Angles, Arcs, and Chords Advanced Geometry Circles Lesson 2.
Complements to Classic Topics of Circles Geometry
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Transcript of 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
1
Ion Patrascu | Florentin Smarandache
Complements to Classic Topics
of Circles Geometry
Ion Patrascu, Florentin Smarandache
2
In the memory of the first author's father
Mihail Patrascu and the second author's
mother Maria (Marioara) Smarandache,
recently passed to eternity...
Complements to Classic Topics of Circles Geometry
3
Ion Patrascu | Florentin Smarandache
Complements to Classic Topics
of Circles Geometry
Pons Editions
Brussels | 2016
Ion Patrascu, Florentin Smarandache
4
Β© 2016 Ion Patrascu & Florentin Smarandache
All rights reserved. This book is protected by
copyright. No part of this book may be
reproduced in any form or by any means,
including photocopying or using any
information storage and retrieval system
without written permission from the
copyright owners.
ISBN 978-1-59973-465-1
Complements to Classic Topics of Circles Geometry
5
Contents
Introduction ....................................................... 15
Lemoineβs Circles ............................................... 17
1st Theorem. ........................................................... 17
Proof. ................................................................. 17
2nd Theorem. ......................................................... 19
Proof. ................................................................ 19
Remark. ............................................................ 21
References. ........................................................... 22
Lemoineβs Circles Radius Calculus ..................... 23
1st Theorem ........................................................... 23
Proof. ................................................................ 23
Consequences. ................................................... 25
1st Proposition. ...................................................... 25
Proof. ................................................................ 25
2nd Proposition. .................................................... 26
Proof. ................................................................ 27
Remarks. ........................................................... 28
3rd Proposition. ..................................................... 28
Proof. ................................................................ 28
Remark. ............................................................ 29
References. ........................................................... 29
Radical Axis of Lemoineβs Circles ........................ 31
1st Theorem. ........................................................... 31
2nd Theorem. .......................................................... 31
1st Remark. ......................................................... 31
1st Proposition. ...................................................... 32
Proof. ................................................................ 32
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Comment. .......................................................... 34
2nd Remark. ....................................................... 34
2nd Proposition. .................................................... 34
References. ........................................................... 34
Generating Lemoineβs circles ............................. 35
1st Definition. ........................................................ 35
1st Proposition. ...................................................... 35
2nd Definition. ....................................................... 35
1st Theorem. .......................................................... 36
3rd Definition. ....................................................... 36
1st Lemma. ............................................................ 36
Proof. ................................................................ 36
Remark. ............................................................ 38
2nd Theorem. ......................................................... 38
Proof. ................................................................ 38
Further Remarks. .............................................. 40
References. ........................................................... 40
The Radical Circle of Ex-Inscribed Circles of a
Triangle ............................................................. 41
1st Theorem. .......................................................... 41
Proof. ................................................................ 41
2nd Theorem. ......................................................... 43
Proof. ................................................................ 43
Remark. ............................................................ 44
3rd Theorem. ......................................................... 44
Proof. ................................................................ 45
References. ........................................................... 48
The Polars of a Radical Center ........................... 49
1st Theorem. .......................................................... 49
Proof. ................................................................ 50
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2nd Theorem. .......................................................... 51
Proof. ................................................................ 52
Remarks. ........................................................... 52
References. ........................................................... 53
Regarding the First Droz-Farnyβs Circle ............. 55
1st Theorem. .......................................................... 55
Proof. ................................................................ 55
Remarks. ........................................................... 57
2nd Theorem. ......................................................... 57
Remark. ............................................................ 58
Definition. ............................................................ 58
Remark. ............................................................ 58
3rd Theorem. ......................................................... 58
Proof. ................................................................ 59
Remark. ............................................................ 60
4th Theorem. ......................................................... 60
Proof. ................................................................ 60
Reciprocally. ..................................................... 61
Remark. ............................................................ 62
References. ........................................................... 62
Regarding the Second Droz-Farnyβs Circle.......... 63
1st Theorem. .......................................................... 63
Proof. ................................................................ 63
1st Proposition. ...................................................... 65
Proof. ................................................................ 65
Remarks. ........................................................... 66
2nd Theorem. ......................................................... 66
Proof. ................................................................ 67
Reciprocally. ..................................................... 67
Remarks. ........................................................... 68
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2nd Proposition. .................................................... 68
Proof. ................................................................ 69
References. ........................................................... 70
Neubergβs Orthogonal Circles ............................. 71
1st Definition. ......................................................... 71
2nd Definition. ....................................................... 72
3rd Definition. ....................................................... 72
1st Proposition. ...................................................... 72
Proof. ................................................................ 73
Consequence. .................................................... 74
2nd Proposition. .................................................... 74
Proof. ................................................................ 74
4th Definition. ....................................................... 75
3rd Proposition. ..................................................... 75
Proof. ................................................................ 75
Reciprocally. ..................................................... 76
4th Proposition. ..................................................... 76
Proof. ................................................................ 77
References. ........................................................... 78
Lucasβs Inner Circles .......................................... 79
1. Definition of the Lucasβs 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 Lucasβs Inner Circles .......... 81
1st Theorem. .......................................................... 81
Proof. ................................................................ 81
2nd Definition. ....................................................... 83
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Remark. ............................................................ 83
2nd Theorem. ......................................................... 84
3rd Definition. ....................................................... 84
3rd Theorem. ......................................................... 84
Proof. ................................................................ 85
Remarks. ........................................................... 86
4th Definition. ....................................................... 87
1st Property. .......................................................... 87
Proof. ................................................................ 87
2nd Property. ......................................................... 88
Proof. ................................................................ 88
References. ........................................................... 88
Theorems with Parallels Taken through a
Triangleβs Vertices and Constructions Performed
only with the Ruler ............................................ 89
1st Problem. ........................................................... 89
2nd Problem........................................................... 89
3rd Problem. .......................................................... 90
1st Lemma. ............................................................ 90
Proof. ................................................................ 91
Remark. ............................................................ 92
2nd Lemma. ........................................................... 92
Proof. ................................................................ 92
3rd Lemma. ........................................................... 93
Solution. ............................................................ 93
Constructionβs Proof. ........................................ 93
Remark. ............................................................ 94
1st Theorem. ......................................................... 95
Proof. ................................................................ 96
2nd Theorem. ....................................................... 97
Ion Patrascu, Florentin Smarandache
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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
Apolloniusβs Circles of kth Rank ......................... 103
1st Definition. ...................................................... 103
2nd Definition. ..................................................... 103
1st Theorem. ........................................................ 103
Proof. .............................................................. 104
3rd Definition. ..................................................... 105
2nd Theorem. ....................................................... 105
Proof. .............................................................. 106
3rd Theorem. ....................................................... 107
Proof. .............................................................. 107
4th Theorem. ....................................................... 108
Proof. .............................................................. 108
1st Remark. ...................................................... 108
5th Theorem. ....................................................... 108
Proof. .............................................................. 109
6th Theorem. ....................................................... 109
Proof. .............................................................. 109
4th Definition. ...................................................... 110
1st Property. ......................................................... 110
2nd Remark. ...................................................... 110
7th Theorem. ........................................................ 111
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Proof. ............................................................... 111
References. .......................................................... 112
Apolloniusβs Circle of Second Rank ................... 113
1st Definition. ....................................................... 113
1st Theorem. ......................................................... 113
1st Proposition. ..................................................... 114
Proof. ............................................................... 114
Remarks. .......................................................... 115
2nd Definition. ...................................................... 115
3rd Definition. ...................................................... 116
2nd Proposition. ................................................... 116
Proof. ............................................................... 116
Remarks. .......................................................... 118
Open Problem. ..................................................... 119
References. ......................................................... 120
A Sufficient Condition for the Circle of the 6 Points
to Become Eulerβs Circle ................................... 121
1st Definition. ....................................................... 121
1st Remark. ....................................................... 122
2nd Definition. ...................................................... 122
2nd Remark. ...................................................... 122
1st Proposition. ..................................................... 122
Proof. ............................................................... 122
3rd Remark. ...................................................... 123
3rd Definition. ...................................................... 123
4th Remark. ..................................................... 124
1st Theorem. ........................................................ 124
Proof. .............................................................. 124
4th Definition. ..................................................... 126
2nd Proposition. .................................................. 126
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1st Proof. .......................................................... 126
2nd Proof. .......................................................... 127
5th Remark. ..................................................... 128
References. ......................................................... 128
An Extension of a Problem of Fixed Point ......... 129
Proof. .............................................................. 130
1st Remark. ....................................................... 132
2nd Remark. ...................................................... 132
3rd Remark. ...................................................... 133
References. .......................................................... 133
Some Properties of the Harmonic Quadrilateral 135
1st Definition. ....................................................... 135
2nd Definition. ...................................................... 135
1st Proposition. ..................................................... 135
2nd Proposition. .................................................. 136
Proof. .............................................................. 136
Remark 1. ......................................................... 137
3rd Proposition. .................................................... 137
Remark 2. ........................................................ 138
Proposition 4. ..................................................... 138
Proof. .............................................................. 139
3rd Remark. ..................................................... 140
5th Proposition. ................................................... 140
6th Proposition. ................................................... 140
Proof. .............................................................. 140
3rd Definition. ..................................................... 142
4th Remark. ..................................................... 142
7th Proposition. ................................................... 143
Proof. .............................................................. 143
5th Remark. ..................................................... 144
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8th Proposition. ................................................... 145
Proof. .............................................................. 145
4th Definition. ..................................................... 146
6th Remark. ..................................................... 146
9th Proposition. ................................................... 146
Proof. .............................................................. 146
10th Proposition. ..................................................147
Proof. ...............................................................147
7th Remark....................................................... 148
11th Proposition. .................................................. 148
Proof. .............................................................. 149
References. ......................................................... 149
Triangulation of a Triangle with Triangles having
Equal Inscribed Circles ..................................... 151
Solution. .............................................................. 151
Building the point D. ........................................... 153
Proof. .............................................................. 154
Discussion. ....................................................... 155
Remark. .......................................................... 156
Open problem. .................................................... 156
An Application of a Theorem of Orthology ........ 157
Proposition. ......................................................... 157
Proof. .............................................................. 158
Remark. .......................................................... 160
Note. ............................................................... 160
References. ......................................................... 162
The Dual of a Theorem Relative to the Orthocenter
of a Triangle ..................................................... 163
1st Theorem. ........................................................ 163
Proof. .............................................................. 164
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Note. ............................................................... 165
2nd Theorem. ....................................................... 165
Proof. .............................................................. 166
References. ......................................................... 168
The Dual Theorem Concerning Aubertβs Line .... 169
1st Theorem. ........................................................ 169
Proof. .............................................................. 169
Note. ................................................................ 171
1st Definition. ....................................................... 171
Note. ................................................................ 171
2nd Definition. ...................................................... 172
2nd Theorem. ........................................................ 172
Note. ................................................................ 172
3rd Theorem. ........................................................ 172
Proof. ............................................................... 173
Notes. ...............................................................174
4th Theorem. ........................................................174
Proof. ...............................................................174
References. ..........................................................176
Bibliography ..................................................... 177
Complements to Classic Topics of Circles Geometry
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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
authorsβ aspiration and attraction, as a poet
writes lyrics about spring according to his
emotions.
Ion Patrascu, Florentin Smarandache
16
Complements to Classic Topics of Circles Geometry
17
Lemoineβs Circles
In this article, we get to Lemoine's circles
in a different manner than the known one.
1st Theorem.
Let π΄π΅πΆ a triangle and πΎ its simedian center. We
take through K the parallel π΄1π΄2 to π΅πΆ, π΄1 β (π΄π΅), π΄2 β
(π΄πΆ); through π΄2 we take the antiparallels π΄2π΅1 to π΄π΅
in relation to πΆπ΄ and πΆπ΅ , π΅1 β (π΅πΆ) ; through π΅1 we
take the parallel π΅1π΅2 to π΄πΆ , π΅2 β π΄π΅; through π΅2 we
take the antiparallels π΅1πΆ1 to π΅πΆ , πΆ1 β (π΄πΆ) , and
through πΆ1 we take the parallel πΆ1πΆ2 to π΄π΅, πΆ1 β (π΅πΆ).
Then:
i. πΆ2π΄1 is an antiparallel of π΄πΆ;
ii. π΅1π΅2 β© πΆ1πΆ2 = {πΎ};
iii. The points π΄1, π΄2 , π΅1 , π΅2, πΆ1, πΆ2 are
concyclical (the first Lemoineβs circle).
Proof.
i. The quadrilateral π΅πΆ2πΎπ΄ is a
parallelogram, and its center, i.e. the
middle of the segment (πΆ2π΄1), belongs to
the simedian π΅πΎ; it follows that πΆ2π΄2 is
an antiparallel to π΄πΆ(see Figure 1).
Ion Patrascu, Florentin Smarandache
18
ii. Let {πΎβ²} = π΄1π΄2 β© π΅1π΅2 , because the
quadrilateral πΎβ²π΅1πΆπ΄2 is a parallelogram;
it follows that πΆπΎβ² is a simedian; also,
πΆπΎ is a simedian, and since πΎ,πΎβ² β π΄1π΄2, it
follows that we have πΎβ² = πΎ.
iii. π΅2πΆ1 being an antiparallel to π΅πΆ and
π΄1π΄2 β₯ π΅πΆ , it means that π΅2πΆ1 is an
antiparallel to π΄1π΄2 , so the points
π΅2, πΆ1, π΄2, π΄1 are concyclical. From π΅1π΅2 β₯
π΄πΆ , β’π΅2πΆ1π΄ β‘ β’π΄π΅πΆ , β’π΅1π΄2πΆ β‘ β’π΄π΅πΆ we
get that the quadrilateral π΅2πΆ1π΄2π΅1 is an
isosceles trapezoid, so the points
π΅2, πΆ1, π΄2, π΅1 are concyclical. Analogously,
it can be shown that the quadrilateral
πΆ2π΅1π΄2π΄1 is an isosceles trapezoid,
therefore the points πΆ2, π΅1, π΄2, π΄1 are
concyclical.
Figure 1
Complements to Classic Topics of Circles Geometry
19
From the previous three quartets of concyclical
points, it results the concyclicity of the points
belonging to the first Lemoineβs circle.
2nd Theorem.
In the scalene triangle π΄π΅πΆ, let K be the simedian
center. We take from πΎ the antiparallel π΄1π΄2 to π΅πΆ ;
π΄1 β π΄π΅, π΄2 β π΄πΆ; through π΄2 we build π΄2π΅1 β₯ π΄π΅; π΅1 β
(π΅πΆ), then through π΅1 we build π΅1π΅2 the antiparallel
to π΄πΆ , π΅2 β (π΄π΅), and through π΅2 we build π΅2πΆ1 β₯ π΅πΆ ,
πΆ1 β π΄πΆ , and, finally, through πΆ1 we take the
antiparallel πΆ1πΆ2 to π΄π΅, πΆ2 β (π΅πΆ).
Then:
i. πΆ2π΄1 β₯ π΄πΆ;
ii. π΅1π΅2 β© πΆ1πΆ2 = {πΎ};
iii. The points π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are
concyclical (the second Lemoineβs circle).
Proof.
i. Let {πΎβ²} = π΄1π΄2 β© π΅1π΅2 , having β’π΄π΄1π΄2 =
β’π΄πΆπ΅ and β’π΅π΅1π΅2 β‘ β’π΅π΄πΆ because π΄1π΄2
Θi π΅1π΅2 are antiparallels to π΅πΆ , π΄πΆ ,
respectively, it follows that β’πΎβ²π΄1π΅2 β‘
β’πΎβ²π΅2π΄1 , so πΎβ²π΄1 = πΎβ²π΅2 ; having π΄1π΅2 β₯
π΅1π΄2 as well, it follows that also πΎβ²π΄2 =
πΎβ²π΅1 , so π΄1π΄2 = π΅1π΅2 . Because πΆ1πΆ2 and
π΅1π΅2 are antiparallels to π΄π΅ and π΄πΆ , we
Ion Patrascu, Florentin Smarandache
20
have πΎ"πΆ2 = πΎ"π΅1; we noted {πΎ"} = π΅1π΅2 β©
πΆ1πΆ2; since πΆ1π΅2 β₯ π΅1πΆ2, we have that the
triangle πΎ"πΆ1π΅2 is also isosceles, therefore
πΎ"πΆ1 = πΆ1π΅2, and we get that π΅1π΅2 = πΆ1πΆ2.
Let {πΎβ²β²β²} = π΄1π΄2 β© πΆ1πΆ2 ; since π΄1π΄2 and
πΆ1πΆ2 are antiparallels to π΅πΆ and π΄π΅ , we
get that the triangle πΎβ²β²β²π΄2πΆ1 is isosceles,
so πΎβ²β²β²π΄2 = πΎβ²β²β²πΆ1 , but π΄1π΄2 = πΆ1πΆ2 implies
that πΎβ²β²β²πΆ2 = πΎβ²β²β²π΄1 , then β’πΎβ²β²β²π΄1πΆ2 β‘
β’πΎβ²β²β²π΄2πΆ1 and, accordingly, πΆ2π΄1 β₯ π΄πΆ.
Figure 2
ii. We noted {πΎβ²} = π΄1π΄2 β© π΅1π΅2 ; let {π} =
π΅2πΆ1 β© π΅1π΄2 ; obviously, π΅π΅1ππ΅2 is a
parallelogram; if πΎ0 is the middle of
(π΅1π΅2), then π΅πΎ0 is a simedian, since π΅1π΅2
is an antiparallel to π΄πΆ, and the middle of
the antiparallels of π΄πΆ are situated on the
Complements to Classic Topics of Circles Geometry
21
simedian π΅πΎ . If πΎ0 β K, then πΎ0πΎ β₯ π΄1π΅2
(because π΄1π΄2 = π΅1π΅2 and π΅1π΄2 β₯ π΄1π΅2 ),
on the other hand, π΅, πΎ0, K are collinear
(they belong to the simedian π΅πΎ ),
therefore πΎ0πΎ intersects π΄π΅ in π΅, which is
absurd, so πΎ0 =K, and, accordingly, π΅1π΅2 β©
π΄1π΄2 = {πΎ} . Analogously, we prove that
πΆ1πΆ2 β© π΄1π΄2 = {πΎ}, so π΅1π΅2 β© πΆ1πΆ2 = {πΎ}.
iii. K is the middle of the congruent
antiparalells π΄1π΄2 , π΅1π΅2 , πΆ1πΆ2 , so πΎπ΄1 =
πΎπ΄2 = πΎπ΅1 = πΎπ΅2 = πΎπΆ1 = πΎπΆ2 . The
simedian center πΎ is the center of the
second Lemoineβs circle.
Remark.
The center of the first Lemoineβs circle is the
middle of the segment [ππΎ], where π is the center of
the circle circumscribed to the triangle π΄π΅πΆ. Indeed,
the perpendiculars taken from π΄, π΅, πΆ on the
antiparallels π΅2πΆ1 , π΄1πΆ2 , π΅1π΄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 π΄, π΅, πΆ). The mediatrix of
the segment π΅2πΆ1 pass though the middle of π΅2πΆ1 ,
which coincides with the middle of π΄πΎ , so is the
middle line in the triangle π΄πΎπ passing through the
middle of (ππΎ) . Analogously, it follows that the
mediatrix of π΄1πΆ2 pass through the middle πΏ1 of [ππΎ].
Ion Patrascu, Florentin Smarandache
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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 MiculiΘa,
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
[Tuckerβs Circle], in Alpha journal, year XV, no. 2/2009.
Complements to Classic Topics of Circles Geometry
23
Lemoineβs Circles Radius
Calculus
For the calculus of the first Lemoineβs
circle, we will first prove:
1st Theorem (E. Lemoine β 1873)
The first Lemoineβs circle divides the sides of a
triangle in segments proportional to the squares of the
triangleβs sides.
Each extreme segment is proportional to the
corresponding adjacent side, and the chord-segment
in the Lemoineβs circle is proportional to the square of
the side that contains it.
Proof.
We will prove that π΅πΆ2
π2 = πΆ2π΅1
π2 = π΅1πΆ
π2 .
In figure 1, πΎ represents the symmedian center
of the triangle π΄π΅πΆ , and π΄1π΄2 ; π΅1π΅2 ; πΆ1πΆ2 represent
Lemoine parallels.
The triangles π΅πΆ2π΄1 ; πΆπ΅1π΄2 and πΎπΆ2π΄1 have
heights relative to the sides π΅πΆ2; π΅1πΆ and πΆ2π΅1 equal
(π΄1π΄2 β₯ π΅πΆ).
Ion Patrascu, Florentin Smarandache
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Hence: π΄πππβ π΅π΄1πΆ2
π΅πΆ2 =
π΄πππβ πΎπΆ2π΄1
πΆ2π΅1 =
π΄πππβ πΆπ΅1π΄2
π΅1πΆ . (1)
Figure 1
On the other hand: π΄1πΆ2 and π΅1π΄2 being
antiparallels with respect to π΄πΆ and π΄π΅, it follows that
Ξπ΅πΆ2π΄1~π₯π΅π΄πΆ and ΞπΆπ΅1π΄2~π₯πΆπ΄π΅ , likewise πΎπΆ2 β₯ π΄πΆ
implies: ΞπΎπΆ2π΅1~π₯π΄π΅πΆ.
