposition.

In an harmonic quadrilateral, the Apollonius’s circle - associated with the vertexes of a diagonal and to the triangles determined by those vertexes to the other diagonal - coincide.

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Radical axis of the Apollonius’s circle is the right determined by the center of the circle circumscribed to the harmonic quadrilateral and by the intersection of its diagonals.

Proof.

Referring to Fig. 5, we observe that the power of O towards the Apollonius’s Circle relative to vertexes B and C of triangles 퐴퐵퐶 and 퐵퐶푈 is: 푂퐵2 = 푂퐶2. So O belongs to the radical axis of the circles. We also have 퐾퐴 ∙ 퐾퐶 = 퐾퐵 ∙ 퐾퐷 , relatives indicating that the point K has equal powers towards the highlighted Apollonius’s circle.

References.

[1] Roger A. Johnson: Advanced Euclidean Geometry, Dover Publications, Inc. Mineola, New York, USA, 2007. [2] F. Smarandache, I. Patrascu: The Geometry of Homological Triangles, The Education Publisher, Inc. Columbus, Ohio, USA, 2012. [2] F. Smarandache, I. Patrascu: Variance on Topics of plane Geometry, The Education Publisher, Inc. Columbus, Ohio, USA, 2013.

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Triangulation of a Triangle with Triangles having Equal Inscribed Circles

In this article, we solve the following problem: Any triangle can be divided by a cevian in two triangles that have congruent inscribed circles.

Solution.

We consider a given triangle 퐴퐵퐶 and we show that there is a point 퐷 on the side (퐵퐶) so that the inscribed circles in the triangles 퐴퐵퐷 , 퐴퐶퐷 are congruent. If 퐴퐵퐶 is an isosceles triangle (퐴퐵 = 퐴퐶), where 퐷 is the middle of the base (퐵퐶), we assume that 퐴퐵퐶 is a non-isosceles triangle. We note 퐼1, 퐼2 the centers of the inscribed congruent circles; obviously,

퐼1퐼2 is parallel to the 퐵퐶. (1) We observe that: 1 푚(∢퐼 퐴퐼 ) = 푚(퐴̂). (2) 1 2 2

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Figure 1.

If 푇1, 푇2 are contacts with the circles of the cevian 퐴퐷 , we have ∆ 퐼1푇1푀 ≡ ∆ 퐼2푇2푀 ; let 푀 be the intersection of 퐼1퐼2 with 퐴퐷, see Figure 1. From this congruence, it is obvious that:

(퐼1푀) ≡ (퐼2푀). (3) Let 퐼 be the center of the circle inscribed in the triangle 퐴퐵퐶; we prove that: 퐴퐼 is a simedian in the triangle 퐼1퐴퐼2. (4)

Indeed, noting 훼 = 푚(퐵퐴̂퐼1) , it follows that

푚(∢퐼1퐴푀) = 훼. From ∢퐼1퐴퐼2 = ∢퐵퐴퐼, it follows that ∢퐵퐴퐼1 ≡ ∢퐼퐴퐼2 , therefore ∢퐼1퐴푀 ≡ ∢퐼퐴퐼2 , indicating that 퐴푀 and 퐴퐼 are isogonal cevians in the triangle

퐼1퐴퐼2. Since in this triangle 퐴푀 is a median, it follows that 퐴퐼 is a bimedian.

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Now, we show how we build the point 퐷, using the conditions (1) – (4), and then we prove that this construction satisfies the enunciation requirements.

Building the point D.

10: We build the circumscribed circle of the given triangle 퐴퐵퐶; we build the bisector of the angle 퐵퐴퐶 and denote by P its intersection with the circumscribed circle (see Figure 2). 20: We build the perpendicular on 퐶 to 퐶푃 and

(퐵퐶) side mediator; we denote 푂1 the intersection of these lines. 0 3 : We build the circle ∁(푂1; 푂1퐶) and denote 퐴’ the intersection of this circle with the bisector 퐴퐼 (퐴’ is on the same side of the line 퐵퐶 as 퐴). 0 4 : We build through 퐴 the parallel to 퐴’푂1 and we denote it 퐼푂1. 0 ′ ′ 5 : We build the circle ∁(푂1; 푂1퐴) and we denote 퐼1, 퐼2 its intersections with 퐵퐼 , and 퐶퐼 respectively. 0 6 : We build the middle 푀 of the segment (퐼1퐼2) and denote by 퐷 the intersection of the lines 퐴푀 and 퐵퐶.

