Four-Fold Symmetry in Universal Triangle Geometry

Home , De Longchamps point, Feuerbach point, Lemoine point, Nagel point, Spieker center, Spieker circle

SCIENTIA

MANU E T MENTE

10th August, 2015

A thesis presented to

The School of Mathematics and Statistics The University of New South Wales

in fulﬁlment of the thesis requirement for the degree of

Doctor of Philosophy

by NGUYEN HONG LE

Supervisor: A/PROF. NORMAN WILDBERGER ORIGINALITY STATEMENT

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School: Mathematics and Statistics Faculty: Science

Title: Four-Fold Symmetry in Universal Triangle Geometry

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We develop a generalized triangle geometry, using an arbitrary bilinear form in an affine plane over a general field. By introducing standardized coordinates we find canonical forms for some basic centers and lines. Strong concurrencies formed by quadruples of lines from the lncenter hierarchy are investigated, including joins of corresponding lncenters, Gergonne, Nagel, Spieker points, Mittenpunkts , Bevan points, midpoints of lncenters and the Centroid, and so on. We identify the resulting centers in Kimberling's list. We also use a Kimberling 6-9-13 triangle to connect the triangle centers we have found with Kimberling's list. The diagrams are taken from Euclidean (blue) geometry and relativistic (green) geometry.

Chromogeometry brings together planar Euclidean geometry, here called blue geometry, and two relativistic geometries, called red and green. We show that if a triangle has four blue lncenters and four red lncenters, then these eight points lie on a green circle, whose center is the green Orthocenter of the triangle, and similarly for the other colours. Tangents to the incenter circles yield interesting additional standard quadrangles and concurrencies. The proofs use the framework of rational trigonometry together with standard coordinates for triangle geometry, while a dilation argument allows us to extend the results also to Nagel and Spieker points.

We investigate the concurrencies of triples of Euler lines (at four points) and the concurrencies of Euler lines and bilines (at 6 points), and prove the pleasant result that these concurrencies (10 points) lie on the Circumcircle. We generalize both the classical definition of the Schiffler point S=X21 and the approach of L. Emelyanov and T. Emelyanov to obtain in fact four Schiffler points. The four-fold Schiffler points give interesting results, for example the four ln-Schiffler lines form a standard quadrilateral s0 s1 s2s3 , and the four ln-Schiffler lines s0 , s1, s2, and s3 are respectively the tangents at 10 , l1, 12, 13 to the In-Circum conic which passes through the Circumcenter and four lncenters.

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Acknowledgements v

1 Introduction 1 1.1 Overview ...... 1 1.2 Background ...... 3 1.2.1 Rational Trigonometry and Universal Geometry ...... 3 1.2.2 Chromogeometry and Euler lines ...... 4 1.2.3 Triangle Geometry ...... 5 1.2.4 The classical Incenter hierarchy ...... 6 1.2.5 Quadrilaterals and quadrangles ...... 8 1.2.6 Construction of a quadrangle containing a given point with a given diagonal triangle ...... 9 1.2.7 Construction of a quadrilateral containing a given line with a given diagonal triangle ...... 10 1.3 Outline of contents ...... 11

2 Basic Universal Triangle Geometry 18 2.1 Affi ne structure and vectors ...... 18 2.2 Metrical structure: quadrance and spread ...... 19 2.3 Triple spread formula ...... 21 2.4 Altitudes and orthocenters ...... 22 2.5 Change of coordinates and an explicit example ...... 23 2.6 Bilines ...... 25 2.7 Standard coordinates and triangle geometry ...... 26 2.7.1 Basic affi ne objects in triangle geometry ...... 28 2.7.2 The Orthocenter hierarchy ...... 29 2.8 Transformations ...... 31 2.8.1 Dilations about the Centroid ...... 32 2.8.2 Reflections and Isogonal conjugates ...... 33 2.8.3 Isotomic conjugates ...... 34 2.9 A Kimberling 6-9-13 triangle ...... 34

3 Universal aﬃ ne triangle geometry and four-fold incenter symmetry 37 3.1 Introduction ...... 37 3.2 Bilines and Incenters ...... 40

i CONTENTS ii

3.3 Strong concurrences ...... 43 3.3.1 Bevan points ...... 43 3.3.2 Sight Lines, Gergonne and Nagel points ...... 45 3.3.3 InMid lines and Mittenpunkts ...... 49 3.3.4 Spieker points ...... 51 3.3.5 Contact triangle and Weill point ...... 53 3.3.6 Summary Table ...... 56

4 Midpoints, four-fold Incenters symmetry and the Euler line 57 4.1 Midpoints of Incenters I and the Centroid G ...... 57 4.2 Midpoints of Incenters I and Circumcenter ...... 60 4.3 Midpoints of Incenters I and Orthocenter H ...... 63 4.4 Midpoints of Incenter I and nine-point center X5 ...... 66 4.5 Midpoints of Incenters I and De-Longchamp point ...... 70 4.5.1 Summary Table ...... 74

5 Incenter circles, chromogeometry, and the Omega triangle 75 5.1 Introduction ...... 75 5.2 The Incenter Circle theorem ...... 76 5.2.1 Equations of Incenter Circles ...... 80 5.2.2 Tangent lines of Incenter Circles ...... 81 5.3 Explicit examples ...... 87 5.3.1 An example over Q √30, √217, √741, √2470, √82 297 . . . . 87 5.3.2 An example over ...... 91 F13 5.4 Spieker circles and Nagel circles ...... 93

6 Euler lines and Schiffl er points 96 6.1 Incenter Euler lines ...... 96 6.1.1 Euler points P0,P1,P2 and P3 ...... 97 6.1.2 Bi-Euler points ...... 101 6.2 The classical Schiffl er point X21, and the Emelyanov’spoint of view . 106 6.3 Four-fold views of Schiffl er points ...... 107 6.3.1 Generalization of the classical Schiffl er point ...... 107 6.3.2 Generalization of L. Emelyanov and T. Emelyanova theorem . 110 6.3.3 In-Schiffl er lines and points, and standard quadrilaterals and quadrangles ...... 113 6.4 The In-Circum conic and its tangent lines ...... 119 6.5 Conclusion and future directions ...... 122 List of Figures

1.1 Euclidean Bilines and Incenters I0,I1,I2 and I3 of A1A2A3 ...... 3 1.2 Blue, red and green Euler lines of A1A2A3 ...... 5 1.3 Euclidean Incenter I0, excenters I1,I2,I3, Gergonne point G = X7 and Nagel point N = X8 ...... 7 1.4 Euclidean Spieker center S and De Longchamps point X20 ...... 8 1.5 A quadrilateral and its opposite quadrangle ...... 9 1.6 Construction of a Quandrangle ...... 10 1.7 Construction of a Quadrilateral ...... 11 1.8 Aspects of the Incenter hierachy in blue geometry ...... 11 1.9 Blue Incenter altitudes, Bevan points L and In-Bevan center C . . . . 12 1.10 Blue Spieker points S, Mittenpunkts D and Mitten-Spieker center H . 13 1.11 Blue MIG-Bevan center X3524 ...... 14 1.12 Blue MIN-Spieker center X3628 ...... 15 1.13 The four blue Incenters of A1A2A3 and red and green Incenter Circles 16 1.14 Blue Circumcircle and Euler points ...... 16

2.1 Euler line in green geometry ...... 24 2.2 Construction of Green bilines b at A1 ...... 26 2.3 The Medial triangle M1M2M3 and Double triangle D1D2D3 ...... 28 2.4 The Euler line of a triangle ...... 30 2.5 The Euler line and X140 in blue geometry ...... 35