We obtain: π΄πππβ π΅πΆ2π΄1
π΄πππβ π΄π΅πΆ =
π΅πΆ22
π2 ;
π΄πππβ πΎπΆ2π΅1
π΄πππβ π΄π΅πΆ=
πΆ2π΅12
π2 ;
π΄πππβ πΆπ΅1π΄2
π΄πππβ π΄π΅πΆ =
πΆπ΅12
π2 . (2)
If we denote π΄πππβ π΄π΅πΆ = π, we obtain from the
relations (1) and (2) that: π΅πΆ2
π2 = πΆ2π΅1
π2 = π΅1πΆ
π2 .
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25
Consequences.
1. According to the 1st Theorem, we find that:
π΅πΆ2 =ππ2
π2+π2+π2; π΅1πΆ =ππ2
π2+π2+π2; π΅1πΆ2 =π3
π2+π2+π2 .
2. We also find that: π΅1πΆ2
π3 = π΄2πΆ1
π3 = π΄1π΅2
π3 ,
meaning that:
βThe chords determined by the first Lemoineβs
circle on the triangleβs sides are proportional to
the cubes of the sides.β
Due to this property, the first Lemoineβs circle is
known in England by the name of triplicate ratio circle.
1st Proposition.
The radius of the first Lemoineβs circle, π πΏ1 is
given by the formula:
π πΏ1
2 =1
4βπ 2(π2+π2+π2)+ π2π2π2
(π2+π2+π2)2, (3)
where π represents the radius of the circle inscribed
in the triangle π΄π΅πΆ.
Proof.
Let πΏ be the center of the first Lemoineβs circle
that is known to represent the middle of the segment
(ππΎ) β π being the center of the circle inscribed in the
triangle π΄π΅πΆ.
Ion Patrascu, Florentin Smarandache
26
Considering C1, we obtain π΅π΅1 =π (π2+π2)
π2+π2+π2 .
Taking into account the power of point π΅ in
relation to the first Lemoineβs circle, we have:
π΅πΆ2 β π΅π΅1 = π΅π2 β πΏπ2,
(π΅π is the tangent traced from π΅ to the first Lemoineβs
circle, see Figure 1).
Hence: π πΏ1
2 = π΅πΏ2 β π΅πΆ2 β π΅π΅1. (4)
The median theorem in triangle π΅ππΎ implies
that:
π΅πΏ2 =2β(π΅πΎ2+π΅π2)βππΎ2
4 .
It is known that πΎ =(π2+π2)βππ
π2+π2+π2 ; ππ =2ππβππ
π2+π2 ,
where ππ and ππ are the lengths of the symmedian
and the median from π΅, and ππΎ2 = π 2 β 3π2π2π2
(π2+π2+π2) , see
(3).
Consequently: π΅πΎ2 =2π2π2(π2+π2)βπ2π2π2
(π2+π2+π2)2 , and
4π΅πΏ2 = π 2 +4π2π2(π2+π2)+π2π2π2
(π2+π2+π2)2 .
As: π΅πΆ2 β π΅π΅1 = π2π2(π2+π2)
(π2+π2+π2)2 , by replacing in (4),
we obtain formula (3).
2nd Proposition.
The radius of the second Lemoineβs circle, π πΏ2, is
given by the formula:
π πΏ2=
πππ
π2+π2+π2 . (5)
Complements to Classic Topics of Circles Geometry
27
Proof.
Figure 2
In Figure 2 above, π΄1π΄2; π΅1π΅2; πΆ1πΆ2 are Lemoine
antiparallels traced through symmedian center πΎ that
is the center of the second Lemoineβs circle, thence:
π πΏ2= πΎπ΄1 = πΎπ΄2.
If we note with π and π the feet of the
symmedian and the median from π΄, it is known that:
π΄πΎ
πΎπ=
π2+π2
π2 .
From the similarity of triangles π΄π΄2π΄1 and π΄π΅πΆ,
we have: π΄1π΄2
π΅πΆ=
π΄πΎ
π΄π .
But: π΄πΎ
π΄π=
π2+π2
π2+π2+π2 and π΄π =2ππ
π2+π2 β ππ.
π΄1π΄2 = 2π πΏ2, π΅πΆ = π, therefore:
π πΏ2=
π΄πΎβπ
2ππ ,
and as π΄πΎ = 2ππ β ππ
π2+π2+π2 , formula (5) is a consequence.
Ion Patrascu, Florentin Smarandache
28
Remarks.
1. If we use π‘ππ =4π
π2+π2+π2, π being the Brocardβs
angle (see [2]), we obtain: π πΏ2= π β π‘ππ.
2. If, in Figure 1, we denote with π1 the middle
of the antiparallel π΅2πΆ1, which is equal to π πΏ2 (due to
their similarity), we thus find from the rectangular
triangle πΏπ1πΆ1 that:
πΏπΆ12 = πΏπ1
2 + π1πΆ12 , but πΏπ1
2 =1
4π2 and π1πΆ2 =
1
2π πΏ2
; it follows that:
π πΏ1
2 =1
4(π 2 + π πΏ2
2 ) =π 2
4(1 + π‘π2π).
We obtain:
π πΏ1=
π
2β β1 + π‘π2π .
3rd Proposition.
The chords determined by the sides of the
triangle in the second Lemoineβs circle are
respectively proportional to the opposing angles
cosines.
Proof.
πΎπΆ2π΅1 is an isosceles triangle, β’πΎπΆ2π΅1 =
β’πΎπ΅1πΆ2 = β’π΄; as πΎπΆ2 = π πΏ2 we have that cos π΄ =
πΆ2π΅1
2π πΏ2
,
deci πΆ2π΅1
cosπ΄= 2π πΏ2
, similary: π΄2πΆ1
cosπ΅=
π΅2π΄1
cosπΆ= 2π πΏ2.
Complements to Classic Topics of Circles Geometry
29
Remark.
Due to this property of the Lemoineβs 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 MiculiΘa,
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.
Ion Patrascu, Florentin Smarandache
30
Complements to Classic Topics of Circles Geometry
31
Radical Axis of Lemoineβs
Circles
In this article, we emphasize the radical
axis of the Lemoineβs circles.
For the start, let us remind:
1st Theorem.
The parallels taken through the simmedian
center πΎ of a triangle to the sides of the triangle
determine on them six concyclic points (the first
Lemoineβs circle).
2nd Theorem.
The antiparallels taken through the triangleβs
simmedian center to the sides of a triangle determine
six concyclic points (the second Lemoineβs circle).
1st Remark.
If π΄π΅πΆ is a scalene triangle and πΎ is its
simmedian center, then πΏ , the center of the first
Lemoineβs circle, is the middle of the segment [ππΎ],
where π is the center of the circumscribed circle, and
Ion Patrascu, Florentin Smarandache
32
the center of the second Lemoineβs circle is πΎ . It
follows that the radical axis of Lemoineβs circles is
perpendicular on the line of the centers πΏπΎ, therefore
on the line ππΎ.
1st Proposition.
The radical axis of Lemoineβs circles is perpendi-
cular on the line ππΎ raised in the simmedian center πΎ.
Proof.
Let π΄1π΄2 be the antiparallel to π΅πΆ taken through
πΎ, then πΎπ΄1 is the radius π πΏ2 of the second Lemoineβs
circle; we have:
π πΏ2=
πππ
π2+π2+π2 .
Figure 1
Complements to Classic Topics of Circles Geometry
33
Let π΄1β² π΄2
β² be the Lemoineβs parallel taken to π΅πΆ;
we evaluate the power of πΎ towards the first
Lemoineβs circle. We have:
πΎπ΄1β² β πΎπ΄2
β² = πΏπΎ2 β π πΏ1
2 . (1)
Let π be the simmedian leg from π΄; it follows
that: πΎπ΄1
β²
π΅π=
π΄πΎ
π΄πβ
πΎπ΄2β²
ππΆ .
We obtain:
πΎπ΄1β² = π΅π β
π΄πΎ
π΄π and πΎπ΄2
β² = ππΆ βπ΄πΎ
π΄π ,
but π΅π
ππΆ=
π2
π2 and π΄πΎ
π΄π=
π2+π2
π2+π2+π2 .
Therefore:
πΎπ΄1β² β πΎπ΄2
β² = βπ΅π β ππΆ β (π΄πΎ
π΄π)2
=βπ2π2π2
(π2 + π2)2;
(π2+π2)2
(π2+π2+π2)2= βπ πΏ2
2 . (2)
We draw the perpendicular in πΎ on the line
πΏπΎ and denote by π and π its intersection to the first
Lemoineβs circle.
We have πΎπ β πΎπ = βπ πΏ2
2 ; by the other hand,
πΎπ = πΎπ (ππ is a chord which is perpendicular to the
diameter passing through πΎ).
It follows that πΎπ = πΎπ = π πΏ2, so π and π are
situated on the second Lemoineβs circle.
Because ππ is a chord which is common to the
Lemoineβs circles, it obviously follows that ππ is the
radical axis.
Ion Patrascu, Florentin Smarandache
34
Comment.
Equalizing (1) and (2), or by the Pythagorean
theorem in the triangle ππΎπΏ, we can calculate π πΏ1.
It is known that: ππΎ2 = π 2 β3π2π2π2
(π2+π2+π2)2 , and
since πΏπΎ =1
2ππΎ, we find that:
π πΏ1
2 =1
4β [π 2 +
π2π2π2
(π2+π2+π2)2].
2nd Remark.
The 1st Proposition, ref. the radical axis of the
Lemoineβs circles, is a particular case of the following
Proposition, which we leave to the reader to prove.
2nd Proposition.
If π(π1, π 1) Εi π(π2, π 2) are two circles such as
the power of center π1 towards π(π2, π 2) is βπ 12, then
the radical axis of the circles is the perpendicular in
π1 on the line of centers π1π2.
References.
[1] F. Smarandache, Ion Patrascu: The Geometry of Homological Triangles, Education Publisher, Ohio, USA, 2012.
[2] Ion Patrascu, F. Smarandache: Variance on Topics of Plane Geometry, Education Publisher, Ohio, USA, 2013.
Complements to Classic Topics of Circles Geometry
35
Generating Lemoineβs circles
In this paper, we generalize the theorem
relative to the first Lemoineβs circle and
thereby highlight a method to build
Lemoineβs 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 π΄π΅πΆ, the cevian π΄π, π β (π΅πΆ), is a
simedian if and only if ππ΅
ππΆ= (
π΄π΅
π΄πΆ)2. For Proof, see [2].
2nd Definition.
It is called a simedian center of a triangle (or
Lemoineβs point) the intersection of triangleβs
simedians.
Ion Patrascu, Florentin Smarandache
36
1st Theorem.
The parallels to the sides of a triangle taken
through the simedian center intersect the triangleβs
sides in six concyclic points (the first Lemoineβs circle
- 1873).
A Proof of this theorem can be found in [2].
3rd Definition.
We assert that in a scalene triangle π΄π΅πΆ the line
ππ, where π β π΄π΅ and π β π΄πΆ, is an antiparallel to π΅πΆ
if β’πππ΄ β‘ β’π΄π΅πΆ.
1st Lemma.
In the triangle π΄π΅πΆ , let π΄π be a simedian, π β
(π΅πΆ). If π is the middle of the segment (ππ), having
π β (π΄π΅) and π β (π΄πΆ), belonging to the simedian π΄π,
then ππ and π΅πΆ are antiparallels.
Proof.
We draw through π and π, ππ β₯ π΄πΆ and ππ β₯ π΄π΅,
π , π β (π΅πΆ) , see Figure 1. Let {π} = ππ β© ππ ; since
ππ = ππ and π΄πππ is a parallelogram, it follows that
π β π΄π.
Complements to Classic Topics of Circles Geometry
37
Figure 1.
Thales's Theorem provides the relations: π΄π
π΄πΆ=
π΅π
π΅πΆ; (1)
π΄π΅
π΄π=
π΅πΆ
πΆπ . (2)
From (1) and (2), by multiplication, we obtain: π΄π
π΄πβπ΄π΅
π΄πΆ=
π΅π
ππΆ . (3)
Using again Thales's Theorem, we obtain: π΅π
π΅π=
π΄π
π΄π , (4)
ππΆ
ππΆ=
π΄π
π΄π . (5)
From these relations, we get π΅π
π΅π=
ππΆ
ππΆ , (6)
or π΅π
ππΆ=
π΅π
ππΆ . (7)
In view of Proposition 1, the relations (7) and (3)
lead to π΄π
π΄π΅=
π΄π΅
π΄πΆ, which shows that βπ΄ππ~βπ΄πΆπ΅ , so
β’π΄ππ β‘ β’π΄π΅πΆ.
Ion Patrascu, Florentin Smarandache
38
Therefore, ππ and π΅πΆ are antiparallels in
relation to π΄π΅ and π΄πΆ.
Remark.
1. The reciprocal of Lemma 1 is also valid,
meaning that if π is the middle of the antiparallel ππ
to π΅πΆ, then π belongs to the simedian from π΄.
2nd Theorem. (Generalization of the 1st Theorem)
Let π΄π΅πΆ be a scalene triangle and πΎ its simedian
center. We take π β π΄πΎ and draw ππ β₯ π΄π΅,ππ β₯ π΄πΆ ,
where π β π΅πΎ, π β πΆπΎ. Then:
i. ππ β₯ π΅πΆ;
ii. ππ,ππ and ππ intersect the sides of
triangle π΄π΅πΆ in six concyclic points.
Proof.
In triangle π΄π΅πΆ, let π΄π΄1, π΅π΅1, πΆπΆ1 the simedians
concurrent in πΎ (see Figure 2).
We have from Thales' Theorem that: π΄π
ππΎ=
π΅π
ππΎ; (1)
π΄π
ππΎ=
πΆπ
ππΎ . (2)
From relations (1) and (2), it follows that π΅π
ππΎ=
πΆπ
ππΎ, (3)
which shows that ππ β₯ π΅πΆ.
Complements to Classic Topics of Circles Geometry
39
Let π , π, π,π, π, π be the intersection points of
the parallels ππ,ππ,ππ of the sides of the triangles to
the other sides.
Obviously, by construction, the quadrilaterals
π΄πππ; πΆπππ; π΅π ππ are parallelograms.
The middle of the diagonal ππ falls on π΄π, so on
the simedian π΄πΎ, and from 1st Lemma we get that ππ
is an antiparallel to π΅πΆ.
Since ππ β₯ π΅πΆ , it follows that ππ and ππ are
antiparallels, therefore the points π, π, π, π are
concyclic (4).
Figure 2.
Analogously, we show that the points π, π, π , π
are concyclic (5). From ππ and π΅πΆ antiparallels, ππ
and π΄π΅ antiparallels, we have that β’πππ΄ β‘ β’π΄π΅πΆ and
β’πππΆ β‘ β’π΄π΅πΆ , therefore: β’πππ΄ β‘ β’πππΆ , and since
Ion Patrascu, Florentin Smarandache
40
ππ β₯ π΄πΆ, it follows that the trapeze ππππ is isosceles,
therefore the points π, π, π, π are concyclic (6).
The relations (4), (5), (6) drive to the
concyclicality of the points π , π, π, π,π, π, and the
theorem is proved.
Further Remarks.
2. For any point π found on the simedian π΄π΄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 π.
4. We get a circle built as in 2nd Theorem from
the first Lemoineβs circle by homothety of pole π and
of ratio π β β.
5. The centers of Lemoineβs circles built as above
belong to the line ππΎ , where π is the center of the
circle circumscribed to the triangle π΄π΅πΆ.
References.
[1] Exercices de Géométrie, par F.G.M., Huitième
Γ©dition, Paris VIe, Librairie GΓ©nΓ©rale, 77, Rue Le Vaugirard.
[2] Ion Patrascu, Florentin Smarandache: Variance on
topics of Plane Geometry, Educational Publishing, Columbus, Ohio, 2013.
Complements to Classic Topics of Circles Geometry
41
The Radical Circle of Ex-
Inscribed Circles of a Triangle
In this article, we prove several theorems
about the radical center and the radical
circle of ex-inscribed circles of a triangle
and calculate the radius of the circle from
vectorial considerations.
1st Theorem.
The radical center of the ex-inscribed circles of
the triangle π΄π΅πΆ is the Spieckerβs point of the triangle
(the center of the circle inscribed in the median
triangle of the triangle π΄π΅πΆ).
Proof.
We refer in the following to the notation in
Figure 1. Let πΌπ , πΌπ , πΌπ 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:
π΄πΉπ = π΄πΈπ = π΅π·π = π΅πΉπ = πΆπ·π = πΆπΈπ = π,
π΅π·π = π΅πΉπ = πΆπ·π = πΆπΈπ = π β π,
Ion Patrascu, Florentin Smarandache
42
πΆπΈπ = πΆπ·π = π΄πΉπ = π΄πΈπ = π β π,
π΄πΉπ = π΄πΈπ = π΅πΉπ = π΅π·π = π β π.
If π΄1 is the middle of segment π·ππ·π , it follows
that π΄1 has equal powers to the ex-inscribed circles
(πΌπ) and (πΌπ). Of the previously set, we obtain that π΄1
is the middle of the side π΅πΆ.
Figure 1.
Also, the middles of the segments πΈππΈπ and πΉππΉπ,
which we denote π and π, have equal powers to the
circles (πΌπ) and (πΌπ).
The radical axis of the circles ( πΌπ ), ( πΌπ ) will
include the points π΄1, π, π.
Because π΄πΈπ = π΄πΉπ and π΄πΈπ = π΄πΉπ, it follows that
π΄π = π΄π and we find that β’π΄ππ =1
2β’π΄, therefore the
Complements to Classic Topics of Circles Geometry
43
radical axis of the ex-inscribed circles (πΉπ) and (πΌπ) is
the parallel taken through the middle π΄1 of the side π΅πΆ
to the bisector of the angle π΅π΄πΆ.
Denoting π΅1 and πΆ1 the middles of the sides π΄πΆ,
π΄π΅, respectively, we find that the radical center of the
ex-inscribed circles is the center of the circle inscribed
in the median triangle π΄1π΅1πΆ1 of the triangle π΄π΅πΆ.
This point, denoted π, is the Spieckerβs point of
the triangle ABC.
2nd Theorem.
The radical center of the inscribed circle (πΌ) and
of the π΅ βex-inscribed and πΆ βex-inscribed circles of
the triangle π΄π΅πΆ is the center of the π΄1 β ex-inscribed
circle of the median triangle π΄1π΅1πΆ1, corresponding to
the triangle π΄π΅πΆ).
Proof.
If πΈ is the contact of the inscribed circle with π΄πΆ
and πΈπ is the contact of the π΅ βex-inscribed circle with
π΄πΆ , it is known that these points are isotomic,
therefore the middle of the segment πΈπΈπ is the middle
of the side π΄πΆ, which is π΅1.
This point has equal powers to the inscribed
circle (πΌ) and to the π΅ βex-inscribed circle (πΌπ), so it
belongs to their radical axis.
Ion Patrascu, Florentin Smarandache
44
Analogously, πΆ1 is on the radical axis of the
circles (πΌ) and (πΌπ).
The radical axis of the circles (πΌ), (πΌπ ) is the
perpendicular taken from π΅1 to the bisector πΌπΌπ.
This bisector is parallel with the internal
bisector of the angle π΄1π΅1πΆ1 , therefore the
perpendicular in π΅1 on πΌπΌπ is the external bisector of
the angle π΄1π΅1πΆ1 from the median triangle.
Analogously, it follows that the radical axis of
the circles (πΌ), (πΌπ) is the external bisector of the angle
π΄1πΆ1π΅1 from the median triangle.
Because the bisectors intersect in the center of
the circle π΄1 -ex-inscribed to the median triangle
π΄1π΅1πΆ1, this point ππ is the center of the radical center
of the circles (πΌ), (πΌπ), (πΌπ).
Remark.
The theorem for the circles (πΌ), (πΌπ), (πΌπ) and (πΌ),
(πΌπ ), (πΌπ ) can be proved analogously, obtaining the
points ππ and ππ.
3rd Theorem.
The radical circleβs radius of the circles ex-
inscribed to the triangle π΄π΅πΆ is given by the formula: 1
2βπ2 + π2, where π is the radius of the inscribed circle.
Complements to Classic Topics of Circles Geometry
45
Proof.
The position vector of the circle πΌ of the
inscribed circle in the triangle ABC is:
ππΌ =1
2π(πππ΄ + πππ΅ + πππΆ ).
Spieckerβs point π is the center of radical circle
of ex-inscribed circle and is the center of the inscribed
circle in the median triangle π΄1π΅1πΆ1, therefore:
ππ =1
π(1
2πππ΄1 +
1
2πππ΅1 +
1
2πππΆ1 ).
Figure 2.
We denote by π the contact point with the π΄-ex-
inscribed circle of the tangent taken from π to this
circle (see Figure 2).
