Harmonic Geometry: Harmonic Involutions, Harmonic Multilaterals and Harmonic Lines

Table of contents :
Copyright
Forword
Terminology
Bibliography
Introduction
Part A. Harmonic Involutions (Collinear and Cocyclic)
1. In General.
A' Harmonic Involutions in Collinear Series of Points
2. in General
B' Symmetrical Pencils and Collinear Harmonic Involutions
3. Analyzing Symmetrical Central Pencils and Collinear Harmonic Involutions
C'. Method of Symmetrical Pencils
4. Method of the Egress from a Line to Its Plane.
5. Method of Symmetrical Central Pencils.
6. Applications of Properties and Methods.
D' Orthooptical Pencils and Orthooptical Points
7. in General
E' Harmonic Involutions of Cocyclic Series of Points
8. In General
9. Applications
Part B. Harmonic Multilaterals and Harmonic Lines
10. In General
A' Harmonic Collinear Multilaterals
11. In General
B' Regular Central Pencils and Collinear Harmonic Groups of n-Points
12. Review of the Closed Regular Central Pencils and Collinear HarmonicGroups of n Series of Points.
C' Method of Regular Central Pencils.
13. In General.
D' Orthooptical Regular Pencils and Orthooptical Points of Closed Harmonic Groups of n.
14. In General.
15. Applications of the Method of Regular Pencils
E' Harmonic Cocyclic Multilaterals
16. In General.
F. Harmonic Closed Central Pencils and their Intersection from a Line and a Circle
17. Analysis of the Closed Harmonic Central Pencils and their Intersectionsrom a Line and a Circle.
G'. Method of the Harmonic Central Pencils.
18. In General.
19. Applications of the Method of Harmonic Pencils and not only that.
H' The Problem of the Vicennium!!!
20. The Problem of the Vicennium
21. The Amazing Method of the Harmonic Transformation
22. Applications of the Method of Harmonic Transformation
I' Pair of Harmonic Collinear and Cocyclic Lines.
23. In General.
24. Pairs of Open Central Flat Harmonic and Regular Pencils of 2n Beams and their Orthooptical Points.
25. Applications Relevant to the Pairs of Collinear and Cocyclic Harmonic Lines and Pencils.
Part C. Supplement of Part A
26. In General.
27. Introducing the Method of Harmonic Transformation in Harmonic lnvolutions
Part D. Supplement of Part B.
28. Generally.
29. Applications of Harmonic Multilaterals, mainly using the Method of Harmonic Transformation.
30. General Conclusions of Harmonic Geometry
Part E. Auxiliary Propositions (Lemmas)
1. In General.
2. Geometrical Auxiliary Propositions (AP).
Epilogue
Figure Index
Contents
It Was Said
Important remark from page 12:
Author

Citation preview

cp=(1/ 5)L=18 1,1oiptc;.

Definitions Methods Propositions Problems Loci

NIKOS D. KYRIAZIS (Researcher Geometrician)

HARMONIC GEOMETRY HARMONIC INVOLUTIONS, HARMONIC MULTILATERALS AND HARMONIC LINES DEFINITIONS-METHODS-PROPOSITIONS-PROBLEMS-LOCI (Research-Investigation) (Our Previous Papers have been honoured from the

ACADEMY OF ATHENS).

Figure of the Construction of a Harmonic Decagon with the new Method of Harmonic Transformation (Construction 155).

Title: Harmonic Geometry

Author: Nikos D. Kyriazis Cover designer: Vassilis loannou Production supervisor: Platon Malliagkas www.mediterrabooks.com

This work was translated from Greek into English by Ylenia Orfanidou (Mathematician) and Harry Nikolaides (Teacher of English) August 2nd, 2017

Copyright 2014 Nikos D. Kyriazis, Trapezountos 9 Kalamaria (Thessaloniki) PC 551 .31 , tel. +30-2310-413.539.

This work as it is published, in whole or in part, or in paraphrasing or adaptation, may not be reproduced, or republished, by any means or by means of (photocopy, printing, microfilm , or other mechanical or electronic method) without the written permission of the Author-Researcher. (N.2121 / 93 No.51 ).

Cover : Implementation of our new " Harmon ic Transformation" Method for the Construction of the Harmonic Decagon from a Normal Decagon (Construction 155)

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This work is dedicated to my parents Aristea and Dimitris, who brought me up and educated me in very hard times.

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The present book is pioneering, since, as far as we know from the bibliography and the internet, we believe that it contains only new elements of Geometry that appear here for the very first time in written presentation, including its title, which is of our own conception and to which its content is perfectly determined. This view is reinforced by the fact that for four years we have been publishing them on the site mathematica. gr, where many thousands of visitors have read it and where we had mentioned that it only contains new information, and no one had any objection. But if any of these are in the bibliography, it is certain that they are differently approached. But it is never too late. We ask even at this very moment, whoever has met something, similar to inform us. It will be highly appreciated. Thus, we believe that a new branch of the Euclidean Geometry has emerged, or if you want a new Geometry, the Harmonic Geometry.

We believe that every following paragraph or following Proposition or following Problem of this book will surprise, move and puzzle the reader friend of Geome!.rv.,__this predominantly Greek Science, which is perfectly combining the grace, the harmony and the logical thinking. Generally, anyone who has got the time, the will to practice his mind, the courage and the strength to study it in depth, will experience pleasant moments and emotions that will be offered generously by the magic of Geometry.

It is aimed at those who really love and are indeed interested in the progress of Geometry; they are thirsty to experience, we believe, new amazing Geometry Elements and to escape the trivial.

The writing of this book started on Tuesday 29th November 2011 and emerged

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after a study and research that was triggered by an exercise that it was given on the internet, specifically on the mathematica.gr website. Since then a great number of those Geometry elements that emerged from the above study-research, we have published them at the same time on the above site, expressing our joy and emotion, for every new important element of Geometry that emerged . We do not agree, with the opinion of friends, who say that we are writing things difficult and very advanced and they cannot fully understand, or that we introduce in Geometry "Subversive Theories".

Today we are delighted to include all of the above mentioned new elements and far more in this book entitled "Harmonic Geometry", as this refers to this field of the Euclidean Geometry, apart from Part E, in which we give a few new auxiliary Geometry Propositions, needed for this work.

For the completeness of this book as "Harmonic Geometry", it would be correct to have included all those harmonic elements of Geometry known to us and exist in the bibliography, for example the well known Harmonic Group of Four Points, the well known Harmonic Quadrilateral with thei r known properties, the well known Apollonian circle, the Newton's Theorem of the Harmonic Group of Four Points with his known evidence, etc. However, we considered that useless, since all these exist in the bibliography and especially in our books [1] and [31], while our own pursuit is for this book to contain only new harmonic elements and because then it would have been unduly huge and would have escaped from its purpose.

With this book, we give our friends a second opportunity to study the above articles in comfort and carefully when they have plenty of time, especially after their retirement, so without the pressure of time to enjoy, all the Greatness of the Harmony of Geometry, much more now, that in this book, the above elements, we believe that they have been logged in a logical mathematical order, fullness and we believe that they are easy to understand and so they will be

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convinced that we do not introduce "Subversive Theories".

The part A refers to the "Geometric Harmonic Involutions", which are collinear and co cyclic, simple and double, which were originated intuitively and then with the application of our new Method of "Symmetrical Central Pencils " 1 which was an important "tool " for proving various relevant Propositions and Problems. The Part B refers to the "Harmonic Multilaterals and Harmonic Lines", namely: Closed and Open Harmonic Series of Points, Collinear and Co cyclic (Harmonic Collinear and Co cyclic Groups of n), in our new method of "Regular Central Pencils", as well as in the new Important Method of the "Harmonic Transformation", which proved to be a powerful "key" for the proof-solution of relevant Propositions-Problems, which until today had remained unsolved. The Parts C and D refer to complementary Geometric data that emerged after the writing of the Parts A and B respectively, as we did not want to alter the order of inventing the reported elements of the Parts A and B for Historical reasons. For the above mentioned reason, obviously I am not happy about the structure of this book. In a future edition of this book, this can be done by us or by someone else who would like to deal with this subject, who even could give to it the perfect logical structure and appearance, especially if he is a talented writer, as

I am not. Furthermore, in the Parts C and D certain elements of parts A and B are dealt with more globally, applying also the New Method of "Harmonic Transformation". In Part E, as we have already mentioned above, we give, I believe, some newly emerging auxil iary Propositions (Lemmas) of Geometry that are needed for this work (which are not harmonic).

Proofs-solutions to Propositions-Problems are usually given in every detail, so that there is no doubt or question of their truth, although all these are verified

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by the accompanying shapes, which have been made meticulously and with great precision. Many times, two or more proofs-solutions are given. In several of these new Propositions, are based on the evidence of others and that is why we give them all and their proofs.

The shapes in this book are often common to various Propositions and Problems that are in different parts. Because this obviously does not make it easier to study, to remove this difficulty, we attach an index of shapes (page 760) and a series of all these figures in volume 3 of the book.

All the elements mentioned in this book, we believe are important and impressive, while they are covering the full range of difficulty (from very simple to very difficult).

We have the ambition in the near future to publish this book in English and to make a second volume, since there is plenty of material. We also believe that Harmonic Geometry will become fashionable and will be studied very much and will grow further.

This book also uses new terms that exist in the terminology section at the beginning of the book.

We underline that this work has been devised and written because of love in Geometry rather and not for commercial reasons, after a serious studyresearch, but in addition to prove that the contemporary Greeks are capable and continue to explore, to invent and to offer in Geometry (and not only ), following the path created by our ancestors.

As the elements of this work we believe to be brand new, it is natural to have some omissions or mistakes. We therefore ask for the lenient judgment of the readers. We will gladly welcome the friendly and well-intentioned comments or suggestions of omissions or mistakes in our address at 9, Trapezountos St.,

Ii'-·-

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Kalamaria , PC 551.31 , tel.2310 413539, in order to introduce the relevant improvements in a new edition of the book. We will consider it a great honor if it is acknowledged that with this work, we have added a small stone to Geometry and if we have contributed further to the search of its friends of the enormously neglected indeed Greek Science of the Euclidean Geometry, whose great merit is undeniable and which has glorified our Homeland worldwide.

This is the seventh of the series, pioneering, avant-garde and fina l book, as the years do not allow us for an additional book unless the Almighty Master may give us a chance. For this reason, we tried to give it the best possible great appearance compared to the previous ones, as it might be my "Swan Song".

This book has been posted and published on the site mathematica.gr in parts and as an entire work.

The entire book, we have the pleasure and the satisfaction to offer it to our friends printed and in full color, w ith our personal work, despite the many and great difficulties we faced , due to financial difficulties we were obliged even to learn computing without an instructor and we believe with great success.

Nikos Kyriazis Author - Researcher. Thessaloniki August 2015

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TERMINOLOGY In General. Some known terms, are reminded below. Additionally, here, new terms are introduced as well, that obviously aim to make the formulation of Propositions-Problems-Exercises easier and in this way they will become more simple, clear and comprehensible. 1. Conjugate Harmonic (CH} Points and Harmonic Group of Four Points. These terms are known from bibliography [11] [17]. We note here, that Conjugate Harmonic Points are symbolized with CH, while the Harmonic Group of Four Points, for example of the ordered points A, B, P,

r

is symbolized with

(APBI:)=1 . 2. Cevians Gergonne of a Triangle. With this term we refer to the three cevians [6] (§11-4.1) of a triangle, are respective to the known Gergonne point [3] (§ 513). 3. Quadrilaterals of a Multilateral with an Even Number of Sides. This term is given for each multilateral separately (Hexagon, Octagon, Decagon, and so on and so forth). 4. Main Diagonals or just Diagonals of Multilaterals with an Even Number of Sides. We use this term to name the diagonals of a multilateral with an even number of sides, those that connect its opposite apexes. 5. First, Second, Third, and so on and so forth, Diagonals of a Multilateral. We use these terms to name the diagonals of a Multilateral, that connect the «one by one», «one by two», «one by three», and so on and so forth, respectively, apexes of the Multilateral (It is obvious that the Quadrilateral has only main diagonals, while the Pentagon has only First Diagonals). 6. Twin Multi laterals of a Multilateral, with an Even Number of Sides. We use this term to characterize the two multilaterals (Collinear or Cocyclic), that their apexes are the «one by one» apexes of a multilateral w ith an even number of sides. 7. Respective Multilaterals, Respective Circle. We use this term to name two multilaterals, when one is inscribed and the other one is circumscribed in the same circle, whose sides intersect the circle at the

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same points. These Multilaterals are also called polar converses [3] (§ 556). Moreover, from these Multilaterals the tangent one is called «tangential» [7] (§ 1177a). (The term Respective Multilaterals is also mentioned in bibliography [20) § 251 note) 8. Harmonic Quadrilateral. As mentioned in book [6] (§ 8-4.10), we call Harmonic Quadrilateral, the quadrilateral that can be inscribed in a circle and whose diagonals are conjugate with respect to its circumscribed circle. Its properties are mentioned in bibliography [1], [6], [26), [31). Because for this quadrilateral the two products of its opposite sides, are proved to be equal and conversely [31) § 1 J.3(38), this criterion is many times used to decide if a quadrilateral is harmonic or not. 9. Harmonic Multilateral. This term is introduced for each Multilateral Separately (Harmonic Pentagon, Hexagon, Octagon, and so on and so forth). 10. Convergent Lines. This term is frequently used instead of the term «Concurrent Lines». 11 . Ordered Points. We have to do w ith a series of points, that lie on a line or a circle in a particular order. 12. Series of Points. It is a series of ordered points, that lie on a line or a circle (Series of points, collinear or cocyclic, respectively, closed or open). 13. Diagonal Triangle of a Complete Quadrilateral ABrllEZ. Triangle HEZ, is called «Diagonal Triangle» of the Complete Quadrilateral ABrllEZ, as H is the intersection of the diagonals in convex quadrilateral ABra and E, Z are the intersections of the diagonals of the concave quadrilaterals AllBr and AB/lr respectively. 14. Rhomboid Quadrilateral.

We use this term to characterize ecery symmetric quadrilateral ABra, with respect to its diagonal Ar or Bil {[7], page 1203}.

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BIBLIOGRAPHY BOOKS OF GEOMETRY

WRITERS

[1]. Geometry. Inscribed-Circumscribed Figures 1993.

Nikos Kyriazis. N.H. Varouchakis.

[2]. Supplement of Geometry 1971 . [3] . Geometry.

Christos Barbastathis.

[4]. Supplement of Geometry 1974.

Nikos Pnevmatikos .

(5]. Geometry. Generalization of the Ceva Theorem 1995.

Nikos Kyriazis.

[6]. Geometry. (Contemporary bookstore publishing).

Giorgios Tsintsifas .

(7]. Geometry Exercises .(A.Karavia publishing 1952).

F.G .-M (Jesuites).

[8]. Planar Geometry (I.Chiotelis publishing).

Christos Tavanlis.

[9]. Theoretical Geometry.

Petros Togas.

(10]. Geometry. Orthocentric Quadrilaterals.

Nikos Kyriazis.

(11]. 11 th Grade Geometry 1986.

Anastasios Skiadas.

(12]. Methods of Solving Geometrical Problems 1976.

Arist. Dimitriou.

(13]. Geometry. (Thessaloniki. His private school 's publishing). P. Vasileiadis. (14]. 11 th Grade Geometry 1987.

Dimitr ios Gouvitsas.

[15]. Supplement of Geometry 1994.

Nikos Kyriazis.

(16]. Theoretical Geometry 10th Grade 1993.

OEDB.

[17]. Theoretical Geometry 11th Grade 1993.

OEDB.

(18]. The candidate's Geometry (KNOSSOS publishing).

A.M.Kourkoulos.

(19]. Methodical Geometry 1971.

Em. And Pol. Georgiakakis.

(20]. Theoretical Geometry 1973.

OEDB (N . Nikolaou).

(21]. Theoretical Geometry 1966.

Georg. Papanikolaou.

(22]. The methodical solution of the Geometric Problem 1978. (23]. Theoretical Geometry 10th Grade 1984. (24]. Theoretical Geometry 11 th Grade 1986. [25]. Geometry (Mathematical Olympics 1987).

G. Lemaire. OEDB. OEDB.

D.G. Kontogiannis.

(26]. Theorems and Problems of Geometry. Ch. Tsarouchis and Nik. Kiskyras. (27]. Euclidean Geometry 1999. (28]. Big Geometry 1971.

OEDB. Aristeidis Pallas.

~

i:- -

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(29). Geometry Lessons.

Christos Tavanlis .

(30]. Euclidean Geometry 1970.

Sp. Kanellos.

(31). New Elements in Geometry.

Nikos Kyriazis.

(32). Geometry Exercises.

G. Papelier.

(33). Coloured Mathematical Encyclopedia (Pagoulatos publishing). German.

MAGAZINES

PUBLISHERS

[A]. EuKAEiliric; A'.

Hellenic Mathematical Society.

[BJ. EuKAEiliric; B'.

Hellenic Mathematical Society.

[CJ. Mathematics in Greek High School. [D]. Appolonios .

Eythymios Kostogiannos . Mathematical Society of lmatheia.

[E]. Mathematical Education.

Charis Vafeiadis.

[F]. . [G]. Diastases.

V.Viskadourakis. Mathematical Society of Thessaloniki.

NOTICE: "Due to incomparable difficulties regarding the figures and the algebraic expressions, of the present book, we were not able to replace the Greek letters (capital and lowercase) in it, with the corresponding English ones. This obviously does not disturb the reader of the present book, as at the end of it ( in page 780 ), the Greek alphabet (capital and lowercase), as well as the translation of some Greek words of the figures can be found because their translation was not feasible in the figures of the book".

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HARMONIC GEOMETRY

RESEARCH (See epilogue page 758).

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HARMONIC GEOMETRY (It is a result of Research)

INTRODUCTION

l•fflfflfOt•JltJ Harmonic Geometry is that field of Euclidean Geometry, that examines only Geometrical Elements w ith Harmonic Properties. Specifically Harmonic Geometry refers: •

To Harmonic Involutions.

These are Harmonic Series of Points, Collinear and Cocyclic with defined Harmonic Properties. •

To Harmonic Multilaterals.

These are Closed Harmonic Groups of n Series of Points with other defined Harmonic Properties. •

To Harmonic Lines.

These are Open Harmonic Series of Points Collinear and Cocyclic with special Harmonic Properties. Subject of Harmonic Geometry. Until now Unhramonic Involutions are known to us (Geometry [7], § 1220 and 2109), Harmonic Groups of Four and Harmonic Quadrilaterals. Here, we extend the above and will refer to Harmonic Involutions (Collinear

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and Cocyclic), to Harmonic Groups of n (six, eight, ten, etc.), to Harmonic Multilaterals (pentagons, hexagons, octagons, decagons, etc.) and to Harmonic Lines (Collinear and Cocyclic). Regarding all these we have not yet found data in bibliography or online, so we faced them in our opinion. For all the above we will provide the relevant definitions, their Constructions (in the proper places), Loci and various relevant Propositions, along with their properties. Used Methods (Keys). To help us prove Geometrical Propositions and solve Geometrical Problems, we invented and use some Methods (keys) which we describe in proper places inside the book. The majority of the Propositions and Problems included in Harmonic Geometry, we believe to be difficult, if not impossible, to be proven or solved, if the above methods are not put into practice together with their Orthooptical Points, for each one of the fields above, whose methods we express below. Specifically: For Collinear Harmonic Involutions, we apply and suggest the application of the Method of Symmetrical Central Pencils . For Collinear Harmonic Multilaterals, we apply and suggest the applicat ion of the Method of Closed Regular Central Pencils . For Collinear Harmonic Lines, we apply and suggest the application of the Method of Open Regular Central Pencils. For all Problems and Propositions in Harmon ic Geometry in general, we apply and suggest the application of the Method of Harmonic Transformation, which connects the Euclidean space with the Harmonic Projective Space and which is the Composition of all the methods above, together with their Orthooptical Points. (To the definitions of the above mentioned terms, we will refer later, in the proper places).

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PART A HARMONIC INVOLUTIONS (Collinear and Cocyclic).

Archimedes (287-212 B.C.).

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HARMONIC INVOLUTIONS (Collinear and Cocyclic). 1. In General. Part A includes Harmonic Involutions (Wrappings) (Collinear and Cocyclic). In addition, this part includes our new «Symmetric Bonds » Method, which is helpful for the solution of problems concerning Harmonic Involutions. Certain components from Part A are more methodically analyzed in Part C, applying our new «Harmonic Transformation» Method, and become more comprehensible, while in Part A they are intuitively approached .

Harmonic Involutions are considered valuable for numerous reasons. A rather important one is their usage in proving many properties of the Harmonic Manifolds , which constitute a special case of Harmonic Involutions, as we will later discover.

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' HARMONIC INVOLUTIONS IN COLLINEAR SERIES OF

flGW4,t4Eil Even though we searched through bibliography relevant to Harmonic Involutions in Collinear Series of Points and asked for relevant information at the mathematica.gr website, no such information has been found until today. We were only able to find some clues regarding Unharmonic Involutions, in the Jesuits book [7] (paragraphs 1220 and 2109). So, at first we approached the matter according to our own view and intuition, while later on we developed and used relevant methods. After thorough study and research, we intuitively discovered that Cocyclic Harmonic Involutions of Series of Points also exist, which we will discuss later. In addition we established that Collinear Harmonic Involutions are an extreme case of Cocylic Harmonic Involutions, during which the circle is degenerated to a line. This means that the circle they belong in has a radius of infinite length (and zero curvature obviously).

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2.1. Definitions and Characteristics. Definitions and Characteristics of Collinear Harmonic Involution of a Closed Series of n-pairs of Points. 2.11 . Definition. The pairs of collinear points A - A', B - B', r - r• , ... ., N - N', that belong to theseries of ordered points M, N, .... ,r , B, A , M', A', B ', r•, ... .,N, are considered to be in Harmonic Involution, with respect to the points M, M' , when all these pairs harmonically divide the segment MM' of the line they belong to.

----------------0 r BAM' A'

M This means that:

r·

B'

MA

MA'

MB

MB'

Mr

Mr'

-AM' =-A'M' -BM' =-B'M' -rM' =-r'M' , .. .... . J

(1) .

J

or (MM'AA')= (MM'BB')= (MMTr')= ...... =1 ,

(2) .

or, according to Newton's Theorem, let O be the middle of the segment MM', we might write:

M0 2 =M'0 2 =0A .OA'= OB.OB'= or.or·= ..... ..

(3).

2.12. Characteristics of the Involution. Point O is called the center of the involution, M02

11

M'0 2 is called the involu-

tion's force. Pairs of points A-A', B-B', r-r·, ... are called conjugate, while M, M' are called double points. It is obvious that we can have only one pair of conjugate points, then we would call it a Harmonic Involution of four, we could also have two conjugate pairs of points, then we would call it a Harmonic Involution of six, etc.

f.fAfl·I4J 2.21 . One-branch. When all t he conjugate pairs of points on the segment MM' are placed on the right (or the left) of point 0. For example, the involution shown in the f igure above is a one branch involution.

2.22. Two-Branch . When conjugate pairs of points on the segment MM' are found both to the left and to the right of point 0 .

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

2.23. Simple. When the involution includes only one pair of double points. 2.24. Double.

When the involution includes two pairs of double points.

Thus, the involution shown in the figure above is a simple, one branch involution. The question then arises: Do Harmonic Involutions really exist? The answer is yes, and it is given in Structures 9, 10, 11, 28, 30. In Figures 6 and 7, that are given below, simple Harmonic Involutions with double points E and E' are given, while in Figure 8 a double involution is given, with two pairs of double points E, E' and Z, Z'. In figures 22 and 23 we can see Harmonic Involutions of n-conjugate pairs of points. Detailed information about these is given in the Propositions they refer to. In figure 6 a Simple One-Branch Harmonic Involution is shown, in figure 7 a Simple Two-Branch Involution is shown, while in figure 8 we can see a Double Involution, which consists of two involutions, an one-branch one and a twobranch one. Figure 24 includes two simple involutions of n-pairs each, the first is an onebranch involution and the second a two-branch , while it is a double involution with M, M' and T, T' the double points pairs. It is obvious that a Collinear Harmonic Involution can include one pair of conjugate points, two pairs, three pairs, etc.

2.3. Middle Harmonic Involution. A Harmonic Involution is called so, when this involution includes two pairs of conjugate points and one of these points is the middle of the distance between two of the other three points.

~

i:- -

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

B' SYMMETRICAL PENCILS AND COLLINEAR HARMONIC INVOLUTIONS 3. Analyzing Symmetrical Central Pencils and Collinear Harmonic Involutions We can find Symmetrical Pencils of one, two, three, ... , n-pairs of beams , as we will later discover, which can consist of one or two branches. Symmetrical Pencils are rather useful , as will be established below, because they help us create Harmonic Collinear Involutions and this is why they will interest us later on. We searched for relevant information in bibliography, online, etc., but we were not able to find anything.

3.1. Simple Symmetrical Central Pencils of One Pair of Beams

l•ffltfliOt•htl Given the Flat Central Pencil of a pair of Lines O(aj.3) (figure

1), where the

pair of beams Oa, 013, are symmetrical to each other relative to a third beam Oy. This pencil is called «Symmetrical Pencil of One Pair of Beams. Given that 15015' is the perpendicular from O to Oy, then Oy and 015 are called «Axes of Symmetry» of this pencil, as we can write : L aOy= L y0j3=cp, L a015'= L'.j.3015= L'.a'Ol5=w.

(1 ).

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

(Where a' is a point on the extension of aO towards 0). Obviously Oy is the internal bisector of angle aOl3, while 15015' is the external bisector of angle aOl3. Moreover 015 can be considered as the internal bisector of angle l30a', while Oy the external one (Oa , 013, are known as isogonal relative to Oy, or 015).

3.2. Simple Harmonic Involution of One Pair of Conjugate Points.

Let (E) be an arbitrary line that intersects the above pencil (Figure 1), and additionally intersects Oa, Oy, Op, 015, at the points A , r , B , 11, respectively and let M be the middle of segment r11, then we could easily establish that the four points A,

r,

B, 11, are Harmonic, therefore accordi ng to Newton's Theorem, we can (2).

write :

Ar Because :

rs=

Al:!,, l:t,,B or

Ar Bl:!,,

rs. l:t,,A =1 ,

or because (2) is true, according to the

relevant definition, and because it is an Elementary Geometric Characteristic, through the repetition of which every Collinear Harmonic Involution is, we say that A, B are in Harmonic Involution, with

r and 11 as double points.

3.3. Simple Symmetrical Pencils of two pairs of Beams. (Simple One-Branch or Two-Branch Pencils).

l•ffltfflOt•J,?i Let there be a Flat Pencil of Two Pairs of Beams O(al313'a') (figure 2), the pairs

• q,

w-

-B

B

'A

a

a-

~

i:- -

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

of beams Oa, Oa' Kai 013, 013' , that belong to which, are symmetrical to each other relative to another beam Oy. This pencil is called Simple Symmetrical Pencil of Two Pairs of Conjugate One-Branch Beams . This pencil includes two pairs of beams, the first pair Oa, Oa', is called «External Pair of Conjugate Beams» and the second one 013, 013', is called «Internal Pair of Conjugate Beams». Moreover, if 06 is the perpendicular from Oto beam Oy, then beams Oy, 06, are called «Axes of symmetry» for the Symmetrical Pencil 's pair. We will establish that later on . We have:

L'.'.aOl3= L'.'. a'Ol3'=0, L'.'.l30y= L'.'.13'0y= w , L'.'.aOy= L'.'.a'Oy=

.

(15).

However due to (14) and (15) ~ LAOB+ L'.'.BOr= LAOr= L fOA'=2w =q> .

(16).

Besides, because of (14) it will also hold that L rOB'=w, because of (16), we have :

L B'OA'= L fOA'- L fOB'=2w-w=w.

(17).

Thus, due to (12), (14), (17) ~ L BOA'= L BOr+ L fOB'+ L B'OA'=w+w+w =3w. Therefore L BOA'=3w =90°, as well as L BOA'=90°, as it was made that way, thus w=30° and q>=2w=60°. Due to all the above, relations (13), (12), (14), (17), lead to: LAOB= L BOr= L fOB'= L B'OA'=w. In addition because rOb.=90°, as it was also made that way, it will hold that: Lb.QA·= L roa- L rOA '=90°-6o 0=30°=q,'=w. Furthermore, let a• be a point on the extension of Ab. towards A, then it will hold that: Finally:

La'OA=180°-5w=180°-150°=30°=q,'=w. La'OA= LAOB= L BOr= L rOB'= L B'OA'= LA'Ob.=w=q,'=30°.

(18).

Pencil O(ABrB'A'b.) , for which relation (18) holds true, is called «Regular of 30° or 6 Beams», as we will later discover (Structure 62 and relevant definition, § 12.3). Additionally, as we will later discover (Part B, § 12.4) the above group of 6 points A , B, r, B' A' a , is a «Collinear Harmonic Group of 6 Points». [Consult also Proposition 36 [Remark (a)] regarding Cocyclic Harmonic Group of 6 or Harmonic Hexagon] (For a Harmonic Group of 8 Points consult Construction 32a). Hence, due to all the above it is originated that from O segments ra, AB' and A'B are seen at right angles. Therefore, 0 is the «Orthooptical Point» in the envolution above, while this pencil is «Orthooptical», and its «Orthooptical point» is O (§ 7.1 below). Conclusion.

il.i·a-___C=o.:..:11.:..:in.:..:e:;.;:ao.:.r-'a""n.:..:d;:....;:::C;..;:o;..;:c'"'""'c.:..:li.;:;c-'H""a"'r'""m:.:.o;;;.:.:.n"'ic'"'l:.:.n:..:v..:;o'lu::::;t:.:.i "" o;:;.n:..:.s =. Because of all the above we conclude that: If we intersect a «Regular Pencil of 30°, or of 6 Beams», with a line the results are a «Collinear Harmonic Group of 6 Points» and a «Collinear Biharmonious Involution of Two Pairs of Conjugate Points».

Finally, if we take a closer look, the Biharmonious Involutions that are formed are actually 3 in total : The one with double points A , B', the one with double points B, A' and the one with double points

r , a.

Furthermore, in the Regular Pencil above we distinguish three «Symmetroharmonical Pencils, with t hese pairs of Axes of Symmetry, OA, OB' and OB, OA' and

or, oa.

As we will later see (Part B), something equivalent happens for all the " Regular Pencils" of 45°, 30°, 22.5°, 18°, etc. (Relevant Structures are 23 and 24, special cases).

3.5. Simple Symmetrical Pencils of Three Pairs of Beams.

j•fflmff!M1H Let there be a flat central Pencil of seven beams O(al3y~y'l3'a')

v· 13'

a'

rxii11a 4.

(figure 4), the pairs of beams Oa, Oa'- 013, 013'- Oy, Oy',that belong to this pencil are symmetrical relative to beam 0~. Pencil O(al3yy'l3'a') is called «Symmetrical Pencil of Three Pairs of Beams». Leto~· be the perpendicular to 0~ from point 0, then 0~, o~·. are called «Axes of Symmetry» of this pencil , as the following are true:

~

i:---

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

L a0i5= L i50a'=o, L l30i5= L i50l3'=cp. L y0i5= L i50y'=w,

(19).

L a'Oi5'= L i5'0a1=0'. L l3'0i5'= L i5'0131=cp', L y'Oi5'= L i5'0y1 =w'.

(20).

(where Oa1, 0131, Oy1 , are the extensions to Oa, 013, Oy, respectively). In addition o+o'=cp+cp'=w+w'=1 L.

(21).

3.6. Simple Harmonic Involution of Three Pairs of Conjugate Points (Figure 4).

l•ffl®Ot•J,?J If we intersect beams Oa, 013, Oy, Oi5, Oy', 013', Oa', Oi5' of the above pencil (figure 4), with a line E, at points A, B, r , 11, r·, B', A', 11' and let M be the midpoint of segment M', then we easily establish that the pairs of points A, A'- B, B'- r, r·, harmonically intersect the same segment M', thus according to known Newton's Theorem, it will hold that: 11M 2=11'M 2=MA.MA'=MB.MB'=Mr.Mr'.

(22).

That, according to the relevant definition, illustrates that the three pairs of points A, A'- B, B'- r, r·, are in «Harmonic Involution», with 11 and 11' as double points. The same happens for every other secant to the beams of this pencil. In addition, it is easily proven that relation (22) is also true when the pencil includes two branches.

3.7. Simple Symmetrical Pencils of n Pairs of Beams. Equivalent to the above, there is the «Simple Symmetrical Pencil of n Pairs of Beams». 3. 71. Simple Harmonic Involution of n Pairs of Conjugate Points. Equivalent to the above, there is the «Harmonic Involution of n Pairs of conjugate Points». All the above lead to the following Conclusion: Conclusion. Every «Simple Symmetrical Pencil of n pairs of Beams», intersected with a line, gives us a «Simple Harmonic Involution of n Pairs of Conjugate Points» (The converse is not true, when generally the center of the pencil is a random point. It is true only when the center of the pencil belongs to the defined locus 1 (§ 3.92), which follows just below).

~ .ia· _ _ _C=o.:..:11.:..:in.:..:e:;.;:ao.:.r-'a""n.:..:d;:....;:::C;..;:o;..;:c'"'""'c.:..:li.;:;c-'H""a"'r'""m:.:.o;;;.:.:.n"'ic'"'l:.:.n:..:v..:;o'lu::::;t:.:.i "" o;:;.n:..:.s =.

This conclusion allows us to easily construct Simple Harmonic Involutions of n Pairs of Conjugate Points (Structures 28 and 29 § 6.22 and 6.23). As we will see in Part C of the book, Harmonic Involutions Collinear or Cocyclic, are easy to develop through inscribed symmetrical n-lateral relative to one of their diagonals (diameter of the circle they are inscribed in). In this case, where a pencil is intersected w ith an arbitrary line, its intersections form a Harmonic Involution, but its center does not belong to the locus 1 that is referred in the conclusion above (§ 3.92), then this pencil , as it is cited in the conclusion, will not be symmetrical , and it will be called «Harmonic lnvoluted Central Pencil».

li' and ordered points A, B, r , B', A' , A on it, from these points pairs A, A' and B, B' are in Harmonic Involution, with double points r, A , then it holds that:

(ABrA')=(A'BTA)

(1).

and conversely, concerning the above ordered points on line l>', if the pairs of points A , A', or B, B' harmonically divide segment rA and relation (1) holds true, then both these pairs of points A, A' and B, B', are in Harmonic Involution w ith r , A as double points [Further analysis in remark (b) below].

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

a . Direct Proof.

1"1 Case Figure 2 . If the involution includes one branch, i.e. its points have the given order. In this case, we draw a circle with diameter the segment ra, and pick a random point O on it. In this way, a pencil O(ABrB'A') is originated, which, according to locus 1 (paragraph 3.92), will be symmetrical relative to or. Therefore, as in paragraph 3. 3 (relation (3)] it is proven that, the above relation ( 1) holds true. Furthermore, Proposition 2 above, would still hold true, even if in relation (1), we use point a instead of r [Consult§ 3.3, relation (3')] .

2nd Case. Fi ure 3

If the involution includes two branches this means that if

the following points be in this order B, r , B', A', a , A , then the case of the Proposition holds true, and we just need to clarify that A is to left of B to proceed with the proof [1 st Special Case§ 3,3 (conclusion 2), relation (7)]. Even here, the above Proposition 2, is also true if instead of point r in relation (1), we use point a [Consult§ 3.3 relation (7')].

{Gllri•Jei94i441 @IH\M§i If the involution includes one branch and the pair of points A, A' harmonically divides segment ra. As it is given that: (ABrA')=(A'BTA) = p AB rA' A'B' rA AB rs· AT (ArBA')=(ATB'A)=1-p Br "A'A = BT "AA'¢> Br "B'A' rA =1 .

(2).

Hence, if in the circle with ra as its diameter we set a random point 0 , then, as a is the Conjugate Harmonic r relative to points A , A', the aforementioned circle is Apollonian and thus or, would actually be the bisector of angle AOA' in triangle OAA'. Therefore, we understand that in triangle OAA', relation (2) holds true and or is the bisector of angle AOA'. That, according to the 1"1 converse of the auxiliary Proposition 2 (§ 2.2 Part E) means that or, would also be the bisector of angle BOB' in triangle OBB'.

Iii·:--

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

Thus, we understand that the two pairs of beams OA, OA' and OB, OB' in pencil O(ABrB'A'), are symmetrical relative to or, therefore this pencil is actually symmetrical relative to or. Consequently, because in triangle 088' line or is the bisector of angle BOB' and because ro -ea, as O belongs to circle of diameter ra, then oa would be the external bisector of the same angle BOB' (in triangle 088'). This means that r , a would be the Conjugate Harmonic of B, B' as well. So we see that both these pairs of points A , A' (hypothesis) and B, B', harmonically intersect with the same segment ra, this, according to the relevant definition of Harmonic Collinear Involutions (§ 2.11 ), means that these two pairs of points are indeed in Harmonic Involution, w ith rand a as double points. It is clear that this Proposition holds true even if it is given that the pair of points B, B' harmonically divides segment ra, instead of the pair A, A '. In that case we proceed to the proof in the same way as above, if we just adjust rela-

BA

rA' BT

tion (2) to its equivalent: Ar . A'B' rB =1 and apply the second converse of the Auxiliary Proposition 2 (Part E).

MWMI

If the involution includes two branches.

The order of the involution's points is the following , B, r , B' , A', a , A which can be seen in figure 3, then obviously pencil O(BB'AA') will include two branches, so Proposition 2 will still hold true, if we were to give to aforementioned points the following order A, B, r , B', A', a , similarly to paragraph 3.3 (1 st special case and conclusion 2). We only need to place point A to the left of point B. In this case, obviously pair of beams OA, OA', would be external , while pair of beams OB, OB' would be external. In that case we proceed to the proof in the same way as above, with the only al-

AB rs· AT

ternation in relation (2), which we transform to its equivalent : Br . 8 , A' rA =1 and apply the second converse of the Auxil iary Proposition 2, Part E.

1;34uFhfl (a). In the special case that: (ArBA') =(ATB'A)=1 , then, as it has been

il.i·a-__...;::C:;..::o;..:;ll""in""e:;.:a,.,_r...::a:.:.an:.;::d'""'C:::;.;o:::;.;c;.,.;:c;.:.:li.;:;.c..:.H.:.;:a,.,_r,.,_m:..;o""n"'ic"-'"'"ln.:.;:v..::;o.:.:lu::.;:t.:.;:io,.,_n~s . proven in paragraph 3.4, the group of six points A , B, r, B', A', A., would be harmonic. (b). Proposition 2 still holds true in the cases that instead of relation (1), we have: (ABA.A')=(A'B'A.A), or (BAA.B')=(B'A'A.B), or (BArB')=(B'ATB),

(3).

Because, as it will be proven below, these three equalities are equivalent to the proven true relation (1). To be able to use these in the following proofs, we will prove that these three equalities are all equivalent to equality (1 ), which we have proven above that both the implication and its converse hold true. Thus, if one of these three equalities in relation (3) holds true and because the pairs of points A, A' and B, B', are in Harmonic Involution with double points r , All

A., it holds that:

/lA'

Ar

=r A'

Bil a nd llB'

Br

=rs• '

(4).

Therefore, from the first equality in (3) and taking into consideration (4), this will apply: (ABA.A')=(A'B'A.A)

Hence, the first equality in (3), is equivalent to equality (1 ). In addition, from the third equality in (3), we will have: Br AB' BT A'B Ar BA' AT B' A (BArB')=(B'ATB)

L'.AKB= L rKa. Hence, because KE, KE' are also bisectors of angles AKB, rKa we easily find that

L

EKE'=1 L, or EK -KE' or EK, KE' are external bisectors of angles rKa,

AKB respectively, thus according to what is known E, E' are the CH points of A, B, but E, E' are the CH points of r, a, or that E, E' harmonically divide both segments AB, ra at the same time. From the way points E, E' are constructed, it occurs that they are unique (fixed). (d). Investigation. Because K, K' are fixed points and can always be constructed because the bisector of angle AKB is unique and can always be constructed then pair of points E, E' is also unique, because something equivalent is true about angle rKa too, the problem has always a solution and actually a uniquely defined one, the pair of points E, E'. It is obvious that if the other intersection K' of the circles above is received we would end up at the solution of the same points E, E', because of the symmetry of K, K', with respect to (E).

IU4uFOd (a). Construction 11 below is an extension to Construction 10, while 10 is relevant to 9. (b). The solutions of Construction 10 above, we believe to be new and firstly appear here. (c). Construction 42 (§ 9.3) is an extension of Construction 10 concerning a circle.

il,i·:-___C=o.:.:ll.:.:in.:.:eaaaa.:.r....:a::.an.:.:d:::...;:;C..;::o..;::c~c.:.:li..a;c...aH..:.;a""'r'""m:.:.o;::;.:..:.n:..:ic'""'l:.:.n:..:v..;::o"'lu:;:;t:.:.io;::.n:.:.s=.

6.5. Double Collinear Harmonic Involution of two Pairs of Conjugate Points each and of two and one Branches . Extension of Constructions 9 and 10 (New Construction). Construction 11 (Figure 8) .

Of a Double Collinear Harmonic Involution of two and one Branch respectively, we are given the two pairs of conjugate points A, r and B, /1 of the first, A, B and r , /1 of the other, and we are asked to define their two pairs of double points E, E' and Z, Z' respectively and prove that points Z', E, Z, E' are a harmonic group group of four, or in an older formation: Line (E) is given and the group of its four ordered points A, r, /1, B. Prove that there is only one pair of points E, E' of the same line (E), that harmonically divide both segments Ar, B/1 and only one pair of points Z, Z' on line (E), that harmonically divide all three segments AB, r/1, EE'.

Solution (Figure 8) .

If we take into consideration Constructions 9 and 10 above, we are led to this Construction below.

liii:-·-

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

b . Construction . Using set segments Afl, Br as diameters we draw circles that intersect at symmetrical points M, M', with respect to (E). If the bisectors of angles AMr, BMfl, intersect line (E) at points E, E' respectively and if the internal and the external bisector of angle AMB, intersect (E) at points 2, 2' respectively, we say that the two pairs of points E, E' and 2, 2', are the desired ones.

ttllkt®I According to the proof of Construction 10 above (2 nd way), points E, E' are the desired ones. Additionally, according to the proof of Construction 9, points 2 , 2', because 2' is the CH of 2 with respect to A, Band r,

fl -

2, 2' harmonically divide AB, rfl.

We are left to prove that 2, 2' harmonically divide segment EE' too. In fact, because ME, ME' from the way they are constructed, are bisectors of angles AMr, BMfl respectively and because Mr, Mfl are isogonal in triangle AMB and angles AMr, BMfl are equal [Proof of Construction 10 (2

nd

way)), we

easily prove that ME, ME' are isogonal in triangle AMB too. Thus the common bisector M2 of angles AMB, rMfl, will be the bisector of angle EME' too, while M2' is the external bisector of the same angle EME', as M2' is perpendicular to M2, 2, 2' will harmonically intersect segment EE' too. Therefore 2, 2', are the desired ones. (d). Investigation Because, according to Constructions 9 and 10 above, these two Problems always reach a solution then Problem 11 always reaches a solution too, while pairs of points E, E' and 2, 2' are unique, as M and ME, ME', M2, M2', are uniquely defined and can always be constructed.

1#3411Filb (a). We would reach the same result if we used point M', because of the symmetry with respect to (E).

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

(b). The Construction above, together with its above solution we believe to firstly appear here, while it is obviously an extension of Constructions 9 and 10, also it obviously constitutes a rather difficult Construction. (c). After Construction 11 above and the definition mentioned i n paragraph 2.1 above, we notice that in figure 8 above, we have the following harmonic involutions:

1/. If K is the midpoint of ZZ', according to Newton's Theorem , it will be: ZK2=Z'K 2= MK 2=KA.KB = Kr.Ka= KE.KE'.

(1) .

This means that here we have a harmonic involution, with three pairs of points A-8, r-a, E-E', center K, double points Z, Z' and force ZK 2 or MK 2 •

2/. If O is the midpoint of EE', according to Newton' s Theorem, it will be: E0 2 =E'0 2= M02=0A.or = OB.ca.

(2).

Additionally, because E, E', harmonically divide ZZ' too, angle OMK is right, as circles 0(0, M) and K(K, M) are orthogonal and K, 0 ~E), it will also be: E0 2 =E'0 2=0Z.OZ'.

(3).

Thus, from (2) and (3), we get: E0 2 =E'0 2 = M0 2=0A.or = 08.0a=OZ.OZ'.

(4).

Relation (4) means that here we have another harmonic involution, with three pairs of points A-r, B-a, Z-Z', center 0 , double points E, E' and force OM 2 or OE 2• (d). We notice that both involutions above, share their two Orthooptical Points

M, M'. Hence segments Aa, rs, EE', ZZ', are seen from M, M' in right angles and thus these are special points, so we call them «Orthooptical Points of a Double Involution of two Pairs of Conjugate Points and of two and one Branches» (§ 7). (e). If on the prolongation of BM towards M, we set a point

13, then

angles J3MZ'

and rMZ are equal, as their sides are perpendicular and they are acute. The same thing applies for angles Z'MA and ZMa and because angles J3MA and rMa are equal , while MZ' and MZ are the bisectors of angles l3MA and rMa, respectively, it will be: L'.J3MZ'= L'.Z'MA= L rMZ= LZMa=cp.

(5).

~

i:- -

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

Additionally, angles AME, f1ME' are equal , because they have perpendicular sides, same for angles EMr and E'MB, and because AMr and f1MB are equal, while ME and ME' are bisectors of angles AMr and BMf1, respectively, it will be: LA ME = L'.'.EMr= Lt..ME'= L'.'.E'MB=w. Moreover it is:

(6) .

LA Mf1= LAME+ L'.'.EMr+ L'.'.rMZ+ LZMf1= w+w+q>+q>=2(w+q>)=1 L,

l'J

(7).

w+q>=45°.

(f). If 4-' , 4-'' are the points of intersection of the circle with diameter MM' with line

(E), then because triangles M4-'4-'', MAa, MrB are orthogonal and Mn is their shared height, it will be: Mn 2=M'n 2 =4-'M 2=4-''n 2 =nA.na=nr.ns.

(8).

This , though, does not allow us to say that in the above way we achieved the construction of Double Poi nts 4-' , 4-'' of a Harmonic Involution, with Pairs of Conjugate Points pairs A , a and r , B. Hence we have not achieved the construction of a Triple Involution of four points A , r, a , B , because A , a, lie on both sides of n , the midpoint of segment 4-'4-''. Similarly for r, B (Pseudoharmonic Involution, § 7.14). (g). From the above and Propositions 30, 31 and 32 a, it occurs that pencil M(Z'EZE') is Regular(§ 12.1). Hence it is: LZ'ME= EMZ= LZME'= =45°. (h). Extension of Construction 11 to a circle is given in Construction 43 (§ 9.4).

(i). Relevant Propositions are 9, 10, 30, 31, 32a, 43, 204.

6.6. Construction of a Point of a Collinear Simple Harmonic Involution. Extension of Constructions 9 1 10 and 11 (New Construction). Construction 12 (Figure 9 and 9a). Line (E) is given and its ordered points A , a , r , B . On line (E) set a point E, for which it will be : Ar Aa rE rs = ar · EB ar

Ar BE ra or rs ·Er· M = 1 ·

Mrs

rE = Ar· BE

ar EB or

rE · Br

rA · Aa = 1 ·

(1 ). (2).

il.i·a-___C=o.:..:11.:..:in.:..:e:;.;:ao.:.r-'a""n.:..:d;:....;:::C;..;:o;..;:c'"'""'c.:..:li.;:;c-'H""a"'r'""m:.:.o;;;.:.:.n"'ic'"'l:.:.n:..:v..:;o'lu::::;t:.:.i "" o;:;.n:..:.s =.

Solution 1st Figure 9 Based on Criterion 4 above (paragraph 3.95).

tQW;ttfflli1tJ Because relation (1) is true, which reminds us Criterion 4 above,

we reach the conclusion that if E is the desired point and if harmonic to

r , with

r•

is the conjugate

respect to A , B, then pairs of points A, B and /1, E, will be in

Harmonic Involution, with double points

r , r· (Converse of Criterion 4).

In this way we are led to the construction of point E below. b . Construction. Because, according to the definition of Harmonic Involution (paragraph 3,4), should be the conjugate har monic of with respect to monic of

r, r·, we

r with respect to A , B

firstly define, in the known way,

r with respect to A , B and consequently E as CH

r·

and E the CH of /1

r · as

conjugate har-

of /1 with respect to

r , r•, that are known.

r·, as mentioned above, E can or2=0/1.0E, after defining point O of rr·.

After constructing tion

be defined also if we use rela-

t&IRI From the above way of constructing points

r·

and E, it occurs that pairs of

points A , B and 11, E, are in Harmonic Involution, with double points vant definition in § 3,4), hence for points A , 11, (Implication of Criterion 4).

r , E, B , r·

r , r·

(rele-

relation (1) will be true

Iii·:--

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

Relation (2), is true because from relation (1) we have proved is true, relation (2) easily occurs, which will obviously be true, as the equivalent to (1) is true. d . Investigation. In equivalent way we construct points r· and E for the cases that the given points have the following order 11, A , r , B, or this order 11, r , B, A too (Harmonic Involution of Two Branches). In these cases too the proof happens in the same way as the proof above, based on the same Criterion 4 again, which is true for these cases as well. It is obvious that there exists a second point E' for which relations (1) and (2) are true. This will obviously be th CH , E' of E with respect to B, r , as for which it

BE' will be E'

BE

Ar BE' rt:i..

r = Er . Hence, (1) becomes rB. E' r. t:i..A = 1.

It is obvious that this Construction happens in the same way as shown here when it concerns another point from A, B, /1 instead of E.

li.t.r, Ar"Br and work in the exact same way as above then we reach a contradiction. So we reached a contradiction because we supposed that another point r• exists in segment B.t., for which (1) is true. Thus, r is a unique point (fixed). If r lies on the prolongation of AE, then in this case too, r1 is fixed and coincides with the intersection of the external bisector of angle AKE with (E), or coincides with the CH, of r with respect to A, E. This is proved as above, but as follows as well: Since we proved that point r is fixed, we conclude that the CH, r1 of r , with respect to A , E will also be fixed . b . Investigation. If in the d irect of Proposition 15, we were given that a relation analogous to (1) was true and that r belongs to segment AB or to .t.E then with similar ways as well it is proven that Proposition 15 is true, but then in way 1, instead of Construction 9, we would use Construction 10 (§ 6.4).

IU4uFUI This Proposition is true in a circle as well [Proposition 49 (§ 9.10)).

6.10. Extension of Locus 1. Locus 16 (Figure 13). We are given a line (E) and its ordered points B, E, Z, r.

ii.I'·-__....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n=s . Find the Locus of variable points A, on the plane of (E) for which pencils A(BEZr), remain symmetrical with respect to its beam At:. [t:. is a point on line (cl], or in an older formation: In triangle ABr, its base Br is fixed and its apex A is variable. Moreover on its side Br we are given the fixed points E, Z (E to the left of Z) in a way that AE, AZ remain isogonals of angle A in triangle ABr. Find the locus of the variable apexes A in triangle ABr.

/).'

(E)

Ixriµa 13.

f:Sfflit•Je?i 1'1 Way. (Figure 13) (Based on Construction 9 and locus 1).

tE1WT®41?J If A is a point in the desired Locus, then pencil A(BEZr) is symmetrical with respect to its beam At:.. Moreover, if the perpendicular to At:. at point A , intersects (c) at t:.', then we notice that we have a pencil A(BEZr) symmetrical, with respect to its beam At:., which is intersected from line (c), at the following pairs of points B, rand E, Z. This, according to the conclusion of paragraph 3.8 above, means that pairs of points B, r and E, Z, will constitute a Simple Harmonic Involution, with double points t:., t:.' , which, can be constructed. (Construction 9 paragraph 6.3). Hence, after constructing the double points of this Involution, we conclude that we have all the features of the Involution, and ask for the Locus of all variable points A on the plane of line (c), for which pencils A(BEZr) are symmetrical with respect to At:..

Iii:-·-

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

Thus we have all the features of Locus 1, that we have given in paragraph 3, 92 above. Consequently, for the construction and proof of the desired Locus, we trace back to the above Locus 1 ( paragraph 3.92).

2 nd Way. (Figure 13) (Older) Based on Construction 13, AP 2 and Proposition

If ABr is one of these triangles and the common bisector of angles BAr, EAZ, intersects Br or (E) at 11, while the perpendicular to A/1 at point A, intersects (E) at /1', then obviously the circle with diameter /1/1' runs through A and is Apollonian, common for both triangles ABr, AEZ. We see then that A lies on the circle with diameter /1/1', which is fixed, as points /1, /1' are fixed. Point /1 is constructed as follows: Because AE, AZ are isogonal from the way they are constructed, for /1, according to Auxiliary Proposition 2 (in Part E), the following relation will be true :

rz ~E ~r·u·Es = 1 ·

B~

(1).

Hence, since (1) is true, /1 is easily constructed and it is fixed (See Construction 13) (§ 6. 7). Thus, /1' is constructed as CH of /1 with respect to B, r or E,

z

[The

fact that /1 is fixed, can occur directly from Proposition 15 above (§ 6.9), as relation (1) is true] . (b). Construction . We construct a point /1 on (E), in such way that relation (1) is true, according to Construction 13. We find, in the known way, the CH /1' of /1 with respect to B, r or E, Z and draw the Apollonian circle with diameter /1/1' , which we say is the desired locus.

tSl:fi®I If A' is an arbitrary point on the circle above, we will prove that A'E, A'Z are isogonal in triangle A'Br, while A'/1 is the common bisector of angles BAT, EA'Z. In fact, since /1 was constructed in such way that (1) holds true (Construction

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

13) and t:., t:.', harmonically divide B, r or E, Z, according to the converse of AP 2 then E, Z or B, r respectively, will harmonically divide points t:., t:.', thus the circle above, due to the way it is constructed, is Apollonian and for both triangles A'Br, A'EZ, A't:. is a common bisector of angles BAT, EA'Z', thus A'E, A'Z' will also be isogonal in triangle A'Br and therefore the circle above is the desired locus. (d . Investigation. Because, according to the investigation of Construction 13, point A, hence point t:. as well are always constructible, then this locus is always constructible as well, that

eventually is the whole circle with diameter M'. Here t:. was

sought in segment EZ, thus t:.' will be located outside of segment Br. We work in a similar way when t:., t:.' are sought in segments BE and rz [Construction 10 (§ 6.4)].

1#3411Fhl According to the definition in paragraph 3.4 above, the following series of points t:.', B, E, t:., Z, B, constitutes a Collinear Harmonic Involution, with conjugate pairs of points B, rand E, z, and double points t:., t:.'.

3rd Way. from Writer and Mathematics Professor Mr. Sotirios Louridas. We will depend on this principal proposition : The ratio of the areas of two triangles, that one of the angles of the first is equal to one of the angles in the other or that one of the first's angles is explementary to one of the angles in the other, is equal to the ratio of their sides that form these angles.

{

Lfl.AZ = L.EAC =;t,

2'.

= 62\

, ..,, .::1li:.4ll L.lJJfli, - L.~A,(f-¾ 11$,.i G

l!li

.'fitJ,

~

~

-

·

,z;c;

~

. H~

~

= n~ . o r,

,rn = . I fH , flr. ~~ . . V~ ~ 7

Hence we are to an Apollonian circle. I feel the need to move on to the process of Composition - Construction of the loci based on the Process of Analysis, due to the completeness since we additionally need the processes of Proof and Investigation. Composition - Construction: In the figure that follows we can see the steps of the construction of the Apollonian circle - locus, that in the figure is shown with the red intermittent one with its diameter segment GR. And this because:

~

i:- -

----=-H;..:..A;..:..R=M""'O::.:N..:.al=C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

S.E.Louridas Attached

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~

/

/

kataskeui.png (53.02 KiB) 445 previews S.E.Louridas

6.11 . Construction of a Point in a Collinear Simple Harmonic Involution. Extension to Constructions 9 to 12 (New Construction). Construction 17 (Figure 9). We are given a line (E) and its ordered points A, 11, r, B. On line (E) define a point E, for which it is:

Ar 2 rB 2

A/1 AE /1B"EB ·

-- =- -

(1 ).

f:fflMit•le?J f®MIMe•Miiini@ii>I Based on Criterion 5 above (paragraph 3.96). tEJW,ffth4t1?J Since relation (1) is true, which reminds us of Criterion 5 above, we conclude that if Eis the desired point and if

r· is the conjugate harmonic

~ .I'· _ _....;C:::;.o""l""li.:.:n.;::;e;.ar:.....;:;:.an:.:..d;:::....;:C'"'o'"'c'"'""c.:..:li""'c'"'"H.:.:a;;.:.r.:.:m.:.:o:;.:.n.:.:.i.;.c""ln:.:..v~o""l..;:::u.;.:ti""'o.:.:n= s.

to

r,

with respect to A , B, then pairs of points A, B and l:J., E, will be in Har-

monic Involution, with double points

r , r· (Converse of Criterion 5).

