Complements to Classic Topics of Circles Geometry

Ion Patrascu | Florentin Smarandache

Complements to Classic Topics of Circles Geometry

Pons Editions Brussels | 2016

Complements to Classic Topics of Circles Geometry

Ion Patrascu | Florentin Smarandache

Complements to Classic Topics of Circles Geometry

1

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...

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Complements to Classic Topics of Circles Geometry

Ion Patrascu | Florentin Smarandache

Complements to Classic Topics of Circles Geometry

Pons Editions Brussels | 2016 3

Ion Patrascu, Florentin Smarandache

Š 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

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Complements to Classic Topics of Circles Geometry

Contents Introduction ....................................................... 15 Lemoine’s Circles ............................................... 17 1st Theorem. ...........................................................17 Proof. .................................................................17 nd 2 Theorem.......................................................... 19 Proof. ................................................................ 19 Remark. ............................................................ 21 References. ........................................................... 22 Lemoine’s Circles Radius Calculus ..................... 23 1st Theorem ........................................................... 23 Proof. ................................................................ 23 Consequences. ................................................... 25 st 1 Proposition....................................................... 25 Proof. ................................................................ 25 nd 2 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

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Ion Patrascu, Florentin Smarandache

Comment........................................................... 34 2nd Remark. ....................................................... 34 nd 2 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 nd 2 Theorem.......................................................... 43 Proof. ................................................................ 43 Remark. ............................................................ 44 rd 3 Theorem. ......................................................... 44 Proof. ................................................................ 45 References. ........................................................... 48 The Polars of a Radical Center ........................... 49 1st Theorem. .......................................................... 49 Proof. ................................................................ 50

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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 nd 2 Theorem.......................................................... 57 Remark. ............................................................ 58 Definition. ............................................................ 58 Remark. ............................................................ 58 rd 3 Theorem. ......................................................... 58 Proof. ................................................................ 59 Remark. ............................................................ 60 th 4 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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Ion Patrascu, Florentin Smarandache

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 nd 2 Proposition. .................................................... 74 Proof. ................................................................ 74 th 4 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 nd 2 Definition. ....................................................... 83

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Complements to Classic Topics of Circles Geometry

Remark. ............................................................ 83 2 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 nd 2 Lemma. ........................................................... 92 Proof. ................................................................ 92 3rd Lemma. ........................................................... 93 Solution. ............................................................ 93 Construction’s Proof. ........................................ 93 Remark. ............................................................ 94 1st Theorem. ......................................................... 95 Proof. ................................................................ 96 2nd Theorem. ....................................................... 97 nd

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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 rd 3 Definition. ..................................................... 105 2nd Theorem........................................................ 105 Proof. .............................................................. 106 rd 3 Theorem. ....................................................... 107 Proof. .............................................................. 107 th 4 Theorem. ....................................................... 108 Proof. .............................................................. 108 1st Remark. ...................................................... 108 5th Theorem. ....................................................... 108 Proof. .............................................................. 109 th 6 Theorem. ....................................................... 109 Proof. .............................................................. 109 4th Definition. ...................................................... 110 1st Property. ......................................................... 110 2nd Remark. ...................................................... 110 7th Theorem. ........................................................ 111

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Complements to Classic Topics of Circles Geometry

Proof. ............................................................... 111 References. .......................................................... 112 Apollonius’s Circle of Second Rank ................... 113 1st Definition. ....................................................... 113 1st Theorem. ......................................................... 113 1st Proposition...................................................... 114 Proof. ............................................................... 114 Remarks. .......................................................... 115 nd 2 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 nd 2 Definition. ......................................................122 2nd Remark. ......................................................122 1st Proposition......................................................122 Proof. ...............................................................122 3rd Remark. ......................................................123 3rd Definition. ......................................................123 4th Remark. ..................................................... 124 st 1 Theorem. ........................................................ 124 Proof. .............................................................. 124 th 4 Definition. ..................................................... 126 2nd Proposition. .................................................. 126

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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 rd 3 Proposition. .................................................... 137 Remark 2......................................................... 138 Proposition 4. ..................................................... 138 Proof. .............................................................. 139 3rd Remark. ..................................................... 140 5th Proposition. ................................................... 140 6th Proposition. ................................................... 140 Proof. .............................................................. 140 rd 3 Definition. ..................................................... 142 4th Remark. ..................................................... 142 7th Proposition. ................................................... 143 Proof. .............................................................. 143 5th Remark. ..................................................... 144

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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 th 11 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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Ion Patrascu, Florentin Smarandache

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

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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.

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Ion Patrascu, Florentin Smarandache

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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 đ??´đ??ľđ??ś a triangle and đ??ž its simedian center. We take through K the parallel đ??´1 đ??´2 to đ??ľđ??ś, đ??´1 ∈ (đ??´đ??ľ), đ??´2 ∈ (đ??´đ??ś); through đ??´2 we take the antiparallels đ??´2 đ??ľ1 to đ??´đ??ľ in relation to đ??śđ??´ and đ??śđ??ľ , đ??ľ1 ∈ (đ??ľđ??ś) ; through đ??ľ1 we take the parallel đ??ľ1 đ??ľ2 to đ??´đ??ś, đ??ľ2 ∈ đ??´đ??ľ; through đ??ľ2 we take the antiparallels đ??ľ1 đ??ś1 to đ??ľđ??ś , đ??ś1 ∈ (đ??´đ??ś) , and through đ??ś1 we take the parallel đ??ś1 đ??ś2 to đ??´đ??ľ, đ??ś1 ∈ (đ??ľđ??ś).

Then: i. đ??ś2 đ??´1 is an antiparallel of đ??´đ??ś; ii. đ??ľ1 đ??ľ2 ∊ đ??ś1 đ??ś2 = {đ??ž }; iii.

The points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are concyclical (the first Lemoine’s circle).

Proof. i.

The quadrilateral đ??ľđ??ś2 đ??žđ??´ is a parallelogram, and its center, i.e. the middle of the segment (đ??ś2 đ??´1 ), belongs to the simedian đ??ľđ??ž; it follows that đ??ś2 đ??´2 is an antiparallel to đ??´đ??ś(see Figure 1).

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Ion Patrascu, Florentin Smarandache

ii.

iii.

Let {đ??žâ€˛} = đ??´1 đ??´2 ∊ đ??ľ1 đ??ľ2 , because the quadrilateral đ??žâ€˛đ??ľ1 đ??śđ??´2 is a parallelogram; it follows that đ??śđ??žâ€˛ is a simedian; also, đ??śđ??ž is a simedian, and since đ??ž, đ??žâ€˛ ∈ đ??´1 đ??´2 , it follows that we have đ??žâ€˛ = đ??ž. đ??ľ2 đ??ś1 being an antiparallel to đ??ľđ??ś and đ??´1 đ??´2 âˆĽ đ??ľđ??ś , it means that đ??ľ2 đ??ś1 is an antiparallel to đ??´1 đ??´2 , so the points đ??ľ2 , đ??ś1 , đ??´2 , đ??´1 are concyclical. From đ??ľ1 đ??ľ2 âˆĽ đ??´đ??ś , âˆ˘đ??ľ2 đ??ś1 đ??´ ≥ âˆ˘đ??´đ??ľđ??ś , âˆ˘đ??ľ1 đ??´2 đ??ś ≥ âˆ˘đ??´đ??ľđ??ś we get that the quadrilateral đ??ľ2 đ??ś1 đ??´2 đ??ľ1 is an isosceles trapezoid, so the points đ??ľ2 , đ??ś1 , đ??´2 , đ??ľ1 are concyclical. Analogously, it can be shown that the quadrilateral đ??ś2 đ??ľ1 đ??´2 đ??´1 is an isosceles trapezoid, therefore the points đ??ś2 , đ??ľ1 , đ??´2 , đ??´1 are concyclical.

Figure 1

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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 đ??´đ??ľđ??ś, let K be the simedian center. We take from đ??ž the antiparallel đ??´1 đ??´2 to đ??ľđ??ś ; đ??´1 ∈ đ??´đ??ľ, đ??´2 ∈ đ??´đ??ś; through đ??´2 we build đ??´2 đ??ľ1 âˆĽ đ??´đ??ľ; đ??ľ1 ∈ (đ??ľđ??ś), then through đ??ľ1 we build đ??ľ1 đ??ľ2 the antiparallel to đ??´đ??ś, đ??ľ2 ∈ (đ??´đ??ľ), and through đ??ľ2 we build đ??ľ2 đ??ś1 âˆĽ đ??ľđ??ś, đ??ś1 ∈ đ??´đ??ś , and, finally, through đ??ś1 we take the antiparallel đ??ś1 đ??ś2 to đ??´đ??ľ, đ??ś2 ∈ (đ??ľđ??ś).

Then: i. đ??ś2 đ??´1 âˆĽ đ??´đ??ś; ii. đ??ľ1 đ??ľ2 ∊ đ??ś1 đ??ś2 = {đ??ž }; iii.

The points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are concyclical (the second Lemoine’s circle).

Proof. i.

Let {đ??žâ€˛} = đ??´1 đ??´2 ∊ đ??ľ1 đ??ľ2 , having âˆ˘đ??´đ??´1 đ??´2 = âˆ˘đ??´đ??śđ??ľ and âˆ˘đ??ľđ??ľ1 đ??ľ2 ≥ âˆ˘đ??ľđ??´đ??ś because đ??´1 đ??´2 Č™i đ??ľ1 đ??ľ2 are antiparallels to đ??ľđ??ś , đ??´đ??ś , respectively, it follows that âˆ˘đ??žâ€˛đ??´1 đ??ľ2 ≥ âˆ˘đ??žâ€˛đ??ľ2 đ??´1 , so đ??žâ€˛đ??´1 = đ??žâ€˛đ??ľ2 ; having đ??´1 đ??ľ2 âˆĽ đ??ľ1 đ??´2 as well, it follows that also đ??žâ€˛đ??´2 = đ??žâ€˛đ??ľ1 , so đ??´1 đ??´2 = đ??ľ1 đ??ľ2 . Because đ??ś1 đ??ś2 and đ??ľ1 đ??ľ2 are antiparallels to đ??´đ??ľ and đ??´đ??ś , we

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Ion Patrascu, Florentin Smarandache

have đ??ž"đ??ś2 = đ??ž"đ??ľ1 ; we noted {đ??ž"} = đ??ľ1 đ??ľ2 ∊ đ??ś1 đ??ś2 ; since đ??ś1 đ??ľ2 âˆĽ đ??ľ1 đ??ś2 , we have that the triangle đ??ž"đ??ś1 đ??ľ2 is also isosceles, therefore đ??ž"đ??ś1 = đ??ś1 đ??ľ2 , and we get that đ??ľ1 đ??ľ2 = đ??ś1 đ??ś2 .

Let {đ??žâ€˛â€˛â€˛} = đ??´1 đ??´2 ∊ đ??ś1 đ??ś2 ; since đ??´1 đ??´2 and đ??ś1 đ??ś2 are antiparallels to đ??ľđ??ś and đ??´đ??ľ , we get that the triangle đ??žâ€˛â€˛â€˛đ??´2 đ??ś1 is isosceles, so đ??žâ€˛â€˛â€˛đ??´2 = đ??žâ€˛â€˛â€˛đ??ś1 , but đ??´1 đ??´2 = đ??ś1 đ??ś2 implies that đ??žâ€˛â€˛â€˛đ??ś2 = đ??žâ€˛â€˛â€˛đ??´1 , then âˆ˘đ??žâ€˛â€˛â€˛đ??´1 đ??ś2 ≥ âˆ˘đ??žâ€˛â€˛â€˛đ??´2 đ??ś1 and, accordingly, đ??ś2 đ??´1 âˆĽ đ??´đ??ś.

Figure 2

ii.

We noted {đ??žâ€˛} = đ??´1 đ??´2 ∊ đ??ľ1 đ??ľ2 ; let {đ?‘‹} = đ??ľ2 đ??ś1 ∊ đ??ľ1 đ??´2 ; obviously, đ??ľđ??ľ1 đ?‘‹đ??ľ2 is a parallelogram; if đ??ž0 is the middle of (đ??ľ1 đ??ľ2 ), then đ??ľđ??ž0 is a simedian, since đ??ľ1 đ??ľ2 is an antiparallel to đ??´đ??ś, and the middle of the antiparallels of đ??´đ??ś are situated on the

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Complements to Classic Topics of Circles Geometry

iii.

simedian đ??ľđ??ž . If đ??ž0 ≠K, then đ??ž0 đ??ž âˆĽ đ??´1 đ??ľ2 (because đ??´1 đ??´2 = đ??ľ1 đ??ľ2 and đ??ľ1 đ??´2 âˆĽ đ??´1 đ??ľ2 ), on the other hand, đ??ľ, đ??ž0 , K are collinear (they belong to the simedian đ??ľđ??ž ), therefore đ??ž0 đ??ž intersects đ??´đ??ľ in đ??ľ, which is absurd, so đ??ž0 =K, and, accordingly, đ??ľ1 đ??ľ2 ∊ đ??´1 đ??´2 = {đ??ž } . Analogously, we prove that đ??ś1 đ??ś2 ∊ đ??´1 đ??´2 = {đ??ž }, so đ??ľ1 đ??ľ2 ∊ đ??ś1 đ??ś2 = {đ??ž }.

K is the middle of the congruent antiparalells đ??´1 đ??´2 , đ??ľ1 đ??ľ2 , đ??ś1 đ??ś2 , so đ??žđ??´1 = đ??žđ??´2 = đ??žđ??ľ1 = đ??žđ??ľ2 = đ??žđ??ś1 = đ??žđ??ś2 .

The simedian center đ??ž is the center of the second Lemoine’s circle.

Remark.

The center of the first Lemoine’s circle is the middle of the segment [đ?‘‚đ??ž ], where đ?‘‚ is the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś. Indeed, the perpendiculars taken from đ??´, đ??ľ, đ??ś on the antiparallels đ??ľ2 đ??ś1 , đ??´1 đ??ś2 , đ??ľ1 đ??´2 respectively pass through O, the center of the circumscribed circle (the antiparallels have the directions of the tangents taken to the circumscribed circle in đ??´, đ??ľ, đ??ś). The mediatrix of the segment đ??ľ2 đ??ś1 pass though the middle of đ??ľ2 đ??ś1 , which coincides with the middle of đ??´đ??ž , so is the middle line in the triangle đ??´đ??žđ?‘‚ passing through the middle of (đ?‘‚đ??ž) . Analogously, it follows that the mediatrix of đ??´1 đ??ś2 pass through the middle đ??ż1 of [đ?‘‚đ??ž ].

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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.

We will prove that

đ??ľđ??ś2 đ?‘?2

=

đ??ś2 đ??ľ1 đ?‘Ž2

=

đ??ľ1 đ??ś đ?‘?2

.

In figure 1, đ??ž represents the symmedian center of the triangle đ??´đ??ľđ??ś , and đ??´1 đ??´2 ; đ??ľ1 đ??ľ2 ; đ??ś1 đ??ś2 represent Lemoine parallels.

The triangles đ??ľđ??ś2 đ??´1 ; đ??śđ??ľ1 đ??´2 and đ??žđ??ś2 đ??´1 have heights relative to the sides đ??ľđ??ś2 ; đ??ľ1 đ??ś and đ??ś2 đ??ľ1 equal (đ??´1 đ??´2 âˆĽ đ??ľđ??ś).

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Hence: đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??ľđ??´1 đ??ś2

=

đ??ľđ??ś2

đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??žđ??ś2 đ??´1 đ??ś2 đ??ľ1

=

đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??śđ??ľ1 đ??´2 đ??ľ1 đ??ś

.

(1)

Figure 1

On the other hand: đ??´1 đ??ś2 and đ??ľ1 đ??´2 being antiparallels with respect to đ??´đ??ś and đ??´đ??ľ, it follows that Δđ??ľđ??ś2 đ??´1 ~đ?›Ľđ??ľđ??´đ??ś and Δđ??śđ??ľ1 đ??´2 ~đ?›Ľđ??śđ??´đ??ľ , likewise đ??žđ??ś2 âˆĽ đ??´đ??ś implies: Δđ??žđ??ś2 đ??ľ1 ~đ?›Ľđ??´đ??ľđ??ś.

We obtain: đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??ľđ??ś2 đ??´1

=

đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??´đ??ľđ??ś đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??žđ??ś2 đ??ľ1 đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??´đ??ľđ??ś

=

đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??śđ??ľ1 đ??´2 đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??´đ??ľđ??ś

=

đ??ľđ??ś22

;

đ?‘?2 đ??ś2 đ??ľ12

;

đ?‘Ž2 đ??śđ??ľ12 đ?‘?2

.

(2)

If we denote đ??´đ?‘&#x;đ?‘’đ?‘Žâˆ† đ??´đ??ľđ??ś = đ?‘†, we obtain from the relations (1) and (2) that: đ??ľđ??ś2 đ?‘?2

=

đ??ś2 đ??ľ1 đ?‘Ž2

=

đ??ľ1 đ??ś đ?‘?2

.

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Consequences.

1. According to the 1st Theorem, we find that: đ??ľđ??ś2 =

đ?‘Žđ?‘? 2 đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

; đ??ľ1 đ??ś =

đ?‘Žđ?‘? 2 đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

; đ??ľ1 đ??ś2 =

đ?‘Ž3 đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

.

2. We also find that: đ??ľ1 đ??ś2 đ?‘Ž3

=

đ??´2 đ??ś1 đ?‘?3

=

đ??´1 đ??ľ2 đ?‘?3

,

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, đ?‘…đ??ż1 is given by the formula: 1 đ?‘…2 (đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )+ đ?‘Ž2 đ?‘? 2 đ?‘? 2 , (đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )2 4

đ?‘…đ??ż21 = ∙

(3)

where đ?‘… represents the radius of the circle inscribed in the triangle đ??´đ??ľđ??ś.

Proof.

Let đ??ż be the center of the first Lemoine’s circle that is known to represent the middle of the segment (đ?‘‚đ??ž) – đ?‘‚ being the center of the circle inscribed in the triangle đ??´đ??ľđ??ś.

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Considering C1, we obtain đ??ľđ??ľ1 =

đ?‘Ž (đ?‘? 2 +đ?‘Ž2 ) đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

.

Taking into account the power of point đ??ľ in relation to the first Lemoine’s circle, we have: đ??ľđ??ś2 ∙ đ??ľđ??ľ1 = đ??ľđ?‘‡ 2 − đ??żđ?‘‡ 2 , (đ??ľđ?‘‡ is the tangent traced from đ??ľ to the first Lemoine’s circle, see Figure 1).

Hence: đ?‘…đ??ż21 = đ??ľđ??ż2 − đ??ľđ??ś2 ∙ đ??ľđ??ľ1 .

(4) The median theorem in triangle đ??ľđ?‘‚đ??ž implies that: đ??ľđ??ż2 =

2∙(đ??ľđ??ž 2 +đ??ľđ?‘‚2 )−đ?‘‚đ??ž 2 4

.

It is known that đ??ž =

(đ?‘Ž2 +đ?‘? 2 )∙đ?‘†đ?‘? đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

; đ?‘†đ?‘? =

2đ?‘Žđ?‘?∙đ?‘šđ?‘? đ?‘Ž2 +đ?‘? 2

,

where đ?‘†đ?‘? and đ?‘šđ?‘? are the lengths of the symmedian and the median from đ??ľ, and đ?‘‚đ??ž 2 = đ?‘…2 −

3đ?‘Ž2 đ?‘? 2 đ?‘? 2 (đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )

, see

(3).

Consequently: đ??ľđ??ž 2 = 4đ??ľđ??ż2 = đ?‘…2 +

2đ?‘Ž2 đ?‘? 2 (đ?‘Ž2 +đ?‘? 2 )−đ?‘Ž2 đ?‘? 2 đ?‘? 2 (đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )2

4đ?‘Ž2 đ?‘? 2 (đ?‘Ž2 +đ?‘? 2 )+đ?‘Ž2 đ?‘? 2 đ?‘? 2

As: đ??ľđ??ś2 ∙ đ??ľđ??ľ1 =

(đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )2 đ?‘Ž2 đ?‘? 2 (đ?‘Ž2 +đ?‘? 2 ) (đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )2

, and

.

, by replacing in (4),

we obtain formula (3).

2nd Proposition.

The radius of the second Lemoine’s circle, đ?‘…đ??ż2 , is given by the formula: đ?‘…đ??ż2 =

đ?‘Žđ?‘?đ?‘? đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

.

(5)

26

Complements to Classic Topics of Circles Geometry

Proof.

Figure 2

In Figure 2 above, đ??´1 đ??´2 ; đ??ľ1 đ??ľ2 ; đ??ś1 đ??ś2 are Lemoine antiparallels traced through symmedian center đ??ž that is the center of the second Lemoine’s circle, thence: đ?‘…đ??ż2 = đ??žđ??´1 = đ??žđ??´2 .

If we note with đ?‘† and đ?‘€ the feet of the symmedian and the median from đ??´, it is known that: đ??´đ??ž đ??žđ?‘†

=

đ?‘? 2 +đ?‘? 2 đ?‘Ž2

.

From the similarity of triangles đ??´đ??´2 đ??´1 and đ??´đ??ľđ??ś, we have:

đ??´1 đ??´2

But:

đ??ľđ??ś đ??´đ??ž đ??´đ?‘†

=

=

đ??´đ??ž

.

đ??´đ?‘€ đ?‘? 2 +đ?‘? 2 đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

and đ??´đ?‘† =

2đ?‘?đ?‘? đ?‘? 2 +đ?‘? 2

∙ đ?‘šđ?‘Ž .

đ??´1 đ??´2 = 2đ?‘…đ??ż2 , đ??ľđ??ś = đ?‘Ž, therefore: đ?‘…đ??ż2 =

đ??´đ??žâˆ™đ?‘Ž

,

2đ?‘šđ?‘Ž 2đ?‘?đ?‘? ∙ đ?‘šđ?‘Ž

and as đ??´đ??ž =

đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

, formula (5) is a consequence.

27

Ion Patrascu, Florentin Smarandache

Remarks.

1. If we use đ?‘Ąđ?‘”đ?œ” =

4đ?‘† đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

, đ?œ” being the Brocard’s

angle (see [2]), we obtain: đ?‘…đ??ż2 = đ?‘… ∙ đ?‘Ąđ?‘”đ?œ”.

2. If, in Figure 1, we denote with đ?‘€1 the middle of the antiparallel đ??ľ2 đ??ś1 , which is equal to đ?‘…đ??ż2 (due to their similarity), we thus find from the rectangular triangle đ??żđ?‘€1 đ??ś1 that: 1

đ??żđ??ś12 = đ??żđ?‘€12 + đ?‘€1 đ??ś12 , but đ??żđ?‘€12 = đ?‘Ž2 and đ?‘€1 đ??ś2 = 4

1 2

đ?‘…đ??ż2 ; it follows that: 1

đ?‘…2

4

4

đ?‘…đ??ż21 = (đ?‘…2 + đ?‘…đ??ż22 ) =

(1 + đ?‘Ąđ?‘”2 đ?œ”).

We obtain: đ?‘…đ??ż1 =

đ?‘… 2

∙ √1 + đ?‘Ąđ?‘”2 đ?œ” .

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. đ??žđ??ś2 đ??ľ1

is

an

isosceles

triangle,

âˆ˘đ??žđ??ś2 đ??ľ1 =

âˆ˘đ??žđ??ľ1 đ??ś2 = âˆ˘đ??´; as đ??žđ??ś2 = đ?‘…đ??ż2 we have that cos đ??´ = deci

đ??ś2 đ??ľ1 cos đ??´

= 2đ?‘…đ??ż2 , similary:

đ??´2 đ??ś1 cos đ??ľ

28

=

đ??ľ2 đ??´1 cos đ??ś

= 2đ?‘…đ??ż2 .

đ??ś2 đ??ľ1 2đ?‘…đ??ż

2

,

Complements to Classic Topics of Circles Geometry

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.

29

Ion Patrascu, Florentin Smarandache

30

Complements to Classic Topics of Circles Geometry

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 đ??ž 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 đ??´đ??ľđ??ś is a scalene triangle and đ??ž is its simmedian center, then đ??ż , the center of the first Lemoine’s circle, is the middle of the segment [đ?‘‚đ??ž ], where đ?‘‚ is the center of the circumscribed circle, and

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Ion Patrascu, Florentin Smarandache

the center of the second Lemoine’s circle is đ??ž . It follows that the radical axis of Lemoine’s circles is perpendicular on the line of the centers đ??żđ??ž, therefore on the line đ?‘‚đ??ž.

1st Proposition.

The radical axis of Lemoine’s circles is perpendicular on the line đ?‘‚đ??ž raised in the simmedian center đ??ž.

Proof.

Let đ??´1 đ??´2 be the antiparallel to đ??ľđ??ś taken through đ??ž, then đ??žđ??´1 is the radius đ?‘…đ??ż2 of the second Lemoine’s circle; we have: đ?‘…đ??ż2 =

đ?‘Žđ?‘?đ?‘? đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

.

Figure 1

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Complements to Classic Topics of Circles Geometry

Let đ??´1′ đ??´â€˛2 be the Lemoine’s parallel taken to đ??ľđ??ś; we evaluate the power of đ??ž towards the first Lemoine’s circle. We have: âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—â€˛ ∙ đ??žđ??´ âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—âƒ—â€˛ = đ??żđ??ž 2 − đ?‘…đ??ż2 . đ??žđ??´ (1) 1

2

1

Let đ?‘† be the simmedian leg from đ??´; it follows that: đ??žđ??´â€˛1 đ??ľđ?‘†

=

đ??´đ??ž đ??´đ?‘†

−

đ??žđ??´â€˛2 đ?‘†đ??ś

.

We obtain: đ??žđ??´1′ = đ??ľđ?‘† ∙ but

đ??ľđ?‘† đ?‘†đ??ś

=

đ?‘?2 đ?‘?2

and

đ??´đ??ž đ??´đ?‘† đ??´đ??ž đ??´đ?‘†

and đ??žđ??´â€˛2 = đ?‘†đ??ś ∙ =

đ?‘? 2 +đ?‘? 2 đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

đ??´đ??ž đ??´đ?‘†

,

.

Therefore: đ??´đ??ž 2 −đ?‘Ž2 đ?‘? 2 đ?‘? 2 ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´1′ ∙ ⃗⃗⃗⃗⃗⃗⃗⃗ đ??žđ??´â€˛2 = −đ??ľđ?‘† ∙ đ?‘†đ??ś ∙ ( ) = 2 ; (đ?‘? + đ?‘? 2 )2 đ??´đ?‘† (đ?‘? 2 +đ?‘? 2 )

2

(đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 )2

= −đ?‘…đ??ż22 .

(2)

We draw the perpendicular in đ??ž on the line đ??żđ??ž and denote by đ?‘ƒ and đ?‘„ its intersection to the first Lemoine’s circle. ⃗⃗⃗⃗⃗⃗ = −đ?‘…đ??ż2 ; by the other hand, We have ⃗⃗⃗⃗⃗ đ??žđ?‘ƒ ∙ đ??žđ?‘„ 2

đ??žđ?‘ƒ = đ??žđ?‘„ (đ?‘ƒđ?‘„ is a chord which is perpendicular to the diameter passing through đ??ž).