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Figure 2.

Proof.

The point 푃 is the middle of the arc 퐵퐶̆ , then 1 푚(푃퐶퐵̂) = 푚(퐴̂). 2 The circle ∁(푂1; 푂1퐶) contains the arc from which points the segment (BC) „is shown” under angle 1 measurement 푚(퐴̂). 2 ′ ′ The circle ∁(푂1; 푂1퐴) is homothetical to the circle ∁(푂1; 푂1퐶) by the homothety of center 퐼 and by 퐼퐴′ the report ; therefore, it follows that 퐼 퐼 will be 퐼퐴 1 2 parallel to the 퐵퐶 , and from the points of circle ′ ′ ∁(푂1; 푂1퐴) of the same side of 퐵퐶 as 퐴, the segment

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1 (퐼 퐼 ) „is shown” at an angle of measure 푚(퐴̂). Since 1 2 2 the tangents taken in 퐵 and 퐶 to the circle ∁(푂1; 푂1퐶) intersect in 푃, on the bisector 퐴퐼, as from a property of simedians, we get that 퐴′퐼 is a simedian in the triangle 퐴′퐵퐶. Due to the homothetical properties, it follows also that the tangents in the points 퐼1, 퐼2 to the circle ′ ′ ∁(푂1; 푂1퐴) intersect in a point 푃’ located on 퐴퐼, i.e. 퐴푃’ contains the simedian (퐴푆) of the triangle 퐼1퐴퐼2, noted ′ {푆} = 퐴푃 ∩ 퐼1퐼2. In the triangle 퐼1퐴퐼2, 퐴푀 is a median, and 퐴푆 is simedian, therefore ∢퐼1퐴푀 ≡ 퐼2퐴푆; on the other hand, ∢퐵퐴푆 ≡ ∢퐼1퐴퐼2; it follows that ∢퐵퐴퐼1 ≡ 퐼2퐴푆, and more: ∢퐼1퐴푀 ≡ 퐵퐴퐼1, which shows that 퐴퐼1 is a bisector in the triangle 퐵퐴퐷; plus, 퐼1, being located on the bisector of the angle 퐵, it follows that this point is the center of the circle inscribed in the triangle 퐵퐴퐷. Analogous considerations lead to the conclusion that 퐼2 is the center of the circle inscribed in the triangle 퐴퐶퐷. Because 퐼1퐼2 is parallel to 퐵퐶, it follows that the rays of the circles inscribed in the triangles 퐴퐵퐷 and 퐴퐶퐷 are equal.

Discussion.

′ ′ ′ The circles ∁(푂1; 푂1퐴 ), ∁(푂1; 푂1퐴) are unique; also, the triangle 퐼1퐴퐼2 is unique; therefore, determined as before, the point 퐷 is unique.

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Remark.

At the beginning of the Proof, we assumed that 퐴퐵퐶 is a non-isosceles triangle with the stated property. There exist such triangles; we can construct such a triangle starting "backwards". We consider two given congruent external circles and, by tangent constructions, we highlight the 퐴퐵퐶 triangle.

Open problem.

Given a scalene triangle 퐴퐵퐶 , could it be triangulated by the cevians 퐴퐷 , 퐴퐸 , with 퐷 , 퐸 belonging to (퐵퐶), so that the inscribed circles in the triangles 퐴퐵퐷, 퐷퐴퐸 and the 퐸퐴퐶 to be congruent?

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An Application of a Theorem of Orthology

In this short article, we make connections between Problem 21 of [1] and the theory of orthological triangles.

The enunciation of the competitional problem is:

Let (푇퐴), (푇퐵), (푇퐶) be the tangents in the peaks 퐴, 퐵, 퐶 of the triangle 퐴퐵퐶 to the circle circumscribed to the triangle. Prove that the perpendiculars drawn from the middles of the

opposite sides on (푇퐴), (푇퐵), (푇퐶) are concurrent and determine their concurrent point. We formulate and we demonstrate below a sentence containing in its proof the solution of the competitional problem in this case.

Proposition.