Ion Patrascu, Florentin Smarandache
46
The radical circleβs radius is given by:
ππ = βππΌπ2 β πΌπ
2
πΌππ =1
2π(ππΌππ΄1
+ ππΌππ΅1 + ππΌππΆ1
).
We evaluate the product of the scales πΌππ β πΌππ ;
we have:
πΌππ2 =
1
4π2 (π2πΌππ΄12 + π2πΌππ΅1
2 + π2πΌππΆ12 + 2πππΌππ΄1
β
πΌππ΅1 + 2πππΌππ΅1
β πΌππΆ1 + 2πππΌππ΄1
β πΌππΆ1 ).
From the law of cosines applied in the triangle
πΌππ΄1π΅1, we find that:
2πΌππ΄1 β πΌππ΅1
= πΌππ΄12 + πΌππ΅1
2 β1
4π2, therefore:
2πππΌππ΄1 β πΌππ΅1
= ππ(πΌππ΄12 + πΌππ΅1
2 β1
4πππ2.
Analogously, we obtain:
2πππΌππ΅1 β πΌππΆ1
= ππ(πΌππ΅12 + πΌππΆ1
2 β1
4π2ππ,
2πππΌππ΄1 β πΌππΆ1
= ππ(πΌππ΄12 + πΌππΆ1
2 β1
4ππ2π.
πΌππ2 =
1
4π2[(π2 + ππ + ππ)πΌππ΄1
2 + (π2 + ππ +
ππ)πΌππ΅12 + (π2 + ππ + ππ)πΌππΆ1
2 βπππ
4(π + π + π)],
πΌππ2 =
1
4π2 [2π(ππΌππ΄12 + ππΌππ΅1
2 + ππΌππΆ12) β 2π ππ],
πΌππ2 =
1
2π(ππΌππ΄1
2 + ππΌππ΅12 + ππΌππΆ1
2) β1
2π π.
From the right triangle πΌππ·ππ΄1, we have that:
πΌππ΄12 = ππ
2 + π΄1π·π2 = ππ
2 + [π
2β (π β π)]
2
=
= ππ2 +
(πβπ)2
4 .
From the right triangles πΌππΈππ΅1 Θi πΌππΉππΆ1 , we
find:
Complements to Classic Topics of Circles Geometry
47
πΌππ΅12 = ππ
2 + π΅1πΈπ2 = ππ
2 + [π
2β (π β π)]
2
=
= ππ2 +
1
4(π + π)2,
πΌππΆ12 = ππ
2 +1
4(π + π)2.
Evaluating ππΌππ΄12 + ππΌππ΅1
2 + ππΌππΆ12, we obtain:
ππΌππ΄12 + ππΌππ΅1
2 + ππΌππΆ12 =
= 2πππ2 +
1
2π(ππ + ππ + ππ) β
1
4πππ.
But:
ππ + ππ + ππ = π2 + π2 + 4π π.
It follows that: 1
2π[ππΌππ΄1
2 + ππΌππ΅12 + ππΌππΆ1
2] = ππ2 +
1
4(π2 + π2) +
1
2π π
and
πΌππ2 = ππ
2 +1
4(π2 + π2).
Then, we obtain:
ππ =1
2βπ2 + π2.
Ion Patrascu, Florentin Smarandache
48
References.
[1] C. Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental Theorems of Triangle Geometry]. Bacau, Romania: Editura
Unique, 2008. [2] I. Patrascu, F. Smarandache: Variance on topics of
plane geometry. Columbus: The Educational
Publisher, Ohio, USA, 2013.
Complements to Classic Topics of Circles Geometry
49
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 π΄π΅πΆ be a given triangle; through the pairs of
points (π΅, πΆ) , (πΆ, π΄) and (π΄, π΅) 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 π΅πΆ, πΆπ΄
and π΄π΅ respectively in three collinear points.
Ion Patrascu, Florentin Smarandache
50
Proof.
We denote by π·, π, πΉ the second point of
intersection of the pairs of circles passing through
(π΄, π΅) and (π΅, πΆ) ; (π΅, πΆ) and (π΄, πΆ) , (π΅, πΆ) and (π΄, π΅)
respectively (see Figure 1).
Figure 1
Let π be the radical center of those circles. In
fact, {π } = π΄πΉ β© π΅π· β© πΆπΈ.
We take from π the tangents π π·1 = π π·2 to the
circle (π΅, πΆ), π πΈ1 = π πΈ2 to the circle (π΄, πΆ) and π πΉ1 =
π πΉ2 to the circle passing through (π΄, π΅). Actually, we
build the radical circle π(π , π π·1) of the given circles.
The polar lines of π to these circles are the lines
π·1π·2, πΈ1πΈ2, πΉ1πΉ2. These three lines cut π΅πΆ, π΄πΆ and π΄π΅
in the points π, π and π , and these lines are
respectively the polar lines of π in respect to the
Complements to Classic Topics of Circles Geometry
51
circles passing through (π΅, πΆ), (πΆ, π΄) and (π΄, π΅) . The
polar lines are the radical axis of the radical circle
with each of the circles passing through
(π΅, πΆ), (πΆ, π΄), (π΄, π΅), respectively. The points belong to
the radical axis having equal powers to those circles,
thereby ππ·1 β ππ·2 = ππΆ β ππ΅.
This relationship shows that the point π has
equal powers relative to the radical circle and to the
circle circumscribed to the triangle π΄π΅πΆ; analogically,
the point π has equal powers relative to the radical
circle and to the circle circumscribed to the triangle
π΄π΅πΆ ; and, likewise, the point π has equal powers
relative to the radical circle and to the circle
circumscribed to the triangle π΄π΅πΆ.
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 π, π and π are
collinear, belonging to the radical axis of the radical
circle and to the circle circumscribed to the given
triangle.
2nd Theorem.
If π is a point in the plane of the triangle π΄π΅πΆ
and the tangents in this point to the circles
circumscribed to triangles πΆ,ππ΄πΆ, ππ΄π΅, respectively,
cut π΅πΆ, πΆπ΄ and π΄π΅, respectively, in the points π, π, π,
then these points are collinear.
Ion Patrascu, Florentin Smarandache
52
Proof.
The point π is the radical center for the circles
(ππ΅πΆ), (ππ΄πΆ), and (ππ΄π΅), and the tangents in π to
these circles are the polar lines to π in relation to
these circles.
If π, π, π are the intersections of these tangents
(polar lines) with π΅πΆ, πΆπ΄, π΄π΅, then they belong to the
radical axis of the circumscribed circle to the triangle
π΄π΅πΆ and to the circle βreducedβ to the point π (ππ2 =
ππ΅ β ππΆ, etc.).
Being located on the radical axis of the two
circles, the points π, π, π 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 π΄π΅πΆ , then they
intersect in Ξ© (the Brocardβs point).
Therefore, we get that the tangents taken in
Ξ© to the adjoin circles cut the sides π΅πΆ , πΆπ΄
and π΄π΅ in collinear points.
Complements to Classic Topics of Circles Geometry
53
References.
[1] C. CoΘniΘΔ: CoordonΓ©es baricentrique [Barycentric Coordinates], Bucarest β Paris: Librairie Vuibert, 1941.
[3] I. PΔtraΘcu: Axe Θi centre radicale ale cercului adjuncte unui triunghi [Axis and radical centers of the adjoin circle of a triangle], in βRecreaΘii
matematiceβ, year XII, no. 1, 2010. [3] C. Mihalescu: Geometria elementelor remarcabile
[The Geometry of Outstanding Elements], Bucharest: Editura Tehnica, 1957.
[4] F. Smarandache, I. Patrascu: Geometry of
Homological Triangle, Columbus: The Educational Publisher Inc., 2012.
Ion Patrascu, Florentin Smarandache
54
Complements to Classic Topics of Circles Geometry
55
Regarding the First Droz-
Farnyβs Circle
In this article, we define the first Droz-
Farnyβs circle, we establish a connection
between it and a concyclicity theorem,
then we generalize this theorem, leading to
the generalization of Droz-Farnyβs circle.
The first theorem of this article was
enunciated by J. Steiner and it was proven
by Droz-Farny (MathΓ©sis, 1901).
1st Theorem.
Let π΄π΅πΆ be a triangle, π» its orthocenter and
π΄1, π΅1, πΆ1 the means of sides (π΅πΆ), (πΆπ΄), (π΄π΅).
If a circle, having its center π», intersects π΅1πΆ1 in
π1, π1 ; πΆ1π΄1 in π2, π2 and π΄1π΅1 in π3, π3 , then π΄π1 =
π΄π1 = π΅π2 = π΅π2 = πΆπ3 = πΆπ3.
Proof.
Naturally, π»π1 = π»π1, π΅1πΆ1 β₯ π΅πΆ, π΄π» β₯ π΅πΆ.
It follows that π΄π» β₯ π΅1πΆ1.
Ion Patrascu, Florentin Smarandache
56
Figure 1.
Therefore, π΄π» is the mediator of segment π1π1;
similarly, π΅π» and πΆπ» are the mediators of segments
π2π2 and π3π3.
Let π1 be the intersection of lines π΄π» and π΅1πΆ1
(see Figure 1); we have π1π΄2 β π1π»
2 = π1π΄2 β π1π»
2. We
denote π π» = π»π1. It follows that π1π΄2 = π π»
2 + (π1π΄ +
π1π»)(π1π΄ β π1π») = π π»2 + π΄π» β (π1π΄ β π1π»).
However, π1π΄ = π1π»1, where π»1 is the projection
of π΄ on π΅πΆ; we find that π1π΄2 = π π»
2 + π΄π» β π»π»1.
It is known that the symmetric of orthocenter π»
towards π΅πΆ belongs to the circle of circumscribed
triangle π΄π΅πΆ.
Denoting this point by π»1β² , we have π΄π» β π»π»1
β² =
π 2 β ππ»2 (the power of point π» towards the
circumscribed circle).
Complements to Classic Topics of Circles Geometry
57
We obtain that π΄π» β π»π»1 =1
2β (π 2 β ππ»2) , and
therefore π΄π12 = π π»
2 +1
2β (π 2 β ππ»2) , where π is the
center of the circumscribed triangle π΄π΅πΆ.
Similarly, we find π΅π22 = πΆπ3
2 = π π»2 +
1
2(π 2 β
ππ»2), therefore π΄π1 = π΅π2 = πΆπ3.
Remarks.
a. The proof adjusts analogously for the
obtuse triangle.
b. 1st Theorem can be equivalently
formulated in this way:
2nd Theorem.
If we draw three congruent circles centered in a
given triangleβs 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 triangleβs orthocenter.
If we denote by π the radius of three congruent
circles having π΄, π΅, πΆ as their centers, we get:
π π»2 = π2 +
1
2(ππ»2 β π 2).
However, in a triangle, ππ»2 = 9π 2 β (π2 + π2 +
π2), π being the radius of the circumscribed circle; it
follows that:
π π»2 = π2 + 4π 2 β
1
2(π2 + π2 + π2).
Ion Patrascu, Florentin Smarandache
58
Remark.
A special case is the one in which π = π , where
we find that π π»2 = π 1
2 = 5π 2 β1
2(π2 + π2 + π2) =
1
2(π 2 +
ππ»2).
Definition.
The circle π(π», π 1), where:
π 1 = β5π 2 β1
2(π2 + π2 + π2),
is called the first Droz-Farnyβs circle of the triangle
π΄π΅πΆ.
Remark.
Another way to build the first Droz-Farnyβs circle
is offered by the following theorem, which, according
to [1], was the subject of a problem proposed in 1910
by V. ThΓ©bault in the Journal de MathΓ©matiques
Elementaire.
3rd Theorem.
The circles centered on the feet of a triangleβs
altitudes passing through the center of the circle
circumscribed to the triangle cut the triangleβs sides
in six concyclical points.
Complements to Classic Topics of Circles Geometry
59
Proof.
We consider π΄π΅πΆ an acute triangle and π»1, π»2, π»3
the altitudesβ feet. We denote by π΄1, π΄2; π΅1, π΅2; πΆ1, πΆ2
the intersection points of circles having their centers
π»1, π»2, π»3 to π΅πΆ, πΆπ΄, π΄π΅, respectively.
We calculate π»π΄2 of the right angled triangle
π»π»1π΄2 (see Figure 2). We have π»π΄22 = π»π»1
2 + π»1π΄22.
Because π»1π΄2 = π»1π , it follows that π»π΄22 =
π»π»12 + π»1π
2 . We denote by π9 the mean of segment
ππ» ; the median theorem in triangle π»1π»π leads to
π»1π»2 + π»1π
2 = 2π»1π92 +
1
2ππ»2.
It is known that π9π»1 is the nine-points circleβs
radius, so π»1π9 =1
2π ; we get: π»π΄1
2 =1
2(π 2 + ππ»2) ;
similarly, we find that π»π΅12 = π»πΆ1
2 =1
2(π 2 + ππ»2) ,
which shows that the points π΄1, π΄2; π΅1, π΅2; πΆ1, πΆ2
belong to the first Droz-Farnyβs circle.
Figure 2.
Ion Patrascu, Florentin Smarandache
60
Remark.
The 2nd and the 3rd theorems show that the first
Droz-Farnyβs 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 π΄1, π΄2; π΅1, π΅2; πΆ1, πΆ2 be the points of
intersection with the sides of triangle π΄π΅πΆ of circles
having their centers in altitudesβ feet π»1, π»2, π»3.
Suppose the points are concyclical; it follows
that their circleβs radical center and of circles centered
in π»2 and π»3 is the point π΄ (sides π΄π΅ and π΄πΆ are
radical axes), therefore the perpendicular from π΄ on
π»2π»3 is radical axis of centers having their centers π»2
and π»3.
Since π»2π»3 is antiparallel to π΅πΆ, it is parallel to
tangent taken in π΄ to the circle circumscribed to
triangle π΄π΅πΆ.
Complements to Classic Topics of Circles Geometry
61
Consequently, the radical axis is the
perpendicular taken in π΄ on the tangent to the
circumscribed circle, therefore it is π΄π.
Similarly, the other radical axis of circles
centered in π»1, π»2 and of circles centered in π»1, π»3 pass
through π, therefore π is the radical center of these
circles.
Reciprocally.
Let π be the radical center of the circles having
their centers in the feet of altitudes. Since π΄π is
perpendicular on π»2π»3 , it follows that π΄π is the
radical axis of circles having their centers in π»2, π»3,
therefore π΄π΅1 β π΄π΅2 = π΄πΆ1 β π΄πΆ2.
From this relationship, it follows that the points
π΅1, π΅2; πΆ1, πΆ2 are concyclic; the circle on which these
points are located has its center in the orthocenter π»
of triangle π΄π΅πΆ.
Indeed, the mediatorsβ chords π΅1π΅2 and πΆ1πΆ2 in
the two circles are the altitudes from πΆ and π΅ of
triangle π΄π΅πΆ, therefore π»π΅1 = π»π΅2 = π»πΆ1 = π»πΆ2.
This reasoning leads to the conclusion that π΅π is
the radical axis of circles having their centers π»1 and
π»3 , and from here the concyclicality of the points
π΄1, π΄2; πΆ1, πΆ2 on a circle having its center in π» ,
therefore π»π΄1 = π»π΄2 = π»πΆ1 = π»πΆ2 . We obtained that
π»π΄1 = π»π΄2 = π»π΅1 = π»π΅2 = π»πΆ1 = π»πΆ2 , which shows
the concyclicality of points π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2.
Ion Patrascu, Florentin Smarandache
62
Remark.
The circles from the 3rd Theorem, passing
through π and having noncollinear centers, admit π
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 douΔ puncte inverse [On some circles that have
as centers two inverse points], in βGazeta Matematicaβ, vol. XXXIII, 1927.
[2] R. Johnson: Advanced Euclidean Geometry. New York: Dover Publication, Inc. Mineola, 2004
[3] Eric W. Weisstein: First Droz-Farnyβs circle. From
Math World β A Wolfram WEB Resurse, http://mathworld.wolfram.com/.
[4] F. Smarandache, I. Patrascu: Geometry of
Homological Triangle. Columbus: The Education Publisher Inc., Ohio, SUA, 2012.
Complements to Classic Topics of Circles Geometry
63
Regarding the Second Droz-
Farnyβs Circle
In this article, we prove the theorem
relative to the second Droz-Farnyβs 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 π΄π΅πΆ passing through its orthocenter π»
intersect the sides π΅πΆ, πΆπ΄ and π΄π΅ respectively in the
points π΄1, π΄2, π΅1, π΅2 and πΆ1, πΆ2, situated on a concentric
circle with the circle circumscribed to the triangle π΄π΅πΆ
(the second Droz-Farnyβs circle).
Proof.
We denote by π1, π2, π3 the middles of π΄π΅πΆ
triangleβs sides, see Figure 1. Because π΄π» β₯ π2π3 and
π» belongs to the circles with centers in π2 and π3, it
follows that π΄π» is the radical axis of these circles,
Ion Patrascu, Florentin Smarandache
64
therefore we have π΄πΆ1 β π΄πΆ2 = π΄π΅2 β π΄π΅1. This relation
shows that π΅1, π΅2, πΆ1, πΆ2 are concyclic points, because
the center of the circle on which they are situated is π,
the center of the circle circumscribed to the triangle
π΄π΅πΆ, hence we have that:
ππ΅1 = ππΆ1 = ππΆ2 = ππ΅2. (1)
Figure 1.
Analogously, π is the center of the circle on
which the points π΄1, π΄2, πΆ1, πΆ2 are situated, hence:
ππ΄1 = ππΆ1 = ππΆ2 = ππ΄2. (2)
Also, π is the center of the circle on which the
points π΄1, π΄2, π΅1, π΅2 are situated, and therefore:
ππ΄1 = ππ΅1 = ππ΅2 = ππ΄2. (3)
The relations (1), (2), (3) show that the points
π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are situated on a circle having the
center in π, called the second Droz-Farnyβs circle.
Complements to Classic Topics of Circles Geometry
65
1st Proposition.
The radius of the second Droz-Farnyβs circle is
given by:
π 22 = 5π2 β
1
2(π2 + π2 + π2).
Proof.
From the right triangle ππ1π΄1, using Pitagoraβs
theorem, it follows that:
ππ΄12 = ππ1
2 + π΄1π12 = ππ1
2 + π1π2.
From the triangle π΅π»πΆ , using the median
theorem, we have:
π»π12 =
1
4[2(π΅π»2 + πΆπ»2) β π΅πΆ2].
But in a triangle,
π΄π» = 2ππ1, π΅π» = 2ππ2, πΆπ» = 2ππ3,
hence:
π»π12 = 2ππ2
2 + 2ππ32 =
π2
4.
But:
ππ12 = π 2 β
π2
4 ;
ππ22 = π 2 β
π2
4 ;
ππ32 = π 2 β
π2
4 ,
where R is the radius of the circle circumscribed to the
triangle π΄π΅πΆ.
We find that ππ΄12 = π 2
2 = 5π 2 β1
2(π2 + π2 + π2).
Ion Patrascu, Florentin Smarandache
66
Remarks.
a. We can compute ππ12 + π1π2 using the
median theorem in the triangle ππ1π» for
the median π1π9 (π9 is the center of the
nine points circle, i.e. the middle of (ππ»)).
Because π9π1 =1
2π , we obtain: π 2
2 =
1
2(ππ2 + π 2). In this way, we can prove
the Theorem computing ππ΅12 and ππΆ1
2.
b. The statement of the 1st Theorem was the
subject no. 1 of the 49th International
Olympiad in Mathematics, held at Madrid
in 2008.
c. The 1st Theorem can be proved in the same
way for an obtuse triangle; it is obvious
that for a right triangle, the second Droz-
Farnyβs circle coincides with the circle
circumscribed to the triangle π΄π΅πΆ.
d. The 1st Theorem appears as proposed
problem in [2].
2nd Theorem.
The three pairs of points determined by the
intersections of each circle with the center in the
middle of triangleβs side with the respective side are
on a circle if and only these circles have as radical
center the triangleβs orthocenter.
Complements to Classic Topics of Circles Geometry
67
Proof.
Let π1, π2, π3 the middles of the sides of triangle
π΄π΅πΆ and let π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 the intersections with
π΅πΆ, πΆπ΄, π΄π΅ respectively of the circles with centers in
π1, π2, π3.
Let us suppose that π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are
concyclic points. The circle on which they are situated
has evidently the center in π, the center of the circle
circumscribed to the triangle π΄π΅πΆ.
The radical axis of the circles with centers π2, π3
will be perpendicular on the line of centers π2π3, and
because A has equal powers in relation to these circles,
since π΄π΅1 β π΄π΅2 = π΄πΆ1 β π΄πΆ2, it follows that the radical
axis will be the perpendicular taken from A on π2π3,
i.e. the height from π΄ of triangle π΄π΅πΆ.
Furthermore, it ensues that the radical axis of
the circles with centers in π1 and π2 is the height
from π΅ of triangle π΄π΅πΆ and consequently the
intersection of the heights, hence the orthocenter π» of
the triangle π΄π΅πΆ is the radical center of the three
circles.
Reciprocally.