Hence we are led to the below construction of point E. b . Construction . Because, according to the definition of Harmonic Involution (paragraph 3,4), should be the conjugate harmonic of with respect to

r, r•, this

conjugate harmonic of

r with

r·

respect to A , B and E the CH of l:J.

is why we firstly define, in the known way,

r•

as the

r with respect to A, B and consequently E as the CH of l:J.

r, r· (r' is already constructed). construction of r·, as mentioned above,

with respect to After the

E is possible to be defined,

using relation Or =0l:J..OE, after we define midpoint O of rr·. 2

{01&1 From the above way of construction of points

r·

and E, it occurs that pairs of

points A , B and l:J., E, are in Harmonic Involution, with double points vant definition § 3,4), hence for points A, l:J.,

r , E,

r, r·

(rele-

B, relation (1) above will be

true (Implication of Criterion 5). (d). Investigation. In an equivalent manner we construct

r• and E, regarding the cases in which the

given points have the following order l:J., A,

r,

B , ot his order l:J.,

r , B,

A as well

(Harmonic Involution of two Branches). In these cases as well the proof is carried through with in the same way as above, based on Criterion 5 again, which is true in these cases as well. It is obvious that a second poitn E' exists for which relation (1) is true and relation (1') below too. This will obviously be the CH , E' of E with respect to A , B, as

AE' for this as well it will be E' B

AE

= EB

Ar 2 rB 2

:

. Hence, (1) becomes :

AA AE' AB. E' B .

(1 ').

It is obvious that this Construction is carried through with in the same ways as it would be if the construction of another point from points A , B, l:J. are desired instead of E.

E i:-·-

----=-H;..:..A;..:..R=M""'O=N-'-'l'""C....;G=E=O=M=ET-'-'Rc..:.Y.:..a,...:.P-=a=rt'-'A -=.

ld4116hl This Construction, is extended to a circle in paragraph 38 (§ 8.75).

Solution 2nd (Figure 9) (Older) (Based on Auxiliary Proposition 3).

We suppose that the desired point is defined and it is point E. If r· is the CH of r with respect to A, B, and if we draw the Apollonian circle with diameter rr•, then for arbitrary point K in this circle, Kr will be the bisector of angleAKB. After all, because (1) is also true, according to the converse of Auxiliary Proposition 3 (in Part E), or of Proposition 61(60) in our book [31] volume 6, KE will be isogonal to Kt.., in triangle KAB, thus we are led to the construction of E shown below. b). Construction. We define the CH, r· of r with respect to A, B and using as diameter segment rr·, we draw the Apollonian circle, on which we define arbitrary point K. We draw KA, KB, Kt.., Kr, Kr' and in triangle KAB the isogonal to Kt., that intersects AB at point E. We say that point Eis the desired one.

Because in triangle KAB, Kr is the bisector of angle AKB and KE is isogonal to Kt.., according to the direct of Auxiliary Proposition 3, or Proposition 113(44) in our book [31] above, and because it is ArJrB=AK/KB, relation (1) will be true, therefore Eis in fact the desired point. (d). Investigation. In figure 9, points r, t.. were received inside segment AB. It is though possible for these, to be received in other locations on (E). It is obvious that the CH, E' of E, with respect to A, B, constitutes another solution to the Problem, since it verifies relation (1) as well, since it is: AE'/E'B=AE/EB.

~ .i:· _ _....;C=-o=-'l""li.:.:n.;:;e=.ar:....::.an:.:.d:....;;:C:..;;o:..;;c;.&.;:c.:.:li.;:;c..:.H.:.:a::.:.r.:.:m.:.:o:..:.n.:..:.i.=.c""'ln:.:.v;:..;o=-'1-=u.;,:ti.;:;o.:.:n= s.

li (1).

(4).

(5).

This shows that r is in fact a solution to

Problem 22. (d . Investigation. Same as the investigation in solution 1.

1#3411FOA (a). It is obvious that for the last two solutions of the above Problem we used the method of «egress from line to plane». (b). It is possible to use some other ways too, from the ones we used in the solution of Constructions 2E(35) to 2E(37) in our book [31] volume 3. (c). According to the definitions in paragraph 3.3 above [remark (c)], pencil K(A/1rMB), (where M midpoint of AB) is a «Symmedianic Pencil», while the series of points A, /1, r , M, B, r · constitutes a «Harmonic Midpoint Involution» (§ 2.3), with conjugate pairs of points the following A, B and /1, M and double points r,

r.

(d). Relevant Propositions are 6, 9, 18, 20.

6.17. Construction of a Symmetroharmonic Pencil(§ 3.3. Special Case 2). Construction 23 (Figure 18). Construct a flat central pencil of five ordered beams O(al3yl3'a'), of which we are given three of its beams and for which it is: O(ayl3a')=O(a'y13'a)=1 .

(1 ).

mBi·:--

---=-H.:.:..A=R..:.:.M:.:.;O::.:N"'"l'-=C....;G::.:E::.:O::.:M=ET-'-'R;..:.Y..:...,' -'P--=a:.:..rt:..:A ..:..:..

M'

~

a

~

a

.

x

~13'

V

13

rxru.1a 1s. Solution Figure 18 . e discern the six following cases:

il4iiifojj 1n this case we are given these three beams Oa, Oy, Oa'. Hence here, we are given beams Oa, Oy, Oa' [The middle beam (Axis of Symmetry) and the two external conjugate ones] and we are asked to construct beams 013, 013' (The two internal conjugate ones).

We suppose that this pencil is constructed and it is O(al3yl3'a'), for which relation (1) is true. Because from (1)

=>

O(ayl3a')=O(a'yl3'a), according to the definition we gave in

paragraph 3.3 above, it occurs that pencil O(al3yl3'a'), is symmetrical. Hence obligatorily Oy should be the bisector of angle aOa'. Additionally, 013, 013' should b isogonal, with respect to Oa, Oa'. Moreover, from (1 ), it occurs that pencils O(ayl3a'), O(a'yl3'a), are harmonic (special case 2). Hence, if from an arbitrary point A, on Oy, we draw a perpendicular to Oa' and

!1191

...;C::;.o;,al"'li"'"n""e.::;.a::...r.::;.a:..:.nd.;:::....;:C'""o'""c-

L'.EKZ'=w=45°.

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cu

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mi-:-

-

---=-H.:.:..A.:.:.R.:.:.M:.:..;O:::.:N"'"l'-=C....;G:::.:E=O:::.:M=ET-'-'R;..:.Y"'""''-'P--=a:.:..rt:..:A ..:..:..

2'0' 2 and pairs of conjugate points

r-r, B-P, A-n , ..., M-M', ... , A'-n', B'-P', r •-r•,

apart from the Cyclic Harmonic Involution with double points M, M', center the midpoint O of MM', force M02 , or M'0"2 and pairs of conjugate points A - A', B -

B', r

- r·,.. ..,N - N', .... , n - n·, P - P', r - r·,... ,T - r.

(d). Easily, in figure 27, we prove that lines AA', BB',

rr·, ... , NN', 22', .. . , rr·,... ,

PP', nn·, .. . ,TT', are concurrent at pole 4' of MM'. The same thing happens w ith

rr,

BP, An, ....,MM', ... , A'n', B'P',

r·r·,... , which,

though , are concurrent here, at

pole of 22', that runs through the intersection K=An'nA 'n . Additionally, we easily prove that, MM ' and 22', run throught the intersection K of An' and A'n, as well , which is the pole of 4', and that BP',

rr·,... ,B'P, r·r,.. .,

also run

through K.

8.21 . Simple. When it includes only one pair of double points (paragraphs 8.62, 8.64 and Constructions 42 and 45). For example the involution in figure 27, whose double points are only the pair M - M', is Simple.

8.22. Double. When it includes two pairs of double points [paragraphs 8.63, 8.65 and Constructions 43 (§ 9.4) and 44 (§ 9.5)]. For example the involution in figure 27, with its double points both pairs M - M' and Z - Z', is Double.

1;34uFhl It is obvious that a Cocyclic Harmonic Involution, can consist of one pair of conj ugate points, two pairs, three pairs, etc. More about the involutions above, will follow in paragraph 8.6.

8.3. Closed Harmonic Cocyclic Midpoint Involution of two Pairs of Points. Using this term we characterize a Cocyclic Harmonic Involution, when this involution includes two pairs of conjugate points, from which the one point of the first pair lies in the middle of the arc the points of the ot her pair form .

!1731

C=o..:.:11..:.:in..:.:e:;.;:a::.:r....:a::.:n..:.:d:::....:::Cc.::oc.::c-

AB M' A' A'B' M' A BM'' A' A -B'M'' AA'



AB M'B' A'M' BM''B'A' " M'A - 1 ·

(3).

Besides, because from the hypothesis A , A' harmonically divide M, M', it will al-

A'M'

so be M' A

A'M

AB M'B' A'M BM' 'B'A' " MA = 1·

= MA , thus (3) will be:

(4).

Relation (4) shows [Auxiliary Proposition 5 (in Part E) (figure Sa)], that the convex

hexagon

ABM'B 'A'M

has

its

diagonals

concurrent.

Hence

it

is

MM'nAB'nA'B=K, or K EMM', then lines MKM' and KM'n share their two points K, M' and thus MKM'n will be a line. Hence, if AA'nBB'=P, because it is AB'nA'B=K and ABnA'B'=n , Kn or MKM'n will be the polar of P, thus si nce AA' and BB', run through pole P of MM' then A , A' and B , B', will harmonically divide pair of points M, M'. This means that A, A ' and B, B' are in Harmonic Involution (Definition§ 8.11), with Double Points MM'. •

If the first in (1) is true and 8 1 B'. harmonically divide points M, M'.

In this case as well , if we work exactly as above, we can easily find that relation (3) is true. Besides, as from the hypothesis B, B' harmonically divide M, M', it will be:

1991

...;C::;.o;,al"'li"'"n""e.::;.a::...r.::;.a:..:.nd.;:::....;:C'""o'""c- (1).

A'A' A'A.

(6).

Hence, if we take into consideration relations (6), then from (1 ), we easily get (2) and (3).

li ap yl>' v'l3' a'l>' ap yl> v'l3' a'l>' 13v . l>y' · wa·. l>a - 1• 13v . l>'y' · wa·. l>'a - 1• 13v . l>y' · wa· . l>'a - 1·

(8)

Relations (8) show that Proposition 7 is true (§ 6.1) [Relations (1), (2) and (2')], but here the proof of this, for the above «Harmonic Collinear Involution», is done in another way. Moreover from the above, we reach the following conclusion : «Relations (1) to (3) of the Cocyclic Harmonic Involution are also, in this way, transferred without alteration to a line as well as a Collinear Harmonic Involution , [relations (8)]» and converse (See also Corollary 168, § 5/).

Additionally, from the above, another way for the construction of such a Collinear Harmonic Involution from a Cocyclic Harmonic Involution occurs, based on a Cocyclic Harmonic Involution, that can be easily constructed. (c). Proposition 40 is obviously an extension to Proposition 7, from a line to a circle.

9.2. Property of the Simple Harmonic Cocyclic Involution of four Pairs of Conjugate Points . Extension to Proposition 8 from a line to a circle. (Application of Propos ition 40 and Criterion 37).

~

i:---

---=-H.:.:..A.:.:.R.:.:.M:.:..;O:::.:N"'"l'-=C....;G:::.:E=O:::.:M=ET-'-'R;..:.Y"'""'' -'P--=a:.:..rt:..:A ..:..:..

Proposition 41 . For every Cocyclic Series of Points of these Ten Points A , B ,

r,

b., E, b.',

r·,

B',

A', E' on circle (o), that constitutes a Simple Harmonic Involution of four pairs of conj ugate points A , A' - B, B', -

r , r·, - b., b.' , with double points E, E', it is :

AB rA EA' r·s· A'E Br . AE . AT'. 8' A'. EA = 1,

(1 ).

AB rA E'A' r·s· A'E' Br . AE'. AT'. 8' A'. E' A = 1,

(2).

AB rA EA' r·s· A'E' Br . AE . AT' . 8' A'. E' A =1-

(3).

(BJ We consider three pairs of points A , A' - B, B' -

r , r•

which will obviously con-

stitute a Harmonic Involution, with double points E, E'. Thus, according to Proposition 40 above, it will be :

AB rE r·s· A'E Br "Er' . B'A'. EA = 1· Similarly, pairs of points

r, r· -

(4).

b., b.' , will constitute a Harmonic Involution as

well, with double points E, E', thus according to Criterion 37, it will be:

rA EA' r'E AE. AT'. Er = 1·

(5).

Therefore from (4). (5), we get:

After all, it is:

AB rE r·s· A'E rA EA' r'E Br . Er' . 8' A'. EA . AE . AT' . Er =1

(1 ).

A'E EA

(6).

A'E' E'A '

A'E EA

A'E' E'A .

Thus, if we take into consideration relations (6), then from (1), we easily get (2) and (3).

ld4eeFhMI (a). Proposition 41 is obviously relevant to Propositions 4, 8, 37 and 40.

~

ii:· _ _...;C::;.o;,al"'li"'"n""e.::;.a::...r.::;.a:..:.nd.;:::....;:C'""o'""c-

(ABZt:,,.)=(t:,,.EZA)=p (AZBt:,,.)=(a2EA) =1-

AZ t:,,.E ZB _ Zt:,,. "EZ" BA - 1

(3').

So we see that relation (3), of the above Construction 48, is equivalent to relation (3) of Construction 47 (§ 9.8). (d). Because here we are given that:

z·

z·

(ABZ't:,,.)=(t:,,.EZ'A)=p (AZ'Bt:,,.)=(a2'EA)

AB li liE A AZ' liE Z'B = 1-p BZ'. l1A = EZ'. Ali Z' li. EZ'. BA

(4').

= 1·

Therefore we see that relation (4), of the above Construction 48, is equivalent to relation (4) of Construction 47 (§ 9.8). From the above we conclude that Construction 48 traces back to Construction 47, as the four equivanelces above are true and because all the rest given data are the same. Solution 2nd (Analytical .(Figure 42

Based on Criterion 36 (paragraph 8.73).

Relations (1) to (4), that are given for Construction 47, remind us of Criterion 36, according to the converse of which, points A , Z,

a, r•,

E, Z', B, r , if they are or-

dered on point (o) and: 1/. If for point r or r· relation (1) or (2) is true, while if r· or r respectively, are chosen in such way that pairs of points r , r· are harmonically divided by pair of points A , B or t:,,., E, then both pairs of points A , B and t:,,., E are in Simple Harmonic Involution with Double Points r , r· (Converse of Criterion 36).

2/. If for point Z or Z' relation (3) or (4) is true, while if Z' or Z respectively, are chosen in such way that pairs of points Z, Z' are harmonically divided by pair of points A , t:,,. or B , E, then both pairs of points A , t:,,. and B, E are in Simple Harmonic Involution with Double Points Z, Z' (Converse of Criterion 36) .

i:- -

~

---=-H.:.:..A.:.:.R.:.:.M:.:..;O:::.:N"'"l'-=C....;G:::.:E=O:::.:M=ET-'-'R;..:.Y"'""'' -'P--=a:.:..rt:..:A ..:..:..

Due to the above we reach the conclusion, that for relations (1) to (4) to be true eight points A , Z, fl,

r·, E, Z ', B , r , should

have the given order on circle (o) and

that then we would have two Simple Harmonic Involutions which combined constitute a Double Harmonic Involution. This means that for the exact place of points

r , r·, Z, Z',

to be defined, the two Pairs of Double Points of this Double

Harmonic Involution need to be constructed, thus we trace back to Construction 43 (§ 9.4) . (b). Construction . For the construction of the above Double Harmonic Involution, we are given the four ordered points A , fl, E, Bon circle (o), pairs of which A , Band fl, E, should constitute the first Simple Harmonic Involution, with Double Points pairs of points A , fl

Kai

r, r · and

B, E, should constitute the other Simple Harmonic Invo-

lution, with Double Points Z, Z'. This construction is carried through with in the exact way mentioned in Construction 43 (§ 9.4) and in this way the exact place of points

r , r·, Z, Z'

is de-

fined . Hence

more

analytically,

we

define

intersections AEnBfl=K, ABnflE=N ,

flnBE=/\, and : 1st

Way. From N and/\ we draw the tangents on circle (o). If

r , r·

and Z, Z' are

the pairs of tacpoints, of these tangents respectively, then we say that these points are the desired ones . 2 nd Way. If Kl\ and KN intersect circle (o), at pairs of points

r , r • and

Z, Z ' re-

spectively, then we say that these points are the desired ones.

ttal#d:ffil We firstly prove in the exact same way as in Construction 43, that the defined as above points

r , r·, z, Z'

in relation to the given points A, fl, E, B, constitute a

Double Harmonic Involution. Consequently, based on the implication of Criterion 36, we prove that for the Simple Harmonic Involution with Pairs of Conjugate Points A , B and fl, E w ith Double Points

r , r·, relations

(1) and (2) are true, while for the Simple Harmonic

Involution with Pairs of Conjugate Points A, fl and B , E w ith Double Points Z, Z', relations (3) and (4) are true.

~ ii:· _ _...;C::;.o;,al"'li"'"n""e.::;.a::...r.::;.a:..:.nd.;:::....;:C'""o'""c-Er

Ar Ar" rB < r"B a nd

=>

Er Er" rA < r" A

=>

Ar Er Ar" Er" - -Br and work in the same way as bove then we reach a contradiction. So we reached a contradiction because we supposed that another point r" exists on arc AE, which does not include points A . Band for which (1) is true. Consequently, r is unique (fixed).

l#J4uFhl It is obvious that Proposition 49 is an extension to Proposition 15, from a line to a circle. Relevant Propositions are 15, 37, 42.

9.11 . Defining a Point on a Simple Cocyclic Harmonic Involution of two Pairs of Conjugate Points. Alteration of Construction 46 (§ 9.7) nd extension to Construction 19 (§ 8.13) from a line to a circle. Construction 50 (Figure 41). On circle (o) we are given the ordered group of four fixed points A, A, r, B. On the same circle (o), define point E for which it would be: (AArB)=(BErA).

(1).

Solution (Figure 41). Tracing back to Construction 46 (paragraph 9. 7). Because we are given that:

(AArB)=(BErA)=p ¢> (ArAB)=(BrEA) =1-p ¢>

AA rs BE r A ¢> Ar BE r A Ar 'BA =Er" · AB rs 'Er' AA - 1

(2).

So we see that relation (1 ), of the Construction 50 above, is equivalent to relation (1) of Construction 46 (§ 9.7), thus Construction 50 traces back to Construction 46, as the rest given data are the same. As for the proof of the Construction, we will use the converse of relation (2) above.

~

ii:· _ _...;C::;.o;,al"'li"'"n""e.::;.a::...r.::;.a:..:.nd.;:::....;:C'""o'""c-=45 µoipcc;

A

(E)

(1).

---K"): B

r

Ixl'J1-1a 44.

~

11

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Pencil O(ABr.t.) is called Closed Regular Central Pencil of Four Beams 1

or of ½ of a right angle, or of 45°.

It is clear that for this pencil this holds true

LA Or= LBO.t.=1 L.

(2).

This pencil , for which relation (2) holds true, is called Orthooptical and its center 0 , Orthooptical Point. 12.2. Closed Collinear Harmonic Group of 4 Points and Harmonic Pancil o

l#•l'ii=1¥iuti If we intersect the above regular pencil O(ABr.t.) (Figure 44), with a random line(£), at points A , 8, r, a , then it is clear that OB is the internal and oa is the external bisector of the angle AOr from the triangle OAr, thus this pencil is harmonic, and therefore the group of 4 points A, r, 8, a , is Harmonic. Moreover, if pencil O(ArBa) is Orthooptical and Harmonic, then it is easily extracted that it is Regular too. Actually then, because L AOr= L BO.t. =90° and the group of 4 points A , r , B, a, is by default harmonic, for triangle OAr, we will have: L AOS= L BOr =45°. In addition, it holds true that : L'.TO.t.= L BO.t.- L BOr=90°-45°=45°, or L'.TO.t.=45°. Besides, let E be a point on the prolongation of AO towards 0, then we can easily understand that: L .t.OE= =45°. Therefore, from the above it is derived that relation (1) holds true, thus pencil O(ABr.t.) is Regular (definition). (See as well Criterion 69, Corollary 2). If the center is a random point P, not on the line (E), then pencil P(ArB.t.) is just a «Harmonic Pencil of Four Beams» [§ 12.13 (13)4/]. 12.3. Closed Regular Central Pencils of Six Beams.

l•ffl®Jit•l,H Let there be a straight angle ZI:Z'=2L (Figure 45), which is divided in six equal parts, thus: LAI:B= Lsrr= Lrra= LaI:E= L EI:Z= LZ'rA= w=Y.L =30°.

(1 ).

Pencil I:(ABr.t.EZ) is called Closed Regular Central Pencil of Six Beams, or of 1/3 of a right angle, or of 30°. It is clear that for this pencil

LAra= LBI:E= LrrZ=1 L.

(2).

(81'-.- -~ H:.:..A=R=M.:..:O=N=l..;:;C-'G=E=O=M=E=T.:..:R..:..Y'-'-P=art'-'-=B. This pencil, for which relation (2) holds true, is called Orthooptical and its center r , Orthoptical Point.

12.4. Closed Collinear Harmonic Group of six Points and Harmonic Pencil

If we intersect the above regular pencil (Figure 45), with an arbitrary line (E), at points A , B, r , /1, E, Z, then it is clear that I:B is the internal and I:E is the external bisector of the angle Arr from the triangle I:Ar, thus this pencil I:(ABrE) is harmonic, and therefore the group of 4 points A , B, r , E, is Harmonic. Moreover, rr is the internal and U is the external bisector of angle BI:/1 of the triangle I:B/1, thus pencil I:(Br/12) is harmonic, and therefore the group of four points B, r, /1, Z, is Harmonic. Moreover, I:/1 is the internal and I:A is the external bisector of angle rrE of the triangle rrE, thus pencil I:(r/1EA) is harmonic, and therefore the group of four points r , /1, E, A , is Harmonic. Let us continue in the same way as above, we will discover that, also these three groups of four points 11, E, Z, B - E, Z, A , r - Z, A , B , 11, are harmonic (See proof of below Criteria 62 and 67). Because of all the above and according to the relevant definition (paragraph 11 .1b), the group of six points A , B, r, 11, E, Z, constitutes a Harmonic Collinear Closed Group of Six Points or just a Harmonic Group of Six, thus this Regular pencil I:(ABr/1EZ) will also be Harmonic, such as

~ . Collinear and Coe clic Harmonic Grou s of n-members.

pencils r(ABrE), r1Braz1, r(r.t.EAl, r(tiEZB), r(EZAn, r(ZABti), would be harmonic (definition§ 12.13 (b) 4/). Inside proof of Proposition 70 (Corollary 2), it is proven that: Every such pencil, that is Orthooptical and Harmonic, will also be Regular and conversely (See also§ 3.4, special case). If the center is a random point P, outside of line (E), then pencil P(ABrtiEZ) would be just a Harmonic Pencil of Six Beams[§ 12.13 (b)4/] . 12.5. Closed Regular Central Pencils of Eight Beams.

l•fflm)Ji[•l,i Let there be a straight angle 0Px'=2L (Figure 46), which is divided in 8 equal parts, then:

L'.APB= LBPr= LrPa= LaPE= L'.EPZ=

L'.ZPH= L HP0= L x'PA = q>=¼L =22,5°.

X~

cp=1/4L=22,5 µofpEr1, trK I') Arz, BrH , rre, arl, ErK will be right. Additionally, since pencil r(apyf>E{1181K) or r(ABrb.EZHelK) is harmonic, according to the relevant definition [§ 12.8 Kai 12.13(P)4/, its ten pencils of four beams below will be harmonic :

r(ABrH), r(Brae), r(rb.EI),

r(b.EZK), r(EZHA), r(ZHeB), r(Helr), r(elKb.), r(IKAE), r(KABZ).

(9).

Hence, since pencil r(ABrH) is harmonic angle BrH is right, according to the converse of Criterion 75, quadrilateral ABr H is symmetrical with respect to BH, therefore

AB=Br.

(10).

Moreover, for the same reason, since pencil r(Brae) is harmonic and angle rre is right, quadrilateral Brae is symmetrical with respect to re, thus Br=ra.

(11 ).

We continue similarly to above and find that: ra=aE, b.E=EZ, EZ=ZH, ZH=He, He=el, el=IK, IK=KA.

(12).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Consequently, from (10), (11 ), (12) it occurs that AB= Br= r .t.= .t.E= EZ= ZH= H0= 01=1K=KA, hence decagon ABr.t.EZH01K ia actually Regular.

This Criterion above is rather significant, as it has many applications, whichi we will find below and because it allows us to give the alternative definition of the Regular Central Pencil of Ten Beams below, apart from the initial definition(§ 12.7): «A Central Pencil of Ten Beams, is called Regular, if we inscribe this pencil in a circle and its intersections with the circle are apexes to a Regular Decagon».

144116hb (a). As we will see below, equivalent Criteria, are true in other Regular Multilaterals as well, such as the Regular Dodecagon etc. that will follow. (b). It is obvious that Criterion 78 is an extension to Criterion 77 to a Regular Decagon. (c). Relevant Propositions are 35, 74, 75, 76, 77.

15.20. General Conclusion. After Criteria 74, 76, 77, 78 above and if we work in the same way as in their proofs, we end up in the general conclusion below: Criterion equivalent to Criteria 74, 76, 77, 78, is also true for any other Regular n-lateral.

15.21. Construction of a Square from a Harmonic Group of Four Points and Conversely. Construction 79 (Figure 57}. Construct a Square, inscribed in a given circle (o), based on a Harmonic Group of Four Points, given to us and conversely, on a given line (E) , construct a Harmonic Group of Four Poins, based on Square given to us.

18=-·__

_,_H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

f:fflMit·l,h tE11•Jlk431 IPl.;ttffll4t1tJ I.e. we are given on a line (E), a Harmonic Group of Four Points a, 13, y , 5 and we are asked , based on this group, to construct a square ABra inscribed in given circle (o). If we take into consideration the converse of Criterion 74 (§ 15.15), we are led to the construction below. 21. Construction.

We draw circles with segments ay and 135 as their diameters, that generally intersect at point r «Orthooptical Point»(§ 14.1). We suppose that r by chance belongs to circle (o). We form pencil r(ayl35), whose beams , if they re-intersect circle (o) at points A, r , B, a , respectively, then we say that quadrilateral ABra is the wanted Square. In the case that the given circle (o) does not run through r, then we draw, in the known way, a circle equal to the given (o), that does run through r and proceed, in the same way as above and after we construct a square ABra, we move it, in the known way, on the given circle (o), which square we say is the desired one.

ktl:ft®I In fact, since the group of four points a, 13, y, 5, is Harmonic then pencil r(al3y5) or r(ABra) would be Harmonic as well and because r is the known , «Orthooptical Point» of it, according to the converse of Criterion 74 (§ 15.15), quadrilateral ABra, will be Regular (square). Thus, square ABra is the desired one, as it is also inscribed in circle (o). (b). Converse.

I.e. we are given in a circle (o), the square inscribed in it ABra and we are aksed, based on that, to construct on a given line (E) a Harmonic Group of Four Points a, 13, y, 5. If we take into consideration the direct of Criterion 74, we are led to the Construction below: /. Construction. On circle (o), we set an arbitrary point rand form pencil r(ArBa) or

~ - Collinear and Coe clic Harmonic Grou s of n-members.

r(al3yl>), whose beams, we suppose that intersect line (E) at points a,

13, y,

l>, respectively. We say that the desired Harmonic Group of Four Points is the following a,

13, V, li.

ktl:ft®I In fact, since Quadrilateral ABrll. is Regular, according to the direct of Criterion 74, pencil r(ABrll.) or r(al3yl>) is harmonic, thus the Group of Four Points a,

13, y, l>

will be orthooptical and harmonic. Hence, it is the desired

one, as it is located on line (E).

134ee6hb (a). The Problem above and its solution, we believe to be new and firstly appear here. (b). Problem 79, the way it is given, it is undefined. If we would like to define it, then sufficient data need to be given. (c). In an equivalent way we can construct on a line Harmonic Collinear Groups of Six, Eight, Ten, etc, or on a circle we can construct Regular Hexagons, Octagons, Decagons, etc. (d). Another way of Constr uction of a Harmonic Group of Four Points, is given in Construction 61. (e). For the Construction above, we used a special way. This happened as this way will help us in the better understanding of our new Method of Harmonic Transformation. (f). Relevant Propositions are 35, 61 , 74, 80, 81 , 99.

15.22. Construction of a Regular Hexagon from a Harmonic Group of Six Points and Conversely. Extension to Construction 79 (§ 15.21) to a Regular Hexagon. Construction 80 (Figure 59). Construct a Regular Hexagon, inscribed in a given circle (o), based on a Harmonic Group of Six Points, given to us and conversely, on a given line (E), construct a Harmonic Group of Six Points, based on a Regular HExagon, given to us.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. f:fflMit·l,h tE11•Jlk431 IPl.;ttffll4t1tJ I.e. we are given on a line (E), the Group of Six Harmonic Points a ,

13, y

, 15, E,

{ and we are asked, based on this group of six, to construct a Regular Hexagon ABraEZ inscribed in given circle (o). If we take into consideration the converse of Criterion 76 (§ 15.17), we are led to the construction below. 21. Construction.

We draw circles with segments al5 and l3E as their diameters, that generally intersect at point r «Orthooptical Point». We suppose that r

by chance belongs to circle (o). We form pencil

r(ayl315E{), whose beams, if they re-intersect circle (o) at points A , r , B, a , E, Z, respectively, then we say that hexagon ABraEZ is the desired Regular one. In the case that the given circle (o) does not run through r, then we draw, in the known way, a circle equal to the given (o), that does run through rand proceed, in the same way as above and after we construct a hexagon ABraEZ, we move it, in the known way, on the given circle (o), which hexagon we say is the desired Regular one.

ktlfil In fact, si nce the group of six points a,

13, y , 15, E, { , is

harmonic and r is its

known, «Orthooptical Point», we prove, as in the proof of Proposition 70, that the Central Pencil r(al3yl5E{) or r(ABraEZ) is Regular, and its equal angles are of measure cp=(1/3)L=30°. Hence, according to the converse of Criterion 76 (§ 15.17), hexagon AsraEZ, will be Regular. Thus, hexagon ABraEZ is the desired one, as it is also inscribed in circle (o). (b). Converse.

I.e. we are given in a circle (o), the Regular Hexagon ABraEZ inscribed in it and we are aksed, based on that, to construct on a given line (E) a Harmonic Group of Six Points a,

13, y , 15, E, {.

If we take into consideration the direct of Criterion 76, we are led to the

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Construction below: 21. Construction.

On circle (o), we set an arbitrary point

r

and form pencil I:(ArBAEZ) or

I:(al3yl5£~), whose beams, we suppose that intersect line (£) at points a ,

13, y,

15, £, ~. respectively. We say that the desired Harmonic Group of Six Points is the following a,

V, 15,

13,

£, ~-

ktl #d!ffi I In fact, since Hexagon ABrAEZ is Regular, according to the direct of Criterion 76, pencil I:(ABrAEZ) or I:(al3yl5£~) is harmonic and

r is its Orthooptical

Point, thus we prove, as in the proof of Construction 62 that the Group of

13, y, 15, £, ~ will be harmonic too. Hence, the Group 13, y, 15, £, ~. is the desired one, as it is located on line(£).

Six Points a, Points a,

of Six

ld4eeFOb (a). The Problem above and its solution, we believe to be new and firstly appear here. (b). Problem 80, the way it is given, it is undefined. If we would like to define it, then sufficient data need to be given. (c). In an equivalent way we can construct on a line Harmonic Collinear Groups of Eight, Ten, etc, or on a circle we can construct Regular Octagons, Decagons, etc. (d). Another way of Construction of a Harmonic Group of Six Points, is given in Construction 62. (e). For the Construction above, we used a special way. This happened as this way will help us in the better understanding of our new Method of Harmonic Transformation. (f). Relevant Propositions are 35, 62, 70, 76, 79, 81, 82, 96, 100.

15.23. Construction of a Regular Octagon from a Harmonic Group of Eight Points and Conversely. Extension to Construction 80 (§ 15.21 l to a Regular Octagon.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. Construction 81 (Figure 60). Construct a Regular Octagon, inscribed in a given circle (o), based on a Harmonic Group of Eight Points, given to us and conversely, on a given line (E), construct a Harmonic Group of Eight Points, based on a Regular Octagon, given to us.

§fflfflh•hi (Ql•Jlk441 lt•½fflili1tJ I.e. we are given on a line (E), the Group of Eight Harmonic Points a,

13, y

, IS,

E, { , 11, 9, and we are asked, based on this group of eight, to construct a Regular Octagon ABrt:..EZH0 inscribed in given circle (o). If we take into consideration the converse of Criterion 77 (§ 15.18), we are led to the construction below. 2/. Construction . We draw circles with segments aE and

13{ as

their diameters , that generally

intersect at point r «Orthooptical Point». We suppose that r

by chance belongs to circle (o). We form pencil

l:(ayl3ISE{l18), whose beams, if they re-intersect circle (o) at points A, r, B,

t:..,

E, Z, H, 0, respectively, then we say that octagon ABrt:..EZH0 is the desired Regular Octagon. In the case that the given circle (o) does not run through r , then we draw, in the known way, a circle equal to the given (o), that does run through rand proceed, in the same way as above and after we construct an octagon A Brt:..EZH0, we move it, in the known way, on the given circle (o), which octagon we say is the desired Regular one.

ktl:ft®I In fact, since the group of Eight points a,

13, y , IS, E,

{ , 11, 9, is Harmonic and

r is its known, «Orthooptical Point», we prove, as in the proof of Proposition 71 , that the Central Pencil l:(al3ylSE{l18) or l:(ABrt:..EZH0) is Regular, and its equal angles are of measure cp=(1/4)L=22.5°. Hence, according to the converse of Criterion 77 (§ 15.18), octagon ABrt:..EZH0, will be Regular. Thus, octagon ABrt:..EZH0 is the desired one, as it is also inscribed in circle (o).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

tGili·lli'Z4?1=1 ;Pl!ttti l'lh tJ I.e. we are given in a circle (o), the Regular Octagon ABraEZH0 inscribed in it and we are aksed, based on that, to construct on a given line (E) a Harmonic Group of Eight Points a,

13, y, l>, E, { ,

11, 8.

If we take into consideration the direct of Criterion 77, we are led to the Construction below: /. Construction. On circle (o), we set an arbitrary point

r

and form pencil l:(ArBaEZH0) or

r(al3yl>E{l18), whose beams, we suppose that intersect line (E) at points a ,

y, l>, E,

{,

13,

11, 8, respectively.

We say that the desired Harmonic Group of Eight Poi nts is the following a,

13, V, l>, E, { ,

11, 8.

ktl:fttt!il In fact , since Octagon ABraEZH0 is Regular, according to the direct of Criterion 77, pencil l:(ABraEZH0) or r(al3yl>E{l18) will be harmonic. Hence, if we take into consideration the relevant definition in paragraph

13, y, l>, E, { , 11, 8, will be harmonic too. Hence, the Group of Eight Points a , 13, y, l>, E, {, 11, 8, is the desired one, as it 12.6 the Group of Eight Points a,

also located on line (E) .

13411FOA (a). The Problem above and its solution , we believe to be new and firstly appear here. (b). Problem 81 , the way it is given, it is undefined. If we would like to define it, then sufficient data need to be given. (c). In an equivalent way we can construct on a line Harmonic Collinear Groups of Ten, Twelve etc, or on a circle we can construct Regular Decagons, Dodecagons, etc. (d). Another way of Construction of a Harmonic Group of Eight Points, is given in Construction 63. (e). For the Construction above, we used a special way. This happened as this way will help us in the better understanding of our new Method of Harmonic Transformation.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-= B.

(f). Relevant Propositions are 35, 63, 68, 71, 77, 80, 82, 101.

15.24. Construction of a Regular Decagon from a Harmonic Group of Ten Points and Conversely. Extension to Construction 81 (§ 15.23) to a Regular Decagon. Construction 82 (Figure 61). Construct a Regular Decagon, inscribed in a given circle (o), based on a Harmonic Group of Ten Points, given to us and conversely, on a given line (E), construct a Harmonic Group of Ten Points, based on a Regular Decagon, given to us.

f:fflfflit•lel tD1•Jli4dl iQl;tmii'lhii I.e. we are given on a line (E), the Group of Ten Harmonic Points a, 13, y , ~. E, { ,

11, 8,

1, K,

and we are asked, based on this group of ten, to construct a

Regular Decagon ABrt.EZH01K inscribed in given circle (o). If we take into consideration the converse of Criterion 78 (§ 15.19), we are led to the construction below. 2/. Construction. We draw circles with segments a{ and 1311 as their diameters, that generally

r «Orthooptical Point». that r by chance belongs

intersect at point We suppose

to circle (o). We form pencil

r(ayl3~E{1181K), whose beams, if they re-intersect circle (o) at points A, r , B,

a, E,

Z, H, 0 , I, K, respectively, t hen we say that decagon ABrt.EZH01K is

the desired Regular Decagon. In the case that the given circle (o) does not run through

r , then we draw, in

the known way, a circle equal to the given (o), that does run through

r and

proceed, in the same way as above and after we construct a decagon ABrt.EZH01K, we move it, in the known way, on the given circle (o), which decagon we say is the desired Regular one.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

ktl#Mffil In fact, since the group of ten points a,

13,

y, li, E, ~. 11, 8, 1, K, is Harmonic

and r is its known, «Orthooptical Point», we prove, as in the proof of Proposition 72, that the Central Pencil I:(al3yl>E~1181K) or I:(ABraEZH01K) is Regular, and its equal angles are of measure cp=(1/5)L=18°. Hence, according to the converse of Criterion 78 (§ 15.19), decagon ABrb.EZH01K, will be Regular. Thus, decagon ABraEZH01K is the desired one, as it is also inscribed in circle (o).

tGili•leW4?14 it•,:.ttffl(i1ii I.e. we are given in a circle (o), the Regular Decagon ABrb.EZH01K inscribed in it and we are aksed, based on that, to construct on a given line (E) a Harmonic Group of Ten Points a ,

13, y , li, E,

~. 11, 8, 1, K.

If we take into consideration the direct of Criterion 78, we are led to the Construction below: /. Construction. On circle (o), we set an arbitrary point rand form pencil I:(ArBaEZH01K) or I:(al3yl>E~1181K), whose beams, we suppose that intersect line (E) at points a,

13, v, li,

E, ~' 8, 1, K, respectively.

We say that the desired Harmonic Group of Ten Points is the following a,

V, li, E,

~.

8, ,,

13,

K.

Ftl#Mffil In fact, since Decagon ABraEZH01K is Regular, according to the direct of Criterion 78, pencil I:(ABrb.EZH01K) or I:(al3yl>E~1181K) will be regular, hence we prove, as in the proof of Construction 64 that the Group of Ten Points a,

13, y, li, 13, y, li,

E, ~' 11, 8,

1,

K, will be harmonic too. Hence, the Group of Ten Points a,

E, ~' 11, 8, 1, K, is the desired one, as it also located on line (E).

ld4eeFhA (a). The Problem above and its solution, we believe to be new and firstly appear here. (b). Problem 82, the way it is given, it is undefined. If we would like to define it, then sufficient data need to be given.

lml=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. (c). In an equivalent way we can construct on a line Harmonic Collinear Groups of Twelve etc, or on a circle we can construct Regular Dodecagons, et c. (d). Another way of Construction of a Harmonic Group of Eight Points, is given in Construction 64. (e). For the Construction above, we used a special way. This happened as this way will help us in the better understanding of our new Method of Harmonic Transformation. (f). Relevant Propositions are 35, 64, 72, 78, 81, 98, 102.

15.25. Construction of a Regular Pentagon from a Harmonic Group of Five Points and Conversely. Extension to Construction 82 (§ 15.24) to a Regular Pentagon. Construction 83 (Figure 61). Construct a Regular Pentagon, inscribed in a given circle (o), based on a Harmonic Group of Five Points, given to us and conversely, on a given line

(E), construct a Harmonic Group of Five Points, based on a Regular Pentagon, given to us.

f:fflfflit•lel [611•112441 itW;fft1l4JhtJ I.e. we are given on a line (E), the Group of Five Harmonic a, y , E, 11, 1, and we are asked , basaed on this group of five, to construct a Regular Pentagon ArEHI inscribed in g iven circle (o). If we take into consideration the definition of the Harmonic Group of Five Points (§ 11.1 e), we conclude that for this Construction, we just need based on the given Harmonic Group of Five a, y, spective Harmonic Group of Ten Points a,

13, y,

l>,

E,

11,

E, ~.

1,

to construct the re-

11, 8, 1, K, thus, based

on it, we can construct the Regular Decagon ABr.t..EZH01K, based on Construction 82 (direct) (§ 15.24), hence simultaneously the desired Regular Pentagon ArEHI occurs .

~ - Collinear and Coe clic Harmonic Grou s of n-members .

. Construction. Since, according to the relevant definition of the Harmonic Group of Ten Points (§ 11.1 d), the group of four a,

13, y,

11, needs to be harmonic and

since we are given its three points a, y, 11, point

f3

is easily constructed in

the known way. Additionally, since the group of four y, l>, E, 1, needs to be harmoncic and since we are given its three points y, E, 1, point l> can be easily constructed in the known way. Similarly, through the Harmonic Group of Four E, ~' 11, a, we define the spot of ~' through thr Harmonic Group of Four 11, 8, 1, y, we define the spot of 8 and through the Harmonic Group of Four 1, K, a, E, we define the spot of K. Thus, after defining the spots of points

13,

l>, ~, 8,

ed the whole Harmonic Group of Ten Points a,

K,

13, y,

we will have construct-

l>,

E, ~,

11, 8, 1, K, hence

we continue as described in the analysis above and we construct the Harmonic Pentagon ArEHI.

ktl#ttMil The proof that A BraEZH01K is Regular, is carried through with as the proof of Construction 82. Hence, since ABraEZH01K is Regular, we can easily find that pentagon ArEHI is regular too.

tl:llri·l,W¼tii it•;ttffli't1~i I.e. we are given the inscribed in circle (o), Regular Pentagon ArEHI and we are asked, based on that, to construct on a given line (E), a Harmonic Group of Five Points a, y, E, 11, 1. If we take into consideration our definition of the Harmonic Group of Five Points (§ 11 .1 e) and Construction 82 (converse), we are led to the construction below: b . Construction . On the circumscribed circle (o) of the given Regular Pentagon ArEHI, we set an arbitrary point r. We draw rA, rr, rE, rH, r1, which we suppose that intersect the given line (E), at points a, y, E, 11, 1, respectively.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

We say that the group of five points a, y,

E,

11,

1,

is the desired Harmonic

one.

ttllHI For the group of five points a, y ,

E,

11,

1,

to be Harmonic, according to the

relevant definition (§ 11.1 e), there needs to be on the line (E), another group of five points 8,

1, K,

13, ~. ~. 8, K,

such that the group of ten points a,

13, y , ~. E,

~.

11,

would be Harmonic.

In fact, if based on the given Regular Pentagon ArEHI, we construct, the Regular Decagon ABraEZH01K. This can be easily achieved, if only we define on circle (o) the spots of points B, a , Z, 0, K too, in the known way or in a way equivalent to the one we used to define points

13, ~. ~.

8, K, on line

(E). We draw rs, ra, rz, re, rK, which intersect line (E) at points

13, ~. ~.

8,

K.

We easily prove, similarly to Construction 82 (converse), that the group of ten points a , 13, y, ~. five points a, y,

E,

E, ~.

11,

1,

11, 8,

1, K,

is Harmonic. This means that the group of

is Harmonic (definition § 11.1 e). Thus this is the de-

sired one, as it is also located on line (E).

@Mulot4iil The Construction above, obviously, the way it is given to us, it is undefined. If we want to make it more specific, sufficient data need to be provided .

1#34116hb (a). In the same way, as above, we prove that the group of five points 8,

K,

13, ~. ~.

is Harmonic too.

(b). Construction 83 we believe to be new and firstly appear here. (c). We notice that using the above new special Method, the construction of the Harmonic Group of Five Points, easily occurs, from a Regular Pentagon, in the exact same way as it does for the Harmonic group of Four, Six, Eight, Ten, etc. (d). Another way of construction of Harmonic Group of Five Points we gave in Construction 65. (e). In the construction above, we used a special way. This happened as this way will help us in the better understanding of our new Method of Harmonic Transformation.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

(f). Relevant Constructions are 64, 65, 82, 102, 103.

15.26. General Conclusion . Due to Criteria 79, 80, 81 , 82, 83 above and if we work as in the Constructions and proofs of these, we reach the following general conclusion: Equivalent Construction, to Constructions 79, 80, 81 , 82, 83, is true for every Regular n-l ateral , as long as this is constructible, or the respective Harmonic Group of n is constr uctible.

15.27. Noteable Property of the Closed Harmonic Collinear Group of Six Points.

Proposition 84 (Figure 45}. For every Harmonic Collinear Group of Six Points A , B,

•.

r , l:J., E, Z, it is:

AB r~ EZ Br . ~E . ZA =1·

(1 ).

,

According to the relevant definition of the Harmon ic Collinear Group of Six Points (§ 11 .1 b), the following groups of four points A , B ,

r , E - r , l:J., E, A

-

E, Z, A , r , are harmonic , from which the relations below occur, respectively:

AB AE r~ rA Br =Er' ~E = AE'

EZ Er ZA = r A .

(2).

Hence, if we multiply the respective parts, in relations (2) and after the proper reductions , we get:

IU4eeEhA (a). This Proposition is not a criterion of harmonic being, as its converse is not true.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-= B. (b). We notice that equivalent Propositions, are true for a Harmonic Group

AB r~

of Four Points A , B, r , t::.., (Br . ~

=1), as well as for a Harmonic Group of

Eight Collinear Points A , B, r , t::.., E, Z, H, 0 (Proposition 85) that follows, etc. (c). This property is true for Simple Harmonic Hexagons too, i.e. when the Group of Six Points A , B , r , t::.., E, Z, is not collinear, but cocyclic (Proposition 119). (d). This Proposition we believe to be new and firstly appear here. (f). Relevant Propositions are 4, 7, 8, 37, 85, 86, 119.

15.28. Significant Property of the Closed Harmonic Collinear Group of Eight Points. Extension to Proposition 84 (§ 15.27) to a Harmonic Collinear Group of Eight Points. Proposition 85 (Figure 52). For every Harmonic Collinear Group of Eight Points A , B, r , t::.., E, Z, H, 0 , it

AB r ~ EZ H0 Br. ~E"ZH"0A

is:

AZ re EB H~ 0E "BH " ~ =1-

= zr.

(1 ).

IM M 1'1 Wa

Based on definition and Criterion 5 .

Since, according to the relevant definition (§ 11.1 c), the Groups of Four Points A , 8 , r , Z and E, Z, H, 8 , are harmonic, or because B, Z harmonically divide A , r and E, H, which means that here we have a Harmonic Involution, of two pairs of Conjugate Points A , r and E, H, with double points B, Z (definition§ 2.11) and thus , according to Criterion 5 and s ince AB/Br=AZ/Zr, EZ/ZH=EB/BH, it would be, respectively:

AB 2 AZ 2 AH AE Br 2 = zr 2 = Hr "Er '

EZ 2 EB 2 Er EA ZH 2 = BH 2 = rH. AH .

(2).

Similarly because, according to the relevant definition (§ 11.1 c) , t he Groups of Four Points r , t::.., E, 0 and H, 0 , A , t::.., are harmonic, or since t::.., 0

~ - Collinear and Coe clic Harmonic Grou s of n-members.

harmonically divider, E and H, A, which means that here as well, we have a Harmonic Involution, of two Pairs of Conjugate Points r , E and H, A, with double points

a, 0 (definition§ 2.11) and thus, according to Criterion 5 and

since rf).Jf).E=r0/0E, H0/0A=Hf).JM, it will be, respectively:

rd2 dE 2

r0 2

rA rH

H0 2 0A 2

=0E =AE. HE ' 2

Hd2

HE Hr

=M =EA . r A .

(3).

2

Hence, from relations (2) and (3), if we multiply the respective parts and after the proper reductions relation (1) occurs. 2nd Way (Based on Construction 21 and Construction 63). According to the second way of construction of Construction 63, we conclude that a Harmonic Group of Eight Points A, B, r , a, E, Z, H, 0, occurs from a Harmonic Group of Four Points A , r , E, H and from its pair of points B, Z, that has harmonically divided segments Ar, EH and from its pair of points

a, 0, that has harmonically divided segments AH , rE, thus, according to what is mentioned in Construction 21 , it will be:

AB 2 Br 2

=

AZ 2 zr 2

=

AH AE Hr· Er '

rd2 dE 2

EZ 2 EB 2 Er EA ZH 2 = BH 2 = rH. AH '

H0 2

r0 2 0E 2

rA rH

=

=AE"HE'

Hd2

0A 2 = M

2

HE Hr = EA . r A .

(4).

Hence, from relations (4), if we multiply the respective parts and after the proper reductions relation (1) occurs.

i;i4116UA (a). This Proposition is not a criterion of harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are true for a Harmonic Group of Four Points A , B, r,

AB rd

a, (Br. M =1), as well as for a Harmonic Group of

Six Collinear Points A , B, r,

AB rd EZ

a, E, Z, (Br . dE. ZA =1) (Proposition 84), etc.

(c). This property is true for Simple Harmonic Octagons too, i.e. when the Group of Eight Points A , B, r ,

a,

E, Z, H, 0, is not collinear, but cocyclic

[See Proposition 120, that we will mention later].

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

(d). The second part of (1), makes it clear for us that Proposition 85 is also true for the Asteroid Harmonic Group of Eight Collinear Points A, 2 , r, 0, E, B, H, 11, that in the Simple Harmonic Octagon, matches to its Asteroid Harmonic Octagon A2r0EBH/1A, as we will later see (Proposition 120). (e). This Proposition, we believe to be new and firstly appear here. (f). Relevant Propositions are 5, 7, 8, 11 , 21 , 63, 84, 86, 120.

15.29. Significant Property of the Closed Harmonic Collinear Group of Ten Points. Extension to Proposition 85 (§ 15.28) to a Harmonic Collinear Group of Ten Points as well. Proposition 86 (Figure 54). For every Harmonic Collinear Group of Ten Points A, B, r, /1, E, 2 , H, 0, I, K, it is:

AB ra EZ H0 IK A0 EB IZ rK H/1 Br. /1E. ZH" 01 . KA - 0E. Bl . zr. KH. 11A - 1-

(1).

IM M 1'1 Way (Based on definition and Criterion 5).

Since, according to the relevant definition (§ 11 .1 d), for the Collinear Harmonic Group of Ten Points A, B, r, 11, E, Z, H, 0, I, K, the Groups of Four Points A , 8 , r , H and Z, H, 0 , 8 , are harmonic, or because 8, H harmonically divide A , r and Z, 0, which means that here we have a Harmonic Involution, of two pairs of Conjugate Points A , rand Z, 0, with double points B, H (definition § 2.11) and thus, according to Criterion 5 and since

AB 2 sr 2

AB/Br=AH/Hr, it would bealso:

=

AH 2 Hr 2

A0 AZ

=er· zr ·

(2).

Moreover, since the following groups of four r , /1, E, I and 0, I, K, /1, are Harmonic as well (definition), in the exact same way we find that it is:

r112 /1E 2

r1 2

rK re

=IE2= KE"0E .

If we continue similarly to above, we also find that:

(3).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

E2 2

EA 2

EB EK

- =- - =- 2H 2 AH 2 BH"KH ' H0 2 Hr 2 Hll HB 01 2 =r12 =T1·01 · IK 2

IE 2

12

Ill

==2 2 KAEA2A. M .

(4).

Hence, if we multiply the respective parts in relations : (2), (3), (4) and after the proper reductions, we get :

AB 2 r112 E2 2 H0 2 IK 2 A0 EB 12 rK Hll sr 2 . llE 2 . 2H 2 . 01 2 . KA 2 0E . Bl . 2r. KH . M AH 2 rl 2 EA 2 Hr 2 IE 2 Hr2 . IE2 . AH2 . rl2 . EA 2 =1,

(5 ).

as the third part of (5) is reducted and becomes equal to 1. Thus, from (5), we can easily get relation (1).

2nd Way (Based on the definition and Criterion 4).

Since, according to the relevant definition (§ 11 .1 d), for the Collinear Harmonic Group of Ten Points A , B, r , fl , E, Z, H, 0, I, K, the Groups of Four Points A , B, r , H and Z, H, 0 , B, are harmonic, or because B , H harmonically divide A , r and Z, 0 , which means that here we have a Harmonic Involution, of two pairs of Conjugate Points A , rand Z, 0 , with double points B , H (definition § 2.11 ), according to Criterion 4 and s ince AB/Br=AH/Hr, it will be :

AB r2 s0 sr ·2s·0A = 1

(6)

(7).

(Here, for our convenience and i n order to apply Criterion 4, we would have to put the points in the following order 0 , A , B, r, Zand 0 , A. H, Moreover, since the groups of four

r , fl , E, I and

r , Z).

0 , I, K, fl , are Harmonic as

well (definition), in the exact same way we will also find that :

rll E0 llK llE. 0/l. Kr=1

(8)

rl E0 llK

a nd

(9).

IE" 011 · Kr = 1·

(Here, for our convenience and i n order to apply Criterion 4, we would have to put the points in the following order K,

r , fl , E, 0

If we continue similarly to above, we also find that:

and K,

r , I, E,

0).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. E2 HK 28 2H"KZ"BE =1 ,

(10)

a nd AH" K2. BE =1,

EA HK 28

H0 18 0A 01 . 80. AH =1,

(12)

a nd n·s0· AH= 1,

IK AA KZ KA. AK·21= 1,

(14)

a nd EA. AK ·21= 1·

Hr 18 0A IE AA K2

(11 ).

(13).

(15).

Hence, if we multiply the respective parts of relations: (6), (8), (10), (12), (14), we get:

AB r A E2 H0 IK Br . AE . 2H . 01 . KA

A0 EB 12 rK HA 0E . Bl . 2r. KH . AA .

(16).

Similarly, from (7), (9), (11), (13), (15), we can easily get:

A0 EB 12 rK HA AH rl EA Hr IE 0E. Bl . 2r. KH . AA - Hr ' IE. AH. rl . EA - 1,

(17).

as the second part of (17) is reducted and becomes equal to 1. Therefore from (16) and (17), we get (1) .

3rd Way (Based on the combination of Propositions 4 and 5). Working similarly to way 1 above, from the first and third part of (2), (3), (4), we get:

AB 2 rA2 E2 2 H0 2 IK 2 A0 EB 12 rK HA Br 2 • AE 2 • 2H 2 • 01 2 • KA 2 = 0E. Bl. 2r'KH . AA.

( 15),

Consequently, working similarly to way 2 above and only from (6), (8), (10), (12), (14), we get (16). Hence, from (16) and (18), we can easily get:

AB 2 rA2 E2 2 H0 2 IK 2 AB rA E2 H0 IK sr 2 • AE 2 • 2H 2 • 01 2 • KA 2 = Br . AE . 2H . 01 . KA ~ AB rA E2 H0 IK Br· AE · 2 H . 01 . KA =1 , thus from this and (16), desired relation (1) occurs.

134eeEhb (a). This Proposition is not a Criterion of Harmonic Being, as its converse is not true. (b). We notice that equivalent Propositions, are true for a Harmonic Group

~ . Collinear and Coe clic Harmonic Grou s of n-members.

of Four Points A, B, r, 11, (For which it is

AB

rt.

Br. f.A =1), as well

as for a Har-

monic Group of Six Collinear Points A, B, r , /1, E, Z, (For which it is

AB

rt. EZ

Br. f.E. ZA =1), as we have proven (Proposition 84), as well as for the Oc-

tagon (Proposition 85). (c). This property is true for Harmonic Collinear Groups of Twelve Points, etc. (d). The second part of (1), makes it clear for us that Proposition 86 is also true for the Asteroid Harmonic Group of Ten A , 0 , E, B, I, Z, r , K, H, /1 (See also Proposition 121). (e). This Proposition, we believe to be new and firstly appear here. (f). This Proposition is also true for Simple Harmonic Decagons , i.e. when points A, B, r, 11, E, Z, H, 0, I, K, are cocyclic (Proposition 121, Harmonic Decagon) . (g). Relevant Propositions are 4, 5, 53, 84, 85, 121 .