The tangential triangle and the median triangle of a given triangle 퐴퐵퐶 are orthological. The orthological centers are 푂 – the center of the circle circumscribed to the triangle 퐴퐵퐶, and 푂9 – the center of the circle of 퐴퐵퐶 triangle’s 9 points.

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Figure 1.

Proof.

Let 푀푎푀푏푀푐 be the median triangle of triangle 퐴퐵퐶 and 푇푎푇푏푇푐 the tangential triangle of the triangle 퐴퐵퐶 . It is obvious that the triangle 푇푎푇푏푇푐 and the triangle 퐴퐵퐶 are orthological and that 푂 is the orthological center.

Verily, the perpendiculars taken from 푇푎, 푇푏, 푇푐 on 퐵퐶; 퐶퐴; 퐴퐵 respectively are internal bisectors in the triangle 푇푎푇푏푇푐 and consequently passing through 푂, which is center of the circle inscribed in triangle

푇푎푇푏푇푐 . Moreover, 푇푎푂 is the mediator of (퐵퐶) and accordingly passing through 푀푎 , and 푇푎푀푐 is perpendicular on 퐵퐶, being a mediator, but also on

푀푏푀푐 which is a median line.

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From orthological triangles theorem, it follows that the perpendiculars taken from 푀푎, 푀푏, 푀푐 on 푇푏푇푐, 푇푐푇푎 , 푇푎푇푏 respectively, are concurrent. The point of concurrency is the second orthological center of triangles 푇푎푇푏푇푐 and 푀푎푀푏푀푐. We prove that this point is 푂9 – the center of Euler circle of triangle 퐴퐵퐶. We take 푀푎푀1 ⊥ 푇푏푇푐 and denote by {퐻1} = 푀푎푀1 ∩ 퐴퐻, 퐻 being the orthocenter of the triangle 퐴퐵퐶. We know that 퐴퐻 = 2푂푀푎; we prove this relation vectorially. From Sylvester’s relation, we have that: 푂퐻⃗⃗⃗⃗⃗⃗ =

푂퐴⃗⃗⃗⃗⃗ + 푂퐵⃗⃗⃗⃗⃗ + 푂퐶⃗⃗⃗⃗⃗ , but 푂퐵⃗⃗⃗⃗⃗ + 푂퐶⃗⃗⃗⃗⃗ = 2푂⃗⃗⃗⃗푀⃗⃗⃗⃗⃗푎 ; it follows that

푂퐻⃗⃗⃗⃗⃗⃗ − 푂퐴⃗⃗⃗⃗⃗ = 2푂⃗⃗⃗⃗푀⃗⃗⃗⃗⃗푎 , so 퐴퐻⃗⃗⃗⃗⃗⃗ = 2푂⃗⃗⃗⃗푀⃗⃗⃗⃗⃗푎 ; changing to module, we have 퐴퐻 = 2푂푀푎 . Uniting 푂 to 퐴, we have 푂퐴 ⊥ 푇푏푇푐 , and because 푀푎푀1 ⊥ 푇푏푇푐 and 퐴퐻 ∥ 푂푀푎 , it follows that the quadrilateral 푂푀푎퐻1퐴 is a parallelogram.

From 퐴퐻1 = 푂푀푎 and 퐴퐻 = 2푂푀푎 we get that 퐻1 is the middle of (퐴퐻), so 퐻1 is situated on the circle of the 9 points of triangle 퐴퐵퐶. On this circle, we find as ′ well the points 퐴 - the height foot from 퐴 and 푀푎 ; ′ 0 since ∢퐴퐴 푀푎 = 90 , it follows that 푀푎퐻1 is the diameter of Euler circle, therefore the middle of

(푀푎퐻1), which we denote by 푂9, is the center of Euler’s circle; we observe that the quadrilateral 퐻1퐻푀푎푂 is as well a parallelogram; it follows that 푂9 is the middle of segment [푂퐻].

In conclusion, the perpendicular from 푀푎 on 푇푏푇푐 pass through 푂9.

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Analogously, we show that the perpendicular taken from 푀푏 on 푇푎푇푐 pass through 푂9 and consequently 푂9 is the enunciated orthological center.

Remark.