It follows that đ??žđ?‘ƒ = đ??žđ?‘„ = đ?‘…đ??ż2 , so đ?‘ƒ and đ?‘„ are situated on the second Lemoine’s circle.

Because đ?‘ƒđ?‘„ is a chord which is common to the Lemoine’s circles, it obviously follows that đ?‘ƒđ?‘„ is the radical axis.

33

Ion Patrascu, Florentin Smarandache

Comment.

Equalizing (1) and (2), or by the Pythagorean theorem in the triangle đ?‘ƒđ??žđ??ż, we can calculate đ?‘…đ??ż1 .

3đ?‘Ž2 đ?‘? 2 đ?‘? 2

It is known that: đ?‘‚đ??ž 2 = đ?‘…2 − (đ?‘Ž2

+đ?‘? 2 +đ?‘? 2 )2

, and

1

since đ??żđ??ž = đ?‘‚đ??ž, we find that: đ?‘…đ??ż21

2 1

đ?‘Ž2 đ?‘? 2 đ?‘? 2

4

+đ?‘? 2 +đ?‘? 2 )2

= ∙ [đ?‘…2 + (đ?‘Ž2

].

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 đ?’ž(đ?‘‚1 , đ?‘…1 ) Ĺ&#x;i đ?’ž(đ?‘‚2 , đ?‘…2 ) are two circles such as the power of center đ?‘‚1 towards đ?’ž(đ?‘‚2 , đ?‘…2 ) is −đ?‘…12 , then the radical axis of the circles is the perpendicular in đ?‘‚1 on the line of centers đ?‘‚1 đ?‘‚2 .

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.

34

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 đ??´đ??ľđ??ś, the cevian đ??´đ?‘†, đ?‘† ∈ (đ??ľđ??ś), is a simedian if and only if

đ?‘†đ??ľ đ?‘†đ??ś

đ??´đ??ľ 2

= ( ) . For Proof, see [2]. đ??´đ??ś

2nd Definition.

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

35

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 đ??´đ??ľđ??ś the line đ?‘€đ?‘ , where đ?‘€ ∈ đ??´đ??ľ and đ?‘ ∈ đ??´đ??ś, is an antiparallel to đ??ľđ??ś if âˆ˘đ?‘€đ?‘ đ??´ ≥ âˆ˘đ??´đ??ľđ??ś.

1st Lemma.

In the triangle đ??´đ??ľđ??ś , let đ??´đ?‘† be a simedian, đ?‘† ∈ (đ??ľđ??ś). If đ?‘ƒ is the middle of the segment (đ?‘€đ?‘ ), having đ?‘€ ∈ (đ??´đ??ľ) and đ?‘ ∈ (đ??´đ??ś), belonging to the simedian đ??´đ?‘†, then đ?‘€đ?‘ and đ??ľđ??ś are antiparallels.

Proof.

We draw through đ?‘€ and đ?‘ , đ?‘€đ?‘‡ âˆĽ đ??´đ??ś and đ?‘ đ?‘… âˆĽ đ??´đ??ľ, đ?‘…, đ?‘‡ ∈ (đ??ľđ??ś) , see Figure 1. Let {đ?‘„} = đ?‘€đ?‘‡ ∊ đ?‘ đ?‘… ; since đ?‘€đ?‘ƒ = đ?‘ƒđ?‘ and đ??´đ?‘€đ?‘„đ?‘ is a parallelogram, it follows that đ?‘„ ∈ đ??´đ?‘†.

36

Complements to Classic Topics of Circles Geometry

Figure 1.

Thales's Theorem provides the relations: đ??´đ?‘ đ??´đ??ś đ??´đ??ľ đ??´đ?‘€

đ??ľđ?‘…

=

đ??ľđ??ś đ??ľđ??ś

=

;

(1)

.

(2)

đ??śđ?‘‡

From (1) and (2), by multiplication, we obtain: đ??´đ?‘

∙

đ??´đ??ľ

đ??´đ?‘€ đ??´đ??ś

=

đ??ľđ?‘… đ?‘‡đ??ś

.

(3)

Using again Thales's Theorem, we obtain: đ??ľđ?‘… đ??ľđ?‘† đ?‘‡đ??ś đ?‘†đ??ś

= =

đ??´đ?‘„ đ??´đ?‘† đ??´đ?‘„ đ??´đ?‘†

,

(4)

.

(5)

From these relations, we get đ??ľđ?‘… đ??ľđ?‘†

đ?‘‡đ??ś

=

đ?‘†đ??ś

,

(6)

.

(7)

or đ??ľđ?‘† đ?‘†đ??ś

=

đ??ľđ?‘… đ?‘‡đ??ś

In view of Proposition 1, the relations (7) and (3) lead to

đ??´đ?‘ đ??´đ??ľ

=

đ??´đ??ľ đ??´đ??ś

, which shows that ∆đ??´đ?‘€đ?‘ ~∆đ??´đ??śđ??ľ , so

âˆ˘đ??´đ?‘€đ?‘ ≥ âˆ˘đ??´đ??ľđ??ś.

37

Ion Patrascu, Florentin Smarandache

Therefore, đ?‘€đ?‘ and đ??ľđ??ś are antiparallels in relation to đ??´đ??ľ and đ??´đ??ś.

Remark.

1. The reciprocal of Lemma 1 is also valid, meaning that if đ?‘ƒ is the middle of the antiparallel đ?‘€đ?‘ to đ??ľđ??ś, then đ?‘ƒ belongs to the simedian from đ??´.

2nd Theorem.

(Generalization of the 1st Theorem)

Let đ??´đ??ľđ??ś be a scalene triangle and đ??ž its simedian center. We take đ?‘€ ∈ đ??´đ??ž and draw đ?‘€đ?‘ âˆĽ đ??´đ??ľ, đ?‘€đ?‘ƒ âˆĽ đ??´đ??ś , where đ?‘ ∈ đ??ľđ??ž, đ?‘ƒ ∈ đ??śđ??ž. Then: i. đ?‘ đ?‘ƒ âˆĽ đ??ľđ??ś; ii. đ?‘€đ?‘ , đ?‘ đ?‘ƒ and đ?‘€đ?‘ƒ intersect the sides of triangle đ??´đ??ľđ??ś in six concyclic points.

Proof.

In triangle đ??´đ??ľđ??ś, let đ??´đ??´1 , đ??ľđ??ľ1 , đ??śđ??ś1 the simedians concurrent in đ??ž (see Figure 2).

We have from Thales' Theorem that: đ??´đ?‘€ đ?‘€đ??ž đ??´đ?‘€ đ?‘€đ??ž

= =

đ??ľđ?‘ đ?‘ đ??ž đ??śđ?‘ƒ

;

(1)

.

(2)

đ?‘ƒđ??ž

From relations (1) and (2), it follows that đ??ľđ?‘ đ?‘ đ??ž

=

đ??śđ?‘ƒ đ?‘ƒđ??ž

,

(3)

which shows that đ?‘ đ?‘ƒ âˆĽ đ??ľđ??ś.

38

Complements to Classic Topics of Circles Geometry

Let đ?‘…, đ?‘†, đ?‘‰, đ?‘Š, đ?‘ˆ, đ?‘‡ be the intersection points of the parallels đ?‘€đ?‘ , đ?‘€đ?‘ƒ, đ?‘ đ?‘ƒ of the sides of the triangles to the other sides.

Obviously, by construction, the quadrilaterals đ??´đ?‘†đ?‘€đ?‘Š; đ??śđ?‘ˆđ?‘ƒđ?‘‰; đ??ľđ?‘…đ?‘ đ?‘‡ are parallelograms.

The middle of the diagonal đ?‘Šđ?‘† falls on đ??´đ?‘€, so on the simedian đ??´đ??ž, and from 1st Lemma we get that đ?‘Šđ?‘† is an antiparallel to đ??ľđ??ś.

Since đ?‘‡đ?‘ˆ âˆĽ đ??ľđ??ś , it follows that đ?‘Šđ?‘† and đ?‘‡đ?‘ˆ are antiparallels, therefore the points đ?‘Š, đ?‘†, đ?‘ˆ, đ?‘‡ are concyclic (4).

Figure 2.

Analogously, we show that the points đ?‘ˆ, đ?‘‰, đ?‘…, đ?‘† are concyclic (5). From đ?‘Šđ?‘† and đ??ľđ??ś antiparallels, đ?‘ˆđ?‘‰ and đ??´đ??ľ antiparallels, we have that âˆ˘đ?‘Šđ?‘†đ??´ ≥ âˆ˘đ??´đ??ľđ??ś and âˆ˘đ?‘‰đ?‘ˆđ??ś ≥ âˆ˘đ??´đ??ľđ??ś , therefore: âˆ˘đ?‘Šđ?‘†đ??´ ≥ âˆ˘đ?‘‰đ?‘ˆđ??ś , and since

39

Ion Patrascu, Florentin Smarandache

đ?‘‰đ?‘Š âˆĽ đ??´đ??ś, it follows that the trapeze đ?‘Šđ?‘†đ?‘ˆđ?‘‰ is isosceles, therefore the points đ?‘Š, đ?‘†, đ?‘ˆ, đ?‘‰ are concyclic (6).

The relations (4), (5), (6) drive to the concyclicality of the points đ?‘…, đ?‘ˆ, đ?‘‰, đ?‘†, đ?‘Š, đ?‘‡, and the theorem is proved.

Further Remarks.

2. For any point đ?‘€ found on the simedian đ??´đ??´1 , 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 đ?‘˜.

4. We get a circle built as in 2nd Theorem from the first Lemoine’s circle by homothety of pole đ?‘˜ and of ratio đ?œ† ∈ â„?.

5. The centers of Lemoine’s circles built as above belong to the line đ?‘‚đ??ž, where đ?‘‚ is the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś.

References.

[1] Exercices de GÊomÊtrie, par F.G.M., Huitième Êdition, Paris VIe, Librairie GÊnÊrale, 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 ExInscribed 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 đ??´đ??ľđ??ś is the Spiecker’s point of the triangle (the center of the circle inscribed in the median triangle of the triangle đ??´đ??ľđ??ś).

Proof.

We refer in the following to the notation in Figure 1. Let đ??źđ?‘Ž , đ??źđ?‘? , đ??źđ?‘? 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: đ??´đ??šđ?‘Ž = đ??´đ??¸đ?‘Ž = đ??ľđ??ˇđ?‘? = đ??ľđ??šđ?‘? = đ??śđ??ˇđ?‘? = đ??śđ??¸đ?‘? = đ?‘?, đ??ľđ??ˇđ?‘? = đ??ľđ??šđ?‘? = đ??śđ??ˇđ?‘? = đ??śđ??¸đ?‘? = đ?‘? − đ?‘Ž,

41

Ion Patrascu, Florentin Smarandache

đ??śđ??¸đ?‘Ž = đ??śđ??ˇđ?‘Ž = đ??´đ??šđ?‘? = đ??´đ??¸đ?‘? = đ?‘? − đ?‘?, đ??´đ??šđ?‘? = đ??´đ??¸đ?‘? = đ??ľđ??šđ?‘? = đ??ľđ??ˇđ?‘? = đ?‘? − đ?‘?.

If đ??´1 is the middle of segment đ??ˇđ?‘? đ??ˇđ?‘? , it follows that đ??´1 has equal powers to the ex-inscribed circles (đ??źđ?‘? ) and (đ??źđ?‘? ). Of the previously set, we obtain that đ??´1 is the middle of the side đ??ľđ??ś.

Figure 1.

Also, the middles of the segments đ??¸đ?‘? đ??¸đ?‘? and đ??šđ?‘? đ??šđ?‘? , which we denote đ?‘ˆ and đ?‘‰, have equal powers to the circles (đ??źđ?‘? ) and (đ??źđ?‘? ).

The radical axis of the circles ( đ??źđ?‘? ), ( đ??źđ?‘? ) will include the points đ??´1 , đ?‘ˆ, đ?‘‰.

Because đ??´đ??¸đ?‘? = đ??´đ??šđ?‘? and đ??´đ??¸đ?‘? = đ??´đ??šđ?‘? , it follows that 1

đ??´đ?‘ˆ = đ??´đ?‘Œ and we find that âˆ˘đ??´đ?‘ˆđ?‘‰ = âˆ˘đ??´, therefore the 2

42

Complements to Classic Topics of Circles Geometry

radical axis of the ex-inscribed circles (đ??šđ?‘? ) and (đ??źđ?‘? ) is the parallel taken through the middle đ??´1 of the side đ??ľđ??ś to the bisector of the angle đ??ľđ??´đ??ś.

Denoting đ??ľ1 and đ??ś1 the middles of the sides đ??´đ??ś, đ??´đ??ľ, respectively, we find that the radical center of the ex-inscribed circles is the center of the circle inscribed in the median triangle đ??´1 đ??ľ1 đ??ś1 of the triangle đ??´đ??ľđ??ś.

This point, denoted �, is the Spiecker’s point of the triangle ABC.

2nd Theorem.

The radical center of the inscribed circle (đ??ź) and of the đ??ľ −ex-inscribed and đ??ś −ex-inscribed circles of the triangle đ??´đ??ľđ??ś is the center of the đ??´1 − ex-inscribed circle of the median triangle đ??´1 đ??ľ1 đ??ś1 , corresponding to the triangle đ??´đ??ľđ??ś).

Proof.

If đ??¸ is the contact of the inscribed circle with đ??´đ??ś and đ??¸đ?‘? is the contact of the đ??ľ −ex-inscribed circle with đ??´đ??ś , it is known that these points are isotomic, therefore the middle of the segment đ??¸đ??¸đ?‘? is the middle of the side đ??´đ??ś, which is đ??ľ1 .

This point has equal powers to the inscribed circle (đ??ź) and to the đ??ľ −ex-inscribed circle (đ??źđ?‘? ), so it belongs to their radical axis.

43

Ion Patrascu, Florentin Smarandache

Analogously, đ??ś1 is on the radical axis of the circles (đ??ź) and (đ??źđ?‘? ).

The radical axis of the circles ( đ??ź ), (đ??źđ?‘? ) is the perpendicular taken from đ??ľ1 to the bisector đ??źđ??źđ?‘? .

This bisector is parallel with the internal bisector of the angle đ??´1 đ??ľ1 đ??ś1 , therefore the perpendicular in đ??ľ1 on đ??źđ??źđ?‘? is the external bisector of the angle đ??´1 đ??ľ1 đ??ś1 from the median triangle.

Analogously, it follows that the radical axis of the circles (đ??ź), (đ??źđ?‘? ) is the external bisector of the angle đ??´1 đ??ś1 đ??ľ1 from the median triangle.

Because the bisectors intersect in the center of the circle đ??´1 -ex-inscribed to the median triangle đ??´1 đ??ľ1 đ??ś1 , this point đ?‘†đ?‘Ž is the center of the radical center of the circles (đ??ź), (đ??źđ?‘? ), (đ??źđ?‘? ).

Remark.

The theorem for the circles (đ??ź), (đ??źđ?‘Ž ), (đ??źđ?‘? ) and (đ??ź), (đ??źđ?‘Ž ), (đ??źđ?‘? ) can be proved analogously, obtaining the points đ?‘†đ?‘? and đ?‘†đ?‘? .

3rd Theorem.

The radical circle’s radius of the circles exinscribed to the triangle đ??´đ??ľđ??ś is given by the formula: 1 2

√đ?‘&#x; 2 + đ?‘?2 , where đ?‘&#x; is the radius of the inscribed circle.

44

Complements to Classic Topics of Circles Geometry

Proof.

The position vector of the circle đ??ź of the inscribed circle in the triangle ABC is: ⃗⃗⃗⃗⃗ + đ?‘?đ?‘ƒđ??ľ ⃗⃗⃗⃗⃗ ). ⃗⃗⃗⃗ = 1 (đ?‘Žđ?‘ƒđ??´ ⃗⃗⃗⃗⃗ + đ?‘?đ?‘ƒđ??ś đ?‘ƒđ??ź 2đ?‘?

Spiecker’s point đ?‘† is the center of radical circle of ex-inscribed circle and is the center of the inscribed circle in the median triangle đ??´1 đ??ľ1 đ??ś1 , therefore: 1 1 ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗1 + 1 đ?‘?đ?‘ƒđ??ľ ⃗⃗⃗⃗⃗⃗⃗1 ). ⃗⃗⃗⃗⃗⃗⃗1 + 1 đ?‘?đ?‘ƒđ??ś đ?‘ƒđ?‘† = ( đ?‘Žđ?‘ƒđ??´ đ?‘? 2

2

2

Figure 2.

We denote by đ?‘‡ the contact point with the đ??´-exinscribed circle of the tangent taken from đ?‘† to this circle (see Figure 2).

45

Ion Patrascu, Florentin Smarandache

The radical circle’s radius is given by: 𝑆𝑇 = √𝑆𝐼𝑎2 − 𝐼𝑎2 1 ⃗​⃗​⃗​⃗​⃗​⃗ ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝐼𝑎 𝑆 = (𝑎𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐴1 + 𝑏𝐼𝑎 𝐵1 + 𝑐𝐼𝑎 𝐶1 ).

2𝑝

⃗​⃗​⃗​⃗​⃗​⃗ We evaluate the product of the scales 𝐼⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝑆 ∙ 𝐼𝑎 𝑆 ; we have: 1 𝐼𝑎 𝑆 2 = 2 (𝑎2 𝐼𝑎 𝐴12 + 𝑏 2 𝐼𝑎 𝐵12 + 𝑐 2 𝐼𝑎 𝐶12 + 2𝑎𝑏𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐴1 ∙ 4𝑝

⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝐼𝑎 𝐵1 + 2𝑏𝑐𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐵1 ∙ 𝐼𝑎 𝐶1 + 2𝑎𝑐𝐼𝑎 𝐴1 ∙ 𝐼𝑎 𝐶1 ).

From the law of cosines applied in the triangle 𝐼𝑎 𝐴1 𝐵1 , we find that: 1 2 2 2 ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 2𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐴1 ∙ 𝐼𝑎 𝐵1 = 𝐼𝑎 𝐴1 + 𝐼𝑎 𝐵1 − 𝑐 , therefore: 4

1 2 2 2 ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 2𝑎𝑏𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐴1 ∙ 𝐼𝑎 𝐵1 = 𝑎𝑏(𝐼𝑎 𝐴1 + 𝐼𝑎 𝐵1 − 𝑎𝑏𝑐 .

4

Analogously, we obtain: 1 2 2 2 ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 2𝑏𝑐𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐵1 ∙ 𝐼𝑎 𝐶1 = 𝑏𝑐(𝐼𝑎 𝐵1 + 𝐼𝑎 𝐶1 − 𝑎 𝑏𝑐, ⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 2𝑎𝑐𝐼⃗​⃗​⃗​⃗​⃗​⃗​⃗​⃗ 𝑎 𝐴1 ∙ 𝐼𝑎 𝐶1 = 𝐼𝑎 𝑆 2 =

1 4𝑝

𝑎𝑐(𝐼𝑎 𝐴12

+

𝐼𝑎 𝐶12

+ (𝑐 2 + 𝑏𝑐 + 𝑎𝑐)𝐼𝑎 𝐶12 −

𝐼𝑎 𝑆 2 =

1 4𝑝2 1 2𝑝

− 𝑎𝑏 2 𝑐.

4

2 2 (𝑏 2 + 𝑎𝑏 + 2 [(𝑎 + 𝑎𝑏 + 𝑎𝑐)𝐼𝑎 𝐴1 +

𝑏𝑐)𝐼𝑎 𝐵12

𝐼𝑎 𝑆 2 =

4 1

𝑎𝑏𝑐 4

(𝑎 + 𝑏 + 𝑐)],

[2𝑝(𝑎𝐼𝑎 𝐴12 + 𝑏𝐼𝑎 𝐵12 + 𝑐𝐼𝑎 𝐶12 ) − 2𝑅𝑆𝑝 ], 1

(𝑎𝐼𝑎 𝐴12 + 𝑏𝐼𝑎 𝐵12 + 𝑐𝐼𝑎 𝐶12 ) − 𝑅𝑟.

2

From the right triangle 𝐼𝑎 𝐷𝑎 𝐴1 , we have that: 2 𝑎 𝐼𝑎 𝐴12 = 𝑟𝑎2 + 𝐴1 𝐷𝑎2 = 𝑟𝑎2 + [ − (𝑝 − 𝑐)] = 2 = 𝑟𝑎2 +

(𝑐−𝑏)2 4

.

From the right triangles 𝐼𝑎 𝐸𝑎 𝐵1 și 𝐼𝑎 𝐹𝑎 𝐶1 , we find:

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Complements to Classic Topics of Circles Geometry

2 𝑏 𝐼𝑎 𝐵12 = 𝑟𝑎2 + 𝐵1 𝐸𝑎2 = 𝑟𝑎2 + [ − (𝑝 − 𝑏)] = 2 1 2 = 𝑟𝑎 + (𝑎 + 𝑐)2 , 4

𝐼𝑎 𝐶12

=

𝑟𝑎2

1

2

+ (𝑎 + 𝑏) .

4

Evaluating 𝑎𝐼𝑎 𝐴12 + 𝑏𝐼𝑎 𝐵12 + 𝑐𝐼𝑎 𝐶12 , we obtain: 𝑎𝐼𝑎 𝐴12 + 𝑏𝐼𝑎 𝐵12 + 𝑐𝐼𝑎 𝐶12 = 1

1

2

4

= 2𝑝𝑟𝑎2 + 𝑝(𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐) − 𝑎𝑏𝑐.

But: 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 = 𝑟 2 + 𝑝2 + 4𝑅𝑟.

It follows that: 1 2𝑝

1

1

4

2

[𝑎𝐼𝑎 𝐴12 + 𝑏𝐼𝑎 𝐵12 + 𝑐𝐼𝑎 𝐶12 ] = 𝑟𝑎2 + (𝑟 2 + 𝑝2 ) + 𝑅𝑟

and 1

𝐼𝑎 𝑆 2 = 𝑟𝑎2 + (𝑟 2 + 𝑝2 ).

4

Then, we obtain: 1

𝑆𝑇 = √𝑟 2 + 𝑝2 .

2

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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.

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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 đ??´đ??ľđ??ś be a given triangle; through the pairs of points (đ??ľ, đ??ś) , (đ??ś, đ??´) and (đ??´, đ??ľ) 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 đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ respectively in three collinear points.

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Ion Patrascu, Florentin Smarandache

Proof.

We denote by đ??ˇ, đ?‘ƒ, đ??š the second point of intersection of the pairs of circles passing through (đ??´, đ??ľ) and (đ??ľ, đ??ś) ; (đ??ľ, đ??ś) and (đ??´, đ??ś) , (đ??ľ, đ??ś) and (đ??´, đ??ľ) respectively (see Figure 1).

Figure 1

Let đ?‘… be the radical center of those circles. In fact, {đ?‘…} = đ??´đ??š ∊ đ??ľđ??ˇ ∊ đ??śđ??¸.

We take from đ?‘… the tangents đ?‘…đ??ˇ1 = đ?‘…đ??ˇ2 to the circle (đ??ľ, đ??ś), đ?‘…đ??¸1 = đ?‘…đ??¸2 to the circle (đ??´, đ??ś) and đ?‘…đ??š1 = đ?‘…đ??š2 to the circle passing through (đ??´, đ??ľ). Actually, we build the radical circle đ?’ž(đ?‘…, đ?‘…đ??ˇ1 ) of the given circles.

The polar lines of đ?‘… to these circles are the lines đ??ˇ1 đ??ˇ2 , đ??¸1 đ??¸2 , đ??š1 đ??š2 . These three lines cut đ??ľđ??ś, đ??´đ??ś and đ??´đ??ľ in the points đ?‘‹, đ?‘Œ and đ?‘? , and these lines are respectively the polar lines of đ?‘… in respect to the

50

Complements to Classic Topics of Circles Geometry

circles passing through (đ??ľ, đ??ś), (đ??ś, đ??´) and (đ??´, đ??ľ) . The polar lines are the radical axis of the radical circle with each of the circles passing through (đ??ľ, đ??ś), (đ??ś, đ??´), (đ??´, đ??ľ), respectively. The points belong to the radical axis having equal powers to those circles, thereby đ?‘‹đ??ˇ1 ∙ đ?‘‹đ??ˇ2 = đ?‘‹đ??ś ∙ đ?‘‹đ??ľ.

This relationship shows that the point đ?‘‹ has equal powers relative to the radical circle and to the circle circumscribed to the triangle đ??´đ??ľđ??ś; analogically, the point đ?‘Œ has equal powers relative to the radical circle and to the circle circumscribed to the triangle đ??´đ??ľđ??ś ; and, likewise, the point đ?‘? has equal powers relative to the radical circle and to the circle circumscribed to the triangle đ??´đ??ľđ??ś.

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 đ?‘‹, đ?‘Œ and đ?‘? are collinear, belonging to the radical axis of the radical circle and to the circle circumscribed to the given triangle.

2nd Theorem.

If đ?‘€ is a point in the plane of the triangle đ??´đ??ľđ??ś and the tangents in this point to the circles circumscribed to triangles đ??ś, đ?‘€đ??´đ??ś, đ?‘€đ??´đ??ľ, respectively, cut đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ, respectively, in the points đ?‘‹, đ?‘Œ, đ?‘?, then these points are collinear.

51

Ion Patrascu, Florentin Smarandache

Proof.

The point đ?‘€ is the radical center for the circles (đ?‘€đ??ľđ??ś), (đ?‘€đ??´đ??ś), and (đ?‘€đ??´đ??ľ), and the tangents in đ?‘€ to these circles are the polar lines to đ?‘€ in relation to these circles.

If đ?‘‹, đ?‘Œ, đ?‘? are the intersections of these tangents (polar lines) with đ??ľđ??ś, đ??śđ??´, đ??´đ??ľ, then they belong to the radical axis of the circumscribed circle to the triangle đ??´đ??ľđ??ś and to the circle “reducedâ€? to the point đ?‘€ (đ?‘‹đ?‘€2 = đ?‘‹đ??ľ ∙ đ?‘‹đ??ś, etc.).

Being located on the radical axis of the two circles, the points đ?‘‹, đ?‘Œ, đ?‘? 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 đ??´đ??ľđ??ś , then they intersect in Ί (the Brocard’s point).

Therefore, we get that the tangents taken in Ί to the adjoin circles cut the sides đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ in collinear points.

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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.

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Ion Patrascu, Florentin Smarandache

54

Complements to Classic Topics of Circles Geometry

Regarding the First DrozFarny’s Circle In this article, we define the first DrozFarny’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 (MathĂŠsis, 1901).

1st Theorem.

Let đ??´đ??ľđ??ś be a triangle, đ??ť its orthocenter and đ??´1 , đ??ľ1 , đ??ś1 the means of sides (đ??ľđ??ś), (đ??śđ??´), (đ??´đ??ľ).

If a circle, having its center đ??ť, intersects đ??ľ1 đ??ś1 in đ?‘ƒ1 , đ?‘„1 ; đ??ś1 đ??´1 in đ?‘ƒ2 , đ?‘„2 and đ??´1 đ??ľ1 in đ?‘ƒ3 , đ?‘„3 , then đ??´đ?‘ƒ1 = đ??´đ?‘„1 = đ??ľđ?‘ƒ2 = đ??ľđ?‘„2 = đ??śđ?‘ƒ3 = đ??śđ?‘„3 .

Proof.