If the circles having the centers in π1, π2, π3
have the orthocenter with the radical center, it follows
that the point π΄, being situated on the height from A
which is the radical axis of the circles of centers π2, π3
Ion Patrascu, Florentin Smarandache
68
will have equal powers in relation to these circles and,
consequently, π΄π΅1 β π΄π΅2 = π΄πΆ1 β π΄πΆ2 , a relation that
implies that π΅1, π΅2, πΆ1, πΆ2 are concyclic points, and the
circle on which these points are situated has π as its
center.
Similarly, π΅π΄1 β π΅π΄2 = π΅πΆ1 β π΅πΆ2 , therefore
π΄1, π΄2, πΆ1, πΆ2 are concyclic points on a circle of center π.
Having ππ΅1 = ππ΅2 = ππΆ1 = ππΆ2 and ππ΄1 β ππ΄2 = ππΆ1 β
ππΆ2 , we get that the points π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are
situated on a circle of center π.
Remarks.
1. The 1st Theorem is a particular case of the
2nd Theorem, because the three circles of
centers π1, π2, π3 pass through π», which
means that π» is their radical center.
2. The Problem 525 from [3] leads us to the
following Proposition providing a way to
construct the circles of centers π1, π2, π3
intersecting the sides in points that
belong to a Droz-Farnyβs circle of type 2.
2nd Proposition.
The circles β (π1,1
2βπ + π2), β (π2,
1
2βπ + π2),
β (π3,1
2βπ + π2) intersect the sides π΅πΆ , πΆπ΄ , π΄π΅
respectively in six concyclic points; π is a conveniently
Complements to Classic Topics of Circles Geometry
69
chosen constant, and π, π, π are the lengths of the sides
of triangle π΄π΅πΆ.
Proof.
According to the 2nd Theorem, it is necessary to
prove that the orthocenter π» of triangle π΄π΅πΆ is the
radical center for the circles from hypothesis.
Figure 2.
The power of π» in relation with
β (π1,1
2βπ + π2) is equal to π»π1
2 β1
4(π + π2) . We
observed that π12 = 4π 2 β
π2
2β
π2
2β
π2
4 , therefore
π»π12 β
1
4(π + π2) = 4π 2 β
π2+π2+π2
4β
1
4π. We use the
same expression for the power of H in relation to the
circles of centers π2, π3, hence H is the radical center
of these three circles.
Ion Patrascu, Florentin Smarandache
70
References.
[1] C. Mihalescu: Geometria elementelor remarcabile [The Geometry of Outstanding Elements]. Bucharest: Editura TehnicΔ, 1957.
[2] I. PΔtraΕcu: Probleme de geometrie planΔ [Some Problems of Plane Geometry], Craiova: Editura Cardinal, 1996.
[3] C. CoΕniΕ£Δ: Teoreme Εi probleme alese de matematicΔ [Theorems and Problems],
BucureΕti: Editura de Stat DidacticΔ Εi PedagogicΔ, 1958.
[4] I. PΔtraΘcu, F. Smarandache: Variance on Topics of
Plane Geometry, Educational Publishing, Columbus, Ohio, SUA, 2013.
Complements to Classic Topics of Circles Geometry
71
Neubergβs 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 Brocardβs point of the triangle π΄π΅πΆ the
point Ξ© with the property: β’Ξ©π΄π΅ = β’Ξ©π΅πΆ = β’Ξ©πΆπ΄ .
The measure of the angle Ξ©π΄π΅ is denoted by π and it
is called Brocard's angle. It occurs the relationship:
ctgπ = ctgπ΄ + ctgπ΅ + ctgπΆ (see [1]).
Figure 1.
Ion Patrascu, Florentin Smarandache
72
2nd Definition.
Two triangles are called equibrocardian if they
have the same Brocardβs angle.
3rd Definition.
The locus of points π from the plane of the
triangle located on the same side of the line π΅πΆ as π΄
and forming with π΅πΆ an equibrocardian triangle with
π΄π΅πΆ, containing the vertex π΄ of the triangle, it's called
A-Neubergβ circle of the triangle π΄π΅πΆ.
We denote by ππ the center of A-Neubergβ circle
by radius ππ (analogously, we define B-Neubergβ and
C-Neubergβ circles).
We get that π(π΅πππΆ) = 2π and ππ =π
2βctg2π β 3
(see [1]).
The triangle ππππππ formed by the centers of
Neubergβs circles is called Neubergβs triangle.
1st Proposition.
The distances from the center circumscribed to
the triangle π΄π΅πΆ to the vertex of Neubergβs triangle
are proportional to the cubes of π΄π΅πΆ triangleβs sides
lengths.
Complements to Classic Topics of Circles Geometry
73
Proof.
Let π be the center of the circle circumscribed to
the triangle π΄π΅πΆ (see Figure 2).
Figure 2.
The law of cosines applied in the triangle ππππ΅
provides: πππ
sin(πππ΅π)=
π
sinπ .
But π(β’πππ΅π) = π(β’πππ΅πΆ) β π(β’ππ΅πΆ) = π΄ β π.
We have that πππ
sin(π΄βπ)=
π
sinπ .
But
sin(π΄βπ)
sinπ=
πΆΞ©
AΞ©=
π
πβ2π sinπ
π
πβ2π sinπ
=π3
πππ=
π3
4π π ,
π being the area of the triangle π΄π΅πΆ.
Ion Patrascu, Florentin Smarandache
74
It follows that πππ =π3
4π , and we get that
πππ
π3 =
πππ
π3 =πππ
π3 .
Consequence.
In a triangle ABC, we have:
1) πππ β πππ β πππ = π 3;
2) ctgπ =πππ
π+
πππ
π+
πππ
π .
2nd Proposition.
If ππππππ is the Neubergβs triangle of the
triangle π΄π΅πΆ, then:
ππππ2 =
(π2 + π2)(π4 + π4) β π2π2π2
2π2π2 + 2π2π2 + 2π2π2 β π4 β π4 β π4 .
(The formulas for ππππ and ππππ are obtained
from the previous one, by circular permutations.)
Proof.
We apply the law of cosines in the triangle
πππππ:
πππ =π3
4π , πππ =
π3
4π ,π(β’πππππ) = 1800 β οΏ½οΏ½.
ππππ2 =
π6 + π6 β 2π3π3cos (1800 β π)
16π2
=π6 + π6 + 2π3π3cosπΆ
16π2 .
Complements to Classic Topics of Circles Geometry
75
But the law of cosines in the triangle π΄π΅πΆ
provides
2cosπΆ =π2 + π2 β π2
2ππ
and, from din Heronβs formula, we find that
16π2 = 2π2π2 + 2π2π2 + 2π2π2 β π4 β π4 β π4.
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 β 18B)
Two circles π(π1, π1), π(π2, π2) are orthogonal if
and only if
π12 + π2
2 = π1π22.
Proof.
Let π(π1, π1), π(π2, π2) be orthogonal (see Figure
3); then, if π΄ is one of the common points, the triangle
π1π΄π2 is a right triangle and the Pythagorean Theorem
applied to it, leads to π12 + π2
2 = π1π22.
Ion Patrascu, Florentin Smarandache
76
Reciprocally.
If the metric relationship from the statement
occurs, it means that the triangle π1π΄π2 is a right
triangle, therefore π΄ is their common point (the
relationship π12 + π2
2 = π1π22 implies π1
2 + π22 >
π1π22), then π1π΄ β₯ π2π΄, so π1π΄ is tangent to the circle
π(π2, π2) because it is perpendicular in π΄ on radius π2π΄,
and as well π2π΄ is tangent to the circle π(π1, π1) ,
therefore the circles are orthogonal.
Figure 3.
4th Proposition.
B-Neubergβs and C-Neubergβs circles associated
to the right triangle π΄π΅πΆ (in π΄) are orthogonal.
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77
Proof.
If π(οΏ½οΏ½) = 900, then ππππ2 =
π6+π6
16π.
ππ =π
2βctg2π β 3 ; ππ =
π
2βctg2π β 3.
But ctgπ βπ2+π2+π2
4π=
π2+π2
2π=
π2
ππ .
It was taken into account that π2 = π2 + π2 and
2π = ππ.
ctg2π β 3 =π4
π2π2β 3 =
(π2 + π2)2 β 3π2π2
π2π2
ctg2π β 3 =π4 + π4 β π2π2
π2π2
ππ2 + ππ
2 =π4 + π4 β π2π2
π2π2(π2 + π2
4)
=(π2 + π2)(π4 + π4 β π2π2)
4π2π2=
π6 + π6
16π2 .
By ππ2 + ππ
2 = ππππ2, it follows that B-Neubergβs
and C-Neubergβs circles are orthogonal.
Ion Patrascu, Florentin Smarandache
78
References.
[1] F. Smarandache, I. Patrascu: The Geometry of Homological Triangles. Columbus: The Educational Publisher, Ohio, USA, 2012.
[2] T. Lalescu: Geometria triunghiului [The Geometry of the Triangle]. Craiova: Editura Apollo, 1993.
[3] R. A. Johnson: Advanced Euclidian Geometry. New
York: Dover Publications, 2007.
Complements to Classic Topics of Circles Geometry
79
Lucasβs Inner Circles
In this article, we define the Lucasβs inner
circles and we highlight some of their
properties.
1. Definition of the Lucasβs Inner Circles
Let π΄π΅πΆ be a random triangle; we aim to
construct the square inscribed in the triangle π΄π΅πΆ ,
having one side on π΅πΆ.
Figure 1.
In order to do this, we construct a square π΄,π΅,πΆ ,π·,
with π΄, β (π΄π΅), π΅,, πΆ , β (π΅πΆ) (see Figure 1).
We trace the line π΅π·, and we note with π·π its
intersection with (π΄πΆ) ; through π·π we trace the
Ion Patrascu, Florentin Smarandache
80
parallel π·ππ΄π to π΅πΆ with π΄π β (π΄π΅) and we project
onto π΅πΆ the points π΄π, π·π in π΅π respectively πΆπ.
We affirm that the quadrilateral π΄ππ΅ππΆππ·π is the
required square.
Indeed, π΄ππ΅ππΆππ·π is a square, because π·ππΆπ
π·,πΆ , =
π΅π·π
π΅π·, =π΄ππ·π
π΄,π·, and, as π·,πΆ , = π΄,π·, , it follows that π΄ππ·π =
π·ππΆπ.
Definition.
It is called A-Lucasβs inner circle of the triangle
π΄π΅πΆ the circle circumscribed to the triangle AAaDa.
We will note with πΏπ the center of the A-Lucasβs
inner circle and with ππ its radius.
Analogously, we define the B-Lucasβs inner circle
and the C-Lucasβs inner circle of the triangle π΄π΅πΆ.
2. Calculation of the Radius of
the A-Lucas Inner Circle
We note π΄ππ·π = π₯, π΅πΆ = π; let βπ be the height
from π΄ of the triangle π΄π΅πΆ.
The similarity of the triangles π΄π΄ππ·π and π΄π΅πΆ
leads to: π₯
π=
βπβπ₯
βπ, therefore π₯ =
πβπ
π+βπ .
From ππ
π =
π₯
π we obtain ππ =
π .βπ
π+βπ . (1)
Complements to Classic Topics of Circles Geometry
81
Note.
Relation (1) and the analogues have been
deduced by Eduard Lucas (1842-1891) in 1879 and
they constitute the βbirth certificate of the Lucasβs
circlesβ.
1st Remark.
If in (1) we replace βπ =2π
π and we also keep into
consideration the formula πππ = 4π π, where π is the
radius of the circumscribed circle of the triangle π΄π΅πΆ
and π represents its area, we obtain:
ππ =π
1+2ππ
ππ
[see Ref. 2].
3. Properties of the Lucasβs Inner Circles
1st Theorem.
The Lucasβs inner circles of a triangle are inner
tangents of the circle circumscribed to the triangle and
they are exteriorly tangent pairwise.
Proof.
The triangles π΄π΄ππ·π and π΄π΅πΆ are homothetic
through the homothetic center π΄ and the rapport: βπ
π+βπ.
Ion Patrascu, Florentin Smarandache
82
Because ππ
π =
βπ
π+βπ, it means that the A-Lucasβs inner
circle and the circle circumscribed to the triangle π΄π΅πΆ
are inner tangents in π΄.
Analogously, it follows that the B-Lucasβs and C-
Lucasβs inner circles are inner tangents of the circle
circumscribed to π΄π΅πΆ.
Figure 2.
We will prove that the A-Lucasβs and C-Lucasβs
circles are exterior tangents by verifying
πΏππΏπ = ππ + ππ. (2)
We have:
ππΏπ = π β ππ;
ππΏπ = π β ππ
and
π(π΄0οΏ½οΏ½) = 2π΅
(if π(οΏ½οΏ½) > 90Β° then π(π΄0οΏ½οΏ½) = 360Β° β 2π΅).
Complements to Classic Topics of Circles Geometry
83
The theorem of the cosine applied to the triangle
ππΏππΏπ implies, keeping into consideration (2), that:
(π β ππ)2 + (π β ππ)
2 β 2(π β ππ)(π β ππ)πππ 2π΅ =
= (ππ + ππ)2.
Because πππ 2π΅ = 1 β 2π ππ2π΅, it is found that (2)
is equivalent to:
π ππ2π΅ =ππππ
(π βππ)(π βππ) . (3)
But we have: ππππ =π 2ππ2π
(2ππ +ππ)(2ππ +ππ) ,
ππ + ππ = π π(π
2ππ +ππ+
π
2ππ +ππ).
By replacing in (3), we find that π ππ 2π΅ =ππ2π
4πππ 2 =
π2
4π2 βΊ sinπ΅ =π
2π is true according to the sines theorem.
So, the exterior tangent of the A-Lucasβs and C-Lucasβs
circles is proven.
Analogously, we prove the other tangents.
2nd Definition.
It is called an A-Apolloniusβs circle of the random
triangle π΄π΅πΆ the circle constructed on the segment
determined by the feet of the bisectors of angle π΄ as
diameter.
Remark.
Analogously, the B-Apolloniusβs and C-
Apolloniusβs circles are defined. If π΄π΅πΆ is an isosceles
triangle with π΄π΅ = π΄πΆ then the A-Apolloniusβs circle
Ion Patrascu, Florentin Smarandache
84
isnβt defined for π΄π΅πΆ , and if π΄π΅πΆ is an equilateral
triangle, its Apolloniusβs circle isnβt defined.
2nd Theorem.
The A-Apolloniusβs circle of the random triangle
is the geometrical point of the points π from the plane
of the triangle with the property: ππ΅
ππΆ=
π
π.
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 circlesβ fascicle is
exterior to them, we say that the fascicle
is of the first type.
b. If the radical axis of the circlesβ fascicle is
secant to the circles, we say that the
fascicle is of the second type.
c. If the radical axis of the circlesβ fascicle is
tangent to the circles, we say that the
fascicle is of the third type.
3rd Theorem.
The A-Apolloniusβs circle and the B-Lucasβs and
C-Lucasβs inner circles of the random triangle π΄π΅πΆ
form a fascicle of the third type.
Complements to Classic Topics of Circles Geometry
85
Proof.
Let {ππ΄} = πΏππΏπ β© π΅πΆ (see Figure 3).
Menelausβs theorem applied to the triangle ππ΅πΆ
implies that: ππ΄π΅
ππ΄πΆ.πΏππ΅
πΏππ.πΏππ
πΏππΆ= 1,
so:
ππ΄π΅
ππ΄πΆ.
ππ
π βππ.π βππ
ππ= 1
and by replacing ππ and ππ, we find that: ππ΄π΅
ππ΄πΆ=
π2
π2.
This relation shows that the point ππ΄ is the foot
of the exterior symmedian from π΄ of the triangle π΄π΅πΆ
(so the tangent in π΄ to the circumscribed circle),
namely the center of the A-Apolloniusβs circle.
Let π1 be the contact point of the B-Lucasβs and
C-Lucasβs circles. The radical center of the B-Lucasβs,
C-Lucasβs circles and the circle circumscribed to the
triangle π΄π΅πΆ is the intersection ππ΄ of the tangents
traced in π΅ and in πΆ to the circle circumscribed to the
triangle π΄π΅πΆ.
It follows that π΅ππ΄ = πΆππ΄ = π1ππ΄, so π1 belongs to
the circle ππ΄ that has the center in ππ΄ and orthogonally
cuts the circle circumscribed in π΅ and πΆ. The radical
axis of the B-Lucasβs and C-Lucasβs circles is ππ΄π1, and
ππ΄π1 is tangent in π1 to the circle ππ΄. Considering the
power of the point ππ΄ in relation to ππ΄, we have:
ππ΄π12 = ππ΄π΅. ππ΄πΆ.
Ion Patrascu, Florentin Smarandache
86
Figure 3.
Also, ππ΄π2 = ππ΄π΅ β ππ΄πΆ ; it thus follows that
ππ΄π΄ = ππ΄π1 , which proves that π1 belongs to the A-
Apolloniusβs circle and is the radical center of the A-
Apolloniusβs, B-Lucasβs and C-Lucasβs circles.
Remarks.
1. If the triangle π΄π΅πΆ is right in π΄ then
πΏππΏπ||π΅πΆ, the radius of the A-Apolloniusβs
circle is equal to: πππ
|π2βπ2|. The point π1 is
the foot of the bisector from π΄. We find
that ππ΄π1 =πππ
|π2βπ2|, so the theorem stands
true.
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87
2. The A-Apolloniusβs and A-Lucasβs circles
are orthogonal. Indeed, the radius of the
A-Apolloniusβs circle is perpendicular to
the radius of the circumscribed circle, ππ΄,
so, to the radius of the A-Lucasβs circle
also.
4th Definition.
The triangle ππ΄ππ΅ππΆ determined by the tangents
traced in π΄, π΅, πΆ to the circle circumscribed to the
triangle π΄π΅πΆ is called the tangential triangle of the
triangle π΄π΅πΆ.
1st Property.
The triangle π΄π΅πΆ and the Lucasβs triangle πΏππΏππΏπ
are homological.
Proof.
Obviously, π΄πΏπ, π΅πΏπ , πΆπΏπ are concurrent in π ,
therefore π, the center of the circle circumscribed to
the triangle π΄π΅πΆ, is the homology center.
We have seen that {ππ΄} = πΏππΏπ β© π΅πΆ and ππ΄ is the
center of the A-Apolloniusβs circle, therefore the
homology axis is the Apolloniusβs line ππ΄ππ΅ππΆ (the line
determined by the centers of the Apolloniusβs circle).
Ion Patrascu, Florentin Smarandache
88
2nd Property.
The tangential triangle and the Lucasβs triangle
of the triangle π΄π΅πΆ are orthogonal triangles.
Proof.
The line ππ΄π1 is the radical axis of the B-Lucasβs
inner circle and the C-Lucasβs inner circle, therefore it
is perpendicular on the line of the centers πΏππΏπ .
Analogously, ππ΅π2 is perpendicular on πΏππΏπ , because
the radical axes of the Lucasβs circles are concurrent
in πΏ, which is the radical center of the Lucasβs circles;
it follows that ππ΄ππ΅ππΆ and πΏππΏππΏπ are orthological and
πΏ is the center of orthology. The other center of
orthology is π the center of the circle circumscribed to
π΄π΅πΆ.
References.
[1] D. BrΓ’nzei, M. MiculiΘa. Lucas circles and spheres. In βThe Didactics of Mathematicsβ, volume 9/1994, Cluj-Napoca (Romania), p. 73-80.
[2] P. Yiu, A. P. Hatzipolakis. The Lucas Circles of a Triangle. In βAmer. Math. Monthlyβ, no.
108/2001, pp. 444-446. http://www.math.fau. edu/yiu/monthly437-448.pdf.
[3] F. Smarandache, I. Patrascu. The Geometry of
Homological Triangles. Educational Publisher, Columbus, Ohio, U.S.A., 2013.
Complements to Classic Topics of Circles Geometry
89
Theorems with Parallels Taken
through a Triangleβs Vertices
and Constructions Performed
only with the Ruler
In this article, we solve problems of
geometric constructions only with the
ruler, using known theorems.
1st Problem.
Being given a triangle π΄π΅πΆ , its circumscribed
circle (its center known) and a point π fixed on the
circle, construct, using only the ruler, a transversal
line π΄1, π΅1, πΆ1, with π΄1 β π΅πΆ, π΅1 β πΆπ΄, πΆ1 β π΄π΅, such that
β’ππ΄1πΆ β‘ β’ππ΅1πΆ β‘ β’ππΆ1π΄ (the lines taken though π
to generate congruent angles with the sides π΅πΆ , πΆπ΄
and π΄π΅, respectively).
2nd Problem.
Being given a triangle π΄π΅πΆ , its circumscribed
circle (its center known) and π΄1, π΅1, πΆ1, such that π΄1 β
Ion Patrascu, Florentin Smarandache
90
π΅πΆ, π΅1 β πΆπ΄, πΆ1 β π΄π΅ and π΄1, π΅1, πΆ1 collinear, construct,
using only the ruler, a point π on the circle
circumscribing the triangle, such that the lines
ππ΄1, ππ΅1, ππΆ1 to generate congruent angles with π΅πΆ,
πΆπ΄ and π΄π΅, respectively.