The triangles 푀푎푀푏푀푐 and 푇푎푇푏푇푐 are homological as well, since 푇푎푀푎 , 푇푏푀푏 , 푇푐푀푐 are concurrent in O, as we already observed, therefore the triangles 푇푎푇푏푇푐 and 푀푎푀푏푀푐 are orthohomological of rank I (see [2]). From P. Sondat theorem (see [4]), it follows that the Euler line 푂푂9 is perpendicular on the homological axis of the median triangle and of the tangential triangle corresponding to the given triangle 퐴퐵퐶.

Note. (Regarding the triangles that are simultaneously orthological and homological)

In the article A Theorem about Simultaneous Orthological and Homological Triangles, by Ion Patrascu and Florentin Smarandache, we stated and proved a theorem which was called by Mihai Dicu in [2] and [3] The Smarandache-Patrascu Theorem of Orthohomological Triangle; then, in [4], we used the term ortohomological triangles for the triangles that are simultaneously orthological and homological. The term ortohomological triangles was used by J. Neuberg in Nouvelles Annales de Mathematiques 160

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(1885) to name the triangles that are simultaneously orthogonal (one has the sides perpendicular to the sides of the other) and homological. We suggest that the triangles that are simultaneously orthogonal and homological to be called ortohomological triangles of first rank, and triangles that are simultaneously orthological and homological to be called ortohomological triangles of second rank.

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References.

[1] Titu Andreescu and Dorin Andrica: 360 de probleme de matematică pentru concursuri [360 Competitional Math Problem], Editura GIL, Zalau, Romania, 2002. [2] Mihai Dicu, The Smarandache-Patrascu Theorem of Orthohomological Triangle, http://www.scrib.com/ doc/28311880/, 2010 [3] Mihai Dicu, The Smarandache-Patrascu Theorem of Orthohomological Triangle, in “Revista de matematică ALPHA”, Year XVI, no. 1/2010, Craiova, Romania. [4] Ion Patrascu, Florentin Smarandache: Variance on topics of Plane Geometry, Educational Publishing, Columbus, Ohio, 2013. [5] Multispace and multistructure neutrosophic transdisciplinarity (100 collected papers of sciences), Vol IV, Edited by prof. Florentin Smarandache, Dept. of Mathematics and Sciences, University of New Mexico, USA - North European Scientific Publishers, Hanco, Finland, 2010.

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The Dual of a Theorem Relative to the Orthocenter of a Triangle

In [1] we introduced the notion of Bobillier’s transversal relative to a point 푶 in the plane of a triangle 푨푩푪; we use this notion in what follows. We transform by duality with respect to a circle ℂ(표, 푟) the following theorem relative to the orthocenter of a triangle.

1st Theorem.

If 퐴퐵퐶 is a nonisosceles triangle, 퐻 its orthocenter, and AA1, BB1, CC1 are cevians of a triangle concurrent at point 푄 different from 퐻, and 푀, 푁, 푃 are the intersections of the perpendiculars taken from 퐻 on given cevians respectively, with 퐵퐶, 퐶퐴, 퐴퐵, then the points 푀, 푁, 푃 are collinear.

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Proof.

We note with 훼 = 푚(∢BAA1); 훽 = 푚(∢CBB1), 훾 = 푚(∢ACC1), see Figure 1. According to Ceva’s theorem, trigonometric form, we have the relation: sin 훼 sin훽 sin훾 ∙ ∙ = 1. (1) sin(퐴−훼) sin(퐵−훽) sin(퐶−훾) We notice that: 푀퐵 Area(푀퐻퐵) 푀퐻∙퐻퐵∙sin sin(푀̂퐻퐵) = = . 푀퐶 Area(푀퐻퐶) 푀퐻∙퐻퐶∙sin(푀퐻퐶̂ ) Because:

∢푀퐻퐵 ≡ ∢퐴1퐴퐶, as angles of perpendicular sides, it follows that 푚(∢푀퐻퐵) = 푚(퐴̂) − 훼. Therewith: 푚(∢푀퐻퐶) = 푚(푀퐻퐵̂ ) + 푚(퐵퐻퐶̂ ) = 1800훼. We thus get that: 푀퐵 sin(퐴 − 훼) 퐻퐵 = ∙ . 푀퐶 sin훼 퐻퐶 Analogously, we find that: 푁퐶 sin(퐵−훽) 퐻퐶 = ∙ ; 푁퐴 sin훽 퐻퐴 푃퐴 sin(퐶−훾) 퐻퐴 = ∙ . 푃퐵 sin훾 퐻퐵 Applying the reciprocal of Menelaus' theorem, we find, in view of (1), that: 푀퐵 퐻퐶 푃퐴 ∙ ∙ = 1. 푀퐶 퐻퐴 푃퐵 This shows that 푀, 푁, 푃 are collinear.