3.1 Aspects of the Incenter hierarchy in green geometry ...... 38 3.2 Green bilines b, Incenters I, Contact points and Incircles ...... 42 3.3 Green Incenter altitudes, Bevan points L and In-New center C . . . . 43 3.4 Green Sight lines, Gergonne points G, In-Gergonne lines and In-Gergonne center X20 ...... 45 3.5 Green Sight lines, Nagel points N, In-Nagel lines and In-Nagel center G = X2 ...... 47 3.6 Green Gergonne-Nagel center X69 and Nagel-Bevan center X20 . . . . 48 3.7 Green InMid lines and Mittenpunkts D ...... 49 3.8 Green Mitten-Bevan center H, Gergonne-Mitten center G and In-Mitten center K ...... 51 3.9 Green Spieker points S and Mitten-Spieker center H ...... 52 3.10 Green contact triangles and Weill points...... 54

iii LIST OF FIGURES iv

4.1 Aﬃ ne ratio of Ij,MIGj, G, Sj, and Nj ...... 58 4.2 Green Incenters, MIG, Centroid, Spiekers and Nagel points ...... 58 4.3 Green MIG-Beven center ...... 59 4.4 Green MIC-Nagel center ...... 60 4.5 Green MIC-Spieker center and MIC-Bevan center ...... 62 4.6 Green MIH-Nagel center ...... 63 4.7 Green MIH-Spieker center and MIH-Bevan center ...... 65 4.8 Green MIN-Nagel center ...... 68 4.9 Green MIN-Bevan center ...... 69 4.10 Green MIN-Spieker center ...... 70 4.11 Green MID-Nagel center ...... 71 4.12 Green MID-Bevan center and MID-Spieker center ...... 73 4.13 Green Euler line...... 74

5.1 Green Incenters and the blue and red Incenter Circles ...... 76 5.2 Three Incenter Circles Cb,Cr and Cg...... 80 5.3 Incenter tangent meets ...... 82 5.4 Quad points and star lines ...... 84 5.5 An example triangle X1X2X3 ...... 88 5.6 Blue, red and green Spieker circles ...... 93 5.7 Blue, red and green Nagel circles ...... 94 5.8 Blue, red, green Orthocenters, Circumcenters and De Longchamps points 95

6.1 Green Euler lines and Euler points on the Circumcircle ...... 99 6.2 Blue Bi-Euler points ...... 101 6.3 Green Bi-Euler points ...... 103 6.4 Green Circumcircle, Euler points and Bi-Euler points ...... 104 6.5 Blue Circumcircle, Euler points and Bi-Euler points ...... 105 6.6 The classical Schiffl er point S = X21 ...... 106 6.7 The Emelyanov characterization of S ...... 107 6.8 Blue Schiffl er points ...... 108 6.9 Green Schiffl er points ...... 109 6.10 Green generalization of L. Emelyanov and T. Emelyanova theorem . . 110 6.11 Blue generalization of L. Emelyanov and T. Emelyanova theorem . . . 111 6.12 Green In-Schiffl er quadrilateral ...... 113 6.13 Blue In-Schiffl er quadrilateral ...... 114 6.14 Green r-Schiffl er standard quadrilaterals ...... 115 6.15 Blue r-Schiffl er standard quadrilaterals ...... 116 6.16 Green r-Schiffl er Quadrangle ...... 117 6.17 Blue r-Schiffl er Quadrangle ...... 118 6.18 Green In-Circum conic and tangent lines ...... 119 6.19 Blue In-Circum conic and tangent lines ...... 121 Acknowledgements

I would like to express my special appreciation and thanks to my supervisor, A. Prof. Norman J. Wildberger, for the patient guidance, encouragement and enlightenment he has provided throughout my time as his student. I have been extremely lucky to have a supervisor who cared so much about my work, and who responded to my questions and queries so promptly. His ﬂexibility, genuine caring and concern, and faith in me during the dissertation process enabled me to attend to life while also earning my Ph.D. He has never judged nor pushed when he knew I needed to juggle priorities. For this, I cannot thank him enough. I am forever grateful. Thank You A. Prof. Norman! I will forever be thankful to my MA research advisor, Professor Matthias Beck. He was and remains my best role model for a mathematician, mentor, and teacher. I still think fondly of my time as an undergraduate student in his classes. He was the reason why I decided to go to pursue a career in research. His enthusiasm and love for teaching is contagious. I would like to express my deep gratitude to the University of New South Wales for funding my study. I would not be able to have such a wonderful opportunity to study in this university without these generous supports. A special thanks to my family. Words cannot express how grateful I am to my parents and sisters for all of the support that you’ve made on my behalf. Your prayer for me was what sustained me thus far. Last, but certainly not least, I must acknowledge with tremendous and deep thanks my husband, Kennard Ngo. There are no words that can express my gratitude and appreciation for all you’ve done and been for me. As I ramble, I still have not found the words that describe or express how I feel for this man and what his presence in my life has meant. He loves me like no one else has and has changed me for the better. Thank you with all my heart and soul. I am forever indebted to you for giving me life, your love, and your heart. Your complete and unconditional love carries me through always.

v Chapter 1

Introduction

1.1 Overview

This thesis aims to study a remarkable four-fold symmetry that occurs in triangle geometry based on the Incenter hierarchy in a very general setting—holdingover general fields and indeed arbitrary bilinear forms. We establish some analogs of results that have appeared first in Universal Hyperbolic Geometry [40]; namely the concurrency of some quadruples of lines associated to the classical Incenters, Gergonne points, Nagel points, Mittenpunkts, Spieker points, Bevan points, Weill points, Schiffl er points, midpoints of Incenters and the Centroid, and other points as well. In particular we show how many of these results can be viewed in the context of chromogeometry, so that they hold not only over the usual Euclidean plane, but also with relativistic geometries. A secondary aim is to provide explicit algebraic formulas for points, lines and transformations of triangle geometry which hold in great generality, over the rational numbers, finite fields, and even the field of complex rational numbers, and with differ- ent bilinear forms determining the metrical structure. This allows us to avoid any use of transcendental quantities. We identify new concurrencies that we find by placing the resulting centers in Kimberling’slist, and show how a strategic use of a particular Kimberling 6-9-13 triangle allows us a practical tool for identifying centers even in more general geometries. Our basic technology is simple but powerful: we propose to replace the affi ne study of a general triangle under a particular bilinear form with the study of a particular triangle under a general bilinear form– analogous to the projective situation as in [40], and using the framework of Rational Trigonometry ([34], [36]). When we study a triangle A1A2A3, we use an affi ne change of coordinates to move it into standard coordinates, so we may assume that

A1 = [0, 0] ,A2 = [1, 0] and A3 = [0, 1] .

1 1. Introduction 2

The various triangle centers and constructions are then expressed in terms of the coeﬃ cients a, b and c of the matrix

a b C ≡ b c! which specifies the new bilinear form (assumed non-degenerate throughout). One advantage here is that since the bilinear form is now arbitrary, we can assume that it was arbitrary to begin with, meaning that our formulas hold not only for Euclidean geometry but for general metrical geometries, including relativistic ones. This allows a systematic augmentation of Kimberling’s Encyclopedia of Triangle Centers ( [18], [19], [20]) to arbitrary quadratic forms and general fields. We get reasonably pleasant and simple formulas for various points, lines, circles and constructions, and our results are in fact very general. Another novelty in our approach is that we acknowledge from the start that the very existence of the Incenter hierarchy is dependent on number-theoretical conditions which end up playing an intimate and ultimately rather interesting role in the theory. While a dynamic geometry package will produce Incenters in all cases, when one looks more carefully at them one over a general field or bilinear form, one sees that there are actually number theoretical conditions that our triangle A1A2A3 needs to satisfy in order to have bilines, and thus incenters. Algebraically it becomes diffi cult to separate the classical incenter I0 from the three closely related excenters I1,I2 and I3 which are also formed from the meets of bilines. The quadratic relations that govern the existence of these four points carry a natural four-fold symmetry between them. This symmetry becomes crucial to simplifying formulas and establishing theorems. So in our framework, there are four Incenters

I0,I1,I2 and I3, not one, and we attempt to extend this four-fold symmetry in other directions. In particular we are interested in constructions that yield groups of four lines which are concurrent, or four points which are collinear. We call this strong concurrency. To showcase the generality of our results, we illustrate theorems not only over the Euclidean plane, but also in the Minkowski plane coming from Einstein’s special theory of relativity in null coordinates, where the metrical structure is determined by the bilinear form

(x1, y1) (x2, y2) x1y2 + y1x2. · ≡ In the language of Chromogeometry ([38] , [39]), this is green geometry, with circles appearing as rectangular hyperbolas with asymptotes parallel to the coordinate axes. Green perpendicularity amounts to vectors being Euclidean reﬂections in these axes, 1. Introduction 3

Figure 1.1: Euclidean Bilines and Incenters I0,I1,I2 and I3 of A1A2A3 while null vectors are parallel to the axes. It is eye-opening to see that triangle geometry is just as rich in such a relativistic setting as it is in the Euclidean one! Chromogeometry brings together planar Euclidean geometry, here called blue geometry, and two relativistic geometries, called red and green. We investigate lovely relations between three coloured Incenter hierarchies of a ﬁxed planar triangle, and standard coordinates give us surprisingly simple and elegant equations for the three coloured Incenter Circles. We also investigate the concurrency of triples of Euler lines and bilines, and extend to a four-fold aspect of Schiﬄ er points, using both a classical approach and a generalization of the approach of L. Emelyanov and T. Emelyanova. We obtain interesting new points on the circumcircle, which we call Euler points and Bi-Euler points.