Naturally, đ??ťđ?‘ƒ1 = đ??ťđ?‘„1 , đ??ľ1 đ??ś1 âˆĽ đ??ľđ??ś, đ??´đ??ť ⊼ đ??ľđ??ś.

It follows that đ??´đ??ť ⊼ đ??ľ1 đ??ś1 .

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Ion Patrascu, Florentin Smarandache

Figure 1.

Therefore, đ??´đ??ť is the mediator of segment đ?‘ƒ1 đ?‘„1 ; similarly, đ??ľđ??ť and đ??śđ??ť are the mediators of segments đ?‘ƒ2 đ?‘„2 and đ?‘ƒ3 đ?‘„3 .

Let đ?‘‡1 be the intersection of lines đ??´đ??ť and đ??ľ1 đ??ś1 (see Figure 1); we have đ?‘„1 đ??´2 − đ?‘„1 đ??ť2 = đ?‘‡1 đ??´2 − đ?‘‡1 đ??ť2 . We 2 denote đ?‘…đ??ť = đ??ťđ?‘ƒ1 . It follows that đ?‘„1 đ??´2 = đ?‘…đ??ť + (đ?‘‡1 đ??´ + 2 đ?‘‡1 đ??ť)(đ?‘‡1 đ??´ − đ?‘‡1 đ??ť) = đ?‘…đ??ť + đ??´đ??ť ∙ (đ?‘‡1 đ??´ − đ?‘‡1 đ??ť).

However, đ?‘‡1 đ??´ = đ?‘‡1 đ??ť1 , where đ??ť1 is the projection 2 of đ??´ on đ??ľđ??ś; we find that đ?‘„1 đ??´2 = đ?‘…đ??ť + đ??´đ??ť ∙ đ??ťđ??ť1 .

It is known that the symmetric of orthocenter đ??ť towards đ??ľđ??ś belongs to the circle of circumscribed triangle đ??´đ??ľđ??ś.

Denoting this point by đ??ť1′ , we have đ??´đ??ť ∙ đ??ťđ??ť1′ = đ?‘…2 − đ?‘‚đ??ť2 (the power of point đ??ť towards the circumscribed circle).

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Complements to Classic Topics of Circles Geometry

1

We obtain that đ??´đ??ť ∙ đ??ťđ??ť1 = ∙ (đ?‘…2 − đ?‘‚đ??ť2 ) , and therefore

đ??´đ?‘„12

=

2 đ?‘…đ??ť

1

+ ∙ 2

(đ?‘…2

− đ?‘‚đ??ť

2 2)

, where đ?‘‚ is the

center of the circumscribed triangle đ??´đ??ľđ??ś.

Similarly,

we

1

2 find đ??ľđ?‘„22 = đ??śđ?‘„32 = đ?‘…đ??ť + (đ?‘…2 − 2

đ?‘‚đ??ť2 ), therefore đ??´đ?‘„1 = đ??ľđ?‘„2 = đ??śđ?‘„3 .

Remarks. a. b.

The proof adjusts analogously for the obtuse triangle.

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 đ?œŒ the radius of three congruent circles having đ??´, đ??ľ, đ??ś as their centers, we get: 1

2 đ?‘…đ??ť = đ?œŒ2 + (đ?‘‚đ??ť2 − đ?‘… 2 ).

2

However, in a triangle, đ?‘‚đ??ť2 = 9đ?‘…2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? đ?‘… being the radius of the circumscribed circle; it follows that: 2 ),

1

2 đ?‘…đ??ť = đ?œŒ2 + 4đ?‘…2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? 2 ).

2

57

Ion Patrascu, Florentin Smarandache

Remark.

A special case is the one in which đ?œŒ = đ?‘…, where 1

1

2

2

2 we find that đ?‘…đ??ť = đ?‘…12 = 5đ?‘… 2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? 2 ) = (đ?‘…2 +

đ?‘‚đ??ť2 ).

Definition.

The circle đ?’ž(đ??ť, đ?‘…1 ), where: 1

đ?‘…1 = √5đ?‘…2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? 2 ), 2

is called the first Droz-Farny’s circle of the triangle đ??´đ??ľđ??ś.

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. ThĂŠbault in the Journal de MathĂŠmatiques 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.

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Complements to Classic Topics of Circles Geometry

Proof.

We consider đ??´đ??ľđ??ś an acute triangle and đ??ť1 , đ??ť2 , đ??ť3 the altitudes’ feet. We denote by đ??´1 , đ??´2 ; đ??ľ1 , đ??ľ2 ; đ??ś1 , đ??ś2 the intersection points of circles having their centers đ??ť1 , đ??ť2 , đ??ť3 to đ??ľđ??ś, đ??śđ??´, đ??´đ??ľ, respectively.

We calculate đ??ťđ??´2 of the right angled triangle đ??ťđ??ť1 đ??´2 (see Figure 2). We have đ??ťđ??´22 = đ??ťđ??ť12 + đ??ť1 đ??´22 .

Because đ??ť1 đ??´2 = đ??ť1 đ?‘‚ , it follows that đ??ťđ??´22 = đ??ťđ??ť12 + đ??ť1 đ?‘‚2 . We denote by đ?‘‚9 the mean of segment đ?‘‚đ??ť ; the median theorem in triangle đ??ť1 đ??ťđ?‘‚ leads to 1

đ??ť1 đ??ť2 + đ??ť1 đ?‘‚2 = 2đ??ť1 đ?‘‚92 + đ?‘‚đ??ť2 .

2

It is known that đ?‘‚9 đ??ť1 is the nine-points circle’s 1

1

2

2 1

radius, so đ??ť1 đ?‘‚9 = đ?‘… ; we get: đ??ťđ??´12 = (đ?‘…2 + đ?‘‚đ??ť2 ) ; similarly, we find that đ??ťđ??ľ12 = đ??ťđ??ś12 = (đ?‘…2 + đ?‘‚đ??ť2 ) , 2

which shows that the points đ??´1 , đ??´2 ; đ??ľ1 , đ??ľ2 ; đ??ś1 , đ??ś2 belong to the first Droz-Farny’s circle.

Figure 2.

59

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 đ??´1 , đ??´2 ; đ??ľ1 , đ??ľ2 ; đ??ś1 , đ??ś2 be the points of intersection with the sides of triangle đ??´đ??ľđ??ś of circles having their centers in altitudes’ feet đ??ť1 , đ??ť2 , đ??ť3 .

Suppose the points are concyclical; it follows that their circle’s radical center and of circles centered in đ??ť2 and đ??ť3 is the point đ??´ (sides đ??´đ??ľ and đ??´đ??ś are radical axes), therefore the perpendicular from đ??´ on đ??ť2 đ??ť3 is radical axis of centers having their centers đ??ť2 and đ??ť3 .

Since đ??ť2 đ??ť3 is antiparallel to đ??ľđ??ś, it is parallel to tangent taken in đ??´ to the circle circumscribed to triangle đ??´đ??ľđ??ś.

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Complements to Classic Topics of Circles Geometry

Consequently, the radical axis is the perpendicular taken in đ??´ on the tangent to the circumscribed circle, therefore it is đ??´đ?‘‚.

Similarly, the other radical axis of circles centered in đ??ť1 , đ??ť2 and of circles centered in đ??ť1 , đ??ť3 pass through đ?‘‚, therefore đ?‘‚ is the radical center of these circles.

Reciprocally.

Let đ?‘‚ be the radical center of the circles having their centers in the feet of altitudes. Since đ??´đ?‘‚ is perpendicular on đ??ť2 đ??ť3 , it follows that đ??´đ?‘‚ is the radical axis of circles having their centers in đ??ť2 , đ??ť3 , therefore đ??´đ??ľ1 ∙ đ??´đ??ľ2 = đ??´đ??ś1 ∙ đ??´đ??ś2 .

From this relationship, it follows that the points đ??ľ1 , đ??ľ2 ; đ??ś1 , đ??ś2 are concyclic; the circle on which these points are located has its center in the orthocenter đ??ť of triangle đ??´đ??ľđ??ś.

Indeed, the mediators’ chords đ??ľ1 đ??ľ2 and đ??ś1 đ??ś2 in the two circles are the altitudes from đ??ś and đ??ľ of triangle đ??´đ??ľđ??ś, therefore đ??ťđ??ľ1 = đ??ťđ??ľ2 = đ??ťđ??ś1 = đ??ťđ??ś2 .

This reasoning leads to the conclusion that đ??ľđ?‘‚ is the radical axis of circles having their centers đ??ť1 and đ??ť3 , and from here the concyclicality of the points đ??´1 , đ??´2 ; đ??ś1 , đ??ś2 on a circle having its center in đ??ť , therefore đ??ťđ??´1 = đ??ťđ??´2 = đ??ťđ??ś1 = đ??ťđ??ś2 . We obtained that đ??ťđ??´1 = đ??ťđ??´2 = đ??ťđ??ľ1 = đ??ťđ??ľ2 = đ??ťđ??ś1 = đ??ťđ??ś2 , which shows the concyclicality of points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 .

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Ion Patrascu, Florentin Smarandache

Remark.

The circles from the 3rd Theorem, passing through đ?‘‚ and having noncollinear centers, admit đ?‘‚ 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.

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Regarding the Second DrozFarny’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 đ??´đ??ľđ??ś passing through its orthocenter đ??ť intersect the sides đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ respectively in the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 and đ??ś1 , đ??ś2 , situated on a concentric circle with the circle circumscribed to the triangle đ??´đ??ľđ??ś (the second Droz-Farny’s circle).

Proof.

We denote by đ?‘€1 , đ?‘€2 , đ?‘€3 the middles of đ??´đ??ľđ??ś triangle’s sides, see Figure 1. Because đ??´đ??ť ⊼ đ?‘€2 đ?‘€3 and đ??ť belongs to the circles with centers in đ?‘€2 and đ?‘€3 , it follows that đ??´đ??ť is the radical axis of these circles,

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Ion Patrascu, Florentin Smarandache

therefore we have đ??´đ??ś1 ∙ đ??´đ??ś2 = đ??´đ??ľ2 ∙ đ??´đ??ľ1 . This relation shows that đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are concyclic points, because the center of the circle on which they are situated is đ?‘‚, the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś, hence we have that: đ?‘‚đ??ľ1 = đ?‘‚đ??ś1 = đ?‘‚đ??ś2 = đ?‘‚đ??ľ2 .

(1)

Figure 1.

Analogously, đ?‘‚ is the center of the circle on which the points đ??´1 , đ??´2 , đ??ś1 , đ??ś2 are situated, hence: đ?‘‚đ??´1 = đ?‘‚đ??ś1 = đ?‘‚đ??ś2 = đ?‘‚đ??´2 .

(2) Also, đ?‘‚ is the center of the circle on which the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 are situated, and therefore: đ?‘‚đ??´1 = đ?‘‚đ??ľ1 = đ?‘‚đ??ľ2 = đ?‘‚đ??´2 .

(3) The relations (1), (2), (3) show that the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are situated on a circle having the center in đ?‘‚, called the second Droz-Farny’s circle.

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Complements to Classic Topics of Circles Geometry

1st Proposition.

The radius of the second Droz-Farny’s circle is given by: 1 đ?‘…22 = 5đ?‘’ 2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? 2 ).

2 Proof.

From the right triangle đ?‘‚đ?‘€1 đ??´1 , using Pitagora’s theorem, it follows that: đ?‘‚đ??´12 = đ?‘‚đ?‘€12 + đ??´1 đ?‘€12 = đ?‘‚đ?‘€12 + đ?‘€1 đ?‘€2 .

From the triangle đ??ľđ??ťđ??ś , using the median theorem, we have: 1 đ??ťđ?‘€12 = [2(đ??ľđ??ť2 + đ??śđ??ť2 ) − đ??ľđ??ś 2 ].

4 But in a triangle, đ??´đ??ť = 2đ?‘‚đ?‘€1 , đ??ľđ??ť = 2đ?‘‚đ?‘€2 , đ??śđ??ť = 2đ?‘‚đ?‘€3 , hence: đ?‘Ž2 đ??ťđ?‘€12 = 2đ?‘‚đ?‘€22 + 2đ?‘‚đ?‘€32 = .

4 But: đ?‘‚đ?‘€12 = đ?‘…2 − đ?‘‚đ?‘€22 = đ?‘…2 − đ?‘‚đ?‘€32 = đ?‘…2 −

đ?‘Ž2 4 đ?‘?2 4 đ?‘?2 4

; ; ,

where R is the radius of the circle circumscribed to the triangle đ??´đ??ľđ??ś.

1

We find that đ?‘‚đ??´12 = đ?‘…22 = 5đ?‘…2 − (đ?‘Ž2 + đ?‘? 2 + đ?‘? 2 ).

2

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Ion Patrascu, Florentin Smarandache

Remarks. a.

We can compute đ?‘‚đ?‘€12 + đ?‘€1 đ?‘€2 using the median theorem in the triangle đ?‘‚đ?‘€1 đ??ť for the median đ?‘€1 đ?‘‚9 (đ?‘‚9 is the center of the nine points circle, i.e. the middle of (đ?‘‚đ??ť)).

1

Because đ?‘‚9 đ?‘€1 = đ?‘… , we obtain: đ?‘…22 = 2

1 2

b.

c.

d.

(đ?‘‚đ?‘€2 + đ?‘…2 ). In this way, we can prove

the Theorem computing đ?‘‚đ??ľ12 and đ?‘‚đ??ś12 .

The statement of the 1st Theorem was the subject no. 1 of the 49th International Olympiad in Mathematics, held at Madrid in 2008.

The 1st Theorem can be proved in the same way for an obtuse triangle; it is obvious that for a right triangle, the second DrozFarny’s circle coincides with the circle circumscribed to the triangle đ??´đ??ľđ??ś.

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.

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Complements to Classic Topics of Circles Geometry

Proof.

Let đ?‘€1 , đ?‘€2 , đ?‘€3 the middles of the sides of triangle đ??´đ??ľđ??ś and let đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 the intersections with đ??ľđ??ś, đ??śđ??´, đ??´đ??ľ respectively of the circles with centers in đ?‘€1 , đ?‘€2 , đ?‘€3 .

Let us suppose that đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are concyclic points. The circle on which they are situated has evidently the center in đ?‘‚, the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś.

The radical axis of the circles with centers đ?‘€2 , đ?‘€3 will be perpendicular on the line of centers đ?‘€2 đ?‘€3 , and because A has equal powers in relation to these circles, since đ??´đ??ľ1 ∙ đ??´đ??ľ2 = đ??´đ??ś1 ∙ đ??´đ??ś2 , it follows that the radical axis will be the perpendicular taken from A on đ?‘€2 đ?‘€3 , i.e. the height from đ??´ of triangle đ??´đ??ľđ??ś.

Furthermore, it ensues that the radical axis of the circles with centers in đ?‘€1 and đ?‘€2 is the height from đ??ľ of triangle đ??´đ??ľđ??ś and consequently the intersection of the heights, hence the orthocenter đ??ť of the triangle đ??´đ??ľđ??ś is the radical center of the three circles.

Reciprocally.

If the circles having the centers in đ?‘€1 , đ?‘€2 , đ?‘€3 have the orthocenter with the radical center, it follows that the point đ??´, being situated on the height from A which is the radical axis of the circles of centers đ?‘€2 , đ?‘€3

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Ion Patrascu, Florentin Smarandache

will have equal powers in relation to these circles and, consequently, đ??´đ??ľ1 ∙ đ??´đ??ľ2 = đ??´đ??ś1 ∙ đ??´đ??ś2 , a relation that implies that đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are concyclic points, and the circle on which these points are situated has đ?‘‚ as its center.

Similarly, đ??ľđ??´1 ∙ đ??ľđ??´2 = đ??ľđ??ś1 ∙ đ??ľđ??ś2 , therefore đ??´1 , đ??´2 , đ??ś1 , đ??ś2 are concyclic points on a circle of center đ?‘‚.

Having đ?‘‚đ??ľ1 = đ?‘‚đ??ľ2 = đ?‘‚đ??ś1 = đ?‘‚đ??ś2 and đ?‘‚đ??´1 ∙ đ?‘‚đ??´2 = đ?‘‚đ??ś1 ∙ đ?‘‚đ??ś2 , we get that the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are situated on a circle of center đ?‘‚.

Remarks.

1.

2.

The 1st Theorem is a particular case of the 2nd Theorem, because the three circles of centers đ?‘€1 , đ?‘€2 , đ?‘€3 pass through đ??ť, which means that đ??ť is their radical center.

The Problem 525 from [3] leads us to the following Proposition providing a way to construct the circles of centers �1 , �2 , �3 intersecting the sides in points that belong to a Droz-Farny’s circle of type 2.

2nd Proposition.

1

The circles ∠(đ?‘€1 , √đ?‘˜ + đ?‘Ž2 ), 2

1

∠(đ?‘€2 , √đ?‘˜ + đ?‘? 2 ), 2

1

∠(đ?‘€3 , √đ?‘˜ + đ?‘? 2 ) intersect the sides đ??ľđ??ś , đ??śđ??´ , đ??´đ??ľ 2 respectively in six concyclic points; đ?‘˜ is a conveniently

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Complements to Classic Topics of Circles Geometry

chosen constant, and đ?‘Ž, đ?‘?, đ?‘? are the lengths of the sides of triangle đ??´đ??ľđ??ś.

Proof.

According to the 2nd Theorem, it is necessary to prove that the orthocenter đ??ť of triangle đ??´đ??ľđ??ś is the radical center for the circles from hypothesis.

Figure 2.

The

power

of

đ??ť

in

1

∠(đ?‘€1 , √đ?‘˜ + đ?‘Ž2 ) is equal to 2 observed

that đ?‘€12 = 4đ?‘… 2 −

đ?‘?2 2

−

1

đ?‘Ž2 +đ?‘? 2 +đ?‘? 2

4

4

đ??ťđ?‘€12 − (đ?‘˜ + đ?‘Ž2 ) = 4đ?‘…2 −

relation 1

đ??ťđ?‘€12 đ?‘?2 2

− (đ?‘˜ + đ?‘Ž −

1

4 đ?‘Ž2

− đ?‘˜.

4

4

,

2)

with . We

therefore

We use the

same expression for the power of H in relation to the circles of centers đ?‘€2 , đ?‘€3 , hence H is the radical center of these three circles.

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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.

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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 đ??´đ??ľđ??ś the point Ί with the property: âˆ˘ÎŠđ??´đ??ľ = âˆ˘ÎŠđ??ľđ??ś = âˆ˘ÎŠđ??śđ??´ .

The measure of the angle Ίđ??´đ??ľ is denoted by đ?œ” and it is called Brocard's angle. It occurs the relationship: ctgđ?œ” = ctgđ??´ + ctgđ??ľ + ctgđ??ś (see [1]).

Figure 1.

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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 đ?‘€ from the plane of the triangle located on the same side of the line đ??ľđ??ś as đ??´ and forming with đ??ľđ??ś an equibrocardian triangle with đ??´đ??ľđ??ś, containing the vertex đ??´ of the triangle, it's called A-Neuberg’ circle of the triangle đ??´đ??ľđ??ś.

We denote by đ?‘ đ?‘Ž the center of A-Neuberg’ circle by radius đ?‘›đ?‘Ž (analogously, we define B-Neuberg’ and C-Neuberg’ circles). đ?‘Ž

We get that đ?‘š(đ??ľđ?‘ đ?‘Ž đ??ś) = 2đ?œ” and đ?‘›đ?‘Ž = √ctg 2 đ?œ” − 3 2

(see [1]).

The triangle đ?‘ đ?‘Ž đ?‘ đ?‘? đ?‘ đ?‘? formed by the centers of Neuberg’s circles is called Neuberg’s triangle.

1st Proposition.

The distances from the center circumscribed to the triangle đ??´đ??ľđ??ś to the vertex of Neuberg’s triangle are proportional to the cubes of đ??´đ??ľđ??ś triangle’s sides lengths.

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Complements to Classic Topics of Circles Geometry

Proof.

Let đ?‘‚ be the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś (see Figure 2).

Figure 2.

The law of cosines applied in the triangle đ?‘‚đ?‘ đ?‘Ž đ??ľ provides: đ?‘‚đ?‘ đ?‘Ž sin(đ?‘ đ?‘Ž đ??ľđ?‘‚)

=

đ?‘… sinđ?œ”

.

But đ?‘š(âˆ˘đ?‘ đ?‘Ž đ??ľđ?‘‚) = đ?‘š(âˆ˘đ?‘ đ?‘Ž đ??ľđ??ś) − đ?‘š(âˆ˘đ?‘‚đ??ľđ??ś) = đ??´ − đ?œ”.

We have that

đ?‘‚đ?‘ đ?‘Ž sin(đ??´âˆ’đ?œ”)

=

đ?‘… sinđ?œ”

.

But sin(đ??´âˆ’đ?œ”) sinđ?œ”

=

đ??śÎŠ AΊ

đ?‘Ž

= đ?‘?đ?‘? đ?‘Ž

∙2đ?‘…sinđ?œ” ∙2đ?‘…sinđ?œ”

=

đ?‘Ž3 đ?‘Žđ?‘?đ?‘?

=

�3 4��

,

đ?‘† being the area of the triangle đ??´đ??ľđ??ś.

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Ion Patrascu, Florentin Smarandache

It follows that đ?‘‚đ?‘ đ?‘Ž = đ?‘‚đ?‘ đ?‘? đ?‘?3

=

đ?‘‚đ?‘ đ?‘? đ?‘?3

�3 4�

, and we get that

đ?‘‚đ?‘ đ?‘Ž đ?‘Ž3

=

.

Consequence.

In a triangle ABC, we have: 1) đ?‘‚đ?‘ đ?‘Ž ∙ đ?‘‚đ?‘ đ?‘? ∙ đ?‘‚đ?‘ đ?‘? = đ?‘…3 ; 2)

ctgđ?œ” =

đ?‘‚đ?‘ đ?‘Ž đ?‘Ž

+

đ?‘‚đ?‘ đ?‘? đ?‘?

+

đ?‘‚đ?‘ đ?‘? đ?‘?

.

2nd Proposition.

If đ?‘ đ?‘Ž đ?‘ đ?‘? đ?‘ đ?‘? is the Neuberg’s triangle of the triangle đ??´đ??ľđ??ś, then: (đ?‘Ž2 + đ?‘? 2 )(đ?‘Ž4 + đ?‘? 4 ) − đ?‘Ž2 đ?‘? 2 đ?‘? 2 đ?‘ đ?‘Ž đ?‘ đ?‘? 2 = 2 2 .

2đ?‘Ž đ?‘? + 2đ?‘? 2 đ?‘? 2 + 2đ?‘? 2 đ?‘Ž2 − đ?‘Ž4 − đ?‘? 4 − đ?‘? 4 (The formulas for đ?‘ đ?‘? đ?‘ đ?‘? and đ?‘ đ?‘? đ?‘ đ?‘Ž are obtained from the previous one, by circular permutations.)

Proof.

We apply the law of cosines in the triangle đ?‘ đ?‘Ž đ?‘‚đ?‘ đ?‘? : đ?‘Ž3 đ?‘?3 đ?‘‚đ?‘ đ?‘Ž = , đ?‘‚đ?‘ đ?‘? = , đ?‘š(âˆ˘đ?‘ đ?‘Ž đ?‘‚đ?‘ đ?‘? ) = 1800 − đ?‘?Ě‚ .

4đ?‘† 4đ?‘† đ?‘Ž6 + đ?‘? 6 − 2đ?‘Ž3 đ?‘? 3 cos(1800 − đ?‘?) đ?‘ đ?‘Ž đ?‘ đ?‘? 2 = 16đ?‘† 2 6 đ?‘Ž + đ?‘? 6 + 2đ?‘Ž3 đ?‘? 3 cosđ??ś = .

16� 2

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Complements to Classic Topics of Circles Geometry

But the law of cosines in the triangle đ??´đ??ľđ??ś provides đ?‘Ž2 + đ?‘? 2 − đ?‘? 2 2cosđ??ś = 2đ?‘Žđ?‘? and, from din Heron’s formula, we find that 16đ?‘† 2 = 2đ?‘Ž2 đ?‘? 2 + 2đ?‘? 2 đ?‘? 2 + 2đ?‘? 2 đ?‘Ž2 − đ?‘Ž4 − đ?‘? 4 − đ?‘? 4 .

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 đ?’ž(đ?‘‚1 , đ?‘&#x;1 ), đ?’ž(đ?‘‚2 , đ?‘&#x;2 ) are orthogonal if and only if đ?‘&#x;12 + đ?‘&#x;22 = đ?‘‚1 đ?‘‚22 .

Proof.

Let đ?’ž(đ?‘‚1 , đ?‘&#x;1 ), đ?’ž(đ?‘‚2 , đ?‘&#x;2 ) be orthogonal (see Figure 3); then, if đ??´ is one of the common points, the triangle đ?‘‚1 đ??´đ?‘‚2 is a right triangle and the Pythagorean Theorem applied to it, leads to đ?‘&#x;1 2 + đ?‘&#x;2 2 = đ?‘‚1 đ?‘‚2 2 .

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Ion Patrascu, Florentin Smarandache

Reciprocally.

If the metric relationship from the statement occurs, it means that the triangle đ?‘‚1 đ??´đ?‘‚2 is a right triangle, therefore đ??´ is their common point (the relationship đ?‘&#x;1 2 + đ?‘&#x;2 2 = đ?‘‚1 đ?‘‚2 2 implies đ?‘&#x;1 2 + đ?‘&#x;2 2 > đ?‘‚1 đ?‘‚2 2 ), then đ?‘‚1 đ??´ ⊼ đ?‘‚2 đ??´, so đ?‘‚1 đ??´ is tangent to the circle đ?’ž(đ?‘‚2 , đ?‘&#x;2 ) because it is perpendicular in đ??´ on radius đ?‘‚2 đ??´, and as well đ?‘‚2 đ??´ is tangent to the circle đ?’ž(đ?‘‚1 , đ?‘&#x;1 ) , therefore the circles are orthogonal.

Figure 3.

4th Proposition.

B-Neuberg’s and C-Neuberg’s circles associated to the right triangle đ??´đ??ľđ??ś (in đ??´) are orthogonal.

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Complements to Classic Topics of Circles Geometry

Proof. đ?‘?

If đ?‘š(đ??´Ě‚) = 900 , then đ?‘ đ?‘? đ?‘ đ?‘?2 = đ?‘?

đ?‘?

2

2 đ?‘? 2 +đ?‘? 2

6 +đ?‘? 6

16�

.

đ?‘›đ?‘? = √ctg 2 đ?œ” − 3 ; đ?‘›đ?‘? = √ctg 2 đ?œ” − 3.

But ctgđ?œ” −

đ?‘Ž2 +đ?‘? 2 +đ?‘? 2 4đ?‘†

=

2�

=

đ?‘Ž2 đ?‘?đ?‘?

.

It was taken into account that đ?‘Ž2 = đ?‘? 2 + đ?‘? 2 and 2đ?‘† = đ?‘?đ?‘?. (đ?‘? 2 + đ?‘? 2 )2 − 3đ?‘? 2 đ?‘? 2 đ?‘Ž4 ctg 2 đ?œ” − 3 = 2 2 − 3 = đ?‘? đ?‘? đ?‘?2đ?‘? 2 4 4 đ?‘? + đ?‘? − đ?‘?2đ?‘? 2 ctg 2 đ?œ” − 3 = đ?‘?2đ?‘? 2 4 4 2 2 2 đ?‘? + đ?‘? − đ?‘? đ?‘? đ?‘? + đ?‘?2 đ?‘›đ?‘?2 + đ?‘›đ?‘?2 = ( ) đ?‘?2đ?‘? 2 4 (đ?‘? 2 + đ?‘? 2 )(đ?‘? 4 + đ?‘? 4 − đ?‘? 2 đ?‘? 2 ) đ?‘? 6 + đ?‘? 6 = = .