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Figure 1.

Note.

1st Theorem is true even if 퐴퐵퐶 is an obtuse, nonisosceles triangle. The proof is adapted analogously.

2nd Theorem. (The Dual of the Theorem 1)

If 퐴퐵퐶 is a triangle, 푂 a certain point in his plan, and 퐴1, 퐵1, 퐶1 Bobillier’s transversals relative to 푂 of 165

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퐴퐵퐶 triangle, as well as 퐴2 − 퐵2 − 퐶2 a certain transversal in 퐴퐵퐶, and the perpendiculars in 푂, and on 푂퐴2, 푂퐵2, 푂퐶2 respectively, intersect the Bobillier’s transversals in the points 퐴3, 퐵3, 퐶3, then the ceviens 퐴퐴3, 퐵퐵3, 퐶퐶3 are concurrent.

Proof.

We convert by duality with respect to a circle ℂ(표, 푟) the figure indicated by the statement of this theorem, i.e. Figure 2. Let 푎, 푏, 푐 be the polars of the points 퐴, 퐵, 퐶 with respect to the circle ℂ(표, 푟). To the lines 퐵퐶, 퐶퐴, 퐴퐵 will correspond their poles {퐴′4 = 푏푛푐; {퐵′4 = 푐푛푎; {퐶′4 = 푎푛푏.

To the points 퐴1, 퐵1, 퐶1 will respectively correspond their polars 푎1, 푏1, 푐1 concurrent in transversal’s pole 퐴1 − 퐵1 − 퐶1. Since 푂퐴1 ⊥ 푂퐴, it means that the polars 푎 and 푎1 ′ ′ ′ are perpendicular, so 푎1 ⊥ 퐵 퐶 , but 푎1 pass through 퐴 , which means that 푄′ contains the height from 퐴′ of ′ ′ ′ 퐴 퐵 퐶 triangle and similarly 푏1 contains the height ′ ′ ′ ′ ′ from 퐵 and 푐1 contains the height from 퐶 of 퐴 퐵 퐶 triangle.

Consequently, the pole of 퐴1 − 퐵1 − 퐶1 transversal is the orthocenter 퐻′ of 퐴′퐵′퐶′ triangle. In the same way, to the points 퐴2, 퐵2, 퐶2 will correspond the polars to 푎2, 푏2, 푐2 which pass respectively through 퐴′, 퐵′, 퐶′ and are concurrent in a point 푄′, the pole of the line 퐴2 − 퐵2 − 퐶2 with respect to the circle ℂ(표, 푟).

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Figure 2.

Given 푂퐴2 ⊥ 푂퐴3, it means that the polar 푎2 and 푎3 are perpendicular, 푎2 correspond to the cevian ′ ′ 퐴 푄 , also 푎3 passes through the the pole of the ′ transversal 퐴1 − 퐵1 − 퐶1, so through 퐻 , in other words ′ ′ ′ 푄3 is perpendicular taken from 퐻 on 퐴 푄 ; similarly, ′ 푏2 ⊥ 푏3, 푐2 ⊥ 푐3, so 푏3 is perpendicular taken from 퐻 ′ ′ on 퐶 푄 . To the cevian 퐴퐴3 will correspond by duality considered to its pole, which is the intersection of the polars of 퐴 and 퐴3, i.e. the intersection of lines 푎 and ′ ′ 푎3 , namely the intersection of 퐵 퐶 with the perpendicular taken from 퐻′ on 퐴′푄′; we denote this

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References.

[1] Ion Patrascu, Florentin Smarandache: The Dual Theorem concerning Aubert’s Line, below. [2] Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles. The Education Publisher Inc., Columbus, Ohio, USA – 2012.

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The Dual Theorem Concerning Aubert’s Line

In this article we introduce the concept of Bobillier’s transversal of a triangle with respect to a point in its plan; we prove the Aubert’s Theorem about the collinearity of the orthocenters in the triangles determined by the sides and the diagonals of a complete quadrilateral, and we obtain the Dual Theorem of this Theorem.