1.2 Background

1.2.1 Rational Trigonometry and Universal Geometry

The new approach to trigonometry developed in [34] by N. J. Wildberger, is called Ra- tional Trigonometry to distinguish it from traditional trigonometry. Rational Trigonom- etry is built on rational arithmetic and algebraic formulas. Distance, d (A1,A2), between points A1 and A2 is replaced with quadrance q (A1,A2), and an angle, θ (l1, l2), between lines l1 and l2 is replaced with the spread S (l1, l2). In the Euclidean case, the 1. Introduction 4 quadrance is the square of the usual distance. But quadrance is a more elementary and fundamental notion than distance, and its algebraic nature makes it ideal for metrical geometry using other bilinear forms. The spread has the value sin2 θ in the Euclidean case where θ is the corresponding angle. Spread is a more algebraic, logical, general and powerful notion than that of angle, and together quadrance and spread provide the foundation for Rational Trigonometry. Our set-up of Rational Trigonometry is based on Cartesian analytic geometry, with a point A defined as an ordered pair [x, y] of rational numbers and a line l defined by an equation ax + by + c = 0 with rational coeffi cients a, b and c. By avoiding use of square root operations (which only give approximate distances) and classical circular functions and their inverses (which only give approximations of angles), we get a more algebraic geometry. Because Rational Trigonometry uses only arithmetic and algebra, many practical problems can now be solved completely in an easier and more accurate than classical trigonometry, which generally provides only approximate solutions. In fact almost all calculations may be done by hand without a calculator in rational trigonometry. Beyond the rational numbers, the laws of Rational Trigonometry apply in any finite field which does not have characteristic 2. In particular they apply to geometry over a finite field. However one should be aware that in this situation there are sometimes null lines with non-zero direction vectors perpendicular to themselves; the same kind of phenomenon familiar to physicists working in special relativity. A natural outgrowth of Rational Trigonometry is Universal Geometry ([34], [36]) which extends Euclidean and non-Euclidean geometries to arbitrary fields and general quadratic forms. By being based only on algebra, other aspects of geometry can be studied in this univeral framework, for example triangle geometry and cubics in [37].

1.2.2 Chromogeometry and Euler lines

The current emphasis with distance and angle as the basis for Euclidean geometry is a historical curiosity contrary to the explicit orientation of Euclid himself, and is a subtle obstacle to understanding the relativistic geometry introduced by Einstein and Minkowski. Universal geometry relates naturally to a remarkable new three-fold symmetry in planar geometry that connects Euclidean and relativistic geometries. This is Chromogeometry ([38], [39]), which brings together planar Euclidean geometry, here called blue geometry based on the symmetric bilinear form x1x2 + y1y2, and two relativistic geometries, called red and green based respectively on the bilinear forms x1x2 y1y2 and x1y2 + y1x2. The three geometries support and interact each − other in a nice and surprising way, and they are ruled by the laws of rational trigonom- 1. Introduction 5 etry. Leonhard Euler (1707-1783) discovered the famous line that bears his name the Euler line (1765) which contains the circumcenter C, the centroid G, and the orthocenter H, always in the respective ratios of 1:2. Later, it was discovered that the nine-point center N also lies on the Euler line. In [38] Wildberger showed that this situation holds quite a lot more generally.

Figure 1.2: Blue, red and green Euler lines of A1A2A3

In Figure 1.2 we see the Centroid G, the blue, red and green Orthocenters respectively Hb,Hr, and Hg, Circumcenters Cb,Cr, and Cg, nine-point centers Nb,Nr, and

Ng, along with the blue, red and green Euler lines eb, er, and eg and also the blue, red and green Circumcircles of A1A2A3. In this thesis, we use the framework of rational trigonometry and universal triangle geometry to study many interesting points, lines and circles associated to a triangle in the plane. Chromogeometry brings out pleasant connections between these triangle centers, and thus we get new perspectives on triangle geometry.

1.2.3 Triangle Geometry

Triangle geometry has a long and cyclical history ([7], [14], [32], [33]). Classical triangle geometry investigates triangle centers like the centroid G = X2, (meet of the medians), circumcenter C = X3 (meet of the perpendicular bisectors), orthocenter H = X4 (meet 1. Introduction 6

of the altitudes), incenter I = X1 (meet of three angle bisectors, or bilines as we prefer to call them), nine-point center N = X5 ([4], [6]) (the center of the circle through the three midpoints of the sides) among others were known to the ancient Greeks, triangle lines such as the Euler line, Brocard line, Lemoine line. The notation here is that of Clark Kimberling, as laid out in the Online Encyclopedia of Triangle Centers ([18], [19], [20]). More than 7000 triangle centers have been as of now classified; each with a format involving X and a subscript number, as in the above examples. In addition to the Incenter, Centroid, Circumcenter, Orthocenter and nine-point center, the symmedian point K = X6 (Lemoine point) ([10]) is the point of concurrence of the symmedians obtained by reflectingthe medians about the corresponding interior angle bisectors. The French mathematician Emile Lemoine proved the existence of the symmedian point in 1873. The De LongChamps point X20 is the orthocenter of the Double (or anti-medial) triangle and is also the reflection of the Orthocenter H in the Circumcenter C ([1], [31]). The De Longchamps point is named after Gaston Albert Gohierre de Longchamps, a French mathematician. Prominent mathematicians like Euler and Gauss contributed to the subject, but it took off mostly in the latter part of the 19th century and the first part of the 20th century, when many new centers, lines, conics, and cubics associated to a triangle were discovered and investigated. Then there was a period when the subject lay somewhat dormant; and now it flourishesonce more– spurred by the power of dynamic geometry packages like GSP, C.a.R., Cabri, GeoGebra, and Cinderella; by the explorations and discussions of the Hyacinthos Yahoo group [15]; and by the heroic efforts of Clark Kimberling in organizing the massive amount of information on Triangle Centers in his Online Encyclopedia. The increased interest in this rich and fascinating subject is great, but there are also questions about the consistency and accessibility of proofs, which have not generally kept up with the greater pace of discoveries. Another diffi culty is that the current framework is modelled on the continuum in terms of real numbers, which often leads synthetic treatments to finesse number-theoretical issues, and is a barrier to investigations over more general fields.

1.2.4 The classical Incenter hierarchy

It is telling that Kimberling decided to list the Incenter as I0 = X1, the ﬁrst point in his list. In fact the Incenter is the centrepiece of a vast hierarchy of tantalizing results, which we refer to as the Incenter hierarchy. Along with the Incenter I0, there are three closely related excenters I1,I2 and I3. The symmetry between these four points is a recurrent tool in this thesis. We now introduce some triangle centers in the Incenter 1. Introduction 7 hierarachy.

Figure 1.3: Euclidean Incenter I0, excenters I1,I2,I3, Gergonne point G = X7 and Nagel point N = X8

The lines from the vertices of a triangle to the points of contact of the opposite sides with the inscribed circle centered at I0 meet at a point called the Gergonne point

G = X7, see ([10], [13]), named after Joseph Diez Gergonne (1771-1859), a French mathematician and geometer. The Nagel point N = X8, see ([10], [17]), is the meet of the lines from the vertices to the points of contact of the opposite sides with the excribed circles centered at I1,I2 and I3. The Nagel point is named after Christian

Heinrich von Nagel (1803-1882), a German mathematician. The Mittenpunkt K = X9 is the symmedian point of the excentral triangle, formed by the excenters of the original triangle. The Mittenpunkt is another point that Nagel studied in 1863 ([8]). The Spieker circle is the incircle of the medial triangle, formed by the midpoints of the sides of a triangle. The center of the Spieker circle is the Spieker center S = X10 ([5], [16]), named in honor of the 19th-century German geometer Theodor Spieker. Some interesting properties of the Spieker center are pointed out in [14] and [26].