15.30. General Conclusion. Due to Propositions 84, 85, 86 above and if we work as in their proofs, we reach the following general conclusion: Equivalent Proposition, to Propositions 84, 85, 86, is true for any other harmonic group of 2n points.

15.31. Noteable Property of the Closed Harmonic Collinear Group of Four Points. Extension to the Known Proposition of Harmonic Quadrilaterals to a Harmonic Group of Four Points as well.

Proposition 87 (Figure 551. For every Harmonic Collinear Group of Four Points A, r, B, /1, it is :

AB.rt.

A/1.Br=Ar.B/1=-2- ·

(1 ).

l8J=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

jitfflliffl

Based on Newton' s and Mac-Learin Theorems trEwµnpia [4]).

Let/\ be the midpoint of segment ra, then according to Mac-Learin Theorem ([4], § 60/4), it will be :

Ar.Aa=AB.A/\, and

(2) .

sr.Ba=AB.BA

(3).

From relations (2), (3), if we multiply the respective parts, we get: Ar.Aa.Br.Ba =AB 2 .A/\.B/\,

Ar

-

After all , since it is:

rs

All.

=-

fl.S

¢,

(4).

Ar.sa=Aa.sr

(5).

Moreover according to Newton' s Theorem ([4] , § 60/2), it will also be: r/\2=M 2 =/\A./\B.

(6).

Hence, relation (4), because of (5), (6), becomes : Ar2 .sa2=Aa 2.sr 2= AB 2.r/\2=AB 2 .M Ar.Ba= Aa.Br = AB.r/\ = AB.M =

2

or

AS(r/\ + .1\/l.) 2

AS.r fl.

=- 2-

~

(1).

i#iM•iel•i'IZfi'J Based on Euler's Theorem trEwµnpia [28], book 3). According to Euler's Theorem ([28] § 295 in book 3) . and concerning point a which lies indise segment AB, it will be: aA.Br= ar.AB - as.Ar or Aa.Br+Ar.Ba = AB.ra. After all , since it is:

Ar

All.

rs = /l.S

¢'

(2). (3) .

Ar.Ba=Aa.Br.

Hence, relation (2), because of (3), becomes: 2Aa.sr=2Ar.sa = AB .ra

r;

AS.rfl.

Aa.Br=Ar.sa = -

2-

.

134ee6hb (a). The so called Euler's Theorem , in one way, can be considered as an extension to Ptolemaios' A ' Theorem about the Harmonic Quadrilateral, but on a line, instead of a circle. (b). It is obvious that Proposition 87, is an extension to a known Proposition ([6], § 8-4.10), from a circle to a line as well. (c). We believe that it is a new Proposition, that firstly appears here. (e). Relevant Propositions are 88, 91 , 92, 144.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

15.32. Significant Property of the Closed Harmonic Collinear Group of Six Points. Extenstion to Proposition 87 about a Harmonic Group of Four to a Harmonic Group of Six Points. Proposition 88 (Figure 56). In every Harmonic Collinear Group of Six Points A, B, r , b., E, Z, it is:

At..BE.rz

AB.rlJ..EZ=Br.tJ.E.ZA= - 8- -

ID.I

(1).

Based on Proposition 87.

According to the relevant definition of the Harmonic Group of Six(§ 11.1 b), the group of four points A, B , r , E, is harmonic, thus if we apply for it Proposition 87, we will have: AB.rE=Br .AE =

BE•Ar 2

(2).

Similarly, since according to the relevant definit ion the group of four points r , b., E, A , is harmonic as well, according to Proposition 87, it will be:

At..rE

rtJ..EA=lJ.E.Ar = -

2- .

(3).

Additionally, since accordi ng to the relevant definition the group of four points E, Z, A , r , is harmonic as well , according to Proposition 87, it will be:

rz.EA 2- .

EZ.Ar=ZA.rE = -

(4).

Hence, from the first and t hird parts of (2), (3), (4) and if we multiply the respective parts, we get:

or

AB.rtJ..EZ.rE.AE.rA=

AB.rtJ..EZ =

BE.Ar.At..rE.rZ.EA 8

At..BE.rz B

(5).

Moreover, from the second and third parts of (2), (3), (4) and if we multiply the respective parts, we get:

18=-·__

_,_H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=a rt '-'-= B.

BE.Ar.A~.rE.rZ.EA

A~.BE.rz

sr.AE.aE.Ar.zA.rE= - - - 8- - - - or sr.aE.ZA = - -8- -

(6).

Consequently, from relations (5), (6):::::(1).

IU4ee6hb (a). We believe that an equivalent Proposition is true for every other harmonic n-lateral, the proof of which can be based on Propositions 87, 88, etc. (b). It is obvious that Proposition 88, is an extension to Proposition 87. (c). We believe that it is a new Proposition. We came up with this Proposition in 1998, we published it in volume 3 of our book [1] (page 1272, Proposition 37 AZ). (d). Extension to Proposition 88, is given with Proposition 144 (§ 19.59), from a line to a circle. (e). Relevant Propositions are 87, 91, 92, 144.

15.33. Construction of a Special Collinear Harmonic Group of Four Points. Generalization of the Geometrical Construction 90. Construction 89 (Figure 55}. The line segment AB=2T, should be harmonically divided from a pair of points r , a (r lies between A , B and a outside of the segment), for which (pair) it should be: ra=2T' (where T, T' are given segments), or in an alternat ive formation: Construct a harmonic group of four points A, B, r , a , such that AB=2T, ra=2T' (where T, T' are given segments).

@fflfflit•leh Since here we have a Harmonic Group of Four, to solve this Problem , we will use the new method of Regular Central Penci ls.

IEiWMfflli1~i To apply th is method, we think in the following way: We suppose that the desired Harmonic Group of Four is constructed and is the following one A, r , B, a. This will have AB=2T and ra=2T'.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

We draw circles with diameters AB=2r or (K) and r.t.=2r' or (A), which intersect at a pair of (Orthooptical Points) points H, H' symmetrical with respect to AB. Pencil H(ArB.t.) is Regular, as it is Orthooptical, since AB, ra are seen from H, H' obviously in right angles and is on the Harmonic Group of Four Points A , r, B, a(§ 12.1, §12.2 and §12.13). Thus, according to Criterion 69 (necessary), if K, I\ are the midpoints of AB, ra, respectively and the circle with diameter Kl\, will run through H, H' and hence it will be LKH/\=1L, while it will also be KH=r, /\H=r'. Thus, Kl\ is constructed , as the hypotenuse of a right triangle with its perpendicular sides the given segments T, r'. Therefore, we are led to the construction below. (b). Construction. After we place on our piece of paper segment AB=2r and after we define midpoint K of segment AB, we also define point I\ (let's say towards B), such that Kl\ would be equal to the hypotenuse of a right triangle with perpendicular sides

T,

r ' (K/\=

,/T 2 + T" 2 ) and consequently we easily define r ,

a, on both sides of/\, as it is 11.r=M=r'. We say that the group of four points A , B , r, a , is the desired harmonic one.

ttal:fGffil From their construction , obviously it is AB=2r

Kai

r.t.=2r', hence we only

need to prove that it is also

(1 ).

We draw circles with diameters AB, ra that intersect at points H, H', thus in triangle HK/\ it is HK=r, H/\=r' and Kl\= .JT2 + T' 2 or Kl\= .JKH 2 + /\H 2 or

KH 2 +H/\2=K/\2.

(2).

1st Way (Based on Criterion 69). Relation (2) means that triangle HK/\ is right at H and thus the circle with diameter Kl\ runs through H. This, according to the sufficient of Criterion 69, means that the group of four A, B, r, a , is harmonic. I.e., relation (1) is in fact true. nd

Way (Analytical).

Relation (2) means that triangle HK/\ is right at H, or that circles (K), (A) are

18=-·__

_,_H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=a rt '-'-= B.

orthogonal. Therefore, since AB is a diameter of circle (K), circle (,\) will harmonically intersect AB at points r , fl , or that r , fl harmonically intersect AB, orthat relation (1) is in fact true.

3rd Way (Based on the Power of a Point with respect to a Circle). From (2), that we proved to be true, it occurs that triangle HK/\ is right at H and thus KH is perpendicular at radius /\H of circle (A) , or that KH is tangent at circle (A). This means that KH 2 or KB 2 or KA 2 or r 2 is the power of point K with respect to circle (A). Thus, it will be KB 2=KA 2 =Kr.Kil. This shows that, r , fl harmonically divide AB (converse of the known Newton's Theorem). Consequently we see that Problem 89, can easily be solved, through tht construction of Center H of the Regular Pencil H(ArBll), which is constructed and proven as we mentioned above (analysis and construction). (d). Investigation. We easily realize that this problem always reaches two solutions. The first is the one we described above and the other occurs, if /\ is received towards the side of A, hence r, will be located between K and A, while fl , will be located to the left of A, while figure 55 will be symmetrical to the new one with respect to the perpendicular line to AB at K.

lil411Fhb (a). It is obvious that Construction 89 is a generalization of Construction 90, that follows. (b). We easily realize that Construction 89, is true for a circle as well (I.e. when A, r, B, fl are cocyclic points) [See Construction 138 (§19.53)]. (c). Construction 89 we believe to be new and firstly appear here. (d). Relevant Constructions are 69, 90, 137, 138.

15.34. Construction of a Special Collinear Harmonic Group of Four Points. Special Case of the Geometrical Construction 89.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Construction 90 (Figures 62 and 63). Line segment AB=2r should be harmonically divided by pair of points r, /1 (r is located between A , B and /1 outside of the segment), for which (pairs) it would be: AB=r/1=2r (r is a given segment), or in another formation: Construct a Harmonic Group of Four Points A, B, r , /1, such that AB=r/1=2r (where

T

given segment).

Moreover, if K, A are the midpoints of AB, ra respectively, measure segment KA according tor and prove that sum Kr+AB is equal to the side of the circumscribed regular octagon in the circle with diameter AB=2r.

(K

K

(E)

First desiratum . Solution 1st (Figure 62). It is obvious that the first desiratum of Construction 90, is a special case of Construction 89. That is why we will work similarly to Construction 89. Since here we have a Harmonic Group of Four, to solve this Problem , we will use the new method of Regular Central Pencils.

ttiW,ttml4JhtJ To apply this method, we think in the following way: We suppose that the desired Harmonic Group of Four is constructed and it is the following one A, r , B, /1. This will have AB=r/1=2r.

18=-·__

_,_H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

We draw circles with diameters AB=2r or (K) and rt1=2r or (A), which intersect at a pair of points symmetrical, with respect to AB, H, H' (Orthooptical Points). Pencil H(ArB/1) is Regular, as it is Orthooptical, since AB, ra are seen from H, H' obviously in right angles and it is on the Harmonic Group of Four Points A , B, r , /1 (§ 12.1 , § 12.2 and § 12.13). Thus, according to Criterion 69 (necessary), if K, /1. are the midpoints of AB, ra, respectively and the circle with diameter K/1., will run through H, H' and hence it will be LKH/\.=1L, while it will also be KH=H/1.=r. Thus, Kl\. is constructed, as the hypotenuse of an isosceles right triangle with its perpendicular sides the given segment T. We would reach the same result if we think analytically as follows: Since each circle from these, harmonically intersects the diameter of the other one, they will be orthogonal. Hence it will be KH .lH/1. and thus KH 2+Hll.2=K/\. 2=2r 2,or K/1.=r ..J2 . Since

T

(1).

is known, Kl\. can be constructed in the known way (as the diagonal

of a square with side r), thus, since K is known, we can easily define /1. as well. As well as r , /1. Thus, we are led to the construction below. (b). Construction . After we place on our piece of paper segment AB=2r and after we define midpoint K of segment AB, we also define point /1. (let's say towards B), such that Kl\. would be equal to the hypotenuse of an isosceles right triangle with perpendicular sides segment r (K/1.=r ..J2 ), thus midpoint /1. of ra is defined and consequently we easily define r, /1, on both sides of/\., as it is 11.r=M=r. We say that the group of four points A , B, r, /1, is the desired harmonic one.

ttll&I From their construction, it obviously is AB=r/1=2r, hence we only need to (2).

prove that it is also

We draw circles with diameters AB, ra that intersect at point H, H', thus in triangle HK/\ it is HK=H/1.=r and Kl\.= .Jy 2 + T 2 or Kl\.= .JKH 2 + /\H 2

or

KH 2 +Hll.2=Kll.2.

(3).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

1st Way {Based on Criterion 69). Relation (3) means that triangle HK/\ is right at H and thus the circle with diameter Kl\ runs through H. This, according to the sufficient of Criterion 69, means that the group of four A, B, r , a , is harmonic. I.e., relation (2) is in fact true.

nd

Way (Analytical).

Relation (3) means that triangle HK/\ is right at H, or that circles {K), (A) are orthogonal . Therefore, since AB is a diameter of circle (K), circle {A) will harmonically intersect AB at points r , a , or that r , a harmonically intersect AB, or that relation (2) is in fact true.

3'd Way Based on the Power of a Point with respect to a Circle).

From (3), that we proved to be true, it occurs that triangle HK/\ is right at H and thus KH is perpendicular at radius /\H of circle {A), or that KH is tangent at circle (A). This means that KH 2 or KB 2 or KA 2 or r 2 is the power of point K with respect to circle (A). Thus, it will be KB 2=KA 2 =Kr.Ka. This shows that, r , a harmonically divide AB (converse of the known Newton's Theorem). Consequently we see that Problem 90, can easily be solved, through the construction of Center H of the Regular Pencil H(ArBl::l.), which is constructed and proven as we mentioned above (analysis and construction). (d). Investigation. We easily realize that this problem always reaches two solutions. The first is the one we described above and the other occurs, if /\ is received towards the side of A, hence r , will be located between K and A, while a , will be located to the left of A, while figure 55 will be symmetrical to the new one with respect to the perpendicular line to AB at K.

ld4eeFhA (a). If 0 is the intersection of HH' and Kl\, then we easily find that from H the s ix segments AK, Kr, re, 08, Bl\, M , are seen in angles with measure equal to w=22,5°.

IBl=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. In fact, since as we saw above pencil H(ArBL'1) is Regular, it will be: LAHr=LrHB=LBHL'1=45°. Additionally, since triangle HKA is right isosceles it will be LHKA= LHAK= LKH0= ..l/\H0=45°, hence also LKAH= LKHA= w= 22,5°.

== LKHr=w=22.5°. LKHr=22.5° == LrH0=w=22.5°.

Moreover, since LAHr=45° and LAHK=22.5° Moreover, since LKH0=45° and

Therefore, LAHK= LKHr= LrH0=w=22,5°.

(4).

Similarly we find that it will also be LBH0=LBHA=LAHL'1=w=22,5°. After what we mentioned above, it occurs that figu re 62, is symmetrical with respect to H0. (b). We easily find that here it also is: (ArKA)=1, (l'1BKA)=1 (hence K, A harmonically

intersect

both

(K/\0 00 )=1 , (hence 0 ,

00 ,

segments

Ar,

8£1),

(Al'10 00 )=1 ,

(Br0oo)=1 ,

harmonically intersect all three segments Al'1, Br,

KA) and (K0rl'1)=1 , (A0BA)=1 . In fact, these are true, as in figure 62 we have the bisectors of five angles of triangles with common apex H that intersect Al'1 at K , A, 0 , r , B and it is KH .lHA, H0 .1H 00 , AH .lHB, rH .1Hl'1.

Ar

All

(c). Moreover, since. rs= !lB , Ar=Bl'1

Solution

== Ar2=Bl'12=Al'1.Br.

2nd ·(Figure

z We suppose that we defined points r , a , which,

harmonically

divide pair of points A , Band it is: AB=ra=21. We draw the circle with diameter

E

.rxr11.1a 63.

AB=2T. (1). We draw the perpendicular diameter EZ to AB at K.

fil1] .

Collinear and Coe clic Harmonic Grou s of n-members.

If Er intersects the circle at H and ZH intersects AB at a•, then since HE is obviously the internal bisector of angle AHB and HZ .1 HE, then HI::,.' will be the external bisector of the same angle AHB. Thus !::,.' would be conjugate harmonic to r with respect to A , B. But also a is conjugate harmonic to r with respect to A, B, due to the implication. This means that /::,.'=/::,.. Right triangles Hra, HZE are equal, since LHEK=LH/::,.K (from cyclic quadrilateral KHI::,.E) and EZ=AB=ra=2r (from the implication). Hence, these will have ZH=Hr,

(2).

And since HA is the bisector of the right angle ZHE, HA will be the perpendicular bisector to rz and since HB .1 AH :::::zrnHB ::::: LZrK= L. HBK.

(3).

Moreover it also is LZrK=LKrE =LHrB or LZrK= LHrB.

(4).

Hence, from (3), (4) ::::: LHrB= LHBr::::: Hr =HB.

(5).

Therefore from (2), (5) "" ZH=HB ::::: arc ZH=arc HB, hence we are led to the construction below. (b) . Construction . After we draw the circle with diameter AB and draw the diameter EZ perpendicular to AB, we define midpoint H of arc 28. If ZH intersects AB at a and EH intersects AB at r, we say that the desired points are r, a .

ttalittffil Since Hr, in the right triangle HAB, is the bisector af angle AHB and EH .1 za, we conclude that r, a harmonically intersect segment AB. We are only left to prove ra=AB=2r. Since LHBA=LHBZ +LZBA =LHZB+LBZE= LHZE or L HBA =LHZE = LHBr :::::LHZE=LHBr.

(6).

Additionally, for cyclic quadrilateral KZHr, it is: LKZH=LHrB or LHrB=LHZE.

(7).

Hence, from (6), (7) :::::LHBr=LHrB,

(8).

and

(9).

Hr=HB.

Consequently we see that right triangles HBA and Hra have Hr=HB and LHra=LHBA [relations (8) and (9)), hence they are equal and thus ra=AB=2r.

lfffiil=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

(d). Investigation. We easily realize that this problem always reaches two solutions. The first is the one we described above and the other occurs, if H is received at the midpoint of arc AZ, hence r , will be located between K and A, while 11, will be located to the left of A, while figure 63 will be symmetrical with respect to line KZ. General Remarks . (a). We easily realize that figure 62 above is a special case of figure 55, for the case where in figure 55, the bisector H0 and the median HZ coincide at the altitude Hn of triangle HBr or when the circles w ith diameters AB, r/1 are equal , or when triangle AH/1 is isosceles or when Ar=B/1 or if the secant (E) of the beams of the symmetrical pencil is perpendicular to its axis K0. (b). We rather easily realized that problem 90 above, is true in a circle as well [Construction 137 (§ 19.52)]. (c). It is obvious that Construction 90, is a special case of Construction 89, through which in Figure 62 line (E) and H0 are perpendicular, or when the altitude or H0 and the median of triangle KHA, coincide. (d) . Relevant Constructions are 89, 137, 138. Second Desiratum . Proof Figures 62 and 63 . 1'1 Way Figure 62 .

For the Harmonic Group of Four A, B, r , /1, according to known Newton' s Theorem it is: KA 2=KB 2 =Kr.K/1 or r 2=Kr(Kr+r/1)=Kr(Kr+2r)=Kr 2+2r.Kr or Kr 2+2r.Kr-r 2 =0. From the solution of this equation according to Kr, we find that: Kr=r( ..f2 -1 ). This, as it is known, shows that Kr is equal to half the side of the circumscribed octagon in circle with diameter AB=2r (rtwµupia [28], book 3, paragraph 244, page 147). But because of the symmetry of figure 62, that we proved in remark (a) of solution 1, it will be Kr=AB, hence Kr+ BA=2Kr= 2r( ..f2 -1). This means that this sum is equal to the side of the circumscribed octagon in circle with diameter AB.

m . Collinear and Coe clic Harmonic Grou

s of n-members.

2nd Way (Figure 63). In figure 63, we draw the tangents at points Z, H that intersect at point I, hence IZ=IH=half the side of the circumscribed regular octagon in circle with diameter AB (BAtm: rEwµupia [28), book 3, paragraph 244 page 147). IZ=I H= r( .vf2 -1 ).

Hence,

Moreover it is Kl lZH and EH lZH

( 1). ~

Kif/Hr.

But, L IHZ= L HEZ = L rEK or L IHZ= L rEK. Hence L IHr = L IHZ + L ZHE = L IHZ+1L = L rEK+1L= L rEK+ L rKE= L KrH or L IHr = L KrH . Therefore quadrialteral rKIH is an isosceles trapezium and thus if we take into consideration (1) as well, it will be :

Kr=IH=r(

.J2 -1 ).

(2).

Since, from the solution of the first desiratum above it is Ar=r/l/2=AB/2, it will be: K/\=Kr+r/\= r( -v'2 -1)+r= r -v'2 -r+r = r .J2 r'J Kl\= r .J2 .

(3).

.J2 -1)=r .J2 -T, then it will be: Br=KB-Kr=r-Kr=r-(T .J2 -r)=r-T .J2 +r=2r-T.J2 , and B/\=K/\-Kr-Br=r .J2 -r .J2 +r+r .J2 -2r=r .J2 -r=r( .J2 -1)=Kr. (4).

Additionally, since because of (2), it is: Kr=r(

Thus, from (2), (4) ~ Kr+B/\=2.r( -v'2 -1)=1N , which is the side of the circumscribed regular octagon , in circle with diameter AB. Final Remarks. (a). We easily realize that segment Br is a side of a square w ith its diagonal the side of the circumscribed regular octagon in circle with diameter AB=2r, as in figure 63 above, IN forms with Br an angle of 45° and Br is the right projection of IN on line Br. (b). Construction 90 we believe to be new and firstly appear here.

15.35. New easy purely Geometrical proof, of the known Newton 's Theorem (condition), using the New Method of «Regular Central Pencils». Criterion of Harmonic being of a Group of Four points.

lml=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Criterion 91 (Figure 55). The sufficient and necessary condition, for four ordered collinear points A, r, B, /1, to constitute a harmonic group of four, is: KA 2=KB 2=Kr.K/1, or

(1). (2).

(where K, A are the midpoints of segments AB, r11, respectively).

own t4ffl('t1tJ Since here we have a Harmonic Group of Four, to prove thi Proposition, we will use the new method of Regular Central Pencils with egress from the line to its planes. To apply this method, we work in the following way: We suppose that the group of four A, B, r, /1, is harmonic and has AB=2r and rt1=2r'. We draw circles with diameters AB=2r or (K) and r/1=2r' or(,\), which intersect at pair of symmetrical, with respect to AB, points H, H' (Orthooptical Points). Pencil H(ArB/1) is Regular, as AB, r/1 are seen from H, H' obviously in right angles and it is on the Harmonic Group of Four Points A, B, r, /1 (§ 12.1, §12.2 and §12.13). Thus, according to Criterion 69 (way 2, necessary), the circle with diameter KA, will run through H, H' and thus it will be LKH/\=1L, while it will also be KH=r, /\H=r'. Therefore, we are led to the proof below.

itl~l434iflh If points A, B, r, /1 constitute a harmonic group of four, we will prove that relation (1) is true. In fact, since, as we saw above triangle HK/\ is right, then KH will be tangent on circle (,\) at point H. Thus KH 2 , will be the power of K with respect to circle (,\) and since KA=KB=KH=r, it will be KA 2 =KB 2=r 2 =KH 2 =Kr.K/1, hence relation (1).

m . Collinear and Coe clic Harmonic Grou

s of n-members.

~-1ff0[3[4,il If relation (1) is true, we will then prove that A, B, r , 11, constitute a harmonic group of four. Here as well, we lay the relevant foundation (analysis) and define H, as mentioned above, but because here we do not know if the following group of four A, B, r , 11, is harmonic, it does not occur that triangle HK/\ is right. We are asked to prove that initially. In

fact,

since

(1)

is

true

and

it

is

KA=KB=KH=r,

it

will

be:

KA 2 =KB 2=r 2 =KH 2=Kr.K/1 or KH 2=Kr.K/1. This shows that KH is tangent on circle (A), at point H. Thus, KH .LAH, hence triangle HK/\ is actually right at H. Therefore, the circle with diameter KA, will run through H, thus, according to the sufficient of way 2 of Criterion 69, the group of four A, B, r , 11, will in fact be harmonic. In the exact similar way, we prove that the Proposition above is true for the case where we would use relation (2) instead of (1 ), too.

liE~rJ81K) is harmonic.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

17.52. Closed Cocyclic Harmonic Group of Five Points or Harmonic Penta-

~ If we inscribe pencil r(ayu11) in circle (o) and if the beams of this pencil intersect circle (o) at points A, r , E, H, I, respectively, while if the beams of pencil r(al3yliE~l')81K), intersect circle (o), at points A, B, r , /1, E, Z, H, 0 , I, K and if these are apexes to a Harmonic Decagon, then this convex pentagon ArEHI is called Harmonic Pentagon.

17.6. Closed Harmonic Central Pencils of n-Beams and Harmonic n-laterals. In equivalent ways, that in the above cases 17.1 to 17.52 of Harmonic Pencils and Harmonic Multilaterals, we gave the relevant definitions, we also define the Closed Harmonic Central Pencils of n-Beams and Harmonic nlaterals.

17.7. General Conclusions. Due to t he above (paragraphs 11, 12, 12.13, 16 and 17), the following occur: a . Regarding the Closed Harmonic Central Pencils. 1/. The construction of Harmonic Pencils, depends on the ability of construction of the respective Harmonic Collinear groups of n (and theirs , as we have already mentioned, from the respective Regular Multilaterals). Th is is after all one of the reasons that we named them «Harmonic Pencils». 2/. Apart from the Closed Harmonic Pencils, Open ones exist as well, to which though we will refer later(§ 23). 3/. As it occurs from the aforementioned, from each intersection of a Closed Harmonic Pencil of n beams with an arbitrary line, a Collinear Closed Harmonic Group of n occurs. 4/. If we inscribe these Pencils, in circles, then their beams that intersect these circles , that they are inscribed in, at four, six, eight, ten, five, etc, points which are the apexes to a Harmonic, quadrilateral, hexagon, octagon, decagon, pentagon, etc, respectively (This property is mainly the other reason why we have given them their name as «Harmonic Pencils»).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

(b). Regarding the Closed Collinear and Cocyclic Harmonic Groups of n-

1/. It is obvious that the Construction of the Harmonic groups of n (collinear and cocyclic), depends on the ability to construct the respective Regular Pencils , which depend on the ability of construction of the respective regular multilaterals.

2/. Apart from the Closed Harmonic Groups of n (collinear and cocyclic), Open ones exist as well , to which though we will refer later(§ 23). 3/. It is obvious and easily proven that if a Pencil Closed Harmonic of n Beams is on a Closed Collinear Harmonic Group of n Points, whose center is not obligatorily its Orthooptical Point, will have n groups of four of its Beams harmonic, as these will be on n Harmonic Groups of Four Points, respectively. Then this pencil obviously is not always Orthooptical and it is called just «Harmonic Closed Central Pencil of n Beams» (non Regular). These pencils, if we inscribe them in circles, then they intersect them at points that are the apexes to Harmonic Multilaterals, as we will later prove, based on Criterion 35 § 8.72. (c). Regarding the Inscribed in Circle Harmonic Pencils. It is obvious that the angles of each Harmonic Pencil of n Beams that are on diagonals (main) of a Harmonic Collinear Group of n, will be on respective diagonals of a Harmonic Cocyclic n-lateral. d . Descri tion-lnter retation of Figures 48 and 49 . Based on the above and paragraphs 11 , 12, 12.13, 16 and 17, we can now understand the description-interpretation of figures 48 and 49 better, that we mentioned in paragraph 12.13 (d), in which we can see many of the above mentioned and that is why we again advise their thorough study.

1#34uFfid (a). We will see and prove the above in Propositions and Problems that will follow, as applications of these. (b). Later, based on figures 48 and 49 and as we mentioned above, we will refer to the method and the interpretation of the method that a Harmonic Multilateral occurs from a Regular (Multilateral), with equal number of

~ . Collinear and Coe clic Harmonic Grou s of n-members.

sides, which also consists the solution to a Problem, that troubled us rather often, for at least twenty years (§ 20). We call this New Method, «Method of the Harmonic Transformation»(§ 21).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

G'. METHOD OF THE HARMONIC CENTRAL PENCILS. 18. In Genera l. We referred to the methods that we came up with and apply in Harmonic Geometry in general in Part A (C' section, paragraphs 4 and 5), while in Part B (C' section , paragraph 13) we referred to the Method of Regular Central Pencils . Below we will only develop the method of Closed Harmonic Central Pencils, while to the method of Open Harmonic Central Pencils (§ 24) and the method of Harmonic Transformation(§ 21), we will refer later.

18.1. Method of Closed Harmonic Central Pencils. It is a general case of Closed Regular Pencils, as Regular Pencils are Harmonic as well , as we mentioned in paragraph 12.13, whose method, tool are the Closed Harmonic Central Pencils (§ 17) and which we apply in Propositions and Problems that concern Collinear and Cocyclic Closed Harmonic Groups of n Series of Points (Collinear Harmonic Multilaterals and mostly in Problems about Cocyclic Harmonic Multilaterals). Their principles and the way of their application, are similar to the ones of the method of Collinear Symmetrical Pencils (§ 5.3) and this is why we will not mention it here. To understand this method, many relevant applications will follow, in the form of Propositions and Construstions.

~ - Collinear and Coe cl ic Harmonic Grou s of n-members.

19. APPLICATIONS OF THE METHOD OF HARMONIC PENCILS AND NOT

[•]~111il#t411 In General. Consequently we will proceed with applications of the method of Harmonic Pencils , which is mentioned in paragraph 18 and not only that, in the form of Geometrical Propositions and Constructions, with their proofs , as we will need them to depend on them the proofs of other Propositions, etc. Moreover on them and the ones in paragraph 15, the New Method of Harmonic Transformation, w ill be based, to which we will refer later(§ 21 ). Simultaneously to the proofs-solutions that we give with the application of the above, often we will give more, that do not include the application of these, mostly in the purpose of comparison and the proof of the huge value of these. As the first Criterion of this paragraph, we whould have placed Criterion 35, which though, due to the needs of this book, was preceded and placed in paragraph 8. 72.

19.1. First Criterion of Harmonic Beaing of an Inscribed Hexagon. Extension to Criterion 35 to a Harmonic Hexagon. Criterion 96 (Figure 69). Every convex inscribed in circle (o) hexagon ABr.t.EZ, is Harmonic, if and only if, pencil I:(ArB.t.EZ) that its center is an ar bitrary point I: of its circumscribed circle (o), is harmonic.

[Eil•Jii431 I.e., if hexagon ABr.t.EZ is harmonic, then pencil I:(ArB.t.EZ) is harmonic (Ter minology§ 16.1b and 17.2). In fact, since hexagon ArB.t.EZ is harmon ic, according to the relevant definition of harmonic being (§ 16.1 b) of the inscribed hexagon , all its six quadrilaterals below are harmonic:

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

ABrE, Br~, raEA, aEZB, EZAr, ZABa.

(1 ).

Hence, according to the direct of criterion 35 (§ 8. 72 or § 17.1 ), the pencils of four beams below will be harmonic, and the above six quadrilaterals are inscribed as well: r(ABrE), r(Br~), r(raEA), r(aEZB), r(EZAr), r(ZABa).

(2).

Thus, if with a line (E) we intersect pencil r(ABraEZ) of six beams, at points a, 13, y, 5, E, ~. respectively, pencil r(al3y5E~) will occur, for which since (2) are true and according to what we know, the groups of four points below will be harmonic: a, 13, y, E - 13, y, 5, ~ - y, 5, E, a - 5, E, ~. 13- E, ~. a, y - ~. a, 13, 5.

(3).

Consequently, according to the relevant definition of harmonic being (§ 11 .1b), the group of six points a, 13, y, 5, E, ~. is harmonic. This, according to the definition of the harmonic central pencil of six beams (§ 17.21), pencil r(al3y5t~), or pencil r(ABraEZ), is harmonic.

tl:lfi•leN4l-i4 I.e., if pencil r(ABraEZ), is harmonic, then hexagon ABraEZ, is harmonic. To prove this, we follow the exactly reverse way. I.e. : Since pencil r(ABraEZ), or pencil r(al3y5t~)) is harmonic, if a line (E) intersects it, according to the relevant definition (§ 17.21), then the collinear group of six points a, 13, y, 5, £, ~. will be harmonic as well, thus according to the relevant definition (§ 11.113) all six groups of collinear points below, will be harmonic: a, 13, y, E - 13, y, 5, ~ - y, 5, E, a - 5, E, ~. 13- E, ~. a, y - ~. a, 13, 5.

(3).

Thus, due to what we know and because of (3) the pencils below, will be harmonic too: r(ABrE), r(Br~), r(raEA), r(aEZB), r(EZAr), r(ZABa).

(2).

Hence, since these six pencils of four beams are harmonic, according to criterion 35 (§ 8.72 or§ 17.1), they will intersect circle (o), at six points A, B, r, a, E, Z, for which, the six groups of four of its cocyclic points below will be harmonic (or that are apexes to six harmonic quadrialterals), as these

~ . Collinear and Coe clic Harmonic Grou s of n-members.

are also inscribed in circle (o): A, B, r, E - B, r , ll, Z - r , ll, E, A - ll, E, z, B - E, z, A , r -

z, A, B, ll.

(1 ).

This, according to the relevant definition of harmonic being (§ 16.1b), means that hexagon ABrllEZ will be harmonic. (c). Investigation. This criterion is also true in the special case where r will coincide with an apex of the hexagon. In fact , if r coincides with for example apex A, then beam rA of pencil r(ABrllEZ) w ill obviously coincide with the tangent x'Ax at point A, thus

L'AI:B = 4:AB , L'.Bff = Li3Ar, .. ., L:'.EI:Z = L'.EAZ , L2I:A = LZ.Ax' . This means that, all the angles formed between the consecutive beams of the two pencils r(ABrllEZ) and A(xBrllEZ) are equal , one by one. Thus, since the first one of the pencils is harmonic, then the second will be too.

The Criterion above is significant, as it has many applications, which we will see below and because it allows us to give the alternative definition of the Harmonic Central Pencil of six beams below, apart from the until now known one(§ 17.2): «A Central Pencil of Six Beams, is called Harmonic, if we inscribe this pencil in a circle and its intersections to it are apexes to a Harmonic Hexagon».

1#411Fhd (a). As we will later see, equivalent Criteria , are true for other Harmonic Multi laterals, as the Harmonic Octagon , Decagon, etc. (b). As we saw in the conclusion above, with this Criterion we extend the definition of the Harmonic Pencils from a line to a circle. (c). Later we will mention other Criteria and Construction of Harmonic Hexagons . (d). This Criterion we believe to be new and firstly appear here.

18=-·__

_,_ H:.:..A=R=M"-'O=N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

(e). The 2 nd, 3rd , 4th , 5th , 6th Criteria of the Harmonic Hexagon, are given in Propositions 109, 112, 139, 140, 141, respectively. (f). Equivalent Criterion, concering a Regular Hexagon, is Criterion 76. (g). Relevant Propositions are 35, 76, 97, 98, 109, 112, 139, 140, 141 .

19.2. Criterion of Harmonic Being of an Inscribed Octagon . Extension to Criterion 96 to a Harmonic Octagon. Criterion 97 (Figure 70). Every convex inscribed in circle (o) octagon ABr.dEZH0, is Harmonic, if and only if, pencil r(ArB.dEZH0) that its center is an arbitrary point r of its circumscribed circle (o), is harmonic.

tD1•Jli4dl I.e., if octagon ABr.dEZH0 is harmonic, then pencil r(ArB.dEZH0) is harmonic (Terminology§ 16.1c and 17.3). In fact, since octagon ArB.dEZH0 is harmonic, according to the relevant definition of harmonic being (§ 16.1 c) of the inscribed octagon, all its eight quadrilaterals below are harmonic: ABrz, Br.dH, r.dE0, .dEZA, EZHB, ZH0r, H0A.d, 0ABE.

(1 ).

Hence, according to the direct of criterion 35, or§ 17.1 , the respective pencils of four beams below will be harmonic too, as the eight quadrilaterals above are inscribed : rcArBZ), r(B.drH), r(rE.d0), r(.dZEA), r(EHZB), r(Z0Hr), r(HA0.d), r(0BAE).

(2).

Thus if we intersect pencil r(ABr.dEZH0) of eight beams with line (E), at points a,

13, y, l'i, E, { , r,,

8, pencil r(al3yl'iE{r,8) will occur, for which according

to what we know and because of relation (2), the groups of four points below will be harmonic: a, 13, V, E, { , r,,

{-13, V, l'i, r, -y, l'i, E, 0 13 - {, r,, 8, y - r,, 8, a, l'i -

l'i, E, {, a 8, a,

13,

E.

(3).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Therefore, according to the relevant definition in § 12.6, the group of eight points a,

13, y , l>, E, ~. 11, 8, is harmonic.

This, according to the definition of the harmonic central pencil of eight beams(§ 17.31), pencil r(al3yl>E~l18), or pencil r(ABraEZH0), is harmonic.

tl:li3•leW4t14 I.e., if pencil r(ABraEZH0), is harmonic, then octagon ABraEZH0, is harmonic too. To prove this , we follow the exactly reverse way. I.e.: Since pencil r(ABraEZH0) is harmonic, if a line (E) intersects it, according to the relevant definition(§ 17.31), then the collinear group of eight points a,

13, y,

l>, E, ~. 11, 8, will be harmonic as well, thus according to the relevant

definition in paragraph 12.6 all eight groups of four collinear points below, will be harmonic: a,

13, v, ~ - 13, v, l>, 11 - v, l>, E, 0 13 - ~. 11, 8, y -11 , 8, a , l> -

E, ~. 11,

l>, E, ~. a 8, a,

13, E.

(3).

Thus, due to what we know and because of (3) the pencils below, w ill be harmonic too: r(A r BZJ, r(sarHJ, r(rEa0J, r(azEAJ, r(EHZB), r(Z0Hr), r(HA0f.), r(0BAE).

(2).

Hence, since these eight pencils of four beams are harmonic, according to the converse of criterion 35, or§ 17.1 , they will intersect circle (o), at eight points A, B, r , a, E, Z, H, 0 , for which, the eight groups of four of its cocyclic points below will be harmonic (or that are apexes to harmonic quadrialterals), as these are also inscribed in circle (o): A, B, r , Z - B, r , a , H E, z, H, B -

z, H, 0 , r

r , a , E, 0

- a , E, Z, A -

- H, 0 , A, a - 0 , A, B, E.

(1 ).

This, according to the relevant definition of harmonic being (§ 16.1c), means that octagon ABraEZH0 will be harmonic. c . Investigation. This criterion is also true in the special case where r coincides w ith an apex of the Octagon.

81=-·__

_,_H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

In fact, if r coincides with for example apex A, then beam rA of pencil r(ABraEZH0) will obviously coincide with the tangent x'Ax at point A, thus

LAI:B = L){AB, &rr

= L'.BAr, .. , L'.HI:0 = L'.HA0,

L0I:a

= L0HA =

LE>Ax'. This means that, all the angles formed between the consecutive beams of the two pencils r(ABraEZH0) and A(xBraEZH0) are equal, one by one. Thus, since the first one of the pencils is harmonic, then the second will be too.

The Criterion above is significant, as it has many applications, which we will see below and because it allows us to give the alternative definition of

the Harmonic Central Pencil of eight beams below, apart from the until now known one(§ 17.3): «A Central Pencil of Eight Beams, is called Harmonic, if we inscribe this pencil in a ci rcle and its intersections to it are apexes to a Harmonic Octagon».

1#34ee60A (a). As we will later see, equivalent Criteria, are true for other Harmonic Multilaterals, as the Harmonic Decagon, etc. (b). As we saw in the conclusion above, with this Criterion we extend the definition of the Harmonic Pencils from a line to a circle. (c). This Criterion we believe to be new and firstly appear here. (d). Equivalent Criterion, concering a Regular Octagon, is Criterion 77. (e). As mentioned and proved in paragraph 16.1c the respective to ABraEZH0, asteroid octagon AaHBE0rZA, is harmonic as well. (f). Relevant Propositions are 35, 77, 81, 96, 98, 101.

19.3. Criterion of Harmonic Being of the Inscribed Decagon . Extension to Criterion 97 to a Harmonic Decagon.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Criterion 98 (Figure 71 ). Every convex inscribed in circle (o) decagon ABraEZH01K, is Harmonic, if and only if, pencil I:(ArBaEZH01K) that its center is an arbitrary point

r of

its circumscribed circle (o), is harmonic. (Equivalent Proposition is true for an Asteroid Decagon AaHKrZHBE0A, Proposition 108).

1/. Regarding the convex Decagon.

[E11•Jlk431 I.e., if decagon ABraEZH01K is harmonic, then pencil I:(ArBaEZH01K) is harmonic (Terminology§ 16.1 d and 17.4). In fact, since decagon ArBaEZH01K is harmonic, according to the relevant definition of harmonic being (§ 16.1d) of the inscribed decagon, all its ten quadrilaterals below are harmonic: ABrH, Brae, rnEI , aEZK, EZHA, (1 ).

ZH0B, H01r, 01Ka, IKAE, KABZ.

Hence, according to the direct of criterion 35 (§ 8.72), or§ 17.1, the respective ten pencils of four beams in relation (1) below will be harmonic too, as the ten quadrilaterals above are inscribed: I:(ABrH), I:(Bra0), I:(raEl),I:(aEZK), I:(EZHA), I:(ZH0B), I:(H01r), I:(01Ka), I:(IKAE), I:(KABZ).

(2).

Thus if we intersect pencil I:(ABraEZH01K) of ten beams with line (E), at points a,

13, y,

6, E, {, 11, 8, ,, K, pencil I:(al3y6E{1181K) will occur, for which ac-

cording to what we know and because of relation (2), the ten groups of four points below will be harmonic: a,

13, V,

{, 11, 8,

11- 13, V, 6, 8 -y, 6, E,

13 -

11, 8,

I,

y - 8,

I,

1-

K, 6 -

6, E, {, I,

K-

E, { , 11, a -

K, a, E - K, a , 13,

{.

(3).

Therefore, according to the relevant definition of harmonic being(§ 11.1d), the group of ten points a,

13, y,

6, E, {, 11, 8, ,, K, will be harmonic.

18=-·__

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

This, according to the definition of the harmonic central pencil of ten beams (§ 17.41), pencil r(al3yl>t~1181K), or pencil r(ABraEZH01K), is harmonic. b). Converse. I.e., if pencil r(ABraEZH01K), is harmonic, then decagon ABraEZH01K, is harmonic too. To prove this, we follow the exactly reverse way. I.e. : Since pencil r(ABraEZH01K), or pencil r(al3yl>t~1181K) is harmonic, if a line

(E) intersects it, according to the relevant definition (§ 17.41), then the collinear group of ten points a,

13,

y, l>, E, ~. 11, 8, 1, K, will be harmonic as well,

thus according to the relevant definition (§ 11.1d) all ten groups of four collinear points below, will be harmonic : a,

13, V, 11 -13, V, l>, 8 - V, l>, E, 1- l>, E, ~. K- E, ~. 11, a 13 - 11, 8, I, y - 8, I, K, l> - I, K ,a, E - K, a, 13, ~-

~. 11, 8,

(3).

Thus, due to what we know and because of (3) the ten pencils below, will be harmonic too: r(ABrH), r(Brn0), r(rnEI), r(aEZK), r(EZHA) r(ZH0B), r(H01r), r(01Ka), r(IKAE), r(KABZ).

(2).

Hence, since these ten pencils of four beams are harmonic, according to the converse of criterion 35 (§ 8. 72), or § 17.1 , their beams will intersect circle (o), at ten points A, B, r , a, E, Z, H, 0, I, K, for which, the ten groups of four of its cocyclic points below will be harmonic (or that are apexes to harmonic quadrialterals), as these are also inscribed in circle (o):

z, H, A, 0, I, K, a-1, K, A, E- K, A, B, z.

A , B, r , H - B, r, a, 0 - r, a, E, I - a, E,

z, H, 0, B -

H, 0, I, r -

z, K -

E,

(1).

This, according to the relevant definition of harmonic being (§ 16.1 d), means that decagon ABraEZH01K, would be harmonic. (c). Investigation. This criterion is also true in the special case where r coincides with an apex of the decagon.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

In fact, if r coincides with for example apex I, then beam r1 of pencil I(ABr.6.EZH01K) will obviously coincide with the tangent x'lx at point I,

LAIB= LAIB , LBir = L'.Blr, LT"I.6. = Lfl.6., ... ,

thus :

L0II = L01X , LIIK = L)('IK, Ll·

Thus, relation (2) according to Auxiliary Proposition 7 (converse) (§ 2.7) in Part E, regarding Hexagon ABrZHe, means that the following relation is true:

AZnBHnre=J\.

(3) .

Moreover, since from the relevant definition this Decagon, will have its quadrilaterals Brae and Heir harmonic too, if we work in the exact same way as above we easily find that BHnrenal=/\.

(4).

Additionally, since from the relevant definition this Decagon, will have its quadrilaterals raEI and e1Ka harmonic too, if we work in the exact same way as above we easily find that rena1nEK=/\.

(5).

Hence, from (3), (4), (5) relation (1) directly occurs.

2nd Way (Based on Auxiliary Proposition 51) . Since the given Decagon is Harmonic (implication), according to the definition (§ 16.1d above), its quadrilaterals ABrH and ZHeB will be harmonic, thus according to Auxiliary Proposition 51 (§ 9.12) (converse), it will be

AZnBHnre=/\.

(6).

Moreover, since due to the definition this Decagon, will have its quadrilaterals Brae and Heir harmonic too, if we work in the exact same way as bove, we easily find that

BHnrenal=/\.

(7).

Additionally, since due to the definition this Decagon , will have its quadrilaterals raEI and e 1Ka harmonic too, if we work in the exact same way as bove, we easily find that

r0na1nEK=/\.

Therefore, from (6), (7), (8) relation (1) directly occurs.

(8).

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

b). Regarding the Asteroid Decagon. It is obvious that the Asteroid Decagon A.6.HKrZIBE0A as well , will share its diagonals with the convex decagon ABr.6.EZH01K, hence for this Asteroid decagon as well relation (1) will actually be true (See Proposition 107). [As it is proven in Proposition 108 below(§ 19.16) the Asteroid Decagon is Harmonic too, as Harmonic is its respective convex Decagon] .

14411EIUA (a). Proposition 106, we believe, to be new and firstly appear here. (b). It is obvious that we would reach the same result, if we consider as harmonic any three other pairs of consecutive quadrilaterals, instead of the above ones, from the quadrialterals of the given Harmonic Decagon. (c). It is obvious that for Proposition 106, the converse is not true. (d). Untill now we have seen that, apart from the Harmonic Decagon, Harmonic Hexagons and Octagons have concurrent diagonals as well. (e). An analytical way of proof of Proposition 106, is given in paragraph 200 (§ 29.1), using the method of Harmonic Transformation. (f). Relevant Propositions are 51, 104, 105, 107, 108, 200.

19.14. Significant Property of the Asteroid Harmonic Decagon. Convergence of the Diagonals of the Asteroid Harmonic Decagon . Proposition 107 (Figure 73). Diagonals (main), of the Asteroid Harmonic Decagons , are concurrent.

1st Way Based on Auxilia

Proposition 7 in Part E.

We consider a Harmonic Asteroid Decagon A.6.HKrZIBE0A . We will provw that it is:

AZnBHnr0na1nEK=/\.

(1).

Since the given Decagon is Harmonic (implication), according to the relevant definition (§ 16.1d, above), its quadrilaterals for example ArE0 and Z0Kr will be harmonic, thus according to our known Theorem, it wil be:

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Ar E0 20 Kr Ar E0 20 Kr rE · 0A =1 and 0K · r2 - 1 == rE · 0A - 0K · r2 ·

(2).

Hence, from relation (2), according to Auxiliary Proposition 7 (converse) (§ 2.7) in Part E, it means that relation is true: AZnr0nEK=A.

(3).

Moreover, since from the relevant definition this Decagon, will have its quadrialterals rEHK and 0KBE harmonic too, if we work in the exact same way as above, we easily find that

r0nEKnHB=A.

(4) .

•.

........•••••••• ..... •••• ••••••• ......... ••••••••• .......... ............. ............ ............ ........... •

Ixrwa 73. K

............ ••••••••••••• ............... •••••••••••••• ................... ................. ••••••••••••••• ................. •••••••••••••••• .................. •••••••••••••••••• •••••••••••••••••• •••••••••••••••••••• •••••••••••••••••••• ...................... •••••••••••••••••••••• •••••••••••••••••••••• ......................... ••••••••••••••••••••••••• .......................... •••••••••••••••••••••••••• ::::::::::::::::::::::::::: •••••••••••••••••••••••••••• ••••••••••••••••••••••••••••• •••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••

Additionally, since from the relevant definition this Decagon, will have its quadrialterals EHIB and KBaH harmonic too, if we work in the exact same way as above, we easily find that

EKnHBnla=A.

Thus, from (3), (4), (5) relation (1) directly occurs.

(5).

IEEil=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

nd

Way Based on Auxilia

Proposition 51 .

Since the given Decagon is Harmonic (implication), according to the definition (§ 16.1 d above), its quadrilaterals for example ArE0 and Z0Kr will be harmonic, thus according to Auxiliary Proposition 51 (§ 9.12, converse), it will be

AznrenEK=/\.

(6).

Moreover, since due to the definition this Decagon, will have its quadrilaterals rEHK and 0KBE harmonic too, if we work in the exact same way as bove, we easily find that

renEKnHB=/\.

(7).

Additionally, since due to the definition this Decagon, will have its quadrilaterals EHIB and KB~H harmonic too, if we work in the exact same way as bove, we easily find that

EKnHBnl~=/\.

(8).

Thus, from (6), (7), (8) relation (1) directly occurs.

lti4eeFiUA (a). Proposition 107, we believe, to be new and firstly appear here. (b). It is obvious that we would reach the same result, if we consider as harmonic any three other pairs of consecutive quadrilaterals, instead of the ab ove ones, from the quadrilaterals of the given Harmonic Decagon . (c). It is obvious that for Proposition 107, the converse is not true. (d). Untill now we have seen that, apart from the Harmonic Asteroid Decagon, Harmonic Hexagons, Octagons, Decagons and Asteroid Harmonic Decagons have concurrent diagonals as well. (e). Relevant Propositions are 51, 104, 105, 106, 108.

19.15. General Conclusion . Due to Propositions, 104, 105, 106, 107, above and if we work s imilarly to their proofs, we reach the follow general conclusion : Equivalent Proposition, to Propositions 104 to 107, can be applied for any Harmonic 2n-lateral, as long as the respective Regular 2n-lateral is constructible (Convex or Asteroid).

~ . Collinear and Coe clic Harmonic Grou s of n-members.

19.16. Property of the Asteroid Harmonic Decagon. Harmonic being of the Asteroid Harmonic Decagons. Criterion 108 (Figure 73). The respective Asteroid Decagon, of each convex Harmonic Decagon , is also Harmonic (§ 16.1 d) and conversely.

itl•JH431 tQWMffll't1tJ I.e., if convex decagon ABrAEZH01K is Harmonic inscribed in circle (o), we will prove that decagon AAHKrZ1BE0A is also harmonic. In fact, if we take into consideration paragraphs§ 16.1d, § 12.8, § 17.4 and Criterion 98 (§ 19.3), then we are led to the proof below.

tGll :fl:ffi I Since, Decagon ABrAEZH01K is Harmonic, according to the direct of Criterion 98, pencil I:(ABrAEZH01K) or I:(al3y~E~1181K) w ill also be Harmonic [where

r is an arbitrary point on circle (o)].

Hence, the group of ten points a,

13, y,

~. E, ~. 11, 8, 1, K, that occurs from the

intersection of the beams of the pencil above, from an arbitrary li ne (E), is harmonic too. Thus, according to paragraph 12.8, it occurs that the group of ten points a, ~. 11, K, y , ~. 1,

13,

E, 8 will be Harmonic too. This means that pencil

I:(a~11KV~1l3t8) or I:(AAHKrZ1BE0) is Harmonic, hence the Asteroid Decagon AAHKrZ1BE0A, is Harmonic (converse of Criterion 98).

J,@i.)1@4141 To prove the converse, we follow the exact reverse way, of the proof above.

134eeFOA (a). Proposition 108, we believe, to be new and firstly appear here. (b). Relevant Propositions are, 98, 107.

81=-·__

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

19.17. Second Criterion of Harmonic Being of a Convex Inscribed Hexagon. Extension to Criterion 67, from a line to a Circle. Criterion 109 (Figure 64). If for a convex inscribed in circle hexagon ABrAEZ its following quadrilaterals are harmonic ABrE, rAEA,EZAr,orBraz,AEZB, ZABA,

(1 ).

then and only then this hexagon is harmonic.

tUl•Hi441 I.e., if the given hexagon ABrAEZ is convex, inscribed in circle (o) and the quadrilaterals of a group of three of its quadrilaterals in (1) are harmonic, then this hexagon is harmonic. We suppose that for example this hexagon has harmonic its quadrilaterals ABrE, rAEA , EZAr. We will prove that it is harmonic. According to known Theorem , regarding harmonic quadrilaterals ABrE, rAEA, EZAr, it will be respectively:

AB AE r.6. rA EZ Er =::::: Br =Er , - = .6.E AE ' ZA rA

AB• r.6. • EZ AB• r.6. • EZ AE•rA•Er 1, or 1. Br• .6.E • ZA Er•AE•rA Br• .6.E• ZA

(2) .

Relation (2), according to the converse of Auxiliary Proposition (Lemma) 5 (§ 2.5) in Part E, means that:

AAnBEnrZ=K.

(3).

[Relation (3), can easily be obtained if we work i n figure 72, similarly to way 3 in the proof of Proposition 104 (§ 19.11 ), after we suppose t h at for the given hexagon ABrAEZ its quadrilaterals ABrE, rAEA, EZAr are harmonic). Hence, s ince (3) is true then quadrilaterals: ABrE, rAEA, EZAr, are harmonic (definition), according to the above Auxiliary Proposition (Lemma) 34 (direct), the respective quadrilaterals AEZB, ZABA, Braz will also be harmonic . Consequently, all the quadrilaterals in (1), are harmonic, thus according to the relevant definition (§ 16.113), given hexagon ABrAEZ is harmonic.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

t613·l,\1§t1=1 We consider a Harmonic Hexagon ABrtiEZ. We will prove that its following quadrilaterals are harmonic: ABrE, raEA, EZAr, Braz, flEZB, ZABfl. According to definition (§ 16.1 b) of Harmonic Hexagons, all the above quadrilaterals in (1) are harmonic too, thus, it is obvious, that this is true.

Due to the Proposition about Harmonic Hexagons about, we reach the following conclusion: Every convex hexagon inscribed in circle (o) is harmonic, if the quadrilaterals of at least one of the groups of quadrilaterals in (1 l are harmonic. Consequently, from now on we can use for Harmonic Hexagons, the following Alternative Definition: «Harmonic hexagon is called each convex inscribed in circle hexagon ABrtiEZ, if the quadrilaterals in at least one of the two groups of three quadrilaterals below are harmonic: ABrE, raEA, EZAr or Braz, flEZB, ZABfl».

134ee60h (a). Criterion 109, we believe, that it is new and firstly appears here. (b). We would obviously reach the same result if we consider that the following quadrilaterals are harmonic Braz, flEZB, ZABfl. (c). The above Criterion of Harmonic being of Inscribed Hexagons, is rather significant, as it allows us to have a simple alternative DEFINITION, from our initial one, for Harmonic Hexagons. (d). The 1•t, 3rd , 4 th , 5th ,

5th,

Criteria about the Harmonic Hexagon, are given

in Propositions 96,112,139,140,141, respectively. (e). Extension to Criterion 109, with Criterion 141. (f). Relevant Propositions are, 94, 96, 110, 111, 112, 139-141.

19.18. Criterion of Harmonic being of a Convex Inscribed Octagon . Extension to Criterion 68 1 from a line to a Circle.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Criterion 110 (Figure 65). Every inscribed in circle convex Octagon, is Harmonic, if and only if, it has concurrent diagonals and four of its eight quadrilaterals are harmonic, those that from these four quadrilaterals, if we receive every two of them, they don't share a diagonal.

tEJl•Jl2441 Let there be the inscribed in circle (o) convex octagon ABraEZH0 for which it is:

AEnsznrHna0=K,

(1).

and four, of its eight quadrilaterals are harmonic, those that no pair of the four quadrilaterals has a shared diagonal with any other of these four quadrilaterals. We discern the cases below: 1/. During the first case four consecutive quadrilaterals of the octagon are harmonic, as for example if the following ones are harmonic: ABrz, sraH , raEe,aEZA.

(2).

In this case, since for example the inscribed hexagon ABrEZH it is AEnBznrH=K and its quadrilateral ABrz is harmonic, according to Proposition 34 (§ 8.71) (direct), quadrilateral EZHB will also be harmonic. We continue with the inscribed hexagon Bra2H0, whose quadrilateral sraH is harmonic and we find , as exactly above, that its quadrilateral ZH0r is harmonic too, as (1) is true too. Moreover, if we continue, as exactly above, we find that quadrilaterals : H0Aa, 0ABE, are harmonic. Thus, from the above we that the quadrilaterals in the group of four quadrilaterals below are harmonic: EZHB, ZH0r, H0Aa, 0ABE.