3rd Problem.
Being given a triangle π΄π΅πΆ inscribed in a circle
of given center and π΄π΄β² a given cevian, π΄β² a point on
the circle, construct, using only the ruler, the isogonal
cevian π΄π΄1 to the cevian π΄π΄β².
To solve these problems and to prove the
theorems for problems solving, we need the following
Lemma:
1st Lemma. (Generalized Simpson's Line)
If π is a point on the circle circumscribed to the
triangle π΄π΅πΆ and we take the lines ππ΄1, ππ΅1, ππΆ1
which generate congruent angles ( π΄1 β π΅πΆ, π΅1 β
πΆπ΄, πΆ1 β π΄π΅) with π΅πΆ, πΆπ΄ and π΄π΅ respectively, then the
points π΄1, π΅1, πΆ1 are collinear.
Complements to Classic Topics of Circles Geometry
91
Proof.
Let π on the circle circumscribed to the triangle
π΄π΅πΆ (see Figure 1), such that:
β’ππ΄1πΆ β‘ β’ππ΅1πΆ β‘ β’ππΆ1π΄ = π. (1)
Figure 1.
From the relation (1), we obtain that the
quadrilateral ππ΅1π΄1πΆ is inscriptible and, therefore:
β’π΄1π΅πΆ β‘ β’π΄1ππΆ. (2).
Also from (1), we have that ππ΅1π΄πΆ1 is
inscriptible, and so
β’π΄π΅1πΆ1 β‘ β’π΄ππΆ1. (3)
Ion Patrascu, Florentin Smarandache
92
The quadrilateral MABC is inscribed, hence:
β’ππ΄πΆ1 β‘ β’π΅πΆπ. (4)
On the other hand,
β’π΄1ππΆ = 1800 β (π΅πΆοΏ½οΏ½ + π),
β’π΄ππΆ1 = 1800 β (ππ΄πΆ1 + π).
The relation (4) drives us, together with the
above relations, to:
β’π΄1ππΆ β‘ β’π΄ππΆ1. (5)
Finally, using the relations (5), (2) and (3), we
conclude that: β’π΄1π΅1πΆ β‘ π΄π΅1πΆ1 , which justifies the
collinearity of the points π΄1, π΅1, πΆ1.
Remark.
The Simsonβs Line is obtained in the case when
π = 900.
2nd Lemma.
If π is a point on the circle circumscribed to the
triangle π΄π΅πΆ and π΄1, π΅1, πΆ1 are points on π΅πΆ , πΆπ΄ and
π΄π΅ , respectively, such that β’ππ΄1πΆ = β’ππ΅1πΆ =
β’ππΆ1π΄ = π, and ππ΄1 intersects the circle a second
time in π΄β, then π΄π΄β² β₯ π΄1π΅1.
Proof.
The quadrilateral ππ΅1π΄1πΆ is inscriptible (see
Figure 1); it follows that:
Complements to Classic Topics of Circles Geometry
93
β’πΆππ΄β² β‘ β’π΄1π΅1πΆ. (6)
On the other hand, the quadrilateral ππ΄π΄β²πΆ is
also inscriptible, hence:
β’πΆππ΄β² β‘ β’π΄β²π΄πΆ. (7)
The relations (6) and (7) imply: β’π΄β²ππΆ β‘ β’π΄β²π΄πΆ,
which gives π΄π΄β² β₯ π΄1π΅1.
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 π(π, π ) , let be a diameter
(π΄π΅)] and let π β π(π, π ). We construct the line π΅π
(see Figure 2). We consider on this line the point π· (π
between π· and π΅). We join π· with π , π΄ with π and
denote π·π β© π΄π = {π}.
We take π΅π and let {π} = π·π΄ β© π΅π. The line ππ
is parallel to π΄π΅.
Constructionβs Proof.
In the triangle π·π΄π΅, the cevians π·π, π΄π and π΅π
are concurrent.
Cevaβs Theorem provides:
Ion Patrascu, Florentin Smarandache
94
ππ΄
ππ΅βππ΅
ππ·βππ·
ππ΄= 1. (8)
But π·π is a median, π·π = π΅π = π .
From (8), we get ππ΅
ππ·=
ππ΄
ππ· , which, by Thales
reciprocal, gives ππ β₯ π΄π΅.
Figure 2.
Remark.
If we have a circle with given center and a
certain line π, we can construct though a given point
π a parallel to that line in such way: we take two
diameters [π π] and [ππ] through the center of the
given circle (see Figure 3).
Complements to Classic Topics of Circles Geometry
95
Figure 3.
We denote π π β© π = {π}; because [π π] β‘ [ππ], we
can construct, applying the 3rd Lemma, the parallels
through π and π to π π which intersect π in πΎ and πΏ ,
respectively. Since we have on the line π the points
πΎ, π, πΏ , such that [πΎπ] β‘ [ππΏ] , we can construct the
parallel through π to π based on the construction
from 3rd Lemma.
1st Theorem. (P. Aubert β 1899)
If, through the vertices of the triangle π΄π΅πΆ, we
take three lines parallel to each other, which intersect
the circumscribed circle in π΄β² , π΅β² and πΆβ² , and π is a
Ion Patrascu, Florentin Smarandache
96
point on the circumscribed circle, as well ππ΄β² β© π΅πΆ =
{π΄1} , ππ΅β² β© πΆπ΄ = {π΅1} , ππΆβ² β© π΄π΅ = {πΆ1}, then π΄1, π΅1, πΆ1
are collinear and their line is parallel to π΄π΄β².
Proof.
The point of the proof is to show that ππ΄1, ππ΅1,
ππΆ1 generate congruent angles with π΅πΆ , πΆπ΄ and π΄π΅ ,
respectively.
π(ππ΄1οΏ½οΏ½) =1
2[π(ποΏ½οΏ½) + π(π΅π΄β² )] (9)
π(ππ΅1οΏ½οΏ½) =1
2[π(ποΏ½οΏ½) + π(π΄π΅β² )] (10)
But π΄π΄β² β₯ π΅π΅β² implies π(π΅π΄β² ) = π(π΄π΅β² ) , hence,
from (9) and (10), it follows that:
β’ππ΄1πΆ β‘ β’ππ΅1πΆ, (11)
π(ππΆ1οΏ½οΏ½) =1
2[π(π΅οΏ½οΏ½) β π(π΄πΆβ² )]. (12)
But π΄π΄β² β₯ πΆπΆβ² implies that π(π΄πΆβ² ) = π(π΄β²οΏ½οΏ½); by
returning to (12), we have that:
π(ππΆ1οΏ½οΏ½) =1
2[π(π΅οΏ½οΏ½) β π(π΄πΆβ² )] =
=1
2[π(π΅π΄β² ) + π(ποΏ½οΏ½)]. (13)
The relations (9) and (13) show that:
β’ππ΄1πΆ β‘ β’ππΆ1π΄. (14)
From (11) and (14), we obtain: β’ππ΄1πΆ β‘
β’ππ΅1πΆ β‘ β’ππΆ1π΄ , which, by 1st Lemma, verifies the
collinearity of points π΄1, π΅1, πΆ1. Now, applying the 2nd
Lemma, we obtain the parallelism of lines π΄π΄β² and
π΄1π΅1.
Complements to Classic Topics of Circles Geometry
97
Figure 4.
2nd Theorem. (MβKensie β 1887)
If π΄1π΅1πΆ1 is a transversal line in the triangle π΄π΅πΆ
(π΄1 β π΅πΆ, π΅1 β πΆπ΄, πΆ1 β π΄π΅), and through the triangleβs
vertices we take the chords π΄π΄β² , π΅π΅β², πΆπΆβ² of a circle
circumscribed to the triangle, parallels with the
transversal line, then the lines π΄π΄β² , π΅π΅β², πΆπΆβ² are
concurrent on the circumscribed circle.
Ion Patrascu, Florentin Smarandache
98
Proof.
We denote by π the intersection of the line π΄1π΄β²
with the circumscribed circle (see Figure 5) and with
π΅1β² , respectively πΆ1
β² the intersection of the line ππ΅β²
with π΄πΆ and of the line ππΆβ² with π΄π΅.
Figure 5.
According to the P. Aubertβs theorem, we have
that the points π΄1, π΅1β² , πΆ1
β² are collinear and that the line
π΄1π΅1β² is parallel to π΄π΄β².
From hypothesis, we have that π΄1π΅1 β₯ π΄π΄β²; from
the uniqueness of the parallel taken through π΄1 to π΄π΄β²,
it follows that π΄1π΅1 β‘ π΄1π΅1β² , therefore π΅1
β² = π΅1 , and
analogously πΆ1β² = πΆ1.
Complements to Classic Topics of Circles Geometry
99
Remark.
We have that: ππ΄1, ππ΅1, ππΆ1 generate congruent
angles with π΅πΆ, πΆπ΄ and π΄π΅, respectively.
3rd Theorem. (Beltrami β 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 π΄π΄β², π΅π΅β², πΆπΆβ² the three parallel lines with a
certain direction (see Figure 6).
Ion Patrascu, Florentin Smarandache
100
Figure 6.
To construct the isogonal of the cevian π΄π΄β², we
take π΄β²π β₯ π΅πΆ, π belonging to the circle circumscribed
to the triangle, having π΅π΄β² β‘ πΆοΏ½οΏ½, it follows that π΄π
will be the isogonal of the cevian π΄π΄β². (Indeed, from
π΅π΄β² β‘ πΆοΏ½οΏ½ it follows that β’π΅π΄π΄β² β‘ β’πΆπ΄π.)
On the other hand, π΅π΅β² β₯ A π΄β² implies π΅π΄β² β‘
π΄οΏ½οΏ½β², and since π΅π΄β² β‘ πΆοΏ½οΏ½ we have that π΄π΅β² β‘ πΆοΏ½οΏ½ ,
which shows that the isogonal of the parallel π΅π΅β² is
π΅π. From πΆπΆβ² β₯ π΄π΄β², it follows that π΄β²πΆ β‘ π΄πΆβ², having
β’π΅β²πΆπ β‘ β’π΄πΆπΆβ², therefore the isogonal of the parallel
πΆπΆβ² is πΆπβ².
Reciprocally.
If π΄π,π΅π, πΆπ are concurrent cevians in π , the
point on the circle circumscribed to the triangle π΄π΅πΆ,
let us prove that their isogonals are parallel lines. To
construct an isogonal of π΄π , we take ππ΄β² β₯ π΅πΆ , π΄β²
belonging to the circumscribed circle. We have ποΏ½οΏ½ β‘
π΅π΄β² . Constructing the isogonal π΅π΅β² of π΅π, with π΅β² on
the circumscribed circle, we will have πΆοΏ½οΏ½ β‘ π΄π΅β² , it
follows that π΅π΄β² β‘ π΄π΅β² and, consequently, β’π΄π΅π΅β² β‘
β’π΅π΄π΄β², which shows that π΄π΄β² β₯ π΅π΅β². Analogously, we
show that πΆπΆβ² β₯ π΄π΄β².
We are now able to solve the proposed problems.
Complements to Classic Topics of Circles Geometry
101
Solution to the 1st problem.
Using the 3rd Lemma, we construct the parallels
π΄π΄β², π΅π΅β², πΆπΆβ² with a certain directions of a diameter of
the circle circumscribed to the given triangle.
We join π with π΄β² , π΅β², πΆβ² and denote the
intersection between ππ΄β² and π΅πΆ, π΄1; ππ΅β² β© πΆπ΄ = {π΅1}
and ππ΄β² β© π΄π = {πΆ1}.
According to the Aubertβs Theorem, the points
π΄1, π΅1, πΆ1 will be collinear, and ππ΄β², ππ΅β², ππΆβ² generate
congruent angles with π΅πΆ, πΆπ΄ and π΄π΅, respectively.
Solution to the 2nd problem.
Using the 3rd Lemma and the remark that follows
it, we construct through π΄, π΅, πΆ the parallels to π΄1π΅1;
we denote by π΄β², π΅β², πΆβ² their intersections with the
circle circumscribed to the triangle π΄π΅πΆ. (It is enough
to build a single parallel to the transversal line π΄1π΅1πΆ1,
for example π΄π΄β²).
We join π΄β² with π΄1 and denote by π the
intersection with the circle. The point π will be the
point we searched for. The constructionβs proof
follows from the MβKensie Theorem.
Ion Patrascu, Florentin Smarandache
102
Solution to the 3rd problem.
We suppose that π΄β² belongs to the little arc
determined by the chord π΅οΏ½οΏ½ in the circle
circumscribed to the triangle π΄π΅πΆ.
In this case, in order to find the isogonal π΄π΄1, we
construct (by help of the 3rd Lemma and of the remark
that follows it) the parallel π΄β²π΄1 to π΅πΆ, π΄1 being on the
circumscribed circle, it is obvious that π΄π΄β² and π΄π΄1
will be isogonal cevians.
We suppose that π΄β² belongs to the high arc
determined by the chord π΅οΏ½οΏ½; we consider π΄β² β π΄οΏ½οΏ½ (the
arc π΄οΏ½οΏ½ does not contain the point πΆ). In this situation,
we firstly construct the parallel π΅π to π΄π΄β², π belongs
to the circumscribed circle, and then through π we
construct the parallel ππ΄1 to π΄πΆ , π΄1 belongs to the
circumscribed circle. The isogonal of the line π΄π΄β² will
be π΄π΄1 . The constructionβs proof follows from 3rd
Lemma and from the proof of Beltramiβs Theorem.
References.
[1] F. G. M.: Exercices de GΓ©ometrie. VIII-e ed., Paris,
VI-e Librairie Vuibert, Rue de Vaugirard,77. [2] T. Lalesco: La GΓ©omΓ©trie 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 TehnicΔ, 1957.
Complements to Classic Topics of Circles Geometry
103
Apolloniusβs Circles
of kth Rank
The purpose of this article is to introduce
the notion of Apolloniusβs circle of k th rank.
1st Definition.
It is called an internal cevian of kth rank the line
π΄π΄π where π΄π β (π΅πΆ), such that π΅π΄
π΄ππΆ= (
π΄π΅
π΄πΆ)π (π β β).
If π΄πβ² is the harmonic conjugate of the point π΄π in
relation to π΅ and πΆ, we call the line π΄π΄πβ² an external
cevian of kth rank.
2nd Definition.
We call Apolloniusβs circle of kth rank with
respect to the side π΅πΆ of π΄π΅πΆ triangle the circle which
has as diameter the segment line π΄ππ΄πβ² .
1st Theorem.
Apolloniusβs circle of kth rank is the locus of
points π from π΄π΅πΆ triangle's plan, satisfying the
relation: ππ΅
ππΆ= (
π΄π΅
π΄πΆ)π.
Ion Patrascu, Florentin Smarandache
104
Proof.
Let ππ΄π the center of the Apolloniusβs circle of kth
rank relative to the side π΅πΆ of π΄π΅πΆ triangle (see
Figure 1) and π, π the points of intersection of this
circle with the circle circumscribed to the triangle π΄π΅πΆ.
We denote by π· the middle of arc π΅πΆ, and we extend
π·π΄π to intersect the circle circumscribed in πβ².
In π΅πβ²πΆ triangle, πβ²π· is bisector; it follows that
π΅π΄π
π΄ππΆ=
πβ²π΅
πβ²πΆ= (
π΄π΅
π΄πΆ)π, so πβ² belongs to the locus.
The perpendicular in πβ² on πβ²π΄π intersects BC on
π΄πβ²β² , which is the foot of the π΅ππΆ triangle's outer
bisector, so the harmonic conjugate of π΄π in relation
to π΅ and πΆ, thus π΄πβ²β² = π΄π
β² .
Therefore, πβ² is on the Apolloniusβs circle of rank
π relative to the side π΅πΆ, hence πβ² = π.
Figure 3
Complements to Classic Topics of Circles Geometry
105
Let π a point that satisfies the relation from the
statement; thus ππ΅
ππΆ=
π΅π΄π
π΄ππΆ ; it follows β by using the
reciprocal of bisector's theorem β that ππ΄π is the
internal bisector of angle π΅ππΆ. Now let us proceed as
before, taking the external bisector; it follows that π
belongs to the Apolloniusβs circle of center ππ΄π. We
consider now a point π on this circle, and we
construct πΆβ² such that β’π΅ππ΄π β‘ β’π΄πππΆβ² (thus (ππ΄π is
the internal bisector of the angle π΅ππΆβ² ). Because
π΄πβ² π β₯ ππ΄π, it follows that π΄π and π΄π
β² are harmonically
conjugated with respect to π΅ and πΆβ² . On the other
hand, the same points are harmonically conjugated
with respect to π΅ and πΆ; from here, it follows that πΆβ² =
πΆ, and we have ππ΅
ππΆ=
π΅π΄π
π΄ππΆ= (
π΄π΅
π΄πΆ)π.
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).
Ion Patrascu, Florentin Smarandache
106
Proof.
Let π΄π΅πΆπ·πΈπΉ a given complete quadrilateral (see
Figure 2). We denote by π»1, π»2, π»3, π»4 respectively the
orthocenters of π΄π΅πΉ, π΄π·πΈ, πΆπ΅πΈ, πΆπ·πΉ triangles, and
let π΄1, π΅1, πΉ1 the feet of the heights of π΄π΅πΉ triangle.
Figure 4
As previously shown, the following relations
occur: π»1π΄.π»1π΄1 β π»1π΅.π»1π΅1 = π»1πΉ.π»1πΉ1; they express
that the point π»1 has equal powers to the circles of
diameters π΄πΆ, π΅π·, πΈπΉ , because those circles contain
respectively the points π΄1, π΅1, πΉ1, and π»1 is an internal
point.
It is shown analogously that the points π»2, π»3, π»4
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
Complements to Classic Topics of Circles Geometry
107
such their centers β which are the middles of the
complete quadrilateral's diagonals β are collinear.
The line formed by the middles of a complete
quadrilateral's diagonals is called Gaussβs line or
Gauss-Newtonβs line.
3rd Theorem.
The Apolloniusβs circle of kth rank of a triangle
are part of a fascicle.
Proof.
Let π΄π΄π , π΅π΅π , πΆπΆπ be concurrent cevians of kth
rank and π΄π΄πβ² , π΅π΅π
β² , πΆπΆπβ² be the external cevians of kth
rank (see Figure 3). The figure π΅πβ² πΆππ΅ππΆπ
β²π΄ππ΄πβ² is a
complete quadrilateral and 2nd theorem is applied.
Figure 5
Ion Patrascu, Florentin Smarandache
108
4th Theorem.
The Apolloniusβs circle of kth rank of a triangle
are the orthogonals of the circle circumscribed to the
triangle.
Proof.
We unite π to π· and π (see Figure 1), ππ· β₯ π΅πΆ
and π(π΄πππ΄πβ² ) = 900, it follows that ππ΄π
β² π΄π = ππ·π΄οΏ½οΏ½ =
πππ΄οΏ½οΏ½.
The congruence ππ΄πβ² π΄π
β‘ πππ΄οΏ½οΏ½ shows that ππ
is tangent to the Apolloniusβs circle of center ππ΄π.
Analogously, it can be demonstrated for the
other Apolloniusβs Circle.
1st Remark.
The previous theorem indicates that the radical
axis of Apolloniusβs circle of kth rank is the perpen-
dicular taken from π to the line ππ΄πππ΅π
.
5th Theorem.
The centers of Apolloniusβs Circle of kth rank of a
triangle are situated on the trilinear polar associated
to the intersection point of the cevians of 2kth rank.
Complements to Classic Topics of Circles Geometry
109
Proof.
From the previous theorem, it results that ππ β₯
πππ΄π, so πππ΄π
is an external cevian of rank 2 for π΅πΆπ
triangle, thus an external symmedian. Henceforth, ππ΄π
π΅
ππ΄ππΆ
= (π΅π
πΆπ)2
= (π΄π΅
π΄πΆ)2π
(the last equality occurs because
π belong to the Apolloniusβs circle of rank π associated
to the vertex π΄).
6th Theorem.
The Apolloniusβs 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 π and π points of intersection of the
Apolloniusβs circle of center ππ΄π with the circle
circumscribed to the π΄π΅πΆ (see Figure 1). We take from
π and π the perpendiculars ππ1 , ππ2 and ππ1, ππ2 on
π΄π΅ and π΄πΆ respectively. The quadrilaterals π΄π΅ππΆ ,
π΄π΅πΆπ are inscribed, it follows the similarity of
triangles π΅ππ1, πΆππ2 and π΅ππ1, πΆππ2, from where we
get the relations: π΅π
πΆπ=
ππ1
ππ2,
ππ΅
ππΆ=
ππ1
ππ2.
Ion Patrascu, Florentin Smarandache
110
But π΅π
πΆπ= (
π΄π΅
π΄πΆ)π,
ππ΅
ππΆ= (
π΄π΅
π΄πΆ)π,
ππ1
ππ2= (
π΄π΅
π΄πΆ)π and
ππ1
ππ2=
(π΄π΅
π΄πΆ)π
, relations that show that π and π belong
respectively to the internal cevian and the external
cevian of rank π + 1.
4th Definition.