Figure 1.4 shows the Double triangle B1B2B3 and the medial triangle M1M2M3 of the triangle A1A2A3, together with the Spieker center S = X10 and the De Longchamps point X20.

The Schiffl er point S = X21 of the triangle A1A2A3 defined by Kurt Schiffl er (1896-

1986) as the intersection of the Euler lines of the four triangles A1A2I, A1A3I, A2A3I, and A1A2A3 where I = X1 is the classical incenter of the triangle A1A2A3, see ([29],

[18], [19]). The Schiﬄ er point lies on the Euler line of the original triangle A1A2A3. In 1. Introduction 8

Figure 1.4: Euclidean Spieker center S and De Longchamps point X20 fact, there are many interesting and remarkable properties of the Schiﬄ er point which have been found over the years: see for example ([12], [9], [24], [30]). The perpendiculars from the excenters to the respective sides meet at a point called the Bevan point X40, named in honor of Benjamin Bevan, an Englishman who proposed the problem of proving that the Circumcenter C was the midpoint of the Incenter I and the circumcenter of the excentral triangle (1806). The problem was solved by John Butterworth in the same year ([2], [3]). One of the main aims of the thesis is to extend the four-fold symmetry between

I0,I1,I2 and I3 to the other points in the incenter hierarchy. So we will introduce four Nagel points, four Gergonne points, four Bevan points, four Mittenpunkt points, four Spieker points, four Weill points and four Schiﬄ er points. In the rest of this section we recall basic facts from the projective geometry of a quadrangle and quadrilateral.

1.2.5 Quadrilaterals and quadrangles

We remind the reader of some basic facts from the projective geometry of a quadrangle (four points) or quadrilateral (four lines), using a visual presentation to avoid the need to introduce notation. In Figure 1.5 we see four blue lines forming a quadrilateral [in this ﬁgure colours are not used in a metrical sense, but only as an aide for explanation]. These four blue lines meet in six points, also in blue. These six blue points determine a further three green diagonal lines, forming the diagonal triangle, in yellow, of the original quadrilateral, whose vertices are three green points. Each green point may be joined via a red line to the two blue points not on either of the two green lines it lies on. This produces six red lines, which somewhat remarkably meet three at a time at four red 1. Introduction 9

Figure 1.5: A quadrilateral and its opposite quadrangle points, giving the opposite quadrangle from the original blue quadrilateral. Note that there is a natural correspondence between the four original blue lines and the four red points. The situation is completely symmetric with regard to points and lines. If we had started out with a quadrangle of four red points, we would join them to form six red lines. These six red lines determine a further three green diagonal points, forming the diagonal triangle of the original quadrangle, whose sides form three green lines. Each green line meets two of the red lines in two new blue points. These six new blue points lie three at a time on four blue lines, giving the opposite quadrilateral from the original red quadrangle. The diagonal green points on a green line are harmonic conjugates with respect to the two blue points on the same line. The diagonal green lines through a green point are harmonic conjugates with respect to the two red lines through the same point. There is another more subtle remark to be made here concerning symmetry: each of the three diagonal points is canonically associated to a subdivision of the four original blue lines into two subsets of two, namely those subsets whose joins meet at that diagonal point. If we start with a triangle, say the yellow triangle in the Figure formed by three green points and three green lines, then any quadrilateral or quadrangle which has that triangle as its diagonal triangle is called standard.

1.2.6 Construction of a quadrangle containing a given point with a given diagonal triangle

Now we construct a quadrangle P0P1P2P3 containing a given point P0, whose diagonal triangle A1A2A3 is given. 1. Introduction 10

Figure 1.6: Construction of a Quandrangle

In Figure 1.6 we start with a triangle A1A2A3 in black, and the red point P0. The three red lines P0A1,P0A2 and P0A3 meet the side lines in three blue points. These three blue points determine three blue lines, and these three blue lines meet the side lines in three green points. We join a green point with the point Ai opposite to the side line that it lies on to form three green lines. We join P0 with three blue points to get three red lines. The three red lines and three green lines meet three at a time (two green lines and a red line) at three other red points. These three red points together with the red point P0 form a quadrangle P0P1P2P3 whose diagonal triangle is A1A2A3.

1.2.7 Construction of a quadrilateral containing a given line with a given diagonal triangle

Now we construct a quadrilateral l0l1l2l3 containing a given line l0, whose diagonal triangle A1A2A3 is given.

In Figure 1.7 we start with a triangle A1A2A3 in black and a blue line l0. The blue line l0 meets three side lines of A1A2A3 at three blue points. The joins of a vertex Ai and a blue point which does not lie on the same side line with Ai form three green lines. These three green lines determine three red points. We join the point Ai and a red point which does not lie on the same green lines with Ai to form three red lines. The three red lines meet three side lines in three green points. The three green points and three blue points determine a further three blue lines. These three blue lines together with the blue line l0 form a quadrilateral whose diagonal triangle A1A2A3. 1. Introduction 11

Figure 1.7: Construction of a Quadrilateral

1.3 Outline of contents

The dissertation consists of six chapters. Chapter 1 is the introduction. Chapter 2 reviews some important aﬃ ne structures, the metrical notions of quadrance and spread and standard coordinates which will be used frequently throughout the dissertation. Chapter 3 and 5 of this thesis are now joint papers with N. J. Wildberger; both have already appeared, see ([23], [22]).

Figure 1.8: Aspects of the Incenter hierachy in blue geometry

Chapter 3 is devoted to our ﬁrst contribution. In this chapter, we reposition and ex- 1. Introduction 12 tend triangle geometry by developing it in the wider framework of Rational Trigonom- etry and Universal Geometry, valid over arbitrary ﬁelds and with general quadratic forms. Our main focus is on strong concurrency results for quadruples of lines associated to the Incenter hierarchy. In this chapter and throughout the thesis, we here refer to all four meets of the vertex bisectors, or bilines, as Incenters, so do not distinguish between the classical incenter and the three excenters.

Associated to any one Incenter Ij is a Gergonne point Gj = X7 (not to be confused with the centroid also labelled G), a Nagel point Nj = X8, a Mittenpunkt Dj = X9, a

Spieker point Sj = X10, a Bevan point Lj = X40, a Weill point Wj = X354 and so on. Figure 1.8 shows just one Incenter and its related hierarchy. Here are some main results in Chapter 3.

The four lines IjLj, j = 0, 1, 2, 3, meet in the Circumcenter C, and in fact C is the midpoint of Ij and Lj. The result is illustrated in Figure 1.9.

Figure 1.9: Blue Incenter altitudes, Bevan points L and In-Bevan center C

The four lines DjSj meet in the Orthocenter H = X4; this is illustrated in Figure 1.10. In Chapter 4, we continue investigating the four-fold incenter symmetry and dis- covering more strong concurrencies formed by quadruples of lines from the Incenter hierarchy which are investigated in the Chapter 3. We focus on the midpoints of Incen- ters and the Centroid, Circumcenter, Orthocenter, Nine-point center, De-Longchamp point, and strong concurrencies formed by quadruples of lines join corresponding these 1. Introduction 13

Figure 1.10: Blue Spieker points S, Mittenpunkts D and Mitten-Spieker center H midpoints and Nagel, Spieker points and Bevan points. Following are some remarkable results.

We introduce the notions of MIG point: the midpoint MIGj between the Incenter

Ij and the Centroid G, and MIN point: the midpoint MINj between the Incenter Ij and the nine-point center N and so on.

The four MIG-Bevan lines MIGjLj are concurrent at MIG-Bevan center, which 4 1 lies on the Euler line. In fact, MIG-Bevan center is X3524 and X3524 = X2 X4; 3 − 4 this is illustrated in Figure 1.11.