4đ?‘? 2 đ?‘? 2 16đ?‘† 2 By đ?‘ đ?‘?2 + đ?‘ đ?‘?2 = đ?‘ đ?‘? đ?‘ đ?‘?2 , it follows that B-Neuberg’s and C-Neuberg’s circles are orthogonal.

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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.

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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 đ??´đ??ľđ??ś be a random triangle; we aim to construct the square inscribed in the triangle đ??´đ??ľđ??ś , having one side on đ??ľđ??ś.

Figure 1.

In order to do this, we construct a square đ??´, đ??ľ, đ??ś , đ??ˇ, with đ??´, ∈ (đ??´đ??ľ), đ??ľ, , đ??ś , ∈ (đ??ľđ??ś) (see Figure 1).

We trace the line đ??ľđ??ˇ , and we note with đ??ˇđ?‘Ž its intersection with (đ??´đ??ś) ; through đ??ˇđ?‘Ž we trace the

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Ion Patrascu, Florentin Smarandache

parallel đ??ˇđ?‘Ž đ??´đ?‘Ž to đ??ľđ??ś with đ??´đ?‘Ž ∈ (đ??´đ??ľ) and we project onto đ??ľđ??ś the points đ??´đ?‘Ž , đ??ˇđ?‘Ž in đ??ľđ?‘Ž respectively đ??śđ?‘Ž .

We affirm that the quadrilateral đ??´đ?‘Ž đ??ľđ?‘Ž đ??śđ?‘Ž đ??ˇđ?‘Ž is the required square.

Indeed, đ??´đ?‘Ž đ??ľđ?‘Ž đ??śđ?‘Ž đ??ˇđ?‘Ž is a square, because đ??ľđ??ˇđ?‘Ž đ??ľđ??ˇ,

=

đ??´đ?‘Ž đ??ˇđ?‘Ž đ??´ , đ??ˇ,

đ??ˇđ?‘Ž đ??śđ?‘Ž đ??ˇ, đ??ś ,

=

and, as đ??ˇ , đ??ś , = đ??´, đ??ˇ, , it follows that đ??´đ?‘Ž đ??ˇđ?‘Ž =

đ??ˇđ?‘Ž đ??śđ?‘Ž .

Definition.

It is called A-Lucas’s inner circle of the triangle đ??´đ??ľđ??ś the circle circumscribed to the triangle AAaDa. We will note with đ??żđ?‘Ž the center of the A-Lucas’s inner circle and with đ?‘™đ?‘Ž its radius.

Analogously, we define the B-Lucas’s inner circle and the C-Lucas’s inner circle of the triangle đ??´đ??ľđ??ś.

2. Calculation of the Radius of the A-Lucas Inner Circle We note đ??´đ?‘Ž đ??ˇđ?‘Ž = đ?‘Ľ, đ??ľđ??ś = đ?‘Ž; let â„Žđ?‘Ž be the height from đ??´ of the triangle đ??´đ??ľđ??ś.

The similarity of the triangles đ??´đ??´đ?‘Ž đ??ˇđ?‘Ž and đ??´đ??ľđ??ś leads to:

đ?‘Ľ đ?‘Ž

=

From

�� �

â„Žđ?‘Ž −đ?‘Ľ â„Žđ?‘Ž

=

, therefore đ?‘Ľ =

đ?‘Ľ đ?‘Ž

we obtain �� =

80

đ?‘Žâ„Žđ?‘Ž đ?‘Ž+â„Žđ?‘Ž đ?‘….â„Žđ?‘Ž đ?‘Ž+â„Žđ?‘Ž

. .

(1)

Complements to Classic Topics of Circles Geometry

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.

If in (1) we replace â„Žđ?‘Ž =

2� �

and we also keep into

consideration the formula đ?‘Žđ?‘?đ?‘? = 4đ?‘…đ?‘†, where đ?‘… is the radius of the circumscribed circle of the triangle đ??´đ??ľđ??ś and đ?‘† represents its area, we obtain: đ?‘™đ?‘Ž =

đ?‘… 1+

2đ?‘Žđ?‘… đ?‘?đ?‘?

[see Ref. 2].

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 đ??´đ??´đ?‘Ž đ??ˇđ?‘Ž and đ??´đ??ľđ??ś are homothetic through the homothetic center đ??´ and the rapport:

81

â„Žđ?‘Ž đ?‘Ž+â„Žđ?‘Ž

.

Ion Patrascu, Florentin Smarandache

Because

�� �

=

â„Žđ?‘Ž đ?‘Ž+â„Žđ?‘Ž

, it means that the A-Lucas’s inner

circle and the circle circumscribed to the triangle đ??´đ??ľđ??ś are inner tangents in đ??´.

Analogously, it follows that the B-Lucas’s and CLucas’s inner circles are inner tangents of the circle circumscribed to đ??´đ??ľđ??ś.

Figure 2.

We will prove that the A-Lucas’s and C-Lucas’s circles are exterior tangents by verifying đ??żđ?‘Ž đ??żđ?‘? = đ?‘™đ?‘Ž + đ?‘™đ?‘? .

(2) We have: đ?‘‚đ??żđ?‘Ž = đ?‘… − đ?‘™đ?‘Ž ; đ?‘‚đ??żđ?‘? = đ?‘… − đ?‘™đ?‘? and Ě‚ ) = 2đ??ľ đ?‘š(đ??´0đ??ś Ě‚ ) = 360° − 2đ??ľ). (if đ?‘š(đ??ľĚ‚) > 90° then đ?‘š(đ??´0đ??ś

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The theorem of the cosine applied to the triangle đ?‘‚đ??żđ?‘Ž đ??żđ?‘? implies, keeping into consideration (2), that: (đ?‘… − đ?‘™đ?‘Ž )2 + (đ?‘… − đ?‘™đ?‘Ž )2 − 2(đ?‘… − đ?‘™đ?‘Ž )(đ?‘… − đ?‘™đ?‘? )đ?‘?đ?‘œđ?‘ 2đ??ľ = = (đ?‘™đ?‘Ž + đ?‘™đ?‘? )2 .

Because đ?‘?đ?‘œđ?‘ 2đ??ľ = 1 − 2đ?‘ đ?‘–đ?‘›2 đ??ľ, it is found that (2) is equivalent to: đ?‘ đ?‘–đ?‘›2 đ??ľ =

đ?‘™đ?‘Ž đ?‘™đ?‘? (đ?‘…−đ?‘™đ?‘Ž )(đ?‘…−đ?‘™đ?‘? )

.

But we have: đ?‘™đ?‘Ž đ?‘™đ?‘? = đ?‘™đ?‘Ž + đ?‘™đ?‘? = đ?‘…đ?‘?(

đ?‘?

2đ?‘Žđ?‘…+đ?‘?đ?‘?

(3) đ?‘…2 đ?‘Žđ?‘? 2 đ?‘?

(2đ?‘Žđ?‘…+đ?‘?đ?‘?)(2đ?‘?đ?‘…+đ?‘Žđ?‘?) đ?‘Ž

+

2đ?‘?đ?‘…+đ?‘Žđ?‘?

,

).

By replacing in (3), we find that đ?‘ đ?‘–đ?‘› 2 đ??ľ = đ?‘?2 4đ?‘Ž2

â&#x;ş sin đ??ľ =

đ?‘?

2đ?‘…

đ?‘Žđ?‘? 2 đ?‘?

4đ?‘Žđ?‘?đ?‘…2

=

is true according to the sines theorem.

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 đ??´đ??ľđ??ś the circle constructed on the segment determined by the feet of the bisectors of angle đ??´ as diameter.

Remark.

Analogously, the B-Apollonius’s and CApollonius’s circles are defined. If đ??´đ??ľđ??ś is an isosceles triangle with đ??´đ??ľ = đ??´đ??ś then the A-Apollonius’s circle

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Ion Patrascu, Florentin Smarandache

isn’t defined for đ??´đ??ľđ??ś , and if đ??´đ??ľđ??ś 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 � from the plane of the triangle with the property:

đ?‘€đ??ľ đ?‘€đ??ś

đ?‘?

= . đ?‘?

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 đ??´đ??ľđ??ś form a fascicle of the third type.

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Proof.

Let {đ?‘‚đ??´ } = đ??żđ?‘? đ??żđ?‘? ∊ đ??ľđ??ś (see Figure 3).

Menelaus’s theorem applied to the triangle đ?‘‚đ??ľđ??ś implies that: đ?‘‚đ??´ đ??ľ đ??ż đ?‘? đ??ľ đ??ż đ?‘? đ?‘‚

.

.

đ?‘‚đ??´ đ??ś đ??ż đ?‘? đ?‘‚ đ??ż đ?‘? đ??ś

= 1,

so: đ?‘‚đ??´ đ??ľ

.

đ?‘™đ?‘?

đ?‘‚đ??´ đ??ś đ?‘…−đ?‘™đ?‘?

.

đ?‘…−đ?‘™đ?‘? đ?‘™đ?‘?

=1

and by replacing đ?‘™đ?‘? and đ?‘™đ?‘? , we find that: đ?‘‚đ??´ đ??ľ đ?‘‚đ??´ đ??ś

=

đ?‘?2 đ?‘?2

.

This relation shows that the point đ?‘‚đ??´ is the foot of the exterior symmedian from đ??´ of the triangle đ??´đ??ľđ??ś (so the tangent in đ??´ to the circumscribed circle), namely the center of the A-Apollonius’s circle.

Let đ?‘ 1 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 đ??´đ??ľđ??ś is the intersection đ?‘‡đ??´ of the tangents traced in đ??ľ and in đ??ś to the circle circumscribed to the triangle đ??´đ??ľđ??ś.

It follows that đ??ľđ?‘‡đ??´ = đ??śđ?‘‡đ??´ = đ?‘ 1 đ?‘‡đ??´ , so đ?‘ 1 belongs to the circle đ?’žđ??´ that has the center in đ?‘‡đ??´ and orthogonally cuts the circle circumscribed in đ??ľ and đ??ś. The radical axis of the B-Lucas’s and C-Lucas’s circles is đ?‘‡đ??´ đ?‘ 1 , and đ?‘‚đ??´ đ?‘ 1 is tangent in đ?‘ 1 to the circle đ?’žđ??´ . Considering the power of the point đ?‘‚đ??´ in relation to đ?’žđ??´ , we have: đ?‘‚đ??´ đ?‘ 1 2 = đ?‘‚đ??´ đ??ľ. đ?‘‚đ??´ đ??ś.

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Figure 3.

Also, đ?‘‚đ??´ đ?‘‚2 = đ?‘‚đ??´ đ??ľ ∙ đ?‘‚đ??´ đ??ś ; it thus follows that đ?‘‚đ??´ đ??´ = đ?‘‚đ??´ đ?‘ 1 , which proves that đ?‘ 1 belongs to the AApollonius’s circle and is the radical center of the AApollonius’s, B-Lucas’s and C-Lucas’s circles.

Remarks.

1.

If the triangle đ??´đ??ľđ??ś is right in đ??´ then đ??żđ?‘? đ??żđ?‘? ||đ??ľđ??ś, the radius of the A-Apollonius’s đ?‘Žđ?‘?đ?‘?

circle is equal to:

|đ?‘? 2 −đ?‘? 2 |

. The point đ?‘ 1 is

the foot of the bisector from đ??´. We find that đ?‘‚đ??´ đ?‘ 1 =

đ?‘Žđ?‘?đ?‘?

, so the theorem stands

|đ?‘? 2 −đ?‘? 2 |

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, đ?‘‚đ??´, so, to the radius of the A-Lucas’s circle also.

4th Definition.

The triangle đ?‘‡đ??´ đ?‘‡đ??ľ đ?‘‡đ??ś determined by the tangents traced in đ??´, đ??ľ, đ??ś to the circle circumscribed to the triangle đ??´đ??ľđ??ś is called the tangential triangle of the triangle đ??´đ??ľđ??ś.

1st Property.

The triangle đ??´đ??ľđ??ś and the Lucas’s triangle đ??żđ?‘Ž đ??żđ?‘? đ??żđ?‘? are homological.

Proof.

Obviously, đ??´đ??żđ?‘Ž , đ??ľđ??żđ?‘? , đ??śđ??żđ?‘? are concurrent in đ?‘‚ , therefore đ?‘‚, the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś, is the homology center.

We have seen that {đ?‘‚đ??´ } = đ??żđ?‘? đ??żđ?‘? ∊ đ??ľđ??ś and đ?‘‚đ??´ is the center of the A-Apollonius’s circle, therefore the homology axis is the Apollonius’s line đ?‘‚đ??´ đ?‘‚đ??ľ đ?‘‚đ??ś (the line determined by the centers of the Apollonius’s circle).

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2nd Property.

The tangential triangle and the Lucas’s triangle of the triangle đ??´đ??ľđ??ś are orthogonal triangles.

Proof.

The line đ?‘‡đ??´ đ?‘ 1 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 đ??żđ?‘? đ??żđ?‘? .

Analogously, đ?‘‡đ??ľ đ?‘ 2 is perpendicular on đ??żđ?‘? đ??żđ?‘Ž , because the radical axes of the Lucas’s circles are concurrent in đ??ż, which is the radical center of the Lucas’s circles; it follows that đ?‘‡đ??´ đ?‘‡đ??ľ đ?‘‡đ??ś and đ??żđ?‘Ž đ??żđ?‘? đ??żđ?‘? are orthological and đ??ż is the center of orthology. The other center of orthology is đ?‘‚ the center of the circle circumscribed to đ??´đ??ľđ??ś.

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.

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

this

article,

geometric

we

solve

constructions

problems

only

with

of the

ruler, using known theorems.

1st Problem.

Being given a triangle đ??´đ??ľđ??ś , its circumscribed circle (its center known) and a point đ?‘€ fixed on the circle, construct, using only the ruler, a transversal line đ??´1 , đ??ľ1 , đ??ś1 , with đ??´1 ∈ đ??ľđ??ś, đ??ľ1 ∈ đ??śđ??´, đ??ś1 ∈ đ??´đ??ľ, such that âˆ˘đ?‘€đ??´1 đ??ś ≥ âˆ˘đ?‘€đ??ľ1 đ??ś ≥ âˆ˘đ?‘€đ??ś1 đ??´ (the lines taken though đ?‘€ to generate congruent angles with the sides đ??ľđ??ś , đ??śđ??´ and đ??´đ??ľ, respectively).

2nd Problem.

Being given a triangle đ??´đ??ľđ??ś , its circumscribed circle (its center known) and đ??´1 , đ??ľ1 , đ??ś1 , such that đ??´1 ∈

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Ion Patrascu, Florentin Smarandache

đ??ľđ??ś, đ??ľ1 ∈ đ??śđ??´, đ??ś1 ∈ đ??´đ??ľ and đ??´1 , đ??ľ1 , đ??ś1 collinear, construct, using only the ruler, a point đ?‘€ on the circle circumscribing the triangle, such that the lines đ?‘€đ??´1 , đ?‘€đ??ľ1 , đ?‘€đ??ś1 to generate congruent angles with đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ, respectively.

3rd Problem.

Being given a triangle đ??´đ??ľđ??ś inscribed in a circle of given center and đ??´đ??´â€˛ a given cevian, đ??´â€˛ a point on the circle, construct, using only the ruler, the isogonal cevian đ??´đ??´1 to the cevian đ??´đ??´â€˛ .

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

1st Lemma.

(Generalized Simpson's Line)

If đ?‘€ is a point on the circle circumscribed to the triangle đ??´đ??ľđ??ś and we take the lines đ?‘€đ??´1 , đ?‘€đ??ľ1 , đ?‘€đ??ś1 which generate congruent angles ( đ??´1 ∈ đ??ľđ??ś, đ??ľ1 ∈ đ??śđ??´, đ??ś1 ∈ đ??´đ??ľ) with đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ respectively, then the points đ??´1 , đ??ľ1 , đ??ś1 are collinear.

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Proof.

Let đ?‘€ on the circle circumscribed to the triangle đ??´đ??ľđ??ś (see Figure 1), such that: âˆ˘đ?‘€đ??´1 đ??ś ≥ âˆ˘đ?‘€đ??ľ1 đ??ś ≥ âˆ˘đ?‘€đ??ś1 đ??´ = đ?œ‘.

(1)

Figure 1.

From the relation (1), we obtain that the quadrilateral đ?‘€đ??ľ1 đ??´1 đ??ś is inscriptible and, therefore: âˆ˘đ??´1 đ??ľđ??ś ≥ âˆ˘đ??´1 đ?‘€đ??ś.

(2).

Also from (1), we have that đ?‘€đ??ľ1 đ??´đ??ś1 is inscriptible, and so âˆ˘đ??´đ??ľ1 đ??ś1 ≥ âˆ˘đ??´đ?‘€đ??ś1 .

(3)

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Ion Patrascu, Florentin Smarandache

The quadrilateral MABC is inscribed, hence: âˆ˘đ?‘€đ??´đ??ś1 ≥ âˆ˘đ??ľđ??śđ?‘€.

(4) On the other hand, Ě‚ + đ?œ‘), âˆ˘đ??´1 đ?‘€đ??ś = 1800 − (đ??ľđ??śđ?‘€ Ě‚ âˆ˘đ??´đ?‘€đ??ś1 = 1800 − (đ?‘€đ??´đ??ś 1 + đ?œ‘).

The relation (4) drives us, together with the above relations, to: âˆ˘đ??´1 đ?‘€đ??ś ≥ âˆ˘đ??´đ?‘€đ??ś1 .

(5) Finally, using the relations (5), (2) and (3), we conclude that: âˆ˘đ??´1 đ??ľ1 đ??ś ≥ đ??´đ??ľ1 đ??ś1 , which justifies the collinearity of the points đ??´1 , đ??ľ1 , đ??ś1 .

Remark.

The Simson’s Line is obtained in the case when đ?œ‘ = 900 .

2nd Lemma.

If đ?‘€ is a point on the circle circumscribed to the triangle đ??´đ??ľđ??ś and đ??´1 , đ??ľ1 , đ??ś1 are points on đ??ľđ??ś , đ??śđ??´ and đ??´đ??ľ , respectively, such that âˆ˘đ?‘€đ??´1 đ??ś = âˆ˘đ?‘€đ??ľ1 đ??ś = âˆ˘đ?‘€đ??ś1 đ??´ = đ?œ‘, and đ?‘€đ??´1 intersects the circle a second time in đ??´â€™, then đ??´đ??´â€˛ âˆĽ đ??´1 đ??ľ1 .

Proof.

The quadrilateral đ?‘€đ??ľ1 đ??´1 đ??ś is inscriptible (see Figure 1); it follows that:

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âˆ˘đ??śđ?‘€đ??´â€˛ ≥ âˆ˘đ??´1 đ??ľ1 đ??ś.

(6) On the other hand, the quadrilateral đ?‘€đ??´đ??´â€˛ đ??ś is also inscriptible, hence: âˆ˘đ??śđ?‘€đ??´â€˛ ≥ âˆ˘đ??´â€˛ đ??´đ??ś.

(7) ′ The relations (6) and (7) imply: âˆ˘đ??´ đ?‘€đ??ś ≥ âˆ˘đ??´â€˛ đ??´đ??ś, which gives đ??´đ??´â€˛ âˆĽ đ??´1 đ??ľ1 .

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 đ?’ž(đ?‘‚, đ?‘…) , let be a diameter (đ??´đ??ľ)] and let đ?‘€ ∈ đ?’ž(đ?‘‚, đ?‘…). We construct the line đ??ľđ?‘€ (see Figure 2). We consider on this line the point đ??ˇ (đ?‘€ between đ??ˇ and đ??ľ ). We join đ??ˇ with đ?‘‚ , đ??´ with đ?‘€ and denote đ??ˇđ?‘‚ ∊ đ??´đ?‘€ = {đ?‘ƒ}.

We take đ??ľđ?‘ƒ and let {đ?‘ } = đ??ˇđ??´ ∊ đ??ľđ?‘ƒ. The line đ?‘€đ?‘ is parallel to đ??´đ??ľ.

Construction’s Proof.

In the triangle đ??ˇđ??´đ??ľ, the cevians đ??ˇđ?‘‚, đ??´đ?‘€ and đ??ľđ?‘ are concurrent.

Ceva’s Theorem provides:

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Ion Patrascu, Florentin Smarandache

đ?‘‚đ??´ đ?‘€đ??ľ

∙

đ?‘‚đ??ľ đ?‘€đ??ˇ

∙

đ?‘ đ??ˇ đ?‘ đ??´

= 1.

(8)

But đ??ˇđ?‘‚ is a median, đ??ˇđ?‘‚ = đ??ľđ?‘‚ = đ?‘….

From (8), we get

đ?‘€đ??ľ đ?‘€đ??ˇ

=

đ?‘ đ??´ đ?‘ đ??ˇ

, which, by Thales

reciprocal, gives đ?‘€đ?‘ âˆĽ đ??´đ??ľ.

Figure 2.

Remark.

If we have a circle with given center and a certain line đ?‘‘, we can construct though a given point đ?‘€ a parallel to that line in such way: we take two diameters [đ?‘…đ?‘†] and [đ?‘ˆđ?‘‰ ] through the center of the given circle (see Figure 3).

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Figure 3.

We denote đ?‘…đ?‘† ∊ đ?‘‘ = {đ?‘ƒ}; because [đ?‘…đ?‘‚] ≥ [đ?‘†đ?‘‚], we can construct, applying the 3rd Lemma, the parallels through đ?‘ˆ and đ?‘‰ to đ?‘…đ?‘† which intersect đ?‘‘ in đ??ž and đ??ż, respectively. Since we have on the line đ?‘‘ the points đ??ž, đ?‘ƒ, đ??ż , such that [đ??žđ?‘ƒ] ≥ [đ?‘ƒđ??ż] , we can construct the parallel through đ?‘€ to đ?‘‘ based on the construction from 3rd Lemma.

1st Theorem.

(P. Aubert – 1899)

If, through the vertices of the triangle đ??´đ??ľđ??ś, we take three lines parallel to each other, which intersect the circumscribed circle in đ??´â€˛ , đ??ľâ€˛ and đ??ś ′ , and đ?‘€ is a

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point on the circumscribed circle, as well đ?‘€đ??´â€˛ ∊ đ??ľđ??ś = {đ??´1 } , đ?‘€đ??ľâ€˛ ∊ đ??śđ??´ = {đ??ľ1 } , đ?‘€đ??ś ′ ∊ đ??´đ??ľ = {đ??ś1 } , then đ??´1 , đ??ľ1 , đ??ś1 are collinear and their line is parallel to đ??´đ??´â€˛ .

Proof.

The point of the proof is to show that đ?‘€đ??´1 , đ?‘€đ??ľ1 , đ?‘€đ??ś1 generate congruent angles with đ??ľđ??ś , đ??śđ??´ and đ??´đ??ľ, respectively.

1 ̆′ Ě‚ ̆ đ?‘š(đ?‘€đ??´ (9) 1 đ??ś ) = [đ?‘š(đ?‘€đ??ś ) + đ?‘š(đ??ľđ??´ )] 2

1 ̆′ Ě‚ ̆ đ?‘š(đ?‘€đ??ľ 1 đ??ś ) = [đ?‘š(đ?‘€đ??ś ) + đ?‘š(đ??´đ??ľ )] 2 ′

(10)

̆′ ) = đ?‘š(đ??´đ??ľ ̆ ′ ) , hence, But đ??´đ??´â€˛ âˆĽ đ??ľđ??ľ implies đ?‘š(đ??ľđ??´ from (9) and (10), it follows that: âˆ˘đ?‘€đ??´1 đ??ś ≥ âˆ˘đ?‘€đ??ľ1 đ??ś, 1 ̆′ Ě‚ ̆ đ?‘š(đ?‘€đ??ś 1 đ??´) = [đ?‘š(đ??ľđ?‘€ ) − đ?‘š(đ??´đ??ś )].

2 ′

(11) (12)

′ đ??ś ); by ̆′ ) = đ?‘š(đ??´Ě† But đ??´đ??´ âˆĽ đ??śđ??ś implies that đ?‘š(đ??´đ??ś ′

returning to (12), we have that: 1 ̆′ Ě‚ ̆ đ?‘š(đ?‘€đ??ś 1 đ??´) = [đ?‘š(đ??ľđ?‘€ ) − đ?‘š(đ??´đ??ś )] = 2 1 ̆ ) + đ?‘š(đ?‘€đ??ś ̆ )]. = [đ?‘š(đ??ľđ??´â€˛ 2

(13)

The relations (9) and (13) show that: âˆ˘đ?‘€đ??´1 đ??ś ≥ âˆ˘đ?‘€đ??ś1 đ??´.

(14) From (11) and (14), we obtain: âˆ˘đ?‘€đ??´1 đ??ś ≥ âˆ˘đ?‘€đ??ľ1 đ??ś ≥ âˆ˘đ?‘€đ??ś1 đ??´ , which, by 1st Lemma, verifies the collinearity of points đ??´1 , đ??ľ1 , đ??ś1 . Now, applying the 2nd Lemma, we obtain the parallelism of lines đ??´đ??´â€˛ and đ??´1 đ??ľ1 .

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Figure 4.

2nd Theorem.

(M’Kensie – 1887)

If đ??´1 đ??ľ1 đ??ś1 is a transversal line in the triangle đ??´đ??ľđ??ś (đ??´1 ∈ đ??ľđ??ś, đ??ľ1 ∈ đ??śđ??´, đ??ś1 ∈ đ??´đ??ľ), and through the triangle’s vertices we take the chords đ??´đ??´â€˛ , đ??ľđ??ľâ€˛ , đ??śđ??ś ′ of a circle circumscribed to the triangle, parallels with the transversal line, then the lines đ??´đ??´â€˛ , đ??ľđ??ľâ€˛ , đ??śđ??ś ′ are concurrent on the circumscribed circle.

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Proof.

We denote by đ?‘€ the intersection of the line đ??´1 đ??´â€˛ with the circumscribed circle (see Figure 5) and with đ??ľ1′ , respectively đ??ś1′ the intersection of the line đ?‘€đ??ľâ€˛ with đ??´đ??ś and of the line đ?‘€đ??ś ′ with đ??´đ??ľ.

Figure 5.

According to the P. Aubert’s theorem, we have that the points đ??´1 , đ??ľ1′ , đ??ś1′ are collinear and that the line đ??´1 đ??ľ1′ is parallel to đ??´đ??´â€˛ .

From hypothesis, we have that đ??´1 đ??ľ1 âˆĽ đ??´đ??´â€˛ ; from the uniqueness of the parallel taken through đ??´1 to đ??´đ??´â€˛ , it follows that đ??´1 đ??ľ1 ≥ đ??´1 đ??ľ1′ , therefore đ??ľ1′ = đ??ľ1 , and analogously đ??ś1′ = đ??ś1 .

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Remark.

We have that: đ?‘€đ??´1 , đ?‘€đ??ľ1 , đ?‘€đ??ś1 generate congruent angles with đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ, 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 đ??´đ??´â€˛ , đ??ľđ??ľâ€˛ , đ??śđ??ś ′ the three parallel lines with a certain direction (see Figure 6).