(3).

Thus, since octagon ABraEZH0 has all its eight quadrilaterals in (2) and (3) harmonic, according to the relevant definition(§ 16.1c), it will be harmonic.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

2/. During the second case the four harmonic quadrilaterals belong to one of the two groups of four below, or to another group of four with equivalent order: ABrz, sraH , H0Aa, 0ABE.

(4).

EZHB, ZH0r, raE0, aEZA.

(5).

Here, if we receive for example the quadrilaterals in (4) as harmonic and we work on hexagon ABrEZH, exactly as above, based on Harmonic Quadrilateral ABrZ then we find that quadrilateral EZHB is harmonic, since (1) is true as well. We continue, similarly to above for hexagon Bra2H0 and based on the harmonic quadrilateral sraH we find that quadrilateral ZH0r is harmon ic too, as (1) is true. Moreover, if we continue as exactly above, we find that quadrilaterals, raE0, aEZA, too are harmonic, thus all the quadrilaterals in (4) and (5), are harmonic and thus the given decagon is harmonic. 3/. Duri ng the third case the four harmonic quadrilaterals belong to one of the two groups of four below, or to another group of four of equ ivalent order:

ABrz, raEe, ZHer, 0ABE.

(6).

EZHB, H0Aa, BraH , aEZA.

(7).

Here, if t he quadrilaterals in for example (6) are received as harmonic and we work in hexagon ABrEZH , as exactly above, based on Harmonic Quadrilateral ABrZ, we find that quadrilateral EZHB is harmonic, as (1) is true too. We continue, as above for hexagon raEH0A as well and based on the harmonic quadri lateral raE0 we find that quadrilateral H0Aa is harmonic too, as (1) is true as well. Moreover, if we continue as exactly above, we find that quadrilaterals BraH , aEZA, are harmonic too, thus all the quadrilaterals in (6) and (7), are harmonic and hence the given decagon is harmonic. (We work in a similar way for any other order of its four given Harmonic Quadrilaterals).

l8J=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

tGlfi•l,fZ4t14 I.e., if the given octagon is harmonic, we will prove that then this will have harmonic, its ten quadrilaterals and (1) will be true. This is obvious, since the given octagon is harmonic, according to the relevant definition(§ 16.1c) and its eight quadrilaterals will be harmonic, while according to Proposition 105, (1) will also be true.

Due to the Proposition about Harmonic Octagons above (Criterion of Harmonic being of inscribed octagons), we reach the following conclusion : Every convex octagon inscribed in circle is harmonic, if its diagonals are concurrent and four of its eight quadrilaterals are harmonic, those that from these four quadrilaterals, if we receive each two of them, they don't share a diagonal. Consequently, from now on we can use for Harmonic Octagons, the following alternatice Definition : «We call Harmonic Octagon, each convex inscribed in circle octagon, if its diagonals are concurrent and four out of its eight quadrilaterals are harmonic, those that four of these quadrilaterals, if we receive any two of them, they don't share a diagonal ».

IU4eeEOb (a). The above Criterion of Harmonic being of Octagons, is rather significant, as it allows us to have a simpler alternative DEFINITION, from the initial one, for Harmonic Octagons, but because it also finds many applications, as we will later find out. (b). We believe that it is a new Criterion and that it firstly appears here. (c). According to the relevant definition(§ 16.1c), this Criterion is true for the Asteroid Octagon Ab.HBE0rZA as well. (d). Relevant Propositions are 34, 68, 109, 111.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

19.19. Criterion of Harmonic Being of a Convex Inscribed Decagon. Criterion 111 (Figure 66). Every inscribed in circle convex Decagon, is Harmonic, if and only if, its diagonals are concurrent and five out of its ten quadrilaterals are harmonic, those that from these five quadrilaterals, if we take each two of them, they don't share a diagonal.

[E11•Jlk431 Let there be the inscribed in circle (o) convex decagon ABraEZH01K for which it is:

AZnBHnr0na1nEK=A,

(1).

and five, out of its ten quadrilaterals are harmonic, those that no pair from these five quadrialterals has a shared diagonal with any other of these five quadrilaterals. We will prove that decagon ABraEZH01K, is harmonic. We discern the following cases:

1/. During the first case five harmonic consecutive quadrilaterals of the decagon are harmonic, as for example if the following are harmonic: ABrH, Brn0, raEI, aEZK, EZHA.

(2).

In this case, since for example in the inscribed hexagon ABrZH0 it is AZnBHnr0=A and its quadrilateral ABrH is harmonic, according to Proposition 34 (§ 8.71) (direct), quadrilateral ZH0B will also be harmonic. We continue with the inscribed hexagon BraH01, whose quadrilateral Brae is harmonic and we find, as exactly above, that quadrilateral H01r is harmonic as well, as relation (1) is true as well. Moreover, if we continue, as exactly above, we find that quadrilaterals: 01Ka, IKAE, KABZ, are harmonic too. Hence, from the above we see that the quadrilaterals in the group of five quadrilaterals below are harmonic:

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

ZH08, H01r, 01Ka, IKAE, KABZ.

(3).

Thus, since decagon ABraEZH01K has all ten quadrilaterals in (2) and (3) harmonic, according to the relevant definition(§ 16.1d), it will be harmonic. 2/. During the second case its five harmonic quadrilaterals are ordered, one by one, thus then we have the two groups of five quadrilaterals below, that at least one of them consists of harmonic quadrilaterals : ABrH, raEI, EZHA, H01r, IKAE,

(4) .

Brae, aEZK, ZH08, 01Ka, KABZ.

(5).

Here, if we consider as harmonic the quadrilaterals in for example (4) and we work in hexagon ABrZH0, as exactly above, based on the Harmonic Quadrilateral ABrH, we find that quadrilateral ZH0B is harmonic too, as relation (1) is true as well. We continue, similarly to above for hexagon raE01K and based on the harmonic quadrilateral raEI, we find that quadrilateral 01Ka is harmonic too, as relation (1) is true. Additionally, if we continue as exactly above, we find that quadrilaterals KABZ, Brae, aEZK, are harmonic as well, thus all the quadrilaterals in (4) and (5), are harmonic and thus the given decagon is harmonic. 3/. During the third case its five harmonic quadrilaterals have for example the order:

KABZ, ABrH, sra0, aEZK, 01Ka and then working as exactly

above we find that the rest five quadrilaterals of the given decagon are harmonic as well, hence this decagon will be harmonic too. 4/. During the fourth case its five harmonic quadrilaterals have the order: KABZ, ABrH , raEI, aEZK, H01r and then working as exactly above we find that the rest five quadrilaterals of the given decagon are harmonic as well, hence this decagon will be harmonic too. (In a similar way, we work for every other order of the given five harmonic quadrilaterals).

tGJli·l,W4ti4 I.e., if the given decagon is harmonic, we will prove that then this will have harmonic its ten decagons and (1) will be true.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

This is obvious, as since the given decagon is harmonic, according to the relevant definition (§ 16.1 d} all its ten quadrilaterals will be harmonic, while according to Proposition 106, (1) will be true.

Gi•lei3.Ut1i•leM Due to the Proposition about Harmonic Decagons above (Criterion of Harmonic being of convex inscribed decagons}, the following conclusion is reached: Each convex decagon inscribed in circle is harmonic, if its diagonals are concurrent and five out of its ten quadrilaterals, those that from these five quadrilaterals, if we receive each two of them, they do not share a diagonal. Consequently, from now on we can use for Harmonic Decagons, the following alternative Definition : «Harmonic decagon is called, each convex inscribed in circle decagon, if its diagonals are concurrent and five out of its ten quadrilaterals are harmonic, those that from these five quadrilaterals, if we receive each two of them, they do not share a diagonal ».

1#34eeFOA (a). The above Criterion of Harmonic being of Decagons, is rather significant, as it allows us to have a simpler alternative DEFINITION, from the initial one, for Harmonic Decagons, but because it also finds many applications, as we w ill later find out. (b). We believe that it is a new Criterion and that it firstly appears here. (c). According to the relevant definition (§ 16.1d), this Criterion is true for the Asteroid Decagon Al'lHKrZIBE0A as well . (d). Relevant Propositions are 34, 109, 110.

19.20. General Conclusion. Due to the Propositions, 109, 110, 111 , above, and if we work sim ilarly to their proofs, we reach the following general conclusion :

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Equivalent Criterion, to Criteria 109 to 111 , can be applied for any Harmonic 2n-lateral.

19.21 . Third Criterion of Harmonic being of a Convex Inscribed Hexagon. Extension to known Theorem about the Lemoine Point of a Triangle to a Hexagon. Does the Harmonic Hexagon have a Lemoine Point? Criterion 112 (Figure 74). If convex inscribed in circle (o) hexagon ABraEZ, has concurrent diagonals (main) and there exists a point on its plane for which the distances from its sides (the hexagon's sides), are analogous to its sides respectively, then and only then this hexagon is harmonic.

tUl•Jl#441 We consider a Harmonic Hexagon ABraEZ, inscribed in circle (o). We will prove that a point K' exists on its plane, for which it is : (1 ). where ua, u~, .... ,

U( ,

are the distances of point K' , from its sides AB=a ,

Br=l3, ... , ZA={, respectively. In fact, since hexagon ABraEZ is harmonic, it will also have AanBEnrZ=K (Proposition about Harmonic Hexagons 104). Moreover, according to the relevant definition (§16.1b) of Harmonic Hexagons, in ABraEZ quadrilaterals ABrE, Braz, ... , , ZABa, will be harmon ic and therefore Aa, BE, rz, will be conjugate to pairs of strings BZ-rE, Ar-

~ . Collinear and Coe clic Harmonic Grou s of n-members.

flZ, Bll-EA, respect ively, thus according to the converse of Auxiliary Proposition (Lemma) 6 in Part E, All, BE, rz, will be symmedians of the pairs of triangles ZAB-rllE, ABr-llEZ, Brll-EZA, respectively. Therefore, All, BE, rz, will be the loci, according which,

to

the

dis-

tances of each of their points, from the pairs of the respective ZA-AB,

sides rll-llE,

AB-Br, llE-EZ, Brr ll,

EZ-ZA,

are

analogous {[6] § 11-14.2}

(direct).

This will obviously be

true

shared

for

the

intersec-

tion K of diago-

rxr)µa 74.

nals All, BE, rz (Propositions 104

about Harmonic Hexagons), as this one belongs to all these three diagonals. This means that point K has all these three properties. Hence for K it will be: (2).

Thus, from the equalities in (2), we can easily find re lation (1 ), that is also true for the intersection K=AllnBEnrz and therefore this means that the desired point K', is actually Kor K'=K. Therefore, the desi red point is the point of convergence of the diagonals (main), of the Harmon ic Hexagon.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

tE91B•iekZ4t14i If for point K=AanBEnrz, relation (1) is true, we will prove that hexagon ABraEZ is harmonic. In fact, from relation (1), we also get that for example:

a =T. This means Ua

Us

that AK or Aa is the symmedian of triangle ZAB (converse of the known Proposition {[6] § 11-14.2} and therefore Aa will be conjugate to 28 with respect to circle (o) [Lemma 6 (direct), in Part E] , thus quadrilateral ZABa is harmonic (definition§ 16.1b). Similarly we find that quadrilaterals Braz, aEZB, are harmonic too, thus according to the simplified alternative definition of Harmonic Hexagons (Criterion 109 of Harmonic being of Inscribed Hexagons), hexagon ABraEZ will be harmonic.

13411609 (a). Two specialized Criteria of Criterion 112, are given in paragraph 81(88) (volume 8) in Geometry [31]. (b). The 1•', 2nd , 4th, 5th, 6th Criteria of the Harmonic Hexagon, are given in Propositions 96, 109, 139, 140, 141 , respectively. (c). We believe that Proposition 112 is new and firstly appears here. (d). Relevant Propositions are 96, 109, 113, 114, 139-141 .

8•h,iui4eil After Criterion 112 of Harmonic Hexagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to wh ich, th is point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and if we depend on the Auxiliary Proposition 6 in Part E, it is easily proven, that t he intersection of the diagonals on the harmonic quadrilateral , has the same, as above, property (See also Exercises of Geometry [32] IV, page 54). Hence the d istance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral. This is why the intersection of the diagonals of the harmonic quadrilateral , was characterized as the Lemoine point of the harmonic quadrilateral.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

(c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, hence we reach the following conclusion:

i•leld.Ut1[•lel Due to the aforementioned and in equivalence to these, the point of convergence of the diagonals of a Harmonic Hexagon is characterized as the LEMOINE POINT, of this Harmonic Hexagon.

Another formation of Criterion 112. After the above, Criterion 112 (Criterion of Harmonic being of convex Inscribed Hexagons), can be stated in the following way too: Every convex Hexagon inscribed in circle, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of its diagonals).

Alternative Definition of the Harmonic Hexagon. Therefore, after the above, from now on we can also use for Harmonic Hexagons, the following alternative Definition: «We call Harmonic Hexagon each convex cyclic Hexagon, if and only if,

i!

has a LEMOINE POINT.

19.22. Criterion of Harmonic being of a Convex Inscribed Octagon . Extension to Criterion 112 to an Octagon. Does the Harmonic Octagon have a Lemoine Point? Criterion 113 (Figure 75). If convex inscribed in circle (o) octagon ABr~EZH0, has concurrent diagonals (main) and there exists a point on its plane for which the distances from its sides (the octagon's sides), are analogous to its sides respectively, then and only then this octagon is harmonic.

81=-·__

_,_H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

tml•Jli441 We consider a Harmonic Octagon ABraEZH0, inscribed in circle (o) (Relevant definition in§ 16.1c). We will prove that a point K' exists on its plane, for which it is: Ua

Up

Uy

U5

u.

u~

u~

Ue

-a =-J3 =-y =-li =-E =-~ =-11 =-8

'

(1 ).

where Ua, up, .... , ue, are the distances of point K', from its sides AB=a , Br=l3, .. ., 0A=9, respectively. In fact, since Octagon ABraEZH0 is harmonic, according to Proposition

B

105 (§ 19.12), it will also have AEnsznrHna0=K, while according to the relevant

definition

(§16.1c), all its following eight quadrilaterals will be harmonic:

rxr11.1a 75.

ABrz, BrnH, rnE0 , ... , 0ABE.

(2).

This means that for example for the harmonic quadrilateral ABrZ its diagonals Ar, BZ, will be conjugate with respect to circle (o), thus according to Auxiliary Proposition 6 (converse) in Part E, for triangle ABr, BZ is its symmedian. Hence, according to the direct of a known Theorem (rtwµnpia (6] § 11-14.2), each point on BZ will be far from AB, Br distance analogous to these sides. This will obviously be true for point K of the convergence of diagonals of this Octagon, which belongs to BZ too, thus for point K, it will be:

(3).

~ . Collinear and Coe clic Harmonic Grou s of n-members.

If we work similarly to above for harmonic quadrilateral Brt.H and more specifically in triangle Bra, we prove that for Kit is also: U~

Uy

J3

V

-= -

(4).

Hence, if we continue as above for the rest six triangles raE, f.EZ, .. .,0AB, we find that for K it is also : (5). Therefore, from relations (3) - (5), we get relation (1 ), which is true for intersection K as well. This means that the desired point K' is Kor K'=K. Hence the desired point, is the point where the diagonals of the Harmonic Octagon coincide.

tGJM·leW4t14i I.e., if the convex inscribed in circle (o) Octagon ABrt.EZH0 has d iagonals concurrent at K and for K relation (1) is true, then this Octagon w ill be harmonic. In fact, since (1) is true, for Kit will be: (3). This, according to the converse of the Theorem in § 11-14.2 in Geometry [6], shows that K belongs to the symmedian matching to the side Ar in triangle ABr and thus, according to the direct of the above Auxiliary Proposition 6 in Part E, BZ and Ar are conjugate w ith respect to circle (o), or quadrilateral ABrZ is harmonic. If we take into consideration, that the rest seven two-part equalities that occur from (1) are true, i.e.: U~

-

J3

Uy

=-

V

Uy

-

V

U5

=-

l> '

U5

-

l>

U,

=-

(6).

£

and if we work as above, we find that all three quadrilaterals in (2) , Brt.H ,

8 ·=-__

_,_H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

rt..E0, .6.EZA, are harmonic, thus, according to Criterion 110 (direct) Octagon ABrt..EZH0 (alternative definition) is actually Harmonic, since it also has concurrent diagonals.

ld411FOb (a). Proposition 113 (criterion), we believ to be new and firstly appear here. (b). Equivalent Proposition (Criterion) we have proved to be true for Harmonic Hexagons as well (Criterion 112). (c). As we will see next, equivalent Criterion is true for Decagon ABrt..EZH01K, etc. (d). Relevant Propositions are 112, 114-118.

8•]11111[4,il After Criterion 113 of Harmonic Octagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to which, this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and easily proven {Criterion 81(87) in our book [31]} , the intersection of the diagonals on the harmonic quadrilateral, has the same, as above, property. Hence the distance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral (See also rEwµupia [32] IV, page 54). This is why the intersection of the diagonals of the harmonic quadrilateral, was characterized as the Lemoine point of the harmonic quadrilateral. (c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, as above, property, thus we reach the following conclusion :

i•lel9i'~1[•Jel After the aforementioned and in equivalence to these, it occurs that: The point of convergence of the diagonals of the Harmonic Octagon, is a LEMOINE POINT, for the Harmonic Octagon.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Another formation of Criterion 113. After the above, Criterion 113 (Criterion of Harmonic being of convex Inscribed Octagons), can be stated in the following way as well : Every convex Octagon inscribed in circle with concurrent diagonals, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of its diagonals).

Alternative Definition of the Harmonic Octagon . Therefore, after the above, from now on we can also use for Harmonic Octagons, the following alternative Definition: «We call Harmonic Octagon each convex cyclic Octagon, if and only if,

i!

has a LEMOINE POINT.

19.23. Criterion of Harmonic being of a Convex Inscribed Decagon. Extension to Criterion 113 to a Decagon . Does the Harmonic Decagon have a Lemoine Point? Criterion 114 (Figure 76). If a convex inscribed in circle (o) decagon ABraEZH01K, has concurrent diagonals (main) and there exists on its plane a point for which the distance from each side (of the decagon), is analogous to the respective side, then and only then, this decagon is harmonic.

ttll•Jli44M We consider a Harmonic Decagon ABraEZH01K, inscribed in circle (o) (Relevant definition in§ 16.1d). We will prove that a point/\' exists on its plane, for which it is:

a' _ j3' _ y' _ li' _ £' _

r_

rJ' _ 8' _ I' aj3yli£{1"18

_ K'

K

(1 ).

18)'-.___,_H:.:..A=R=M.:...:O=N=l..=cC-'G=E=O=M=E=T.:...:R...:..Y'-'-P=art:...:..::..:.B.

where a',

13', .. .. ,

K' , is the distance of point /\', from each side AB=a,

Br=l3, ... , KA=K, respectively.

In fact, since Decagon ABraEZH01K is harmonic, according to Proposition 106 (§ 19.13), it will have AZnBHnr0na1nEK=/\, while according to the relevant definition(§ 16.1d), all its ten quadrilaterals will be harmonic: ABrH, Brn0, rnEI, ... , KABZ.

(2).

This means that for example for the harmonic quadrilateral ABrH its diagonals Ar, BH, will be conjugate with respect to circle (o), hence according to the Auxiliary Proposition 6 (converse) in Part E, for triangle ABr, BH is its symmedian. Thus, according to the direct of a known Theorem (rEwµETpfa [6] § 11-14.2) , each point on BH will be far from AB, Br distance analogous

~ - Collinear and Coe clic Harmonic Grou s of n-members.

to these sides. This will obviously be true for the point of convergence of the diagonals of the Decagon I\, which too belongs to BH, thus for point /1., it will be :

a' 13' - =-

(3).

If we work similarly to above for the harmonic quadrilateral Brae and more specifically in triangle Bra, we prove that for I\ it is also:

13'

y'

- =p V

(4).

Hence, if we continue as above for the rest eight triangles raE, aEZ, ... ,KAB, we find that for I\ is also is:

v· y

K' li' li , ... ., K

a' a

(5).

Therefore, from relations (3) - (5), we get relation (1 ), which is true for intersection I\. This means that the desired point I\' is /1. or /\'=.I\. Hence the desired point, is the point where the diagonals of the Decagon coincide.

tGifi•l,l'Z4t14i I.e., if the convex inscribed in circle (o) Decagon ABraEZH01K has d iagonals concurrent at I\ and for I\ relation (1) is true, then this Decagon will be harmonic. In fact, si nce (1) is true, for /1. it will be :

a'

13'

- =a J3.

(3).

This, according to the converse of the Theorem in § 11-14.2 in Geometry [6], shows that I\ belongs to the symmedian matching to the side Ar i n triangle ABr and thus, according to the direct of the above Auxiliary Proposition 6 in Part E, BH and Ar are conjugate with respect to circle (o) , or quadrilateral ABrH is harmonic. If we take into consideration, that the rest nine two-part equalities that occur from (1) are true, i.e :

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

J3' v·

-=J3 V

(6).

and if we work as above, we find that the rest nine quadrilaterals in (2), Br~0, r~EI, ~EZK, EZHA, etc, are harmonic, thus, according to the definition of the Simple Harmonic Decagon (§ 16.1d), then Decagon ABr~EZH01K is in fact Harmonic.

ld41160A (a). Proposition 114 (criterion), we believ to be new and firstly appear here. (b). Equivalent Propositions (Criteria) we have proved to be true for Harmonic Hexagons (Criterion 112) and Octagons (Criterion 113) as well. (c). As we will see next, equivalent Criterion is true for the Asteroid non regular Decagon A~HKrZIBE0A, etc. I.e. that this is also harmonic, etc. (d). Relevant Propositions are 112,113, 115-118.

Gi•i11l11i4,il After Criterion 114 of Harmonic Decagons , we also notice that: (a). Point Lemoine in a triangle, has the property, according to which , this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and easily proven {Criterion 81(87) in our book (31] and Exercises (32] page 54}, the intersection of the diagonals on each harmonic quadrilateral, has the same, as above, property. Hence the distance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral. This is why the intersection of the diagonals of the harmonic quadrilateral, was characterized as the Lemoine point of the harmonic quadrilateral. (c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, as above, property.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

(e). According to Criterion 114 above for every Harmonic Decagon, the intersection of its diagonals (main), has the same, as above, property, thus we reach the following conclusion:

Gi•lei3i't1t•i,M After the aforementioned and in equivalence to these, it occurs that : The point of convergence of the diagonals of the Harmonic Decagon , is a LEMOINE POINT, for the Harmonic Decagon. Another formation of Criterion 114. After the above, Criterion 114 (Criterion of Harmonic being of convex Inscribed Decagons), can be stated in the following way as well: Every convex Decagon inscribed in circle with concurrent diagonals, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of its diagonals). Alternative Definition of the Harmonic Decagon. Therefore, after the above, from now on we can also use for Harmonic Octagons, the following alternative Definition: «We call Harmonic Decagon each convex cyclic Octagon, if and only if ,

i!

has a LEMOINE POINT.

19.24. Criterion of Harmonic being of an Asteroid Inscribed Octagon. Extension to Criterion 113 and 114, to aa Asteroid Octagon. Does the Asteroid Harmonic Octagon have a Lemoine Point? Criterion 115 (Figure 77). If an asteroid inscribed in circle (o) octagon Af.HBE0rZA, has concurrent diagonals (main) and there exists on its plane a point for which the distance from each side (of the Asteroid Octagon), is analogous to the respective side, then and only then, this octagon is harmonic.

181=-·- -~ H:.:..A=-=-R=M"-'O=N=l..;:;C....:G=E=O=M=E=T"-'R-=-Y'-'-P=art~B.

ttll•Jli44M We consider a Harmonic Asteroid Octagon AaHBE0rZA, inscribed in circle (o) (Relevant definition in § 16.1c). We will prove that a point K' exists on its plane, for which itis: Ua

u~

Uy

a

~

V

u~ l>

u~ u~ Ue - =- ={ 8 ' 11 where

u, E

(1 ).

Ua, Up, .. .. , Ue,

is

the distance of point K', from each side Aa=a,

aH=l3,

HB=y, ... ,rz=11,

ZA=8,

respectively, of the Asteroid Octagon. In fact, since Octagon AaHBE0rZAis

har-

monic, according to the relevant definition (§16.1c), all its following eight quadrilaterals will be harmonic: AaH0, aHBr, HBEZ, BE0A, E0ra, 0rzH, rZAB, ZAaE,

(2).

Which are obviously the quadrilaterals of the convex octagon ABraEZH0, thus this convex Octagon will also be harmonic. Hence, it will have AEnsznrHna0=K (Proposition 105, § 19.12). This means that for example for the harmonic quadrilateral AaH0 its diagonals AH , a0, will be conjugate with respect to circle (o), thus according to Auxiliary Proposition 6 (converse) in Part E, for triangle AaH , a0 is its symmedian. Hence, according to the direct of a known Theorem (rtwµnpfa [6] § 11-14.2), each point on a0 will be far from Aa, aH distance analogous

~ - Collinear and Coe clic Harmonic Grou s of n-members.

to these sides. This will obviously be true for point K of the convergence of diagonals of this Octagon, which belongs to /10 too, thus for point K , it will be : (3). If we work similarly to above for harmonic quadrilateral /1HBr and more specifically in triangle /1HB, we prove that for K it is also: Up

-

13

Uy

=-

(4).

V

Hence, if we continue as above for the rest six triangles HBE, BE0, E0r, 0rz, rZA, ZA/1, we find that it is also : U~

-

ri

Ue

=-

Ue

-

8 ' 8

Ua

=-

a

(5).

Therefore, from relations (3) - (5), we get relation (1 ), which is true for intersection K as well. This means that the desired point K' is Kor K'=K. Hence the desired point, is the point where the diagonals of the Harmonic Asteroid Octagon coincide, which are the same as the ones for the respective Harmonic Octagon. b). Converse. I.e., if the convex inscribed in circle (o) Asteroid Octagon A/1HBE0rZA has diagonals concurrent at K and for K relation (1) is true, then this Octagon will be harmonic. In fact, since (1) is true, for Kit will be :

(3). This , according to the converse of the Theorem in § 11-14.2 in Geometry [6], shows that K belongs to the symmedian matching to the side AH in triangle A/1H and thus, according to the direct of the above Auxiliary Proposition 6 in Part E, /10 and AH are conjugate with respect to circle (o), or quadrilateral A/1H0 is harmonic.

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

If we take into consideration, that the rest seven two-part equalities that occur from (1) are true, i.e.: Up

Uy

Uy

U 1,

U 1,

u,

J3=y, y=f, , f>=E

u,

u~

u~

u~

E ={ , {=11

(6).

and if we work as above, we find that the rest seven quadrilaterals in (2), b.HBr, HBEZ, BE0A, E0ra, erzH , rZAB, ZAb.E, are harmonic, thus, according to the definition of the Simple Asteroid Harmonic Octagon (§ 16.1c), Octagon Ab.HBE0rZA is in fact Harmonic, as it is also inscribed in circle (o).

1«41,Fiih (a). Proposition 115 (criterion), we believe to be new and firstly appear here. (b). Equivalent Propositions (Criteria) we have proved to be true for Harmonic Hexagons (Criterion 112) and Octagons (Criterion 113), etc. (c). As we will see next, equivalent Criterion 116 is true for the Asteroid non regular Decagon Ab.HKrZIBE0A, etc. I.e. that this is also harmonic, etc. (d). Relevant Propositions are 112-114, 116-118.

i·hll11I4hl After Criterion 115 of Asteroid Harmonic Octagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to which , this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and easily proven {Criterion 81(87) in our book [31]} , the intersection of the diagonals on each harmonic quadrilateral, has the same, as above, property (See also [32) IV, page 54). Hence the distance of this intersection from the sides of the quadrilateral, are analogous , to the respective s ides of t he quadrilateral. This is why the intersection of the diagonals of the harmonic quadri lateral, was characterized as the Lemoine point of the harmon ic quadrilateral. (c). Accord ing to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this

~ . Collinear and Coe clic Harmonic Grou s of n-members.

is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, property. (e). According to Criterion 114 above for every Harmonic Decagon, the intersection of its diagonals (main), has the same, property. (f). According to Criterion 115 above for every Asteroid Harmonic Octagon, the intersection of its diagonals (main), has the same, as above, property, thus we reach the following conclusion:

i•lll4Ut1i•l,M After the aforementioned and in equivalence to these, it occurs that : The point of convergence of the diagonals of the Asteroid Harmonic Octagon, is a LEMOINE POINT, for the Harmonic Octagon .

Another formation of Criterion 115. After the above, Criterion 115 (Criterion of Harmonic being of concave Inscribed Asteroid Octagons), can be stated in the following way as well: Every Asteroid Octagon inscribed in circle with concurrent diagonals, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of its diagonals).

Alternative Definition of the Harmonic Asteroid Octagon. Therefore, after the above, from now on we can also use for Asteroid Harmonic Octagons, the following alternative Definition: ««We call Asteroid Harmonic Octagon each asteroid cyclic Octagon, if and only if, it has a LEMOINE POINT».

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

19.25. Criterion of Harmonic being of an Asteroid Inscribed Decagon . Extenstion to Criterion 115 to an Asteroid Decagon. Does the Asteroid Harmonic Decagon have a Lemoine Point? Criterion 116 (Figure 78). If an asteroid inscribed in circle (o) decagon AilHKrZIBE0A, has concurrent diagonals (main) and there exists on its plane a point for which the distance from each side (of the Asteroid Decagon), is analogous to the respective side, then and only then, this decagon is harmonic.

rxr11-1a 78.

[Ql•Jl2491 We consider a Harmonic Asteroid Decagon AilHKrZIBE0A, inscribed

~ . Collinear and Coe clic Harmonic Grou s of n-members.

in circle (o) (Relevant definition in§ 16.1d). We will prove that a point A' exists on its plane, for which it is : Ua

Up

Uy

U5

U,

Ui

U~

Ue

U,

U•

-a------- --{ --., --8 -- --K ' l3yl>E

(1 ).

where Ua , up, .. .. , UK, is the distance of point A', from each side of the Asteroid Decagon A/1=a, /1H=l3, HK=y, ... ,E0=1, 0A=K, respectively. In fact, since decagon A/1HKr218E0A is harmonic, according to Proposition 107 (§ 19.14), it will have A2n8Hnr0n111nEK=/\, while according to the relevant definition (§ 16.1 d), all its following ten quadrilaterals will be harmonic: A/1HI, /1HK8, HKrE, Kr20, r21A , 218/1, 18EH, 8E0K, E0Ar, 0A/12.

(2).

This means that for example for the harmonic quadrilateral A/1HI its diagonals AH , /11, will be conjugate with respect to circle (o), thus according to Auxiliary Proposition 6 (converse) in Part E, for triangle A/1H , /11 is its symmedian. Hence, according to the direct of a known Theorem (rEwµupia [6] § 11-14.2), each point on /11 will be far from A/1, /1H distance analogous to these s ides. This will obviously be true for point /\ of the convergence of diagonals of this Octagon, which belongs to /11 too, thus for point /\, it will be:

(3).

If we work similarly to above for harmonic quadrilateral /1HK8 and more specifically in triangle /1HK, we prove that for K it is also:

~=~

13

V

(4).

Hence, if we continue as above for t he rest eight triangles HKr, Kr2, r21, 218, 18E, 8E0, E0A, 0A/1, we find that for/\ it is also:

u~

Ue

Ue

u,

-= - =11 8 ' 8

(5).

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Therefore, from relations (3) - (5), we get relation (1 ), which is true for intersection I\ as well. This means that the desired point I\' is /1. or /\'=.I\. Hence the desired point, is the point where the diagonals of the Asteroid Decagon coincide.

(GJfi•l,@4?14 I.e., if the convex inscribed in circle (o) Asteroid Decagon Ab.HKrZIBE0A has diagonals concurrent at /1. and for /1. relation (1) is true, then this Decagon will be harmonic . Ua

Up

a=Jf·

In fact, since (1) is true, for /1. it will be:

(3).

This, according to the converse of the Theorem in § 11-14.2 in Geometry [6], shows that I\ belongs to the symmedian matching to the side AH in triangle Ab.H and thus, according to the direct of the above Auxiliary Proposition 6 in Part E, b.l and AH are conjugate with respect to circle (o), or quadrilateral Ab.HI is harmonic. If we take into consideration, that the rest nine two-part equalities that occur from (1) are true, i.e.: Up

-

13

Uy

Uy

=- -

u~

y'y

u~

u~

U5

=-

U5

-

l)'l)

Ue

Ue

U,

=-

E '

u,

U, E

u,

u. u.

Ua

T=~·~=a·a=,.,=K·K=a

(6).

and if we work as above, we find that the rest nine quadrilaterals in (2), b.HKB, HKrE, KrZ0, rZIA, ZIBb., IBEH, BE0K, E0Ar, 0Aa2, are harmonic, thus, according to the definition of the Simple Asteroid Harmonic Decagon (§ 16.1d), Decagon Ab.HKrZIBE0A is in fact Harmonic, as it is also inscribed in circle (o). [The construction of the Asteriod Harmonic Decagon (§ 16.1 d), as we will later see (§ 19,65, Construction 150), is easy . This is carried through with based on the Construction of the convex Harmonic Decagon ABrb.EZH01K and Proposition 108].

~ . Collinear and Coe clic Harmonic Grou s of n-members.

14411EOA (a). Proposition 116 (criterion), we believe to be new and firstly appear here. (b). Equivalent Propositions (Criteria) we have proved to be true for Harmonic Hexagons (Criterion 112) and Octagons (Criterion 113), etc. (c). As we will see next, equivalent Criterion is true for the convex Harmonic pentagon ArEHI and for the Asteroid non regular Pentagon ArEHIA, etc. (d). Relevant Propositions are 108, 112-115, 117,118,150.

i·h,iui§,il After Criterion 116 of Asteroid Harmonic Decagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to which, this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle's). (b). As it is known and easily proven {Criterion 81(87) in our book [31]} , the intersection of the diagonals on each harmonic quadrilateral, has the same, as above, property (See also [32) IV, page 54). Hence the distance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral. This is why the intersection of the diagonals of the harmonic quadrilateral, was characterized as the Lemoine point of the harmonic quadrilateral. (c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, property.

(e). According to Criterion 114 above for every Harmonic Decagon, the intersection of its diagonals (main), has the same, property. (f). According to Criterion 115 above for every Asteroid Harmonic Octagon, the intersection of its diagonals (main), has the same, as above, property. (g). According to Criterion 116 above for every Asteroid Harmonic Decagon, the intersection of its diagonals (main), has the same, as above, property,

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

thus we reach the following conclusion:

After the aforementioned and in equivalence to these, it occurs that: The point of convergence of the diagonals of the Asteroid Harmonic Decagon, is a LEMOINE POINT, for the Harmonic Octagon.

Another formation of Criterion 116. After the above, Criterion 116 (Criterion of Harmonic being of convex Inscribed Asteroid Decagons), can be stated in the following way as well: Every Asteroid Decagon inscribed in circle with concurrent diagonals, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of its diagonals).

Alternative Definition of the Harmonic Asteroid Decagon. Therefore, after the above, from now on we can also use for Asteroid Harmonic Decagons, the following alternative Definition: ««We call Asteroid Harmonic Decagon each asteroid cyclic decagon, if and only if, it has a LEMOINE POINT».

19.26. Criterion of Harmonic being of an Inscribed Pentagon. Extension to Criterion 116 to a Pentagon. Does the Harmonic Pentagon have a Lemoine Point? Criterion 117 (Figure 79). If for convex inscribed in circle (o) Pentagon ArEHI, there exists on its plane a point for which its distance from its side (of the Pentagon), is analogous to the respective side, then and only then this Pentagon is Harmonic.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Ixr'iµa 79.

We consider a Harmonic Pentagon ArEHI , inscribed in circle (o) (Relevant terminology in§ 16.1 e). We will show that there is a point A' on its plane, for which it is : Ua

Uy

U,

U~

U1

-a --- ---V £ 11 • '

(1 ).

where ua, Uy, u ,, u~, u ,, is the distance of point A', from each of its sides Ar=a , rE=y, EH=E, Hl=r,, IA=1, respectively. In fact, since pentagon ArEHI is Harmonic, according to the relevant definition(§ 16.1e), there will be points B, 11, Z, 0 , K, on circle (o), such that decagon ABrt::.EZH01K, would be Harmonic, thus this will have concurrent

81=-·__

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

diagonals, at point let us say A (AZnBHnr0na1nEK=A) ([Proposition 106 (§

19.13). Moreover, since the convex decagon ABrb.EZH01K ia Harmonic, according to the direct of Proposition 108 (§ 19.16), the Asteroid Decagon Ab.HKrZI BE0A will also be Harmonic, which shares its diagonals with the above convex one, hence they will share point A too, thus for this, according to the relevant definition (§ 16, 1 d), all its ten following quadrilaterals will be Harmonic: ArE0, Bb.21 , rEHK, a20A, EHIB, Z0Kr, HIAb., 0KBE, IArz, KBb.H .

(2).

This means that for example quadrilateral EHIB is harmonic, then BH , El, will be conjugate with respect to circle (o), thus according to Auxiliary Proposition 6 (converse) in Part E, for triangle EHi, HB is its symmedian and hence, according to the direct of a known Theorem (rEwµupia [6] § 11-

14.2), each point on HB will be far from sides EH , HI , distance analogous to these sides. Hence from point A too. Thus for point A, it will be:

~=~ E '1

(3).

Similarly, for A and the Harmonic Quadrilaterals ArE0, rEHK, HIAb., IArZ, Ua

Uy

Uy

U,

U~

U,

U,

Ua

1

a

wefindthatalso respectively: . - = - - = - - = - - = - .

'

a

y'y

t ' ri

1 '

(4).

Therefore, from (3), (4), relation (1) occurs , which is true for A. This means that the desired point A' is A or A'=A. Hence the desired point, is the point where the diagonals of the Harmonic Decagon coincide, from which it is produced.

tGlfi•l,fJ4t141 I.e., if for the convex inscribed in circle (o) pentagon ArEHI a point A exists, for which relation (1) is true, then we will prove that this pentagon is Harmonic. We define the intersections Z, 0 , K, B, a , of AA, rA, EA, HA, IA, with circle (o) respectively. Hence, since relation (1) is true, for A, it will be for example:

~ . Collinear and Coe clic Harmonic Grou s of n-members.

u,

-

E

u~ 11

=-

(3).

This, according to the converse of the Theorem in § 11-14.2 in Geometry [6], means that A belongs to the symmedian that matches to the side El of triangle EHi and thus, according to the direct of the above Auxiliary Proposition 6, in Part E, AH or BH and El are conjugate with respect to circle (o) [where B is the intersection of HA with circle (o)], or that quadrilateral EHIB is Harmonic. If we take into to consideration, the fact that the rest four two-part equalities in (1) are true, i.e.: Ua

-

a

Uy

=-

U,

Ua

- =a,

V

(4).

and if we work similarly to above, we find that the rest four quadrilaterals in (2), ArE0, rEHK, HIAa, IArz, are Harmonic. Hence,

since

for

example

quadrilateral

ArE0,

is

Harmonic

and

AZnrenEK=A, according to Auxiliary Proposition 34, quadrilateral Z0Kr will also be Harmonic. In a similar way we find that the followi ng quadrilaterals are harmonic too: B~I, ~0A, 0KBE, KBaH.

(5).

Thus, all the quadrilaterals in (2) are harmonic, hence the Asteroid Decagon AaHKrZIBE0A (definition§ 16.1d), will be harmonic and thus ABraEZH01K will also be harmonic (Proposition 108, converse). Hence, according to the relevant definition (§ 16.1e), Pentagon ArEHI will in fact be harmonic, as it is also inscribed in circle (o).

ld4eeFOb (a). Proposition 117 (criterion), we believe to be new and firstly appear here. (b). Equivalent Propositions (Criteria) we have proved to be true for Harmonic Hexagons (Criterion 112), Octagons (Criterion 113), etc. (c). As we already saw, equivalent Criterion is true for the Asteroid non regular Decagon AaHKrZIBE0A [Proposition 116 (§ 19.25)] . (d). As we will later see, equivalent Criterion is also true for the Asteroid non Regular Pentagon AElrHA (Proposition 118), etc. (e). Relevant Propositions are 112-116, 118.

8 ·=-__

_,_H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

i•l11l1114,il After Criterion 117 of Harmonic Pentagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to which , this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and easily proven {Criterion 81(87) in our book [31]} , the intersection of the diagonals on each harmonic quadrilateral, has the same, as above, property (See also [32] IV, page 54). Hence the distance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral. This is why the intersection of the diagonals of the harmonic quadrilateral, was characterized as the Lemoine point of the harmonic quadrilateral. (c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, property. (e). According to Criterion 114 above for every Harmonic Decagon, the intersection of its diagonals (main), has the same, property. (f). According to Criteria 115 and 116 above for every Asteroid Harmonic Octagon, as well as for every Asteroid Harmonic Decagon the intersection of its diagonals (main), has the same, as above, property. (g). According to Criterion 117 above for every convex Harmonic Pentagon, the intersection of the diagonals (main) of the decagon from which this is created, has the same, as above, property, thus we reach the following conclusion:

i•lel3i't1i•lel After the aforementioned and in equivalence to these, it occurs that: The point of convergence of the diagonals of every Harmonic Decagon, is a LEMOINE POINT for its twin Harmonic Pentagons as well.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Another formation of Criterion 117. After the above, Criterion 117 (Criterion of Harmonic being of convex Inscribed Pentagons), can be stated in the following way as well: Every pentagon inscribed i n circle, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of the diagonals of the Harmonic Decagon, from which it is created).

Alternative Definition of the Harmonic Pentagon. Therefore, after the above, from now on we can also use for Harmonic Pentagons, the following alternative Definition : «We call Harmonic Pentagon each cyclic pentagon, if and only if, it has a LEMOINE POINT».

19.27. Criterion of Harmonic being of an Inscribed Asteroid Pentagon. Extension to Criterion 117 to an Asteroid Pentagon. Does the Asteroid Harmonic Pentagon have a Lemoine Point? Criterion 118 (Figure 80). If for convex inscribed in circle (o) Asteroid Pentagon AElrHA, there exists on its plane a point for which its distance from its side (of the Asteroid Pentagon), is analogous to the respective side, then and only then this Pentagon is Harmonic.

tm1,11z4;1 We consider the Asteroid Harmonic Pentagon AElrHA, inscribed in circle (o) (Relevant terminology in§ 16.1 e). We will show that there is a point A' on its plane, for which it is: Ua

-

a

U,

=-

£

U,

Uy

=- =-

V

=

U~

., '

(1 ).

(Bi.:-- --=-H=A=R=M:.:..O=N=l:..=C;....;G=E=O=-M=E...:..;TR~Y""-'-P-=a=-=rt-=B'"'". where ua, u,, u, , uv, u~, is the distance of point /\', from each of its sides AE=a, El=E, Ir=,, rH=y, HA=ri, respectively.

rxrwa 80. Since from the implication, the Asteroid Pentagon AElrHA is harmonic, according to the relevant definition [§ 16.1 e (b)], between its consecutive apexes, there will be one point of the circle (o), from the group of five points 0 , B, Z, K, fl, respectively, such that the Asteroid Decagon AllHKZIBE0A, would be harmonic. Thus, since the Asteroid Decagon AllHKZIBE0A, is harmonic, then the convex decagon ABrllEZH01K will also be harmonic (converse of Proposit ion 108), while these will also have AZnBHnr0na1nEK=/\ ([Propositions

106 (§ 19.13) and 107 (§ 19.14).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Thus, since decagon ABraEZH01K is Harmonic, according to the relevant definition(§ 16.1d), all its ten quadrilaterals will be harmonic: ABrH , srne, rnEI, aEZK, EZHA, ZH0B, H01r, 01Ka, IKAE, KABZ.

(2).

Therefore, since for example quadrilateral IKAE is Harmonic, its diagonals KE, IA will be conjugate with respect to circle (o). This means that EK is symmedian of the triangle AEI [Auxiliary Proposition 6 (converse) in Part E. Thus all the points on EK will be far from AE, El analogous distance Ua, u , (Theorem in§ 11-14.2 in Geometry [6]). Thus for point /1. as well. Hence for I\ it will be:

Ua

U,

a

E

(3).

In the exact same way, for/\, from the harmonic quadrilaterals raEI , H01r, ABrH, EZHA, respectively, we also find that: U,

U,

E-1 '

U,

,

Uy

Uy

U~

-y ' y- 11 '

.,-a

U~

Ua

(4).

Thus, from relations (3) , (4), relation (1) occurs, which is true for point /1. too. This means that the desired point I\' is /1. or I\'=/\. Hence the desired point, is the point where the diagonals of the Asteroid Decagon coincide, from which the pentagon is created.

tGJM•hW4i#4 I.e., if for the inscribed in circle (o) Asteroid Pentagon AElrHA a point /1. exist s, for which relation (1) is true, then we will prove that this pentagon is Harmonic. In fact, if for /1. relation (1) is t rue and A/\, El\, I/\, r/1., HI\, re-intersect circle (o) at points Z, K, a , 0, B, respectively, then because of (1 ), it will be for example:

Ua

U,

a

E

(3).

This means that in triangle EAi , EK is symmedian (Theorem in§ 11 -14.2 in Geometry [6]), thus EK, IA will be conj ugate with respect to circle (o) [Auxiliary Proposition 6 (direct) in Part E] and thus quadrilateral IKAE is Harmonic.

8l=-·__

_,_H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

In the exact similar way, if we take into consideration relation (1), we also find that quadrilaterals: ABrH , raEI, EZHA, H01r, are Harmonic. Therefore, since for example quadrilateral IKAE is harmonic and it is Azna1nEK=/\, quadrilateral aEZK will also be harmonic (Criterion 34). In a similar way we find that the following quadrilaterals are harmonic too: ZH08, 01Ka, KABZ, Brae. Thus all ten quadrilaterals, in relation (2) are harmonic. This means that the convex decagon ABraEZH01K is also harmonic, thus the Asteroid Decagon AaHKZIBE0A is also Harmonic (Proposition 108). This shows that, since points Z, K, a , 0 , B exist and this Asteroid Decagon is harmonic, then the Asteroid Pentagon AElrHA will be harmonic too (definition § 16.1 e).

1341160A (a). Proposition 118 (criterion), we believe to be new and firstly appear here. (b). Equivalent Propositions (Criteria) we have proved to be true for Harmonic Hexagons (Criterion 112), Octagons (Criterion 113), etc. (c). As we already saw, equivalent Criterion is true for the Asteroid non regular Pentagon ArEHI [Proposition 117 (§ 19.26)]. (d). Relevant Propositions are 34, 106-108, 112-117.

i•l11l11I4,il After Criterion 118 of Asteroid Harmonic Pentagons, we also notice that: (a). Point Lemoine in a triangle, has the property, according to which, this point is far from the sides of the triangle, distances analogous to its respective sides (the triangle' s). (b). As it is known and easily proven {Criterion 81(87) in our book [311}, the intersection of the diagonals on each harmonic quadrilateral, has the same, as above, property. Hence the distance of this intersection from the sides of the quadrilateral, are analogous, to the respective sides of the quadrilateral. This is why the intersection of the diagonals of the harmonic quadrilateral, was characterized as the Lemoine point of the harmonic quadrilateral.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

(c). According to Criterion 112 above in the Harmonic Hexagon, the intersection of its diagonals (main), has the same, as above, property, and this is why we named it the Lemoine point of the Harmonic Hexagon. (d). According to Criterion 113 above for every Harmonic Octagon as we saw, the intersection of its diagonals (main), has the same, property. (e). According to Criterion 114 above for every Harmonic Decagon, the intersection of its diagonals (main), has the same, property. (f). According to Criteria 115, 116 and 117 above for every Asteroid Harmonic Octagon, every Asteroid Harmonic Decagon , as well as for every Harmonic Pentagon, respectively, the intersection of its diagonals (main), has the same, as above, property. (g). According to Criterion 118 above for every convex Asteroid Harmonic Pentagon, the intersection of the diagonals (main) of the decagon from which this is created, has the same, as above, property, thus we reach the following conclusion:

(ri•lei4i'i1Ithh After the aforementioned and in equivalence to these, it occurs that: The point of convergence of the diagonals of every Harmonic Decagon, is a LEMOINE POINT for its twin Asteroid Harmonic Pentagons as well.

Another formation of Criterion 118. After the above, Criterion 118 (Criterion of Harmonic being of Asteroid Inscribed Pentagons), can be stated in the following way as well: Every asteroid pentagon inscribed in circle, is harmonic, if and only if, it has a LEMOINE POINT. (This coincides with the convergence point of the diagonals of the Harmonic Decagon, from which it is created).

Alternative Definition of the Harmonic Asteroid Pentagon . Therefore, after the above, from now on we can also use for Asteroid Harmonic Pentagons, the following alternative Definition:

18=-·__

_,_H:.:..A=R=M"-'O=N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

«We call Asteroid Harmonic Pentagon each cyclic asteroid pentagon, if and only if, it has a LEMOINE POINT».

19.28. General Conclusion of Harmonic Decagons and their Twin Pentagons. We have proved in Criterion 114, that the convex Harmonic Decagon has a Lemoine point which coincides with the intersection of its diagonals (main). Additionally we have proved in Criterion 116, that the respective Asteroid Harmonic Decagon of this Harmonic Decagon, has a Lemoine point that also coincides with tha above intersection of the diagonals (main) of the respective convex Harmonic Decagon (they obviously have main diagonals). Moreover we have proven in Criteria 117 and 118 above, that both the two convex Harmonic Pentagons, (twin), as well as their respective Asteroid Harmonic Pentagons (twin), that its apexes are the one by one apexes of a convex Harmonic Decagon, have Lemoine points that also coincide with the above intersection, of the diagonals (main) of the Harmonic Decagon, from which they are created . Hence, each convex Harmonic Decagon, its twin convex Harmonic Pentagons, its Asteroid Harmonic Decagon and its twin Asteroid Harmonic Pentagons, that their sides are diagonals (not main) of this Harmonic Decagon, have a common Lemoine point that coincides, for all these, with the intersection of the diagonals (main) of the convex Harmonic Decagon.

19.29, General Conclusion concerning Multilaterals. If we take into consideration Criteria 112-118 and if we continue for a Harmonic Dodecagon, etc, we will realize that there are true equivalent conclusions to the ones in Criteria 112-118 as well, thus we reach the following General Conclusion: The convergence point of the diagonals of every Harmonic Multilateral, is a LEMOINE POINT for this Harmonic Multilateral.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

19.30. General Formation of Criteria 112-118 for Multilaterals as well. If we take into consideration the above, as well as Criteria 112-118, we easily realize that the following Criterion about Multilaterals is true as well: Each convex or Asteroid Multilateral inscribed in circle, with concurrent diagonals (main), is Harmonic, if and only if, is has a LEMOINE POINT, which coincides with the convergence point of its diagonals as well, or of the diagonals of the convex Multilateral to which this one belongs (As for example the Pentagon, Heptagon, etc, that don't have main diagonals, but the convex Decagon, Tetradecagon, etc, respectively, do have, to which they belong as twins).

19. 31 . Alternative Definition of the Harmonic Multilateral. Consequently, due to the above, from now on we can use for Harmonic Multilaterals as well, the alternative Definition below: «We call Harmonic Multilateral each convex or asteroid cycl ic Multilateral, if and only if, this has a LEMOINE POINT.

19.32. Significant Property of the Simple Harmonic Hexagon. Extension to Proposition 84. from a line to a circle . Proposition 119 (Figure 64).

AB rb EZ

For every Simple Harmonic Hexagon ABrt.EZ, it is': Br. bE . ZA =1.

(1) .

1"1 Way (Based on Auxiliary Proposition 5 in Part E). Since, according to Proposition 104 (§ 19.11), every Harmonic Hexagon has

18l=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

concurrent diagonals (main), according to the direct of Auxiliary Proposition 5 (§ 2.5) in Part E, relation (1) above will in fact be true. 2nd Way Based on the relevant Definition . Since this hexagon is harmonic, its quadrilaterals ABrE, raEA , EZAr will be harmonic too (definition§ 16.1 b), for which it will be respectively:

AB Br

AE

=Er '

rb bE

rA

=AE '

EZ ZA

Er

=r A.

(2).

Thus, from relations (2), if we multiply the respective parts and after the proper reductions, we get:

AB r b EZ Br. bE . ZA

AE r A Er

=Ef · AE. rA =1 ==< 1l·

13411Ehb (a). Proposition 119, we believe to be new and firstly appear here. (b). As we already saw in paragraph 16.1 a, equivalent Proposition, is true

AB rb

for the Harmonic Quadrilateral ABra as well (For which it is Br. M

=1),

while Equivalent Propositions we will prove to be true for Harmonic Octagons, Decagons, etc (Propositions 120, 121 , etc, respectively). (c). This Proposition is not a Criterion of Harmonic being, as its converse is not true. (d). This Property is true for Harmonic Collinear Groups of Six Points as well (See Proposition 84). (e). Relevant Propositions are 84, 104, 120, 121.

19.33. Significant Property of the Simple Harmonic Octagon. Extension to Proposition 85, from a line to a circle . Proposition 120 (Figure 65). For every Simple Harmonic Octagon ABraEZH0 and its respective Asteroid Octagon AZr0EBHaA, it is:

AB r b EZ H0 AZ re EB Hb Br . bE. ZH . 0A - zr . 0E . BH. M - 1·

(1 ).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

Since, according to the relevant Definition(§ 16.1c), for the Harmonic Octagon ABraEZH0, for example quadrialterals ABrZ, EZHB, are Harmonic, the pair of points B, Z, harmonically divides the two pairs of points A , r and E, H. Thus, according to Proposition 38 (§ 8.75) and since it is AB/Br=AZ/Zr, it will also be:

AB2

A-Z:-

AH AE

- - = -2- = - -

er2 zr

Hr· Er·

(2).

Moreover, since quadrilaterals raE0, H0Aa, are Harmonic too (definition) and since it is rataE=r0/0E, in the exact same way we find that it is:

r112 r0 2 rA rH -- = -=-/1E2 0E 2 AE. HE .

(3),

If we continue as above, we also find that:

EZ 2 EB 2 Er EA ZH 2 =BH 2 =rH "AH '

H0 2 H/12 HE Hr 0A 2 = /iA 2 = EA . r A .

(4).

Thus, from relations (2)-(4), if we multiply their respective parts and after the proper reductions we have relation (1 ).

134116hd (a). This Proposition is not a Criterion of Harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are true for the Harmonic Quadrilateral ABra as well (For which it is

AB r/1 er· /iA =1) and for Harmenic HExa-

gon ABraEZ too (Proposition 119). (c). This Property is true for Harmonic Collinear Groups of Eight Points as well [See Proposition 85 (§ 15.28)]. (d). The second part in (1 ), shows us that Proposition 120, is true for the Asteroid Harmonic Octagon AZr0EBHM. (e).This Proposition we believe to be new and firstly appear here. (f). Relevant Propositions are 38, 85, 119, 121.

8)'-.___,_H:.:..A"-R=M.:..:O=N..:.:l..;:;C....;:G=E=O=M=E= T.:..:R...:..Y'-'-P=art'-'-='-B.

19.34. Significant Property of the Simple Harmonic Decagon .. Extension to Proposition 86 1 from a line to circle. Proposition 121 (Figure 81 }. For every Simple Harmonic Decagon ABr~EZH01K and its matching Asteroid Decagon A0EBIZrKHM, it is:

AB rt1 EZ H0 IK A0 EB 12 rK H/1 Br. l1E. ZH. 01 . KA= 0E. Bl . zr. KH . M =1 ·

'"'

·"-

~ ~~ ~ '..y'

AJ3'. >..l5' =1 and from

m . Collinear and Coe clic Harmonic Grou

s of n-members.

Since for example it is LAKB= LrKt. and LKAB= LKt.r triangles KAB, Kt.r are similar, thus according to what we know, it will be :

a a' - = -y' . V

(3).

In the exact same way, we also find that:

-13 = -13' 15 15' .

(4).

Thus, from relations (3), (4), if we multiply their respective parts, we get:

a y

a' y'

13. 15 = 13'. 15' .

(5).

Thus, if we take into consideration the known Criterion 213(21) in book [31], then (5) becomes:

a y

a' y'

ji°"l5 = 13,. 15, =1

or

a' y'

13, . 15, =1,

.

.

which 1s the proven rela-

tion (1).

tl:lti•l,W4i#41 I.e., if (1) is true, we will prove that quadrilateral A Bra, is harmonic. We find, as in the direct of way 2 above, that relations (3) and (4) are true, from which we get (5). Hence, since (1) is true as well, then (5), becomes:

a y

a' y'

l3, 15 - 13, . 15, -1,

or

a y

l3 . 15 -1.