The four MIN-Spieker lines MINjSj are concurrent at MIN-Spieker center, which 15 1 lies on the Euler line. In fact, MIN-Spieker center is X3628 and X3628 = 16 X2 + 16 X4; this is illustrated in Figure 1.12. The Chapter 5 investigates a surprising connection between three closely related Incenter hierarchies of a ﬁxed planar triangle. The framework here is that of Ratio- nal Trigonometry which allows a consistent universal triangle geometry valid for any symmetric bilinear form, together with the three-fold symmetry of chromogeometry. By working with the rational notions of quadrance and spread instead of the transcendental notions of distance and angle, the main laws of Rational Trigonometry allow metrical geometry, and thus triangle geometry, to be developed in each of these three geometries in a parallel fashion, with mostly identical formulas and theorems. The results in this chapter have already appeared in [22]. 1. Introduction 14

Figure 1.11: Blue MIG-Bevan center X3524

The ﬁrst results of the Chapter 5 concern the four Incenters of a planar triangle in one of the three geometries, and were announced in [23]. If a triangle A1A2A3 has four b b b b blue Incenters I0,I1,I2 and I3, then all four points lie both on a red incenter circle b b with center the red Orthocenter Hr, and on a green incenter circle with center Cr Cg the green Orthocenter Hg; this is illustrated in Figure 1.13. Similarly, if a triangle has r red Incenters, then these lie both on a green incenter circle g with center Hg, and a r C blue incenter circle Cb with center the blue Orthocenter Hb. If a triangle has green g Incenters, these lie both on a blue incenter circle b with center Hb, and on a red g C incenter circle r with center Hr. Furthermore, if both red and green Incenters exist, C then they lie on the same blue incenter circle, and similarly for the other colours. The proofs are algebraic, and rely on non-obvious simpliﬁcations found by the help of a computer. So the Omega triangle formed by the three Orthocenters Ω HbHrHg, ≡ introduced in [38], has an intimate connection with the Incenter hierarchies. These facts relate also to elegant classical properties of quadrangles. In [11] Haskell showed that if two quadrangles have the same diagonal triangle, then all eight points of these quadrangles lie on a single conic; and in [42] Woods found a synthetic derivation of the same result. Now it is obvious that the four Incenters of a triangle, with respect to any bilinear form, will form a standard quadrangle in this sense, meaning that the 1. Introduction 15

Figure 1.12: Blue MIN-Spieker center X3628 diagonal triangle coincides with the original triangle. As a consequence, if blue and red Incenters exist, then they must lie on a conic. Our assertion is that this conic is g g actually a green circle = r g with center Hg. Cb C ≡ C In the case of blue Incenters, the four tangent lines to the red incenter circle b at Cr the blue Incenters form a standard quadrilateral, implying that they meet in six points b Rij, which lie two at a time on the three lines of A1A2A3, where they are harmonic conjugates with respect to A1,A2 and A3; and similarly the four tangent lines to the green incenter circle b at the blue Incenters meet in six points Gb on the three lines. Cg ij This is also seen on the Figure 1.13. Similarly there is a corresponding result when we look at red Incenters, and at green Incenters. b The six lines AkRij, for i, j, k distinct, are the lines of a complete quadrangle, so b b they meet three at a time at four quad points Qrj. Similarly, the six lines AkGij meet b b b b three at a time at points Qgj. Somewhat remarkably, the four star lines sj QrjQgj b b b b ≡ form a standard quadrilateral s0s1s2s3. In Chapter 6, we explore interesting concurrencies of Euler lines, and we extend the classical Schiﬄ er point S = X21 and the point of view of L. Emelyanov and T.

Emelyanova in [9] to four-fold Schiffl er points, S0,S1,S2 and S3, in the language of standard coordinates. We also introduce the In-Schiffl er line sj IjSj, the In-Schiffl er ≡ point Rij sisj, and standard quadrilaterals and standard quadrangles relate to In- ≡ 1. Introduction 16

Figure 1.13: The four blue Incenters of A1A2A3 and red and green Incenter Circles

Schiﬄ er lines and In-Schiﬄ er points. Here are some interesting results of this chapter.

Figure 1.14: Blue Circumcircle and Euler points

The ﬁrst result of Chapter 6 states that the Euler lines of A2A3I1, A1A3I2 and

A1A2I3 are concurrent and meet at the Euler point P0, where A1,A2 and A3 are vertices of A1A2A3, and I0,I1,I2 and I3 are Incenters of A1A2A3. Similarly, the Euler lines of A2A3I0, A1A2I2 and A1A3I3 are concurrent and meet at P1; the Euler lines of A1A3I0, A1A2I1 and A2A3I3 are concurrent and meet at P2; and the Euler lines of A1A2I0, A1A3I1 and A2A3I2 are concurrent and meet at P3. There are four Euler 1. Introduction 17 points and each associated to an Incenter.

In Figures 1.14, the line eji is the Euler line of AkAlIj where k, l = i, for k, l, i = 6 1, 2, 3,and j = 0, 1, 2, 3. Later in this chapter, we will see that the blue points S0,S1,

S2 and S3 are the Schiﬄ er points which are other concurrencies of Euler lines. The second result states that the four Euler points lie on the Circumcircle of

A1A2A3. Chapter 2

Basic Universal Triangle Geometry

2.1 Aﬃ ne structure and vectors

We begin with some terminology and concepts for elementary affi ne geometry in a linear algebra setting, following [34]. Fix a field F, of characteristic not two, whose elements will be called numbers. We work in a two-dimensional affi ne space A2 over F, with V2 the associated two-dimensional vector space. A point is then an ordered pair A [x, y] of numbers enclosed in square brackets, typically denoted by capital ≡ letters, such as A, B, C etc. A vector of V2 is an ordered pair v (x, y) of numbers ≡ enclosed in round brackets, typically u, v, w etc. Any pair of points A and B determines a vector v = −−→AB; so for example if A [2, 1] and B [5, 1] , then v = −−→AB = (3, 2), ≡ − ≡ and this is the same vector v = −−→CD determined by C [4, 1] and D [7, 3]. ≡ ≡ The non-zero vectors v1 (x1, y1) and v2 (x2, y2) are parallel precisely when ≡ ≡ one is a non-zero multiple of the other, this happens precisely when

x1y2 x2y1 = 0. − Vectors may be scalar-multiplied and added component-wise, so that if v and w are vectors and α, β are numbers, the linear combination αv +βw is defined. For points A and B and a number λ, we may define the affi ne combination C = (1 λ) A+λB − either by coordinates or by interpreting it as the sum A + λ−−→AB. An important special case is when λ = 1/2; in that case the point C A/2 + B/2 is the midpoint of AB, ≡ a purely affi ne notion independent of any metrical framework. Once we fix an origin O [0, 0], the affi ne space A2 and the associated vector space ≡ V2 are naturally identified: to every point A [x, y] there is an associated position ≡ vector a = −→OA = (x, y). So points and vectors are almost the same thing, but not

18 2. Basic Universal Triangle Geometry 19 quite. The choice of a distinguished point also allows us to have a useful notational shortcut: we agree that for a point A [x, y] and a number λ we write ≡ λ [x, y] (1 λ) O + λA = [λx, λy] . (2.1) ≡ − A line is a proportion l a : b : c where a and b are not both zero. The point ≡ h i A [x, y] lies on the line l a : b : c , or equivalently the line l passes through ≡ ≡ h i the point A, precisely when ax + by + c = 0.