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Figure 6.

To construct the isogonal of the cevian đ??´đ??´â€˛ , we take đ??´â€˛ đ?‘€ âˆĽ đ??ľđ??ś, đ?‘€ belonging to the circle circumscribed ̆ ′ ≥ đ??śđ?‘€ ̆ , it follows that đ??´đ?‘€ to the triangle, having đ??ľđ??´ will be the isogonal of the cevian đ??´đ??´â€˛ . (Indeed, from ̆ ′ ≥ đ??śđ?‘€ ̆ it follows that âˆ˘đ??ľđ??´đ??´â€˛ ≥ âˆ˘đ??śđ??´đ?‘€.) đ??ľđ??´ ̆ ≥ On the other hand, đ??ľđ??ľâ€˛ âˆĽ A đ??´â€˛ implies đ??ľđ??´â€˛ ̆ ′ ≥ đ??śđ?‘€ ̆ ′ ≥ đ??śđ?‘€ ̆ ′ , and since đ??ľđ??´ ̆ we have that đ??´đ??ľ ̆ , đ??´đ??ľ which shows that the isogonal of the parallel đ??ľđ??ľâ€˛ is đ??ľđ?‘€. From đ??śđ??ś ′ âˆĽ đ??´đ??´â€˛ , it follows that đ??´â€˛ đ??ś ≥ đ??´đ??ś ′ , having âˆ˘đ??ľâ€˛ đ??śđ?‘€ ≥ âˆ˘đ??´đ??śđ??ś ′ , therefore the isogonal of the parallel đ??śđ??ś ′ is đ??śđ?‘€â€˛ .

Reciprocally.

If đ??´đ?‘€, đ??ľđ?‘€, đ??śđ?‘€ are concurrent cevians in đ?‘€, the point on the circle circumscribed to the triangle đ??´đ??ľđ??ś, let us prove that their isogonals are parallel lines. To construct an isogonal of đ??´đ?‘€ , we take đ?‘€đ??´â€˛ âˆĽ đ??ľđ??ś , đ??´â€˛ ̆ ≥ belonging to the circumscribed circle. We have đ?‘€đ??ś ′ ′ ̆ . Constructing the isogonal đ??ľđ??ľ of đ??ľđ?‘€, with đ??ľâ€˛ on đ??ľđ??´ ̆ ′ , it ̆ ≥ đ??´đ??ľ the circumscribed circle, we will have đ??śđ?‘€ ̆ ′ ≥ đ??´đ??ľ ̆ ′ and, consequently, âˆ˘đ??´đ??ľđ??ľâ€˛ ≥ follows that đ??ľđ??´ âˆ˘đ??ľđ??´đ??´â€˛ , which shows that đ??´đ??´â€˛ âˆĽ đ??ľđ??ľâ€˛ . Analogously, we show that đ??śđ??ś ′ âˆĽ đ??´đ??´â€˛ .

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 đ??´đ??´â€˛ , đ??ľđ??ľâ€˛ , đ??śđ??ś ′ with a certain directions of a diameter of the circle circumscribed to the given triangle.

We join đ?‘€ with đ??´â€˛ , đ??ľâ€˛ , đ??ś ′ and denote the intersection between đ?‘€đ??´â€˛ and đ??ľđ??ś, đ??´1 ; đ?‘€đ??ľâ€˛ ∊ đ??śđ??´ = {đ??ľ1 } and đ?‘€đ??´â€˛ ∊ đ??´đ?‘‰ = {đ??ś1 }.

According to the Aubert’s Theorem, the points đ??´1 , đ??ľ1 , đ??ś1 will be collinear, and đ?‘€đ??´â€˛ , đ?‘€đ??ľâ€˛ , đ?‘€đ??ś ′ generate congruent angles with đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ, respectively.

Solution to the 2nd problem.

Using the 3rd Lemma and the remark that follows it, we construct through đ??´, đ??ľ, đ??ś the parallels to đ??´1 đ??ľ1 ; we denote by đ??´â€˛ , đ??ľâ€˛ , đ??ś ′ their intersections with the circle circumscribed to the triangle đ??´đ??ľđ??ś. (It is enough to build a single parallel to the transversal line đ??´1 đ??ľ1 đ??ś1 , for example đ??´đ??´â€˛ ).

We join đ??´â€˛ with đ??´1 and denote by đ?‘€ the intersection with the circle. The point đ?‘€ 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 đ??´â€˛ belongs to the little arc ̆ in the circle determined by the chord đ??ľđ??ś circumscribed to the triangle đ??´đ??ľđ??ś.

In this case, in order to find the isogonal đ??´đ??´1 , we construct (by help of the 3rd Lemma and of the remark that follows it) the parallel đ??´â€˛ đ??´1 to đ??ľđ??ś, đ??´1 being on the circumscribed circle, it is obvious that đ??´đ??´â€˛ and đ??´đ??´1 will be isogonal cevians.

We suppose that đ??´â€˛ belongs to the high arc ̆ ; we consider đ??´â€˛ ∈ đ??´đ??ľ ̆ (the determined by the chord đ??ľđ??ś ̆ does not contain the point đ??ś). In this situation, arc đ??´đ??ľ we firstly construct the parallel đ??ľđ?‘ƒ to đ??´đ??´â€˛ , đ?‘ƒ belongs to the circumscribed circle, and then through đ?‘ƒ we construct the parallel đ?‘ƒđ??´1 to đ??´đ??ś , đ??´1 belongs to the circumscribed circle. The isogonal of the line đ??´đ??´â€˛ will be đ??´đ??´1 . The construction’s proof follows from 3rd Lemma and from the proof of Beltrami’s Theorem.

References.

[1] F. G. M.: Exercices de GĂŠometrie. VIII-e ed., Paris, VI-e Librairie Vuibert, Rue de Vaugirard,77.

[2] T. Lalesco: La GĂŠomĂŠtrie 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 k t h rank.

1st Definition.

It is called an internal cevian of kth rank the line đ??´đ??´đ?‘˜ where đ??´đ?‘˜ ∈ (đ??ľđ??ś), such that

đ??ľđ??´ đ??´đ?‘˜ đ??ś

đ??´đ??ľ đ?‘˜

= ( ) (đ?‘˜ ∈ â„?). đ??´đ??ś

đ??´â€˛đ?‘˜

If is the harmonic conjugate of the point đ??´đ?‘˜ in relation to đ??ľ and đ??ś, we call the line đ??´đ??´â€˛đ?‘˜ an external cevian of kth rank.

2nd Definition.

We call Apollonius’s circle of kth rank with respect to the side đ??ľđ??ś of đ??´đ??ľđ??ś triangle the circle which has as diameter the segment line đ??´đ?‘˜ đ??´â€˛đ?‘˜ .

1st Theorem.

Apollonius’s circle of kth rank is the locus of points đ?‘€ from đ??´đ??ľđ??ś triangle's plan, satisfying the relation:

đ?‘€đ??ľ đ?‘€đ??ś

đ??´đ??ľ đ?‘˜

=( ) . đ??´đ??ś

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Proof.

Let đ?‘‚đ??´đ?‘˜ the center of the Apollonius’s circle of kth rank relative to the side đ??ľđ??ś of đ??´đ??ľđ??ś triangle (see Figure 1) and đ?‘ˆ, đ?‘‰ the points of intersection of this circle with the circle circumscribed to the triangle đ??´đ??ľđ??ś.

We denote by đ??ˇ the middle of arc đ??ľđ??ś, and we extend đ??ˇđ??´đ?‘˜ to intersect the circle circumscribed in đ?‘ˆ ′ .

In đ??ľđ?‘ˆ ′ đ??ś triangle, đ?‘ˆ ′ đ??ˇ is bisector; it follows that đ??ľđ??´đ?‘˜ đ??´đ?‘˜ đ??ś

=

đ?‘ˆâ€˛đ??ľ đ?‘ˆâ€˛đ??ś

đ??´đ??ľ đ?‘˜

= ( ) , so đ?‘ˆ ′ belongs to the locus. đ??´đ??ś

The perpendicular in đ?‘ˆ ′ on đ?‘ˆ ′ đ??´đ?‘˜ intersects BC on đ??´â€˛â€˛ đ?‘˜ , which is the foot of the đ??ľđ?‘ˆđ??ś triangle's outer bisector, so the harmonic conjugate of đ??´đ?‘˜ in relation ′ to đ??ľ and đ??ś, thus đ??´â€˛â€˛ đ?‘˜ = đ??´đ?‘˜ .

Therefore, đ?‘ˆ ′ is on the Apollonius’s circle of rank đ?‘˜ relative to the side đ??ľđ??ś, hence đ?‘ˆ ′ = đ?‘ˆ.

Figure 3

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Let đ?‘€ a point that satisfies the relation from the statement; thus

đ?‘€đ??ľ đ?‘€đ??ś

=

đ??ľđ??´đ?‘˜ đ??´đ?‘˜ đ??ś

; it follows – by using the

reciprocal of bisector's theorem – that đ?‘€đ??´đ?‘˜ is the internal bisector of angle đ??ľđ?‘€đ??ś. Now let us proceed as before, taking the external bisector; it follows that đ?‘€ belongs to the Apollonius’s circle of center đ?‘‚đ??´đ?‘˜ . We consider now a point đ?‘€ on this circle, and we construct đ??ś ′ such that âˆ˘đ??ľđ?‘ đ??´đ?‘˜ ≥ âˆ˘đ??´đ?‘˜ đ?‘ đ??ś ′ (thus (đ?‘ đ??´đ?‘˜ is ̂′ ). Because the internal bisector of the angle đ??ľđ?‘ đ??ś đ??´â€˛đ?‘˜ đ?‘ ⊼ đ?‘ đ??´đ?‘˜ , it follows that đ??´đ?‘˜ and đ??´â€˛đ?‘˜ are harmonically conjugated with respect to đ??ľ and đ??ś ′ . On the other hand, the same points are harmonically conjugated with respect to đ??ľ and đ??ś; from here, it follows that đ??ś ′ = đ??ś, and we have

đ?‘ đ??ľ đ?‘ đ??ś

=

đ??ľđ??´đ?‘˜ đ??´đ?‘˜ đ??ś

đ??´đ??ľ đ?‘˜

=( ) . đ??´đ??ś

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 đ??´đ??ľđ??śđ??ˇđ??¸đ??š a given complete quadrilateral (see Figure 2). We denote by đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 respectively the orthocenters of đ??´đ??ľđ??š, đ??´đ??ˇđ??¸, đ??śđ??ľđ??¸, đ??śđ??ˇđ??š triangles, and let đ??´1 , đ??ľ1 , đ??š1 the feet of the heights of đ??´đ??ľđ??š triangle.

Figure 4

As previously shown, the following relations occur: đ??ť1 đ??´. đ??ť1 đ??´1 − đ??ť1 đ??ľ. đ??ť1 đ??ľ1 = đ??ť1 đ??š. đ??ť1 đ??š1 ; they express that the point đ??ť1 has equal powers to the circles of diameters đ??´đ??ś, đ??ľđ??ˇ, đ??¸đ??š , because those circles contain respectively the points đ??´1 , đ??ľ1 , đ??š1 , and đ??ť1 is an internal point.

It is shown analogously that the points đ??ť2 , đ??ť3 , đ??ť4 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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such their centers – which are the middles of the complete quadrilateral's diagonals – are collinear.

The line formed by the middles of a complete quadrilateral's diagonals is called Gauss’s line or Gauss-Newton’s line.

3rd Theorem.

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

Proof.

Let đ??´đ??´đ?‘˜ , đ??ľđ??ľđ?‘˜ , đ??śđ??śđ?‘˜ be concurrent cevians of kth rank and đ??´đ??´â€˛đ?‘˜ , đ??ľđ??ľđ?‘˜â€˛ , đ??śđ??śđ?‘˜â€˛ be the external cevians of kth rank (see Figure 3). The figure đ??ľđ?‘˜â€˛ đ??śđ?‘˜ đ??ľđ?‘˜ đ??śđ?‘˜â€˛ đ??´đ?‘˜ đ??´â€˛đ?‘˜ 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 đ?‘‚ to đ??ˇ and đ?‘ˆ (see Figure 1), đ?‘‚đ??ˇ ⊼ đ??ľđ??ś ′ ′ 0 Ě‚ Ě‚ and đ?‘š(đ??´Ě‚ đ?‘˜ đ?‘ˆđ??´ ) = 90 , it follows that đ?‘ˆđ??´ đ??´đ?‘˜ = đ?‘‚đ??ˇđ??´đ?‘˜ = đ?‘˜

đ?‘˜

Ě‚đ?‘˜ . đ?‘‚đ?‘ˆđ??´ ′ Ě‚ Ě‚ The congruence đ?‘ˆđ??´ đ?‘˜ đ??´đ?‘˜ ≥ đ?‘‚đ?‘ˆđ??´đ?‘˜ shows that đ?‘‚đ?‘ˆ is tangent to the Apollonius’s circle of center đ?‘‚đ??´đ?‘˜ .

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 đ?‘‚ to the line đ?‘‚đ??´đ?‘˜ đ?‘‚đ??ľđ?‘˜ .

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 đ?‘‚đ?‘ˆ ⊼ đ?‘ˆđ?‘‚đ??´đ?‘˜ , so đ?‘ˆđ?‘‚đ??´đ?‘˜ is an external cevian of rank 2 for đ??ľđ??śđ?‘ˆ triangle, thus an external symmedian. Henceforth, đ?‘‚đ??´đ?‘˜ đ??ľ đ?‘‚đ??´đ?‘˜ đ??ś

=(

đ??ľđ?‘ˆ 2

đ??´đ??ľ 2đ?‘˜

đ??śđ?‘ˆ

đ??´đ??ś

) =(

)

(the last equality occurs because

đ?‘ˆ belong to the Apollonius’s circle of rank đ?‘˜ associated to the vertex đ??´).

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 đ?‘ˆ and đ?‘‰ points of intersection of the Apollonius’s circle of center đ?‘‚đ??´đ?‘˜ with the circle circumscribed to the đ??´đ??ľđ??ś (see Figure 1). We take from đ?‘ˆ and đ?‘‰ the perpendiculars đ?‘ˆđ?‘ˆ1 , đ?‘ˆđ?‘ˆ2 and đ?‘‰đ?‘‰1 , đ?‘‰đ?‘‰2 on đ??´đ??ľ and đ??´đ??ś respectively. The quadrilaterals đ??´đ??ľđ?‘‰đ??ś , đ??´đ??ľđ??śđ?‘ˆ are inscribed, it follows the similarity of triangles đ??ľđ?‘‰đ?‘‰1 , đ??śđ?‘‰đ?‘‰2 and đ??ľđ?‘ˆđ?‘ˆ1 , đ??śđ?‘ˆđ?‘ˆ2 , from where we get the relations: đ??ľđ?‘‰ đ?‘‰đ?‘‰1 đ?‘ˆđ??ľ đ?‘ˆđ?‘ˆ1 = , = . đ??śđ?‘‰ đ?‘‰đ?‘‰2 đ?‘ˆđ??ś đ?‘ˆđ?‘ˆ2

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But

đ??ľđ?‘‰ đ??śđ?‘‰

=(

đ??´đ??ľ đ?‘˜ đ?‘ˆđ??ľ đ??´đ??ś

) ,

đ?‘ˆđ??ś

đ??´đ??ľ đ?‘˜ đ?‘‰đ?‘‰1

=( ) , đ??´đ??ś

��2

đ??´đ??ľ đ?‘˜

đ?‘ˆđ?‘ˆ1

đ??´đ??ś

đ?‘ˆđ?‘ˆ2

= ( ) and

=

đ??´đ??ľ đ?‘˜

( ) , relations that show that đ?‘‰ and đ?‘ˆ belong đ??´đ??ś

respectively to the internal cevian and the external cevian of rank đ?‘˜ + 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 đ?‘Šđ?‘˜ , đ?‘Šđ?‘˜â€˛ ).

1st Property.

If đ?‘Šđ?‘˜ , đ?‘Šđ?‘˜â€˛ are isodynamic centers of kth rank, then: đ?‘Šđ?‘˜ đ??´. đ??ľđ??ś đ?‘˜ = đ?‘Šđ?‘˜ đ??ľ. đ??´đ??ś đ?‘˜ = đ?‘Šđ?‘˜ đ??ś. đ??´đ??ľđ?‘˜ ; đ?‘Šđ?‘˜â€˛ đ??´. đ??ľđ??ś đ?‘˜ = đ?‘Šđ?‘˜â€˛ đ??ľ. đ??´đ??ś đ?‘˜ = đ?‘Šđ?‘˜â€˛ đ??ś. đ??´đ??ľđ?‘˜ .

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 đ?‘˜ = 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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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 đ??´ of the triangle đ??´đ??ľđ??ś the circle constructed on the segment determined on the simedians’ feet from đ??´ on đ??ľđ??ś as diameter.

1st Theorem.

The Apollonius’s circle of second rank relative to the vertex đ??´ of the triangle đ??´đ??ľđ??ś intersect the circumscribed circle of the triangle đ??´đ??ľđ??ś 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 đ??´ of the triangle đ??´đ??ľđ??ś intersects the circumscribed circle in two points đ?‘„ and đ?‘ƒ (đ?‘„ on the same side of đ??ľđ??ś as đ??´). Then, (đ?‘„đ?‘† is a bisector in the triangle đ?‘„đ??ľđ??ś, S is the simedian’s foot from đ??´ of the triangle đ??´đ??ľđ??ś.

Proof. đ?‘„ belongs to the Apollonius’s circle of second rank, therefore: đ?‘„đ??ľ đ?‘„đ??ś

=(

đ??´đ??ľ 2 đ??´đ??ś

) .

(1)

Figure 1.

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

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đ?‘†đ??ľ đ?‘†đ??ś

đ??´đ??ľ 2

=( ) .

(2)

đ??´đ??ś

From relations (1) and (2), we note that đ?‘„đ??ľ đ?‘„đ??ś

=

đ?‘†đ??ľ đ?‘†đ??ś

,

a relation showing that đ?‘„đ?‘† is bisector in the triangle đ?‘„đ??ľđ??ś.

Remarks.

1.

2.

The Apollonius’s circle of second relative to the vertex đ??´ of the triangle đ??´đ??ľđ??ś (see Figure 1) is an Apollonius’s circle for the triangle đ?‘„đ??ľđ??ś. Indeed, we proved that đ?‘„đ?‘† is an internal bisector in the triangle đ?‘„đ??ľđ??ś, and since đ?‘†â€˛, the external simedian’s foot of the triangle đ??´đ??ľđ??ś , belongs to the Apollonius’s Circle of second rank, we have đ?‘š(âˆ˘đ?‘†â€˛đ?‘„đ?‘†) = 900 , therefore đ?‘„đ?‘†â€™ is an external bisector in the triangle đ?‘„đ??ľđ??ś. đ?‘„đ?‘ƒ is a simedian in đ?‘„đ??ľđ??ś . Indeed, the Apollonius’s circle of second rank, being an Apollonius’s circle for đ?‘„đ??ľđ??ś, intersects the circle circum-scribed to đ?‘„đ??ľđ??ś after đ?‘„đ?‘ƒ, 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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one of them is tangent to the triangle’s side. We denote (đ??ľđ??´Ě…) the adjoint circle that passes through đ??ľ and đ??´, and is tangent to the side đ??´đ??ś in đ??´.

About the circles (đ??ľđ??´Ě…) and (đ??śđ??´Ě…), we say that they are adjoint to the vertex đ??´ of the triangle đ??´đ??ľđ??ś.

3rd Definition.

It is called the second Brocard’s triangle the triangle đ??´2 đ??ľ2 đ??ś2 whose vertices are the projections of the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś on triangle’s simedians.

2nd Proposition.

The Apollonius’s circle of second rank relative to the vertex đ??´ of triangle đ??´đ??ľđ??ś and the adjoint circles relative to the same vertex đ??´ intersect in vertex đ??´2 of the second Brocard’s triangle.

Proof.

It is known that the adjoint circles (đ??ľđ??´Ě…) and (đ??śđ??´Ě…) intersect in a point belonging to the simedian đ??´đ?‘†; we denote this point đ??´2 (see [3]).

We have: âˆ˘đ??ľđ??´2 đ?‘† = âˆ˘đ??´2 đ??ľđ??´ + âˆ˘đ??´2 đ??´đ??ľ, but: âˆ˘đ??´2 đ??ľđ??´ ≥ âˆ˘đ??ľđ??´2 đ?‘† = âˆ˘đ??´2 đ??´đ??ľ + đ??´2 đ??´đ??ś = âˆ˘đ??´.

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Analogously, âˆ˘đ??śđ??´2 đ?‘† = âˆ˘đ??´, therefore (đ??´2 đ?‘† is the bisector of the angle đ??ľđ??´2 đ??ś. The bisector’s theorem in this triangle leads to: đ?‘†đ??ľ đ?‘†đ??ś

=

đ??ľđ??´2 đ??śđ??´2

,

but: đ?‘†đ??ľ đ?‘†đ??ś

đ??´đ??ľ 2

=( ) , đ??´đ??ś

consequently: đ??ľđ??´2 đ??śđ??´2

đ??´đ??ľ 2

=( ) , đ??´đ??ś

so đ??´2 is a point that belongs to the Apollonius’s circle of second rank.

Figure 2.

We prove that đ??´2 is a vertex in the second Brocard’s triangle, i.e. đ?‘‚đ??´2 ⊼ đ??´đ?‘†, O the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś.

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Ě‚ We pointed (see Figure 2) that đ?‘š(đ??ľđ??´ 2 đ??ś ) = 2đ??´, if Ě‚ ) = 2đ??´ , âˆ˘đ??´ is an acute angle, then also đ?‘š(đ??ľđ?‘‚đ??ś therefore the quadrilateral đ?‘‚đ??śđ??´2 đ??ľ is inscriptible. Ě‚ ) = 900 − đ?‘š(đ??´Ě‚) , it follows that Because đ?‘š(đ?‘‚đ??śđ??ľ 0 Ě‚ Ě‚ đ?‘š(đ??ľđ??´ 2 đ?‘‚ ) = 90 + đ?‘š(đ??´).

0 Ě‚ Ě‚ On the other hand, đ?‘š(đ??´đ??´ 2 đ??ľ ) = 180 − đ?‘š(đ??´), so 0 Ě‚ Ě‚ đ?‘š(đ??ľđ??´ 2 đ?‘‚ ) + đ?‘š(đ??´đ??´2 đ??ľ ) = 270 and, consequently, đ?‘‚đ??´2 ⊼

đ??´đ?‘†.

Remarks.

1.

2.

3.

If đ?&#x2018;&#x161;(đ??´Ě&#x201A;) < 900 , then four remarkable circles pass through đ??´2 : the two circles adjoint to the vertex đ??´ of the triangle đ??´đ??ľđ??ś, the circle circumscribed to the triangle đ??ľđ?&#x2018;&#x201A;đ??ś (where đ?&#x2018;&#x201A; is the center of the circumscribed circle) and the Apolloniusâ&#x20AC;&#x2122;s circle of second rank corresponding to the vertex đ??´.

The vertex đ??´2 of the second Brocardâ&#x20AC;&#x2122;s triangle is the middle of the chord of the circle circumscribed to the triangle đ??´đ??ľđ??ś containing the simedian đ??´đ?&#x2018;&#x2020;.

The points đ?&#x2018;&#x201A; , đ??´2 and đ?&#x2018;&#x2020;â&#x20AC;&#x2122; (the foot of the external simedian to đ??´đ??ľđ??ś) are collinear.

Indeed, we proved that đ?&#x2018;&#x201A;đ??´2 â&#x160;Ľ đ??´đ?&#x2018;&#x2020;; on the other hand, we proved that (đ??´2 đ?&#x2018;&#x2020; is an internal bisector in the triangle đ??ľđ??´2 đ??ś, and

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since đ?&#x2018;&#x2020;â&#x20AC;˛đ??´2 â&#x160;Ľ đ??´đ?&#x2018;&#x2020;, the outlined collinearity follows from the uniqueness of the perpendicular in đ??´2 on đ??´đ?&#x2018;&#x2020;.

Open Problem.

The Apolloniusâ&#x20AC;&#x2122;s circle of second rank relative to the vertex đ??´ of the triangle đ??´đ??ľđ??ś intersects the circle circumscribed to the triangle đ??´đ??ľđ??ś in two points đ?&#x2018;&#x192; and đ?&#x2018;&#x201E; (đ?&#x2018;&#x192; and đ??´ apart of đ??ľđ??ś).

We denote by đ?&#x2018;&#x2039; the second point of intersection between the line đ??´đ?&#x2018;&#x192; and the Apolloniusâ&#x20AC;&#x2122;s circle of second rank.

What can we say about đ?&#x2018;&#x2039;?

Is đ?&#x2018;&#x2039; a remarkable point in triangleâ&#x20AC;&#x2122;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 triangleâ&#x20AC;&#x2122;s vertex are called isogonal cevians if they create congruent angles with triangleâ&#x20AC;&#x2122;s bisector taken through the same vertex.

2nd Remark.

In the Figure 1 we represented the bisector đ??´đ??ˇ and the isogonal cevians đ??´đ??´1 and đ??´đ??´2 . The following relations take place: Ě&#x201A; đ??´Ě&#x201A; 1 đ??´đ??ˇ â&#x2030;Ą đ??´2 đ??´đ??ˇ ; Ě&#x201A;1 â&#x2030;Ą đ??śđ??´đ??´ Ě&#x201A;2 . đ??ľđ??´đ??´

1st Proposition.

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

Proof.

Let đ??´đ??ľđ??ś an acute-angled triangle with the height đ??´đ??´â&#x20AC;˛ and the radius đ??´đ?&#x2018;&#x201A; (see Figure 2).

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Figure 2.

The angle đ??´đ?&#x2018;&#x201A;đ??ś is a central angle, and the angle Ě&#x201A; = 2đ??´đ??ľđ??ś Ě&#x201A; . It follows đ??´đ??ľđ??ś is an inscribed angle, so đ??´đ?&#x2018;&#x201A;đ??ś Ě&#x201A; = 900 â&#x2C6;&#x2019; đ??ľĚ&#x201A; . that đ??´đ?&#x2018;&#x201A;đ??ś Ě&#x201A; = 900 â&#x2C6;&#x2019; đ??ľĚ&#x201A; , so đ??´đ??´â&#x20AC;˛ and On the other hand, đ??ľđ??´đ??´â&#x20AC;˛ đ??´đ?&#x2018;&#x201A; are isogonal cevians.

The theorem can be analogously proved for the obtuse triangle.

3rd Remark.

One can easily prove that in a triangle, if đ??´đ?&#x2018;&#x201A; is circumscribed circleâ&#x20AC;&#x2122;s radius, its isogonal cevian is the triangleâ&#x20AC;&#x2122;s height from vertex đ??´.

3rd Definition.

Two points đ?&#x2018;&#x192;1 , đ?&#x2018;&#x192;2 in the plane of triangle đ??´đ??ľđ??ś are called isogonals (isogonal conjugated) if the ceviansâ&#x20AC;&#x2122; pairs (đ??´đ?&#x2018;&#x192;1 , đ??´đ?&#x2018;&#x192;2 ), (đ??ľđ?&#x2018;&#x192;1 , đ??ľđ?&#x2018;&#x192;2 ), (đ??śđ?&#x2018;&#x192;1 , đ??śđ?&#x2018;&#x192;2 ), are composed of isogonal cevians.