.

.

This, according to the converse of the known

Criterion 213(21), means that quadrilateral ABrt., is actually harmonic.

134eeEOb (a). Proposition 122 is the thirteenth criterion of harmonic being of a quadrilateral. We have come up with the first ten criteria of harmonic being of the harmonic quadrilateral, earlier and they are included in book [31]. Criteria 11 , 12, 14, are the Propositions 33, 35, 133, respectively. (b). As we will see below, equivalent Propositions, are true for Hexagon

.

. .

ABft.EZ, (For which 1t 1s:

a' y'

£'

l3' . 15, · f-=1) , for Octagon, for Decagon,

are not Criteria of Harmonic being though.

etc, that

18=-·__

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

(c). This Proposition we believe to be a new Criterion and it firstly appears here. (d). Relevant Propositions are 123-127.

19.37. Significant Property of the Lemoine Point of a Simple Harmonic Hexagon . Extension to Proposition 122 to a Harmonic Hexagon. Proposition 123 (Figure 74}. For every Simple Harmonic Hexagon ABraEZ, it is: Ua Uy u. - . - . - =1, Up Ui; U(

(1 ).

where ua, up, Uy, .. ., U(, is the distance of the convergence point K of the diagonals (main) of this hexagon, from its side AB=a, Br=J3, ra=y, ... ,ZA=\, respectively.

1st Way (Based on Propositions 104, 112 and 119). Since hexagon ABraEZ is harmonic, according to Proposition 104 (§ 19.11), it will have concurrent diagonals at poi nt K (Lemoine Point of the Hexagon) for which, according to Criterion 112 (§ 19.21 ), it will be: Ua

Up

a

J3

-= -

Uy

=-

V

Ui;

U•

U(

ij

E

{

=-=-= -

1

=-

A

or

a=A.ua, f3=,\.up, y=A.Uy, i5=,\.ui;, t=,\.u., \=A.U(. After all, according to Proposition 119 (§ 19.32), it is:

to (2), becomes:

i·¼·f

(2). =1 , which due

A.U a A.Uy A.u • .---- . .---- . . - - =1 and from which we get the proven re/\,Up /\,U i; /\,U (

lation (1 ). 2nd Way Based on Propositions 104 and 119.. Since Hexagon ABraEZ is Harmonic, according to Proposition 104, it will have concurrent diagonals at point K (Lemoine Point of the Hexagon).

filzj .

Collinear and Coe clic Harmonic Grou s of n-members.

Since for example it is LAKB= Lt.KE and LKAB= LKE.t. triangles KAB, KE.ti are similar, thus according to what we know, it will be: 0

=

l>

Ua

(3).

U5

In the exact same way, we also find that:

£ u, - =-

J3

V

Uy

- =-

(4).

Up

Hence, from relations (3), (4), if we multiply their respective parts , we get:

a V

£

Ua

u,

Uy

p·~-~ = ~ - ~ - ~

(5).

Thus, if we take into consideration Proposition 119 (§ 19.32), according to which it is

i. f. f

=1 , then from (5) we get:

Ua

-

Up

Uy

.-

U5

U,

.-

=1, which is the

U;

proven relation (1 ).

3rd Way (Based on Propositions 104 and 119). Since Hexagon ABr.t.EZ is Harmonic, according to Proposition 104, it will have concurrent diagonals at point K (Lemoine Point of the Hexagon). Since for example it is LAKB=Lt.KE, according to known Theorem, for triangles KAB, K.t.E, it will be:

KA.KB ~.KE

(KAB)

a .ua

KA.KB

=(Kf.E) = l>.u 5

or Kf..KE

=

O.U a

(6).

l>.u 5 .

In the exact same way, we also find that:

KE.KZ KB.Kr

(KEZ)

£.u ,

=(KBr) = J3.up

Similarly we find that:

KE.KZ £.u , = KB.Kr J3.u p .

or

Kr.Kt. KZ.KA

(7).

(8).

Thus, from relations (6), (7), (8) if we multiply the respective parts and after the proper reductions , we get:

a V

£

Ua

Uy

u,

1=p·~-~- ~ -~ -~

-

(9).

Thus, if we take into consideration Proposition 119 (§ 19.32) according to

Bl=-·__ a V

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

E

which it is n·s·7=1, then (9), becomes: I-'

Ua

-

Up

Uy U,

.-

U15

.-

u

.,

Ua

Uy

u,

Up

U15

U(

1=1.-.-.- or

=1 , which is the proven relation (1).

U(

IU4ufhh (a). Proposition 123 is not a Criterion of Harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are also true for the Harmonic Quadrilateral ABr.6. (For which, according to Proposition Ua

-

Up

Uy

.-

122, it is

=1 ), for the Harmonic Octagon, for the Harmonic Decagon, etc.

U15

(c). Proposition 123 is also true for the Asteroid Harmonic Octagon, Decagon, etc. (e). This Proposition we believe is a new Proposition and firstly appears here. (f). Relevant Propositions are 104, 112, 119, 122, 124-127.

19.38. Significant Property of the Lemoine Point of a Simple Harmonic Octagon. Extension to Proposition 123 to the Harmonic Octagon. Proposition 124 (Figure 75). For every convex Simple Harmonic Octagon ABr.6.EZH0, it is: Ua

Uy

U,

U~

Up U15

U(

Ue

-

. - . - . -=1 ,

(1 ).

where Ua, Up, Uy, . . . , u e, is the distance of the convergence point K of the diagonals (main) of this octagon, from its side AB=a, Br=l3, ra=y, ... ,0A=8, respectively.

[ml Since octagon ABr.6.EZH0 is Harmonic, according to Proposition 105 (§ 19.12), it will have concurrent diagonals at point K (Lemoine point of the octagon) for which, according to Criterion 113 (§ 19.22), it will be:

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Ua

-

a

Up

=-

~

Uy

=-

V

U5

=-

l>

U, U( U~ Ue 1 = - = - = - = - = - or E

{

11

8

A

a=A.ua, f3=A.up, y=A.uy, li=A.u5, E=A.u,, ~=A.u(, 11=>..u~, 9=A.ua.

(2).

After all , according to Proposition 120 (§ 19.33), it is: i·f·t-i=1 , which

A.Ua A.U y A.u ,

due to (2), becomes :

A.u~

.

.

.---- . .---- . .---- . .---- =1 and from whrch we directly

/\.U p /\.U 5 "·U ( /\.U e

get proven relation (1).

li

0 ,-13'-y'-l>'_

E(1181K

E' _f_ri·- 8 ,-1-~ -A

or

a=l\.a', J3=.\.J3', y=.\.y',

l>=l\.l>', r.=l\.r.', ~=A.~', ri=l\.r)', 8=1\.8', 1=1\.1', K=A.K'.

(2).

After all, according to Proposition 121 (§ 19.34), it is:

i·¼-r

i-°iZ=1 , which

l\a' Ay' 11.E' All' Al' . due to (2), becomes: 11.13·. 11.l>'. 11.r. 11.8'. AK' =1 and from which we get the proven relation (1 ).

2n d Way (Based on Propositions 106 and 121). Since Decagon ABr~PZH01K is Harmonic, according to Proposition 106 (§ 19.13), it will have concurrent diagonals at point /\ (Lemoine Point of the decagon). Since for example it is LA/\B= LH/\Z and L/\AB= L/\HZ triangle /\AB, /\HZ are similar, thus according to what we know, it will be :

a

a'

(3).

In the exact same way, we also find that:

11

V

.,.

y'

1'

E

E'

-l3 =-l3' ' -8 =-8' ' -l> =-l>' ' -K =-K' .

(4).

Hence, from relations (3), (4), if we multiply their respective parts , we get:

a y

E

11

a' y' r.'

1

l3. l> . (. 8 . K

=

11'

1'

l3' . l>'. r. 8' . K' .

(5).

Thus, if we take into consideration Proposition 121 (§ 19.34), according to

.

. .

which 1t 1s

a y

E 11

I

J3. ~. ~. 8 .K=1 , then (5) becomes:

the proven relation (1).

a' y'

E' .,.

1'

.

.

13' "l>'. f . 8, . K' =1 , which 1s

~ . Collinear and Coe clic Harmonic Grou s of n-members.

3rd Way (Based on Propositions 106 and 121). Since Decagon ABraPZH01K is Harmonic, according to Proposition 106 (§ 19.13), it will have concurrent diagonals at point /\ (Lemoine Point of the decagon). Since for example it is LA/\B= LH/\2, according to known Theorem , for tri-

AA.AB

angles /\AB, /\HZ, it will be:

(AAB)

J\Z.AH = (J\ZH) =

a.a'

~-r .

(6).

In the exact same way, we also find that:

AB.Ar _ (fl.Br) _ l3.13' . AH.fl.0 _ rJ.11' AH.fl.0 - (AH0) - l').rJ' 11 AB.Ar - 13-13'

(7).

.

Similarly we find that :

Ar.M fl.0.AI

y.y'

= 8.8' '

Al.AK M.AE

1••1'

= l>.l>' '

AE./\Z. AK.AA

E.E'

= K.K' .

( 5),

Thus, from relations (6), (7), (8) if we multiply the respective parts and after the proper reductions , we get:

_ a y £ 11 1 a' y' £" 11' 1' 1-13 ° l>, ~. 8. K · 13• l>', 8'. K' 0

r.

(9).

.

Thus, if we take into consideration Proposition 121 (§ 19.34) according to

.

. .

Which It IS

a' y'

£"

a y £ 11 1 p. ~.~.8 .~ =1 , then

11' 1'

.

.

(9), becomes: .

13'. l>'. {' . 8 , . K ' =1 , which 1s the proven relation (1).

134eeEhd (a). Proposition 125 is not a Criterion of Harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are also true for the Harmonic Quadrilateral ABra (For which, according to

a' y'

Proposition

122, it is

13' ' l>' =1 ), for the Harmonic Hexagon A BraEZ (For which, according to

l8J=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. Proposition 123, it is:

a' y' E'

13'. l>' . f' =1 ), for the

.

. .

Harmonic Octagon (Propos1t1on

124), etc. (c). Equivalent Proposition to Proposition 125, is also true for the Asteroid Harmonic Decagon At.HKrZIBE0A (Proposition 127). (d). This Proposition we believe is a new Proposition and firstly appears here. (f). Relevant Propositions are 106, 114, 121-124, 126, 127.

19.40. Significant Property of the Lemoine Point of an Asteroid Harmonic Octagon. Extension to Proposition 124 to the Asteroid Harmonic Octagon . Proposition 126 (Figure 77). For every Asteroid Harmonic Octagon At.HBE0rZA, it is: Ua Uy U, U~ - . - . - . - =1 , Up Ui; U( Ue

(1 ).

where ua, up, Uy, ..., ue, is t he distance of the convergence point K of the diagonals (main) of this octagon, from its side At.=a , t.H=J3, HB=y, ...,ZA=8, respectively.

lnl Since octagon At.HBE0rZA is Harmonic, according to Proposition 105 (§ 19.12), it will have concurrent diagonals at point K (Lemoine point of the octagon) for which, according to Criterion 115 (§ 19.24), it will be: Ua

Up

Uy

Ui;

U,

U(

U~

Ue

1

- -- -- -- -- -- -- -- -- ~ a J3 y l> E ~ 11 8 A a=A.Ua, l3=A.up, y=A.uy, l>=A.ui;, E=A.u,, ~=A.U(, ri=A.u~, 8=A.ue.

(2).

After all , according to Proposition 120 (§ 19.33), it is: i·f·t-i=1 , which

due to (2), becomes : get proven relation (1).

A..U a A.U y A..u , A..u ~ . . . -.. -. . - - . . - - =1 and from which we directly /\.Up /\.U i; "·U ( /\.Ue

~ . Collinear and Coe clic Harmonic Grou s of n-members.

1341,EiUA (a). Proposition 126 is not a Criterion of Harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are also true for the Harmonic

u

u

Quadrilateral ABrA, for which , according to Proposition 122, it is ~-_y_ =1 , Up Uo for the Harmonic Hexagon, for the Harmonic Decagon, etc. (c). Proposition 126, is also true for the Asteroid Harmonic Decagon (Proposition 127), etc. (d). This Proposition we believe is a new Proposition and firstly appears here. (f). Relevant Propositions are 105, 115, 120, 122- 125, 127, etc.

19.41 . Significant Property of the Lemoine Point of an Asteroid Harmonic Decagon. Extension to Proposition 125 to the Asteroid Harmonic Decagon . Proposition 127 (Figure 78). For every Asteroid Harmonic Decagon AAHKrZ1BE0A, it is : Ua Uy U, U~ U, --- - =1 Up · Uo · u~ · Ue · u. '

(1 ).

where Ua, up, Uy, ... , u., is the distance of the convergence point/\ of the diagonals (main) of this decagon, from its side AA=a, AH=f3, HK=y, .. . ,E0=1, 0A=K, respectively.

1'1 Way (Based on Propositions 106, 116 and 121). Since decagon AAHKfZ1BE0A is harmonic, according to Proposition 106 (§ 19.13), it will have concurrent diagonals at point /\ (Lemoine Point of the decagon) for which, according to Criterion 116 (§ 19.25), it will be :

_ __ __ __ __ __ __ __ ____ __ Ua

Up

Uy

Uo

u,

u~

U~

Ue

a

p

y

l5

E

{

rJ

9

u,

u.

1

K

A

~

a=A.ua, f3=A.up, y=A.uy, l>=A.uo, t=A.u,, ~=A.u~, ri=A.u~, 8=A.ue, 1=>..u,, K=A.u •.(2).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

After all, according to Proposition 121 (§ 19.34), it is: A.ua

A.Uy

A.u ,

A.u~

A.u,

/\.Up

/\.U15

/\,U{

/\,U e

/\.u .

i·t-ri-

~=1, which .

due to (2), becomes: .---- . .---- . .--- . .--- . .---- =1 and from which we get the proven relation (1). nd

Way (Based on Propositions 106 and 121).

Since Decagon AaHKrZIBE0A is Harmonic, according to Proposition 106 (§ 19.13), it will have concurrent diagonals at point/\ (Lemoine Point of the decagon). Since for example it is LAM=Ll/1.Z and L/1.Aa=L/I.IZ triangles /\Aa, /\IZ are Ua

Q

similar, thus according to what we know, it will be:

(3).

In the exact same way, we also find that :

.!l=~ y=~ - =~ ~=~ l3

Up '

8

i5

Ue '

U11 ' K

(4).

u.

Hence, from relations (3), (4), if we multiply their respective parts, we get:

a V £ ., I Ua Uy u , u~ u, l3. i5 • (. 8 . K = Up • U11 • U{ , Ue • U•

(5).

Thus, if we take into consideration Proposition 121 (§ 19.34), according to ..

.

which ,t 1s

Ua

OVE:111 p. ~.~. 8 ."iZ =1 , then

-

(5) becomes:

Up

Uy U,

.-

U15

.-

U(

U~

.-

Ue

U1

.-

=1 ,

U•

which is the proven relation (1). 3rd Way (Based on Propositions 106 and 121). Since Decagon AaHKrZIBE0A is Harmonic, according to Proposition 106 (§ 19.13), it will have concurrent diagonals at point /\ (Lemoine Point of the decagon). Since for example it is LAM= Ll/1.Z, according to known Theorem, for tri-

fl.A.M AZ.A.I

angles /\Aa, /\IZ, it will be :

(AA~)

a,u a

= (A.ZI) = {.u{ .

(6).

In the exact same way, we also find that:

/\~.AH fl.I.AB

(MH) A.IB

l3.u p

,

A.I.AB M.fl.H

- - = -(- ) = - - ., - - = 11.u~

11-U~ l3.u p .

(7).

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Similarly we find that:

/\H./\K /\8./\E

/\E./\0

y.uy

/\r.J\Z /\0./\A

I.U ,

= 8.u a ' /\K./\r =

l>.U 5 '

E.U ,

=

K.U .

(8).

Thus, from relations (6), (7), (8) if we multiply the respective parts and after the proper reductions, we get:

a y E 11 I U a Uy U , U n U, 1- - .- .- .- .- - .- .- . - . - .

l3

l) ~ 8 K

Up

U5

U;

Ua

(9).

U•

Thus, if we take into consideration Proposition 121 (§ 19.34) according to

. . . OVEll

which 1t 1s ~. ~. ~. Ua

1=1. -

Uy

.-

Up

U5

U,

.-

U;

Un

.-

Ua

1

9 .K=1 , then (9), becomes: U,

.-

or

U•

Ua

-

Up

Uy

.-

U,

.-

U5

U;

Un

.-

Ua

U,

.-

=1 , which is the proven re-

U•

lation (1 ).

1;34u6hff (a). Proposition 127 is not a Criterion of Harmonic being, as its converse is not true. (b). We notice that equivalent Propositions, are also true for the Harmonic Quadrilateral ABrt.., for which, according to Proposition 122, it is ;: . ~: =1, for the Harmonic Hexagon ABrt..EZ, for which, according to Proposition

. . a' y' E' . 123, 1t 1s: 13 , . l>' • =1 , for the Harmonic Octagon , etc.

f

(c). Equivalent Proposition to Proposition 127, is also true for the Asteroid Harmonic Octagon (Proposition 126). (d). This Proposition we believe is a new Proposition and firstly appears here. (f). Relevant Propositions are 106, 114-116, 121-126.

19.42. General Conclusion. Due to Propositions 122-127 above and if we work in the same way as in their proofs, we reach the following result:

18l=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. Equivalent Proposition, to Propositions 119-127, can be made and proven to be true for any other Harmonic Multilateral convex or asteroid, with an even number of sides.

19.43. Criterion of Harmonic being of a Simple Inscribed Pentagon. Extension to Criterion 117. (Theorem analogous to Gergonne Theorem for the Harmonic 5-gon). Criterion 128 (Figure 83). We consider the inscribed in circle (o), convex pentagon ArEHI and its respective (its polar converse), circumscribed (tangential) pentagon ayEr11 (where a the intersection of the tangents at points E, H, where y the intersection of the tangents at points H, I, and so on and so forth). Pentagon ArEHI is Harmonic, if and only if, Aa, rv, EE, HI'), II , coincide at point let's say A.

~

Ix,jµa 83.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

(611•111431 I.e., if it is

(1 ).

AanrvnEEnHr,nl1 =A ,

We will prove that ArEHI is harmonic. According to the relevant definition, for ArEHI to be Harmonic(§ 16.1e), on circle (o), we need to have points B, /1, Z, 0 , K, such that the convex Decagon ABr/1EZH01K, would be Harmonic. In fact, if Aa, rv, EE, Hr,, 11 , intersect circle (o), at points Z, 0 , K, B, 11, respectively, then since for example AZ runs through pole a of EH , then quadrilateral EZHA would be Harmonic. Similarly we find that the rest four quadrilaterals of the pentagon are Harmonic , i.e. H01r, IKAE, ABrH, r/1EI, are Harmonic, then, according to Proposition 111 (§19.19), Decagon ABr/1EZH01K is Harmonic, s ince, from the implication, relation (1) is also true. Thus, since for the intersections B, 11, Z, 0 , K, Decagon ABr/1EZH01K is harmonic, then according to the relevant definition (§ 16.1e) pentagon ArEHI, is Harmonic as well.

tGiti•leN§?i4i I.e., if pentagon ArEHI is Harmonic, we will prove that Aa, rv, EE, Hr,, II, coincide (Theorem analogous to Theorem Gergonne for the Harmonic 5-gon) . Since pentagon ArEHI is Harmonic, according to the relevant definition (§ 16.1 e), in circle (o), we will have points B', /1', Z', 0', K', for which the convex decagon ABT/1'EZ'H0'1K', is Harmonic. Therefore, since ABT/1'EZ'H0'1K', would be Harmonic, then its ten quadrilaterals will be harmonic too:

ABTH , BT/1'0', r/1'EI,

/1'EZ'K' , EZ'HA, Z'H0'B', H0'1r, 0'1K'/1', IK'AE, K'AB' Z' .

(2).

Thus, since for example quadrilateral EZ'HA would be Harmonic, then AZ', will be conjugate to EH , hence AZ' will run through pole a of EH. Th is means that then

z· is the intersection of Aa with circle (o). But the intersec-

tion of Aa with circle (o) is called Z. This means that Z'=Z. In the same way we also find that :

B'=B, 11'=/1, 0'=0, K'=K.

Therefore, since B', /1', Z', 0 ', K', were received in such way that decagon

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

ABTt.'EZ'H0'1K', would be Harmonic, then ABrt.EZH01K would also be harmonic, hence its diagonals AZ, re, EK, HB, It., or Aa, rv, EE, Hri, h, will coincide at point let's say/\ (Proposition 106, § 19.13).

Due to the above Criterion of Harmonic Pentagons (Criterion of Harmonic being of convex inscribed pentagons), we reach the following conclusion: Each convex pentagon ArEHI inscribed in circle {ol, is harmonic, if and only if, for it and for its respective {its polar converse), circumscribed (tangential) pentagon ayEQI (where a is the intersection of the tangents at points E, H, where y is the intersection of the tangents at points H, I, and so on and so forth), diagonals Aa, ry, EE. Hn, h, of the decagon AnriEaHylE, are concurrent. Therefore, from now on we can use for Harmonic pentagons, the following alternative Definition: «Harmonic Pentagon is called, each convex pentagon ArEHI inscribed in circle (o), if for it and its respective (its polar converse), circumscribed (tangential) pentagon ayEl"ll (where a is the intersection of tangents at points E, H, where y is the intersection of tangents at points H, I, and so on and so forth) , diagonals Aa, rv, EE, Hri, h, of the decagon ArinEaHylE, are concurrent».

134eeEiUA (a). The above Criterion of Harmonic being of Pentagons, is rather significant, as it allows us to have an alternativer simpler DEFINITION, from our initial definition, about Harmonic Pentagons, as well as because it has many appl ications, as we will realize from now on. (b). This Criterion we believe is a new Proposition and firstly appears here. (c). It is obvious that the converse of Criterion 128, is a Theorem analogous to Gergonne Theorem for the Harmonic 5-gon . (d). Relevant Propositions are 109, 110, 111 , 129, 130.

19.44. Criterion of Harmonic being of a Simple Inscribed Pentagon. Extension to Criteria 117 and 128.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Criterion 129 {Figure 83). We consider the inscribed in circle (o) convex pentagon ArEHI. If on circle (o) we have points B, fl., Z, 0 , K, for which the five quadrilaterals: ABrH , rll.EI , EZHA, H01r, IKAE,

(1).

are Harmonic and Decagon ABrb.EZH01K is convex with concurrent diagonals (main), then and only then pentagon ArEHI is Harmonic.

[Ql•Jli491 I.e., if for the convex decagon ABrb.EZH01K, it is: AZnr0nEKnHBnlb.

=/\,

(2)

And if the quadrilaterals in relation (1 ), are Harmonic, we will prove that pentagon ArEHI is Harmonic too.

1 st Way (Based on Proposition 128). If the 5-gon ayt111 is the respective circumscribed one of the 5-gon ArEHI (where a is the intersection of the tangents at points E, H, where y is the intersection of the tangents at points H, I, and so on and so forth), since, for example quadrilateral EZHA, is Harmonic from the hypothesis, then AZ, will run through pole a of EH. Sim ilarly we also find that re, EK, HB, lb., will run through poles y , t , 11, 1, of HI, IA, Ar, rE, respectively, hence relation (2) becomes:

AanrvnEtnH11nh =/\.

(3).

Hence, since (3) is true, according to the direct of Criterion 128 (§ 19.43), pentagon ArEH , is Harmonic.

nd

Way (Based on Proposition 111 ).

Since Decagon ABrb.EZH01K, has concurrent diagonals and relation (1) is true, it will be harmonic (Direct of Criterion 111 ). Hence, we see that for Pentagon ArEHI there are points B , fl., Z, 0 , K, for which decagon ABrb.EZH01K, is harmonic. This, according to the relevant definition (§ 16.1 e), means that pentagon ArEHI is harmonic too.

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. tl:ll3·l,W4t14 I.e., if Pentagon ArEHI is Harmonic, we will prove that on circle (o), there are points B, a ,

z, 0,

K, for which on decagon ABraEZH01K, AZ, r0, EK,

HB, II::,,., coincide and the five quadrilaterals in relation (1), are Harmonic. 1st Way (Based on Proposition 128). Since pentagon ArEHI is harmonic, according to the converse of Criterion 128, Aa, rv, E£, Hri, h, will be concurrent (where a a is the intersection of the tangents at points E, H, where y is the intersection of the tangents at points H, I, and so on and so forth). Hence, if Z, 0 , K, B, a , are the intersections Aa, rv, E£, Hri, h, with the circle (o), then for Z, 0 , K, B, a , decagon ABfl::,,.EZH01K will in fact have AZ, r0, EK, HB, II::,,. concurrent and the five quadrilaterals in (1) are harmonic, as AZ, r0, EK, HB, 11::,,., run through poles a, y, £, ri,

1,

of EH, HI, IA, Ar, rE respec-

tively. nd

Way (Based on Proposition 106).

Actually, since pentagon ArEHI, is Harmonic, according to the relevant definit ion (§ 16.1e), there will be points B, a , Z, 0 , K, for which decagon ABraEZH01K, will be Harmonic, thus, according to Proposition 106 (§ 19.13), its diagonals will be concurrent. Hence it will be AZnr0nEKnHBnll::,,. =A Moreover, since for points B, a , Z, 0 , K, decagon ABraEZH01K, is Harmonic, according to the relevant definition of the Harmonic Decagon(§ 16.1d), it will have Harmonic all its ten quadrilaterals, among which there are also the five quadrilaterals in relation (1 ). Hence the five quadrilaterals in relation (1 ), are in fact Harmonic.

(+i•ll[411i1t•leh Due to the above Criterion of Harmonic Pentagons (Criterion of Harmonic being of convex inscribed pentagons), this conclusion is reached: Every convex pentagon AfEHI inscribed in circle (0) 1 is harmonic, if and only if, on circle (ol there are points B, a, Z, 0, K, for which the following five quadrilaterals ABrH, fl::,,.EI, EZHA, H01r, IKAE, are Harmonic and decagon ABfl::,,.EZH01K is convex with concurrent diagonals (main). Alternative Definition of the Harmonic Pentagon.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Consequently, from now on we can use for Harmonic pentagons as well, the alternative Definition below: «Harmonic Pentagon is called, every convex pentagon ArEHI inscribed in circle (o), if and only if, on circle (o) there are points B, I!, Z, 0, K, for which the five following quadrilaterals ABrH , rl!EI , EZHA, H01r, IKAE, are Harmonic and decagon ABrl!EZH01K is convex with concurrent diagonals (main)».

1.141,Fiid (a). The above Criterion of Harmonic being of Pentagons, is rather significant. as it allows us to have an alternative simpler DEFINITION, from our initial definition, for Harmonic Pentagons, as well as because it finds many applications, as we will later realize. (b). This Criterion we believe to be new and firstly appear here. (c). Relevant Propositions are 106,111 , 117,128,130,131.

19.45. Criterion of Harmonic being of the Inscribed Pentagon. Extension to Criteria 117, 128 and 129. Does the Harmonic Pentagon have an Analogous Gergonne Point? Criterion 130 (Figure 83). Every Harmonic Pentagon and its respective tangential or its polar converse pentagon, are homologous, or in another formation: The tangential or polar converse pentagon, of every Harmonic Pentagon, has an analogous Gergonne points, and conversely: If the tangential or polar converse pentagon, inscribed in circle of a pentagon, has an analogous Gergonne point, then the second pentagon is Harmonic.

tEll•Jlk431 I.e. if ArEHI is a Harmonic Pentagon inscribed in circle (o) and ayE111 (where

IB=-·__

_,_ H:.:..A=R=M"-'O =N.:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

a is the intersection of the tangents at E, H, where y is the intersection of the tangents at H, I, etc), is its tangential or polar converse pentagon, we will prove that it is:

AanrvnEtnHrinh=/\,

(1)

(/\ is analogous to the Gergonne point of the triangle). Since pentagon ArEHI is Harmonic, according to the converse of Criterion

128 (§ 19.43), relation (1) will in fact be true. [In which relation /\ is also the convergence

point

of

the

diagonals

of

the

Harmonic

Decagon

ABr.l'lEZH01K, which is also the Lemoine point of the Harmonic Decagon, as well of the Harmonic Pentagon ArEHI (Criteria 114 and 117)].

tGJ13·l,W§i-i4i I.e., if for the convex inscribed in circle (o) pentagon ArEHI and for its tangential Pentagon ayEl')I, relation (1) is true, we will then prove that ArEHI is Harmonic. Since from the hypothesis relation (1) is true, according to the direct of Criterion 128 (§ 19.43) , pentagon ArEHI is in fact Harmonic.

Due to the above and since the convergence point of diagonals/\ of the respective Harmonic Decagon is also the Lemoine point of this Harmonic Decagon [Criterion 114 (§ 19.23)], as well as of its twin respective Harmonic Pentagons [Criterion 117 (§ 19.26)], the following conclusion is reached : «The tangential or the Polar converse pentagon of ever Harmonic Pentagon, has a point analogous to the Gergonne point of a triangle, which coincides with the common Lemoine Point of the Harmonic Pentagon, as well as its respective Harmonic Decagon, which is a convergence point of the diagonals of this Harmonic Decagon».

134uEOA (a). We believe that equivalent Propositions, are true for Triangle, etc. (b). This Proposition, we believe to be a new Proposition and firstly appear here.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

(c). Actually Propositions 128 and the above 130 are the same, but in another formation. (d). Relevant Propositions are 114,117,128, 129, 131.

19.46. Significant Property of the Lemoine Point of a Simple Harmonic Pentagon. Extension to Propositions 117, 128-130. Proposition 131 (Figure 84). We consider a Harmonic Pentagon ArEHI inscribed in circle (o) and the Lemoine point/\ of it. If A/\, r/\, El\, HI\, 1/\ re-intersect circle (o), at points Z, 0 , K , B, fl, respectively, then pentagon B/120K will be Harmonic t oo and it shares its Lem oine point/\ with ArEHI.

rxrjµa 84.

81=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Since pentagon ArEHI is Harmonic, according to Criterion 117 (§ 19.26), it will in fact have a Lemoine point /1.. If ua, uy, u,, u~, u,, is the distance of /1. from side Ar=a, rE=y, EH=E, Hl=ri, IA=,, respectively, according to Criterion 117 (direct)(§ 19.26), it will be: Ua

Uy

U,

U~

a

y

E

11

Thus it will also be for example:

u, E

U,

u~ 11

(1 ).

(2).

Therefore in triangle EHi , H/1. or HB is the symmedian (converse of Theorem

11-14.2 in Geometry [6]). Hence, according to the direct of the Auxiliary Proposition 6 (§ 2.6) in Part E, El, HB, will be conjugate, thus quadrilateral BEHi will be Harmonic. Consequently, since quadrilateral BEHi is Harmonic and from the implication it is also KE

n BH n .61=/I.,

according to Proposition 34 (§ 8. 71) quadri-

lateral KB.6H, will be Harmonic too. Thus, since quadrilateral KB.6H is Harmonic, then .6K and BH will be conjugate, hence BH will be the symmedian of .6K in triangle KB.6 [Proposition 6 (§ 2.6) in Part E] (converse)). In this way, all the points on BH, hence /1. as well, will be far from sides BK=K, B.6=13 distance,

u.,

Up analogous to these sides respectively (direct of

Theorem 11-14.2 in rEwµETpia [6)). Hence, it will be:

u.

Up

~=13·

(3).

In the exact same way we prove that the distance of /1. from the pairs of sides .6B-.6.Z, Z.6-20, 0Z-0K, K0-KB, are analogous to these sides. Thus the following relations are true:

(4). From relations (3), (4) it occurs that relation (1) is true, i.e. that the distance

~ - Collinear and Coe clic Harmonic Grou s of n-members.

of /\ from each side of the pentagon B.t.20K is analogous to its respective sides. This means that pentagon B.t.20K is Harmonic and /\ is a Lemoine point for this pentagon as well [converse of Proposition 117 (§ 19.26) and the comment after its conclusion].

IU411Fhb (a). Equivalent Propositions can be formed for hexagons as well (Corollary

3 of Criterion 139), etc. This time as well, the proofs are carried through with in similar ways, as for pentagons ArEHI and B.t.20K. (b). Proposition 131 , we believe to be new and firstly appear here. (c). Relevant Propositions are 34, 114, 117, 128, 129, 130, 139.

19.47. Significant Property of the Simple Harmonic Decagon . Proposition 132 (Figure 84). The twin pentagons, of every Harmonic Decagon, are harmonic.

ml I.e., if the inscribed in circle (o) decagon ABrAEZH01K, is harmonic, we will prove that its twin pentagons ArEHI and B.t.20K, are harmonic as well. In fact, pentagon ArEHI is harmonic, as, according to the relevant definition (§ 16.1 e), there on circle (o) points B, A, Z, 0 , K, for which decagon A-

BrAEZH01K, is harmonic. In the exact same way we prove that pentagon B.t.20K, is harmonic as well.

liE~l18), the beams of which re-intersect circle (o) at points A', B', r'

a·, E', Z', H', 0', respectively. We say that the desired Harmonic Octagon is A'BT'a'E'Z'H'0'.

[01:/t#t!il If we work similarly to the proof of the converse of Construction 81 , we find that the Group of Eight Points a ,

13, y, l>,

E, ~.

11, 8, is harmonic, as octagon

ABraEZH0, is Regular, while if we work afterwards, as in the direct of the proof of Construction 101 and based on this group of eight, we find that octagon A 'BT'a 'E'Z'H'0 ', is harmonic, as it is inscribed in circle (o') and then the group of eight points a,

13, y, l>, E, ~.

11, 8, is harmonic.

Comment. It is obvious that the method above is another way of construction of Harmonic Group of Eight and of a Harmonic Octagon, apart from the ones we used in Constructions 63 (§ 15.3) and 149 (§ 19.64).

;w.u@Yt11 For the converse we follow the opposite way.

tm11-ttm,11i1 I.e. here we are given the Harmonic Octagon A'BT'a'E'Z'H'0', which is inscribed in circle (o') and we are asked, based on that and with proper projective transformations to construct a Regular Octagon ABraEZH0, inscribed in circle (o).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

If we take again into consideration Propositions 101 and 81 and we £2!!!.: bine them properly, we are led to the construction below (For our convenience and for a better understanding of the solution, we just need to properly combine figures 70 and 60, in such way that figure 49 occurs): b . Construction. We work similarly to Constructions 101 (converse) and 81 (direct), respectively. I.e., using an arbitrary point T as center of circle (o'), we form the Central Pencil T(A'BT'Jl'E'Z'H'0') whose beams, we suppose that intersect a random line (t), at points a, 13, y, ~. £, ~. r,, 8, respectively. We find the Orthooptical Point r (§ 14), of the group of eight a, 13, y, ~. £, ~.

r,, 8, from which we suppose that circle (o) runs through and we define on circle (o) the intersections A, B, r, Jl, E, Z, H, 0, of pencil I:(al3y~t~r,8) [If circle (o), does not run through

r,

then we work as in the direct of Con-

struction 81 (§ 15.23,(a) 2/)]. We say that octagon ABrt.EZH0 is the desired Regular one.

t01=tt®I If we work similarly to the proof of the converse of Construction 101, we find that the Group of Eight Points a, 13, y, ~. £, ~. 11, 8, 11, is harmonic, as octagon A'BT'Jl'E'Z'H'0' is harmonic, while if we work afterwards, as in the direct of the proof of Construction 81, we find that octagon ABrt.EZH0, is a Regular Octagon as it is inscribed in circle (o), the group of eight points a,

13, y, ~.

£, ~.

11, 8, is harmonic and r, which belongs to circle (o), is the Or-

thooptical Point of the Harmonic group of eight points a, 13, y, ~. £, ~. 11, 8, from the way it is constructed.

lti4u6UA (a). Problem 154, the way it is given, it is undefined. If we want to define it, then we need to be given sufficient data. (b). The Problem above and its solution, we believe to be new and firstly appear here. (c). We notice that with this method, firstly the harmonic collinear group of eight occurs and then the desired harmonic octagon.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

I.e. using the New Method of the Harmonic Transformation, we simultaneously achieve the Construction of the Harmonic Group of Eight as well. (d). Relevant Propositions are 35, 63, 77, 81 , 97, 101 , 149.

22.4 . Construction of a Harmonic Decagon from a Regular Decagon. (How from a Regular Decagon a Harmonic Decagon occurs). MAGNIFICENT CONSTRUCTION . Construction 155 (Figure 99). Based on proper projective transformations and intersections from a line and circles, construct a Harmonic Decagon based on the Regular Decagon, given to us and conversely, similarly and based on a Harmonic Decagon, given to us construct a Regular Decagon.

q,=(1/5)L=18 µoipE. • ya = 1·

(10).

A'B' r' .a• ET' BT' . .a'E' . r• A' = 1·

(11 ).

From relations (9)-(11 ), comes the invariant of this case. Relation (9) is obvious, as it is AB=.aE, Br=r.a, Ar=rE, relation (10) is true due to Criterion 4 [relation (1 )] and (11) if we take into consideration Criterion 37 [relation (2')].

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

[Since angle

yr~ is right and

in triangle l:aE,

r13

and Hi are isogonal, as it is

aj3 yf, El:

known, it will also be:

(10')].

J3v · flt · ra =1

4/. According to Criteria 5 and 38 (Figure 107). We will only refer to Harmonic Involutions of two Pairs of Conjugate Points, as only then Corollary 168 is true.

rxrum 101.

It is:

Ar 2 rE 2

AZ 2

AB A/1

=ZE 2 =BE. /1E.

ay 2 a{ 2 aj3 afJ yE2 = ~£2 = j3E. fJE.

AT' A'Z' A'B' A' /1' --=--=---r'E'2 Z'E' 2 B'E'. /1'E'. 2

(12).

(13).

2

From relations (12)-(14), comes the invariant of this case.

(14).

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Relation (12) is obvious, as AB=b.E, Ab.=BE, Ar=rE, AZ=ZE, relation (13) is true according to Criterion 5 [relation (1)] and (14) if we take into consideration Criterion 38 [relations (1) and (1 '). [Since angle

yr~ is right and

in t ri angle

raE, r13

and

r~

are isogonal , as it is

known (Theorem of isogonals) ,it will also be:

ra 2 ay 2

a( 2

ap a~

rE 2 = yE 2 = (E 2 = PE· ~E ·

(13')].

5/. According to Propositions 8 and 41 (Figures 100 and 102). We will only refer to Harmonic Involutions of one and four Pairs of Conjugate Points, as in an equivalent way it is proved that the same applies to n Pairs of conjugate points. al. Of one pair (Figure 100). The above relation (1 ), can also be written in the following way:

AB r~ ap y~ A'B' r· ~· Br.~ =1 , py. ~a =1 , BT' . ~· A' =1 ,

(15).

From relation (15), comes the invariant of this case.

AB

r~

The fact that Br.~ =1 , is obvious as AB=Br and Aa=ar (The rema ining two equalities, are known to us). b/. Of four pai rs (Figure 102). It is:

AB r~ EZ H0 IK Br . ~E. ZH" 01 . KA = 1·

ap

y~

E(

.,a

IK

(16).

Pv · ~E · )=(AT'B'.6')=1.

(1 ).

From relation (1 ), comes the invariant of this case. The fact that (ArB.6)=1 , is obvious as AB=Br=r.6=.6A (The rest second and third, are known to us, § 11.1 a and 16.1 a, respectively). b/. Decagon (Figure 99). It is :

(ArBH)= (B.6r0)=(rE.61)= ... =(KBAZ)=1 .

(2).

(avl311)= (l3l>y8)=(y£l>1)= ... =(Kl3a{)=1 .

(3).

(AT'B'H')= (B '.6T'0 ')=(r'E'.6'1')= .. .=(K'B'A'Z')=1.

(4).

From relations (1 )-(3), comes the invariant of this case. Relation (2) is obvious as AB=Br=r.6= ... Kai AH=B0=rl=...

, relations (3)

and (4) are true due to the relevant definitions § 11.1 d and 16.1 d, respectively. [It will obviously be :

r(ayl311)= r(J3l>y8)=r(yt:l>1)= ... =r(Kl3a{)=1.

(3')].

2/. According to Criteria 86 and 121 (Figures 97 and 99). We will only refer to Harmonic Quadrilaterals and Decagon, as in an equivalent way it is proven that the same applies to 2n-l aterals. a/. Quadrilateral (Figure 97). The above relation (1), is also w ritten in the followi ng way :

AB rA aJ3 yfJ A'B' r'A' Br . M = 1 ' J3y . 7:Ja = 1 ' BT' . A' A' = 1'

(4').

From relation (4'), comes the invariant of t his case.

AB rA

The fact that Br . M =1 , is obvious as AB=Br=r.6=AA, the second and third parts, are known to us. b/. Decagon (Figure 99). It is :

AB rA EZ H0 IK Br . AE . ZH" 01 . KA = 1,

(5).

al3 yf, £~ 118 IK 13v · fJE · ~11 · 81 · Ka =1•

(6).

A'B' r'A' E'Z' H'0' l'K' BT' " A'E' . Z'H' 0'1' "K'A'= 1·

(7).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

From relations (5)-(7), comes the invariant of this case. Relation (5) is obvious, as AB=Br=ra= ... , relation (6) and (7) are true due to Propositions 86 and 121, respectively.

3/. According to Propositions 87, 88 and 144 (Figures 97 and 98). We will only refer to Harmonic Quadrilaterals and Decagon , as in an equivalent way it is proven that the same applies to 2n-laterals. a/. Quadrilateral (Figure 97).

Ar.Bf.

It is: AB.rll=Br.M= - 2-

·

ay.l3l> AT' .B' f.' , A'B'.r'Jl'=BT'.Jl'A'= al3 .yl>=l3y.5a=-2. 2

(8).

From relation (8), comes the invariant of this case. The second relation in (8) is true due to Proposition 87, while the first and third one are true due to the known Proposition in [32] page 53 IV, as quadrilaterals ABra and A 'BT'Jl', are harmonic (Proposition 151). b/. Hexagon (Figure 98). It is:

AB.rll.EZ=Br.aE.ZA-

al3.y6.E~=l3y.6£.~a=

Af..BE.rZ 8

al>.l3E.y~

8

,

A' f.'.B'E'.r'Z'

A'B'.r'Jl'.E'Z'=BT'.Jl'E'.Z'A'= - - - - 8

(9).

(10).

(11 ).

From relations (9)-(11 ), comes the invariant of this case. Relation (10) is true due to Proposition 88, while (9) and (10) are true due to Proposition 144, as hexagons ABrllEZ and A'BT'Jl'E'Z' are harmonic (Proposition 151 ).

4/. According to Criterion 114. We will only refer to Harmonic Quadrilaterals and Decagon , as in an equivalent way it is proven that the same applies to 2n-laterals. a/. Quadrilateral (Figure 97). It is:

(12).

8 =-·__

_,_ H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y'-'-P=art'-'-=B.

a''

J3''

y''

~··

-=-=-=(Where a',

13',

(13).

y', '6', is the distance of 0, from each side AB=a, Br=l3, rt::..=y,

t::..A='6, respectively and a", 13", y", '6", the distance of intersection K of AT' and B't::..', from each side A'B'=a1, BT'=l31, r·t::..·=v1, t::..'A'='61, respectively). From relations (12) and (13), comes the invariant of this case. It is obvious that relation (12), is true as it is a'=J3'=y'='6' and a=l3=y='6, while relation (13) is true due the Proposition in page 54 IV in [32]. b/. Decagon {Figure 99).

13' y' a f3 V a'' J3'' y'' - =- =a'

It is:

(Where a',

13',

K'

(14).

K

K"

(15).

y', ... ,K', is the distance of 0 , from each side AB=a, Br=13,

rt::..=y, ... , KA=K, respectively and a", 13", y", ... , K", the distance of intersection/\ of A'Z', B'H' , ... from each side A'B'=a1, BT'=l31, r·t::..·=v1, ... , K'A'=K1, respectively). From relations (14) and (15), comes the invariant of this case. It is obvious that relation (14) is true, as it is a'=J3'=y'= ... =K' and a=l3=y= ... =K, while relation (15) is true due to Criterion 114.

5/. According to Criteria 122 and 125. We will only refer to Harmonic Quadrilaterals and Decagon, as in an equivalent way it is proven that the same applies to 2n-laterals. a/. Quadrilateral (Figure 97).

a' y'

It is:

13'. l)' =1, a" y" jr-·~=1.

(Where a',

13', y', '6',

(16).

(17).

is the distance of 0 , from each side AB=a, Br=l3, rt::..=y,

t::..A='6, respectively and a", 13", y", '6", the distance of intersection K of AT'

i§Q1J .

Collinear and Coe clic Harmonic Grou s of n-members.

and B'.6.', from each side A 'B'=a,, BT'=j3,, r·.ti.·=v, , .6.'A'='i:J,, respectively). From relation (16) and (17), comes the invariant of this case. It is obvious that relation (16) is true, as it is a'=J3'=y'='i:J', while relation (17) is true due to Criterion 122. b/. Decagon (Figure 99).

a' y' ~

It is:

rf ~ -

13' . l)' . (' . 8' . K' -1 , ~

r_

~ ~

~-

J3". ij". ~". 8". K" - 1·

(18).

(19).

(Where a', 13', y', ... , K', is the distance of 0, from each side AB=a, Br=13, r.ti.=y, ... , KA=K, respectively and a",

13", y",

... , K", the distance of intersec-

tion /\ of A'Z', B'H' , ... from each side A'B'=a,, BT'=J31, r·.ti.·=v1 , ... , K'A'=K1, respectively). From relation (18) and (19), comes the invariant of this case. It is obvious that relation (18) is true, as it is a'=j3'=y'= ... =K', while relation (19) is true due to Criterion 125.

3. For Harmonic Lines Collinear and Cocyclic . (a). According to Relevant Definitions (§ 23.1, 23.2 and 23.3) ( Figure 111 ). It is: (ArBB') =(B.ti.rr')=(rEM')= ... = (AT'B'B) =(B'.6.TT)=(r'E'.6.'.6.)= . .. =1.

(1).

( ayl313') =(j3'i:Jyy')=(yE'i:J'i:J')= ... =( a 'y'J3'13) =(j3''i:J'y'y)=(y'E''i:J''i:J)= .. . =1 .

(2).

(A1r 1B1B'1)=(B1.6.1r1r'1)=(r1E1.6.1.6.'1)= ... =(A'1r'1B'1B1)=(B'1.6.'1r·1r1)=(r'1E'1.6.'1.6.1) .. .. =1.

(3).

From relations (1) - (3), comes the invariant of this case. Relation (1) above it obvious, as it is for example AB=Br and AB'=BT, etc (definition§ 23.3). Relations (2) and (3) are true due to the relevant definitions § 23.1 and § 23.2, respectively. (b). According to Criterion 178 (Figure 111 ). If TT, p, o , T, .. • , is the distance of O from each one of the following AB=a,

l8J=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Br=j3, ra=v, aE=l>, ... , if rr', p', a', r', ... is the distance of O from each one of the following A'B'=a', BT'=j3', r ·a·=v', a'E'=l>', ... and if TT1 , p1, 1 .. . , av rr'1 , p'1 , '1... , respectively, then it will be: TT

p

a

T

TT'

p'

a'

T'

-a =-13 =-v =-l> - ... - -a' =-13' =-y' =-l>' - ....

(4).

(5). From relations (4) and (5), comes the invariant of this case. It is obvious that relation (4) is true, as it is rr=p=a=r= ... rr'=p'=a'=r'= ... and a=j3=y=l>= ... a'=j3'=y'=l>'= .. ., while (5) is true due to Criterion 178.

j;(4116hb (a). It is obvious that during the Harmonic Transformation, the circumscribed circle of the Regular or Harmonic Multilateral stays invariant as well (Except of course of the size of the circle, which we choose at will). (b). Relevant Propositions are 4, 5, 8, 33, 37, 38, 41 , 86-89, 114, 121, 122,

125,144, 151 , 178.

~ - Collinear and Coe clic Harmonic Grou s of n-members.

H' PAIR OF HARMONIC COLLINEAR ND COCYCLIC LINES. 3. In General. It has to do with Pairs of Open Harmonic Collinear and Cocyclic Series of Points, n points of each series of points, or Pairs of Polygonal Harmonic Lines, which have some harmonic properties, to which we will refer below. We could not find features, regarding the Harmonic Lines, in bibliography, even though we asked for relevant information at the mathematica.gr website, no such information has been found until today (except for Harmonic Groups of Four Points and Harmonic Quadrilaterals, that we have forementioned as known to us). So, at first we approached the matter according to our own view and intuition, while later on we developed and used relevant methods Through the passing of time and after research and study, we realized that the Pairs of Open Collinear Groups of n, is an extreme case of Pairs of Cocyclic Harmonic Groups of n, during which their circumscribed circle is degenerate to a line. Hence the circle to which they belong has a radius of infin ite length (With zero curvature obviously).

8 =-·__

_,_H:.:..A=R=M"-'O =N .:..:l..::; C....;:G=E=O =M =E =-T -'-'R ..a.Y '-'-P=art'-'-=B.

23.1. Pairs of Open Harmonic Collinear Series of Points of 2n Points. (Harmonic Polygonal Line degenerate to a line). Definitions. We consider, on line (E), the group of n ordered points A , B, r , 11, E,... ,N (Figure 108) and consequently, the group of n ordered points A', B' , r·, 11',

E',... ,N'. 11'

(E)

E'

A

The set of the above 2n points will be called «Harmonic Open Collinear Series of Points of 2n Points or more simply Harmonic Polygonal Collinear Line, or Harmonic Collinear Line», if all the groups of four points below, are harmonic: A , B, r, B', - B, r, /1, r', - r, /1, E, /1', ,n-2 groups A', B', r•, B - B', r•, /1', r, - r•, /1', E', 11,.., n-2 -//-

}

or 2(n-2) groups, (1).

the above 2(v-2) groups of four points in relation (1), we will call them « Group of 2(n-2) Groups of Four Points, of the Harmonic Open Collinear Series of Points of 2n Points».

The following line segments AB, Br, r11,

/1E, ... and A'B', BT', r· /1', /1' E', ... we will call them sides and points A , B , r, 11, E, ... N, .. . A', B', r·, 11', E', . .. N', . .. we will call them apexes of this Harmonic Open Series of Points of 2n Points, while the n segments, AA', BB',rr•,

M', EE', .... ,NN', we will call them «Main Diagonals» of it. Remarks. (a). We named the above group of 2n points «Collinear», since, as we will see below, there is a matching Harmonic Open «Cocyclic» Series of Points of 2n Points. (b). Each such series of points of 2n points, consists of two series of points of n points each .

~ - Collinear and Coe clic Harmonic Grou s of n-members.

(c). The above collinear group of 2n points, can be considered as degenerate «Cocyclic» Series of Points of 2n Points, whose the circumscribed circle is degenerate to a line (£), as it has a radius of infinite length (zero curvature) . (d). Below we will also mention various properties that each Harmonic Open Collinear Series of Points of 2n Points has, as well as some Criteria of its Harmonic being. Moreover we will refer to the ways of construction of an arbitrary Harmonic Open Collinear Series of Points of 2n Points (Construction 171 , etc), among those also using the Method of Harmonic Transformation (Construction 179). 23.2.

Pair of a Harmonic Open Convex Cocyclic Series of Points of 2n Points (Apexes). (Harmonic Open Polygonal Line). Definitions .

r·

We consider on circle (o), the group of n ordered points A, B, r,

E'

\

t,., E, ... ,N (Figure 109) and consequently the group

\

J

(o

N'

A

of

points A',

n

ordered

B', r• , t,.' ,

E', ... ,N'. The set of the above 2n points, we will call «Pair

of

Harmonic

Open Convex Cocyclic

r rx~µa 109.

Series of Points of 2n Points, or more simply Harmonic

Polygonal

Cocyclic Line, or Harmonic Cocyclic Line», if all its 2(n-2) quadrilaterals below are harmonic : ABrB', Brar·, rtJ.EtJ.', ..... .,n-2 quadri laterals } A'BT'B, BT'tJ.T, r"tJ.'E'.t., ... ,n-2 quadrilaterals

2(n-2) quadrilaterals, (2).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

The above quadrilaterals in relation (2), we will call them «2(n-2l Quadrilaterals of the Harmonic Open Cocyclic Series of Points of 2n Points». The line segments AB, Br, ra, aE, ... and A'B', BT', r·a·, a'E', ... we will call them sides, the points A, B, r, a, E, ... ,N, ... , - A', B', r·, a·, E', ... , N', ... we will call them apexes of the Harmonic Open Cocyclic Series of Points of 2n Points, while the n segments, AA' , BB' , rr·, M', EE', .... ,NN', we will name them its «Main Diagonals». Remarks. (a). We called the above group of 2n points «Cocyclic», since as we saw above, there is a matching Open Harmonic «Collinear» Group of 2n Points. (b). Each such series of points of 2n points (apexes), consists of two series of points of n points each. (c). It is obvious that if the radius of the circumscribed circle of this Open Harmonic group of 2n, is of infinite length or if the circle has zero curvature, then this group of 2n is degenerate to an Open Harmonic «Collinear» group of 2n Points. (d). Below we will also mention various properties that each Open Harmonic Cocyclic Series of Points of 2n Points has, in the form of Propositions and Criteria of its Harmonic being. For example the convergence of its diagonals (Proposition 177), etc. Moreover we will refer to the ways of construction of an arbitrary such group of 2n points, among which also using the Method of Harmonic Transformation.

23.3.

Pair of a Regular Open Cocyclic Series of Points of 2n Points.(Regular Open Polygonal Cocyclic Line).

Definitions. We consider on circle (o), the group of n ordered points A, B, r, a, E, ... ,N (Figure 110) for which it is:

AB=Br=ra=aE= ... =1,

(3).

and the group of then points A', B', r•, a•, E',... ,N', symmetrical to the above first group of n with respect to a center, the center O of circle (o) respectively.

i§Q] .

Collinear and Coe clic Harmonic Grou s of n-members.

Here as well, it will obviously be:

A'B'= BT'= r·a·= .6.'E'=... = T,

(4).

and A', B', r·, .6.', E', ... ,N', will belong to circle (o) . The set of the above 2n-points we would call «Pair of a Regular Open Cocyclic Series of Points of 2n Points or more simply Regular Polygonal Cocyclic Line, or Regular Cocyclic Line». We easily realize that all the 2(n-2) quadrilaterals in (5) below, that belong to the group of 2n above, are rhomboid and therefore they are harmonic, as for example the rhomboid quadrilateral ABrB' has AB.rB'=Br.B'A, thus it will be Harmonic. In the same way we also find that all the rhomboid quadrilaterals in (5) below, are Harmonic.

ABrB', Brar•, raE.6.', ..... ., n-2 quadrilaterals} A'BT'B, BT'.6.T, r'.6.'E'.6., ... ,n-2 quadrilaterals The line segments AB , Br,

ra, aE, .. . and A'B', BT', r·a·,

2(n-2) quadrilaterals,

(5).

r·

.6.'E', ...we will call them sides,

points

A,

E, ... N, ... - A', B',

B, r , a, r·, a·, E', ...

N', ... we will call them apexes of the Har monic Open Cocyclic Series of Points of 2n Points, while the n segents, AA', BB',

rr·, M', EE'

, .. .. , NN', we will call them its «Main Diagonals» [here they are also diameters of circle

r

(o)].

Comments. (a). In all three cases of lines above, the pairs of points B-8', r-r', .6.-.6.', ... harmonically divide the pairs of segments Ar-AT', B.6.-8'.6.', rE-r'E', .. ., respectively (In the Third case they are obviously perpendicular).

18l=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

(b). It is obvious that in all three cases of series of points above, some apexes of each series of points can be located among other apexes of the same series of points, or all of them together to not cover the whole circle (Hence the term «open»). (c). The construction of the above series of points is easy, in comparison to the construction of the Harmonic Multilaterals, as the above series of points are open and thus the difficulty in closing them does not exist (Construction 172, etc).

~ - Collinear and Coe clic Harmonic Grou s of n-members.

24. Pairs of O en Central Flat Harmonic and Regular Pencils of 2n Beams and their Orthooptical Points. 24.1. In General. Below we will refer to the Harmonic and Regular Pairs of Open Flat Central Pencils of 2n Beams, as well as to their Orthooptical Points.

24.2. Harmonic Pairs of an Open Flat Central Pencil of 2n Beams (Figure

11.11 a. I•ffl mfl mfii. Based on what we know so far and from all we have mentioned here (for example paragraphs 12.13, 23.1 , etc), we can give the definition of these pencils below: We call Pair of a Harmonic Open Central Pencil of 2n Beams, that pencil whose intersections to an arbitrary line, consist a Pair of Harmonic Open Collinear Series of Points of 2n Points» graph 23.1 ).