For any two distinct points A1 [x1, y1] and A2 [x2, y2], there is a unique line ≡ ≡ l A1A2 which passes through them both; namely the join ≡

A1A2 = y1 y2 : x2 x1 : x1y2 x2y1 . (2.2) h − − − i

In vector form, this line has parametric equation l : A1 + λv, where v = −−−→A1A2 =

(x2 x1, y2 y1) is a direction vector for the line, and λ is a parameter. The − − direction vector of a line is unique up to a non-zero multiple. The line l a : b : c ≡ h i has a direction vector v = ( b, a). − Two lines are parallel precisely when they have parallel direction vectors. For every point P and line l, there is then precisely one line m through P parallel to l, namely m : P + λv, where v is any direction vector for l. For any two lines l1 ≡ a1 : b1 : c1 and l2 a2 : b2 : c2 which are not parallel, there is a unique point h i ≡ h i A l1l2 which lies on them both; using (2.1) we can write this meet as ≡ b1c2 b2c1 c1a2 c2a1 1 A l1l2 = − , − = (a1b2 a2b1)− [b1c2 b2c1, c1a2 c2a1] . ≡ a1b2 a2b1 a1b2 a2b1 − − − − − (2.3)

Three points A1 = [x1, y1] ,A2 = [x2, y2] ,A3 = [x3, y3] are collinear precisely when they lie on a common line, which amounts to the condition

x1y2 x1y3 + x2y3 x3y2 + x3y1 x2y1 = 0. − − −

Three lines l1 = a1 : b1 : c1 , l3 = a2 : b2 : c2 and l3 = a3 : b3 : c3 are concur- h i h i h i rent precisely when they pass through the same point P l1l2l3, which amounts to ≡ the condition

a1b2c3 a1b3c2 + a2b3c1 a3b2c1 + a3b1c2 a2b1c3 = 0. − − −

2.2 Metrical structure: quadrance and spread

We now introduce a metrical structure, which is determined by a non-degenerate symmetric 2 2 matrix × 2. Basic Universal Triangle Geometry 20

a b C = (2.4) b c! with entries a, b, c in the fixed field F over which we work. This matrix defines a symmetric bilinear form on vectors, regarded as row matrices, by the formula

v u = vu = vCuT . · Here non-degenerate means det C = 0, and it implies that if v u = 0 for all vectors u 6 · then v = 0. Note our introduction of the simpler notation v u = vu, so that also v v = v2. · · There should be no confusion with matrix multiplication, even if v and u are viewed as 1 2 matrices. Since C is symmetric, v u = vu = uv = u v. × · · Two vectors v and u are perpendicular precisely when v u = 0. Since the matrix · C is non-degenerate, for any vector v there exists, up to a scalar, exactly one vector u which is perpendicular to v. The bilinear form determines the main metrical quantity: the quadrance of a vector v is the number 2 Qv v v = v . ≡ · 2 A vector v is null precisely when Qv = v v = v = 0, in other words precisely when · v is perpendicular to itself. The quadrance between the points A and B is

Q(A, B) Q . ≡ −→AB This is the replacement in Rational Trigonometry for the notion of distance d(A, B). Two lines l and m are perpendicular precisely when they have perpendicular direction vectors. A line is null precisely when it has a null direction vector (in which case all direction vectors are null). We now make the important observation that the aﬃ ne notion of parallelism may also be recaptured via the bilinear form. (This result also appears with the same title in [41].)

Theorem 1 (Parallel vectors) Vectors v and u are parallel precisely when

2 QvQu = (vu) .

Proof. Using (2.4) for C, v = (x, y) and u = (z, w) , then an explicit computation shows that

4 2 2 2 (xw yz) ac b QvQu (vu) = − − . 2 2 2 2 2 2 − −(ax + 2bxy + cy ) (az + 2bzw + cw ) 2. Basic Universal Triangle Geometry 21

Since the quadratic form is non-degenerate, ac b2 = 0, so we see that the left hand − 6 side is zero precisely when xw yz = 0, in other words precisely when v and u are − parallel. This motivates the following measure of the non-parallelism of two vectors; the spread between non-null vectors v and u is the number (vu)2 s (v, u) 1 . ≡ − QvQu This is the replacement in rational trigonometry for the transcendental notion of angle θ. The spread s (v, u) is unchanged if either v or u are multiplied by a non-zero number, and so we define the spread between any non-null lines l and m with direction vectors v and u to be s (l, m) s (v, u). From the Parallel vectors theorem, the spread ≡ between parallel lines is 0. Two non-null lines l and m are perpendicular precisely when the spread between them is 1. A circle is given by an equation of the form Q (A, X) = K for some fixed point A called the center, and a number K called the quadrance. Note that it is not required that a circle have any points X lying on it: in this case by enlarging the field to a quadratic extension we can guarantee that it does. The three particular planar geometries we are most interested in come from the blue, red and green bilinear forms given by the respective matrices 1 0 1 0 0 1 Cb Cr and Cg . ≡ 0 1! ≡ 0 1! ≡ 1 0! − The corresponding formulas for the blue, red and green quadrances between points

A1 [x1, y1] and A2 [x2, y2] are ≡ ≡ 2 2 Qb (A1,A2) = (x2 x1) + (y2 y1) − − 2 2 Qr (A1,A2) = (x2 x1) (y2 y1) − − − Qg (A1,A2) = 2 (x2 x1)(y2 y1) . − −

2.3 Triple spread formula

We now derive one of the basic formulas in the subject: the relation between the three spreads made by three (coplanar) vectors, and give a linear algebra proof, following the same lines as the papers [36] and [41].

Theorem 2 (Triple spread formula) Suppose that v1, v2, v3 are (planar) non-null vectors with respective spreads s1 s (v2, v3), s2 s (v1, v3) and s3 s (v1, v2) . Then ≡ ≡ ≡ 2 2 2 2 (s1 + s2 + s3) = 2 s1 + s2 + s3 + 4s1s2s3. (2.5) 2. Basic Universal Triangle Geometry 22

Proof. We may assume that at least two of the vectors are linear independent, as otherwise all spreads are zero and the relation is trivial. So suppose that v1 and v2 linearly independent, and v3 = kv1 + lv2. Suppose the bilinear form is given by the matrix a b C = b c! with respect to the ordered basis v1, v2. Then in this basis v1 = (1, 0) , v2 = (0, 1) and v3 = (k, l) and we may compute that

ac b2 l2 ac b2 k2 ac b2 s = − s = − s = − . 3 ac 2 a (ak2 + 2bkl + cl2) 1 b (ak2 + 2bkl + cl2) Then (2.5) is an identity, satisﬁed for all a, b, c, k and l. We now mention three consequences of the Triple spread formula, taken from

[34]. The Equal spreads theorem asserts that if s1 = s2 = s, then s3 = 0 or 2 2 2 2 2 s3 = 4s (1 s). This follows from the identity (s + s + s3) 2 s + s + s3 4s s3 = − 2 − − s3 s3 4s + 4s . The Complementary spreads theorem asserts that if s3 = 1 − − thens1 + s2 = 1. This follows by rewriting the Triple spread formula in the form 2 (s3 s1 s2) = 4s1s2 (1 s3). − − − And the Perpendicular spreads theorem asserts that if v and u are non-null planar vectors with perpendicular vectors v⊥ and u⊥, then s (v, u) = s v⊥, u⊥ . This follows from the Complementary spreads theorem, since if s v, v⊥ = s u, u⊥ = 1, then s v , u = 1 s v , u = 1 (1 s (v, u)) = s (v, u). ⊥ ⊥ − ⊥ − − 2.4 Altitudes and orthocenters

Given a line l and a point P, there is a unique line n through P which is perpendicular to the line l; it is the line n : P +λw, where w is a perpendicular vector to the direction vector v of l. We call n the altitude to l through P. Note that this holds true even if l is a null line; in this case a direction vector v of l is null, so the altitude to l through P agrees with the parallel to l through P. We use the following conventions: a set A, B of two distinct points is a side and { } is denoted AB, and a set l, m of two distinct lines is a vertex and is denoted lm. { } A set A1,A2,A3 of three distinct non-collinear points is a triangle and is denoted { } A1A2A3. The triangle A1A2A3 has lines l3 A1A2, l2 A1A3 and l1 A2A3 (by ≡ ≡ ≡ assumption no two of these are parallel), sides A1A2, A1A3 and A2A3, and vertices l1l2, l1l3 and l2l3.

The triangle A1A2A3 also has three altitudes n1, n2, n3 passing through A1,A2,A3 and perpendicular to the opposite lines A2A3,A1A3,A1A2 respectively. The following 2. Basic Universal Triangle Geometry 23 holds both for aﬃ ne and projective geometries: we give a short and novel proof here for the general aﬃ ne case.

Theorem 3 (Orthocenter) For any triangle A1A2A3 the three altitudes n1, n2, n3 are concurrent at a point H.

Proof. Suppose that a1, a2, a3 are the associated position vectors to A1,A2,A3 respectively. Since no two of the lines of the triangle A1A2A3 are parallel, the Perpendicular spreads theorem implies that no two of the three altitude lines are parallel. Deﬁne H to be the meet of n1 and n2, with h the associated position vector. In the identity

(h a1)(a3 a2) + (h a2)(a1 a3) = (h a3)(a1 a2) − − − − − − the left hand side equals 0 by assumption, so the right hand is also equal to 0, implying that h a3 is perpendicular to the line a1a2. Therefore, the three altitude lines n1, n2, n3 − are concurrent at the point H.