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4th Remark.

In a triangle, the orthocenter and circumscribed circleâ&#x20AC;&#x2122;s center are isogonal points.

1st Theorem.

(The 6 points circle)

If đ?&#x2018;&#x192;1 and đ?&#x2018;&#x192;2 are two isogonal points in triangle đ??´đ??ľđ??ś , and đ??´1 , đ??ľ1 , đ??ś1 respectively đ??´2 , đ??ľ2 , đ??ś2 their projections on the sides đ??ľđ??ś , đ??śđ??´ and đ??´đ??ľ of triangle, then the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 concyclical.

the are the are

Proof.

The mediator of segment [đ??´1 đ??´2 ] passes through the middle đ?&#x2018;&#x192; of segment [đ?&#x2018;&#x192;1 , đ?&#x2018;&#x192;2 ] because the trapezoid đ?&#x2018;&#x192;1 đ??´1 đ??´2 đ?&#x2018;&#x192;2 is rectangular and the mediator of [ đ??´1 đ??´2 ] contains its middle line, therefore (see Figure 3), we have: đ?&#x2018;&#x192;đ??´1 = đ?&#x2018;&#x192;đ??´2 (1). Analogously, it follows that the mediators of segments [đ??ľ1 đ??ľ2 ] and [đ??ś1 đ??ś2 ] pass through đ?&#x2018;&#x192;, so đ?&#x2018;&#x192;đ??ľ1 = đ?&#x2018;&#x192;đ??ľ2 (2) and đ?&#x2018;&#x192;đ??ś1 = đ?&#x2018;&#x192;đ??ś2 (3). We denote by đ??´3 and đ??´4 respectively the middles of segments [đ??´đ?&#x2018;&#x192;1 ] and [ đ??´đ?&#x2018;&#x192;2 ]. We prove that the triangles đ?&#x2018;&#x192;đ??´3 đ??ś1 and 1

đ??ľ2 đ??´4 đ?&#x2018;&#x192; are congruent. Indeed, đ?&#x2018;&#x192;đ??´3 = đ??´đ?&#x2018;&#x192;2 (middle 2

1

line), and đ??ľ2 đ??´4 = đ??´đ?&#x2018;&#x192;2 , because it is a median in the 2

rectangular triangle đ?&#x2018;&#x192;2 đ??ľ2 đ??´ , so đ?&#x2018;&#x192;đ??´3 = đ??ľ2 đ??´4 ; analogously, we obtain that đ??´4 đ?&#x2018;&#x192; = đ??´3 đ??ś1 .

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Figure 3.

We have that: Ě&#x201A; Ě&#x201A; Ě&#x201A; Ě&#x201A; Ě&#x201A; đ?&#x2018;&#x192;đ??´ 3 đ??ś1 = đ?&#x2018;&#x192;đ??´3 đ?&#x2018;&#x192;1 + đ?&#x2018;&#x192;1 đ??´3 đ??ś1 = đ?&#x2018;&#x192;1 đ??´đ?&#x2018;&#x192;2 + 2đ?&#x2018;&#x192;1 đ??´đ??ś1 = = đ??´Ě&#x201A; + đ?&#x2018;&#x192;Ě&#x201A; 1 đ??´đ??ľ ; Ě&#x201A; Ě&#x201A; Ě&#x201A; đ??ľĚ&#x201A; đ??´ đ?&#x2018;&#x192; = đ??ľ đ??´ đ?&#x2018;&#x192; + đ?&#x2018;&#x192;đ??´ đ?&#x2018;&#x192; = đ?&#x2018;&#x192;Ě&#x201A; 2 4 2 4 2 4 2 1 đ??´đ?&#x2018;&#x192;2 + 2đ?&#x2018;&#x192;2 đ??´đ??ľ2 = = đ??´Ě&#x201A; + đ?&#x2018;&#x192;Ě&#x201A; 2 đ??´đ??ś . Ě&#x201A; But đ?&#x2018;&#x192;Ě&#x201A; 1 đ??´đ??ľ = đ?&#x2018;&#x192;2 đ??´đ??ś , because the cevians đ??´đ?&#x2018;&#x192;1 and Ě&#x201A; Ě&#x201A; đ??´đ?&#x2018;&#x192;2 are isogonal and therefore đ?&#x2018;&#x192;đ??´ 3 đ??ś1 = đ??ľ2 đ??´4 đ?&#x2018;&#x192; . Since â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ??´3 đ??ś1 = â&#x2C6;&#x2020;đ??ľ2 đ??´4 đ?&#x2018;&#x192;, it follows that đ?&#x2018;&#x192;đ??ľ2 = đ?&#x2018;&#x192;đ??ś1 (4).

Repeating the previous reasoning for triangles đ?&#x2018;&#x192;đ??ľ3 đ??ś1 and đ??´2 đ??ľ4 đ?&#x2018;&#x192; , where đ??ľ3 and đ??ľ4 are respectively the middles of segments (đ??ľđ?&#x2018;&#x192;1 ) and (đ??ľđ?&#x2018;&#x192;2 ), we find that they are congruent and it follows that đ?&#x2018;&#x192;đ??ś1 = đ?&#x2018;&#x192;đ??´2 (5).

The relations (1) - (5) lead to đ?&#x2018;&#x192;đ??´1 = đ?&#x2018;&#x192;đ??´2 = đ?&#x2018;&#x192;đ??ľ1 = đ?&#x2018;&#x192;đ??ľ2 = đ?&#x2018;&#x192;đ??ś1 = đ?&#x2018;&#x192;đ??ś2 , which shows that the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are located on a circle centered in đ?&#x2018;&#x192;,

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the middle of the segment determined by isogonal points given by đ?&#x2018;&#x192;1 and đ?&#x2018;&#x192;2 .

4th Definition.

It is called the 9 points circle or Eulerâ&#x20AC;&#x2122;s circle of the triangle đ??´đ??ľđ??ś the circle that contains the middles of triangleâ&#x20AC;&#x2122;s sides, the triangle heightsâ&#x20AC;&#x2122; feet and the middles of the segments determined by the orthocenter with triangleâ&#x20AC;&#x2122;s vertex.

2nd Proposition.

If đ?&#x2018;&#x192;1 , đ?&#x2018;&#x192;2 are isogonal points in the triangle đ??´đ??ľđ??ś and if on the circle of their corresponding 6 points there also are the middles of the segments (đ??´đ?&#x2018;&#x192;1 ), (đ??ľđ?&#x2018;&#x192;1 ), ( đ??śđ?&#x2018;&#x192;1 ), then the 6 points circle coincides with the Eulerâ&#x20AC;&#x2122;s circle of the triangle đ??´đ??ľđ??ś.

1st Proof.

We keep notations from Figure 3; we proved that the points đ??´1 , đ??´2 , đ??ľ1 , đ??ľ2 , đ??ś1 , đ??ś2 are on the 6 points circle of the triangle đ??´đ??ľđ??ś, having its center in đ?&#x2018;&#x192;, the middle of segment [đ?&#x2018;&#x192;1 đ?&#x2018;&#x192;2 ].

If on this circle are also situated the middles đ??´3 , đ??ľ3 , đ??ś3 of segments (đ??´đ?&#x2018;&#x192;1 ), (đ??ľđ?&#x2018;&#x192;1 ), (đ??śđ?&#x2018;&#x192;1 ), then we have đ?&#x2018;&#x192;đ??´3 = đ?&#x2018;&#x192;đ??ľ3 = đ?&#x2018;&#x192;đ??ś3 .

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We determined that đ?&#x2018;&#x192;đ??´3 is middle line in the 1

triangle đ?&#x2018;&#x192;1 đ?&#x2018;&#x192;2 đ??´ , therefore đ?&#x2018;&#x192;đ??´3 = đ??´đ?&#x2018;&#x192;2 , analogously 2

1

1

2

2

đ?&#x2018;&#x192;đ??ľ3 = đ??ľđ?&#x2018;&#x192;2 and đ?&#x2018;&#x192;đ??ś3 = đ??śđ?&#x2018;&#x192;2 , and we obtain that đ?&#x2018;&#x192;2 đ??´ = đ?&#x2018;&#x192;2 đ??ľ = đ?&#x2018;&#x192;2 đ??ś, consequently đ?&#x2018;&#x192; is the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś, so đ?&#x2018;&#x192;2 = đ?&#x2018;&#x201A;.

Because đ?&#x2018;&#x192;1 is the isogonal of đ?&#x2018;&#x201A;, it follows that đ?&#x2018;&#x192;1 = đ??ť, therefore the circle of 6 points of the isogonal points đ?&#x2018;&#x201A; and đ??ť is the circle of 9 points.

2nd Proof.

Because đ??´3 đ??ľ3 is middle line in the triangle đ?&#x2018;&#x192;1 đ??´đ??ľ, it follows that â&#x2C6;˘đ?&#x2018;&#x192;1 đ??´đ??ľ â&#x2030;Ą â&#x2C6;˘đ?&#x2018;&#x192;1 đ??´2 đ??ľ3 .

(1) Also, đ??´3 đ??ś3 is middle line in the triangle đ?&#x2018;&#x192;1 đ??´đ??ś, and đ??´3 đ??ś3 is middle line in the triangle đ?&#x2018;&#x192;1 đ??´đ?&#x2018;&#x192;2 , therefore we get â&#x2C6;˘đ?&#x2018;&#x192;đ??´3 đ??ś3 â&#x2030;Ą â&#x2C6;˘đ?&#x2018;&#x192;2 đ??´đ??ś.

(2) The relations (1), (2) and the fact that đ??´đ?&#x2018;&#x192;1 and đ??´đ?&#x2018;&#x192;2 are isogonal cevians lead to: â&#x2C6;˘đ?&#x2018;&#x192;1 đ??´2 đ??ľ3 â&#x2030;Ą đ?&#x2018;&#x192;đ??´3 đ??ś3 .

(3) The point đ?&#x2018;&#x192; is the center of the circle circumscribed to đ??´3 đ??ľ3 đ??ś3 ; then, from (3) and from isogonal ceviansâ&#x20AC;&#x2122; properties, one of which is circumscribed circle radius, it follows that in the triangle đ??´3 đ??ľ3 đ??ś3 the line đ?&#x2018;&#x192;1 đ??´3 is a height, as đ??ľ3 đ??ś3 â&#x2C6;Ľ đ??ľđ??ś, we get that đ?&#x2018;&#x192;1 đ??´ is a height in the triangle ABC and, therefore, đ?&#x2018;&#x192;1 will be the orthocenter of the triangle

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đ??´đ??ľđ??ś , and đ?&#x2018;&#x192;2 will be the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś.

5th Remark.

Of those shown, we get that the center of the circle of 9 points is the middle of the line determined by triangleâ&#x20AC;&#x2122;s orthocenter and by the circumscribed circleâ&#x20AC;&#x2122;s center, and the fact that Eulerâ&#x20AC;&#x2122;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] CÄ&#x192;tÄ&#x192;lin Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental theorems of triangleâ&#x20AC;&#x2122;s geometry]. BacÄ&#x192;u: 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 đ?&#x203A;ž, and the case for the problem for the ex-inscribed circle đ??ś. 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: â&#x20AC;&#x153;We fix an angle đ?&#x203A;ž â&#x2C6;&#x2C6; (0, 900 ) and the line đ??´đ??ľ which divides the plane in two half-planes đ?&#x203A;ž and đ?&#x203A;žĚ&#x2026; . The point đ??ś in the half-plane đ?&#x203A;ž is situated Ě&#x201A; ) = đ?&#x203A;ž. The circle inscribed in the such that đ?&#x2018;&#x161;(đ??´đ??śđ??ľ triangle đ??´đ??ľđ??ś with the center đ??ź is tangent to the sides đ??´đ??ś and đ??ľđ??ś in the points đ??š and đ??¸ , respectively. The point đ?&#x2018;&#x192; is located on the segment line (đ??źđ??¸ , the point đ??¸ between đ??ź and đ?&#x2018;&#x192; such that đ?&#x2018;&#x192;đ??¸ â&#x160;Ľ đ??ľđ??ś and đ?&#x2018;&#x192;đ??¸ = đ??´đ??š. The point đ?&#x2018;&#x201E; is situated on the segment line (đ??źđ??š, such that đ??š is between đ??ź and đ?&#x2018;&#x201E; ; đ?&#x2018;&#x201E;đ??š â&#x160;Ľ đ??´đ??ś and đ?&#x2018;&#x201E;đ??š = đ??ľđ??¸ . Prove

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that the mediator of segment đ?&#x2018;&#x192;đ?&#x2018;&#x201E; passes through a fixed point.â&#x20AC;? (Stanislav Chobanov) Proof.

Firstly, it is useful to note that the point đ??ś varies in the half-plane đ?&#x153;&#x201C; on the arc capable of angle đ?&#x203A;ž; we Ě&#x201A; ) = 900 + đ?&#x203A;ž, so đ??ź varies on the know as well that đ?&#x2018;&#x161;(đ??´đ??źđ??ľ 2 0

đ?&#x203A;ž

arc capable of angle of measure 90 + situated in the 2

half-plane đ?&#x153;&#x201C;.

Another useful remark is about the segments đ??´đ??š and đ??ľđ??¸ , which in a triangle have the lengths đ?&#x2018;? â&#x2C6;&#x2019; đ?&#x2018;&#x17D;, respectively đ?&#x2018;? â&#x2C6;&#x2019; đ?&#x2018;? , where đ?&#x2018;? is the half-perimeter of the triangle đ??´đ??ľđ??ś with đ??´đ??ľ = đ?&#x2018;? â&#x20AC;&#x201C; constant; therefore, we have â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ??¸đ??ľ â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??šđ?&#x2018;&#x201E; with the consequence đ?&#x2018;&#x192;đ??ľ = đ?&#x2018;&#x201E;đ??´.

Considering the vertex đ??ś of the triangle đ??´đ??ľđ??ś the middle of the arc capable of angle đ?&#x203A;ž built on đ??´đ??ľ, we observe that đ?&#x2018;&#x192;đ?&#x2018;&#x201E; is parallel to đ??´đ??ľ ; more than that, đ??´đ??ľđ?&#x2018;&#x192;đ?&#x2018;&#x201E; is an isosceles trapezoid, and segment đ?&#x2018;&#x192;đ?&#x2018;&#x201E; mediator will be a symmetry axis of the trapezoid, so it will coincide with the mediator of đ??´đ??ľ, which is a fixed line, so we're looking for the fixed point on mediator of đ??´đ??ľ.

Let đ??ˇ be the intersection of the mediators of segments đ?&#x2018;&#x192;đ?&#x2018;&#x201E; and đ??´đ??ľ , see Figure 1, where we considered đ?&#x2018;&#x161;(đ??´Ě&#x201A;) < đ?&#x2018;&#x161;(đ??ľĚ&#x201A;) . The point đ??ˇ is on the mediator of đ??´đ??ľ, so we have đ??ˇđ??´ = đ??ˇđ??ľ; the point đ??ˇ is also on the mediator of đ?&#x2018;&#x192;đ?&#x2018;&#x201E;, so we have đ??ˇđ?&#x2018;&#x192; = đ??ˇđ?&#x2018;&#x201E;; it

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follows that: â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;&#x201E;đ??´đ??ˇ, a relation from where we get that â&#x2C6;˘đ?&#x2018;&#x201E;đ??´đ??ˇ = â&#x2C6;˘đ?&#x2018;&#x192;đ??ľđ??ˇ.

Figure 1

Ě&#x201A; ) = đ?&#x2018;Ľ and đ?&#x2018;&#x161;(đ??ˇđ??´đ??ľ Ě&#x201A; ) = đ?&#x2018;Ś , we If we denote đ?&#x2018;&#x161;(đ?&#x2018;&#x201E;đ??´đ??š 0 Ě&#x201A; = đ?&#x2018;Ľ + đ??´ + đ?&#x2018;Ś, đ?&#x2018;&#x192;đ??ľđ??ˇ Ě&#x201A; = 360 â&#x2C6;&#x2019; đ??ľ â&#x2C6;&#x2019; đ?&#x2018;Ś â&#x2C6;&#x2019; (900 â&#x2C6;&#x2019; đ?&#x2018;Ľ). have đ?&#x2018;&#x201E;đ??´đ??ˇ From đ?&#x2018;Ľ + đ??´ + đ?&#x2018;Ś = 3600 â&#x2C6;&#x2019; đ??ľ â&#x2C6;&#x2019; đ?&#x2018;Ś â&#x2C6;&#x2019; 900 + đ?&#x2018;Ľ, we find that đ??´ + đ??ľ + 2đ?&#x2018;Ś = 2700 , and since đ??´ + đ??ľ = 1800 â&#x2C6;&#x2019; đ?&#x203A;ž, we find that 2đ?&#x2018;Ś = 900 â&#x2C6;&#x2019; đ?&#x203A;ž, therefore the requested fixed point đ??ˇ is the vertex of triangle đ??ˇđ??´đ??ľ , situated in đ?&#x153;&#x201C;Ě&#x2026; Ě&#x201A; ) = 900 â&#x2C6;&#x2019; đ?&#x203A;ž. such that đ?&#x2018;&#x161;(đ??´đ??ˇđ??ľ

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1st Remark.

If đ?&#x203A;ž = 900 , we propose to the reader to prove that the quadrilateral đ??´đ??ľđ?&#x2018;&#x192;đ?&#x2018;&#x201E; is a parallelogram; in this case, the requested fixed point does not exist (or is the point at infinity of the perpendicular lines to đ??´đ??ľ).

2nd Remark.

If đ?&#x203A;ž â&#x2C6;&#x2C6; (900 , 1800 ), the problem applies, and we find that the fixed point đ??ˇ is located in the half-plane đ?&#x153;&#x201C; , such that the triangle đ??ˇđ??´đ??ľ is isosceles, having Ě&#x201A; ) = đ?&#x203A;ž â&#x2C6;&#x2019; 900 . đ?&#x2018;&#x161;(đ??´đ?&#x2018;&#x201A;đ??ľ We suggest to the reader to solve the following problem: We fix an angle đ?&#x203A;ž â&#x2C6;&#x2C6; (00 , 1800 ) and the line AB which divides the plane in two half-planes, đ?&#x153;&#x201C; and đ?&#x153;&#x201C;Ě&#x2026;. The point đ??ś in the half-plane đ?&#x153;&#x201C; is located Ě&#x201A; ) = đ?&#x203A;ž . The circle đ??ś â&#x20AC;&#x201C; exsuch that đ?&#x2018;&#x161;(đ??´đ??śđ??ľ inscribed to the triangle đ??´đ??ľđ??ś with center đ??źđ?&#x2018;? is tangent to the sides đ??´đ??ś and đ??ľđ??ś in the points đ??š and đ??¸, respectively. The point đ?&#x2018;&#x192; is located on the line segment (đ??źđ?&#x2018;? đ??¸ , đ??¸ is between đ??źđ?&#x2018;? and đ?&#x2018;&#x192; such that đ?&#x2018;&#x192;đ??¸ â&#x160;Ľ đ??ľđ??ś and đ?&#x2018;&#x192;đ??¸ = đ??´đ??š . The point đ?&#x2018;&#x201E; is located on the line segment (đ??źđ?&#x2018;? đ??š such that đ??š is between đ??ź and đ?&#x2018;&#x201E; , đ?&#x2018;&#x201E;đ??š â&#x160;Ľ đ??´đ??ś and đ?&#x2018;&#x201E;đ??š = đ??ľđ??¸ . Prove that the mediator of the segment đ?&#x2018;&#x192;đ?&#x2018;&#x201E; passes through a fixed point.

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3rd Remark.

As seen, this problem is also true in the case đ?&#x203A;ž = 90 , more than that, in this case, the fixed point is the middle of đ??´đ??ľ. Prove!

0

References.

[1] Ion Patrascu: Probleme de geometrie planÄ&#x192; [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

Apolloniusâ&#x20AC;&#x2122;s

Circle.

1st Definition.

A convex circumscribable quadrilateral đ??´đ??ľđ??śđ??ˇ having the property đ??´đ??ľ â&#x2C6;&#x2122; đ??śđ??ˇ = đ??ľđ??ś â&#x2C6;&#x2122; đ??´đ??ˇ is called harmonic quadrilateral.

2nd Definition.

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

1st Proposition.

In the triangle đ??´đ??ľđ??ś, the cevian đ??´đ??´1 , đ??´1 â&#x2C6;&#x2C6; (đ??ľđ??ś) is a simedian if and only if

đ??ľđ??´1 đ??´1 đ??ś

property, see infra.

135

đ??´đ??ľ 2

= ( ) . For Proof of this đ??´đ??ś

Ion Patrascu, Florentin Smarandache

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 đ??´đ??ľđ??śđ??ˇ be an harmonic quadrilateral and {đ??ž } = đ??´đ??ś â&#x2C6;Š đ??ľđ??ˇ (see Figure 1). We prove that đ??ľđ??ž is simedian in the triangle đ??´đ??ľđ??ś.

From the similarity of the triangles đ??´đ??ľđ??ž and đ??ˇđ??śđ??ž, we find that: đ??´đ??ľ đ??ˇđ??ś

=

đ??´đ??ž đ??ˇđ??ž

=

đ??ľđ??ž đ??śđ??ž

.

(1)

From the similarity of the triangles đ??ľđ??śđ??ž Ĺ&#x;i đ??´đ??ˇđ??ž, we conclude that:

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đ??ľđ??ś đ??´đ??ˇ

=

đ??śđ??ž đ??ˇđ??ž

=

đ??ľđ??ž đ??´đ??ž

.

(2)

From the relations (1) and (2), by division, it follows that: đ??´đ??ľ đ??´đ??ˇ

.

đ??ľđ??ś đ??ˇđ??ś

=

đ??´đ??ž đ??śđ??ž

.

(3)

But đ??´đ??ľđ??śđ??ˇ is an harmonic quadrilateral; consequently, đ??´đ??ľ đ??´đ??ˇ = ; đ??ľđ??ś đ??ˇđ??ś substituting this relation in (3), it follows that: đ??´đ??ľ 2 đ??´đ??ž ( ) = ; đ??ľđ??ś đ??śđ??ž As shown by Proposition 1, đ??ľđ??ž is a simedian in the triangle đ??´đ??ľđ??ś. Similarly, it can be shown that đ??´đ??ž is a simedian in the triangle đ??´đ??ľđ??ˇ, that đ??śđ??ž is a simedian in the triangle đ??ľđ??śđ??ˇ, and that đ??ˇđ??ž is a simedian in the triangle đ??´đ??ˇđ??ś.

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.

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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 đ??´đ??ľđ??ś be considered; we construct the simedian of A, be it đ??´đ??ž, and we denote

by D the intersection of the simedian đ??´đ??ž with the circle. The quadrilateral đ??´đ??ľđ??śđ??ˇ is an harmonic quadrilateral.

Proposition 4.

In a triangle đ??´đ??ľđ??ś, the points of the simedian of A are situated at proportional lengths to the sides đ??´đ??ľ and đ??´đ??ś.

Figura 2.

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Proof.

We have the simedian đ??´đ??´1 in the triangle đ??´đ??ľđ??ś (see Figure 2). We denote by đ??ˇ and đ??¸ the projections of đ??´1 on đ??´đ??ľ, and đ??´đ??ś respectively.

We get: đ??ľđ??´1 đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D;â&#x2C6;&#x2020; (đ??´đ??ľđ??´1 ) đ??´đ??ľ â&#x2C6;&#x2122; đ??´1 đ??ˇ = = . đ??śđ??´1 đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D;â&#x2C6;&#x2020; (đ??´đ??śđ??´1 ) đ??´đ??ś â&#x2C6;&#x2122; đ??´1 đ??¸ Moreover, from 1st Proposition, we know that đ??ľđ??´1 đ??´đ??ľ 2 =( ) . đ??´1 đ??ś đ??´đ??ś Substituting in the previous relation, we obtain that: đ??´1 đ??ˇ đ??´đ??ľ = . đ??´1 đ??¸ đ??´đ??ś On the other hand, đ??ˇđ??´1 = đ??´đ??´1 . From đ??ľđ??´đ??´1 and Ě&#x201A;1 , hence: đ??´1 đ??¸ = đ??´đ??´1 â&#x2C6;&#x2122; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??śđ??´đ??´ đ??´1 đ??ˇ đ??´1 đ??¸

=

Ě&#x201A;1 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??ľđ??´đ??´ Ě&#x201A;1 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??śđ??´đ??´

=

đ??´đ??ľ đ??´đ??ś

.

(4)

If M is a point on the simedian and đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;1 and đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;2 are its projections on đ??´đ??ľ , and đ??´đ??ś respectively, we have: Ě&#x201A;1 , Ě&#x201A;1 , đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;1 = đ??´đ?&#x2018;&#x20AC; â&#x2C6;&#x2122; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??ľđ??´đ??´ đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;2 = đ??´đ?&#x2018;&#x20AC; â&#x2C6;&#x2122; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??śđ??´đ??´ hence: Ě&#x201A;1 đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;1 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??ľđ??´đ??´ = . Ě&#x201A;1 đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;2 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ??śđ??´đ??´ Taking (4) into account, we obtain that: đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;1 đ??´đ??ľ = . đ?&#x2018;&#x20AC;đ?&#x2018;&#x20AC;2 đ??´đ??ś

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3rd Remark.

The converse of the property in the statement above is valid, meaning that, if đ?&#x2018;&#x20AC; 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 đ??´đ??ľđ??śđ??ˇ be an harmonic quadrilateral and đ?&#x2018;&#x20AC; any point within.

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We denote by đ?&#x2018;Ľ, đ?&#x2018;Ś, đ?&#x2018;§, đ?&#x2018;˘ the distances of đ?&#x2018;&#x20AC; to the đ??´đ??ľ , đ??ľđ??ś , đ??śđ??ˇ , đ??ˇđ??´ sides of lenghts đ?&#x2018;&#x17D;, đ?&#x2018;?, đ?&#x2018;?, and đ?&#x2018;&#x2018; (see Figure 3).

Figure 3.

Let đ?&#x2018;&#x2020; be the đ??´đ??ľđ??śđ??ˇ quadrilateral area.

We have: đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?đ?&#x2018;Ś + đ?&#x2018;?đ?&#x2018;§ + đ?&#x2018;&#x2018;đ?&#x2018;˘ = 2đ?&#x2018;&#x2020;.

This is true for đ?&#x2018;Ľ, đ?&#x2018;Ś, đ?&#x2018;§, đ?&#x2018;˘ and đ?&#x2018;&#x17D;, đ?&#x2018;?, đ?&#x2018;?, đ?&#x2018;&#x2018; real numbers.