If the Apolloniusβs 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 ππ ,ππβ²).
1st Property.
If ππ ,ππβ² are isodynamic centers of kth rank,
then:
πππ΄. π΅πΆπ = πππ΅. π΄πΆπ = πππΆ. π΄π΅π;
ππβ²π΄. π΅πΆπ = ππ
β²π΅. π΄πΆπ = ππβ²πΆ. π΄π΅π.
The proof of this property follows immediately
from 1st Theorem.
2nd Remark.
The Apolloniusβs circle of 1st rank is the
investigated Apolloniusβs circle (the bisectors are
cevians of 1st rank). If π = 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
Complements to Classic Topics of Circles Geometry
111
in triangleβs vertices to the circumscribed circle. In
this case, for the Apolloniusβs circle of 2nd rank, the 3rd
Theorem becomes:
7th Theorem.
The Apolloniusβs 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.
Ion Patrascu, Florentin Smarandache
112
References.
[1] N. N. MihΔileanu: LecΘii complementare de geometrie [Complementary Lessons of Geometry], Editura DidacticΔ Θi PedagogicΔ,
BucureΘti, 1976. [2] C. Mihalescu: Geometria elementelor remarcabile
[The Geometry of Outstanding Elements],
Editura TehnicΔ, BucureΘti, 1957. [3] V. Gh. VodΔ: Triunghiul β ringul cu trei colΘuri [The
Triangle-The Ring with Three Corners], Editura Albatros, BucureΘti, 1979.
[4] F. Smarandache, I. PΔtraΘcu: Geometry of
Homological Triangle, The Education Publisher Inc., Columbus, Ohio, SUA, 2012.
Complements to Classic Topics of Circles Geometry
113
Apolloniusβs Circle of Second
Rank
This article highlights some properties of
Apolloniusβs circle of second rank in
connection with the adjoint circles and the
second Brocardβs triangle.
1st Definition.
It is called Apolloniusβs circle of second rank
relative to the vertex π΄ of the triangle π΄π΅πΆ the circle
constructed on the segment determined on the
simediansβ feet from π΄ on π΅πΆ as diameter.
1st Theorem.
The Apolloniusβs circle of second rank relative to
the vertex π΄ of the triangle π΄π΅πΆ intersect the
circumscribed circle of the triangle π΄π΅πΆ in two points
belonging respectively to the cevian of third rank
(antibisectorβs isogonal) and to its external cevian.
The theoremβs proof follows from the theorem
relative to the Apolloniusβs circle of kth rank (see [1]).
Ion Patrascu, Florentin Smarandache
114
1st Proposition.
The Apolloniusβs circle of second rank relative to
the vertex π΄ of the triangle π΄π΅πΆ intersects the
circumscribed circle in two points π and π (π on the
same side of π΅πΆ as π΄). Then, (ππ is a bisector in the
triangle ππ΅πΆ , S is the simedianβs foot from π΄ of the
triangle π΄π΅πΆ.
Proof.
π belongs to the Apolloniusβs circle of second
rank, therefore:
ππ΅
ππΆ= (
π΄π΅
π΄πΆ)2. (1)
Figure 1.
On the other hand, π being the simedianβs foot,
we have:
Complements to Classic Topics of Circles Geometry
115
ππ΅
ππΆ= (
π΄π΅
π΄πΆ)2. (2)
From relations (1) and (2), we note that ππ΅
ππΆ=
ππ΅
ππΆ ,
a relation showing that ππ is bisector in the triangle
ππ΅πΆ.
Remarks.
1. The Apolloniusβs circle of second relative
to the vertex π΄ of the triangle π΄π΅πΆ (see
Figure 1) is an Apolloniusβs circle for the
triangle ππ΅πΆ. Indeed, we proved that ππ
is an internal bisector in the triangle ππ΅πΆ,
and since πβ², the external simedianβs foot
of the triangle π΄π΅πΆ , belongs to the
Apolloniusβs Circle of second rank, we
have π(β’πβ²ππ) = 900 , therefore ππβ is an
external bisector in the triangle ππ΅πΆ.
2. ππ is a simedian in ππ΅πΆ . Indeed, the
Apolloniusβs circle of second rank, being
an Apolloniusβs circle for ππ΅πΆ, intersects
the circle circum-scribed to ππ΅πΆ after ππ,
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
Ion Patrascu, Florentin Smarandache
116
one of them is tangent to the triangleβs side. We denote
(π΅οΏ½οΏ½) the adjoint circle that passes through π΅ and π΄,
and is tangent to the side π΄πΆ in π΄.
About the circles (π΅οΏ½οΏ½) and (πΆοΏ½οΏ½), we say that they
are adjoint to the vertex π΄ of the triangle π΄π΅πΆ.
3rd Definition.
It is called the second Brocardβs triangle the
triangle π΄2π΅2πΆ2 whose vertices are the projections of
the center of the circle circumscribed to the triangle
π΄π΅πΆ on triangleβs simedians.
2nd Proposition.
The Apolloniusβs circle of second rank relative to
the vertex π΄ of triangle π΄π΅πΆ and the adjoint circles
relative to the same vertex π΄ intersect in vertex π΄2 of
the second Brocardβs triangle.
Proof.
It is known that the adjoint circles (π΅οΏ½οΏ½) and (πΆοΏ½οΏ½)
intersect in a point belonging to the simedian π΄π; we
denote this point π΄2 (see [3]).
We have:
β’π΅π΄2π = β’π΄2π΅π΄ + β’π΄2π΄π΅,
but:
β’π΄2π΅π΄ β‘ β’π΅π΄2π = β’π΄2π΄π΅ + π΄2π΄πΆ = β’π΄.
Complements to Classic Topics of Circles Geometry
117
Analogously, β’πΆπ΄2π = β’π΄ , therefore (π΄2π is the
bisector of the angle π΅π΄2πΆ. The bisectorβs theorem in
this triangle leads to: ππ΅
ππΆ=
π΅π΄2
πΆπ΄2 ,
but:
ππ΅
ππΆ= (
π΄π΅
π΄πΆ)2,
consequently:
π΅π΄2
πΆπ΄2= (
π΄π΅
π΄πΆ)2,
so π΄2 is a point that belongs to the Apolloniusβs circle
of second rank.
Figure 2.
We prove that π΄2 is a vertex in the second
Brocardβs triangle, i.e. ππ΄2 β₯ π΄π, O the center of the
circle circumscribed to the triangle π΄π΅πΆ.
Ion Patrascu, Florentin Smarandache
118
We pointed (see Figure 2) that π(π΅π΄2οΏ½οΏ½) = 2π΄, if
β’π΄ is an acute angle, then also π(π΅ποΏ½οΏ½) = 2π΄ ,
therefore the quadrilateral ππΆπ΄2π΅ is inscriptible.
Because π(ππΆοΏ½οΏ½) = 900 β π(οΏ½οΏ½) , it follows that
π(π΅π΄2οΏ½οΏ½) = 900 + π(οΏ½οΏ½).
On the other hand, π(π΄π΄2οΏ½οΏ½) = 1800 β π(οΏ½οΏ½), so
π(π΅π΄2οΏ½οΏ½) + π(π΄π΄2οΏ½οΏ½) = 2700 and, consequently, ππ΄2 β₯
π΄π.
Remarks.
1. If π(οΏ½οΏ½) < 900 , then four remarkable
circles pass through π΄2 : the two circles
adjoint to the vertex π΄ of the triangle π΄π΅πΆ,
the circle circumscribed to the triangle
π΅ππΆ (where π is the center of the
circumscribed circle) and the Apolloniusβs
circle of second rank corresponding to the
vertex π΄.
2. The vertex π΄2 of the second Brocardβs
triangle is the middle of the chord of the
circle circumscribed to the triangle π΄π΅πΆ
containing the simedian π΄π.
3. The points π , π΄2 and πβ (the foot of the
external simedian to π΄π΅πΆ) are collinear.
Indeed, we proved that ππ΄2 β₯ π΄π; on the
other hand, we proved that (π΄2π is an
internal bisector in the triangle π΅π΄2πΆ, and
Complements to Classic Topics of Circles Geometry
119
since πβ²π΄2 β₯ π΄π , the outlined collinearity
follows from the uniqueness of the
perpendicular in π΄2 on π΄π.
Open Problem.
The Apolloniusβs circle of second rank relative to
the vertex π΄ of the triangle π΄π΅πΆ intersects the circle
circumscribed to the triangle π΄π΅πΆ in two points π and
π (π and π΄ apart of π΅πΆ).
We denote by π the second point of intersection
between the line π΄π and the Apolloniusβs circle of
second rank.
What can we say about π?
Is π a remarkable point in triangleβs geometry?
Ion Patrascu, Florentin Smarandache
120
References.
[1] I. Patrascu, F. Smarandache: Cercurile Apollonius de rangul k [The Apolloniusβs Circle of kth rank]. In: βRecreaΘii matematiceβ, Anul XVIII, nr.
1/2016, IaΘi, RomΓ’nia, p. 20-23. [2] I. Patrascu: Axe Θi centre radicale ale cercurilor
adjuncte ale unui triunghi [Axes and Radical
Centers of Adjoint Circles of a Triangle]. In: βRecreaΘii matematiceβ, Anul XVII, nr. 1/2010,
IaΘi, RomΓ’nia. [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.
Complements to Classic Topics of Circles Geometry
121
A Sufficient Condition for
the Circle of the 6 Points
to Become Eulerβs 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 triangleβs points, we obtain the
circle of 9 points (the Eulerβs C ircle).
1st Definition.
It is called cevian of a triangle the line that
passes through the vertex of a triangle and an opposite
sideβs point, other than the two left vertices.
Figure 1.
Ion Patrascu, Florentin Smarandache
122
1st Remark.
The name cevian was given in honor of the
italian geometrician Giovanni Ceva (1647 - 1734).
2nd Definition.
Two cevians of the same triangleβs vertex are
called isogonal cevians if they create congruent angles
with triangleβs bisector taken through the same vertex.
2nd Remark.
In the Figure 1 we represented the bisector π΄π·
and the isogonal cevians π΄π΄1 and π΄π΄2. The following
relations take place:
π΄1π΄οΏ½οΏ½ β‘ π΄2π΄οΏ½οΏ½;
π΅π΄π΄1 β‘ πΆπ΄π΄2.
1st Proposition.
In a triangle, the height and the radius of the
circumscribed circle corresponding to a vertex are
isogonal cevians.
Proof.
Let π΄π΅πΆ an acute-angled triangle with the height
π΄π΄β² and the radius π΄π (see Figure 2).
Complements to Classic Topics of Circles Geometry
123
Figure 2.
The angle π΄ππΆ is a central angle, and the angle
π΄π΅πΆ is an inscribed angle, so π΄ποΏ½οΏ½ = 2π΄π΅οΏ½οΏ½. It follows
that π΄ποΏ½οΏ½ = 900 β οΏ½οΏ½.
On the other hand, π΅π΄π΄β² = 900 β οΏ½οΏ½ , so π΄π΄β² and
π΄π are isogonal cevians.
The theorem can be analogously proved for the
obtuse triangle.
3rd Remark.
One can easily prove that in a triangle, if π΄π is
circumscribed circleβs radius, its isogonal cevian is the
triangleβs height from vertex π΄.
3rd Definition.
Two points π1, π2 in the plane of triangle π΄π΅πΆ are
called isogonals (isogonal conjugated) if the ceviansβ
pairs (π΄π1, π΄π2), (π΅π1, π΅π2), (πΆπ1, πΆπ2), are composed
of isogonal cevians.
Ion Patrascu, Florentin Smarandache
124
4th Remark.
In a triangle, the orthocenter and circumscribed
circleβs center are isogonal points.
1st Theorem. (The 6 points circle)
If π1 and π2 are two isogonal points in the
triangle π΄π΅πΆ , and π΄1, π΅1, πΆ1 respectively π΄2, π΅2, πΆ2 are
their projections on the sides π΅πΆ , πΆπ΄ and π΄π΅ of the
triangle, then the points π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are
concyclical.
Proof.
The mediator of segment [π΄1π΄2] passes through
the middle π of segment [π1, π2] because the trapezoid
π1π΄1π΄2π2 is rectangular and the mediator of [π΄1π΄2]
contains its middle line, therefore (see Figure 3), we
have: ππ΄1 = ππ΄2 (1). Analogously, it follows that the
mediators of segments [π΅1π΅2] and [πΆ1πΆ2] pass through
π, so ππ΅1 = ππ΅2 (2) and ππΆ1 = ππΆ2 (3). We denote by
π΄3 and π΄4 respectively the middles of segments [π΄π1]
and [π΄π2 ]. We prove that the triangles ππ΄3πΆ1 and
π΅2π΄4π are congruent. Indeed, ππ΄3 =1
2π΄π2 (middle
line), and π΅2π΄4 =1
2π΄π2, because it is a median in the
rectangular triangle π2π΅2π΄ , so ππ΄3 = π΅2π΄4 ;
analogously, we obtain that π΄4π = π΄3πΆ1 .
Complements to Classic Topics of Circles Geometry
125
Figure 3.
We have that:
ππ΄3πΆ1 = ππ΄3π1
+ π1π΄3πΆ1 = π1π΄π2 + 2π1π΄πΆ1 =
= οΏ½οΏ½ + π1π΄οΏ½οΏ½;
π΅2π΄4οΏ½οΏ½ = π΅2π΄4π2 + ππ΄4π2
= π1π΄π2 + 2π2π΄π΅2 =
= οΏ½οΏ½ + π2π΄οΏ½οΏ½.
But π1π΄οΏ½οΏ½ = π2π΄οΏ½οΏ½ , because the cevians π΄π1 and
π΄π2 are isogonal and therefore ππ΄3πΆ1 = π΅2π΄4οΏ½οΏ½. Since
βππ΄3πΆ1 = βπ΅2π΄4π, it follows that ππ΅2 = ππΆ1 (4).
Repeating the previous reasoning for triangles
ππ΅3πΆ1 and π΄2π΅4π , where π΅3 and π΅4 are respectively
the middles of segments (π΅π1) and (π΅π2), we find that
they are congruent and it follows that ππΆ1 = ππ΄2 (5).
The relations (1) - (5) lead to ππ΄1 = ππ΄2 = ππ΅1 =
ππ΅2 = ππΆ1 = ππΆ2 , which shows that the points
π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are located on a circle centered in π,
Ion Patrascu, Florentin Smarandache
126
the middle of the segment determined by isogonal
points given by π1 and π2.
4th Definition.
It is called the 9 points circle or Eulerβs circle of
the triangle π΄π΅πΆ the circle that contains the middles
of triangleβs sides, the triangle heightsβ feet and the
middles of the segments determined by the
orthocenter with triangleβs vertex.
2nd Proposition.
If π1, π2 are isogonal points in the triangle π΄π΅πΆ
and if on the circle of their corresponding 6 points
there also are the middles of the segments (π΄π1), (π΅π1),
(πΆπ1 ), then the 6 points circle coincides with the
Eulerβs circle of the triangle π΄π΅πΆ.
1st Proof.
We keep notations from Figure 3; we proved that
the points π΄1, π΄2, π΅1, π΅2, πΆ1, πΆ2 are on the 6 points circle
of the triangle π΄π΅πΆ, having its center in π, the middle
of segment [π1π2].
If on this circle are also situated the middles
π΄3, π΅3, πΆ3 of segments (π΄π1), (π΅π1), (πΆπ1), then we have
ππ΄3 = ππ΅3 = ππΆ3.
Complements to Classic Topics of Circles Geometry
127
We determined that ππ΄3 is middle line in the
triangle π1π2π΄ , therefore ππ΄3 =1
2π΄π2 , analogously
ππ΅3 =1
2π΅π2 and ππΆ3 =
1
2πΆπ2, and we obtain that π2π΄ =
π2π΅ = π2πΆ, consequently π is the center of the circle
circumscribed to the triangle π΄π΅πΆ, so π2 = π.
Because π1 is the isogonal of π, it follows that
π1 = π», therefore the circle of 6 points of the isogonal
points π and π» is the circle of 9 points.
2nd Proof.
Because π΄3π΅3 is middle line in the triangle π1π΄π΅,
it follows that
β’π1π΄π΅ β‘ β’π1π΄2π΅3. (1)
Also, π΄3πΆ3 is middle line in the triangle π1π΄πΆ, and
π΄3πΆ3 is middle line in the triangle π1π΄π2, therefore we
get
β’ππ΄3πΆ3 β‘ β’π2π΄πΆ. (2)
The relations (1), (2) and the fact that π΄π1 and
π΄π2 are isogonal cevians lead to:
β’π1π΄2π΅3 β‘ ππ΄3πΆ3. (3)
The point π is the center of the circle
circumscribed to π΄3π΅3πΆ3 ; then, from (3) and from
isogonal ceviansβ properties, one of which is
circumscribed circle radius, it follows that in the
triangle π΄3π΅3πΆ3 the line π1π΄3 is a height, as π΅3πΆ3 β₯ π΅πΆ,
we get that π1π΄ is a height in the triangle ABC and,
therefore, π1 will be the orthocenter of the triangle
Ion Patrascu, Florentin Smarandache
128
π΄π΅πΆ , and π2 will be the center of the circle
circumscribed to the triangle π΄π΅πΆ.
5th Remark.
Of those shown, we get that the center of the
circle of 9 points is the middle of the line determined
by triangleβs orthocenter and by the circumscribed
circleβs center, and the fact that Eulerβs circle radius is
half the radius of the circumscribed circle.
References.
[1] Roger A. Johnson: Advanced Euclidean Geometry. New York: Dover Publications, 2007.
[2] CΔtΔlin Barbu: Teoreme fundamentale din geometria triunghiului [Fundamental theorems of triangleβs geometry]. BacΔu: Editura Unique,
2008.
Complements to Classic Topics of Circles Geometry
129
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 πΎ, and the case
for the problem for the ex-inscribed circle
πΆ. 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:
βWe fix an angle πΎ β (0, 900) and the line
π΄π΅ which divides the plane in two half-planes πΎ
and οΏ½οΏ½. The point πΆ in the half-plane πΎ is situated
such that π(π΄πΆοΏ½οΏ½) = πΎ. The circle inscribed in the
triangle π΄π΅πΆ with the center πΌ is tangent to the
sides π΄πΆ and π΅πΆ in the points πΉ and πΈ ,
respectively. The point π is located on the
segment line (πΌπΈ , the point πΈ between πΌ and π
such that ππΈ β₯ π΅πΆ and ππΈ = π΄πΉ. The point π is
situated on the segment line (πΌπΉ, such that πΉ is
between πΌ and π ; ππΉ β₯ π΄πΆ and ππΉ = π΅πΈ . Prove
Ion Patrascu, Florentin Smarandache
130
that the mediator of segment ππ passes through
a fixed point.β (Stanislav Chobanov)
Proof.
Firstly, it is useful to note that the point πΆ varies
in the half-plane π on the arc capable of angle πΎ; we
know as well that π(π΄πΌοΏ½οΏ½) = 900 +πΎ
2, so πΌ varies on the
arc capable of angle of measure 900 +πΎ
2 situated in the
half-plane π.
Another useful remark is about the segments π΄πΉ
and π΅πΈ , which in a triangle have the lengths π β π ,
respectively π β π , where π is the half-perimeter of
the triangle π΄π΅πΆ with π΄π΅ = π β constant; therefore,
we have βππΈπ΅ β‘ βπ΄πΉπ with the consequence ππ΅ = ππ΄.
Considering the vertex πΆ of the triangle π΄π΅πΆ the
middle of the arc capable of angle πΎ built on π΄π΅, we
observe that ππ is parallel to π΄π΅ ; more than that,
π΄π΅ππ is an isosceles trapezoid, and segment ππ
mediator will be a symmetry axis of the trapezoid, so
it will coincide with the mediator of π΄π΅, which is a
fixed line, so we're looking for the fixed point on
mediator of π΄π΅.
Let π· be the intersection of the mediators of
segments ππ and π΄π΅ , see Figure 1, where we
considered π(οΏ½οΏ½) < π(οΏ½οΏ½) . The point π· is on the
mediator of π΄π΅, so we have π·π΄ = π·π΅; the point π· is
also on the mediator of ππ, so we have π·π = π·π; it
Complements to Classic Topics of Circles Geometry
131
follows that: βππ΅π· β‘ βππ΄π·, a relation from where we
get that β’ππ΄π· = β’ππ΅π·.
Figure 1
If we denote π(ππ΄οΏ½οΏ½) = π₯ and π(π·π΄οΏ½οΏ½) = π¦ , we
have ππ΄οΏ½οΏ½ = π₯ + π΄ + π¦ , ππ΅οΏ½οΏ½ = 3600 β π΅ β π¦ β (900 β π₯).
From π₯ + π΄ + π¦ = 3600 β π΅ β π¦ β 900 + π₯, we find
that π΄ + π΅ + 2π¦ = 2700, and since π΄ + π΅ = 1800 β πΎ, we
find that 2π¦ = 900 β πΎ, therefore the requested fixed
point π· is the vertex of triangle π·π΄π΅ , situated in οΏ½οΏ½
such that π(π΄π·οΏ½οΏ½) = 900 β πΎ.