1st Theorem. (E. Bobillier)

Let 퐴퐵퐶 be a triangle and 푀 a point in the plane of the triangle so that the perpendiculars taken in 푀, and 푀퐴, 푀퐵, 푀퐶 respectively, intersect the sides 퐵퐶, 퐶퐴 and 퐴퐵 at 퐴푚, 퐵푚 şi 퐶푚. Then the points 퐴푚, 퐵푚 and 퐶푚 are collinear.

Proof.

퐴푚퐵 aria (퐵푀퐴푚) We note that = (see Figure 1). 퐴푚퐶 aria (퐶푀퐴푚)

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Figure 1.

1 Area (퐵푀퐴푚) = ∙ 퐵푀 ∙ 푀퐴푚 ∙ sin(퐵푀퐴푚̂ ). 2 1 Area (퐶푀퐴푚) = ∙ 퐶푀 ∙ 푀퐴푚 ∙ sin(퐶푀퐴푚̂ ). 2 Since 3휋 푚(퐶푀퐴푚̂ ) = − 푚(퐴푀퐶̂ ), 2 it explains that sin(퐶푀퐴푚̂ ) = − cos(퐴푀퐶̂ ); 휋 sin(퐵푀퐴푚̂ ) = sin (퐴푀퐵̂ − ) = − cos(퐴푀퐵̂ ). 2 Therefore: 퐴푚퐵 푀퐵 ∙ cos(퐴푀퐵̂ ) = (1). 퐴푚퐶 푀퐶 ∙ cos(퐴푀퐶̂ ) In the same way, we find that:

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퐵푚퐶 푀퐶 cos(퐵푀퐶̂ ) = ∙ (2); 퐵푚퐴 푀퐴 cos(퐴푀퐵̂ ) 퐶푚퐴 푀퐴 cos(퐴푀퐶̂ ) = ∙ (3). 퐶푚퐵 푀퐵 cos(퐵푀퐶̂ ) The relations (1), (2), (3), and the reciprocal Theorem of Menelaus lead to the collinearity of points 퐴푚, 퐵푚, 퐶푚.

Note.

Bobillier's Theorem can be obtained – by converting the duality with respect to a circle – from the theorem relative to the concurrency of the heights of a triangle.

1st Definition.

It is called Bobillier’s transversal of a triangle 퐴퐵퐶 with respect to the point 푀 the line containing the intersections of the perpendiculars taken in 푀 on 퐴푀, 퐵푀, and 퐶푀 respectively, with sides 퐵퐶, CA and 퐴퐵.

Note.

The Bobillier’s transversal is not defined for any point 푀 in the plane of the triangle 퐴퐵퐶, for example, where 푀 is one of the vertices or the orthocenter 퐻 of the triangle.

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2nd Definition.

If 퐴퐵퐶퐷 is a convex quadrilateral and 퐸, 퐹 are the intersections of the lines 퐴퐵 and 퐶퐷 , 퐵퐶 and 퐴퐷 respectively, we say that the figure 퐴퐵퐶퐷퐸퐹 is a complete quadrilateral. The complete quadrilateral sides are 퐴퐵, 퐵퐶, 퐶퐷, 퐷퐴 , and 퐴퐶, 퐵퐷 and 퐸퐹 are diagonals.

2nd Theorem. (Newton-Gauss)

The diagonals’ middles of a complete quadrilateral are three collinear points. To prove 2nd theorem, refer to [1].

Note.

It is called Newton-Gauss Line of a quadrilateral the line to which the diagonals’ middles of a complete quadrilateral belong.

3rd Theorem. (Aubert)

If 퐴퐵퐶퐷퐸퐹 is a complete quadrilateral, then the orthocenters 퐻1, 퐻2, 퐻3, 퐻4 of the traingles 퐴퐵퐹, 퐴퐸퐷, 퐵퐶퐸, and 퐶퐷퐹 respectively, are collinear points.

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Proof.