Following Cauchy-Buniakowski-Schwarz Inequality, we get: (đ?&#x2018;&#x17D;2 + đ?&#x2018;? 2 + đ?&#x2018;? 2 + đ?&#x2018;&#x2018; 2 )(đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 + đ?&#x2018;§ 2 + đ?&#x2018;˘2 ) â&#x2030;Ľ (đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?đ?&#x2018;Ś + đ?&#x2018;?đ?&#x2018;§ + đ?&#x2018;&#x2018;đ?&#x2018;˘)2 , and it is obvious that: 4đ?&#x2018;&#x2020; 2 đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 + đ?&#x2018;§ 2 + đ?&#x2018;˘2 â&#x2030;Ľ 2 . đ?&#x2018;&#x17D; + đ?&#x2018;?2 + đ?&#x2018;? 2 + đ?&#x2018;&#x2018;2 We note that the minimum sum of squared distances is:

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4đ?&#x2018;&#x2020; 2 = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ą. đ?&#x2018;&#x17D;2 + đ?&#x2018;? 2 + đ?&#x2018;? 2 + đ?&#x2018;&#x2018; 2 In Cauchy-Buniakowski-Schwarz Inequality, the equality occurs if: đ?&#x2018;Ľ đ?&#x2018;Ś đ?&#x2018;§ đ?&#x2018;˘ = = = . đ?&#x2018;&#x17D; đ?&#x2018;? đ?&#x2018;? đ?&#x2018;&#x2018; Since {đ??ž } = đ??´đ??ś â&#x2C6;Š đ??ľđ??ˇ is the only point with this property, it ensues that đ?&#x2018;&#x20AC; = đ??ž, so đ??ž has the property of the minimum in the statement.

3rd Definition.

We call external simedian of đ??´đ??ľđ??ś triangle a cevian đ??´đ??´1 â&#x20AC;&#x2122; corresponding to the vertex đ??´, where đ??´1 â&#x20AC;&#x2122; is the harmonic conjugate of the point đ??´1 â&#x20AC;&#x201C; simedianâ&#x20AC;&#x2122;s foot from đ??´ relative to points đ??ľ and đ??ś.

4th Remark.

In Figure 4, the cevian đ??´đ??´1 is an internal simedian, and đ??´đ??´1 â&#x20AC;&#x2122; is an external simedian.

We have: đ??´1 đ??ľ đ??´1 â&#x20AC;˛đ??ľ = . đ??´1 đ??ś đ??´1 â&#x20AC;˛đ??ś In view of 1st Proposition, we get that: đ??´1 â&#x20AC;˛đ??ľ đ??´đ??ľ 2 =( ) . đ??´1 â&#x20AC;˛đ??ś đ??´đ??ś

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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 đ?&#x2018;&#x192; be the intersection of a tangent taken in đ??ˇ to the circle circumscribed to the harmonic quadrilateral đ??´đ??ľđ??śđ??ˇ with đ??´đ??ś (see Figure 4).

Figure 4.

Since triangles PDC and PAD are alike, we conclude that: đ?&#x2018;&#x192;đ??ˇ đ?&#x2018;&#x192;đ??´

=

đ?&#x2018;&#x192;đ??ś đ?&#x2018;&#x192;đ??ˇ

=

đ??ˇđ??ś đ??´đ??ˇ

.

(5)

From relations (5), we find that: đ?&#x2018;&#x192;đ??´ đ?&#x2018;&#x192;đ??ś

=(

đ??´đ??ˇ 2 đ??ˇđ??ś

) .

(6)

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This relationship indicates that P is the harmonic conjugate of K with respect to A and C, so đ??ˇđ?&#x2018;&#x192; is an external simedian from D of the triangle đ??´đ??ˇđ??ś.

Similarly, if we denote by đ?&#x2018;&#x192;â&#x20AC;&#x2122; the intersection of the tangent taken in B to the circle circumscribed with đ??´đ??ś, we get: đ?&#x2018;&#x192;â&#x20AC;˛đ??´ đ?&#x2018;&#x192;â&#x20AC;˛đ??ś

đ??ľđ??´ 2

=( ) .

(7)

đ??ľđ??ś

From (6) and (7), as well as from the properties of the harmonic quadrilateral, we know that: đ??´đ??ľ đ??´đ??ˇ = , đ??ľđ??ś đ??ˇđ??ś which means that: đ?&#x2018;&#x192;đ??´ đ?&#x2018;&#x192;â&#x20AC;˛đ??´ = , đ?&#x2018;&#x192;đ??ś đ?&#x2018;&#x192;â&#x20AC;˛đ??ś hence đ?&#x2018;&#x192; = đ?&#x2018;&#x192;â&#x20AC;&#x2122;.

Similarly, it is shown that the tangents taken to A and C intersect at point Q located on the diagonal đ??ľđ??ˇ.

5th Remark. a. The points P and Q are the diagonal poles of đ??ľđ??ˇ and đ??´đ??ś 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 đ??´đ??ľđ??śđ??ˇ 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 {đ??ž } = đ??´đ??ś â&#x2C6;Š đ??ľđ??ˇ, then the orthocenter of triangle đ?&#x2018;&#x192;đ??žđ?&#x2018;&#x201E; is O. Proof.

From the properties of tangents taken from a point to a circle, we conclude that đ?&#x2018;&#x192;đ?&#x2018;&#x201A; â&#x160;Ľ đ??ľđ??ˇ and đ?&#x2018;&#x201E;đ?&#x2018;&#x201A; â&#x160;Ľ đ??´đ??ś. These relations show that in the triangle đ?&#x2018;&#x192;đ??žđ?&#x2018;&#x201E;, đ?&#x2018;&#x192;đ?&#x2018;&#x201A; and đ?&#x2018;&#x201E;đ?&#x2018;&#x201A; are heights, so đ?&#x2018;&#x201A; is the orthocenter of this triangle.

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4th Definition.

The Apolloniusâ&#x20AC;&#x2122;s circle related to the vertex A of the triangle đ??´đ??ľđ??ś is the circle built on the segment [đ??ˇđ??¸ ] in diameter, where D and E are the feet of the internal, respectively external, bisectors taken from A to the triangle đ??´đ??ľđ??ś.

6th Remark.

If the triangle đ??´đ??ľđ??ś is isosceles with đ??´đ??ľ = đ??´đ??ś , the Apolloniusâ&#x20AC;&#x2122;s circle corresponding to vertex A is not defined.

9th Proposition.

The Apolloniusâ&#x20AC;&#x2122;s circle relative to the vertex A of the triangle đ??´đ??ľđ??ś 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 đ??´đ??ľđ??ś with đ??ľđ??ś (see Figure 6).

Assuming that đ?&#x2018;&#x161;(đ??ľĚ&#x201A;) > đ?&#x2018;&#x161;(đ??śĚ&#x201A; ), we find that: Ě&#x201A; ) = 1 [đ?&#x2018;&#x161;(đ??ľĚ&#x201A;) + đ?&#x2018;&#x161;(đ??śĚ&#x201A; )]. đ?&#x2018;&#x161;(đ??¸đ??´đ??ľ 2

Ě&#x201A; Oa being a tangent, we find that đ?&#x2018;&#x161;(đ?&#x2018;&#x201A;Ě&#x201A; đ?&#x2018;&#x17D; đ??´đ??ľ ) = đ?&#x2018;&#x161;(đ??ś ).

Withal, 1 đ?&#x2018;&#x161;(đ??¸đ??´đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; ) = [đ?&#x2018;&#x161;(đ??ľĚ&#x201A;) â&#x2C6;&#x2019; đ?&#x2018;&#x161;(đ??śĚ&#x201A; )] 2

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and 1 đ?&#x2018;&#x161;(đ??´đ??¸đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; ) = [đ?&#x2018;&#x161;(đ??ľĚ&#x201A; ) â&#x2C6;&#x2019; đ?&#x2018;&#x161;(đ??śĚ&#x201A; )].

2

It results that đ?&#x2018;&#x201A;đ?&#x2018;&#x17D;đ??¸ = đ?&#x2018;&#x201A;đ?&#x2018;&#x17D;đ??´; onward, đ??¸đ??´đ??ˇ being a right angled triangle, we obtain: đ?&#x2018;&#x201A;đ?&#x2018;&#x17D;đ??´ = đ?&#x2018;&#x201A;đ?&#x2018;&#x17D;đ??ˇ; hence Oa is the center of Apolloniusâ&#x20AC;&#x2122;s circle corresponding to the vertex đ??´.

Figura 6.

10th Proposition.

Apolloniusâ&#x20AC;&#x2122;s circle relative to the vertex A of triangle đ??´đ??ľđ??ś cuts the circle circumscribed to the triangle following the internal simedian taken from A. Proof.

Let S be the second point of intersection of Apolloniusâ&#x20AC;&#x2122;s Circle relative to vertex A and the circle circumscribing the triangle đ??´đ??ľđ??ś.

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Because đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; đ??´ is tangent to the circle circumscribed in A, it results, for reasons of symmetry, that đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; đ?&#x2018;&#x2020; will be tangent in S to the circumscribed circle.

For triangle đ??´đ??śđ?&#x2018;&#x2020; , đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; đ??´ and đ?&#x2018;&#x201A;đ?&#x2018;&#x17D; đ?&#x2018;&#x2020; are external simedians; it results that đ??śđ?&#x2018;&#x201A;đ?&#x2018;&#x17D; is internal simedian in the triangle đ??´đ??śđ?&#x2018;&#x2020; , furthermore, it results that the quadrilateral đ??´đ??ľđ?&#x2018;&#x2020;đ??ś is an harmonic quadrilateral.

Consequently, đ??´đ?&#x2018;&#x2020; is the internal simedian of the triangle đ??´đ??ľđ??ś 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 Apolloniusâ&#x20AC;&#x2122;s circle relative to the vertex A for the triangle đ??´đ??ľđ??ˇ. This circle (of center Q and radius QC) is also an Apolloniusâ&#x20AC;&#x2122;s circle relative to the vertex C of the triangle đ??ľđ??śđ??ˇ.

Similarly, the Apolloniusâ&#x20AC;&#x2122;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 Apolloniusâ&#x20AC;&#x2122;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 Apolloniusâ&#x20AC;&#x2122;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 Apolloniusâ&#x20AC;&#x2122;s Circle relative to vertexes B and C of triangles đ??´đ??ľđ??ś and đ??ľđ??śđ?&#x2018;&#x2C6; is: đ?&#x2018;&#x201A;đ??ľ2 = đ?&#x2018;&#x201A;đ??ś 2 .

So O belongs to the radical axis of the circles.

We also have đ??žđ??´ â&#x2C6;&#x2122; đ??žđ??ś = đ??žđ??ľ â&#x2C6;&#x2122; đ??žđ??ˇ , relatives indicating that the point K has equal powers towards the highlighted Apolloniusâ&#x20AC;&#x2122;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

cevian

in

can

two

be

divided

triangles

that

by

a

have

congruent inscribed circles.

Solution.

We consider a given triangle đ??´đ??ľđ??ś and we show that there is a point đ??ˇ on the side (đ??ľđ??ś) so that the inscribed circles in the triangles đ??´đ??ľđ??ˇ , đ??´đ??śđ??ˇ are congruent. If đ??´đ??ľđ??ś is an isosceles triangle (đ??´đ??ľ = đ??´đ??ś), where đ??ˇ is the middle of the base (đ??ľđ??ś), we assume that đ??´đ??ľđ??ś is a non-isosceles triangle. We note đ??ź1 , đ??ź2 the centers of the inscribed congruent circles; obviously, đ??ź1 đ??ź2 is parallel to the đ??ľđ??ś.

(1) We observe that: 1 đ?&#x2018;&#x161;(â&#x2C6;˘đ??ź1 đ??´đ??ź2 ) = đ?&#x2018;&#x161;(đ??´Ě&#x201A;).

(2) 2

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Figure

1.

If đ?&#x2018;&#x2021;1 , đ?&#x2018;&#x2021;2 are contacts with the circles of the cevian đ??´đ??ˇ , we have â&#x2C6;&#x2020; đ??ź1 đ?&#x2018;&#x2021;1 đ?&#x2018;&#x20AC; â&#x2030;Ą â&#x2C6;&#x2020; đ??ź2 đ?&#x2018;&#x2021;2 đ?&#x2018;&#x20AC; ; let đ?&#x2018;&#x20AC; be the intersection of đ??ź1 đ??ź2 with đ??´đ??ˇ, see Figure 1.

From this congruence, it is obvious that: (đ??ź1 đ?&#x2018;&#x20AC;) â&#x2030;Ą (đ??ź2 đ?&#x2018;&#x20AC;).

(3) Let đ??ź be the center of the circle inscribed in the triangle đ??´đ??ľđ??ś; we prove that: đ??´đ??ź is a simedian in the triangle đ??ź1 đ??´đ??ź2 .

(4) Ě&#x201A;1 ) , it follows that Indeed, noting đ?&#x203A;ź = đ?&#x2018;&#x161;(đ??ľđ??´đ??ź đ?&#x2018;&#x161;(â&#x2C6;˘đ??ź1 đ??´đ?&#x2018;&#x20AC;) = đ?&#x203A;ź. From â&#x2C6;˘đ??ź1 đ??´đ??ź2 = â&#x2C6;˘đ??ľđ??´đ??ź, it follows that â&#x2C6;˘đ??ľđ??´đ??ź1 â&#x2030;Ą â&#x2C6;˘đ??źđ??´đ??ź2 , therefore â&#x2C6;˘đ??ź1 đ??´đ?&#x2018;&#x20AC; â&#x2030;Ą â&#x2C6;˘đ??źđ??´đ??ź2 , indicating that đ??´đ?&#x2018;&#x20AC; and đ??´đ??ź are isogonal cevians in the triangle đ??ź1 đ??´đ??ź2 .

Since in this triangle đ??´đ?&#x2018;&#x20AC; is a median, it follows that đ??´đ??ź is a bimedian.

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Now, we show how we build the point đ??ˇ, using the conditions (1) â&#x20AC;&#x201C; (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 đ??´đ??ľđ??ś; we build the bisector of the angle đ??ľđ??´đ??ś and denote by P its intersection with the circumscribed circle (see Figure 2).

0 2 : We build the perpendicular on đ??ś to đ??śđ?&#x2018;&#x192; and (đ??ľđ??ś) side mediator; we denote đ?&#x2018;&#x201A;1 the intersection of these lines.

0 3 : We build the circle â&#x2C6; (đ?&#x2018;&#x201A;1 ; đ?&#x2018;&#x201A;1 đ??ś) and denote đ??´â&#x20AC;&#x2122; the intersection of this circle with the bisector đ??´đ??ź (đ??´â&#x20AC;&#x2122; is on the same side of the line đ??ľđ??ś as đ??´).

0 4 : We build through đ??´ the parallel to đ??´â&#x20AC;&#x2122;đ?&#x2018;&#x201A;1 and we denote it đ??źđ?&#x2018;&#x201A;1 .

0 5 : We build the circle â&#x2C6; (đ?&#x2018;&#x201A;1â&#x20AC;˛ ; đ?&#x2018;&#x201A;1â&#x20AC;˛ đ??´) and we denote đ??ź1 , đ??ź2 its intersections with đ??ľđ??ź , and đ??śđ??ź respectively.

0 6 : We build the middle đ?&#x2018;&#x20AC; of the segment (đ??ź1 đ??ź2 ) and denote by đ??ˇ the intersection of the lines đ??´đ?&#x2018;&#x20AC; and đ??ľđ??ś.

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Figure 2.

Proof. Ě&#x2020; , then The point đ?&#x2018;&#x192; is the middle of the arc đ??ľđ??ś Ě&#x201A; ) = 1 đ?&#x2018;&#x161;(đ??´Ě&#x201A;). đ?&#x2018;&#x161;(đ?&#x2018;&#x192;đ??śđ??ľ 2

The circle â&#x2C6; (đ?&#x2018;&#x201A;1 ; đ?&#x2018;&#x201A;1 đ??ś) contains the arc from which points the segment (BC) â&#x20AC;&#x17E;is shownâ&#x20AC;? under angle 1 measurement đ?&#x2018;&#x161;(đ??´Ě&#x201A;).

2

The circle â&#x2C6; (đ?&#x2018;&#x201A;1â&#x20AC;˛ ; đ?&#x2018;&#x201A;1â&#x20AC;˛ đ??´) is homothetical to the circle â&#x2C6; (đ?&#x2018;&#x201A;1 ; đ?&#x2018;&#x201A;1 đ??ś) by the homothety of center đ??ź and by the report

đ??źđ??´â&#x20AC;˛ đ??źđ??´

; therefore, it follows that đ??ź1 đ??ź2 will be

parallel to the đ??ľđ??ś , and from the points of circle â&#x2C6; (đ?&#x2018;&#x201A;1â&#x20AC;˛ ; đ?&#x2018;&#x201A;1â&#x20AC;˛ đ??´) of the same side of đ??ľđ??ś as đ??´, the segment

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(đ??ź1 đ??ź2 ) â&#x20AC;&#x17E;is shownâ&#x20AC;? at an angle of measure

1 2

đ?&#x2018;&#x161;(đ??´Ě&#x201A;). Since

the tangents taken in đ??ľ and đ??ś to the circle â&#x2C6; (đ?&#x2018;&#x201A;1 ; đ?&#x2018;&#x201A;1 đ??ś) intersect in đ?&#x2018;&#x192;, on the bisector đ??´đ??ź, as from a property of simedians, we get that đ??´â&#x20AC;˛ đ??ź is a simedian in the triangle đ??´â&#x20AC;˛ đ??ľđ??ś.

Due to the homothetical properties, it follows also that the tangents in the points đ??ź1 , đ??ź2 to the circle â&#x2C6; (đ?&#x2018;&#x201A;1â&#x20AC;˛ ; đ?&#x2018;&#x201A;1â&#x20AC;˛ đ??´) intersect in a point đ?&#x2018;&#x192;â&#x20AC;&#x2122; located on đ??´đ??ź, i.e. đ??´đ?&#x2018;&#x192;â&#x20AC;&#x2122; contains the simedian (đ??´đ?&#x2018;&#x2020;) of the triangle đ??ź1 đ??´đ??ź2 , noted {đ?&#x2018;&#x2020;} = đ??´đ?&#x2018;&#x192;â&#x20AC;˛ â&#x2C6;Š đ??ź1 đ??ź2 .

In the triangle đ??ź1 đ??´đ??ź2 , đ??´đ?&#x2018;&#x20AC; is a median, and đ??´đ?&#x2018;&#x2020; is simedian, therefore â&#x2C6;˘đ??ź1 đ??´đ?&#x2018;&#x20AC; â&#x2030;Ą đ??ź2 đ??´đ?&#x2018;&#x2020;; on the other hand, â&#x2C6;˘đ??ľđ??´đ?&#x2018;&#x2020; â&#x2030;Ą â&#x2C6;˘đ??ź1 đ??´đ??ź2 ; it follows that â&#x2C6;˘đ??ľđ??´đ??ź1 â&#x2030;Ą đ??ź2 đ??´đ?&#x2018;&#x2020;, and more: â&#x2C6;˘đ??ź1 đ??´đ?&#x2018;&#x20AC; â&#x2030;Ą đ??ľđ??´đ??ź1 , which shows that đ??´đ??ź1 is a bisector in the triangle đ??ľđ??´đ??ˇ; plus, đ??ź1 , being located on the bisector of the angle đ??ľ, it follows that this point is the center of the circle inscribed in the triangle đ??ľđ??´đ??ˇ.

Analogous considerations lead to the conclusion that đ??ź2 is the center of the circle inscribed in the triangle đ??´đ??śđ??ˇ. Because đ??ź1 đ??ź2 is parallel to đ??ľđ??ś, it follows that the rays of the circles inscribed in the triangles đ??´đ??ľđ??ˇ and đ??´đ??śđ??ˇ are equal.

Discussion.

The circles â&#x2C6; (đ?&#x2018;&#x201A;1 ; đ?&#x2018;&#x201A;1 đ??´â&#x20AC;˛ ), â&#x2C6; (đ?&#x2018;&#x201A;1â&#x20AC;˛ ; đ?&#x2018;&#x201A;1â&#x20AC;˛ đ??´) are unique; also, the triangle đ??ź1 đ??´đ??ź2 is unique; therefore, determined as before, the point đ??ˇ is unique.

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Remark.

At the beginning of the Proof, we assumed that đ??´đ??ľđ??ś 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 đ??´đ??ľđ??ś triangle.

Open problem.

Given a scalene triangle đ??´đ??ľđ??ś , could it be triangulated by the cevians đ??´đ??ˇ , đ??´đ??¸ , with đ??ˇ , đ??¸ belonging to (đ??ľđ??ś), so that the inscribed circles in the triangles đ??´đ??ľđ??ˇ, đ??ˇđ??´đ??¸ and the đ??¸đ??´đ??ś 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 (đ?&#x2018;&#x2021;đ??´ ), (đ?&#x2018;&#x2021;đ??ľ ), (đ?&#x2018;&#x2021;đ??ś ) be the tangents in the peaks đ??´, đ??ľ, đ??ś of the triangle đ??´đ??ľđ??ś to the circle circumscribed to the triangle. Prove that the perpendiculars drawn from the middles of the opposite sides on (đ?&#x2018;&#x2021;đ??´ ), (đ?&#x2018;&#x2021;đ??ľ ), (đ?&#x2018;&#x2021;đ??ś ) 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 đ??´đ??ľđ??ś are orthological. The orthological centers are đ?&#x2018;&#x201A; â&#x20AC;&#x201C; the center of the circle circumscribed to the triangle đ??´đ??ľđ??ś, and đ?&#x2018;&#x201A;9 â&#x20AC;&#x201C; the center of the circle of đ??´đ??ľđ??ś triangleâ&#x20AC;&#x2122;s 9 points.

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Figure 1.

Proof.

Let đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? be the median triangle of triangle đ??´đ??ľđ??ś and đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? the tangential triangle of the triangle đ??´đ??ľđ??ś . It is obvious that the triangle đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and the triangle đ??´đ??ľđ??ś are orthological and that đ?&#x2018;&#x201A; is the orthological center.

Verily, the perpendiculars taken from đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; , đ?&#x2018;&#x2021;đ?&#x2018;? , đ?&#x2018;&#x2021;đ?&#x2018;? on đ??ľđ??ś ; đ??śđ??´; đ??´đ??ľ respectively are internal bisectors in the triangle đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and consequently passing through đ?&#x2018;&#x201A;, which is center of the circle inscribed in triangle đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? . Moreover, đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x201A; is the mediator of (đ??ľđ??ś) and accordingly passing through đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; , and đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;? is perpendicular on đ??ľđ??ś , being a mediator, but also on đ?&#x2018;&#x20AC;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? which is a median line.

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From orthological triangles theorem, it follows that the perpendiculars taken from đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; , đ?&#x2018;&#x20AC;đ?&#x2018;? , đ?&#x2018;&#x20AC;đ?&#x2018;? on đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? , đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; , đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? respectively, are concurrent. The point of concurrency is the second orthological center of triangles đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? . We prove that this point is đ?&#x2018;&#x201A;9 â&#x20AC;&#x201C; the center of Euler circle of triangle đ??´đ??ľđ??ś. We take đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;1 â&#x160;Ľ đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and denote by {đ??ť1 } = đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;1 â&#x2C6;Š đ??´đ??ť, đ??ť being the orthocenter of the triangle đ??´đ??ľđ??ś. We know that đ??´đ??ť = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; ; we prove this relation vectorially.

From Sylvesterâ&#x20AC;&#x2122;s relation, we have that: â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; đ?&#x2018;&#x201A;đ??ť = â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; + đ?&#x2018;&#x201A;đ??ľ â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; + đ?&#x2018;&#x201A;đ??ś â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; , but đ?&#x2018;&#x201A;đ??ľ â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; + đ?&#x2018;&#x201A;đ??ś â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC; â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;đ?&#x2018;&#x17D; ; it follows that đ?&#x2018;&#x201A;đ??´ â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;đ?&#x2018;&#x17D; , so â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;đ?&#x2018;&#x17D; ; changing to module, đ?&#x2018;&#x201A;đ??ť â&#x2C6;&#x2019; â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014;â&#x192;&#x2014; đ?&#x2018;&#x201A;đ??´ = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC; đ??´đ??ť = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC; we have đ??´đ??ť = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; . Uniting đ?&#x2018;&#x201A; to đ??´ , we have đ?&#x2018;&#x201A;đ??´ â&#x160;Ľ đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? , and because đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;1 â&#x160;Ľ đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and đ??´đ??ť â&#x2C6;Ľ đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; , it follows that the quadrilateral đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ??ť1 đ??´ is a parallelogram.

From đ??´đ??ť1 = đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; and đ??´đ??ť = 2đ?&#x2018;&#x201A;đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; we get that đ??ť1 is the middle of (đ??´đ??ť), so đ??ť1 is situated on the circle of the 9 points of triangle đ??´đ??ľđ??ś. On this circle, we find as well the points đ??´â&#x20AC;˛ - the height foot from đ??´ and đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; ; since â&#x2C6;˘đ??´đ??´â&#x20AC;˛ đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; = 900 , it follows that đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ??ť1 is the diameter of Euler circle, therefore the middle of (đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ??ť1 ), which we denote by đ?&#x2018;&#x201A;9 , is the center of Eulerâ&#x20AC;&#x2122;s circle; we observe that the quadrilateral đ??ť1 đ??ťđ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x201A; is as well a parallelogram; it follows that đ?&#x2018;&#x201A;9 is the middle of segment [đ?&#x2018;&#x201A;đ??ť].

In conclusion, the perpendicular from đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; on đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? pass through đ?&#x2018;&#x201A;9 .

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Analogously, we show that the perpendicular taken from đ?&#x2018;&#x20AC;đ?&#x2018;? on đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? pass through đ?&#x2018;&#x201A;9 and consequently đ?&#x2018;&#x201A;9 is the enunciated orthological center.

Remark.

The triangles đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? and đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? are homological as well, since đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; , đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? , đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? are concurrent in O, as we already observed, therefore the triangles đ?&#x2018;&#x2021;đ?&#x2018;&#x17D; đ?&#x2018;&#x2021;đ?&#x2018;? đ?&#x2018;&#x2021;đ?&#x2018;? and đ?&#x2018;&#x20AC;đ?&#x2018;&#x17D; đ?&#x2018;&#x20AC;đ?&#x2018;? đ?&#x2018;&#x20AC;đ?&#x2018;? are orthohomological of rank I (see [2]).

From P. Sondat theorem (see [4]), it follows that the Euler line đ?&#x2018;&#x201A;đ?&#x2018;&#x201A;9 is perpendicular on the homological axis of the median triangle and of the tangential triangle corresponding to the given triangle đ??´đ??ľđ??ś.

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

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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

Bobillierâ&#x20AC;&#x2122;s transversal relative to a point đ?&#x2018;ś in the plane of a triangle đ?&#x2018;¨đ?&#x2018;Šđ?&#x2018;Ş; we use this notion in what follows.

We transform by duality with respect to a circle

â&#x201E;&#x201A;(đ?&#x2018;&#x153;, đ?&#x2018;&#x;)

the

following

theorem

relative to the orthocenter of a triangle.

1st Theorem.

If đ??´đ??ľđ??ś is a nonisosceles triangle, đ??ť its orthocenter, and AA1 , BB1 , CC1 are cevians of a triangle concurrent at point đ?&#x2018;&#x201E; different from đ??ť, and đ?&#x2018;&#x20AC;, đ?&#x2018; , đ?&#x2018;&#x192; are the intersections of the perpendiculars taken from đ??ť on given cevians respectively, with đ??ľđ??ś, đ??śđ??´, đ??´đ??ľ, then the points đ?&#x2018;&#x20AC;, đ?&#x2018; , đ?&#x2018;&#x192; are collinear.

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Proof.