We call H the orthocenter of the triangle A1A2A3.

2.5 Change of coordinates and an explicit example

If we change coordinates via either an affi ne transformation in the original affi ne space A2, or equivalently a linear transformation in the associated vector space V2, then the matrix for the form changes in the familiar fashion. Suppose φ : V V is a linear → transformation given by an invertible 2 2 matrix M, so that φ (v) = vM = w, with × inverse matrix N, so that wN = v. Define a new bilinear form by ◦ T T T w1 w2 (w1N) (w2N) = (w1N) C (w2N) = w1(NCN )w . (2.6) ◦ ≡ · 2 So the matrix C for the original bilinear form becomes the matrix D NCN T for · ≡ the new bilinear form . ◦ Example 4 We illustrate these abstractions in a concrete example that will be used throughout in our diagrams. Our basic Triangle shown in Figure 2.1 has points A1 ≡ [3, 1], A2 [4, 4] and A3 [47/5, 29/5], and lines A1A2 = 3 : 1 : 8 , A1A3 = ≡ ≡ h− i 3 : 4 : 5 and A2A3 = 1 : 3 : 8 . The bilinear form we will consider is that of green h− i h − i geometry in the language of chromogeometry ([38], [39]), determined by the symmetric 0 1 matrix Cg = and corresponding quadrance Q(x,y) = 2xy. After translation by 1 0! 2. Basic Universal Triangle Geometry 24

Figure 2.1: Euler line in green geometry

( 3, 1) we obtain A1 = [0, 0], A2 = [1, 3], A3 = [32/5, 24/5] . The matrix N and its − − inverse M e e e 1 5 1 3 1 3 24 N = 32 24 M = N − = −4 5 ! ! 5 5 9 − 72 send [1, 0] and [0, 1] to A2 and A3, and A2 and A3 to [1, 0] and [0, 1] respectively. So the eﬀect of translation followed by multiplication by M is to send the original triangle e e e e to the standard triangle with points [0, 0] , [1, 0] and [0, 1]. T The bilinear form in these new standard coordinates is given by the matrix NCgN which is, up to a multiple,

1 4 1 a b C = 64 = . 1 25 ! b c!

We will shortly see that the Orthocenter in standard coordinates is

2 1 ac b − [b (c b) , b (a b)] . − − − In our example this would be the point 13 , 25 , and to convert that back into the − 3 12 original coordinates, we would multiply by N to get

13 25 N = 9 3 − 3 12 − h i h i and translate by (3, 1) to get the original orthocenter H = [12, 2] . This is shown − in Figure 2.1, along with the Centroid G = [82/15, 18/5] and the Circumcenter C = [11/5, 32/5]– we will meet these points shortly. 2. Basic Universal Triangle Geometry 25

2.6 Bilines

A biline of the non-null vertex l1l2 is a line b which passes through l1l2 and satisﬁes s(l1, b) = s(l2, b). The existence of bilines depends on number-theoretical considera- tions of a particularly simple kind.

Theorem 5 (Vertex bilines) If v and u are linearly independent non-null vectors, then there is a non-zero vector w with s (v, w) = s (u, w) precisely when 1 s (v, u) is − a square. In this case we may renormalize v and u so that Qv = Qu, and then there are exactly two possibilities for w up to a multiple, namely v + u and v u, and these − are perpendicular.

Proof. Since v and u are linearly independent, any vector can be written uniquely as w = kv + lu for some numbers k and l. The condition s (v, w) = s (u, w) amounts to

2 2 (vw) (uw) 2 2 = u2 kv2 + lvu = v2 kvu + lu2 QvQw QuQw ⇐⇒ 2 2 u2 k2 v2 + 2lkv2 (vu) + l2 (vu )2 = v2 k2 (vu)2 + 2lku2 (vu) + l2 u2 ⇐⇒ 2 2 k2u2 v2 + l2u2 (vu)2 = k2v2 (vu)2+ l2v2 u2 ⇐⇒ v2u2 (vu)2 k2v2 l2u2 = 0. ⇐⇒ − − Since v and u are by assumption not parallel, the ﬁrst term is non-zero by the Parallel vector theorem, and so the condition s (v, w) = s (u, w) is equivalent to k2v2 = l2u2. Since v, u are non-null, v2 and u2 are non-zero, so k and l are also, since by assumption w = kv + lu is non-zero. So if s (v, w) = s (u, w) then we may renormalize v and u so that v2 = u2 (by for example setting v = kv and u = lu, and then replacing v, u by v, u again), and then 1 s(v, u) = (vu)2 / v2 2 is a square. There are then two solutions: w = v + u and − w = v u, correspondinge to le= k. Since (v + u)(v u)e =ev2 u2 = 0, these vectors − ± − − are perpendicular. The converse is straightforward along the same lines.

Example 6 In our example triangle of Figure 2.1, v1 = −−−→A2A3 = (27/5, 9/5) , v2 =

−−−→A1A3 = (32/5, 24/5) and v3 = −−−→A1A2 = (1, 3) , so

T 2 v2Cgv 25 s (v , v ) = 1 3 = 2 3 T T 16 − v2Cgv2 v3Cgv3 T is a square, so the vertex at A1 has bilines. Since Qv2 = v2Cgv2 = 1536/25 and Qv3 = T v3Cgv3 = 6, we can renormalize v2 by scaling it by 5/16 to get u2 = −−→A1B = (2, 3/2) so that now Qu = Qv . This means that u2 + v3 = −−−→A1C1 and u2 v3 = −−−→A1C2 are the 2 3 − 2. Basic Universal Triangle Geometry 26

Figure 2.2: Construction of Green bilines b at A1

direction vectors for the bilines of the vertex at A1. These are shown in Figure 2.2, along with three of the four Incenters I (the other two vertices also have bilines, and they are mutually concurrent). Naturally this triangle has been chosen carefully to ensure that Incenters do exist. In green geometry, a vertex formed from a light-like line and a time-like line will not have bilines, not even approximately over the rational numbers.

2.7 Standard coordinates and triangle geometry

Our principle strategy to study triangle geometry is to apply an aﬃ ne transformation to move a general triangle to standard position:

A1 = [0, 0] A2 = [1, 0] and A3 = [0, 1] . (2.7)

With this convention, A1A2A3 will be called the (standard) Triangle, with Points

A1,A2,A3. The Lines of the Triangle are then

l1 A2A3 = 1 : 1 : 1 l2 A1A3 = 1 : 0 : 0 l3 A2A1 = 0 : 1 : 0 . ≡ h − i ≡ h i ≡ h i All further objects that we define with capital letters refer to this standard Triangle, and coordinates in this framework are called standard coordinates. In general the standard coordinates of points and lines in the plane of the original triangle depend on the choice of affi ne transformation– we are, in principle, free to permute the vertices– but triangle centers and central lines will have well-defined standard coordinates independent of such permutations. Since we have performed an affi ne transformation, whatever metrical structure we started with has changed as in (2.6). So we will assume that the new metrical structure, 2. Basic Universal Triangle Geometry 27 in standard coordinates, is determined by a bilinear form with generic symmetric matrix a b C . (2.8) ≡ b c! We assume that the form is non-degenerate, so that the determinant

∆ det C = ac b2 ≡ − is non-zero. Another important number is the mixed trace

d a + c 2b. ≡ − It will also be useful to introduce the closely related secondary quantities

a c b b a c c a b ≡ − ≡ − ≡ − to simplify formulas. For example d = a + c.