Let 퐴1, 퐵1, 퐹1 be the feet of the heights of the triangle 퐴퐵퐹 and 퐻1 its orthocenter (see Fig. 2). Considering the power of the point 퐻1 relative to the circle circumscribed to the triangle ABF, and given the triangle orthocenter’s property according to which its symmetrics to the triangle sides belong to the circumscribed circle, we find that:

퐻1퐴 ∙ 퐻1퐴1 = 퐻1퐵 ∙ 퐻1퐵1 = 퐻1퐹 ∙ 퐻1퐹1.

Figure 2.

This relationship shows that the orthocenter 퐻1 has equal power with respect to the circles of diameters [퐴퐶], [퐵퐷], [퐸퐹]. As well, we establish that the orthocenters 퐻2, 퐻3, 퐻4 have equal powers to these circles. Since the circles of diameters [퐴퐶], [퐵퐷], [퐸퐹] have collinear centers (belonging to the Newton- 173

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Gauss line of the 퐴퐵퐶퐷퐸퐹 quadrilateral), it follows that the points 퐻1, 퐻2, 퐻3, 퐻4 belong to the radical axis of the circles, and they are, therefore, collinear points.

Notes.

1. It is called the Aubert Line or the line of the complete quadrilateral’s orthocenters the line to which the orthocenters 퐻1, 퐻2, 퐻3, 퐻4 belong. 2. The Aubert Line is perpendicular on the Newton-Gauss line of the quadrilateral (radical axis of two circles is perpendicular to their centers’ line).

4th Theorem. (The Dual Theorem of the 3rd Theorem)

If 퐴퐵퐶퐷 is a convex quadrilateral and 푀 is a point in its plane for which there are the Bobillier’s transversals of triangles 퐴퐵퐶 , 퐵퐶퐷 , 퐶퐷퐴 and 퐷퐴퐵 ; thereupon these transversals are concurrent.

Proof.

Let us transform the configuration in Fig. 2, by duality with respect to a circle of center 푀. By the considered duality, the lines 푎, 푏, 푐, 푑, 푒 şi 푓 correspond to the points 퐴, 퐵, 퐶, 퐷, 퐸, 퐹 (their polars). It is known that polars of collinear points are concurrent lines, therefore we have: 푎 ∩ 푏 ∩ 푒 = {퐴′},

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푏 ∩ 푐 ∩ 푓 = {퐵′} , 푐 ∩ 푑 ∩ 푒 = {퐶′} , 푑 ∩ 푓 ∩ 푎 = {퐷′} , 푎 ∩ 푐 = {퐸′}, 푏 ∩ 푑 = {퐹′}. Consequently, by applicable duality, the points 퐴′, 퐵′, 퐶′, 퐷′, 퐸′ and 퐹′ correspond to the straight lines 퐴퐵, 퐵퐶, 퐶퐷, 퐷퐴, 퐴퐶, 퐵퐷.

To the orthocenter 퐻1 of the triangle 퐴퐸퐷 , it corresponds, by duality, its polar, which we denote ′ ′ ′ 퐴1 – 퐵1 − 퐶1, and which is the Bobillier’s transversal of the triangle 퐴’퐶’퐷’ in relation to the point 푀. Indeed, the point 퐶’ corresponds to the line 퐸퐷 by duality; to the height from 퐴 of the triangle 퐴퐸퐷, also by duality, ′ it corresponds its pole, which is the point 퐶1 located ̂ ′ 0 on 퐴’퐷’ such that 푚(퐶′푀퐶1) = 90 . To the height from 퐸 of the triangle 퐴퐸퐷 , it ′ ̂ ′ corresponds the point 퐵1 ∈ 퐴′퐶′ such that 푚(퐷′푀퐵1) = 900. ′ Also, to the height from 퐷, it corresponds 퐴1 ∈ ′ ′ 0 퐶′퐷’ on C such that 푚(퐴 푀퐴1) = 90 . To the orthocenter 퐻2 of the triangle 퐴퐵퐹, it will correspond, ′ by applicable duality, the Bobillier’s transversal 퐴2 − ′ ′ ′ ′ ′ 퐵2 − 퐶2 in the triangle 퐴 퐵 퐷 relative to the point 푀. To the orthocenter 퐻3 of the triangle 퐵퐶퐸 , it will ′ correspond the Bobillier’s transversal 퐴3′ − 퐵 3 − 퐶3′ in the triangle 퐴′퐵′퐶′ relative to the point 푀, and to the orthocenter 퐻4 of the triangle 퐶퐷퐹 , it will ′ ′ ′ correspond the transversal 퐴4 − 퐵4 − 퐶4 in the triangle 퐶′퐷′퐵′ relative to the point 푀.