We note with đ?&#x203A;ź = đ?&#x2018;&#x161;(â&#x2C6;˘BAA1 ); đ?&#x203A;˝ = đ?&#x2018;&#x161;(â&#x2C6;˘CBB1 ), đ?&#x203A;ž = đ?&#x2018;&#x161;(â&#x2C6;˘ACC1 ), see Figure 1.

According to Cevaâ&#x20AC;&#x2122;s theorem, trigonometric form, we have the relation: sin đ?&#x203A;ź

â&#x2C6;&#x2122;

sinđ?&#x203A;˝

sinđ?&#x203A;ž

â&#x2C6;&#x2122;

sin(đ??´â&#x2C6;&#x2019;đ?&#x203A;ź) sin(đ??ľâ&#x2C6;&#x2019;đ?&#x203A;˝) sin(đ??śâ&#x2C6;&#x2019;đ?&#x203A;ž)

= 1.

(1)

We notice that: đ?&#x2018;&#x20AC;đ??ľ đ?&#x2018;&#x20AC;đ??ś

=

Area(đ?&#x2018;&#x20AC;đ??ťđ??ľ) Area(đ?&#x2018;&#x20AC;đ??ťđ??ś)

=

Ě&#x201A;) đ?&#x2018;&#x20AC;đ??ťâ&#x2C6;&#x2122;đ??ťđ??ľâ&#x2C6;&#x2122;sin sin(đ?&#x2018;&#x20AC;đ??ťđ??ľ Ě&#x201A; đ?&#x2018;&#x20AC;đ??ťâ&#x2C6;&#x2122;đ??ťđ??śâ&#x2C6;&#x2122;sin(đ?&#x2018;&#x20AC;đ??ťđ??ś )

.

Because: â&#x2C6;˘đ?&#x2018;&#x20AC;đ??ťđ??ľ â&#x2030;Ą â&#x2C6;˘đ??´1 đ??´đ??ś, as angles of perpendicular sides, it follows that đ?&#x2018;&#x161;(â&#x2C6;˘đ?&#x2018;&#x20AC;đ??ťđ??ľ) = đ?&#x2018;&#x161;(đ??´Ě&#x201A;) â&#x2C6;&#x2019; đ?&#x203A;ź.

Therewith: Ě&#x201A; ) = 1800 đ?&#x203A;ź. Ě&#x201A; ) + đ?&#x2018;&#x161;(đ??ľđ??ťđ??ś đ?&#x2018;&#x161;(â&#x2C6;˘đ?&#x2018;&#x20AC;đ??ťđ??ś) = đ?&#x2018;&#x161;(đ?&#x2018;&#x20AC;đ??ťđ??ľ We thus get that: đ?&#x2018;&#x20AC;đ??ľ sin(đ??´ â&#x2C6;&#x2019; đ?&#x203A;ź) đ??ťđ??ľ = â&#x2C6;&#x2122; . đ?&#x2018;&#x20AC;đ??ś sinđ?&#x203A;ź đ??ťđ??ś Analogously, we find that: đ?&#x2018; đ??ś đ?&#x2018; đ??´ đ?&#x2018;&#x192;đ??´ đ?&#x2018;&#x192;đ??ľ

= =

sin(đ??ľâ&#x2C6;&#x2019;đ?&#x203A;˝) đ??ťđ??ś

â&#x2C6;&#x2122;

sinđ?&#x203A;˝ đ??ťđ??´ sin(đ??śâ&#x2C6;&#x2019;đ?&#x203A;ž) đ??ťđ??´ sinđ?&#x203A;ž

â&#x2C6;&#x2122;

đ??ťđ??ľ

; .

Applying the reciprocal of Menelaus' theorem, we find, in view of (1), that: đ?&#x2018;&#x20AC;đ??ľ đ??ťđ??ś đ?&#x2018;&#x20AC;đ??ś

â&#x2C6;&#x2122;

â&#x2C6;&#x2122;

đ?&#x2018;&#x192;đ??´

đ??ťđ??´ đ?&#x2018;&#x192;đ??ľ

= 1.

This shows that đ?&#x2018;&#x20AC;, đ?&#x2018; , đ?&#x2018;&#x192; are collinear.

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Figure 1.

Note.

1st Theorem is true even if đ??´đ??ľđ??ś is an obtuse, nonisosceles triangle.

The proof is adapted analogously.

2nd Theorem.

(The Dual of the Theorem 1)

If đ??´đ??ľđ??ś is a triangle, đ?&#x2018;&#x201A; a certain point in his plan, and đ??´1 , đ??ľ1 , đ??ś1 Bobillierâ&#x20AC;&#x2122;s transversals relative to đ?&#x2018;&#x201A; of

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đ??´đ??ľđ??ś triangle, as well as đ??´2 â&#x2C6;&#x2019; đ??ľ2 â&#x2C6;&#x2019; đ??ś2 a certain transversal in đ??´đ??ľđ??ś, and the perpendiculars in đ?&#x2018;&#x201A;, and on đ?&#x2018;&#x201A;đ??´2 , đ?&#x2018;&#x201A;đ??ľ2 , đ?&#x2018;&#x201A;đ??ś2 respectively, intersect the Bobillierâ&#x20AC;&#x2122;s transversals in the points đ??´3 , đ??ľ3 , đ??ś3 , then the ceviens đ??´đ??´3 , đ??ľđ??ľ3 , đ??śđ??ś3 are concurrent.

Proof.

We convert by duality with respect to a circle â&#x201E;&#x201A;(đ?&#x2018;&#x153;, đ?&#x2018;&#x;) the figure indicated by the statement of this theorem, i.e. Figure 2. Let đ?&#x2018;&#x17D;, đ?&#x2018;?, đ?&#x2018;? be the polars of the points đ??´, đ??ľ, đ??ś with respect to the circle â&#x201E;&#x201A;(đ?&#x2018;&#x153;, đ?&#x2018;&#x;). To the lines đ??ľđ??ś, đ??śđ??´, đ??´đ??ľ will correspond their poles {đ??´â&#x20AC;˛ 4 = đ?&#x2018;?đ?&#x2018;&#x203A;đ?&#x2018;?; {đ??ľâ&#x20AC;˛ 4 = đ?&#x2018;?đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;; {đ??ś â&#x20AC;˛ 4 = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;?.

To the points đ??´1 , đ??ľ1 , đ??ś1 will respectively correspond their polars đ?&#x2018;&#x17D;1 , đ?&#x2018;?1 , đ?&#x2018;?1 concurrent in transversalâ&#x20AC;&#x2122;s pole đ??´1 â&#x2C6;&#x2019; đ??ľ1 â&#x2C6;&#x2019; đ??ś1 .

Since đ?&#x2018;&#x201A;đ??´1 â&#x160;Ľ đ?&#x2018;&#x201A;đ??´, it means that the polars đ?&#x2018;&#x17D; and đ?&#x2018;&#x17D;1 are perpendicular, so đ?&#x2018;&#x17D;1 â&#x160;Ľ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ , but đ?&#x2018;&#x17D;1 pass through đ??´â&#x20AC;˛ , which means that đ?&#x2018;&#x201E;â&#x20AC;˛ contains the height from đ??´â&#x20AC;˛ of đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ triangle and similarly đ?&#x2018;?1 contains the height from đ??ľâ&#x20AC;˛ and đ?&#x2018;?1 contains the height from đ??ś â&#x20AC;˛ of đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ triangle.

Consequently, the pole of đ??´1 â&#x2C6;&#x2019; đ??ľ1 â&#x2C6;&#x2019; đ??ś1 â&#x20AC;˛ transversal is the orthocenter đ??ť of đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ triangle. In the same way, to the points đ??´2 , đ??ľ2 , đ??ś2 will correspond the polars to đ?&#x2018;&#x17D;2 , đ?&#x2018;?2 , đ?&#x2018;?2 which pass respectively through đ??´â&#x20AC;˛ , đ??ľâ&#x20AC;˛ , đ??ś â&#x20AC;˛ and are concurrent in a point đ?&#x2018;&#x201E;â&#x20AC;˛ , the pole of the line đ??´2 â&#x2C6;&#x2019; đ??ľ2 â&#x2C6;&#x2019; đ??ś2 with respect to the circle â&#x201E;&#x201A;(đ?&#x2018;&#x153;, đ?&#x2018;&#x;).

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Figure 2.

Given đ?&#x2018;&#x201A;đ??´2 â&#x160;Ľ đ?&#x2018;&#x201A;đ??´3 , it means that the polar đ?&#x2018;&#x17D;2 and đ?&#x2018;&#x17D;3 are perpendicular, đ?&#x2018;&#x17D;2 correspond to the cevian đ??´â&#x20AC;˛ đ?&#x2018;&#x201E;â&#x20AC;˛ , also đ?&#x2018;&#x17D;3 passes through the the pole of the transversal đ??´1 â&#x2C6;&#x2019; đ??ľ1 â&#x2C6;&#x2019; đ??ś1 , so through đ??ťâ&#x20AC;˛ , in other words đ?&#x2018;&#x201E;3 is perpendicular taken from đ??ťâ&#x20AC;˛ on đ??´â&#x20AC;˛ đ?&#x2018;&#x201E;â&#x20AC;˛ ; similarly, đ?&#x2018;?2 â&#x160;Ľ đ?&#x2018;?3 , đ?&#x2018;?2 â&#x160;Ľ đ?&#x2018;?3 , so đ?&#x2018;?3 is perpendicular taken from đ??ťâ&#x20AC;˛ on đ??ś â&#x20AC;˛ đ?&#x2018;&#x201E;â&#x20AC;˛ . To the cevian đ??´đ??´3 will correspond by duality considered to its pole, which is the intersection of the polars of đ??´ and đ??´3 , i.e. the intersection of lines đ?&#x2018;&#x17D; and đ?&#x2018;&#x17D;3 , namely the intersection of đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ with the perpendicular taken from đ??ťâ&#x20AC;˛ on đ??´â&#x20AC;˛ đ?&#x2018;&#x201E;â&#x20AC;˛ ; we denote this

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point by đ?&#x2018;&#x20AC;â&#x20AC;˛ . Analogously, we get the points đ?&#x2018; â&#x20AC;˛ and đ?&#x2018;&#x192;â&#x20AC;˛ .

Thereby, we got the configuration from 1st Theorem and Figure 1, written for triangle đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ of orthocenter đ??ťâ&#x20AC;˛ . Since from 1st Theorem we know that đ?&#x2018;&#x20AC;â&#x20AC;˛ , đ?&#x2018; â&#x20AC;˛ , đ?&#x2018;&#x192;â&#x20AC;˛ are collinear, we get the the cevians đ??´đ??´3 , đ??ľđ??ľ3 , đ??śđ??ś3 are concurrent in the pole of transversal đ?&#x2018;&#x20AC;â&#x20AC;˛ â&#x2C6;&#x2019; đ?&#x2018; â&#x20AC;˛ â&#x2C6;&#x2019; đ?&#x2018;&#x192;â&#x20AC;˛ with respect to the circle â&#x201E;&#x201A;(đ?&#x2018;&#x153;, đ?&#x2018;&#x;), and 2nd Theorem is proved.

References.

[1] Ion Patrascu, Florentin Smarandache: The Dual Theorem concerning Aubertâ&#x20AC;&#x2122;s Line, below.

[2] Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles. The Education Publisher Inc., Columbus, Ohio, USA â&#x20AC;&#x201C; 2012.

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The Dual Theorem Concerning Aubertâ&#x20AC;&#x2122;s Line In this article we introduce the concept of Bobillierâ&#x20AC;&#x2122;s transversal of a triangle with respect to a point in its plan ; we prove the Aubertâ&#x20AC;&#x2122;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 đ??´đ??ľđ??ś be a triangle and đ?&#x2018;&#x20AC; a point in the plane of the triangle so that the perpendiculars taken in đ?&#x2018;&#x20AC;, and đ?&#x2018;&#x20AC;đ??´, đ?&#x2018;&#x20AC;đ??ľ, đ?&#x2018;&#x20AC;đ??ś respectively, intersect the sides đ??ľđ??ś, đ??śđ??´ and đ??´đ??ľ at đ??´đ?&#x2018;&#x161;, đ??ľđ?&#x2018;&#x161; Ĺ&#x;i đ??śđ?&#x2018;&#x161;. Then the points đ??´đ?&#x2018;&#x161;, đ??ľđ?&#x2018;&#x161; and đ??śđ?&#x2018;&#x161; are collinear.

Proof.

We note that

đ??´đ?&#x2018;&#x161;đ??ľ đ??´đ?&#x2018;&#x161;đ??ś

=

aria (đ??ľđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161;) aria (đ??śđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161;)

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Figure 1.

1

Ě&#x201A; ).

Area (đ??ľđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161;) = â&#x2C6;&#x2122; đ??ľđ?&#x2018;&#x20AC; â&#x2C6;&#x2122; đ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; â&#x2C6;&#x2122; sin(đ??ľđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; 2 1

Ě&#x201A; ).

Area (đ??śđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161;) = â&#x2C6;&#x2122; đ??śđ?&#x2018;&#x20AC; â&#x2C6;&#x2122; đ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; â&#x2C6;&#x2122; sin(đ??śđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; 2

Since Ě&#x201A; ) = 3đ?&#x153;&#x2039; â&#x2C6;&#x2019; đ?&#x2018;&#x161;(đ??´đ?&#x2018;&#x20AC;đ??ś Ě&#x201A; ), đ?&#x2018;&#x161;(đ??śđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; 2

it explains that Ě&#x201A; ) = â&#x2C6;&#x2019; cos(đ??´đ?&#x2018;&#x20AC;đ??ś Ě&#x201A; ); sin(đ??śđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; đ?&#x153;&#x2039; Ě&#x201A; ) = sin (đ??´đ?&#x2018;&#x20AC;đ??ľ Ě&#x201A; â&#x2C6;&#x2019; ) = â&#x2C6;&#x2019; cos(đ??´đ?&#x2018;&#x20AC;đ??ľ Ě&#x201A; ). sin(đ??ľđ?&#x2018;&#x20AC;đ??´đ?&#x2018;&#x161; 2 Therefore: Ě&#x201A;) đ??´đ?&#x2018;&#x161;đ??ľ đ?&#x2018;&#x20AC;đ??ľ â&#x2C6;&#x2122; cos(đ??´đ?&#x2018;&#x20AC;đ??ľ (1). = Ě&#x201A;) đ??´đ?&#x2018;&#x161;đ??ś đ?&#x2018;&#x20AC;đ??ś â&#x2C6;&#x2122; cos(đ??´đ?&#x2018;&#x20AC;đ??ś In the same way, we find that:

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Ě&#x201A;) đ??ľđ?&#x2018;&#x161;đ??ś đ?&#x2018;&#x20AC;đ??ś cos(đ??ľđ?&#x2018;&#x20AC;đ??ś (2); = â&#x2C6;&#x2122; Ě&#x201A;) đ??ľđ?&#x2018;&#x161;đ??´ đ?&#x2018;&#x20AC;đ??´ cos(đ??´đ?&#x2018;&#x20AC;đ??ľ Ě&#x201A;) đ??śđ?&#x2018;&#x161;đ??´ đ?&#x2018;&#x20AC;đ??´ cos(đ??´đ?&#x2018;&#x20AC;đ??ś (3). = â&#x2C6;&#x2122; Ě&#x201A;) đ??śđ?&#x2018;&#x161;đ??ľ đ?&#x2018;&#x20AC;đ??ľ cos(đ??ľđ?&#x2018;&#x20AC;đ??ś The relations (1), (2), (3), and the reciprocal Theorem of Menelaus lead to the collinearity of points đ??´đ?&#x2018;&#x161;, đ??ľđ?&#x2018;&#x161;, đ??śđ?&#x2018;&#x161;.

Note.

Bobillier's Theorem can be obtained â&#x20AC;&#x201C; by converting the duality with respect to a circle â&#x20AC;&#x201C; from the theorem relative to the concurrency of the heights of a triangle.

1st Definition.

It is called Bobillierâ&#x20AC;&#x2122;s transversal of a triangle đ??´đ??ľđ??ś with respect to the point đ?&#x2018;&#x20AC; the line containing the intersections of the perpendiculars taken in đ?&#x2018;&#x20AC; on đ??´đ?&#x2018;&#x20AC;, đ??ľđ?&#x2018;&#x20AC;, and đ??śđ?&#x2018;&#x20AC; respectively, with sides đ??ľđ??ś, CA and đ??´đ??ľ.

Note.

The Bobillierâ&#x20AC;&#x2122;s transversal is not defined for any point đ?&#x2018;&#x20AC; in the plane of the triangle đ??´đ??ľđ??ś, for example, where đ?&#x2018;&#x20AC; is one of the vertices or the orthocenter đ??ť of the triangle.

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2nd Definition.

If đ??´đ??ľđ??śđ??ˇ is a convex quadrilateral and đ??¸, đ??š are the intersections of the lines đ??´đ??ľ and đ??śđ??ˇ , đ??ľđ??ś and đ??´đ??ˇ respectively, we say that the figure đ??´đ??ľđ??śđ??ˇđ??¸đ??š is a complete quadrilateral. The complete quadrilateral sides are đ??´đ??ľ, đ??ľđ??ś, đ??śđ??ˇ, đ??ˇđ??´ , and đ??´đ??ś, đ??ľđ??ˇ and đ??¸đ??š are diagonals.

2nd Theorem.

(Newton-Gauss)

The diagonalsâ&#x20AC;&#x2122; 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 diagonalsâ&#x20AC;&#x2122; middles of a complete quadrilateral belong.

3rd Theorem.

(Aubert)

If đ??´đ??ľđ??śđ??ˇđ??¸đ??š is a complete quadrilateral, then the orthocenters đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 of the traingles đ??´đ??ľđ??š, đ??´đ??¸đ??ˇ, đ??ľđ??śđ??¸, and đ??śđ??ˇđ??š respectively, are collinear points.

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Proof.

Let đ??´1 , đ??ľ1 , đ??š1 be the feet of the heights of the triangle đ??´đ??ľđ??š and đ??ť1 its orthocenter (see Fig. 2).

Considering the power of the point đ??ť1 relative to the circle circumscribed to the triangle ABF, and given the triangle orthocenterâ&#x20AC;&#x2122;s property according to which its symmetrics to the triangle sides belong to the circumscribed circle, we find that: đ??ť1 đ??´ â&#x2C6;&#x2122; đ??ť1 đ??´1 = đ??ť1 đ??ľ â&#x2C6;&#x2122; đ??ť1 đ??ľ1 = đ??ť1 đ??š â&#x2C6;&#x2122; đ??ť1 đ??š1 .

Figure 2.

This relationship shows that the orthocenter đ??ť1 has equal power with respect to the circles of diameters [đ??´đ??ś ], [đ??ľđ??ˇ], [đ??¸đ??š ] . As well, we establish that the orthocenters đ??ť2 , đ??ť3 , đ??ť4 have equal powers to these circles. Since the circles of diameters [đ??´đ??ś ], [đ??ľđ??ˇ], [đ??¸đ??š ] have collinear centers (belonging to the Newton-

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Gauss line of the đ??´đ??ľđ??śđ??ˇđ??¸đ??š quadrilateral), it follows that the points đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 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 quadrilateralâ&#x20AC;&#x2122;s orthocenters the line to which the orthocenters đ??ť1 , đ??ť2 , đ??ť3 , đ??ť4 belong.

2. The Aubert Line is perpendicular on the Newton-Gauss line of the quadrilateral (radical axis of two circles is perpendicular to their centersâ&#x20AC;&#x2122; line).

4th Theorem.

(The Dual Theorem of the 3rd Theorem)

If đ??´đ??ľđ??śđ??ˇ is a convex quadrilateral and đ?&#x2018;&#x20AC; is a point in its plane for which there are the Bobillierâ&#x20AC;&#x2122;s transversals of triangles đ??´đ??ľđ??ś , đ??ľđ??śđ??ˇ , đ??śđ??ˇđ??´ and đ??ˇđ??´đ??ľ ; thereupon these transversals are concurrent.

Proof.

Let us transform the configuration in Fig. 2, by duality with respect to a circle of center đ?&#x2018;&#x20AC;.

By the considered duality, the lines đ?&#x2018;&#x17D;, đ?&#x2018;?, đ?&#x2018;?, đ?&#x2018;&#x2018;, đ?&#x2018;&#x2019; Ĺ&#x;i đ?&#x2018;&#x201C; correspond to the points đ??´, đ??ľ, đ??ś, đ??ˇ, đ??¸, đ??š (their polars).

It is known that polars of collinear points are concurrent lines, therefore we have: đ?&#x2018;&#x17D; â&#x2C6;Š đ?&#x2018;? â&#x2C6;Š đ?&#x2018;&#x2019; = {đ??´â&#x20AC;˛ },

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đ?&#x2018;? â&#x2C6;Š đ?&#x2018;? â&#x2C6;Š đ?&#x2018;&#x201C; = {đ??ľ â&#x20AC;˛ } , đ?&#x2018;? â&#x2C6;Š đ?&#x2018;&#x2018; â&#x2C6;Š đ?&#x2018;&#x2019; = {đ??ś â&#x20AC;˛ } , đ?&#x2018;&#x2018; â&#x2C6;Š đ?&#x2018;&#x201C; â&#x2C6;Š đ?&#x2018;&#x17D; = {đ??ˇ â&#x20AC;˛ } , đ?&#x2018;&#x17D; â&#x2C6;Š đ?&#x2018;? = {đ??¸ â&#x20AC;˛ }, đ?&#x2018;? â&#x2C6;Š đ?&#x2018;&#x2018; = {đ??š â&#x20AC;˛ }.

Consequently, by applicable duality, the points â&#x20AC;˛ â&#x20AC;˛ â&#x20AC;˛ đ??´ , đ??ľ , đ??ś , đ??ˇâ&#x20AC;˛ , đ??¸ â&#x20AC;˛ and đ??š â&#x20AC;˛ correspond to the straight lines đ??´đ??ľ, đ??ľđ??ś, đ??śđ??ˇ, đ??ˇđ??´, đ??´đ??ś, đ??ľđ??ˇ.

To the orthocenter đ??ť1 of the triangle đ??´đ??¸đ??ˇ , it corresponds, by duality, its polar, which we denote đ??´1â&#x20AC;˛ â&#x20AC;&#x201C; đ??ľ1â&#x20AC;˛ â&#x2C6;&#x2019; đ??ś1â&#x20AC;˛ , and which is the Bobillierâ&#x20AC;&#x2122;s transversal of the triangle đ??´â&#x20AC;&#x2122;đ??śâ&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122; in relation to the point đ?&#x2018;&#x20AC;. Indeed, the point đ??śâ&#x20AC;&#x2122; corresponds to the line đ??¸đ??ˇ by duality; to the height from đ??´ of the triangle đ??´đ??¸đ??ˇ, also by duality, it corresponds its pole, which is the point đ??ś1â&#x20AC;˛ located Ě&#x201A; â&#x20AC;˛ ) = 900 . on đ??´â&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122; such that đ?&#x2018;&#x161;(đ??śâ&#x20AC;˛đ?&#x2018;&#x20AC;đ??ś 1

To the height from đ??¸ of the triangle đ??´đ??¸đ??ˇ , it Ě&#x201A; â&#x20AC;˛) = corresponds the point đ??ľâ&#x20AC;˛ â&#x2C6;&#x2C6; đ??´â&#x20AC;˛đ??śâ&#x20AC;˛ such that đ?&#x2018;&#x161;(đ??ˇâ&#x20AC;˛đ?&#x2018;&#x20AC;đ??ľ 1

1

900 .

Also, to the height from đ??ˇ, it corresponds đ??´1â&#x20AC;˛ â&#x2C6;&#x2C6; đ??śâ&#x20AC;˛đ??ˇâ&#x20AC;&#x2122; on C such that đ?&#x2018;&#x161;(đ??´â&#x20AC;˛ đ?&#x2018;&#x20AC;đ??´1â&#x20AC;˛ ) = 900 . To the orthocenter đ??ť2 of the triangle đ??´đ??ľđ??š, it will correspond, by applicable duality, the Bobillierâ&#x20AC;&#x2122;s transversal đ??´â&#x20AC;˛2 â&#x2C6;&#x2019; đ??ľ2â&#x20AC;˛ â&#x2C6;&#x2019; đ??ś2â&#x20AC;˛ in the triangle đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ˇâ&#x20AC;˛ relative to the point đ?&#x2018;&#x20AC;.

To the orthocenter đ??ť3 of the triangle đ??ľđ??śđ??¸ , it will correspond the Bobillierâ&#x20AC;&#x2122;s transversal đ??´3 â&#x20AC;˛ â&#x2C6;&#x2019; đ??ľâ&#x20AC;˛ 3 â&#x2C6;&#x2019; đ??ś3 â&#x20AC;˛ in the triangle đ??´â&#x20AC;˛ đ??ľâ&#x20AC;˛ đ??ś â&#x20AC;˛ relative to the point đ?&#x2018;&#x20AC;, and to the orthocenter đ??ť4 of the triangle đ??śđ??ˇđ??š , it will correspond the transversal đ??´â&#x20AC;˛4 â&#x2C6;&#x2019; đ??ľ4â&#x20AC;˛ â&#x2C6;&#x2019; đ??ś4â&#x20AC;˛ in the triangle đ??ś â&#x20AC;˛ đ??ˇâ&#x20AC;˛ đ??ľâ&#x20AC;˛ relative to the point đ?&#x2018;&#x20AC;.

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The Bobillierâ&#x20AC;&#x2122;s transversals đ??´â&#x20AC;˛đ?&#x2018;&#x2013; â&#x2C6;&#x2019; đ??ľđ?&#x2018;&#x2013;â&#x20AC;˛ â&#x2C6;&#x2019; đ??śđ?&#x2018;&#x2013;â&#x20AC;˛ , đ?&#x2018;&#x2013; = Ě&#x2026;Ě&#x2026;Ě&#x2026;Ě&#x2026; 1,4 Ě&#x2026;Ě&#x2026;Ě&#x2026;Ě&#x2026; correspond to the collinear points đ??ťđ?&#x2018;&#x2013; , đ?&#x2018;&#x2013; = 1,4.

These transversals are concurrent in the pole of the line of the orthocenters towards the considered duality.

It results that, given the quadrilateral đ??´â&#x20AC;&#x2122;đ??ľâ&#x20AC;&#x2122;đ??śâ&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122;, the Bobillierâ&#x20AC;&#x2122;s transversals of the triangles đ??´â&#x20AC;&#x2122;đ??śâ&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122; , đ??´â&#x20AC;&#x2122;đ??ľâ&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122; , đ??´â&#x20AC;&#x2122;đ??ľâ&#x20AC;&#x2122;đ??śâ&#x20AC;&#x2122; and đ??śâ&#x20AC;&#x2122;đ??ˇâ&#x20AC;&#x2122;đ??ľâ&#x20AC;&#x2122; relative to the point đ?&#x2018;&#x20AC; are concurrent.

References.

[1]

Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles. The Education Publisher Inc., Columbus, Ohio, USA â&#x20AC;&#x201C; 2012.

[2] Ion Patrascu, Florentin Smarandache: Variance on Topics of Plane Geometry. The Education Publisher Inc., Columbus, Ohio, USA â&#x20AC;&#x201C; 2013.

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180

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 authorsâ&#x20AC;&#x2122; aspiration and attraction, as a poet writes lyrics about spring according to his emotions.