(Relevant Terminology in para-

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

b. Q,t4,h4J These Open Pencils have many and significant properties, some of which we will mention below: 1/. The Harmonic Open Central Flat Pencil of 2n Beams T(al3yl>E ... v, a'l3'y'l>'E' .. .v') in figure 111], if we take into consideration paragraph 23.1, we easily find that it includes 2(n-2) Harmonic Central Flat Pencils of four beams. These are: T(ayl3!3'), T(J3l>yy'), ... , T(a'y'J3'13), T(J3'l>'y'y), .. ..

2/. If we intersect this pencil with a line it gives us 2n points, that constitute an Open Harmonic Collinear Series of Points of 2n Points, or a Collinear Harmonic Line (Relevant terminology in paragraph 23.1). 3/. A Harmonic Open Central Pencil of 2n Beams, if we inscribe them in circle (o'), then tis beams re-intersect the circle at points that obviously are the points of a Harmonic Open Cocyclic Series of Points of 2n Points. Thus from the above Harmonic Pencil T(al3yl>E ... v,

a'l3'y'l>'E' ... v') in figure 111,

occurred the Harmonic Open Cocycl ic Series of Points of 2n Points A 1, 81 , r1, .6.1 , E1, ... ,N1 ,

A'1, 8'1, r·1, .6.'1, E'1, ... ,N'1, which for the random point T of

circle (o') has 2(n-2) Harmonic Pencils of fou r of its beams: T(A1r 1B181°), T(B1.6.1r1r'1), ... , T(A'1r'1B'1B1), T(B'1.6.'1r'1r 1), ....

24.3.

Pair of a Regular Open Central Flat Pencil of 2n Beams (Figure 111).

a. I•ffi fflfl fflffl. «We call Regular, a Pair of Open Central Flat Pencil of 2n Beams I:(al3yl>E .. .v ,

a' J3'y'l>'E' .. .v') , if the angles of this pencil, that are formed be-

tween the consecutive beams, of each pencil of its above n beams, are equal to each other, with random measure cp and f for these pencils, the pairs of their homologous beams, are perpendicularly intersected.

I.e., for the Pair of this Regular Pencil above : I:(al3yl>E ... v, - a'l3'y'l>'E' ... v') it is:

LaI:!3= LJ3I:y= LyI:l>= Ll>I:E= LEI:~= ...

= La'I:13'= L13'I:y'= Ly'I:l>'= Ll>'I:E'= LE'I:{'= .. .=cp,

(1 ).

@ .

Collinear and Coe clic Harmonic Grou s of n-members.

and Lara'= Lj3I:13'= Lvrv·= Ll>I:l>'= LEI:E' = ... ... = LvI:v'=1 L.

(2).

(The question is raised. Does such a pencil exist? The answer is positive, as it is constructed easily, if we depend in relations (1) and (2). This after all can also be seen in figure 111 . Proof in Construction 179). Due to relation (2), these pencils can also be called Orthooptical.

Each such Regular Pencil , is also Orthooptical , as relation (2) is true. b.

d•l4,h4i

These Open Pencils have many and significant properties, some of which we will mention below: 1/. The Regular Open Central Flat Pencil of 2n Beams I:(aj3yl>E ...v, a'l3'y'l>'E' ...v') in figure 111 , has 2(n-2) Regular Pencils of four beams. These are : I:(aj3yl3'), I:(j3yl>y'), .. . I:(a'l3'y'j3), I:(13'y'l>'y), ... .

2/. It is easily proven (Proposition 169) that, if we intersect this pencil with a line, it gives us 2n points, that are a Pair of a Harmonic Collinear Series of Points of 2n Points (Relevant terminology in paragraph 23.1). This is the reason why we call this pencil Harmon ic. Hence, if we also take into consideration the definition of Closed Harmonic Pencils in paragraph 12.13, we reach the conclusion that: «All these Regular Pencils, are also Harmonic». Relevant constructions are 61-64, 79, 97 and 171 , that will follow. 3/. A Regular Open Pencil of 2n beams, if we inscribe them in circle (o), then its beams re-intersect the circle at points which are the points of a Regular Open Cocyclic Series of Poi nts of 2n Points (Criterion 173). Hence from the above Regular Penci l I:(aj3yl>E ...v , a'l3'y'l>'E' ... v') in figure 111 , occured the Regular Open Cocyclic Series of Points of 2n Poi nts A, B, r , t:,,., E, ... ,N, - A' B' r· t:,,.• E' ... ,N', which for point r of circle (o), has 2(n-2) Regular Pencils of four beams : I:(ABrB'), I:(Brt:,,.r'), .. . I:(A'BT'B), I:(BT't:,,.T), .... 4/. Additionally, when we intersect this pencil w ith a line, then its center r is an «Orthooptical Point» for some segments (diagonals) at which the line is

lml=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

intersected, while if we intersect it with a circle that runs through its center

r, then

its center is an «Orthooptical Point» for some diameters of the circle

in which it is inscribed.

r, is an «Orthooptical Point», for segments (diagonals) 1313', yy', lili' .... , as well as for diameters AA' , BB', rr·, M', ...

Thus, in figure 111 aa',

24.4.

Pair of Orthooptical Points, of a Pair of Open Collinear Harmonic Lines (Figure 111 ).

As we will later prove, each Collinear Harmonic Line, has two symmetrical with respect to this line Orthooptical Points, from which its diagonals, are seen in right angles. Such points are these we mentioned above (§ b 4/).

24.5. Description-Analysis of Figure 111 . Based on the above, we can easily describe figure 111. In this figure we can see the Open Regular Flat Central Pencil of 2n Beams I:(al3yli£ ... v, - a'l3'y'li'£' ... v') with equal angles of random measure cp, which intersects line (£) at the Open Harmonic Collinear Series of Points of 2n Points a,

13, y,

li, £, ... v, - a',

13', y', li',

£', .. . ,v', but which also intersects circle

(o) at points A, B, r, /1, E, ... N, - A', B', r·, /1', E' , ... ,N', which are obviously apexes of an Open Regular Cocyclic Series of Points of 2n Points. Here quadrilaterals ABrB', Brar·, ... , A'BT'B, BT'/1T ,.. , are rhomboids with axes of symmetry the diameters BB', rr• .... , that are obviously harmonic. Additionally, in this figure, we can see the Open Central Harmonic Pencil of 2n Beams T(al3yl5£ ... v, - a'j3'y'l5"£' ... v'), which intersects line(£) at th Open Harmonic Series of Points of 2n Points a,

13,

y, 15, £, ... ,v, - a',

13', y',

15', £',

... ,v' (known from the previous). Moreover, if we set an arbitrary point T on a random circle (o'), then pencil I:(al3yl5£ ... v, - a'j3'y'l5'£' ... v') re-intersects circle (o') at the Open Harmonic

@ .

Collinear and Coe clic Harmonic Grou s of n-members.

Cocyclic Series of Points of 2n Points A1, B1, r1, l::..1, E1 , ... ,N1 , - A'1 , B'1, r 0 1, l::..'1, E'1, ... ,N'1, which are apexes to the Open Harmonic Cocyclic Series of Points of 2n Points. It is obvious that, if the sides of the Open Regular Cyclic Series of Points of 2n Points, are many and in comparison to the circle big, then some apexes can be located among other of its apexes. Hence, for example in figure 111, apex 0 lies between A' and B', while apex 0' lies between A and B.

134ee6hb (a). We will prove the above in Propositions and Problems that will follow or that have already been mentioned. (b). Below based on figure 111, as we mentioned above, we will give the way and the interpretation of the way that an Open Collinear and a Cocyclic Harmonic Series of Points of 2n Points occurs, from an Open Cocyclic Regular Series of Points of 2n Points, with an equal number of sides, using the new Method of the «Harmonic Transfomation» (Construction 179), that is the solution to a Problem that troubled us often, for at least twenty years. (A first taste of this method, was given above in paragraph 24.5).

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

25. APPLICATIONS RELEVANT TO THE PAIRS OF COLLINEAR AND COCYCLIC HARMONIC LINES AND PENCILS. In General. Consequently we will proceed to applications of the method of Harmonic Pencils , that is mentioned in paragraphs 5 and 14, adjusted though here to the pencils of paragraph 24, in the form of Geometrical Propositions and Constructions, with their proofs, as we will need them to base on them the proofs of other Propositions, etc. Moreover on these and those in paragraphs 13, 17 and 18, we will also depend the New Method of the Harmonic Transformation(§ 21), to which we will refer later, adjusted though here to the pencils of paragraph 24.

25.1. Intersection of a Pair of a Regular Open Flat Pencil of 2n Beams, from

a Line. Proposition 169 (Figure 112). Each line that intersects a Pair of an Open Regular Flat Pencil of 2n Beams, is intersected at 2n points, which constitute a Pair of a Harmonic Open Collinear Series of Points of 2n Points.

[ml Let there be the Pair of the Open Regular Flat Pencil of 2n Beams T(ABraE ... N -A'BT'a'E' ... N'),

(1).

which is intersected from line (E) at points:

A, B,

r, a, E, ... ,N - A', B', r·, a·,

E', ... ,N',

(2).

lfil]] .

Collinear and Coe clic Harmonic Grou s of n-members.

respectively. We will prove that the pair of the series of points (2), is Harmonic(§ 23.1). Since the given pencil (1) is Regular, according to the relevant definition (§ 24.3a), it will have: LATB= LBTr= LrTa= LaTE= LETZ= ... ,=LA'TB'= LB'Tr'= Lr'Ta'= La'TE'= LE'TZ'= .. . =cp,

(3)

and LATA'=LBTB'=LrTr'=LaTa'=LETE' = ... =LNTN ' =1L.

(4).

Hence, since TB is the bisector of LA Tr and TB' is perpendicular to TB, then B, B' will harmonically divide A and r or it will be (ArBB')=1 . Additionally, since Tr is the bisector of angle BTa and Tr' is perpendicular to Tr, then r , r• will harmonically divide B, a, or it will be (Barr')=1. We continue in a similar way and we find that : (rEM')=1 , (a2EE')=1 , and so on and so forth . Similarly, since TB' is the bisector of angle A'Tr' and TB is perpendicular to TB', then B', B will harmonically divide A', r·, or it will be: (AT'B'B)=1. In a similar way we conti nue and find that it is: (B'aTT)=1, (r'E'a'a)=1 , (a'Z'E'E)=1, and so on and so forth. Due to the above and according to the relevant definitions(§ 23.1), the pair of the series of points A, B, r , a , E, .. . ,N - A ', B', r· , a·, E', ... ,N', is actually harmonic.

Gi•lei3.Ut1I·UM From the above Proposition and the relevant definition of the Pair of the Open Regular Flat Pencil of 2n Beams (§ 24), we reach the following conclusion: Each Pair of an Open Regular Flat Pencil of 2n Beams, is Harmonic as well.

Thus, to easily deifne, on a line, an arbitrary Pair of an Open Regular Flat Series of Points of 2n Points, we just need to intersect this line with a Pair of an Open Regular Flat Pencil of 2n Beams.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

14411609 (a). Proposition 169 we believe to be new and firstly appear here. (b). The above Central Pencil in figure 112, is symmetrical too, with respect to the bisector of angle ATN' , or A'TN, and so on and so forth. (c). Relevant Constructions are 11, 32a, 61-65.

25.2. Orthooptical Points of a Pair of a Harmonic Open Series of Points of 2n Points. (Significant Property. Extenstion to Corollary 1 of Proposition 72). Proposition 170 (Figure 113). We are given the Pair of the Harmonic Open Collinear Series of Points of 2n Points A, B,

r,

/1, E, ... ,N, - A', B',

r·, /1',

that the circles with diameters AA', BB',

E', ... ,N' and we are asked to prove

rr·,

M', EE', ... , NN', are concur-

rent, hence they are a pencil of circles of the first kind.

tlbl#dfflil Let (E) be the line of the above points A , 8 ,

r,

/1, E, ...,N,

- A', B',

r·,

/1' ,

E',... ,N' .

According to the relevant definition of the Harmonic Collinear Series of Points of 2n Points (§ 23.1) , this should have harmonic the following groups of four points:

r, B'- B, r, /1, r•- r , /1, E, /1'-, ...... .. ,n-2 groups} A', B', r·, B- B', r, 11', r - r·, 11' , E', /1-, . .. ,n-2 groups A, B,

2(n-2) groups. (1) .

To prove Proposition 170, we use the new Method of «Symmetrical Central

@ .

Collinear and Coe clic Harmonic Grou s of n-members.

Pencils» (§ 5), adjusted to the pencils in paragraph 24, and taking into consideration the definition of the Pairs of Regular Open Central Pencils of 2n Beams(§ 24.3), that are harmonic too. To apply this method, we draw circles (o) , (K) with diameters BB' and rr· respectively, that intersect at points T, T', symmetrical with respect to (E). We draw TA, TB, Tr, Th., TE, TZ, ...... , TA', TB', Tr', Th.', TE', TZ', ...... If we take into consideration relation (1) above and we suppose that for triangle A Tr since (ArBB')=1 and LBTB'=90°, circle (o) is Apollonian, then LA TB= LBTr.

(2).

Moreover, in triangle BTh., since it is (Barr')=1 and LrTr'=90°, circle (K) is Apollonian, hence

LBTr=LrTa.

(3).

Thus, from the relations in (2), (3), it occurs that:. LA TB= LBTr= LrTh.=cp.

(4).

We continue as above and if we consider triangle A'Tr' for which since (r'A'B'B)=1 and LBTB'=90°, circle (o) is Apollonian, then LA'TB'=LB'Tr'.

(5).

Moreover, for triangle B'Th.', since it is (h.'BTT)=1 and LrTr'=90°, circle (K) is Apollonian, then

LB'Tr'= Lr'Th.'.

(6).

Hence, from the relations in (5), (6), it occurs that: LA'TB'=LB'Tr'= Lr'Th.'=w. After all it is

LBTr=cp= LB'Tr'=w, or w=cp,

(7). (8).

w~ ouµrrAripwµara rri~ ywvia~ rTB'.

Arr6 Tl~ oxt.oE1~ (4), (7), (8) rrpoKUTTTEI 6r1:. LA TB= LBTr= LrTh.= LA'TB'= LB'Tr"= Lr'Th.'=cp=w.

(9).

Since it is LrTr'=90° and since, due to (9), it also is LrTa=Lr'Th.', it will be LrTr-LrTa+ Lr'Th.'=90° or LaTh.'=90° , thus then circle (A), with diameter M', will run through T, as well as through T', its symmetrical with respect to (E). Additionally, for triangle rTE, since it is (rEM')=1 and since Lh.Th.'=90°, circle (,\) is Apollonian, thus due to (9), it will be: Lrrn= LaTE=cp.

(10).

Similarly, for triangle r'TE', since it is (r'E'h.'h.)=1 and since La'Th.=90°, circle (,\) is Apollonian, thus due to (9), it will be:

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

Lr'T11'= L/1'TE'=q>.

(11 ).

Since it is L/1T/1'=90° and due to (10), (11) it isL/1TE=L/1'TE', and it will also be L/1T/1'-L/1TE+L/1'TE'=90° or LETE'=90° , thus circle (v), with diameter EE' , will run through T, as well as through T', its symmetrical with respect to (E). In the exact same way we continue and we find that circle (rr) with diameter

ZZ' , runs through T, T', as well, and so on and so forth , until circle with diameter MM'. But when we try to prove that the circle w ith diameter NN' runs through T, T' we encounter difficulty. To achieve this we add in the end of each one of the two given series of points one more point = , =' respectively, such that it would be (M=NN')=1 , (M'='N'N)=1 and we work as above. This can easily happen as points M, N, N', exist, so we define in the known way point=, while since points M', N', N, exist, we define in the known way='· Finally since we have not proven that circle (µ) with diameter AA', runs through T, T', we will prove that too. To prove that we work in the following way: Since it is LBTB'=90° and since, due to (9), it is LA TB= LA'TB', it will be LBTB' -LA'TB'+ LA TB=90° = L.ATA' or L.ATA'=90°, hence circle (µ) , w ith diameter AA', will also run through T, as well as through T', its symmetrical with respect to (E).

(a). From the above it finally occurs that: LATA'=LBTB'=LrTr"=L/1T/1'=LETE'= .....=LNTN '=90°.

(12).

This means that the circles with diameters AA'=BB'=rr'=M'=EE'= .. ...=NN', do in fact run through the pair of points T, T'. This means that they constitute a pencil of circles of the first kind and since these segments, are seen in right angles from T, T', they are «Orthooptical Points» of the given harmonic series of points. (b). Additionally from (9), (10), (11) it also occurs that : LATB=LBTr=LrT/1=L/1TE= .... = LA'TB'= LB'Tr "= L r'T/1'= L/1'TE'= ... .=q>.

(13).

lfil]] .

Collinear and Coe clic Harmonic Grou s of n-members.

(c). Hence, since (12), (13), are true, pencil T(ABr.llE ...N - A'BT'.ll'E' ... N'), is regular (definition§ 24.3).

1;3411609 (a). Proposition 170, is obviously an extension to Corollary 1 of Proposition 72, which refers to a Harmonic Group of Ten Collinear Points. (b). The Proposition above we believe to be new and firstly appear here. (c). We have to do w ith a significant Proposition , as it will be useful in many cases below. (d.) Relevant Constructions are 32a, 63. 68, 70-72, 169, 171.

25.3. Easy Construction of a Pair of a Harmonic Open Collinear Series of Points of 2n Points. (Harmonic Open Polygonal Collinear Line). Construction 171 (Figure 113). Construct an arbitrary Pair of a Harmonic Open Collinear Series of Points of 2n Points, on a given line (E).

§fflfflii•llh 1st Way (Analytical).

We suppose that the desired series of points is constructed on line (E) and it is this one A, B, r , .ll, E,... ,N, - A', B', r•, .ll', E' ,... ,N'. If we take into consideration the definitions of Regular Central Pencils (§ 12.13 and 24, especially their paragraphs a1/ and 24.3a/, respectively) and the definition of the Pair of a Harmonic Open Collinear Series of Points of 2n Points (§ 23.1), we are easily led to the construction below. (b) . Construction . Outside of line (E), we set a point T, from which we draw a line that intersects (E), at let's say point A. Consequently from T we draw a second line, that intersects (E) at let's say B and forms with TA an acute angle

E, ... v , -

a'l3'y'l>'E', ...,v'), the beams of which, we suppose re-intersect circle (o) at points A, B,

r , a, E, ...,N, - A', B', r·, a·, E', .. . ,N' , respectively [If relation

(1) is

not given to us, we construct it ourselves (Construction 177)). We say that the series of points A, B,

r, a, E, ... ,N,

- A', B',

r·, a·, E', ... ,N',

is

the desired Open Harmonic one.

ktl:/t®I Since the Open Collinear Series of Points of 2n Points, a,

13', y',

13, y,

l>,

E, . . . v,

- a',

l>', E', . .. ,v', is Harmonic, according to the relevant definition (§ 24.2),

Central Pencil T(al3yl>E .. .v- a'l3'y'l>'E' ... v'), or T(ABraE ... N - A'BT'.6.'E' ... N') will also be Harmonic.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. Thus, according to Criterion 174 (converse) the cocyclic series of points A, B,

r,

1::,.,

E, ... ,N,

A', 8',

r•,1::,.•, E', ... ,N' , will also be Harmonic, thus it is the

desired one, as it belongs to circle (o).

tiDfi•l,W4t14 it#!fflt1(41tJ I.e. we are given on circle (o), the Pair of the Harmonic Open Cocyclic Series of Points of 2n Points : A, 8 ,

r , 1::,., E, ... ,N, - A', 8' , r·,1::,.·, E', ... ,N' and we

are asked to construct on a given line (£) the Pair of a Harmonic Open Collinear Series of Points of 2n Points: a,

J3,

V, c5,

£ , ... v,

- a', P', y', c5', E', ... ,v',

(1 ).

If we take into consideration Criterion 174 and the relevant definitions (§ 23.1, 23.2 and 24.2) , we are led to the construction below. 2/. Construction. On circle (o), we set an arbitrary point T and form pencil T(A 8 A' 8'

r·

I::,.'

points a,

r I::,. E ... N

-

E' ... N'), the beams of which , we suppose intersect line (t) at

13, y,

15, £, ... v, - a',

13',

y', 15', t', ... ,v', respectively [If the Harmonic

Open Cocyclic Series of Points of 2n Points : A, B,

r , 1::,., E, ... ,N, - A', 8', r·,1::,.·,

E', .. . ,N', is not given to us we can construct it in the same way as it is mentioned in construction 172 (§ 25.4)] . We say that the desired Harmonic Open Collinear Series of Points of 2n Points is the following: a,

13, y, 15, £, ... v,

- ,a',

13', y', 15', t', .. .,v'.

ktl :fttffi I Since the Pair of the Open Cocyclic Series of Points of 2n Points: A, 8 , r, 1::,., E, ... ,N, .. -... A' , 8', 8'

r

I::,.'

E' ... N'),

r•, !::,.', E', .. .,N', ri T(al3yl5£ ... v,

is Harmonic then Pencil T(A 8

r

I::,.

E ... N - A'

a'j3'y'l5't' .. . v'), will also be Harmonic (Crite-

rion 174, direct). Hence, according to the relevant definition (§ 24.2), since the Central Pencil T(al3yl5£ ... v, - a'l3'y'i5't' ... v'). is harmonic, then the following Open Collinear Series of Points of 2n Points will also be harmonic : a ,

13,

y , 15, £, ... v, - a',

13',

y', c5', E.' , ..• ,v'. Hence it is the desired one, as it is located on line(£).

1#34uEiUA (a). The Problem above and its solution, we believe to be new and firstly appear here.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

(b). Problem 176, the way it is given, it is undefined. If we want to define it we need to be given sufficient data. (c). In an equivalent way we have constructed on a line harmonic groups of four, eight, ten, etc, or on a circle harmonic quadrilaterals, hexagons , octagons, etc (Constructions 99, 100, 101 , 102, etc, respectively). (d). Relevant Propositions are 35, 61-66, 96 - 103, 145, 171 , 172, 174, 175, 179.

25.9. Converse of Diagonals of a Convex Pair of a Harmonic Open Cocyclic Series of Points of 2n Points (Significant Property. Extension to Proposition 106). (Pair of a Convex Harmonic Cocyclic Polygonal Line). Proposition 177 (Figure 109). The diagonals, of each Pair of a Convex Harmonic Open Cocyclic Series of Points of 2n Points, are concurrent.

We consider the Pair of a Convex Harmonic Open Cocyclic Series of Points of 2n Points: We will prove that:

A, B, r, l:J., E, ... ,N - A', B', r•, l:J.', E', ... ,N'. AA'nBB'nrr·nM'nEE'n ... nNN'=:L

(1).

(2).

1 st Way [Based on the Auxiliary Proposition 7 (in Part E)]. Since the given Pair of a Harmonic Open Cocyclic Series of Points of 2n Points is Harmonic (implication), according to the definition (§ 23.2), it will have harmonic for example quadrilaterals ABrB' and A'BT'B, thus according to known Theorem, it will be:

AB rs· A'B' r·s AB rs· A'B' r·s Br . B' A = 1 a nd BT' . BA' = 1 ~ Br . B' A= BT' . BA' .

(3).

Thus, from relation (3), according to Auxi liary Proposition 7 (converse) (in Part E) , it means that the following relation is true: AA'nss·nrr· =r.

(4).

Moreover, since from the definition the given Pair, will have harmonic quad

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

rilaterals Brt,r and BT't..T as well, if we work in the exact same way as above, we easily find that:

BB'nrr•nt..t..' =:L

(5).

Moreover, since from the definition the given Pair, will have harmonic quadrilaterals rt.Et.' and r't..'E't.., as well, if we work in the exact same way as above, we easily find that

rr·nt..t..'nEE'=r.

(6).

We continue in the exact same way as above and finally reach the harmonic quadrilaterals AMNM' and A'M'N'M, from which we get: M'nMM'nNN'=r.

(,\).

Thus, from (4), (5), (6), .... , (,\), relation (2) occurs.

2nd Way (Based on Proposition 51). Since the given Pair of a Harmonic Open Cocyclic Series of Points of 2n Points is Harmonic (implication), according to the definition (§ 23.2), will have harmonic for example quadrilaterals ABrB' and A'BT'B, thus according to Proposition 51 (converse)(§ 9.12), it will be

AA'nBB'nrr·=r.

(4).

Moreover, since from the definition the given Pair, will have harmonic quadrilaterals Brar· and BT't..T as well, if we work in the exact same way as above, we easily find that:

BB'nrr·nt..t..' =r.

(5).

Moreover, since from the definition the given Pair, will have harmonic quadrilaterals rt.Et..' and r't..'E't.., as well, if we work in the exact same way as above, we easily find that

rr·nt..t..'nEE'=r.

(6).

We continue in the exact same way as above and finally reach the harmonic quadrilaterals AMNM' and A'M'N'M, from which we get: M'nMM'nNN'=r.

(A).

Thus, from (4), (5), (6), .... , (,\), relation (2) occurs.

J rd Wa

Based on Construction 172 .

If we take into consideration the way of construction of a Pair of a Harmonic Open Cocyclic Series of Points of 2n Points 172 (§ 25.4), it occurs that this pair, has concurrent diagonals. Hence relation (2) is true.

1#34uEiUA (a). Proposition 177, we believe to be new and firstly appear here. (b). It is obvious that for Proposition 177, the converse is not true.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

(c). Until now we have seen that, apart from the Pair of a Harmonic Open Cocyclic Series of Points of 2n Points, concurrent diagonals have the following Harmonic hexagons, octagons, decagons, etc (Propositions 104, 105, 106, respectively). (d). Relevant Propositions are 51 , 104-106, 172.

25.10. Criterion of Harmonic being of a Pair of a Convex Open Cocyclic Series of Points of 2n Points (Pair of a Convex Harmonic Cocyclic Polygonal Line) . Existence of a Lemoine Point on a Pair of a Harmonic Open Cocyclic Series of Points of 2n Points (Significant Property. Extension to Criterion 114). Criterion 178 (Figure 109). If a Pair of a Convex Open Cocyclic Series of Points of 2n Points, has concurrent diagonals and there exists a point on its plane for which its distance from each side of it, is analogous to the matching side, then and only then this Pair of a Convex Open Cocyclic Series of 2n Points, is Harmonic.

tml•Jli44M We consider the inscribed in circle (o) Harmonic Pair of an Open Convex Series of Points of 2n Points:

A, B,

r, 11, E, .. ,N - A', B', r·, 11', E', ... ,N'.

(1 ).

(Relevant terminology in § 23.2). We will prove that a point r' exists on its plane, for which it is: Ua

a -

Up

J3 -

Uy

U5

U,

Uµ

U'a

V - IS - E - . ... - µ - a' -

U'p

J3' -

U'y

U'5

u',

U'µ

y' - IS' - E' - .... - µ' -'A. ( 2)-

where ua, up, .•. . , Uµ, - u'a, u'p, ... •, u'µ, is the distance of point

r·,

from its side

AB=a, Br=l3, ... , MN=µ, A'B'=a', BT'=j3', ... , M'N'=µ', respectively. In fact, since the convex inscribed in circle (o) Pair of an Open Cocyclic Series of Points of 2n Points, in (1), is harmonic, according to Proposition 177

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

(§ 25.9) , it will have AA'nBB'n rr•naa'nEE'n ... n NN'=I:, while according to

the relevant definition (§ 23.2), its following quadrilaterals will be harmonic: ABrB', Brar·, rt.Ea', .. .... ,n-2 quadri laterals

l

A'BT'B, BT't.T, r't.'E't., .. ..,n-2 quadrilateralsJ 2(n-2) quadrilaterals.

(3).

This means that for example in quadrilateral ABrB' its diagonals Ar, BB', will be conjugate with respect to circle (o), thus according to auxiliary Proposition 6 (converse) in Part E, for triangle ABr BB' is its symmedian. Thus, according to known Theorem (rEwµETpia [6] § 11-14.2), each point on BB' will be far from AB, Br distance analogous to each of these sides. This will obviously be true for the point

r of convergence of the diagonals of the

given inscribed in circle (o) Pair of Open Cocyclic Series of Points of 2n Points, which belongs to BB' as well, hence for point

r, it will

be: (4).

If we work as above for the harmonic quadrilateral Brar· and more specifically for triangle Bra, we prove that for U~

-

~

r it is as well:

Uy

=-

(5).

V

Additionally, if we continue as exactly above for the rest triangles raE, t.EZ, ... ,/\MN , we find that for Uy

r it is as well: Ui;

U i;

U,

~=~

V =~ ' ~=E ,.... , A

(6).

µ

Moreover, from (3) it occurs that for example for quadrilateral A'BT'B its diagonals AT', B'B, will be conjugate with respect to circle (o), thus according to auxiliary Proposition 6 (converse) in Part E, for triangle A'BT' B'B is its symmedian. Thus, according to known Theorem (rEwµETpia [6] § 1114.2), each point on BB' will be far from A 'B', BT' distance analogous to each of these sides. This will obviously be true for the point

r

of conver-

gence of the diagonals of the given inscribed in circle (o) Pair of Open Cocyclic Series of Points of 2n Points, which belongs to BB' as well , hence for point

r , it will

be:

~ . Collinear and Coe clic Harmonic Grou s of n-members.

u' a u'p - =-

(7).

W.

a'

If we work as above for the harmonic quadrilateral BT'LI.T and more specifically for triangle BT'LI.', we prove that for r it is as well: (8).

Additionally, if we continue as exactly above for the rest triangles r'Ll.'E', Ll.'E'Z', ... ,/\'M'N', we find that for r it is as well:

u'v

u'6

u'6

u',

y'=~, ~=7,····,

~=~

A'

µ' .

(9).

After all, for example from the similar triangles ArB, B'rA', it occurs that: Ua

a

U1 a

a'·

(10).

Consequently, from relations (4) - (10) above, we easily get relation (2), which is true for intersection r. This means that the desired point r' is r or r·=r. This means that the desired point is point r where the diagonals of the given inscribed in circle (o) Pair of an Open Series of Points of 2n Points in (1 ).

tl:lfi•ltl'Z4t14 I.e., if the inscribed in circle (o) Pair of an Open Convex Cocyclic Series of Points of 2n Points in (1), has concurrent diagonals at point rand for r relation (2) is true, then we will prove that this pair will be harmonic .. In fact, since (2) is true, for r it will be : (4).

This, according to the converse of the Theorem in§ 11-14.2 in rtwµnpfa [6], shows that r belongs to the symmedian of side Ar in triangle ABr and thus, according to the direct of the above auxiliary Proposition 6 (in Part E), BB' and Ar are conjugate with respect to circle (o), or quadrilateral ABrB' is harmonic.

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

If we take into consideration, that the rest two-part equalities that occur from (2) are true, i.e. the following ones: Ull

-

13

Uy

=-

V (11).

and if we work as exactly above, we find that the rest quadrilaterals in (3), Brar·, raEa', aEZE', EZHZ', ..... , A'BT'B, BT'aT, r·a'E'a, ... , are harmonic, hence, according to the definition of the Pair of an Open Cocyclic Series of Points of 2n Points (§ 23.2) then the given Pair of an Open Cocyclic Series of Points of 2n Points in (1), is in fact Harmonic too.

IU411EOA (a). Proposition 178 (criterion), we believe to be new and firstly appear here. (b). We have proven earlier here that an equivalent Proposition (Criterion), is also true for Harmonic Hexagons, Octagons, Decagons, etc (Criteria 112, 113, 114, etc, respectively). (c). Relevant Propositions are 112, 113-115, 177.

(i•J11l11i4hl Due to the above Criterion 178 of the Pair of a Harmonic Open Cocyclic Series of Points of 2n Points, we notice that: (a). The Lemoine Point of every triangle, has the property, according to which, this point is far from each side of the triangle, distance analogous to the respective side (of the triangle). (b). Equivalent Property is true for Harmonic quadrilaterals, Hexagons, Octagons, Decagons, etc (Criteria 112-114, etc). This is the reason why the intersection of the diagonals of the Harmonic Polygonal Lines, is called Lemoine Point of these Harmonic Polygonal Lines, so we reach the following conclusion. Due to what we have mentioned before and in equivalence to these, it occurs that:

~ . Collinear and Coe clic Harmonic Grou s of n-members.

The point of convergence of the diagonals of every Harmonic Pair of an Open Cocyclic Series of Points of 2n Points, is a LEMOINE POINT, for this Harmonic pair.

Due to the above Criterion 178, we reach another conclusion as well: Every convex Harmonic Pair of an Open Cocyclic Series of Points of 2n Points, w ith concurrent diagonals, is harmonic, if and only if, it has a LEM OINE POINT, which coincides with the point of convergence of its diagonals.

Alternative Definition of a Harmonic Pair of an Open Cocyclic Series of Points of 2n Points. Consequently, after the second conclusion above, from now on we can use for the Harmonic Pair of an Open Cocyclic Series of Points of 2n Points , the Alternative Definition below: «Harmonic Pair of an Open Cocyclic Series of Points of 2n Points (Harmonic Cocyclic Line), is called every convex Pair of an Open Cocyclic Series of Points of 2n Points , then and only then, if it has a LEMOINE POINT.

25.11. Construction of a Pair of a Harmonic Open Cocyclic Series of Points of 2n Points, using our New Method of the «HARMONIC TRANSFORMATION». (Pair of a Convex Harmonic Cocyclic Polygonal Line). How from a Regular Pair of an Open Cocyclic Series of Points of 2n Points Occurs a Harmonic one (Significant Construction) .

lntri4,i4Eil We have always believed that a Harmonic Line (Collinear and Cocyclic), comes from a Regular Cocyclic Line. But how though?

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B.

The solution to the Construction of a Pair of a Harmonic Open Cocyclic Series of Points of 2n Points, based on a Pair of a Regular Open Cocyclic Series of Points of 2n Points, though we tried, we could not achieve until we came up with the Method of the Harmonic Transformation. Another of construction of the Harmonic Cocyclic Line, is given in Construction 172 (§ 25.4), but not based on a Regular Cocyclic Line. Finally this Construction, is easily achieved applying our New Method of the «Harmonic Transformation», (See relevantly§ 21 and the Problem of the Vicennium 157, § 22.6), which we will also apply here given that this method can be easily applied to all relevant problems, but here based on a Regular Cocyclic Line. Using this Method we firstly achieve the construction of the Collinear and consequently of the Cocyclic Harmonic Line. A first taste of this application is given in paragraph 24.5. The application of this method in the construction of a Collinear and Cocyclic Line is complex and achieved through the proper SYNTHESIS of Constructions 175 and 176 (Synthesis of figures 116 and 115). For Problems 152-155 we have applied this method for quadrilaterals, hexagons, octagons, decagons and now below for a Pair of a Harmonic Open Collinear Series of Points of 2n Points.

General Construction 179 (Figure 111 }. Based on proper projective transformations and intersections from a line and circles, construct a Pair of a Harmonic Convex Open Cocyclic Series of Points of 2n Points, based on a Regular respective pair, and conversely in a similar way and based on a Harmonic Cocyclic Pair construct the respective Regular one.

~ . Collinear and Coe clic Harmonic Grou s of n-members.

f§fflfflii·llh itl•Jli441 EtW;®i't1tJ Here we are given the Pair of a Regular Open Cocyclic Series of Points of 2n Points:

A, 8,

r, 11, E, ... ,N -A',

8' , r·, /1', E', ... ,N',

(1).

which is inscrcibed in circle (o) and we are asked, based on that and with proper projective transformations to construct a Pair of a Harmonic Open Cocyclic Series of Points of 2n Points: (2). If we take into consideration Constructions 175 (§ 25.7) and 176 (§ 25.8) and we combine them properly, we are led to the construction below. b/. Construction. Initially, we work in the exact same way as in Construction 175 (converse) and after we define an arbitrary point

r

on circle (o), we form the regular

pencil r(Ar8/1E ... N - A'8T'/1'E' ... N'), the beams of which intersect an arbitrary line (E), at points a,

13, y,

l>, E, ... ,v - a',

13', y', l>', E', ... ,v',

respectively.

Consequently, we work in the exact same way as in Construction 176 (direct) and after we define an arbitrary point T on a random circle (o'), we form Pencil T(al3yl>E ... v

- a'l3'y'l>'E' ... v'), whose beams re-intersect circle

We say that the Pair of an Open Cocyclic Series of Points of 2n Points A1,

r:11:ft®I Initially we work in the exact same way as in the proof of Construction 175 (converse) and we prove, similarly to there, that the Pair of the Regular Open Collinear Series of Points of 2n Points a,

13,

y, l>, E, ... ,v - a',

13', y',

l>',

t', ... , v', is Harmonic.

Consequently we work, in the exact same way as in the proof of Construction 176 (direct) and we prove that the Pair of the Open Cocyclic Series of Points of 2n Points A1, 81, r1, /11, E1, ... ,N1 - A'1 , 8'1, r·1, /1'1 , E'1, ... ,N'1, is Harmonic.

18l=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=8.

rW•l,@4t14i For the converse we follow the exactly reverse route.

I.e., here we are given the Pair of the Harmonic Open Cocyclic Series of Points of 2n Points: A1 , 81 , r1 , A1 , E1, ... , N1 -A'1, 8'1, r·1 , A'1 , E'1 1 ••• ,N'1, which is inscribed in circle (o') and we ask, based on that and with proper projective transformations to construct a Pair of a Regular Open Cocyclic Series of Points of 2n Points: A, 8, r, A, E, ... ,N - A', 8', r•, A', E', ... ,N'. If we take into consideration here as well Constructions 176 and 175 and we combine them properly, we are led to the Construction below. b/. Construction. Initially, we work in the exact same way as in Construction 176 (converse) and after we define an arbitrary point T on circle (o'), we form Pencil T(A181r1A1E1 ... N1 -A'18'1r'1A'1E'1 ... N'1), the beams of which intersect an arbitrary line (E), at points a,

13, y,

15, E, ... ,v - a',

13', y',

15', E', ... ,v', respectively.

Consequently, we work in the exact same way as in Construction 175 (direct) and after we define at least one «Orthooptical Point» section of circles with diameters for example segments

r,

1313',

(as the inter-

yy'), of the Pair

of the Harmonic Open Collinear Series of Points of 2n Points: a, E, ... ,v - a',

13', y',

13,

y, 15,

15', E', ... ,v', we draw an arbitrary circle (o), that runs through

r. If A, 8, r,

a,

E, ... ,N - A', 8', r·,

a·,

E', ... ,N'. are the points where Pencil

T(al3yl5E ... v - a'J3'y'l5'E' ... v') whose beams re-intersect circle (o), respectively, we say that the Pair of the Open Cocyclic Series of Points of 2n Points A, 8, r,

a,

a·, E', ... ,N'. is the desired Regular one. and does not run through r, then we work in the exact

E, ... ,N - A', 8', r·,

[If circle (o) is given

same way as mentioned in Construction 175 (direct, § (a)2/)].

M=tt®I Initially we work in the exact same way as in the proof of Construction 176 (converse) and we prove, similarly to there, that the Pair of an Open Collinear Series of Points of 2n Points: a, Harmonic.

13,

y, 15, E, .. .,v - a',

13',

y', 15', E', ... ,v', is

~ . Collinear and Coe clic Harmonic Grou s of n-members.

Consequently we work in the exact same way as in the proof of Construction 175 (direct) and we prove, similarly to there, that Pencil r(al3yl>t ...v ' a'l3'y'IS't' ... v'), or r(ABraE ... N - A'BT'a'E' ... N'), is Regular, thus the Pair of a Harmonic Open Cocyclic Series of Points of 2n Points, A, B, A', B',

r·, a·, E', ... ,N', is Regular too and thus it

r , a, E, ... ,N

-

is the desired one.

134u&iib (a). Problem 179, the way it is given , it is undefined. If we want to define it we need to be given sufficient data. (b). The Problem above and its solution, we believe to be new and firstly appear here. (c). We notice that this way above, is applied to a large number of relevant construction . This means that it is a method, which we have named «Harmonic Transformation» (§ 21 ). The detailed interpretation of this method, we tried to practically give in the proof of Problem 179 above. (d). We notice that using this method, the Harmonic Collinear Group of 2n occurs and then the Cocyclic one. (e). The application of the above Method, is seen rather thoroughly and clearly in the above exact figure 111 , which obviously occurs through the synthesis of figures 115 and 116. (f). Relevant Propositions are 35, 74, 98, 152-157, 173-176.

25.12. General Conclusion of the new Method of Harmonic Transfomation. The Method of the Harmonic Transformation is applied to the solution of Problems regarding : (a). Harmonic Involutions (Collinear and Cocyclic) Simple, for which we use inscribed multilaterals, symmetrical w ith respect to one of their diagonals (main) (Corollaries 158, 163, 165) and Double, for which we use inscribed multilaterals, symmetrical with respect to two perpendicular diagonals (main) or theirs (Corollary 167). (b). Harmonic Multilaterals and Harmonic Lines(Collinear and Cocyclic)

181=-·___,_H:.:..A=R=M"-'O=N.:..:l..::;C....;:G=E=O=M=E=-T-'-'R..a.Y'-'--P=art'-'-=B. for which we use Regular Multilaterals, that are obviously symmetrical with respect to all their diagonals (main) (Constructions 152-157) and Pairs of Regular inscribed Lines (Construction 179).

(In Part D, we give other relevant complementary features of the Harmonic Multilaterals and Harmonic Lines, mostly applying the Method of the Harmonic Transformation).

\

v'

PARTC SUPPLEMENT OF PART A

Applications of Harmonic Involutions, Mostly using the New Method of Harmonic Transformation.

Carl Friedrich Gauss (1777-1855).

Ela.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:-:..P=art~C.

SUPPLEMENT OF PART A

Applications of Harmonic Involutions, Mostly using the New Method of Harmonic Transformation. 26. In General. Part C is actually an extension of Part A' and includes some additional information about Harmonic Involutions (Collinear and Cocyclic), that came up after having written Part B. These data will obviously be inserted in Part A in a new edition of the book. Mostly though it includes applications of Harmonic Involutions using our New Method of Harmonic Transformation, which we have developed in paragraph 21 (although here it is adjusted to Harmonic Involutions) as while writing Part A, we had not invented this method yet. Please note here that using this important Method of «Harmonic Transformation», we can methodically, easily and spherically solve the majority of the Problems in Harmonic Geometry, even the ones that have been impossible to solve so far. Especially for the application of this method in Harmonic Involutions, we will use the Rules (Corollaries), we have given with Corollaries 158 till 168. As we will later discover, while applying the Harmonic Transformation, on Harmonic Involutions :

lement of Collinear - Coe clic Harmonic Involutions.

a. Simple Harmonic Involutions (Collinear and Cocyclicl, occur from inscribed 2n-laterals in circle, symmetrical respective to one of their diagonals, or from rhomboids inscribed in circle that have a common diagonal respective to which they are symmetrical. b. Double Harmonic Involution (Collinear and Cocyclicl, occur from 2nlaterals inscribed in circle, symmetrical respectively to two perpendicular diagonals, or from rectangles inscribed in circle that have their two axes of symmetry in common .

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:-:.. P=art~C.

27. Introducing the Method of Harmonic Transformation in Harmonic lnvo-

27.1 . Simple Harmonic Involutions (Collinear and Cocylic). Just like we have mentioned earlier, Simple Harmonic Involutions (Collinear and Cocylic), occur from 2n-laterals inscribed in circle symmetrical respective to one of their main diagonals, or from rhomboids inscribed in circle which share their diagonal, the one that is their axis of symmetry.

27.2. Applications of Simple Harmonic Involutions. 27.21. Construction of a Simple Harmonic Involution with One Pair of Conjugate Points (See§ 3.2, 8.11, 8.61 and Construction 135). (How a Simple Harmonic Cocyclic Involution occurs from an Inscribed Rhomboid) . Construction 180 (Figure 100). Based on appropriate projective transformations and intersections with line and circles, construct a Simple Cocyclic Harmonic Involution with One Pair of Conjugate Points, based on an Inscribed Rhomboid, given to us, and conversely, sim ilarly based on a Simple Cocylic Harmonic Involution of One Pair of Conjugate Points, given to us, construct an Inscribed Rhomboid.

f:fflMh•lel itl•Jlk441 tEIIEittthi1 t-1 Here we have as given the rhomboid ABr~ inscribed in circle (o)

and ,

based on that and appropriate projective transformations, we are asked to find a Simple Cocyclic Harmonic Involution with double points B', ~· and a pair of Conjugate Points A', r ·. Taking into consideration Corollary 158 (§ 22.61 A), Criteria 35 and 75, as well as paragraphs 3.2, 8.1 1, 8.61 , are led to the construction below:

lement of Collinear - Coe clic Harmonic Involutions.

b . Construction. We work in the same way as in the implication (first case) of Corollary 158, we firstly construct the Symmetrical Pencil I:(ABra) or I:(al3y6) and then its intersection with line (E), thus the Collinear Harmon ic Involution of Double Points 13, 6 and of One Pair of Conjugate Points a, y appears. Consequently, we construct the Harmonic Pencil T(ayl36) or T(AT'B'a ") [where TE(o')], or the Simple Cocyclic on circle (o') Harmonic Involution with B', a • double points and of One Pair of Conjugate Points A', r•, which we say is the demanded one.

tSliMffil According to the relevant definition (§ 3.1), we can easily find that pencil I:(ABra) or I:(al3y6) is symmetrical respectively to rs and ra. Thus the points of its i ntersection with line (E) a, y , will be in Harmonic Involution, with 13, 6 its Double Points, or that the Group of Four Points a , 13, y , 6 , is harmonic, therefore pencil T(ayl36) or T(AT'B'a'), will be harmonic (Criterion 35). This means that A ' and

r,

harmonically divide B', a ·, thus accord-

ing to the definition in paragraph 8.11, points A', r ·, are in Harmonic Involution with Double Points B', a ·.

~)ti•lefJ4t14 For the converse we have to follow the opposite direction.

So, we are given the Harmonic Involution of One Pair of Conjugate Points A', r· and of Double Points B' , a· , in a circle (o'), and we are asked to inscribe in circle (o) a rhomboid ABra, with Ba its axis of symmetry. Taking into consideration Corollary 158, Criteria 35 and 75, as well as paragraphs 8.11 , 8.61 , 3.2, we are led to the construction below. (b). Construction. We work similarly to the converse (first case) of Corollary 158, we firstly construct pencil T(AT'B'a') or T(ayl36) which intersects with line (E) at points a, 13, y , 6. Because we want the quadrilateral ABra, that will occur to be symmetrical with respect to Ba, this is why we set a point I: on the circle of diameter 136 and on circle (o). This means that we set as r t he point of intersection of these two circles.

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:-:.. P=art~C.

Consequently, we construct pencil I:(ayl3li), which also intersects circle (o) at points A, B, r, /1 and we say that quadrilateral ABrt:.., is the wanted rhomboid, symmetrical with respect to Ba. [In the case that the circle of diameter l3li does not intersect with circle (o), then we use an auxiliary circle, equal to circle (o), with which it intersects and after we construct ABra, we move it to circle (o)].

tSlitt®I In fact, because points A', B', r·, a· are in Harmonic Involution, from the implication, with Double Points B', /1' and Pair of Conjugate Points A ', r• , quadrilateral AT'B '/1' is harmonic, thus according to the converse of Criterion 35, pencil r(AT'B'a') or r(ayl3~) would be harmonic. Therefore the group of four points a , y, 13, li , would be harmonic, or these points will be in Harmonic Involution, with Double Points 13, li and Pair of Conjugate Points a, y. Moreover, because the group of four points a, y, 13, li is harmonic and angle 13rli is right, in triangle ray, r13 will be the bisector of angle ary. Thus rs is the bisector of angle Arr, therefore AB=Br and because angle Bra is right, Ba is a diameter of circle (o). As a result, we can easily find that quadrilateral ABrt:.. is in fact symmetrical with respect to Ba, this means that it is the wanted rhomboid. (d . Investigation. In the special case that for the rhomboid A Bra , Ar is also a diameter, then this will be a square, so for the implication we work similarly to paragraph 1/ above. Concerning the converse to find square ABra we have to set the intersection of circles of diameters l3li and ay as

:r

(Orthooptical Point)

(See Construction 152).

l.34uFIUA (a). Problem 180, the way it is given is undefined. If we want to define it, we need sufficient data. (b). The above Problem and its solution, we believe to be new and firstly appear here. (c). We see that for the solution of the above Construction, where we used

lement of Collinear - Coe clic Harmonic Involutions.

the Method of Harmonic Transformation , we did not use the Theory about Poles and Polar Coordinates (§ 8.61 ). (d). We notice that with this method , firstly a Harmonic Collinear Involution occurs and then the wanted Harmonic Involution. This means that w ith the New Method of Harmonic Transformation, we can simultaneously achieve the Construction of the Harmonic Collinear Involution. (e). Two other solutions of Problem 180, can be found in Construction 135). (f). Other relevant Propositions are 35, 61 , 74, 75, 79, 99,135, 158.

27.22. Construction of the Simple Cocyclic Harmonic Involution of two Pairs of Conjugate Points (Relevant paragraphs 3.4 and 8.62). (How a Simple Harmonic Cocylic Involution occurs from a Convex Inscribed Symmetrical Hexagon). Construction 181 (Figure 103). Based on appropriate projective transformations and intersections with line and circles , construct a Simple Cocyclic Harmonic Involution of two Pairs of Conjugate Points, based on a Inscribed Convex Symmetrical with respect to one of its Diagonals Hexagon, which is given to us, and con versely, similarly and based on a Simple Cocyclic Involution of two Pairs of Conjugate Points, which is given to us, construct an Inscribed Convex Symmetrical with respect to one of its Diagonals Hexagon.

f®Mh•leh 1'1 Way (Based on Construction 180).

ltl•Jli431 tEJWfflft1~i Here a hexagon ABrtiEZ inscribed in circle (o) convex symmetrical with respect to BE is given and we are asked , based on that and with appropriate projective transformations to find a Simple Cocyclic Harmonic Involution with double points B', E' and two pairs A' , r• and fl', Z' of Conjugate Points. Taking into consideration Construction 180 (§ 27.21), Criteria 35 and 75, just like in paragraphs 3.2, 8.11 , 8.62, we are led to the construction below:

8la.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:-:..P=art~C. (b). Construction. We notice that the symmetrical with respect to BE hexagon ABrt.EZ, is analyzed in two rhomboids ABrE and /1EZB, with common diagonal BE. This way, based on these rhomboids and Construction 180, we create on line (E) the Collinear Harmonic Involution of one pair of Conjugate Points a, y and Double points

13, E

and afterwards in circle (o') we construct the

Cocyclic Harmonic Involution of one pair of Conjugate Points A', r· and Double points 8', E'. Consequently, working just like above, we construct line (E) the Collinear Harmonic Involution of one pair of Conjugate Points, 15, Points

~

and Double

13, E and afterwards on circle (o') we construct the Cocyclic Harmon-

ic Involution of one pair of Conjugate Points /1', Z' and Double Points B', E'. We say that the desired involution, consists of two Pairs of Conjugate Points A', r· and /1', Z', and Double Points B', E'.

{cll:fflffil In fact, if we combine together the above constructed involutions of one Pair of Conjugate Points each one of line (E), we will have the Harmonic Involution of two Pairs of Conjugate Points a, y and 15, ~' with Double Points

13, E.

In addition, if we combine together the above constructed involutions of one Pair of Conjugate Points each one of circle (o'), we will have the Harmonic Involution of two Pairs of Conjugate Points A', r· and /1', Z' and with Double Points 8', E', which is in fact t he desired one.

~./fi•llt'l4t14i For the converse we follow the opposite direction.

{EiW4t®ft1ii Here we are given the Harmonic Involution, w ith two Pairs of Conjugate Points A', r· and /1', Z' and with Double Points 8', E' and we are asked, based on that and with the appropriate projective transformations to find a hexagon ABr/1EZ inscribed in circle (o) convex and symmetrical with respect to BE. Taking into consideration Construction 180 (§ 27.2), Criteria 35 and 75, as well as paragraphs 3.2, 8.11, 8.62, we are led to the construction below :

lement of Collinear - Coe clic Harmonic Involutions.

(b). Construction. Based on Construction 180, we construct figure 103. In which we notice that the Harmonic Involution of two Pairs of Conjugate Points A', r· and A', Z' and with Double Points B', E', is analyzed in two Harmonic Involutions with common Double Points B', E' and each one of one Pair of Conjugate Points, which are A', r· for one and A', Z' for other, thus based on Construction 180 (converse), the two Involutions of one Pair of Conjugate Points each occur on line (E) respectively, from which with the appropriate combination a collinear involution of two Pairs of Conjugate Points occurs . Thus, from the two involutions above and based on Construction 180 (converse), we realize that the two quadrilaterals ABrE and AEZB are rhomboids that have BE as a common axis of symmetry and from which the desired symmetrical hexagon ABrAEZ occurs.

{tal@ffil In fact, if we combine together the two involutions of line (E) above the Pair of Conjugate Points of the first is, a, y and of the other c5, ~. and with Pair of Double Points

13, £ , then

gate Points a, y and 15,

~

the Harmonic Involution of two Pairs of Conju-

and with Double Points

13,

E

occur.

Moreover, we easily realize that if we combine the two rhomboids ABrE and ABZE, that share their axis of symmetry BE, hexagon ABrAEZ occurs, which we easily realize that it is symmetrical with respect to BE.

2nd Way (Based on Corollary 159).

itl•Jli431 tmumn1~1 Taking into consideration Corollary 159, we are led to the construction below: (b . Construction. Based on Construction 180, we construct figure 103, in which we say that the two pairs of points A', r• and A', Z', are in Harmonic Involution, with its Double Points B', E' and that it is the desired involution.

B a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:-:.. P=art~C.

[010'®1 Because hexagon ABrt.EZ is symmetrical with respect to BE, Ar and t.Z. will be perpendicular to BE, thus, according to Corollary 159 (direct) these, transform to the conjugate lines AT' and /1'2', with B'E' both. Therefore, according to known facts, the two pairs of points A', r· and /1', 2', are in Harmonic Involution, with Double Points B', E' and therefore this is in fact the one desired.

For the converse we follow the opposite direction.

Taking into consideration Corollary 159, we are led to following construction: (b). Construction. Based on Construction 180, we construct figure 103. in which we say that hexagon ABrt.E2, is the desired symmetrical with respect to BE.

ttalitt®I Because pairs of points A', r· and /1', 2', are in Harmonic Involution, with Double Points B', E', AT' and /1'2', are conjugate with respect to B'E'. Thus, according to the converse of Corollary 159, AT' and /1'2', transform into the lines Ar and t.Z. perpendicular to diameter BE, therefore it easily occurs that hexagon ABrt.E2, is in fact the desired symmetric with respect to BE one.

ld411Elih (a). Problem 181, the way it is given, it is undefined. If we want to define it, then we need sufficient data. (b). The above Problem and its solutions, we believe are new and firstly appear here. (c). We notice that with this method, the Harmonic Collinear Involution occurs

first

and

then

the

desired

Harmonic

Cocyclic

Involution.

This means that with the New Method of Harmonic Transformation, we simultaneously achieve the Construction of the Harmonic Collinear Involution.

lement of Collinear - Coe clic Harmonic Involutions.

(d). In figure 103, the involution of line (E) is a two-branch one, as point

r

was set on the arc AZ which is placed between the two parallel lines Ar and

/j,Z,

Would we want it to be a one-branch one, we should place point

r on arcs

ABroraEZ. (e). Construction 181 and its proof, can be based on Corollary 162, or Corollary 163, or paragraphs 3.3, 3.4 and 8.62 (Simple Harmonic Involute Pencil of three pairs of beams), etc. (f). Other relevant Propositions are 35, 42, 45, 75, 135, 158, 159, 162, 163, 180.

27.23. Construction of a Simple Cocyclic Harmonic Involution of three Pairs of Conjugate Points (Relevant§ 8.64). (How a Simple Harmonic Cocyclic Involution occurs from a Convex Inscribed Symmetrical Octagon). Construction 182 (Figure 104). Based on appropriate projective transformations and intersections with line and circles, construct a Simple Cocyclic Harmonic Involution of three Pairs of Conjugate Points, based on an Inscribed Convex Symmetrical with respect to its Diagonal Octagon, which is given to us, and conversely, similarly and based on a Simple Cocyclic Harmonic Involution of three Pairs of Conjugate Points, which is given to us, construct an Inscribed Convex Symmetrical with respect to its Diagonal Octagon .

f#fflMh•leh Based on Corollary 164 (§22.61G), we construct figure 104. Hence, based on this figure and through analyzing it to and combining it with figures equivalent to figures 100 and 103, working similarly to Constructions 180 and 181, we can easily solve this Problem as well , taking into consideration paragraph 8.64).

li 47, 14 --> 48, 20 --> 52, 21 --> 53, etc .