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′ ′ ′ ̅̅̅̅ The Bobillier’s transversals 퐴𝑖 − 퐵𝑖 − 퐶𝑖 , 𝑖 = 1,4 correspond to the collinear points 퐻𝑖, 𝑖 = 1̅̅,̅4̅. These transversals are concurrent in the pole of the line of the orthocenters towards the considered duality. It results that, given the quadrilateral 퐴’퐵’퐶’퐷’, the Bobillier’s transversals of the triangles 퐴’퐶’퐷’ , 퐴’퐵’퐷’ , 퐴’퐵’퐶’ and 퐶’퐷’퐵’ relative to the point 푀 are concurrent.

References.

[1] Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles. The Education Publisher Inc., Columbus, Ohio, USA – 2012. [2] Ion Patrascu, Florentin Smarandache: Variance on Topics of Plane Geometry. The Education Publisher Inc., Columbus, Ohio, USA – 2013.

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Bibliography

[1] C. Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental Theorems of Triangle Geometry]. Bacau, Romania: Editura Unique, 2008. [2] N. Blaha: Asupra unor cercuri care au ca centre două puncte inverse [On some circles that have as centers two inverse points], in “Gazeta Matematica”, vol. XXXIII, 1927. [3] D. Brânzei, M. Miculița. Lucas circles and spheres. In „The Didactics of Mathematics”, volume 9/1994, Cluj-Napoca (Romania), p. 73- 80. [4] C. Coșniță: Coordonées baricentrique [Barycentric Coordinates]. Bucarest – Paris: Librairie Vuibert, 1941. [5] C. Coşniţă: Teoreme şi probleme alese de matematică [Theorems and Problems], Bucureşti: Editura de Stat Didactică şi Pedagogică, 1958. [6] D. Efremov: Noua geometrie a triunghiului [The New Geometry of the Triangle], translation from Russian into Romanian by Mihai Miculița. Zalau, Romania: Cril Publishing House, 2010.

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[7] Roger A. Johnson: Advanced Euclidean Geometry. New York: Dover Publications, 2007. [8] T. Lalescu: Geometria triunghiului [The Geometry of the Triangle]. Craiova: Editura Apollo, 1993. [9] N. N. Mihaileanu: Lecții complementare de geometrie [Complementary Lessons of Geometry], Editura Didactică și Pedagogică, București, 1976. [10] Gh. Mihalescu: Geometria elementelor remarcabile [The Geometry of the Outstanding Elements]. Bucharest: Tehnica Publishing House, 1957. [11] Ion Patrascu: Probleme de geometrie plană [Planar Geometry Problems]. Craiova: Editura Cardinal, 1996. [12] I. Patrascu: Axe și centre radicale ale cercului adjuncte unui triunghi [Axis and radical centers of the adjoin circle of a triangle], in “Recreații matematice”, year XII, no. 1, 2010. [13] Ion Patrascu, Gh. Margineanu: Cercul lui Tucker [Tucker’s Circle], in “Alpha journal”, year XV, no. 2/2009. [14] I. Patrascu, F. Smarandache, Variance on Topics of Plane Geometry, Educational Publisher, Ohio, USA, 2013.

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[15] F. Smarandache, I. Patrascu, The Geometry of Homological Triangles. The Education Publisher, Ohio, USA, 2012. [16] V. Gh. Voda: Triunghiul – ringul cu trei colțuri [The Triangle-The Ring with Three Corners]. Bucharest: Editura Albatros, 1979. [17] Eric W. Weisstein: First Droz-Farny’s circle. From Math World – A Wolfram WEB Resurse, http://mathworld.wolfram.com/. [18] P. Yiu, A. P. Hatzipolakis. The Lucas Circles of a Triangle. In “Amer. Math. Monthly”, no. 108/2001, pp. 444-446. http://www.math.fau. edu/yiu/monthly437-448.pdf.

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We approach several themes of classical geometry of the circle and complete them with some original results, showing that not everything in traditional math is revealed, and that it still has an open character. The topics were chosen according to authors’ aspiration and attraction, as a poet writes lyrics about spring according to his emotions.