Ion Patrascu, Florentin Smarandache
132
1st Remark.
If πΎ = 900, we propose to the reader to prove that
the quadrilateral π΄π΅ππ is a parallelogram; in this case,
the requested fixed point does not exist (or is the point
at infinity of the perpendicular lines to π΄π΅).
2nd Remark.
If πΎ β (900, 1800), the problem applies, and we
find that the fixed point π· is located in the half-plane
π , such that the triangle π·π΄π΅ is isosceles, having
π(π΄ποΏ½οΏ½) = πΎ β 900.
We suggest to the reader to solve the following
problem:
We fix an angle πΎ β (00, 1800) and the line
AB which divides the plane in two half-planes, π
and οΏ½οΏ½. The point πΆ in the half-plane π is located
such that π(π΄πΆοΏ½οΏ½) = πΎ . The circle πΆ β ex-
inscribed to the triangle π΄π΅πΆ with center πΌπ is
tangent to the sides π΄πΆ and π΅πΆ in the points πΉ
and πΈ, respectively. The point π is located on the
line segment (πΌππΈ , πΈ is between πΌπ and π such
that ππΈ β₯ π΅πΆ and ππΈ = π΄πΉ . The point π is
located on the line segment (πΌππΉ such that πΉ is
between πΌ and π , ππΉ β₯ π΄πΆ and ππΉ = π΅πΈ . Prove
that the mediator of the segment ππ passes
through a fixed point.
Complements to Classic Topics of Circles Geometry
133
3rd Remark.
As seen, this problem is also true in the case πΎ =
900, more than that, in this case, the fixed point is the
middle of π΄π΅. Prove!
References.
[1] Ion Patrascu: Probleme de geometrie planΔ [Planar
Geometry Problems]. Craiova: Editura Cardinal, 1996.
Ion Patrascu, Florentin Smarandache
134
Complements to Classic Topics of Circles Geometry
135
Some Properties of the
Harmonic Quadrilateral
In this article, we review some properties
of the harmonic quadrilateral related to
triangle simedians and to Apolloniusβs
Circle.
1st Definition.
A convex circumscribable quadrilateral π΄π΅πΆπ·
having the property π΄π΅ β πΆπ· = π΅πΆ β π΄π· is called
harmonic quadrilateral.
2nd Definition.
A triangle simedian is the isogonal cevian of a
triangle median.
1st Proposition.
In the triangle π΄π΅πΆ, the cevian π΄π΄1, π΄1 β (π΅πΆ) is
a simedian if and only if π΅π΄1
π΄1πΆ= (
π΄π΅
π΄πΆ)2. For Proof of this
property, see infra.
Ion Patrascu, Florentin Smarandache
136
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 π΄π΅πΆπ· be an harmonic quadrilateral and
{πΎ} = π΄πΆ β© π΅π· (see Figure 1). We prove that π΅πΎ is
simedian in the triangle π΄π΅πΆ.
From the similarity of the triangles π΄π΅πΎ and
π·πΆπΎ, we find that: π΄π΅
π·πΆ=
π΄πΎ
π·πΎ=
π΅πΎ
πΆπΎ . (1)
From the similarity of the triangles π΅πΆπΎ Εi π΄π·πΎ,
we conclude that:
Complements to Classic Topics of Circles Geometry
137
π΅πΆ
π΄π·=
πΆπΎ
π·πΎ=
π΅πΎ
π΄πΎ . (2)
From the relations (1) and (2), by division, it
follows that: π΄π΅
π΅πΆ.π΄π·
π·πΆ=
π΄πΎ
πΆπΎ . (3)
But π΄π΅πΆπ· is an harmonic quadrilateral;
consequently, π΄π΅
π΅πΆ=
π΄π·
π·πΆ ;
substituting this relation in (3), it follows that:
(π΄π΅
π΅πΆ)2
=π΄πΎ
πΆπΎ;
As shown by Proposition 1, π΅πΎ is a simedian in
the triangle π΄π΅πΆ. Similarly, it can be shown that π΄πΎ is
a simedian in the triangle π΄π΅π·, that πΆπΎ is a simedian
in the triangle π΅πΆπ·, and that π·πΎ is a simedian in the
triangle π΄π·πΆ.
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.
Ion Patrascu, Florentin Smarandache
138
Remark 2.
From 2nd and 3rd Propositions above, it results a
simple way to build an harmonic quadrilateral.
In a circle, let a triangle π΄π΅πΆ be considered; we
construct the simedian of A, be it π΄πΎ, and we denote
by D the intersection of the simedian π΄πΎ with the
circle. The quadrilateral π΄π΅πΆπ· is an harmonic
quadrilateral.
Proposition 4.
In a triangle π΄π΅πΆ, the points of the simedian of A
are situated at proportional lengths to the sides π΄π΅
and π΄πΆ.
Figura 2.
Complements to Classic Topics of Circles Geometry
139
Proof.
We have the simedian π΄π΄1 in the triangle π΄π΅πΆ
(see Figure 2). We denote by π· and πΈ the projections
of π΄1 on π΄π΅, and π΄πΆ respectively.
We get:
π΅π΄1
πΆπ΄1=
π΄πππβ(π΄π΅π΄1)
π΄πππβ(π΄πΆπ΄1)=
π΄π΅ β π΄1π·
π΄πΆ β π΄1πΈ .
Moreover, from 1st Proposition, we know that
π΅π΄1
π΄1πΆ= (
π΄π΅
π΄πΆ)2
.
Substituting in the previous relation, we obtain
that: π΄1π·
π΄1πΈ=
π΄π΅
π΄πΆ .
On the other hand, π·π΄1 = π΄π΄1. From π΅π΄π΄1 and
π΄1πΈ = π΄π΄1 β π πππΆπ΄π΄1, hence: π΄1π·
π΄1πΈ=
π πππ΅π΄π΄1
π πππΆπ΄π΄1=
π΄π΅
π΄πΆ . (4)
If M is a point on the simedian and ππ1 and ππ2
are its projections on π΄π΅ , and π΄πΆ respectively, we
have:
ππ1 = π΄π β π πππ΅π΄π΄1, ππ2 = π΄π β π πππΆπ΄π΄1,
hence:
ππ1
ππ2=
π πππ΅π΄π΄1
π πππΆπ΄π΄1
.
Taking (4) into account, we obtain that: ππ1
ππ2=
π΄π΅
π΄πΆ .
Ion Patrascu, Florentin Smarandache
140
3rd Remark.
The converse of the property in the statement
above is valid, meaning that, if π 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 π΄π΅πΆπ· be an harmonic quadrilateral and π
any point within.
Complements to Classic Topics of Circles Geometry
141
We denote by π₯, π¦, π§, π’ the distances of π to the
π΄π΅ , π΅πΆ , πΆπ· , π·π΄ sides of lenghts π, π, π, and π (see
Figure 3).
Figure 3.
Let π be the π΄π΅πΆπ· quadrilateral area.
We have:
ππ₯ + ππ¦ + ππ§ + ππ’ = 2π.
This is true for π₯, π¦, π§, π’ and π, π, π, π real numbers.
Following Cauchy-Buniakowski-Schwarz Ine-
quality, we get:
(π2 + π2 + π2 + π2)(π₯2 + π¦2 + π§2 + π’2)
β₯ (ππ₯ + ππ¦ + ππ§ + ππ’)2 ,
and it is obvious that:
π₯2 + π¦2 + π§2 + π’2 β₯4π2
π2 + π2 + π2 + π2 .
We note that the minimum sum of squared
distances is:
Ion Patrascu, Florentin Smarandache
142
4π2
π2 + π2 + π2 + π2= ππππ π‘.
In Cauchy-Buniakowski-Schwarz Inequality, the
equality occurs if: π₯
π=
π¦
π=
π§
π=
π’
π .
Since {πΎ} = π΄πΆ β© π΅π· is the only point with this
property, it ensues that π = πΎ, so πΎ has the property
of the minimum in the statement.
3rd Definition.
We call external simedian of π΄π΅πΆ triangle a
cevian π΄π΄1β corresponding to the vertex π΄, where π΄1β
is the harmonic conjugate of the point π΄1 β simedianβs
foot from π΄ relative to points π΅ and πΆ.
4th Remark.
In Figure 4, the cevian π΄π΄1 is an internal
simedian, and π΄π΄1β is an external simedian.
We have:
π΄1π΅
π΄1πΆ=
π΄1β²π΅
π΄1β²πΆ .
In view of 1st Proposition, we get that:
π΄1β²π΅
π΄1β²πΆ= (
π΄π΅
π΄πΆ)2
.
Complements to Classic Topics of Circles Geometry
143
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 π be the intersection of a tangent taken in π·
to the circle circumscribed to the harmonic
quadrilateral π΄π΅πΆπ· with π΄πΆ (see Figure 4).
Figure 4.
Since triangles PDC and PAD are alike, we
conclude that: ππ·
ππ΄=
ππΆ
ππ·=
π·πΆ
π΄π· . (5)
From relations (5), we find that:
ππ΄
ππΆ= (
π΄π·
π·πΆ)2. (6)
Ion Patrascu, Florentin Smarandache
144
This relationship indicates that P is the harmonic
conjugate of K with respect to A and C, so π·π is an
external simedian from D of the triangle π΄π·πΆ.
Similarly, if we denote by πβ the intersection of
the tangent taken in B to the circle circumscribed with
π΄πΆ, we get:
πβ²π΄
πβ²πΆ= (
π΅π΄
π΅πΆ)2. (7)
From (6) and (7), as well as from the properties
of the harmonic quadrilateral, we know that: π΄π΅
π΅πΆ=
π΄π·
π·πΆ ,
which means that:
ππ΄
ππΆ=
πβ²π΄
πβ²πΆ ,
hence π = πβ.
Similarly, it is shown that the tangents taken to
A and C intersect at point Q located on the diagonal π΅π·.
5th Remark.
a. The points P and Q are the diagonal poles of
π΅π· and π΄πΆ 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.
Complements to Classic Topics of Circles Geometry
145
Figure 5.
8th Proposition.
Let π΄π΅πΆπ· 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 {πΎ} = π΄πΆ β© π΅π·, then the orthocenter of
triangle ππΎπ is O.
Proof.
From the properties of tangents taken from a
point to a circle, we conclude that ππ β₯ π΅π· and ππ β₯
π΄πΆ. These relations show that in the triangle ππΎπ, ππ
and ππ are heights, so π is the orthocenter of this
triangle.
Ion Patrascu, Florentin Smarandache
146
4th Definition.
The Apolloniusβs circle related to the vertex A of
the triangle π΄π΅πΆ is the circle built on the segment [π·πΈ]
in diameter, where D and E are the feet of the internal,
respectively external, bisectors taken from A to the
triangle π΄π΅πΆ.
6th Remark.
If the triangle π΄π΅πΆ is isosceles with π΄π΅ = π΄πΆ ,
the Apolloniusβs circle corresponding to vertex A is not
defined.
9th Proposition.
The Apolloniusβs circle relative to the vertex A of
the triangle π΄π΅πΆ 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 π΄π΅πΆ with π΅πΆ (see Figure 6).
Assuming that π(οΏ½οΏ½) > π(οΏ½οΏ½), we find that:
π(πΈπ΄οΏ½οΏ½) =1
2[π(οΏ½οΏ½) + π(οΏ½οΏ½)].
Oa being a tangent, we find that π(πππ΄οΏ½οΏ½) = π(οΏ½οΏ½).
Withal,
π(πΈπ΄ππ) =1
2[π(οΏ½οΏ½) β π(οΏ½οΏ½)]
Complements to Classic Topics of Circles Geometry
147
and
π(π΄πΈππ) =1
2[π(οΏ½οΏ½) β π(οΏ½οΏ½)].
It results that πππΈ = πππ΄; onward, πΈπ΄π· being a
right angled triangle, we obtain: πππ΄ = πππ·; hence Oa
is the center of Apolloniusβs circle corresponding to
the vertex π΄.
Figura 6.
10th Proposition.
Apolloniusβs circle relative to the vertex A of
triangle π΄π΅πΆ cuts the circle circumscribed to the
triangle following the internal simedian taken from A.
Proof.
Let S be the second point of intersection of
Apolloniusβs Circle relative to vertex A and the circle
circumscribing the triangle π΄π΅πΆ.
Ion Patrascu, Florentin Smarandache
148
Because πππ΄ is tangent to the circle circum-
scribed in A, it results, for reasons of symmetry, that
πππ will be tangent in S to the circumscribed circle.
For triangle π΄πΆπ , πππ΄ and πππ are external
simedians; it results that πΆππ is internal simedian in
the triangle π΄πΆπ , furthermore, it results that the
quadrilateral π΄π΅ππΆ is an harmonic quadrilateral.
Consequently, π΄π is the internal simedian of the
triangle π΄π΅πΆ 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
Apolloniusβs circle relative to the vertex A for the
triangle π΄π΅π·. This circle (of center Q and radius QC)
is also an Apolloniusβs circle relative to the vertex C of
the triangle π΅πΆπ·.
Similarly, the Apolloniusβs 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 Apolloniusβs
circle - associated with the vertexes of a diagonal and
to the triangles determined by those vertexes to the
other diagonal - coincide.
Complements to Classic Topics of Circles Geometry
149
Radical axis of the Apolloniusβs 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 Apolloniusβs Circle relative to vertexes
B and C of triangles π΄π΅πΆ and π΅πΆπ is:
ππ΅2 = ππΆ2.
So O belongs to the radical axis of the circles.
We also have πΎπ΄ β πΎπΆ = πΎπ΅ β πΎπ· , relatives
indicating that the point K has equal powers towards
the highlighted Apolloniusβs 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.
Ion Patrascu, Florentin Smarandache
150
Complements to Classic Topics of Circles Geometry
151
Triangulation of a Triangle
with Triangles having Equal
Inscribed Circles
In this article, we solve the following
problem:
Any triangle can be divided by a
cevian in two triangles that have
congruent inscribed circles.
Solution.
We consider a given triangle π΄π΅πΆ and we show
that there is a point π· on the side (π΅πΆ) so that the
inscribed circles in the triangles π΄π΅π· , π΄πΆπ· are
congruent. If π΄π΅πΆ is an isosceles triangle (π΄π΅ = π΄πΆ),
where π· is the middle of the base (π΅πΆ), we assume
that π΄π΅πΆ is a non-isosceles triangle. We note πΌ1, πΌ2 the
centers of the inscribed congruent circles; obviously,
πΌ1πΌ2 is parallel to the π΅πΆ. (1)
We observe that:
π(β’πΌ1π΄πΌ2) =1
2π(οΏ½οΏ½). (2)
Ion Patrascu, Florentin Smarandache
152
Figure 1.
If π1, π2 are contacts with the circles of the cevian
π΄π· , we have β πΌ1π1π β‘ β πΌ2π2π ; let π be the
intersection of πΌ1πΌ2 with π΄π·, see Figure 1.
From this congruence, it is obvious that:
(πΌ1π) β‘ (πΌ2π). (3)
Let πΌ be the center of the circle inscribed in the
triangle π΄π΅πΆ; we prove that: π΄πΌ is a simedian in the
triangle πΌ1π΄πΌ2. (4)
Indeed, noting πΌ = π(π΅π΄πΌ1) , it follows that
π(β’πΌ1π΄π) = πΌ. From β’πΌ1π΄πΌ2 = β’π΅π΄πΌ, it follows that
β’π΅π΄πΌ1 β‘ β’πΌπ΄πΌ2 , therefore β’πΌ1π΄π β‘ β’πΌπ΄πΌ2 , indicating
that π΄π and π΄πΌ are isogonal cevians in the triangle
πΌ1π΄πΌ2.
Since in this triangle π΄π is a median, it follows
that π΄πΌ is a bimedian.
Complements to Classic Topics of Circles Geometry
153
Now, we show how we build the point π·, using
the conditions (1) β (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 π΄π΅πΆ; we build the bisector of the
angle π΅π΄πΆ and denote by P its intersection
with the circumscribed circle (see Figure
2).
20: We build the perpendicular on πΆ to πΆπ and
(π΅πΆ) side mediator; we denote π1 the
intersection of these lines.
30: We build the circle β(π1; π1πΆ) and denote π΄β
the intersection of this circle with the
bisector π΄πΌ (π΄β is on the same side of the
line π΅πΆ as π΄).
40: We build through π΄ the parallel to π΄βπ1 and
we denote it πΌπ1.
50: We build the circle β(π1β² ; π1
β²π΄) and we denote
πΌ1, πΌ2 its intersections with π΅πΌ , and πΆπΌ
respectively.
60: We build the middle π of the segment (πΌ1πΌ2)
and denote by π· the intersection of the
lines π΄π and π΅πΆ.
Ion Patrascu, Florentin Smarandache
154
Figure 2.
Proof.
The point π is the middle of the arc π΅οΏ½οΏ½ , then
π(ππΆοΏ½οΏ½) = 1
2π(οΏ½οΏ½).
The circle β(π1; π1πΆ) contains the arc from
which points the segment (BC) βis shownβ under angle
measurement 1
2π(οΏ½οΏ½).
The circle β(π1β² ; π1
β²π΄) is homothetical to the
circle β(π1; π1πΆ) by the homothety of center πΌ and by
the report πΌπ΄β²
πΌπ΄; therefore, it follows that πΌ1πΌ2 will be
parallel to the π΅πΆ , and from the points of circle
β(π1β² ; π1
β²π΄) of the same side of π΅πΆ as π΄, the segment
Complements to Classic Topics of Circles Geometry
155
(πΌ1πΌ2) βis shownβ at an angle of measure 1
2π(οΏ½οΏ½). Since
the tangents taken in π΅ and πΆ to the circle β(π1; π1πΆ)
intersect in π, on the bisector π΄πΌ, as from a property
of simedians, we get that π΄β²πΌ is a simedian in the
triangle π΄β²π΅πΆ.
Due to the homothetical properties, it follows
also that the tangents in the points πΌ1, πΌ2 to the circle
β(π1β² ; π1
β²π΄) intersect in a point πβ located on π΄πΌ, i.e. π΄πβ
contains the simedian (π΄π) of the triangle πΌ1π΄πΌ2, noted
{π} = π΄πβ² β© πΌ1πΌ2.
In the triangle πΌ1π΄πΌ2, π΄π is a median, and π΄π is
simedian, therefore β’πΌ1π΄π β‘ πΌ2π΄π; on the other hand,
β’π΅π΄π β‘ β’πΌ1π΄πΌ2; it follows that β’π΅π΄πΌ1 β‘ πΌ2π΄π, and more:
β’πΌ1π΄π β‘ π΅π΄πΌ1, which shows that π΄πΌ1 is a bisector in the
triangle π΅π΄π·; plus, πΌ1, being located on the bisector of
the angle π΅, it follows that this point is the center of
the circle inscribed in the triangle π΅π΄π·.
Analogous considerations lead to the conclusion
that πΌ2 is the center of the circle inscribed in the
triangle π΄πΆπ·. Because πΌ1πΌ2 is parallel to π΅πΆ, it follows
that the rays of the circles inscribed in the triangles
π΄π΅π· and π΄πΆπ· are equal.
Discussion.
The circles β(π1; π1π΄β²), β(π1
β² ; π1β²π΄) are unique;
also, the triangle πΌ1π΄πΌ2 is unique; therefore,
determined as before, the point π· is unique.
Ion Patrascu, Florentin Smarandache
156
Remark.
At the beginning of the Proof, we assumed that
π΄π΅πΆ 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 π΄π΅πΆ triangle.
Open problem.
Given a scalene triangle π΄π΅πΆ , could it be
triangulated by the cevians π΄π· , π΄πΈ , with π· , πΈ
belonging to (π΅πΆ), so that the inscribed circles in the
triangles π΄π΅π·, π·π΄πΈ and the πΈπ΄πΆ to be congruent?
Complements to Classic Topics of Circles Geometry
157
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 (ππ΄), (ππ΅), (ππΆ) be the tangents in the
peaks π΄, π΅, πΆ of the triangle π΄π΅πΆ to the circle
circumscribed to the triangle. Prove that the
perpendiculars drawn from the middles of the
opposite sides on (ππ΄), (ππ΅), (ππΆ) 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 π΄π΅πΆ are orthological. The
orthological centers are π β the center of the circle
circumscribed to the triangle π΄π΅πΆ, and π9 β the center
of the circle of π΄π΅πΆ triangleβs 9 points.
Ion Patrascu, Florentin Smarandache
158
Figure 1.
Proof.
Let ππππππ be the median triangle of triangle
π΄π΅πΆ and ππππππ the tangential triangle of the triangle
π΄π΅πΆ . It is obvious that the triangle ππππππ and the
triangle π΄π΅πΆ are orthological and that π is the
orthological center.
Verily, the perpendiculars taken from ππ, ππ , ππ
on π΅πΆ ; πΆπ΄; π΄π΅ respectively are internal bisectors in
the triangle ππππππ and consequently passing through
π, which is center of the circle inscribed in triangle
ππππππ . Moreover, πππ is the mediator of (π΅πΆ) and
accordingly passing through ππ , and ππππ is
perpendicular on π΅πΆ , being a mediator, but also on
ππππ which is a median line.