27.51 . Construction of a Simple Cocyclic Harmonic Involution of two Pairs of Conjugate Points. Solution to the Problems 9, 10 and 42 (§ 9.3), using the Method of Harmonic Transformation. Construction 187 (Figure 118). On a circle (o'), there is an ordered group of points A', r·, /1', Z' given. If pairs of points A', r· and /1', Z' are the two Pairs of Conjugate Points of a Simple Cocyclic Harmonic Involution, we are asked to define the pair of Double Points B', E' of the involution using the Method of Harmonic Transformation ,

or to state it otherwise:

Let there be a circle (o') and the group of four ordered points A', r·, 11', Z' on it. Prove, using the Method of Harmonic Transformation , that only one pair of points B', E' on the same circle (o') exists, which harmonically divides both pairs of points A '-r' and /1'-Z '.

f§fflfflh•hi [f;lltittfflff1~j Let us assume that the desired Double Points of the involution are set and are B', E'. If we take into consideration Construction 181 (§ 27.22) (converse) and locus 16 (§ 6.10), we can easily construct figure 118, as shown below: We set a random point Ton circle (o') and create pencil T(A'BT'/1'E'Z') or T(al3y~E{). If we intersect this pencil with a random line (E), according to construction 181 (converse), the Simple Harmonic Involution with Double Points and two Pairs of Conjugate Points a, y and ~. { will occur. Let us draw arcs of circles in correspondence to chords a~ and y{ and

13,

E

Ela.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:-:..P=art~C. equal acute (preferably) angles, towards the same side of (E) and if these arcs intersect at point r, then the pencil r(af3yl>E~) or r(ABrAEZ) is formed.

Because it is true that L arl>= Lyr~~Lar~= Lyrl>, hence pencil r(a~l>y) is symmetrical with respect to rE and thus , according to locus 16, point r will belong to the Apollonian circle of diameter j3E. Let (o) be a random circle that runs through point r, then pencil r(af3yl>E~) re-intersects circle (o) at points A, B, r , A, E, Z. Since as well L'.'.ArA= Lrrz quadrilateral ArAZ is a trapezium with Ar // AZ. But, we can easily find that: LArB= LBra= L j3ry= LBrr= LBAr or LArB= LBAr~AB=Br. In a similar way we also find that AE=EZ, thus hexagon ABrAEZ, Eiva1 is symmetrical w ith respect to BE, which will also be a diameter of circle (o), as it is also true that L j3rE= LBrE=1 L. From all the above, we realize that hexagon ABrAEZ is constructed, so we are led to the construction below. (b). Construction. On circle (o'), we set a random point T. Pencil T(AT'A'Z') intersects a random line (E) at points a, y, l>, ~- We write

lement of Collinear - Coe clic Harmonic Involutions.

arcs with chords al> and yl, and equal acute angles, that intersect at point

r. Because L'.arl,=Lyrl>==LArz=Lrra, we will have AZ=ra and therefore Ar

II az. Let us draw the diameter that is perpendicular to Ar and

az, that intersect

circle (o) at points Band E. Based on B and E we can easily define intersections p and E on (E) and ultimately intersections B' and E' on circle (o'), which we say are the desired Double Points of the involution, with Pairs of Conjugate Points A', r• and b.',Z.'

tcll:ft®I Because hexagon ABrb.EZ is symmetrical with respect to BE, if we work in a way similar to the implication of Construction 181, we prove that

p, y , 6, E, l, and A', B', r·, a·, E', Z', constitute Simple Harmonic Involutions of two Pairs of Conjugate Points each and Double Points p, E for points a,

the first and B', E' for the second . This means we have defined the desired Double Points B', E', of the involution.

lt3411Fiid (a). From Construction 187 it occurs that the Double Points B', E', of the involution were defined, without using poles and polar coordinates, that we had used in Construction 42. Here, as we already saw, we used the method of Harmonic Transformation. (b). The Problem above and its solution, we believe to be new and firstly appear here. (c). We notice that using this method, firstly the Double Points

p,

E

of the

Harmonic Collinear Involution occur and then the desired Double Points B', E', of the Harmonic Cocyclic Involution (Construction 42). This means that with this New Method of Harmonic Transformation, we simultaneously achieve the Construction of Double Points and of the Harmonic Collinear Involution (Constructions 9 and 10). (d). The proof of Construction 187, can be based on Corollaries of paragraph 22.61. (e). If angles arl> and yrl, are drawn to be right, then it easily occurs that Ab. and rz that will appear, would be diameters of circle (o).

S a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:-:.. P=art~C.

(f). In the above proof of Constructions 180-186 we tried to show the detailed way of applying this method in Simple Involutions. (g). The execution of the Method above, is descriptively and unambiguously developed in figures 118 above. (h). Taking into consideration what was mentioned in paragraph 8.62 and in some of Corollaries 158-168, or in Construction 42, then we can easily solve Problem 9 (§ 6.3), based on the Theory of Poles and Polar Coordinates. In fact, in figure 118, if on line (E), points a, y, 15, Points

13,

E

~

are given and Double

of the involution are demanded, then we set a random point T

on a random circle (o') and intersections A',

r·,

/1', Z', we work similarly to

Constructions 42 and after we define intersections P, , K

=A' /1' nr·z·, we

can easily define intersections B', E' and through these the desired points

13, E. (i). Other relevant Propositions are 1, 9. 10, 16, 42, 180-186, 188.

27.52. Constructions of a Double Cocyclic Harmonic Involution of two Pairs of Conjugate Points . Solution to Problems 11 (§ 6.5) and 43 (§ 9.4) , using the Method of Harmonic Transformation . Construction 188 (Figure 117).

r·,

On circle (o'), we are given the ordered set of points A', B',

/1', which

constitute a Double Cocyclic Harmonic Involution, with two Pairs of Conjugate Points A', /1' and B',

r· of one simple

conjugate points A', B' - H', 0' and

involution and three pairs of

r·, /1' of the other simple involution and

we are asked to define their two pairs of Double Points H', 0' and E', Z', respectively, using the Method of Harmonic Transformation, or to state it differently: On circle (o') we are given an ordered set of its points A', B',

r·,

/1'. Prove

that only one pair of points H', 0' on the same circle (o') exists, which harmonically define the two pairs of points A', /1'

Kai

B',

r·

and only one

pair of points E', Z' on the same circle (o'), which harmonically divide the three pairs of points A', B' - H', 0' and

r·, /1'.

lement of Collinear - Coe clic Harmonic Involutions.

f#fflfflh•leh {Gift®kt1~J We assume that the two Pairs of Double Points of the Double Involution, we constr ucted and are H', 0 ' and E', Z'. According to Construction 184 (converse), we can easily construct firstly, on line (E), the Double Harmonic Involution of two Pairs of Conjugate Points a, 13 and y, l5 with their Double Points the following pairs 11, 9 and£, ~. and consequently, after we define the Orthooptical Point r (§ 7.14), we draw in circle (o) that runs through r , the Rectangle ABra, as well as pairs of points H, 0 and E, Z. For the rectangle H0 and EZ are its two axes of symmetry, which are obviously perpendicular and diameters of circle (o). We notice that points A, B, r , a are fixed on circle (o) and constructed as points A', B', r ·, a·, are given. In this way, rectangle ABra is constructed, thus its axes of symmetry H0 and EZ are constructed as well, so we are led to Construction below. b . Construction. On circle (o') we set a random point T and let pencil T(A'BT'a ') intersect line (E) at points a, 13, y , l:i, respectively. Using diameters ay and l3l:i we draw circles that intersect each other at point r . If pencil r(al3yl:i) re-intersects circle (o) at points A, B, r , a , we say that ABra is a rectangle, whose axes of symmetry H0 and EZ are easily constructed. Consequently, based on rectangle ABra and points of intersection H, 0 and E, Z of the axes of symmetry and the circle (o), easily, as in the direct of Construction 184 we set on line (E) points a, 13, y , l:i, £, ~. 11, 9 and afterwards on circle (o') points A', B', r·, a·, E', Z', H', 0', which we say constitute a Double Harmonic Involution, of which the two desired Pairs of Double Points are H', 0' and E', Z'.

[01:ftMii Because angles arv and 13rl:i or Arr and Bra are right, then Ar and Ba are diameters of the circle (o). Hence quadrilateral ABra is actually a rectangle and its two axes of symmetry H0 and EZ are diameters of the circle (o).

B a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:-:.. P=art~C.

Thus, as points A', B',

r,

r·, fl',

E', Z', H', 0', were drawn based on points A, B,

fl, E, Z, H, 0, according to the proof of the implication in Construction

184 and points A', B',

r·, fl',

E', Z', H', 0', would constitute a Double Har-

monic Involution of two Pairs of Conjugate Points A, B and

r,

fl and two

Pairs of Double Points H', 0' and E', Z', this means that H', 0' and E', Z', are the desired Double Points of the Involution, as they also belong in circle (o'), in addition H'0' and E'Z' are conjugate, as they stem from H0, EZ which are perpendicular to each other [Corollary 159 (implication)]. [It is possible for us to have another simple solution to Problem 188, if we apply twice Construction 187. Once to define Double Points H', 0' and another one to define Double Points E', Z'. Attention is needed though when constructing point

r,

which should be Orthooptical (§ 7.14), and is defined

from the intersection of circles with diameters

E~

and 118].

[;(§11tlid (a). From Construction 188 it occurs that Double Points H', 0'

Kai

E', Z', of

the involution, were defined without using poles and polar coordinates, that we used in Construction 43. Here, as seen, we achieved the proof using the method of Harmonic Transformation. (b). The Problem above and its solution, we believe to be new and firstly appear here. (c). We notice that using the method of Harmonic Transformation, firstly the Double Points 11, 8 and E,

~

of the Harmonic Collinear Involution occur

and then the desired Double Points H', 0' and E', Z', of the Harmonic Cocyclic Involution. This means that with this New Method of Harmonic Transformation, we simultaneously achieve the Construction of Double Points and of the Harmonic Collinear Involution. (d). The proof of Construction 188, can be based on Corollaries of paragraph 22.61. (e). We tried to provide the detailed way of applying this method on Double Involutions, in the proofs of Constructions 184-188 above. (f). Execution of the above Method, is shown step by step, descriptively and unambiguously in figure 117 above.

lement of Collinear - Coe clic Harmonic Involutions.

(g). Taking into consideration what was mentioned in paragraph 8.63 and in some of Corollaries 158-168, or in Construction 43, then we can easily solve Problem 11 (§ 6.5), based on the Theory of Poles and Polar Coordinates. In fact, in figure 117, if on line (E), we are given points a, asked for Double Points 11, 8 and E,

~

13, y,

l> and are

of the involution, then we set a ran-

dom point Ton a random circle (o') and intersections A', B',

r·, 11', we work

similarly to Construction 43 and after we define intersections P, , K

=A'

r· ns·a·, then intersections H', 0' and E', Z' are easily defined, and through these the desired points 11, 8, and E, ~(h). If we work in the same way as in Propositions 180-188, we can obviously prove or solve many other Propositions and Problems mentioned in Part A. (i). Other relevant Propositions are 1, 11, 16, 43, 180-187.

B a.. .·__..;.H:..;..;A:.:..;R=M=O.:..:.Nl:..:::C...;:G=E=O=M=E'-'-'TR'-''-P'-'a=rt;....:C -Y =.

I

PARTD SUPPLEMENT OF PART B Applications of Harmonic Multilaterals, Mainly using the New Method of Harmonic Transformation.

Euclid of Alexandria.

S a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

COMPLEMENT OF PART B

Applications of Harmonic Multilaterals, Mainly using the New Method of Harmonic Transformation. 28. Generall . Part D is actually an extension to Part B and includes some additional data about Harmonic Multilaterals (Collinear and Cocyclic), that came up after having written Part C. These data will obviously be included in Part B in a new edition of the book. This did not happen here, so as to not change the sequence of their conception, mostly for historic purposes, something we did not want to happen. Mainly though, it includes applications of Harmonic Multilaterals, using our New Method of Harmonic Transformation, which we have developed in paragraph 21. We note that using this significant method of «Harmonic Transformation», we can solve methodically, easily and spherically most of the Problems of Harmonic Geometry, even those that were impossible to solve until now.

lement of Harmonic Multilaterals.

29. Applications of Harmonic Multilaterals, mainly using the Mathod o Harmonic Transformation .

29.1. Application of the Method of Harmonic Transformation, Significant Property of the Harmonic Decagon. Another proof to Proposition 106. Convergence of Diagonals in Harmonic Decagons. Proposition 200 (Figure 120). The diagonals (main), of Harmonic Decagons, are concurrent.

cp=(115)L=18 µoip., BH nAr=µ, the above Collinear Series of Points, constitutes a Simple Collinear Harmonic Involution, with Double Points B, Hand these Pairs of Conjugate Points: I, { -

a, E - 13, 6 - K, y - " · r,' -J\', 8' - >., {' - a', E'

-13', 6' - µ , y'.

(1 ).

Proof (Figures 99 and 121 ). (a). Analysis (Figure 99). Here we will remember our new Method of «Harmonic Transformation» and more specifically, the construction way of a Harmonic Decagon through a Regular Decagon and conversely the way of constructing a Regular Decagon using a Harmonic Decagon, which we have already mentioned here in Construction 155 (§ 22.4) and mainly the converse, according to which, from the given Harmonic Decagon A'BT"a'E'Z'H '0'1'K', we get a Regular Decagon ABraEZH01K, through the intervention of a Harmonic Collinear Group of Ten Points a, 13, seen in figure 99.

y, 6, E, { ,

r,, 8, 1, K and its Orthooptical Point

r , as

B i·:-

- ---'-' H""'A"-'R=M=O=N=IC::;.....;:::G=E=O=M=E:...:.T.:...:R..:..Y..:..P-=a:..;:rt..::..:.D.

rxr11.ia 121. 8'

More specifically, if we work in figure 99 and based on the given Harmonic Decagon A'BT'a'E'Z'H'0'1'K' and an arbitrary point T of its circumscribed circle (o'), we construct on an arbitrary line (E) the Harmonic Group of Ten Points a,

13, y ,

l>, E, ~. 11, 8, 1, K and after we define its one «Orthooptical

Point» r , we draw a random circle (o), that runs through rand in this way we get the Regular Pencil r(al3yl>E~1181K) of ten beams with cp=U5=18°, whose beams intersect circle (o), at t he apexes of the Regular Decagon ABraEZH01K.

lement of Harmonic Multilaterals.

After this construction, we notice that in triangle l:£1, rr, is the bisector of angle

El:1

( LEl:r,= Lr,l:1=2.18°=36°)

and

that

angle

r,r~

is

right

( Lr,r~=S.18°=90°). This means that Pencil l:(~r,£1) is Harmonic, hence Pencil T(~r,£1) or T(B'H'E'I') is also Harmonic or that the quadrilateral B'E'H'I' is harmonic and thus B'H' and E'I' are conjugate with respect to circle (o'). If we work in the exact same way, we also find that B'H' and a'K' are conjugate with respect to circle (o'), as in triangle l:l>K, rr, is the bisector of angle lil:K ( Llil:r,= Lr,l:K=3.18°=54°) and angle ~rr, is right ( L~rr,=5.18°=90°), thus Pencil l:(~r,l>K) is Harmonic too, and so on and so forth. (b). Proof (Figure 121). Based on what is mentioned above and if we now move to figure 121 , we see that for the Harmonic Decagon ABraEZH01K too, aK and El are conjugate to BH. After all , from the definition of the Simple Harmonic Decagon (§ 16.1d), quadrilaterals ABrH, 2H0B are Harmonic too and Ar, 20 are conjugate, to BH. Consequently all the following Ar, aK, El, 20 are conjugate to BH, with respect to circle (o). Hence, because Ar and BH are conjugate with respect to circle (o), BH will run through pole

v· of Ar or that v· EBH.

Similarly, because 02 and BH are also conjugate with respect to circle (o), BH will run through pole y of 20 or that y EBH. Additionally, because aK, El are conjugate to BH too, with respect to circle (o), BH will run through poles

r. {of aK, El , respectively, or that~·. {

EBH.

Additionally, because aK, 20 are also conjugate to BH, pair of points B, H will harmonically divide pairs of points a , K and 2, 0. Thus, according to Proposition 51 (§ 9.12), intersections a0 nzK=a and

az n0K=E,

will belong

to BH. Similarly, because Ar, El are also conjugate to BH, pair of points B, H will

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

harmonically divide pairs of points A , r and E, I. Thus, according to Proposition 51 (§ 9.12), intersections AE nn=a' and rE nAl=E', will belong to BH. Moreover, because El and 20 are conjugate to BH , intersections

13

and

13'

and

6EBH . Additionally, because Ar and .t.K are conjugate BH, intersections 6'EBH. Similarly, because Ar and 20 are conjugate to BH , intersections AZ nr0=A and A0 nrz=11· will belong to BH. Finally, because .t.K and El are, too, conjugate to BH, intersections .t.l nEK=A', .t.E nKl=8' will belong to BH. From the above we see that in fact the following sixteen points y', y , ~. r . a, E, a', E',

13, 6, 13', 6', A, 11', A', 8',

belong to diagonal BH.

After all, because diagonals of the Simple Harmonic Decagon are concurrent (Proposition 106, paragraph 19.13), it will be A'=A, hence it will be 8'=11', such as (HBA8')=1, (HBA11')=1

~

8'=11· too. This means that, intersections

and points, that are collinear, while in reality are 18 in number, because it is A'=A, 8'=11', in fact these points are 16. From the above it occurs that the series points above, is a Simple Harmonic Involution of Two-Branches with Double Points B, H and Pairs of Conjugate Points these of relation (1). Hence, if M is the midpoint of BH , then it will also be: MB 2=MH 2 = M1.M~ = Ma.ME = Ml3.M6 = MK.My = MA.M11 ' = MA'.M8' =Mh.Mr=Ma'. ME' =Mj3'.M6' =Mµ.My'.

(2).

(c). Investigation. Diagonal BH , because it was received as arbitrary, a Proposition equivalent to 201 , will be true for each one of the rest diagonals AZ, re, .t.l, EK, of Decagon ABr.t.EZH01K and thus we actually have the properties of a complex of a complete Harmonic Decagon, the tangential of the complete Decagon, the complete Decagon of the first, the second and the third diagonals

lement of Harmonic Multilaterals.

of the Harmonic Decagon (terminology § 5). This complex that will occur, will be equivalent to figures 3.2.6.(4).(b).1/ and 4.2.5.(5).(b).5/, in our book [1], which refer to a Complex of a Harmonic Hexagon and a Harmonic Octagon, respectively (See also General Remarks, § 29.10 below).

1#341160A (a). The above Proposition and its proof, we believe to be new and firstly appear here. (b). Analogous Propositions are true for all other Harmonic Multilaterals as well. (c) . Relevant Propositions are 202-204.

29.3. Another Discussion on Proposition 201, Mainly with the Simple Application of Corollaries 158 to 168. Remarkable Properties of Harmonic Decagons. Proposition 202 (Figures 122). Every Regular Decagon ABrt.EZH01K, is Harmonically transformed, to a Harmonic Decagon A'BT't.'E'Z'H'0'1'K', which, concerning its arbitrary diagonal B'H' (main), has the following properties: AT', K't.', l'E', 0'2', are concurrent at the intersection of the tangents at points B', H'. Intersections E'Z'nl'0'=l>,

A'Z'nr'0'=J\, t.'Z'nK'0'=E,

t.'l'nE'K'=J\', K'l'nt.'E'=8,

t.'0'nK'Z'=a, A'E'nr'l'=a',

E'0'nl'Z'=J3, A't.'nr'K'=J3',

A'K'nr't.'=l>', A'l'nr'E'=E', A'0'nr'Z'=rl', t.'E'nK'l'=8' and intersections y , ~. y' {', of the tangents at the following pairs of points 0'- Z', I'- E', A'- r·, K'- t.' , respectively, lie on diagonal B'H', while J\=J\' , 8'= rf=P e B'H'. Quadrilaterals A'BT'H', Z'H'0'B', B't.'H'K', B'E'H 'I', are Harmonic. Moreover, if B'H' nE'l'=1 , B'H' n2'0'=K, B'H' nt.'K'=A, B'H' nAT'=µ, the above Collinear Series of Points of line B'H', constitutes a Simple Collinear Harmonic Involution, with Double Points B', H' and Pairs of Conjugate Points: 1,

~

- a, E - 13, l> - K, y - J\, 11' -/\', 8' - A, {' - a', E' -

13', l>' - µ , y'.

(1 ).

Additionally, the Cocyclic Series of Points A', B', r•, t.', E', Z', H', 0', I', K', constitutes a Simple Cocyclic Harmonic Involution, with Double Points B', H' and Pairs of Conjugate Points: A',

r• - t.' , K' - E', I' - Z', 0',

(2).

Sia. .·- ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D. If M is the midpoint of B'H', then it will also be:

MB' 2 =MH' 2 = M1 .M~ =

Ma.ME= Ml3.Mc5 = MK.My= M/\.Mrl' = M/\'.M9' = MA.Mr=Ma'.Mt' = Ml3'.Mc5' = Mµ.My'= MA'.Mr'= Ma'.MK'= ME'.MI'= MZ'.M0'.

(3).

Properties analogous to the above, are true for every other diagonal of the Harmonic Decagon A'BT'a'E'Z'H'0'1'K', and conversely, every Harmonic Decagon A'BT'a'E'Z'H'0'1'K', is possible to be Harmonically transformed, to a Regular Decagon ABraEZH01K, which has the following properties : Ar, Ka, IE, 02, are concurrent at infinity (are parallel). It is symmetric with respect to BH. Intersections of these pairs of chords AZ -re, al-EK, a0-KZ, E0-IZ, EZ-10, az-K0, Kl-aE, AE-rl, Aa-rK, AK-ra, Al-rE, A0-rz, aE-IK, such as intersections of tangents at these pairs of symmetric points 0- Z, I- E, A- r, K- a respectively, will lie on BH, while chords AZ, re, a1, EK, BH (diagonals), would be concurrent at center O of circle (o) and A0, BH, rz, aE, IK, will be concurrent at infinity (they are parallel). The two above involutions would be true [Collinear (1) and cocyclic (2)], such as relation (3), properly adjusted. Quadrilaterals ABrH, ZH0B, BaHK, BEHi , would be symmetric, with respect to BH. Analogous properties, to the above, are true for every other diagonal of the Regular Decagon ABraEZH01K.

Proof (Figure 122).

[EJl•lli441 Since Decagon A'BT'a'E'Z'H'0'1'K' is Harmonic, according to Construction

155 (direct) (§ 22.4), this will have occurred from a Regular Decagon ABraEZH01K, with the application of the New Method of Harmonic Transformations (§ 21 and Construction 155). Hence, if we receive as axis for example diagonal BH of ABraEZH01K and to this draw a perpendicular to its center 0, then we have a second axis MN [where M, N are the points of intersection of this axis and circle (o)].

lement of Harmonic Multilaterals.

&

Ela.. ·- ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D. It easily occurs that, quadrilaterals AfZ0 and aEIK are rectangles and share their axes of symmetry, diameters BH and MN. This, according to Corollary 167 (direct) (§ 22.61 I), means that these rectangles are Harmonically Transformed to Harmonic Quadrilaterals AT'Z'0' and a'E'l'K' whose diagonals A'Z', r·e·, a'I' E'K', are concurrent at the intersection of chords B'H' and M'N', that have occurred from axes BH and MN respectively, according to the above Harmonic Transformation. Hence it will be: B'H"nM'N"nA·z·nr·e·=/\=l'a'nE'K'=A' r'1 A'=/\=B'H"nM'N'.

(4).

Moreover, according to the same Corollary 167 (direct)(§ 22.61 I), It will also be :

AT'nK'a'nM'N"nl"E"n0'Z'=.

(5).

Additionally, according to the above Corollary 167 (direct), it will be: B'H'nA'0'nr·z·=rJ'=K'l'na·E'=8'=P EB'H' r'1 8'=rJ'=P EB'H'.

(6).

After all, since Decagon A'BT'a'E'Z'H'0'1'K', comes from Regular Decagon ABraEZH01K, which is symmetrical with respect to BH, according to Corollary 165 (direct)(§ 22.61 H), intersections: /\= B'H'nM'N', P=8'=rl', a , 13, y, li, E, ~. a',

13', y', li' , E',

~· .

all in fact lie on diag-

onal B'H'. Similarly, according to the same Corollary 165 (direct), quadrilaterals A'BT'H', Z'H'0'B', B'a'H'K', B'E'H'I', are Harmonic. Analogous properties, to the above, Harmonic Decagon A'BT"a"E'Z'H'0'1'K' has concerning every other diagonal of it too, as this Harmonic Decagon, occurs from Regular Decagon ABraEZH01K, which is obviously symmetric with respect to every other of its diagonals, since BH , was randomly received. From the above, we easily realize that the above series of points of line B'H', constitutes a Simple Collinear Harmonic Involution of two Branches, with Double Points B', H' and Pairs of Conjugate Points those in relation (1), (definition§ 2.11). In fact, because for example Z'0' is the polar of y, with respect to circle (o'), it will be (yK B'H')=1 . Additionally it is (lil3 B'H')=1 , and so on and so forth. Moreover, because chords AT', a'K', E'I', Z'0' are conjugate to line B'H', the Cocyclic Series of Points A', B', r·, a·, E', Z', H', 0', I', K', will constitute a

lement of Harmonic Multilaterals.

Simple Cocyclic Involution, with Double Points also B', H' and Pairs of Conjugate Points those in relation (2), (definition§ 8.11), as it will be for example (a'K'B'H')=1, (AT'B'H')=1, and so on and so forth. Therefore, from the two Involutions above, it occurs that the above relation (3) is true (§ 8, 11 ).

tl:lli·J,VJ4t14 If Decagon ABraEZH01K has been Harmonically transformed, from the Harmonic Decagon A'BT'a"E'Z'H'0'1'K' in the known way [Construction 155 (converse) (§ 22.4)), in such way that it is Regular, then in this all what is desired in Proposition 202 will be true (Converses of Corollaries 165 and 167). These, besides, are rather easy to prove, if we depend on Theorem 253 in volume A' of book [28] and take into consideration that ever other diagonal of the regular Decagon is an axis of symmetry for it. Moreover because Decagon ABraEZH01K is Regular and because of its symmetry with respect to BH, we easily realize that here as well, analogous Involutions occur (collinear and cocyclic), to those that are mentioned in the direct, consequently for those too, a relation analogous to (3) will be true, properly adjusted, (without the marks on the letters). For example, from the Cocyclic Involution of points A, B, r, a, E, Z, H, 0, I, K, easily and directly it occurs that: 08 2 =0H 2 =R 2 = OA.Or=oa.OK=OI.OE=OZ.00. (c). Investigation. We easily prove that, the five intersections of the opposite sides (of their carriers) of the Harmonic Decagon (figure 122) are collinear, Ka8wc; as these belong to the polar of A (point of convergence of the main diagonals of the Harmonic Decagon). This means that, intersections A'B'nZ'H', BT"nH'0', r·a·ne'I', a'E'nl'K', E'Z"nK'A', are collinear (equivalent to Pascal's Theorem). From these five intersections, some of its diagonals run through too, main and not, uniformly distributed (three from every intersection), in a way that

Ela.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D.

they are completely in order and constitute a Harmonic Complex of lines and Central Pencils of five beams each (this figure can count for an exercise for those interested). Moreover, the carriers of the first (terminology §5) diagonals of the Harmonic Decagon, constitute two Complete Harmonic Pentagons, while the second and third diagonals of this Decagon, respectively constitute two Complete Harmonic Asteroid Decagons. After all , except for the above mentioned five collinear intersections, that belong to the polar of intersection /\, we also have intersections of sides and diagonals that intersect in pairs on the main diagonals {See above (figure 122), which though we have approached in alternative ways and extensively, in our book [1]}. Such complexes, of Complete Harmonic Quadrilaterals, Hexagons, Octagons and not only that, are mentioned in our book of five volumes [1] 1990 (See also General Annotations § 29.10 below). There, we have approached these matters from another point of view, while here we use the respective circumscribed Complete Multilaterals (polar converses). After making up the Method of Harmonic Transformation(§ 21), Corollaries from 158 to 168 and its applications , in Harmonic Multilaterals, the General Conclusion below occurs:

General Conclusion. A ll Harmonic Multilaterals occur from the respective Regular Multilaterals (with an equal amount of sides), applying the Method of Harmonic Transformation and conversely. Every apex, side, diagonal of a Regular Multilateral matches to an apex, side, diagonal of its respective Harmonic Multilateral. In general, every center of a pencil of a Regular Multilateral, matches to a center of a pencil, of an equal amount of beams, of its respective Harmonic Multilateral with an equal amount of s ides and every line of a Regular Multilateral matches to a line of its respective Harmonic Multilateral.

lement of Harmonic Multilaterals.

Moreover, to each diagonal {main) of the Regular Multilateral, belongs an involutory series of points that consists of the convergence of equal number of its sides or its diagonals. The exact same thing happens in the respective Harmon ic Multilateral.

l;i4uFO@ (a). Proposition 202 above is actually Proposition 201 (paragraph 29.2), in another point of view, extended and with more data desired, while its converse is also given. (b). This Proposition, is significant as it provides us the whole image of every Complete Harmonic Decagon, as with it various and impressive properties of the Harmonic Decagon are revealed, which no other multilateral has, some of which will be mentioned below in Construction 203. (c). Axis BH was randomly chosen. We could have chosen any other diagonal (main) of ABraEZH01K. (d). We notice that in Harmonic Transformations, multilaterals (Regular and its respective Harmonic), have an equal number of sides and angles, while every point on one multilateral matches a point on the other. The same happens for rectilinear segments, and by extension for lines. (e) . The Proposition above and its proof, we believe to be new and firstly appear here. (f). Analogous Propositions are true in the rest Harmonic Multilaterals. (g). Relevant Propositions are 152, 155, 165,167, 201 , 203.

29.4. Gigantic Problem (in Size and in Difficulty). Big Challenge. The Big, Rather Difficult and lmpresive Problem. Construction of a Special Convex Inscribed in Circle Decagon, with Various and Impressive Properties. Construction 203 (Figures 123-126). In a given circle (o), we inscribe a special convex Decagon ABraEZH01K (not regular), which has the main properties below: (a). Its main diagonals (terminology§ 4), are concurrent at point A. I.e. it is: AZnBHnr0na1nEK=A.

(1).

S a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

AB r~ EZ H0 IK Br. ~E. ZH" 01 . KA = 1·

(b). Additionally, it is:

(2).

(c). If a , 13, y, ... .. K, are the distances of A. from sides AB, Br, ra, ... , KA respectively, it would be:

a AB

13 _ Br

V -~-___!_ _ _J_ _ _B_ _ _!_ __ r~ ~E EZ ZH H0 01 IK

a y £ 8

_~ KA .

1

P. ~.{ ·;, ·tc=1 .

(d). Moreover, it will be :

(3).

(4).

(e). The five intersections of its opposite sides, are collinear. Hence intersections: ran01=n, aEnJK=P, EZnKA=I, ZHnAB =T, H0nBr=Y,

(5).

belong on a line (E), while every other of its main diagonals, runs through an intersection from these. I.e. it is: nEAZ, PEBH, Ier0, TEal, YEEK.

(6).

(f). Each one from the five intersections of its second opposite diagonals

(terminology § 5), coincides to one of the above collinear intersections. Hence it is: BEnHK=n, rznA0=P, aHnBl=I, E0nrK=T, ZlnaA=Y.

(7).

(g). The ten intersections KrnBl=a', AanrK=l3', BEnaA=y', .. . ,IBnA0=K',

(8)

of its second diagonals, lie on its main diagonals: AZ, BH , r0, ... ,KE, respectively (two intersections on each diagonal) . (h). The ten intersections KanB0=a", AEnrl=l3", BznaK=v", ... ,1rnAH=K",

(9).

of its third diagonals, lie on its main diagonals : AZ, BH, r0, ... ,KE, respectively (two intersections on each diagonal). (i). The ten intersections IKnBr=a, KAnra=j3, ABnaE=y, ... ,01nAB=K,

(10).

of its one by two sides , lie on its main diagonals: AZ, BH , re, .. . , KE, respectively (two intersections on each diagonal).

(j). The ten intersections 0KnBa=a, , IAnrE=131, KBnaz=y,, ... ,HlnAr=K1,

(11 ).

of its first diagonals, lie on its main diagonals : AZ, BH, r0, ... ,KE, respectively (two intersections on each diagonal).

lement of Harmonic Multilaterals.

(k). If the tangents of circle (o), at pairs of its opposite apexes /1 and I, E and K, Zand A, H and B, 0 and r , intersect at points n·, P', r·, T', Y', respectively, then these intersections lie on line (E) above. (I). If the tangents of circle (o), at pairs of its apexes K and B, A and r , B and /1, .... ,I and A, intersect at points A', B', r·, ... , K', respectively, then these intersections lie on its main diagonals AZ, BH, r0, .. .. ,KE, respectively (two intersections on each diagonal). (m). If the tangents of circle (o), at pairs of points r and I, /1 and K, E and A, .... ,0 and B, intersect at points A1, 81 , r1 , ... , K1 , respectively, then these intersections lie on its main diagonals AZ, BH , r0, .. .. ,KE, respectively (two intersections on each diagonal). (n). Each one of the five intersections of its opposite first diagonals, coincides to one of the above collinear intersections n •, P',

r , T', Y'.

Hence it is:

rEnK0=n·, /1ZnlA=P', EHnBK=r·, z0nrA=T', B/1nHl=Y'.

(12).

(o). Each one of the five intersections of its opposite third diagonals, coincides to one of the above collinear intersections n', P', r, T', Y'. Hence it is: BZnHA=n·, rHnB0=P', 110nn=r·, Elnl1K=T', ZKnAE=Y'.

(13).

(p). The eighteen intersections above that lie on each of its main diagonals, constitute a Simple Collinear Harmonic Involution, with Double Points the pairs of apexes of the decagon that lie on this same diagonal of it. Moreover, every eight apexes of the decagon, each time, constitute a Simple Cocyclic Harmonic Involution with Double Points the two apexes remaining. For example, on diagonal AZ, as we saw above, intersections n, A1, a1 , a, A', A=BKnAZ, a', a", v=rlnAZ, /\, TT=l10nAZ, t", r, µ=EHnAZ, Z', t, t1, Z1 exist, which constitute a Simple Collinear Harmonic Involution, with Double Points the pair of its apexes A and Z. Additionally, apexes B, r , 11, E, H, 0, I, K, constitute a Simple Cocyclic Harmonic Involution, with Double Points the pair of apexes A and Z aswell. This means that, let M be the midpoint of AZ, it will be: MA 2= MZ 2 = MA.MA'= Ma'.Ma= Ma".Ma1= Mv.MA1= MAMn= MTT.MZ1= Mt".Mt1= Mr.Mt= Mµ.MZ'= MB.MK= Mr.Ml= M/1.M0= ME.MH.

(14).

(q). The ten intersections n , n·, P, P', r, r·, T, T', Y, Y', that belong to line (E), constitute a Collinear Harmonic Group of Ten Points. (Its properties are endless. But the above are enough for us here).

B a.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D.

rxrn.1a 123. Solution (Figures 123-125). [We highlight the fact that figures 123-125, regularly should be one uniform

lement of Harmonic Multilaterals.

S a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

figure and circle (o), be of the same size, in all figures . In that case, though, the design paper would need to be huge. For this reason we separated it in three parts and with circle (o), in various sizes, depending on our needs. Thus the reader should each time go to the figure that includes the geometrical figures that interest him or her).

We suppose that the desired Decagon is constructed and is the ABrt..EZH01K (Figure 123). Thus its main diagonals would be concurrent at point A. [relation (1)], for which, according to the data of the Problem, relation (3) will also be true. This, according to the converse of Criterion 114 (§ 19.23), means that the desired Decagon needs to be Harmonic, whose Construction can easily be made with the application of our new Method of Harmonic Transformation [This is possible to depend on the converse of Criterion 98 (§19.3), as for example pencil l(n'PP'H'TT'YY'n) or l(IKABrt..EZH0), would be harmonic). (b) . Construction . For the Constructions of the Harmonic Decagon, we work in the same way as in Construction 155 (§ 22.4) (implication). Hence, if the Decagon that occurs is ABrt..EZH01K, then we say that this is the desired one.

tell :ft£ffi I For the constructed, as above, to be the desired Harmonic Decagon, we need to prove that for it all properties mentioned in Construction 203 above are true, paragraphs (a) to (q). In fact, since Decagon ABrt..EZH01K, is Harmonic: 1/. According to Proposition 106 (§ 19.13), relation (1) will be true.

2/. According to Proposition 121 (§ 19.34), relation (2) will be true. 3/. According to Criterion 114 (§ 19.23) (w8u), relation (3) will be true. 4/. According to Proposition 125 (§ 19.39), relation (4) will also be true.

lement of Harmonic Multilaterals.

Ela.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D.

5/. We easily realize that intersections

n, P, r,

T, Y are conjugate to point /1.,

with respect to circle (o), thus according to what is known, all these will belong to the polar (E) of point A After all , according to Proposition 201 (§ 29.2) or Proposition 202 (§ 29.3), relation (6) will also be true. 6/. Additionally, according to Proposition 201 (§ 29.2) or Proposition 202 (§

29.3), relation (7) will also be true. 7/. According to Proposition 201 (§ 29.2) or Proposition 202 (§ 29.3), all relations from (8) to (11) will be true. 8/. We easily realize that intersections

n·, P', r·, T', Y', are conjugate to point

/1., with respect to circle (o), thus according to what we know, all these would belong on the polar (E) of point A Hence all ten points

n,

P, r, T, Y,

n·, P', r·, T', Y', lie on line (E).

9/. According to Proposition 201 (§ 29.2) or Proposition 202 (§ 29.3), all the desired i n paragraph (I) would be true .

10/ According to Proposition 201 (§ 29.2) or Proposition 202 (§ 29.3), the desired of paragraph (m) would be true.

11 /. Because for example rE is the polar of b.' and K0 is the polar of I', b.'I' is the polar of the intersection of rE and K0. Hence, because b.'I ', and b.l coincide as we saw, their poles will coincide as well. This means that the intersection of rE and K0 will coincide with

n·.

Similarly we find that the

rest of what is desire in (12) is also true.

12/. Because for example BZ is the polar of b.1 and AH is the polar of 11 , b.1'1 will be the polar of the intersection of BZ and AH . Hence, because b.11 1 and b.l, coincide as we saw, their poles will also coincide. This means that the intersection of BZ and AH will coincide with

n•. Similarly, we

find that the

rest of what is desired in (13) are true.

13/. According to Proposition 201 (§ 29.2) or Proposition 202 (§ 29.3), all of what is desired in paragraph (p) are true.

14/. We notice that for example the central pencil 0(ABrb.EZH01K) of ten beams (with the tangent at 0 included), will be harmonic(§ 12.8), as it is

lement of Harmonic Multilaterals.

inscribed in circle (o) and Decagon ABraEZH01K will be Harmonic because of the way it is constructed (Criterion 98, § 19.3, direct). This means that this pencil is intersected from line (E) at the following ten points

r, r·,

n, n·, P,

P',

T, T', Y, Y', these points will in fact constitute a Harmonic Group of

Ten (§ 11.1d and§ 12.8). (d) . Description - Interpretation of Figures 123-126. Figures 123-125, constitute a uniform figure, as each one completes the other two or is part of the other two. We made them into three figures though, in order to avoid using big size paper.

Figure 123. In this figure intersections of diagonals of the Harmonic Decagon that lie inside circle (o) are included. Hence, we notice that with these intersections the two following decagons are formed

a'l3' ... K' and a"l3" ... K", from which, the first one is formed from

intersections of the second diagonals of the Harmonic Decagon and the second one from intersections of its third diagonals. Figure 124. In this figure intersections of sides and diagonals of the Harmonic Decagon and its tangential are included, which are located outside of circle (o), but are relatively close to it. Hence, we notice that with these intersections the following decagons are formed: 1/ From the intersections of the sides and the first diagonals the following decagons are formed

al3 ...K and a,l3, ... K1, respectively.

2/. From the intersections of the sides of the tangential (polar converse) of the Harmonic Decagon, decagons A'B' ... K' and A,B, ... K, are formed.

Figure 125. In this figure intersections of the sides and the diagonals of the Harmonic Decagon and its respective tangential are included, that are located outside of circle (o) and actually are far away from circle (o).

8 '-·__

-:..:. H=-:A.:..:R=M-=O=N=IC:;...,;G=EO=M=E"'-TR:...:.Y -'-'P=art'-'-='-D.

lement of Harmonic Multilaterals.

In this figure we notice that ten central pencils are formed , whose centers n , n·, P, P', r , r·, T, T', Y, Y', are collinear and constitute a Harmonic Group of Ten. From these : 1/. Five w ith centers n, P, r , T, Y, are formed from the opposite sides, main diagonals and the opposite second diagonals of the Harmonic Decagon.

2/. Five with centers n·, P', r·, T', Y', are formed from the opposite sides of the tangential of the Harmonic Decagon , from the opposite first and third diagonals of the Harmonic Decagon . It is obvious that, these three figures above constitute a Uniform Harmonic Set, that has analogous characteristics, to the Regular Decagon, which is (regular) and Harmonic (Proposition 151) and from which it is after all originated , through a Harmonic Transformation (Construction 155).

Figure 126. This figure is independent from figures 123-125 and its purpose is to present the form of a finished Complete Harmonic Decagon ABr ... K aJ3y ... K

A'BT' ... K' A1B1r1...K1 nPrTY, which consists of the initial Simple Harmonic Decagon ABr ... K, the Primary Decagon aJ3y ... K, the Secondary Decagon

A'BT' ... K', the Tertiary Decagon A1B1r1 .. . K1 and the quaternary collinear Group of Five Points n , P, r, T, Y, which constitutes a Harmonic Pentagon, degenerate to a line (E).

i;i4uEUA (a). Problem 203, the way it is given, is undefined (does not constit ute a specific Harmonic Decagon). If we want to define it, then we need to be given sufficient data. (b). The above Problem and its solution, we believe to be new and firstly appear here. (c). Regular Decagon, as all Regular Multilaterals are Harmonic too (Proposition 151 , § 19.66), but constitutes a special case of the harmonic. Therefore the regular, has itself all properties of the harmonic, but in a special Regular form (for example the line of the intersections of the opposite sides, is found at infinity).

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

(d). Constructions equivalent to 203, are possible to be manufactured for all other Harmonic Multilaterals. (e). As we will realize from the above, Problem 203 is gigantic, as its formulation is rather long and mostly because this decagon has various propert ies and because for its solution we need to depend on many Propositions and Constructions, that cover almost all material in this book. it is an application covering almost the whole outline of Harmonic Geometry. We believe that there is no other way of solving it and for this reason it is a great challenge for Geometry fans. (f). If we include, in a uniform figure, the Complete Decagons that are formed from the sides of the Simple Harmonic Decagon (as in figure 126), from the Tangential Complete Decagon and the three Complete Decagons that each one is formed from the first, second and third diagonals of the Harmonic Decagon, then a complex with various properties will occur. Such complexes are mentioned in our book [1], concerning Harmonic Quadrilaterals [Figures 2.2.2.(1) .(a).3/ and 4/ pages 696 and 697], Hexagons [Figure 3.2.6.(4).(b).1/, page 1285] and Octagons [Figure 4.2.5.(5).((3).3/ page 1746], which we reached though in another way, different from the one we used here (See also General Annotations § 29.10 below). (g). Some more properties of the decagon in Construction 203, are the following: The main diagonals of its tangential decagon are concurrent at A (This is an extension to Newton' s Theorem , and Brianhon's Theorem as well. See Proposition 206). Many intersections of the sides and diagonals of the Harmonic Decagon, and its tangential , belong to every main diagonal of its tangential. Hence something equivalent to Proposition 201 or 202 is happening. Each one of the main d iagonals of the tangential, runs through one of the intersections

n·, P', r , T',Y '.

(h). For more features of Harmonic Complexes, Quadrilaterals, Hexagons and Octagons, we cite book [1]. (i). Relevant Propositions are 98, 106, 114, 121,125, 151 , 155, 201 , 202.

lement of Harmonic Multilaterals.

29.5. Notable Property of Harmonic Groups of Four. Extension to Constructions 11 , 32a and 63. Proposition 204 (Figure 53). On a line (E) a group of eight points A, B, r , 11, E, Z, H, 0 is given, for which the group of four points A, r , E, H is harmonic and for which group of four, the two pairs of points A, r and E, H, are in Harmonic Involution with Double Points B, Z, while the two pairs of points A , H and r , E are in Harmonic Involution with Double Points /1, 0 [Hence A, r, E, H. Are in Double Harmonic Involution (definition§ 2.24)]. Prove that the following group of eight points A, B, r , /1, E, Z, H, 0 , will be harmonic.

1"1.Way Based on Construction 32 a . Accord ing to Construction 32a (§ 7.22), there will be a point P from which segments AB, Br, r/1, /1E, EZ, ZH , H0, will be seen in angles equal and of measure U4 r'J q,=22.5°. This means that pencil P(ABr/1EZH0) will be Closed Regular of ¼ of a right(§ 12.5), such as also LAP0'=22,5°, thus the group of eight points A, B, r , /1, E, Z, H, 0 , will also be harmonic(§ 12.6 and general conclusion of§ 15.6).

2nd .Way (Based on Construction 63). From what is given in Proposition 204, it occurs that the four below «one by one» groups of four points in (1), are harmonic : A, B,

r, Z - r , 11, E, 0

- E, z, H, B - H, 0 , A , /1.

(1 ).

In addition, from the hypothesis, the group of points A, r , E, H, is also harmonic, thus, if we work in the same way as in the proof of way 3 in Construction 63, it will easily occur that the group of eight points A , B, r , 11, E, Z, H, 0 , will in fact be harmonic.

i•Jeidl 1~11•1eh Regarding Propos ition 204 and if we take into consideration Constructions 11 , 32a and 63, the conclusion below occurs:

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

If, regarding Proposition 204, the group of four points A, r , E, H, is not harmonic then the group of eight points A, B, r ,

t:.., E, Z, H, 0 , will

also not

be harmon ic, while if the group of four points A, r , E, H, is harmonic then the group of eight points A, B, r , t:.., E, Z, H, 0 , will also be harmonic.

IU411Fhb (a). The Proposition above and its solution, we believe to be new and firstly appear here. (b). We reach an equivalent result also if it is given that the group of four points B,

t:..,

Z, 0 , is harmonic and the pairs of points B,

t:..

and Z, 0 , are in

Harmonic Involution with Double Points r , H and pairs of points B, 0 and

t:..,

Z, are in Harmonic Involution with Double Points A, E. Extension to a circle in Proposition 205. (c) . Relevant Propositions are 11 , 32a, 63, 149, 205.

29.6. Notable Property of Harmonic Quadrilaterals . Extension of Proposition 204, from a line to a circle. Proposition 205 {Figure 96). In circle (o), its inscribed convex octagon ABrt:..EZH0 is given whose diagonals are concurrent and for which quadrilateral

ArEH is harmonic, the

two pairs of points A, r and E, H, are in Harmonic Involution with Double Points B, Z, while the two pairs of points A, H and r , E are in Harmonic Involution with Double Points

t:.., 0

[Hence points A, r , E, H. Are in Double

Harmonic Involution (definition§ 8.22)]. Prove that octagon ABrt:..EZH0, is harmonic.

tE1W4®i't1tJ If we take into consideration Construction 149 (§ 29.64), we easily reach the proof below.

lement of Harmonic Multilaterals.

tml&I In figure 96, using Construction 149 [§ (b)] , we have constructed : The random Simple Harmonic Quadrilateral ArEH . The two pairs of points a, 0 and B, Z which harmonically divide pairs of points A, H - r , E, and A, r - E, H, respectively and which are constructed in such way that AEnsznrHnae=K. We will prove that octagon ABraEZH0, is harmonic. In fact, if we work in the exact same way, as in the proof of Construction

149 [§ (c)], we prove, based on Criterion 110 (§ 19.18) too, that octagon A sraEZH0, is harmonic.

Regard ing Proposition 205, if we take into consideration Construction 149 and Proposition 204, the conclusion below occurs: If, regarding Proposition 205, quadri lateral ArEH , is not harmonic then octagon ABraEZH0, will not be harmonic as well, while if quadrilateral ArEH , is har monic then octagon ABraEZH0, will be harmonic.

1#34uFhb (a). The Proposition above and its solution , we believe to be new and firstly appear here. (b). We would reach an equivalent result also if it was given that quadrilateral Ba20 is harmonic and pai rs of points 8, a and Z, 0 , are in Harmonic Involution with Double Points r, H and pairs of points B, 0 and , a , Z, are in Harmonic Involution with Double Points A, E. (c). Relevant Propositions are 63, 110, 149, 204.

29. 7. Convergence of Diagonals of the Tangential Decagon of each Simple Harmonic Decagon .(Equivalent to Point Brianchon and Newton's Theorem, for a Harmonic Decagon). Extension of Proposition 146 to a Harmonic Decagon.

Sia.·. - ---'-'H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y.:....:..P=art:...:..=D.

Proposition 206 (Figure 127). The diagonals (main), of the tangential or Polar Converse decagon, of every Simple Harmonic Decagon, are concurrent or to state it otherwise : The tangential or Polar Converse decagon, of every Simple Harmonic Decagon , has an equivalent Brianchon point.

p

Ixri11a 121.

~ Let there be a Harmonic Decagon ABraEZH01K and A'B"r'a'E'Z'H'0'1'K' (where A' is the intersection of the tangents at points A, B - B' is the intersection of the tangents at points B, r , and so on and so forth) its respective tangential or the polar converse of this Harmonic Decagon, with its operator circle (o) (Geometry [26], paragraph 246a). We will prove that it is:

A'Z'nB'H'nr·e·na'l'nE'K'=N.

(1).

lement of Harmonic Multilaterals.

1st Way (Analytical). Let the tangents at points r , I intersect at point n and the tangents at points /1, 0 , intersect at point P. In this way, the circumscribed quadrilateral r•p0·n occurs in quadrilateral ra01, for which, according to known Newton' s Theorem (Geometry [26], § 236a), they have r0na1nr•0•nnP=/\, where /\= AZnBHnr0na1nEK [Criterion 106 (§ 19.15)]. Hence diagonal r•0• runs through /\, which is the point of convergence of the diagonals of the Harmonic Decagon ABr/1EZH01K. In the exact same way we work to find that A'Z', B'H' , /1'1', E'K', run through I\.

Hence it finally is: AZnBHnr0na1nEKnA'Z'nB'H'nr·0·na'l'nE'K'=/\=N.

2nd Way [Based on Proposition H> 1)] . There is a known Newton' s Theorem according to which, two polar converse quadrilaterals have concurrent diagonals. We have generalized this Theorem for polar converse 2n-laterals, in Proposition 16(1) in our book «Geometry [31] (volume 1), which Proposition is always true for 2n-laterals, as long as the diagonals of their inscribed 2nlateral are concurrent. Therefore, if we consider known and depend on Proposition Hi(1 ), then because the Harmonic Decagon ABr/1EZH01K has always diagonals concurrent at point/\ [Criterion 106 (§ 19.15)], at the same point/\ diagonals of the decagon A'BT'/1'E'Z'H'0'1'K' will be concurrent. Hence it will be: AZnBHnr0na1nEK=A'Z'nB'H'nr·0·na'l'nE'K'=/\=N.

Conclusion. Through the above and because the Lemoine point of the Harmonic Decagon coincides with the convergence point of its diagonals/\ [Criterion 114 (§ 19.23)], this conclusion is reached :

«The diagonals (main), of the tangential or Polar Converse decagon, of every Simple Harmonic Decagon, are concurrent at the convergence point of the diagonals of this Harmonic Decagon, which is not only the Lemoine point of this Harmonic Decagon, but also the equivalent to the Brianchon

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

and Newton point, of the tangential decagon, or their Lemoine, Newton and Brianchon points coincide at the convergence point of the Harmonic Decagon's diagonals ».

ld4uFOA (a). In analogous ways we can prove that equivalent Propositions and Conclusions, are true for the Quadrilateral , the Harmonic Hexagon (Proposition 146), for the Harmonic Octagon , etc. (b). This Proposition, we believe to be new and firstly appear here. (c). It is obvious that Proposition 206 is an extension to Newton's Theorem (Geometry [26], § 236a), which is known to be true only for quadrilateral Polar Converses. (d). A Proposition equivalent to Proposition 201, is proven to be true for every diagonal of the tangential decagon, that matches to a Harmonic Decagon. (e). Proof of Proposition 206, is possible to be achieved using the Method of Harmonic Transformation, as done in Propositions 200, 202, based on Corollaries 158-168. (f). Relevant Propositions are 106, 114, 146.

29.8. Auxiliary Proposition (Lemma), New Discussion on Criterion 34 (§ 8.71). Criterion 207 (Figure 128). In

every

convex

inscribed

in

circle

(o)

hexagon

ABraEZ

with

AanBEnrZ=K, if one of its quadrilaterals: ABrE, Braz, raEA, b.EZB, EZAr, ZABb.,

(1 ).

Is harmonic, then every other of the quadrilaterals in (1) will also be harmonic, following the equivalences below: ABrE- aEZB, raEAZABb., EZAr- sraz,

(2).

And conversely, if one pair, of the pairs of quadrilaterals in relation (2) are harmoniv, then it will also be AanBEnrZ=K (and not only that).

lement of Harmonic Multilaterals.

1'1 Way (Based on Lemma 5).

tEll•Jli431 We suppose for example that the harmonic quadrilateral is ABrE. We will prove that its respective quadrilateral AEZB is harmonic, if AAnBEnrZ=K.

Ia

p

\~ rxrUJa 12s.

In fact, according to Lemma 5 (5~ Auxiliary Proposition, in Part E), regarding hexagon ABrAEZ, for which it is AAnBEnrZ=K (hypothesis), it will also be :

AB fJi EZ Bf"JiE"ZA = 1 ·

(3).

Additionally, according to Lemma 5 and for the concave hexagon ZAErAB, because it shares its diagonals (main) with hexagon ABrAEZ it will have: AAnBEnrZ=K, thus it will be:

ZA Er JiB AE . r Ji . 82 = 1 ·

(4).

B a. ·

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

AB EZ Er AB ::::: Br . AE . AE . BZ = 1·

From relations (3), (4)

(5).

Moreover, because quadrilateral ABrE is harmonic (hypothesis), according to known Theorem {[6] § 8-4.1 O}, it will be:

AB.rE=Br.EA.

Hence, relation (5), because of (6), will be: .6.E.ZB=B.6..EZ.

(6).

(7).

The latter relation (7), according to known Theorem [6], means that the inscribed quadrilateral .6.EZB is harmonic.

tG>M·llfJ4ti4i Hence, if the given convex hexagon ABr.6.EZ, has harmonic for example the two quadrilaterals ABrE and .6.EZB, then it will also be A.6.nBEnrZ=K. In fact, because quadrilaterals ABrE and .6.EZB are harmonic from the hy-

az are conjugate to BE and thus if n is the pole of BE, Ar and az will run through n. Hence, intersection n· of Ar and az coincides with pole n of BE. pothesis, Ar and

But intersection

n· of Ar and BZ, as it is known {[4], § 81}, has KP as its po-

lar (where P=AZnra). So we see that the poles of BE and KP coincide, thus their polars will coincide too. Hence it is BE=KP or BE runs through K and P, therefore it is in fact AanBEnrZ=K (but also AZnBEnra=P). For the proof of the rest pairs of quadrilaterals in relation (2) we work in a similar way.

2nd Way Based on Lemma 7 .

(£11•J12441 We suppose again that ABrE is the harmonic quadrilateral. We will prove that its respective quadrilateral .6.EZB is harmonic, if AannBEnrZ=K. In fact, according to Lemma 7 (7th Auxiliary Proposition, Part E), concerning Central Pencil K(ABrE) or K(.6.EZB), it is:

AB rE AE ZB Br . EA = EZ . BA .

(8).

After all, because from hypothesis quadrilateral ABrE, is harmonic, it will also be:

AB rE

AB.rE=Br.EA or Br. EA =1.

(9).

lement of Harmonic Multilaterals .

Therefore, relation (8), because of (9), becomes:

.t.E ZB EZ . B.t. =1 or

b.E.ZB=EZ.Bb.. The latter relation means that quadrilateral b.EZB is harmonic etc.

tGlli•1,W4t¼4i I.e., if the given convex hexagon ABraEZ, has harmonic for example two quadrilaterals ABrE and b.EZB, then it will also be AanBEnrZ=K. In fact, because quadrilaterals ABrE and b.EZB are harmonic from the hypothesis, they will have: AB rE .t.E ZB AB rE .t.E ZB Br . EA = 1• EZ . B.t. =1 ::::: Br . EA= EZ . B.t. . The last relation, according to the converse of the above Auxiliary Proposition 7 (Part E), means that it is: AanBEnrZ=K.

Similarly we work for the proof of the rest pairs of quadrilaterals in relation (2). 3rd Way (Using the Method of Harmonic Transformation .