Complements to Classic Topics of Circles Geometry
159
From orthological triangles theorem, it follows
that the perpendiculars taken from ππ , ππ , ππ on ππππ,
ππππ , ππππ respectively, are concurrent. The point of
concurrency is the second orthological center of
triangles ππππππ and ππππππ. We prove that this point
is π9 β the center of Euler circle of triangle π΄π΅πΆ. We
take πππ1 β₯ ππππ and denote by {π»1} = πππ1 β© π΄π», π»
being the orthocenter of the triangle π΄π΅πΆ. We know
that π΄π» = 2πππ; we prove this relation vectorially.
From Sylvesterβs relation, we have that: ππ» =
ππ΄ + ππ΅ + ππΆ , but ππ΅ + ππΆ = 2πππ ; it follows that
ππ» β ππ΄ = 2πππ , so π΄π» = 2πππ
; changing to module,
we have π΄π» = 2πππ . Uniting π to π΄ , we have ππ΄ β₯
ππππ , and because πππ1 β₯ ππππ and π΄π» β₯ πππ , it
follows that the quadrilateral ππππ»1π΄ is a
parallelogram.
From π΄π»1 = πππ and π΄π» = 2πππ we get that π»1
is the middle of (π΄π»), so π»1 is situated on the circle of
the 9 points of triangle π΄π΅πΆ. On this circle, we find as
well the points π΄β² - the height foot from π΄ and ππ ;
since β’π΄π΄β²ππ = 900 , it follows that πππ»1 is the
diameter of Euler circle, therefore the middle of
(πππ»1), which we denote by π9, is the center of Eulerβs
circle; we observe that the quadrilateral π»1π»πππ is as
well a parallelogram; it follows that π9 is the middle
of segment [ππ»].
In conclusion, the perpendicular from ππ on ππππ
pass through π9.
Ion Patrascu, Florentin Smarandache
160
Analogously, we show that the perpendicular
taken from ππ on ππππ pass through π9 and
consequently π9 is the enunciated orthological center.
Remark.
The triangles ππππππ and ππππππ are
homological as well, since ππππ , ππππ , ππππ are
concurrent in O, as we already observed, therefore the
triangles ππππππ and ππππππ are orthohomological of
rank I (see [2]).
From P. Sondat theorem (see [4]), it follows that
the Euler line ππ9 is perpendicular on the homological
axis of the median triangle and of the tangential
triangle corresponding to the given triangle π΄π΅πΆ.
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
Complements to Classic Topics of Circles Geometry
161
(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.
Ion Patrascu, Florentin Smarandache
162
References.
[1] Titu Andreescu and Dorin Andrica: 360 de probleme de matematicΔ 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 βRevista de matematicΔ ALPHAβ, 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.
Complements to Classic Topics of Circles Geometry
163
The Dual of a Theorem
Relative to the Orthocenter
of a Triangle
In [1] we introduced the notion of
Bobillierβs transversal relative to a point
πΆ in the plane of a triangle π¨π©πͺ; we use
this notion in what follows.
We transform by duality with respect to a
circle β(π, π) the following theorem
relative to the orthocenter of a triangle.
1st Theorem.
If π΄π΅πΆ is a nonisosceles triangle, π» its
orthocenter, and AA1, BB1, CC1 are cevians of a triangle
concurrent at point π different from π», and π,π, π are
the intersections of the perpendiculars taken from π»
on given cevians respectively, with π΅πΆ, πΆπ΄, π΄π΅, then
the points π,π, π are collinear.
Ion Patrascu, Florentin Smarandache
164
Proof.
We note with πΌ = π(β’BAA1); π½ = π(β’CBB1), πΎ =
π(β’ACC1), see Figure 1.
According to Cevaβs theorem, trigonometric form,
we have the relation: sinπΌ
sin(π΄βπΌ)β
sinπ½
sin(π΅βπ½)β
sinπΎ
sin(πΆβπΎ)= 1. (1)
We notice that: ππ΅
ππΆ=
Area(ππ»π΅)
Area(ππ»πΆ)=
ππ»βπ»π΅βsin sin(ππ»οΏ½οΏ½)
ππ»βπ»πΆβsin(ππ»οΏ½οΏ½) .
Because:
β’ππ»π΅ β‘ β’π΄1π΄πΆ,
as angles of perpendicular sides, it follows that
π(β’ππ»π΅) = π(οΏ½οΏ½) β πΌ.
Therewith:
π(β’ππ»πΆ) = π(ππ»οΏ½οΏ½) + π(π΅π»οΏ½οΏ½) = 1800πΌ.
We thus get that:
ππ΅
ππΆ=
sin(π΄ β πΌ)
sinπΌβπ»π΅
π»πΆ.
Analogously, we find that: ππΆ
ππ΄=
sin(π΅βπ½)
sinπ½βπ»πΆ
π»π΄ ;
ππ΄
ππ΅=
sin(πΆβπΎ)
sinπΎβπ»π΄
π»π΅ .
Applying the reciprocal of Menelaus' theorem,
we find, in view of (1), that: ππ΅
ππΆβπ»πΆ
π»π΄βππ΄
ππ΅= 1.
This shows that π,π, π are collinear.
Complements to Classic Topics of Circles Geometry
165
Figure 1.
Note.
1st Theorem is true even if π΄π΅πΆ is an obtuse,
nonisosceles triangle.
The proof is adapted analogously.
2nd Theorem. (The Dual of the Theorem 1)
If π΄π΅πΆ is a triangle, π a certain point in his plan,
and π΄1, π΅1, πΆ1 Bobillierβs transversals relative to π of
Ion Patrascu, Florentin Smarandache
166
π΄π΅πΆ triangle, as well as π΄2 β π΅2 β πΆ2 a certain
transversal in π΄π΅πΆ, and the perpendiculars in π, and
on ππ΄2, ππ΅2, ππΆ2 respectively, intersect the Bobillierβs
transversals in the points π΄3, π΅3, πΆ3, then the ceviens
π΄π΄3, π΅π΅3, πΆπΆ3 are concurrent.
Proof.
We convert by duality with respect to a circle
β(π, π) the figure indicated by the statement of this
theorem, i.e. Figure 2. Let π, π, π be the polars of the
points π΄, π΅, πΆ with respect to the circle β(π, π). To the
lines π΅πΆ, πΆπ΄, π΄π΅ will correspond their poles {π΄β²4 =
πππ; {π΅β²4 = πππ; {πΆβ²4 = πππ.
To the points π΄1, π΅1, πΆ1 will respectively
correspond their polars π1, π1, π1 concurrent in
transversalβs pole π΄1 β π΅1 β πΆ1.
Since ππ΄1 β₯ ππ΄, it means that the polars π and π1
are perpendicular, so π1 β₯ π΅β²πΆβ², but π1 pass through π΄β²,
which means that πβ² contains the height from π΄β² of
π΄β²π΅β²πΆβ² triangle and similarly π1 contains the height
from π΅β² and π1 contains the height from πΆβ² of π΄β²π΅β²πΆβ²
triangle.
Consequently, the pole of π΄1 β π΅1 β πΆ1
transversal is the orthocenter π»β² of π΄β²π΅β²πΆβ² triangle. In
the same way, to the points π΄2, π΅2, πΆ2 will correspond
the polars to π2, π2, π2 which pass respectively through
π΄β², π΅β², πΆβ² and are concurrent in a point πβ², the pole of
the line π΄2 β π΅2 β πΆ2 with respect to the circle β(π, π).
Complements to Classic Topics of Circles Geometry
167
Figure 2.
Given ππ΄2 β₯ ππ΄3, it means that the polar π2 and
π3 are perpendicular, π2 correspond to the cevian
π΄β²πβ², also π3 passes through the the pole of the
transversal π΄1 β π΅1 β πΆ1, so through π»β², in other words
π3 is perpendicular taken from π»β² on π΄β²πβ²; similarly,
π2 β₯ π3, π2 β₯ π3, so π3 is perpendicular taken from π»β²
on πΆβ²πβ². To the cevian π΄π΄3 will correspond by duality
considered to its pole, which is the intersection of the
polars of π΄ and π΄3, i.e. the intersection of lines π and
π3 , namely the intersection of π΅β²πΆβ² with the
perpendicular taken from π»β² on π΄β²πβ²; we denote this
Ion Patrascu, Florentin Smarandache
168
point by πβ². Analogously, we get the points πβ² and πβ².
Thereby, we got the configuration from 1st Theorem
and Figure 1, written for triangle π΄β²π΅β²πΆβ² of orthocenter
π»β². Since from 1st Theorem we know that πβ², πβ², πβ² are
collinear, we get the the cevians π΄π΄3, π΅π΅3, πΆπΆ3 are
concurrent in the pole of transversal πβ² β πβ² β πβ² with
respect to the circle β(π, π), and 2nd Theorem is proved.
References.
[1] Ion Patrascu, Florentin Smarandache: The Dual Theorem concerning Aubertβs Line, below.
[2] Florentin Smarandache, Ion Patrascu: The
Geometry of Homological Triangles. The Education Publisher Inc., Columbus, Ohio, USA
β 2012.
Complements to Classic Topics of Circles Geometry
169
The Dual Theorem Concerning
Aubertβs Line
In this article we introduce the concept of
Bobillierβs transversal of a triangle with
respect to a point in its plan ; we prove the
Aubertβs 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 π΄π΅πΆ be a triangle and π a point in the plane
of the triangle so that the perpendiculars taken in π,
and ππ΄,ππ΅,ππΆ respectively, intersect the sides
π΅πΆ, πΆπ΄ and π΄π΅ at π΄π, π΅π Εi πΆπ. Then the points π΄π,
π΅π and πΆπ are collinear.
Proof.
We note that π΄ππ΅
π΄ππΆ=
aria (π΅ππ΄π)
aria (πΆππ΄π) (see Figure 1).
Ion Patrascu, Florentin Smarandache
170
Figure 1.
Area (π΅ππ΄π) =1
2β π΅π β ππ΄π β sin(π΅ππ΄οΏ½οΏ½).
Area (πΆππ΄π) =1
2β πΆπ β ππ΄π β sin(πΆππ΄οΏ½οΏ½).
Since
π(πΆππ΄οΏ½οΏ½) =3π
2β π(π΄ποΏ½οΏ½),
it explains that
sin(πΆππ΄οΏ½οΏ½) = β cos(π΄ποΏ½οΏ½);
sin(π΅ππ΄οΏ½οΏ½) = sin (π΄ποΏ½οΏ½ βπ
2) = β cos(π΄ποΏ½οΏ½).
Therefore:
π΄ππ΅
π΄ππΆ=
ππ΅ β cos(π΄ποΏ½οΏ½)
ππΆ β cos(π΄ποΏ½οΏ½) (1).
In the same way, we find that:
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171
π΅ππΆ
π΅ππ΄=
ππΆ
ππ΄βcos(π΅ποΏ½οΏ½)
cos(π΄ποΏ½οΏ½) (2);
πΆππ΄
πΆππ΅=
ππ΄
ππ΅βcos(π΄ποΏ½οΏ½)
cos(π΅ποΏ½οΏ½) (3).
The relations (1), (2), (3), and the reciprocal
Theorem of Menelaus lead to the collinearity of points
π΄π,π΅π, πΆπ.
Note.
Bobillier's Theorem can be obtained β by
converting the duality with respect to a circle β from
the theorem relative to the concurrency of the heights
of a triangle.
1st Definition.
It is called Bobillierβs transversal of a triangle
π΄π΅πΆ with respect to the point π the line containing
the intersections of the perpendiculars taken in π on
π΄π, π΅π, and πΆπ respectively, with sides π΅πΆ, CA and
π΄π΅.
Note.
The Bobillierβs transversal is not defined for any
point π in the plane of the triangle π΄π΅πΆ, for example,
where π is one of the vertices or the orthocenter π» of
the triangle.
Ion Patrascu, Florentin Smarandache
172
2nd Definition.
If π΄π΅πΆπ· is a convex quadrilateral and πΈ, πΉ are the
intersections of the lines π΄π΅ and πΆπ· , π΅πΆ and π΄π·
respectively, we say that the figure π΄π΅πΆπ·πΈπΉ is a
complete quadrilateral. The complete quadrilateral
sides are π΄π΅, π΅πΆ, πΆπ·, π·π΄ , and π΄πΆ, π΅π· and πΈπΉ are
diagonals.
2nd Theorem. (Newton-Gauss)
The diagonalsβ 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 diagonalsβ middles of a complete
quadrilateral belong.
3rd Theorem. (Aubert)
If π΄π΅πΆπ·πΈπΉ is a complete quadrilateral, then the
orthocenters π»1, π»2, π»3, π»4 of the traingles π΄π΅πΉ, π΄πΈπ·,
π΅πΆπΈ, and πΆπ·πΉ respectively, are collinear points.
Complements to Classic Topics of Circles Geometry
173
Proof.
Let π΄1, π΅1, πΉ1 be the feet of the heights of the
triangle π΄π΅πΉ and π»1 its orthocenter (see Fig. 2).
Considering the power of the point π»1 relative to the
circle circumscribed to the triangle ABF, and given the
triangle orthocenterβs property according to which its
symmetrics to the triangle sides belong to the
circumscribed circle, we find that:
π»1π΄ β π»1π΄1 = π»1π΅ β π»1π΅1 = π»1πΉ β π»1πΉ1.
Figure 2.
This relationship shows that the orthocenter π»1
has equal power with respect to the circles of
diameters [π΄πΆ], [π΅π·], [πΈπΉ]. As well, we establish that
the orthocenters π»2, π»3, π»4 have equal powers to these
circles. Since the circles of diameters [π΄πΆ], [π΅π·], [πΈπΉ]
have collinear centers (belonging to the Newton-
Ion Patrascu, Florentin Smarandache
174
Gauss line of the π΄π΅πΆπ·πΈπΉ quadrilateral), it follows
that the points π»1, π»2, π»3, π»4 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 quadrilateralβs orthocenters the line to
which the orthocenters π»1, π»2, π»3, π»4 belong.
2. The Aubert Line is perpendicular on the
Newton-Gauss line of the quadrilateral (radical axis of
two circles is perpendicular to their centersβ line).
4th Theorem. (The Dual Theorem of the 3rd Theorem)
If π΄π΅πΆπ· is a convex quadrilateral and π is a
point in its plane for which there are the Bobillierβs
transversals of triangles π΄π΅πΆ , π΅πΆπ· , πΆπ·π΄ and π·π΄π΅ ;
thereupon these transversals are concurrent.
Proof.
Let us transform the configuration in Fig. 2, by
duality with respect to a circle of center π.
By the considered duality, the lines π, π, π, π, π Εi
π correspond to the points π΄, π΅, πΆ, π·, πΈ, πΉ (their polars).
It is known that polars of collinear points are
concurrent lines, therefore we have: π β© π β© π = {π΄β²},
Complements to Classic Topics of Circles Geometry
175
π β© π β© π = {π΅β²} , π β© π β© π = {πΆβ²} , π β© π β© π = {π·β²} , π β©
π = {πΈβ²}, π β© π = {πΉβ²}.
Consequently, by applicable duality, the points
π΄β², π΅β², πΆβ², π·β², πΈβ² and πΉβ² correspond to the straight lines
π΄π΅, π΅πΆ, πΆπ·, π·π΄, π΄πΆ, π΅π·.
To the orthocenter π»1 of the triangle π΄πΈπ· , it
corresponds, by duality, its polar, which we denote
π΄1β² β π΅1
β² β πΆ1β², and which is the Bobillierβs transversal of
the triangle π΄βπΆβπ·β in relation to the point π. Indeed,
the point πΆβ corresponds to the line πΈπ· by duality; to
the height from π΄ of the triangle π΄πΈπ·, also by duality,
it corresponds its pole, which is the point πΆ1β² located
on π΄βπ·β such that π(πΆβ²ππΆ1β² ) = 900.
To the height from πΈ of the triangle π΄πΈπ· , it
corresponds the point π΅1β² β π΄β²πΆβ² such that π(π·β²ππ΅1
β² ) =
900.
Also, to the height from π·, it corresponds π΄1β² β
πΆβ²π·β on C such that π(π΄β²ππ΄1β² ) = 900. To the
orthocenter π»2 of the triangle π΄π΅πΉ, it will correspond,
by applicable duality, the Bobillierβs transversal π΄2β² β
π΅2β² β πΆ2
β² in the triangle π΄β²π΅β²π·β² relative to the point π.
To the orthocenter π»3 of the triangle π΅πΆπΈ , it will
correspond the Bobillierβs transversal π΄3β² β π΅β²3 β πΆ3β²
in the triangle π΄β²π΅β²πΆβ² relative to the point π, and to
the orthocenter π»4 of the triangle πΆπ·πΉ , it will
correspond the transversal π΄4β² β π΅4
β² β πΆ4β² in the triangle
πΆβ²π·β²π΅β² relative to the point π.
Ion Patrascu, Florentin Smarandache
176
The Bobillierβs transversals π΄πβ² β π΅π
β² β πΆπβ² , π = 1,4
correspond to the collinear points π»π, π = 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 π΄βπ΅βπΆβπ·β,
the Bobillierβs transversals of the triangles π΄βπΆβπ·β ,
π΄βπ΅βπ·β , π΄βπ΅βπΆβ and πΆβπ·βπ΅β relative to the point π are
concurrent.
References.
[1] Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles. The
Education Publisher Inc., Columbus, Ohio, USA β 2012.
[2] Ion Patrascu, Florentin Smarandache: Variance on Topics of Plane Geometry. The Education Publisher Inc., Columbus, Ohio, USA β 2013.
Complements to Classic Topics of Circles Geometry
177
Bibliography
[1] C. Barbu: Teoreme fundamentale din geometria
triunghiului [Fundamental Theorems of
Triangle Geometry]. Bacau, Romania: Editura
Unique, 2008.
[2] N. Blaha: Asupra unor cercuri care au ca centre
douΔ puncte inverse [On some circles that have
as centers two inverse points], in βGazeta
Matematicaβ, vol. XXXIII, 1927.
[3] D. BrΓ’nzei, M. MiculiΘa. Lucas circles and
spheres. In βThe Didactics of Mathematicsβ,
volume 9/1994, Cluj-Napoca (Romania), p. 73-
80.
[4] C. CoΘniΘΔ: CoordonΓ©es baricentrique
[Barycentric Coordinates]. Bucarest β Paris:
Librairie Vuibert, 1941.
[5] C. CoΕniΕ£Δ: Teoreme Εi probleme alese de
matematicΔ [Theorems and Problems],
BucureΕti: Editura de Stat DidacticΔ Εi
PedagogicΔ, 1958.
[6] D. Efremov: Noua geometrie a triunghiului
[The New Geometry of the Triangle],
translation from Russian into Romanian by
Mihai MiculiΘa. Zalau, Romania: Cril Publishing
House, 2010.
Ion Patrascu, Florentin Smarandache
178
[7] Roger A. Johnson: Advanced Euclidean
Geometry. New York: Dover Publications, 2007.
[8] T. Lalescu: Geometria triunghiului [The
Geometry of the Triangle]. Craiova: Editura
Apollo, 1993.
[9] N. N. Mihaileanu: LecΘii complementare de
geometrie [Complementary Lessons of
Geometry], Editura DidacticΔ Θi PedagogicΔ,
BucureΘti, 1976.
[10] Gh. Mihalescu: Geometria elementelor
remarcabile [The Geometry of the Outstanding
Elements]. Bucharest: Tehnica Publishing
House, 1957.
[11] Ion Patrascu: Probleme de geometrie planΔ
[Planar Geometry Problems]. Craiova: Editura
Cardinal, 1996.
[12] I. Patrascu: Axe Θi centre radicale ale cercului
adjuncte unui triunghi [Axis and radical
centers of the adjoin circle of a triangle], in
βRecreaΘii matematiceβ, year XII, no. 1, 2010.
[13] Ion Patrascu, Gh. Margineanu: Cercul lui
Tucker [Tuckerβs Circle], in βAlpha journalβ,
year XV, no. 2/2009.
[14] I. Patrascu, F. Smarandache, Variance on
Topics of Plane Geometry, Educational
Publisher, Ohio, USA, 2013.
Complements to Classic Topics of Circles Geometry
179
[15] F. Smarandache, I. Patrascu, The Geometry of
Homological Triangles. The Education
Publisher, Ohio, USA, 2012.
[16] V. Gh. Voda: Triunghiul β ringul cu trei colΘuri
[The Triangle-The Ring with Three Corners].
Bucharest: Editura Albatros, 1979.
[17] Eric W. Weisstein: First Droz-Farnyβs circle.
From Math World β A Wolfram WEB Resurse,
http://mathworld.wolfram.com/.
[18] P. Yiu, A. P. Hatzipolakis. The Lucas Circles of
a Triangle. In βAmer. Math. Monthlyβ, no.
108/2001, pp. 444-446. http://www.math.fau.
edu/yiu/monthly437-448.pdf.
Ion Patrascu, Florentin Smarandache
180
We approach several themes of clas-
sical 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
authorsβ aspiration and attraction, as a
poet writes lyrics about spring according to
his emotions.