(Ql•Jli441 We suppose for example that the harmonic quadrilateral is ABrE. We will prove that its respective quadri lateral b.EZB is also harmonic if: Ab.nBEnrZ=K.

(10).

In fact, since quadrilateral ABrE is harmonic, Ar and BE will be conjugate with respect to circle (o). This means that during the Harmonic Transformation, Ar, BE will have come from two perpendicular chords AT', B'E' of a circle (o'), from which B'E' is the diameter of circle (o') (Converse of Corollary 159), thus B'E' is the perpendicular bisector to AT'. Additionally, during the above Harmonic Transformation, because relation (10) is true, it will be A'a·nr'Z'=PEB'E' and A'b.', r·z·, will be symmetrical chords , with respect to diameter B'E' of circle (o') (Converse of Corollary 161).

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

Therefore A'K', r'K' will also be symmetrical, with respect to B'K' and because B'E' is the perpendicular bisector to AT', triangle K'A' r· is isosceles thus LA'AT'=O.T'A', but it also is LA'ZT'=LA't:,,.T'. Hence, triangles A'ZT', r't:,,.'A' are similar and because they share s ide AT', they will be equal , thus A't:,,.'=r·z·. Therefore AT't:,,.'Z' is an isosceles trapezium and because B'E' lAT', it will also be B'E' .l.t:,,.'Z'. Thus, from the Harmonic Transformation and because B'E' .l.t:,,.'Z', B'E', l:,,.'Z', are transformed to the conjugate chords BE, l:,,.Z with respect to circle (o) (Direct of Corollary 159). Therefore, since BE, l:,,.Z are conjugate then quadrilateral t:,,.EZB, will be harmonic (definition).

tG>fi•l,fJ4i#i I.e., if the given convex hexagon ABrt:,,.EZ, has harmonic for example these two quadrilaterals ABrE and t:,,.EZB, then it will also be At:,,.nBEnrZ=K. In fact , then Ar and l:,,.Z will be conjugate, with respect to circle (o), to BE (definition). This means that during the Harmonic Transformation, Ar, l:,,.Z came the chords AT', t:,,.'Z' perpendicular to diameter B'E' of circle (o') (converse of the Corollary 159), Hence, AT't:,,.'Z' is an isosceles trapezium, as B'E' is the perpendicular bisector to AT', t:,,.'Z'. Therefore chords A'Z' and r•t:,,.• are symmetrical to diameter B'E', thus intersection A't:,,.'nr'Z'=K' EB'E'. Through all the above, during the Harmonic Transformation (Implication of Corollary 161), A't:,,.', r·z· are transformed to chords At:,., rz of circle (o), which intersect at chord BE, thus relation (10) will be true (but also AZnBEnrn=P).

We work similarly for the proof of the rest pairs of quadrilaterals in relation (2).

ld4u6hd (a). Proposition 207, is stated otherwise and proven in a diffirent way in Criterion 34 (§ 8.71). (b). We notice that quadrilaterals ABrE, t:,,.EZB, share diagonal BE.

lement of Harmonic Multilaterals.

(c). It is obvious that Criterion 207 is true and if instead of Harmonic Quadrilateral ABrE, quadrilateral aEZB is given as the harmonic one. This is proved in the same way as for ABrE above . (d). Relevant propositions are 34, 159, 161 .

29.9. Another Proof of Criterion 51, using the method of Harmonic Transformation . Criterion 208 (Figure 128). Every convex hexagon ABraEZ inscribed in circle (o) with concurrent diagonals (main) and whose diagonal for example BE runs through intersection P of its two opposite sides AZ and ra, has its two quadrilaterals ABrE and aEZB harmonic, and conversely.

[£11•Jli441 In fact, because from hypothesis we have

AanBEnrZ=K,

(1 ).

and

AZnBEnra=P,

(2) .

hexagon ABraEZ, during the Harmonic Transformation (Converse of Corollary 163 (Rule 6)], has come from a hexagon A'BT'a'E'Z', inscribed in circle (o') , which besides everything else, is symmetrical with respect to its diagonal B'E', with respect to which its quadrilaterals A'BT'E' and a'E'Z'B' are obviously symmetrical too. Therefore, symmetrical hexagon A 'BT'a'E'Z' with respect to its diagonal B'E', during the Harmonic Transformation (Direct of Corollary 163), this will be transformed to a hexagon ABraEZ, inscribed in circle (o), which besides everything else, has harmonic its quadrilaterals ABrE, aEZB.

tUlfi•lel'Z4t14i I.e., if the given convex hexagon ABraEZ, has harmonic its two quadrilaterals from example ABrE and aEZB, then it will be AanBEnrZ=K and AZnBEnra=P. To prove this we will depend on Corollaries 159 and 161.

B a.·.

- ---'-' H'""'A'--'R=M.;::O""'N""'IC:;...,:G=EO=M=E..:..TR:...:.Y .:....:.. P=art:...:..=D.

In fact, because quadrilaterals ABrE and1 aEZB are harmonic from the hypothesis, Ar and t:iZ. will be conjugate to BE (definition), with respect to circle (o). This means that during the Harmonic Transformation, Ar, t:iZ. came from perpendicular chords AT', a ·z· to diameter B'E' of circle (o') (converse of Corollary 159), Thus, AT'a'Z' is an isosceles trapezium, since B'E' is the perpendicular bisector to AT', a·z·. Therefore chords A'Z' and r·a· are symmetrical to diameter B 'E', hence intersections A'a'nr'Z'=K' and A'Z'nr·a·=P', will belong to B'E'. Due to the above, during the Harmonic Transformation (Impl ication of Corollary 161), A'Z', r•a• are transformed to chords AZ, ra of circle (o), which will intersect on the chord BE. This means that relation (2) is true. Moreover, for the same reason , relation (1 ).

We work similarly to prove the other two pairs of quadrilaterals Brt:iZ.- EZAr and raEA- ZABa.

li LAEE' (where l> a point on the extension of A/1 and towards A). Then we easily find that LAAE' = LE'AA'=q,' , q>+q,'= LEAE'=1 L.

B Ixr11.1a 3.

Because triangles ABE, AEr have equal angles LBAE, LEAr it will hold true that:

(ABE) AB•AE (ArE) = AE• Ar. Moreover, if L/1'Ar=w, because L/1'Ar + LBA/1 =w+ LBA/1=2L, LAAE

(5).

m F-.

- --'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

+ LEAa'= LaAE+q,=2L, regarding the two pairs of triangles Ara·, ABa and AaE, Aa'E, we similarly find that:

(ArA') Ar•AA' (ABA) = AB•AA '

(AAE) AA•AE (AA'E)= AE•AA''

(6).

respectively. Thus, from relations (5), (6)~(1 ). Thus in this case also relation (1) is true. Therefore in all three cases above (Figures 1-3), relation (1) is true.

2/. Proof of relation (2). 1st Case (Figure 1). If points a, a· lie on side Br. Because for triangles ABE' and ArE' it is true that: LE'Ar+LE'AB = LE'AE- w -q>+ LE'AE+w+q> = 2LE'AE=2L, it will be:

(ABE') AB• AE' (ArE') = AE'•Ar .

(7).

Additionally because, LrAa"= LBAa=w, LaAE'+ La'AE'=2L, we similarly find that:

(ArA') Ar• AA' (ABA) = AB• AA'

(AAE') (AA'E')

AA• AE' AE'•AA'.

(8).

Thus, from relations (7), (8)~2).

2n d Case (Figure 2) . If points a, a· are outside Br and lie on both its sides. Because in triangle AM' AB, Ar are isogonal and if we apply relationship (2) from above, then for triangle AM', it will be:

lmJ, - - -A'-'-=u'""xi=li=a"-'ry'--'-P'""'roc.. LAEE' (where li is a point on the extension of A/1 and towards A). Then we easily find that L/1AE' =LE'A/1'=q>', q>+q>'=LEAE'=1L. Because for triangles ABE', ArE': LBAE'+ LE'Ar= LBAE'+w+q>' =2L,

(ABE') AB• AE' (ArE') = AE'•Ar .

It will be:

Additionally, because L/1'Ar+ LBA/1 =w+ LBA/1=2L,

(10).

L/1AE'= LE'A/1'=q>',

regarding the two pairs of triangles Ar/1', AB/1 and A/1E', A/1'E', we similarly find that:

(Arfl') (ABfl)

Ar. Afl' AB• Afl'

(AflE') Afl. AE' (Afl'E') - AE'•Afl''

(11 ).

respectively. Thus, through relationships (10), (11)==(2). Hence relation (2) is true in this case too. Therefore in all three cases above (Figures 1-3), relation (2) is true.

B'. Let us work in triangle AM', with AB, Ar isogonal : If we use triangle AM' with AB and Ar isogonal and work just as we did above, we prove that relations (1) and (2) are still true. (It is possible to have the same result, if we depend on what is proven above).

- --'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

~

F-.

lt:Jii•l,VJ4ti4i A'. If we work in triangle ABr, with At:., At:.' isogonal : 1'1 Case (Figures 1-3). If relation (1 l is true.

In all three cases of the implication above, where relation (1) is true, we work in the following way: We consider that in triangle ABr - At:., At:.' are not isogonal, then we draw At:.1 isogonal to At:., hence:

(ABE) (Ar ll.1) (All.E) (ArE). {All.1E). (ABll.) =1 ·

(12).

Thus, as relations (1) and (12) are true, it easily occurs through them that:

(Ar ll.1) (Ar fl.') (All. 1E) = (All.'E). Hence, as all its four triangles, share the same altitude, rll.1

it will also be ll. 1E

rll.'

= fl.' E

~

t:.1=t:.' and because At:.1 is isogonal to At:., At:.'

will be isogonal too. 2nd Case (Figures 1-3). If relation (2) is true. In all three cases where relation (2) is true above, we work similarly to case 1 just above (of the converse) and easily prove that here as well At:.' is in fact isogonal to At:., in triangle ABr, using relation (2) and that its four triangles that will occur, are of the same altitude.

B'. If we work in triangle AM', with AB, Ar isogonal : 1st Case (Figures 1-3). If relation (1) is true. In all three cases of the direct proof above, where relation (1) is true, we

lmJ, - - -A'-'-=u'""xi=li=a"-'ry'--'-P'""'roc.. LAEE' (where ~ a point on the extension of A/1 and towards A). Then we easily find that also Ll1AE' =LE'Al1'=q>', q>+q>'=LEAE'=1L. According to Auxiliary Proposition (AP) 1 (Case 3), relation (1) will be true here as well, this means that:

(ABE) (Ar.6.') (A.6.E) (ArE) . (A.6.' E). (AB.6.) = 1 ·

(6).

~

F-.-

--'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

Hence, because the six triangles in (6) have the same altitude it will be:

BE r~· E~ B~ E~' rE Er . ~· E. B~ = 1 or ~E· ~T-EB =1·

(7).

This means that the second equality of (1) holds true here as well.

But, because AE is the bisector of BAr, it will be

AB

BE

Ar = Er , therefore if we

take into consideration (7), we receive the triple equality of relation (1). Therefore, in all three cases above (Figures 1-3), the whole relation (1) is true.

Proof of relation (1 '). We observe that easily, from the second part of relation (1 ), we can derive the second part of (1'). This means that relation (1') (second part) is equivalent to the second part of (1). This means that in all three cases above where the second part of (1) is true, the second part of (1') will also be true. After all, because for all three cases above (figures 1-3), AE is the bisector of angle Mt:.' and AE' is the external bisector of the same angle, it will be

~E

M

E~' =A~', we easily derive the first part of (1'), from the second part of (1 '). Therefore, in all three cases above (Figures 1-3), the whole relation (1') is true.

Proof of relation (1"). To prove relation (1") for all three cases in figures 1 to 3, we work in the

~

,---A'-'-=u'""xi=li=a"-'ry'--'-P'""'roc.. E 1r and (9).

We realize that we have reached a contradiction, because then (8) is not true, but (9) is. We have reached a contradiction, because we set as bisector line AE1 and not AE. We will also reach a contradiction if we suppose that for point E1 it holds that BE1>BE, thus ~E1>~E, etc. Therefore, AE will in fact be the bisector of angle BAr. 3 rd Way (Computational) (Figure 1-3). a). Direct Proof. Because A~, A~', in triangle ABr, are isogonal , according to known Theorem [Geometry Book [6] § 11 -12.1), it will be :

AB 2

BE 2

BA BA'

-- = -- = --Ar2 Er 2 Ar• AT

0

(1).

After all because, in triangle AM', AB, Ar are isogonal too, for the same reason, it will be:

AA2 AE 2 AB Ar A' A 2 = A'E 2 =BA'. rt.·.

(2).

lmJ1----'A"'u"-'x= i l=ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'('-=L=e=m=m-'-'-=a=s)'"."

Hence, from relations (1), (2), we receive:

BE 2 t.E 2 t.B Bt.' t.B t.r Bt.2 Bt. BE t.E Er 2 't.'E 2 = t.r·rt.•' Bt.''rt.·= t.T 2 or -t.,-r =-E-r .-Et.-' or BE

Bt. t.'E

-=----

or

BE Bt. Et.' AB Er = t.E. t.T =Ar. This means that relation (1) is

proven true.

Proof of the Direct of Relation (1') of the 2 nd and 3 rd Way of Proof (Figure 1-

~ To prove relation (1 '), we work similarly to the proof of relation (1) above, or in a more simple way, as shifting appropriately the segments in relation (1'), we receive relation (1 ), we conclude that relations (1) and (1 ') are equivalent and because (1) is true then (1') will also be true. Auxiliary Proposition 2 is proven in analogous ways, and is true when for example Ar, AB are symmetrical with respect to AE', which is perpendicular to AE at point A.

Proof of the Direct of Relation (1") of the 2 nd and 3rd Way of Proof of (1") (Figure 1-3). It is obvious that relations (1) and (1 ') above are true and for the Conjugate Harmonic E' of E, with respect to B, rand /1, /1', as it is:

BE BE' t.' E t.' E' Er = ET a nd Et. = E' t. .

(a).

Thus, from relation (1) and (a) we can easily find that:

AB Ar

BE Er

BE' ET

Bt. Et.' t.E . t.T .

(b).

~

F-.-

--'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

From relation (b), we can easily find that:

Bil Ell' rE' llE. llT. E'B =1,

llE llT E'B Ell'. rE' . Bil = 1·

(c).

This means that the first and the second equality in relation (1") are true. After all, from the second equality in (c) and (a), we can easily derive the following:

Bil E' fl' rE' llE'. llT . E'B = 1·

Hence the third equality in (1") is also true.

b . Converses of the 3rd Wa of Proof. The proof of the converse here as well, can be performed, similarly to the converses of the 2nd Way above, or the other ways above.

8·lui11I4,il According to relevant definitions, in paragraph 3.3 (of Part A), pencil A(Bl1E/1TE') is a «flat symmetrical pencil of six beams, with respect to an axis», as pencils A(Bl1Ef) and A(f/1'EB) are symmetrical with respect to their beam AE and /1E', while the following four collinear points B, 11, E, /1', constitute a «Simple Harmonic Involution of Two Pairs of Conjugate Points with Double Points E, E' (paragraph 3.4, of the main part of this book). In paragraph 3.9 the properties of the above pencil and the involution are mentioned, in the form of Propositions (Criteria, with their proofs included).

lrnJ1----'-A"'u"-'xil= = ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'('-=L=e=m=m-'-'-=a=s)'"."

BA Ar

-

From the pt Part of relation (1) we have:

BA EA' AE"AT.

=---

This means that the above Proposition 2 is true in triangle ABr. Additionally, from the 1st Part of relation (1) we have:

BE

-

BA EA'

=- - -

This means that Proposition 2 is true in a triangle degenerate into a line. Thus it is also true in a special collinear series of points which we have defined as «Simple Collinear Harmonic Involution» (See Criterion 4, in Part A) .

2.3. lsogonal Property of a Triangle. (Auxiliary Proposition). This Proposition exists in Bibliography [6] (§11 -12.1), but only its implication. Here, besides its direct proof, we refer to its converse as well. Moreover, we mention some of its special cases, which are essential to this book. Proposition 3. (Figures 1-3). If two isogonal lines with respect to angle A in triangle ABr, intersect its side Br at points 11, /1' and if AE is the internal , AE' the external bisector of angle A in this triangle or in triangle AM', then we will have:

BA 2 BE 2 BE' 2 BA BA' -- = -- = -- = -Ar2 Er 2 ET 2 Ar• AT•

(1 ).

AA 2 AE 2 AE' 2 AB Ar 2 2 AA' = EA' = E' A' 2 =BA'. rA'.

(2).

F-.-

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And conversely : If AE is the internal bisector of angle BAr, in triangle ABr and AE' the external bisector of the same angle and if relation (1) is true, then AE is the bisector of angle /1A/1' too, or if AE is the internal bisector of angle /1A/1', in triangle AM' and AE' the external bisector of the same angle and if relation (2) is true, then AE is the bisector of angle BAr too.

(1 ). Direct Proof. (a). Based on Triangle ABr.

Proof of relation (1 ). 1st Case (Figure 1). If points /1 1 /1' lie on side er. Because in triangles AB/1 and Ar/1' angles BA/1, rA/1' are equal , it will be:

(AB/1) AB• A/1 (Ar/1') = Ar•A/1''

(3).

Moreover because, LBA/1'=LrA/1, it will also hold that :

(AB/1') AB• A/1' (Ar/1) = Ar•A/1.

(4).

Thus, through relations

(3), (4)~

(AB/1) (AB/1') AB• A/1 AB• A/1' = (Ar /1') . (Ar /1) Ar• A/1' . Ar• A/1

BA 2 Ar 2

•

(5).

After all , because the four triangles in (5) are of the same altitude, it will be:

(AB/1) (AB/1') 8/1 8/1' 8/1 8/1' (Ar11·)" (Ar11) =11T · 11r =11r · 11T ·

(6).

~

,----'A'-'-u=x=i= l ia=ry....___,_P'-'-r-=-oc..po=s=i=tio=n'-"s=--i.: (L=e=m=m=a=s=).

Therefore, from (5) , (6) and because AE is the internal and AE' the external BA BE BE' bisector of angle BAr, it is: Ar = Er =ET , wanted relation (1) occurs easily.

2 nd Case (Figure 2). If points /1 1 /1' are outside Br and lie on both its sides .. Again, because in triangles AB/1 and Ara· angles BA/1, rA/1' are equal , it

(ABA) AB•AA (ArA') = Ar•AA' .

will be:

(7).

Additionally because, LBA/1'= LrA/1, it wi II also be:

(ABA') AB• AA' (ArA) = Ar•AA.

(8).

Hence, from relations

(7), (8)~

(ABA) (ABA') AB• AA AB• AA' BA 2 (Ar A'). (Ar A) = Ar• AA'. Ar• AA = Ar 2

•

(9).

After all, because the four triangles in (9) are of the same altitude, it will be:

(ABA) (ABA')

BA BA'

BA BA'

(Ar A') . (Ar A) =AT . Ar =Ar. AT .

(10).

Therefore, from (9) , (10) and because AE is the internal and AE' the external BA BE BE' bisector of angle BAr, it is : Ar= Er =ET , wanted relation (1) easily occurs. 3rd Case (Figure 3). If points /1 1 /1' are outside Br and towards the same side of M midpoint of segment EE'. This case will obviously occur, if q,= Li5AE= LEA/1'> LAEE' (where i5 is a point on the extension of A/1 towards A). Then we can easily find that also L/1AE' =LE'A/1'=q,', q,+q,'=LEAE'=1L.

F-.-

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Additionally

because

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in

triangles

AB/1

and

Ara·

we

have:

LBA/1+ LrA/1'= L BA/1+w= L BA/1+ L BA(> =2L,

it will be:

{ABA)

AB•AA

{Ar A')

Ar. AA' .

(1 1).

Moreover because, LrA/1+ LBA/1'= (w+2 0

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It is obvious that the last relations can occur directly from (3) or even (2)]. (2). Converse (Figure Sy). If for example relation (2) is true, we will prove that : AllnBEnrZ=K. In fact, if All, BE, rz are not concurrent and we set: AllnBE=a, BEnrZ=13, rznAll=y (it is possi ble that a, 13, y , are outside of the c i rcle), then from the pairs of similar triangles ABa-Ella, Br 13 - ZEl3, rE13 - 8213, EAa - llBa, we get :

AB EA Ba= Aa '

J3B 132 Br= ZE '

rE BZ EJ3 = ZJ3 '

aE aA EA= AB.

(4).

From relations (4), if we multiply their analogous sides, it appears that:

AB.rE.BJ3.Ea Br.EA.Ba.EJ3

AE.ZB EZ.BA .

(5).

Relation (5), because of (2), which is given to us , becomes : : ; = : : ::= a=l3. This means that rz, that passes through 13, will also pass through a the intersection of All and BE, hence: AllnBEnrz = K =a=l3=y.

144uE09 (a). This Proposition was established in our book [31) [§ 2a(45) which can also occur from Proposition 2a(44) in book [31), or Proposition 50lr (page 1582) in book [1]. Both these firstly appearing Propositions are important, as their converses are also true and mostly because they expand the known Pappus 's Theorem , which has to do with intersections of a flat central pencil from lines, while here we have intersections of a flat central pencil of 3 and 4 beams respectively, intersected and re-int ersected by a circle.

lmJ1----'A"'u"-'x= i l=ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'('-=L=e=m=m-'-'-=a=s)'"."

Both these Propositions, have also been published in the magazine «Mathematics in High School (issue 13 page 36), with some of their applications. (b). The above Proposition 7 was proven valuable for the proof of Propositions and particularly for Propositions concerning Harmonic Hexagons.

2.8. Important Auxiliary Proposition. Prolongation of Newton 's Theorem in Hexagons {Special Case of Proposition Hi(1) in Geometry [31]}. This Auxiliary Proposition (Lemma) , is mentioned here, as we will depend on it to prove other Propositions. Proposition 8 (Figures 9). If the main diagonals of a hexagon inscribed in circle are concurrent at a point, then this point coincides with point Brianchon {[6] (§ 11-11 )}, of the respective circumscribed (tangential).

If i.e. we are given two respective hexagons ABrAEZ inscribed in circle (0, R) and A'BT'A'E'Z' circumscribed (tangential) in the same circle, as to AEA'Z', BEA'B', reBT', ... and it is: AAnBEnrZ=K, we will prove that also A'A"nB'E"nr'Z'=K. Hence K coincides with point Brianchon, of hexagon A'BT'A'E'Z'. In fact, according to known Newton 's Theorem {Books [7] § 1274 and [31] Proposition 113(2)}, where a new proof is given}, for the respective

~

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quadrilaterals ABaE

in-

scribed

and

A'H'a'0'

cir-

cumscri bed (where H'=A'B"nr·a· , 0'=A'Z'na"E'), their

d iago-

nals will pass through the intersection Aa

and

of BE,

which is point K. Hence, to state it otherwise A"a" goes through K (as does H'0'). In the exact same way we prove that B'E', r•z• go through K. Consequently we see that diagonals A'a' , B'E', r·z· go through K , which is known to be point Brianchon {[6]( § 11 -11)}, og hexagon A'BT'a'E'Z'.

134uEOb (a). We can easily realize that the converse of Lemma 8 is not true, as it is known, that circumscribed in circle hexagons, always have concurrent diagonals at point Brianchon. (b). Generalization of Proposition 8, for n-laterals, is given in § 1l>(1) of book [31]. Both these Pr opositions are obviously generalizations of Newton 's Theorem , which is true only in quadrilaterals. (c). Proposition 8 is also true in concave hexagons, (See Geometry [31], Structures 1l>(2) and 1l>(3).

lmJ1----'-A"'"'u"-'x=-i'-' li=-ary'-'--'-P-'-ro=.,p"-o=-s=-i'-'-ti'"'o'"'"n'"'s'--'('-=L'"'e'"'"m'"'"m'-'-'-'a"'s"-'-.)

(d). Lemma 8 above was proven valuable for the proof of Propositions concerning Harmonic Hexagons, whose diagonals are concurrent. (e). Proposition 146 in Part B is also relevant, which is about Harmonic Hexagons and Proposition 206 about Harmonic Decagons.

~

F-.-

--'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

EnlAOrOI Here is the end of the "SKY" ! ! ! We started our research in the " sky" about four years ago with a high cubic capacity motorcycle [as it is shown in the picture at the beginning of the book (page 13)] and after crossing difficult paths and "Symplegades Rocks"( " Clashing Rocks " ), which I assure you were Exciting, enchanting, delightful and effective, I arrived happy and healthy at the end of the writing of my new book "ARMONIC GEOMETRY". Everything has its end!!! But that does not mean that the work is over. There is still a lot of work to be done on complements, error corrections (where applicable) and improvements. I thank God who gave me the strength, the ability and the days, to finish my new book "Harmonic Geometry". I also thank all the friends who followed the entire or part of the journey I had and I believe they also enjoyed it. I also would like to thank my own people, who tolerated and helped me morally. Finally, I am extremely indebted to the site mathematica.gr, which hosted me and gave me the opportunity to present all the work to my friends. We have been working methodically, with passion, patience and perseverance, and we believe that our realistic but not ambitious goal has been achieved. However, this is enough for us for the moment as we believe that the work is of HIGH QUALITY.

~

1----'-A"'"'u"-'x=-i'-' li=-ary'-'--'-P-'-ro=.,p"-o=-s=-i'-'-ti'"'o'"'"n'"'s'--'('-=L'"'e'"'"m'"'"m'-'-'-'a"'s"-'-.)

Thus, we believe that a new Euclidean geometry branch has emerged, or if you want a new Geometry, the Harmonic Geometry. Anyone who will carefully study this book, I am convinced, that he will really enjoy the Magic, the Harmony and all the Greatness of Geometry, and will find in it Geometric Treasures that were hidden till today. There are many and great challenges in this book, and we have given all the relevant tools to those who want to expand it. Finally, we believe that we built (stone by stone) a building of a great Architectural value and of great Strength. It is proved that way once again that Geometry is alive and grows, against the stubbornness of its enemies. Whoever wants to really get involved seriously with the development of hard Geometry needs to do further research work but also to study the new elements of Geometry that come to light.

RESEARCH = Enjoyment+ Discoveries+ Intense Emotions "' Creation

== Progress == Civilization

If, with my previous frequent encouragement, I managed to persuade even a friend of Geometry to follow this difficult, but attractive and constructive way of research, it would be a great satisfaction for me. We believe that Harmonic Geometry and more specifically the New Method of the Harmonic transformation, can be introduced to other fields of Mathematics as well , such as Complex Analysis, etc. This is stated as an idea for research , for those who are interested.

8 F-·

- --'-' H'--'-AR'""""M"'-O=N=l-=-C-=G=E=O=M=ET-'-'-R-'-'Y__,_P--=a"-'rt-='-E.

FIGURE INDEX

Equivalence of the Figure Number and the Page it is on. A. Harmonic Geometry.

1+--+22, 2+--+23

KOi

35, 20+--+29, 3+--+26, 4+--+31 , 5+--+42, 6+--+58,

7+--+61 , 8+--+65, 9+--+69, 90+--+ 72, 10+--+76, 11 +--+86, 12+--+87, 13+--+89, 14+--+95, 15+--+101 , 16+--+102, 17+--+ 114, 18+--+118, 19+--+123, 20+--+124, 21 +--+129, 22+--+133, 23+--+138, 24+--+141 , 25+--+152, 26+--+155, 260+--+163, 27+--+171 , 28+--+174, 29+--+177, 30+--+179, 31 +--+181 , 32+--+185, 33+--+187, 34+--+192, 35+--+194, 36+--+197, 37+--+208, 38+--+212, 39+--+215, 40+--+220, 41 +--+233, 42+--+237, 43+--+244, 430+--+247, 44+--+264, 45+--+266, 46+--+267, 47+--+269, 48+--+274, 49+--+275, 50+--+280, 51 +--+283, 52+--+290, 53+--+292, 54+--+296, 55+--+313, 56+--+317, 57+--+328, 58+--+330, 59+--+333, 60+--+336, 61 +--+340, 62+--+367, 63+--+370, 64+--+382, 65+--+383, 66+--+385, 67+--+386, 68+--+389, 69+--+390, 70+--+391 , 71 +--+393, 72+--+420, 73+--+425, 74+--+437, 75+--+440, 76+--+444, 77+--+448, 78+--+452, 79+--+457, 80+--+462, 81+--+470, 82+--+474, 83+--+486, 84+--+493, 85+--+498, 86+--+500, 87+--+503, 88+--+505, 89-509,90+--+513, 91 +--+520,92+--+523,93+--+526,94+--+528,

Auxiliary Propositions (Lemmas).

95+-+531 , 96+-+535, 97+-+546, 98+-+549, 99+-+555, 100+-+566, 101 +-+573,

102+-+575,

103+-+578,

104+-+582,

105+-+586,

106+-+591 ,

107+-+596,

108+-+604,

109+-+605,

110+-+607,

111 +-+609,

112+-+614,

113+-+616,

114+-+622,

115+-+625,

116+-+627,

117+-+660,

118+-+668,

120+-+677,

121 +-+680,

122+-+685,

123+-+692,

124+-+693,

125+-+695,

126+-+698,

127+-+704, 128+-+707, 3.2.6.(4).(J3).1/+-+714. B. Auxiliary Propositions.

1+-+722, 2+-+722, 3+-+723, 4+-+743, 5+-+746, 6+-+747, 7+-+750, 8+-+753, 9+-+756.

m F-.

- --'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

CONTENTS Paragraphs

Pages

4 TERMINOLOG

8

BIBLIOGRAPHY

11

INTRODUCTION

14

l•fflrnflit•l,?1

14

):tJ,jif-1 HARMONIC INVOLUTIONS (Collinear and Cocyclic). 1. In General. ' HARMONIC INVOLUTIONS OF COLLINEAR SERIES OF POINTS

17, 18 18 19

2. In General.

19

2.1. Definitions and Features.

20

2.2. Types.

21

B' SYMMETRICAL PENCILS & COLLINEAR HARMONIC INVOLUTIONS 22 3. Analysis of the Symmetrical Central Pencils and the Collinear Harmonic Involutions. 22 3.1 . Simple Symmetrical Central Pencils of One Pair of Beams.

22

3.2. Simple Harmonic Involution of One Pair of Points.

23

3.3. Simple Symmetrical Pencils of Two Pairs of Beams.

23

3.4. Simple Harmonic Involution of Two Pairs of Points.

28

3.5. Simple Symmetrical Pencils of Three Pairs of Beams.

31

~

1----'A"'u"-'x=il=ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'('-=L=e=m=m-'-'-=a=s)'"."

3.6. Simple Harmonic Involution of Three Pairs of Points.

32

3.7. Simple Symmetrical Pencils of n Pairs of Beams.

32

3.8. Double Symmetrical Pencils of two, four, six, eight, ... ,2n Pairs of Beams. 33 3.9. Properties of Symmetrical Pencils and Harmonic Involutions.

34

C'. METHOD OF SYMMETRICAL PENCILS

51

4. Method of egress from line to its plane.

51

5. Method of symmetrical central pencils.

52

5.1. Historical Background.

52

5.2. Name.

52

5.3. Description of the Method and Comments.

52

5.4. Other Types of Central Pencils.

54

5.5. Value of Research.

54

6. Applications of the properties and the methods.

55

D' ORTHOOPTICAL PENCILS AND ORTHOOPTICAL POINTS.

151

7. In General.

151

7.1 . Orthooptical Symmetrical Pencils and Orthooptical Points of Harmonic Involutions. 151 7.2. Applications.

160

E' HARMONIC INVOLUTIONS OF COCYCLIC SERIES OF POINTS

169

8. In General.

169

8.1. Definitions and Features.

170

8.2. Types.

172

8.3. Harmonic Cocyclic Involution of two Pairs of Points of Midpoint.

172

8.4. Historical Background.

173

8.5. Extension to Newton's Theorem to a Circle.

174

8.6. Analysis of Cocyclic Harmonic Involutions.

177

8.7. Properties of the Harmonic Cocyclic Involutions.

192

m F-.

- --'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

9. Applications.

219

114:Jil:) HARMONIC MULTILA TERALS AND HARMONIC LINES (Collinear and Cocyclic). 256 10. In General. 'HARMONIC COLLINEAR MUL TILATERALS 11 . In General.

256 257 257

11.1. Definitions of Closed Harmonic Collinear groups of n Series of Points.

258

12. Analysis of Closed Regular Central Pencils and Collinear Harmonic Groups of n Series of Points. 264 12.1. Closed Regular Central Pencils of Four Beams.

264

12.2. Closed Collinear Harmonic Group of Four Points and Harmonic Pencil of Four Beams. 265 12.3. Closed Regular Central Pencils of Six Beams.

265

12.4. Closed Collinear Harmonic Group of Six Points and Harmonic Pencil of Six Beams. 266 12.5. Closed Regular Central Pencils of Eight Beams.

267

12.6. Closed Collinear Harmonic Group of Eight Points and Harmonic Pencil of Eight Beams. 267 12.7. Closed Regular Central Pencils of Ten Beams.

268

12.8. Closed Collinear Harmonic Group of Ten Points and Harmonic Pencils of Ten Beams. 269 12.9. Closed Regular Central Pencils of Five Beams.

270

lmJ1----'A"'u"-'x=il=ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'(=L=e=m=m-'-'-=a=s)'"."

12.10. Closed Collinear Harmonic Group of Five Points and Pencil of Five Beams. 271 12.11. Closed Regular Central Pencils of n Beams.

271

12.13. Conclusions.

271

C' . METHOD OF THE REGULAR CENTRAL PENCILS.

277

13. In General.

277

13.1. Method of Closed Regular Central Pencils.

277

14. In General.

278

14.1. Orthooptical Regular Pencils and Orthooptical Points of Closed Collinear Harmonic Groups of n. 278 15. Applications of the method of regular pencils.

280

E' HARMONIC COCYCLIC MULTILATERALS

380

16. In General.

380

16.1. Definitions of Simple Harmonic Multilaterals.

381

17. Analysis of Closed Harmonic Central Pencils and their Intersections from a Line and a Circle.

389

17.1 . Closed Harmonic Central Pencils of Four Beams.

389

17.2. Closed Harmonic Central Pencils of Six Beams.

390

17.3. Closed Harmonic Central Pencils of Eight Beams.

391

17.4. Closed Harmonic Central Pencils of Ten Beams.

392

17.5. Closed Harmonic Central Pencils of Five Beams.

394

17.6. Closed Harmonic Central Pencils of n-Beams and Harmonic n-laterals.

395

17.7. General Conslusions.

395

~

F-.-

--'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

G'. METHOD OF HARMONIC PENCILS.

398

18. In General.

398

18.1. Method of Closed Harmonic Central Pencils.

398

19. Applications of the method of harmonic pencils and not only that. 399

H'. THE PROBLEM OF THE VICENNIUM !!! ND THE METHOD OF THE HARMONIC TRANSFORMATION

541

20. The problem of the vicennium.

541

21 . The method of the Harmonic Transformation .

542

22. Applications of the method of the Harmonic Transformation.

545

I' PAIR OF HARMONIC COLLINEAR AND COCYCLIC LINES.

603

23. In General.

603

23.1. Pairs of Open Harmonic Collinear Series of Points of 2n Points. 604 23.2.

Pair of Harmonic Open Convex Cocyclic Series of Points of 2n Points (Apexes). 605

23.3.

Pair of Regular Open Cocyclic Series of Points of 2n Points. 606

24. Pairs of Open Central Flat Harmonic and Regular Pencils of 2n Beams and their Orthooptical Points.

609

24.1. In General.

609

24.2. Harmonic Pair of Open Flat Central Pencil of 2n Beams.

609

24.3.

Pair of a Regular Open Central Flat Pencil of 2n Beams.

610

24.4.

Pair of Orthooptical Points, of a Pair of Open Collinear Harmonic Lines .

612

25. Applications relevant to the Pairs of Collinear and Cocyclic Harmonic Lines and Pencils.

614

lmJ1--- -'A"'u"-'x=il=ia=ry-'-'-P....,_r=o=p=o=s= it= i o=n=s'-'('-=L=e=m=m-'-'-=a=s)'"."

SUPPLEMENT OF PART A

648

26. In General.

648

27. Introduction of the Method of Harmonic Transformation for the Harmonic Involutions. 650 27.1 . Simple Harmonic Involutions (Collinear and Cocyclic).

650

27.2. Applications of Simple Harmonic Involutions.

650

27.3. Double Harmonic Involutions (Collinear and Cocyclic).

659

27.4. Applications of Double Harmonic Involutions.

660

27.5 Various Constructions of Collinear and at the same time Cocyclic Involutions, using the Method of the Harmonic Transformation. 666

SUPPLEMENT OF PART B

676

28. In General.

676

29. Applications of Harmonic Multi laterals, mainly using the Method of Harmonic Transformation. 677 30. General conclusions of Harmonic Geometry.

UXILIARY PROPOSITIONS

716

720

1. In General.

720

2. Geometrical Auxiliary Propositions (AP) .

721

l:1~11•©113

758

FIGURES INDEX

760 762 768

~

F-.-

--'-' HA'-'-'--'R=M'-=O"-N'"'""IC=--=G=E=O=M=E"-'T'"'"R-'-'Y_,P'"""'a=rt-=-='-E.

ITWAS SAID (See at Google entry Nikos Kyriazis a distinguished member).

I would like to welcome to Mathematica the exceptional expert in geometry Nikos Kyriazis , whose proofs-solutions and suggestions we have read in several magazines/ books both Greek and foreign! Alexandros Kongelakis (Professor of Mathematics, Mathematica Site Administra tor).

I would also like to officially welcome the eminent and multi award-winning expert in geometry Nikos Kyriazis. He is the author of the well known multi-volume Geometry, which should be acquired by all of us, in addition to a great number of articles in magazines. His presence here is an honor for all of us. I wish him good health and to provide us with new and original solutions to the issues we are discussing. Babis Stergiou (Professor of Mathematics - Author).

At this very moment, I really felt completely ignorant, as I did not know (even though I say that I am involved in Geometry) about this great work of Nikos Kyriazis! Thanks to Mathematica, I was informed and I thank above all Nikos for this. I hope to find and study thorough his work. Nikos, even delayed, warm congratulations (please forgive me of my ignorance). Andreas Varverakis (Professor of Mathematics).

I was completely ignorant about it before it was mentioned here, so when I visited the Mathematical Library (the bookshop of Haris Vafeiadis) a few days later, I asked for and saw some of the volumes of Nikos ... and I was really stunned. For me it is already too late to wander in such a lengthy work - even though I consider myself an expert in geometry (or rather •a man of symmetry •) - apart from the topics (and discussions) provoked by such endeavors regarding the possibility of further originality in sectors which are considered studied thoroughly are indeed provocative! (In addition of course , an additional big issue is when and why a mathematical sector could be considered as an "exhausted one"--deep waters ... ) Giorgos Baloglou (Professor of Mathematics at the University of New York.

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I am stunned facing the beauty of the problem and the sensational simplicity of Nikos' solution! If this construction is universally unknown, I would suggest that the term "Kyriazis 's hexagon" should be internationally adopted. Giorgos Baloglou (Professor of Mathematics at the University of New York).

Many Happy returns to Nikos Kyriazis, a loyal lover of geometry, with a rich and original work. Leonidas Tharralidis (Professor of Mathematics).

Nikos, a wonderful and absolutely basic solution! It may very well be used in the classroom as a spontaneous adventure with the students as a part of the quadrilaterals course. With a circle, everything is over ............. and at the same time everything is restarted! Thank you for the inspired solution. Babis Stergiou (Professor of Mathematics - Author).

Nikos congratulations on the honorary distinction and thank you once more for honoring us with your membership at our Club. Mihalis Lambrou, General Coordinator (Professor of Mathematics, University of Crete).

Nikos Kyriazis (whom I personally met during the summer and admired his liveliness, zeal, passion and love, I would say, he shows for Geometry). He does not deal with the foundation of Geometry. He has fully grasped and entirely respects the work of Euclid. What he does, based on this work, is to discover the qualities of the various shapes and then he is trying to prove them in as many ways as possible. Nowhere did he claim that these should be taught at high school. He has said several times that these are for the lovers of geometry or for spiritual satisfaction only. Is there anything better and purer? I am sure that, if there had been an opportunity for Euclid to study his work, he would have congratulated him. Antonis Kyriakopoulos (Professor of Mathematics - Author).

Nikos, you are getting on our nerves with your criteria. We will all be suffering from complexes. Andreas Poulios (Doctor of Pedagogy, Professor of Mathematics - Writer).

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Nikos, although I think I do not know you personally, I admire indeed the tremendous will you possess to accomplish your work!! Good follow-up. I shall always follow the progress of your work. loannis Dimitris (Mathematician, lstiaia Evia)

My friend Nikos, I was deeply touched with your "GREAT THANK YOU". Your progress so far guarantees that you will offer us more, because our astronomical age is of no importance. The importance is in the age of our soul and you have proved to us that your soul is one of a teenager. Yannis Kerassarides (Professor of Mathematics).

Dear Nikos Kyriazis, looking once more at the shapes in your article, I remember the despair (and the dizziness ... ) I felt when I received them by fax from Nikos losifidis, when we had to type and layout the text! Keep well and continue your EXCELLENT WORK. Georgios Rizos, Coordinator (Professor of Mathematics).

Moreover, everybody bows in front of a zis .

giant of geometry such as Nikos Kyria-

Stathis Koutras (Professor of Mathematics).

My tireless friend Nikos Your suggestions are inconceivable and your extensions are magnificent. I would like to publicly confess that at the age of 52, the emotion I get from you concerning the greatness of geometry is unprecedented. Your new proposal is incredible; fantastic as well are the shapes that you have created the beauty of "symmetry" in all its grandeur. I could not fail to react at your new challenge (and I confess that I have lost my sleep because of it). Stathis Koutras (Professor of Mathematics).

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Dear friend Nikos, Your love and your efforts for "Hardcore Geometry" are worthy of admiration and respect, especially for your deeds at that beautiful but rather neglected branch of Mathematics. Giorgos Apokis (Professor of Mathematics).

Niko, Best wishes and keep well for the future, continue to impress us. You are an example to us (I will not stop saying it) we draw strength from your appetite and energy, I am 33 years old and I already feel tired and disappointed by many things in our field ... It is a good refuge for us and that's the reason we are here. Concerning your books, I am ashamed to admit that I haven't bought them yet, but at the first opportunity I will acquire them and I will bring them to our first meeting, for your signing them all - we would need at least three hours!! Keep well. Makis Hatzopoulos (Professor of Mathematics).

Mr. Nikos, keep well and have in mind that we, the younger ones, admire you and have you as an example to follow. Makis Hatzopoulos (Professor of Mathematics).

Mr. Nikos, I very often follow your discoveries and admire you. Andreas Poulos (Doctor of Pedagogy, Professor of Mathematics - Author) .

Nikos, I do not know if you have realized, you have become the first of your league and the honored one. Andreas Poulos (Doctor of Pedagogy, Professor of Mathematics - Author) .

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Nikos, the right movement is not for us to accept your gratitude but for you to accept our admiration. Continue to produce work on the Eternal Building of Euclid and let us admire you. Nikos Mavroyiannis, General Coordinator (Professor of Mathematics).

I hope that you will continue your excellent work with the same passion, so that all the following generations (and mine amongst them) shall inherit something great in the field of Geometry. Vassilis Papaspilotopoulos (Electrical Engineering Student at N.M.P.).

Mr Nikos, I would simply like to personally thank you for the subjects, you tirelessly post at the site. I hope to find the time to personally work with the problems you set, which are numerous and excellent so far. Achilleas Sinefakopoulos

Mr. Nikos good m orning We have to voice a huge Thank You for a great number of reasons : -for your being one of us - for giving us generously all that from your personal work-for teach ing us so many things - for being able to teach us even more if we had more time to work upon it -I wish you the best -1 am glad to see you through your publications - all the best for the future -With love and unlimited respect Fotini Kaldi (Professor of Mathematics).

Nikolas, although I do not know you personally, I wish you to be long lived and keep working on mathematics (although I have to admit, I do not always fully grasp the issues you deal with). Omega Man

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Many Happy returns from me to all those people having their birthday and above all to the tireless man of geometry Nikos Kyriazis whom I hope to meet personally in the future. Marantidis Fotis (Professor of Mathematics).

My Friend Nikos, good evening . It is a beautiful proposal indeed the one you have suggested. I would like to meet you in person. I feel the need to tell you that you are GREAT and please simply accept it. Stathis Koutras (Professor of Mathematics).

I would like to present a nice problem by a friend of mine Nikos Kyriazis from Thessaloniki - Hellas. Kostas Vittas (Architect at NMP).

File comment: Kyriazis's problem of concurrency. Kostas Vittas (Architect at NMP).

Dear Nikos, Be always well , strong, in good health, with inspirations and discoveries! Rekoumis Konstantanos (Professor of Mathematics).

Dear Nikos, I follow with great interest your research studies - so to speak - in geometry and I am not hesitating to voice my thoughts, I admire the methodology of your work and the depth of your research. Keep well! With my best wishes for good health, every success, progress, prosperity and inspiration! Rekoumis Konstantanos (Professor of Mathematics).

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Dear Nikos. I want to meet you in person so as to think together of our school years. In one of your messages you mentioned that when we were pupils, you admired me. Now I have to tell you that I admire you for your excellent work in Geometry (Honestly, why you did not become a mathematician?) Antonis Kyriakopoulos (Professor of Mathematics - Author)

My Friend Nikos. I'm following what you write about geometry in " mathematica" and I'm proud of being your friend . Antonis Kyriakopoulos (Professor of Mathematics - Author).

Dear friends and colleagues, the work dealing with this subject is excellent. Of course , congratulations to Mr. Nikos Kyriazis - the tire less and always creative man - who started this project. I too believe that all this informative material should be exploited in multiple ways Andreas Poulos (Doctor of Pedagogy, Professor of Mathematics - Writer).

Thank you for sharing with us your thoughts and concerns on Geometry! Thanks to you we've learned a lot of new things! Keep well and continue what you are doing! I w ish you all the best! Nikos Katsipis (Professor of Mathematics).

My particular GEOMETRY w ishes to the kind

GIANT of Geometry NIKO KIRIAZI.

Nikos, be always strong and be next to us with your new and amazing discoveries. Stathis Koutras (Professor of Mathematics).

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Dear Nikos ............ You made me feel proud, when you said that I, even without knowing it, became the reason for you to work w ith mathematics and become the most important researcher in Euclidean Geometry and to gift us with your great and significant work (Over 30 volumes!!! w ith honorable distinctions from the Academy of Athens and awards obtained by many other important associations).

The torrent of the original proposals you articulate and prove in your books and in your articles about the Euclidean Geometry, place you among the great men of geometry of all time. Nikos, keep well and continue to surprise us with the results of your research in this exceptional branch of mathematics . Antonis Kyriakopoulos (Mathematician-Author).

I would like to thank Antonis Kyriakopoulos and Nikos Kyriazis for their contribution to Mathematics! The least we can do for them , we, the total number of the coordinators at mathematica.gr, have decided to add them in the team of t he Distinguished Members as the New Regulation foresees! Both of them, each one from a different stand, are trying to improve and promote Mathematics in our country ... Thank you very much and we urge the younger generation to follow you r example! Alexandros Sygelakis (Professor of Mathematics-Administrator).

Antonis Kyriakopoulos , Nikos Kyriazis, allow us to nominate you as Distinguished Members of the Greek Mathem atical, web community, as a minimum recognition of your contribution to the mathematical education in this country. Grigoris Kostakos (Professor of Mathematics - Administrator).

Antonis Kyriakopoulos - Nikos Kyriazis thank you for your long-term contribution to the mathematical commun ity. We've gained a lot of mat h knowledge from you and we hope to gain even more in the future . The education of Mathematics in our country is in need of people of your kind . With respect and appreciation toward you , I ask for your permission to call you my masters. Kardamitsis Spiros (Professor of Mathematics - Coordinator).

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It is our honor, if anything else, that people of this magnitude at mathematics are close to us. Above all I wish you good health. Christos Kyriazis (Professor of Mathematics).

I will also congratulate the excellent masters, Antonis Kyriakopoulos and Nikos Kyriazis , for their enormous contribution in the science of Mathematics. Stavros Stavropoulos (Professor of Mathematics).

Mathematics with us, feel very proud of you. Congratulations from my part as well! Tilegrafos Kostas (Professor of Mathematics).

Antonis and Nikos I think your inclusion on that list is yours by right, not to say officially! We are proud to have you next to us, you beautify and ornament our mathematics, you make them simpler and more glamorous, consistent and accurate and with substance and purpose ... You are our masters in mathematica.gr. Makis Hatzopoulos (Professor of Mathematics).

We thank our colleagues Antonis and Nikos, who are examples for us, the younger mathematicians. Sukaras Efthimios (Professor of Mathematics).

Nikos Kyriazis. I congratulate you for your contribution to Geometry. It is great honor to have you with us. X.K. (Professor of Mathematics)

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A QUESTION. DOES ANY UNIVERSITY PROFESSOR VISIT THE MATEMA TICA WEBSITE? A GREAT NUMBER OF ORDINARY PEOPLE WERE NAMED HONORARY DOCTORS; NO ONE WAS MOVED BY THE GREATNESS OF THE EXCEPTIONAL WORK OF MISTER NIKOS? I WONDER IF MISTER NIKOS WAS HONORED BY THE ACADEMIC WORLD. Mr Nikos, I apologize for my message, the greatness of your modesty would not have wanted me to send it out ... but I would have exploded if I hadn't written it. Finally, I'm sorry I do not have more time to work on it any longer ... and I will not have because I simply do not hope to get a pension. Leonidas 1

I have to congratulate you on your work or better on your valuable contribution. I hope you will be an example to others and especially Mathematicians. I also wish for the continuation of your efforts. I offer you all of my Geometry books (six volumes), as well as my five research papers, because I'm sure they will be of a value to you. Nikos Kiskyras (Mathematician) .

. .. I congratulate you on your appeal to deal with such passion for Geometry, but also for your achievements, which are giving the impression not only for your complete knowledge of Geometry but, above all, a deeply complex and unified thought. I wish you well and to keep working, with similar great success, on the most beautiful of sciences, on this the purely Greek science. I also wish you to keep well and have every success in the divine paths of research and knowledge. Michael Diakomanolis (Mathematician).

The main Theorem of your article is very interesting. I have not seen it anywhere and it can be a powerful tool for proving other Theorems like the ones of Steiner, Vecten, Morley, Brocard, Kariya, convergence of triangle heights, etc Assessor in Geometry, the periodical Mathematics and Informatics Quarterly.

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Your massive and multi-page work in Euclidean Geometry has been a pleasant surprise for me in terms of its content. This work is definitely the fruit of many years of long term , deeply successful or better intelligent research. Mr Nikolaos Kyriazis , a lover and admirer of Geometry, with this extraordinary written work , emerges as a brilliant man of geometry, a remarkable researcher in Geometric issues and in any case all praise is deserved. Zissis Balpouzis (Mathematician).

THE ATHENS ACADEMY WAS OFFERED THE HONOR TO ATTRIBUTE TO NIKOLAO KYRIAZI CONGRATULATIONS CONCERN ING HIS WORK AND SUCCESSFUL CONTRIBUTION ON THE SCIENCE OF BASIC GEOMETRY. THE HONORS TOOK PLACE IN THE MEETING OF THE MONTH DECEMBER THE SEVENTH OF THE YEAR TWO THOUSAND TWO. Athens Academy (This honor was obviously awarded for his work prior to 27-122002).

Nikos, a work of such importance will stay immortal and you as well with it! Keep well, live long and enjoy the fruits and glories of this unique effort and contribution! I hope that among all those that will honor you , there will be with additional love the Greek scientific Associations , the institutions and the Universities! Babis Stergiou (Professor of Mathematics - Author).

Mr. Nikos Kyriazis , I would like you to allow me to thank you for this unforgettable gift that touches the spirit of Euclid and you generously offered it to us. I sincerely wish you to continue such scientific interventions of progress and substantial advancement. Best wishes to you and your family . (*) ... Personally, I strongly believe that the message of this work is clearly wider and it is an essential response to the challenges of our time having several recipients .... Sotiris E. Louridas (Professor of Mathematics - Author).

Personally, as I have already written , Nikos thank you again, I admire you for what you are doing even if I do not understand you. Parmenides51 (Professor of Mathematics).

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In certain research work a thank you is not enough!!! We will "gulp" it, Nikos at every opportunity !!! Sotiris D. Hasapis (Professor of Mathematics).

I admire Mr. Nikos Kyriazis for his courage to engage in Euclidean Geometry. When a person has got a professional career that is irrelevant to mathematics and manages to have so great results at an age when others lose interest in what was once their main activity, then this man is enviable. I really would like to meet and to him ... I have so many things to ask him ... I would like him so many things to tell me ... What I'm writing is not exaggeration, it's from the bottom of my heart ... Mr. Nikos, I do not know if we will ever meet, I want you to know that you are an example to be followed. Keep well and continue to work - with your own way - with the Euclidean Geometry. You offer and help ... And you show what a retired person can do when he keeps in mind and heart something he loved from the time he was a student

Tilemachos Baltsavias. A simple Professor of Mathematics. Nikos, this is my greatest honor ever. I feel like I do not deserve it, but because it is from you , I accept it humbly. I hope for you the best, keep your spirit always in a creative mood and give to all of us these geometric diamonds. Good day and every happiness to you! Babis Stergiou. (Professor of Mathematics, Mathematics Bookwriter).

Nikos, thank you very much for your enormous contribution to this field of geometry, which may be for the young people who shall wish to study the publications a real treasure. Keep always good and I think we will be amazed again with your new ideas. DIMITRIS IOANNOU (Professor of Mathematics).

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Important remark from page 12: 1. Greek alphabet, in which we can see the sequence (order) of the Greek letters, that interests the researcher. (a). CAPITAL LETTERS. A-alpha,

I-iota,

P-rho,

8-beta ,

K-kappa,

r-sigma,

r-gamma,

L-lambda,

T-tau,

.6-delta,

M-mu,

Y-upsilon,

E-epsilon,

N-nu,

-phi ,

Z-zeta,

::-xi ,

X-chi,

H-eta,

0-omicron,

4'-psi,

0-theta,

n-pi,

n-omega

(bl. LOWERCASE LETTERS. a-alpha,

1-iota,

p-rho,

13-beta,

K-kappa,

a-sigma,

y-gamma,

A-lambda,

T-tau,

~-delta,

µ-mu,

u-upsilon,

E-epsilon,

v-nu,

q>-phi,

~-zeta,

~-xi,

x -chi,

ri-eta,

o-omicron,

ljl-psi ,

8-theta,

TT-pi,

w-omega

2. Translation in English of some words that appear in the book in Greek. rxr;µa - Figure, MoipE~ - Degrees, rwvia - Angle, npof3o,\r') - Projection, T6~o - Arc, ruvriµµtvo - Attached, MtyE8o~ - Size.

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HARMONIC GEOMETRY Part E.

NIKOS KYRIAZIS is a retired high-ranking officer of the Greek Army. Because of his great love for mathematics, he devoted himself to studies and research in Euclidean geometry after he retired. The fruit of the above-mentioned research is the creation of many important new elements in geometry and he recently formulated Harmonic Geometry. All this work has been documented in five books, twentytwo (22) volumes so far, of which thirteen (13) have been released, available at cost price for geometry lovers. He has also written fifty-seven (57) articles in Greek and foreign magazines and newspapers, some of which have been published in two additional volumes (the 23rd and 24th). In his works, he only records new elements, which have not been encountered before and which arose from his research. The value of his books do not derive from their perfect appearance and their elegant presentation, but from their original content, which has significantly changed what we know about geometry, and will surprise the reader. A part of the work is secondary-school level , while the rest is of a higher level, suitable for the Olympiads, the Balkaniade, etc. The comments and reviews of mathematics experts are excellent and favourable (Page 768). The press wrote numerous glowing comments about him and his work. All his books and articles prior to 27-12-2002 were honoured by the ACADEMY OF ATHENS at its annual celebratory session on 27-12-2002. The Academy hailed the whole work as "outstanding and useful " when presenting it with the award. For his work and his Academy Award, he was honoured by many other institutions in Greece. In 2011, he was unanimously named Distinguished Member in geometry by the website mathematica.gr. In 1957, Nikos Kyriazis graduated from the Hellenic Army Cadet School as a Second Lieutenant of Engineering, in 1966, from the School of Civil Engineering in Athens, in 1975, from STAMAM (Military School, similar to the School of Civil Engineering of the Polytechnic School) and in 1978, from the Supreme War College. He served his country continuously for 33 years and retired from his honourable service in 1987, with the rank of Brigadier. He has received numerous medals of honour during his military service. Nikos Kyriazis was born in Thouria, Kalamata (Greece) in 1934. He permanently resides and creates in Kalamaria (Trapezoundos 9, tel. +30 2310-413.539). He is married and has two children and three grandchildren.