Theorem 7 (Standard triangle quadrances and spreads) The quadrances and spreads of A1A2A3 are

Q1 Q(A2,A3) = d Q2 Q(A1,A3) = c Q3 Q(A1,A2) = a ≡ ≡ ≡ and ∆ ∆ ∆ s1 s (A1A2,A1A3) = s2 s (A2A3,A2A1) = s3 s (A3A1,A3A2) = . ≡ ac ≡ ad ≡ cd Furthermore b2 (c)2 (a)2 1 s1 = 1 s2 = 1 s3 = . − ac − ad − cd Proof. Using the deﬁnition of quadrance,

T Q1 Q (A2,A3) = Q = ( 1, 1) C ( 1, 1) = a + c 2b = d ≡ −−−→A2A3 − − − and similarly for Q2 and Q3. Using the deﬁnition of spread,

2 (1, 0) C (0, 1)T s1 s (A1A2,A1A3) = s ((1, 0) , (0, 1)) = 1 ≡ − (1, 0) C(1, 0)T (0, 1)C (0, 1)T 1 ∆ = 1 b2 = − ac ac and similarly for s2 and s3. 2. Basic Universal Triangle Geometry 28

2.7.1 Basic aﬃ ne objects in triangle geometry

We now write down some basic central objects which ﬁgure prominently in triangle geometry, all with reference to the standard triangle A1A2A3 in the form (2.7). The derivations of these formulas are mostly immediate using the two basic operations of joins (2.2) and meets (2.3). We begin with some purely aﬃ ne notions, independent of the bilinear form. The Midpoints of the Triangle are 1 1 1 1 M = , M = 0, M = , 0 . 1 2 2 2 2 3 2 The Medians are d1 A1M1 = 1 : 1 : 0 d2 A2M2 = 1 : 2 : 1 d3 A3M3 = 2 : 1 : 1 . ≡ h − i ≡ h − i ≡ h − i The Centroid is the common meet of the Medians 1 1 G = , . 3 3

Figure 2.3: The Medial triangle M1M2M3 and Double triangle D1D2D3

The Circumlines are the lines of the Medial triangle M1M2M3, these are b1 M2M3 = 2 : 2 : 1 b2 M3M1 = 2 : 0 : 1 b3 M1M2 = 0 : 2 : 1 . ≡ h − i ≡ h − i ≡ h − i

The Double triangle of A1A2A3 (usually called the anti-medial triangle) is formed from lines through the Points parallel to the opposite Lines. This is D1D2D3 where

D1 = [1, 1] D2 = [ 1, 1] D3 = [1, 1] . − −

The lines of D1D2D3 are

D2D3 = 1 : 1 : 0 D1D3 = 1 : 0 : 1 D1D2 = 0 : 1 : 1 . h i h − i h − i Figure 2.3 shows these objects for our example Triangle. 2. Basic Universal Triangle Geometry 29

2.7.2 The Orthocenter hierarchy

We now introduce some objects involving the metrical structure, and so the entries a, b, c of C from (2.8). Recall that a c b and c = a b. ≡ − − The Altitudes of A1A2A3 are the lines

n1 = c : a : 0 n2 = b : c : b n3 = a : b : b . h − i h − i h − i Theorem 8 (Orthocenter formula) The three Altitudes meet at the Orthocenter b b H = [c b, a b] = [a, c] . (2.9) ac b2 − − ∆ − Proof. We know that the altitudes meet from the Orthocenter theorem. We check that n1 passes through H by computing

1 b∆− (ac ac) = 0. −

Also n2 passes through H since b b (ba + cc) b = b (c b) + c (a b) ac + b2 = 0 ∆ − ∆ − − − and similarly for n3.

The Midlines m1, m2 and m3 are the lines through the midpoints M1,M2 and

M3 perpendicular to the respective sides– these are usually called perpendicular bisectors. They are also the altitudes of M1M2M3:

m1 = 2c : 2a : b m2 = 2b : 2c : c m3 = 2a : 2b : a . − h − i h − i

Theorem 9 (Circumcenter) The Midlines m1, m2, m3 meet at the Circumcenter 1 1 C = [c (a b) , a (c b)] = [cc, aa] . 2 (ac b2) − − 2∆ − Proof. We check that m1 passes through C by computing 1 1 2c2c + 2a2a + b = 2 (a b)2 c + 2 (c b)2 a + (a c) = 0 2∆ − 2 (ac b2) − − − − − and similarly for m2 and m3. As Gauss realized, this is also a consequence of the Orthocenter theorem applied to the Medial triangle M1M2M3, since the altitudes of the Medial triangle are the Midlines of the original Triangle.

The three altitudes of the Double triangle D1D2D3 are

t1 = c : a : b t2 = b : c : a t3 = a : b : c . − − h − i h − i

2. Basic Universal Triangle Geometry 30

Figure 2.4: The Euler line of a triangle

Theorem 10 (Double orthocenter formula) The three altitudes of the Double triangle meet in the De Longchamps point

1 2 2 X20 b 2bc + ac, b 2ab + ac . ≡ ∆ − − Proof. We check that t1 passes through X20 by computing 1 c b2 2bc + ac a b2 2ab + ac b ∆ − − − − 1 = (a b) b2 2bc +ac (c b) b2 2ab + ac (a c) ac b2 = 0 ∆ − − − − − − − − and similarly for t2 and t3. The existence of an Euler line in relativistic geometries was established in [39], here we extend this to the general case.

Theorem 11 (Euler line) The points H,C and G are collinear, and satisfy G = 1 H + 2 C. The Euler line e CH is 3 3 ≡ e = ∆ 3bc : ∆ + 3ba : bb . − −

Proof. Using the formulas above for H and C, we see that

1 2 1 1 2 1 1 1 H + C = [ba, bc] + [cc, aa] = ac b2, ac b2 = [1, 1] = G. 3 3 3 ∆ 3 2∆ 3∆ − − 3 Computing the equation for the Euler line CH is straightforward. In Figure 2.4 we illustrate the situation with our basic example triangle with the Altitudes, Medians and Midlines meeting to form the Orthocenter H, Centroid G and Circumcenter C respectively on the Euler line e. 2. Basic Universal Triangle Geometry 31

The bases of altitudes of M1M2M3 are: 1 1 1 E = [c, a] E = [c, a] E = [c, a] . 1 2d 2 2c 3 2a

The joins of Points and corresponding bases of altitudes of M1M2M3 are

A1E1 = a : c : 0 A2E2 = a : c : a A3E3 = a : c : c . h − i h − i h − i

Theorem 12 (Medial base perspectivity) The three lines A1E1,A2E2,A3E3 meet at the point 1 X = [c, a] . 69 a + c b − Proof. Straightforward. The Nine-point center is the midpoint of Circumcenter and Orthocenter

1 1 (∆ + ba) (∆ + bc) N = H + C = , = X . 2 2 4∆ 4∆ 5 Theorem 13 For any triangle

Q (A1N) + Q (A2N) + Q (A3N) = 3Q (A1C) Q (C,N) − Proof. We have a2c + 6abc + ac2 8b3 9a2c 8ab2 10abc + ac2 + 8b3 Q (A N) = − Q (A N) = − − 1 16 (ac b2) 2 16 (ac b2) − − a2c 10abc + 9ac2 + 8b3 8b2c (a 2b + c) ac Q (A N) = − − Q (A C) = − and 3 16 (ac b2) 1 4 (ac b2) − − a2c + 8ab2 10abc + ac2 8b3 + 8b2c Q (C,N) = − − 16 (ac b2) − therefore

Q (A1N) + Q (A2N) + Q (A3N) = 3Q (A1C) Q (C,N) − 11a2c 8ab2 14abc + 11ac2 + 8b3 8b2c = − − − . 16 (ac b2) − 2.8 Transformations

Important classical transformations of points associated to a triangle include dilations in the centroid, and the isogonal and isotomic conjugates. It is useful to have general formulae for these mappings in our standard coordinates. 2. Basic Universal Triangle Geometry 32

2.8.1 Dilations about the Centroid

The dilation δ of factor λ centered at the origin takes [x, y] to λ [x, y] . This also acts on vectors by scalar multiplying, and in particular it leaves spreads unchanged and multiplies any quadrance by a factor of λ2. Similarly the dilation centered at a point A takes a point B to A + λ−−→AB. Any dilation preserves directions of lines, so preserves spreads, and changes quadrances between points proportionally.

Given our Triangle A1A2A3 with centroid G, deﬁne the central dilation δ 1/2 − to be the dilation by the factor 1/2 centered at G. It takes the three Points of the − Triangle to the midpoints M1,M2,M3 of the opposite sides. This medial triangle

M1M2M3 then clearly has lines which are parallel to the original triangle.

Since the central dilation preserves spread, the three altitudes of A1A2A3 are sent by δ 1/2 to the three altitudes of the medial triangle, which are the midlines/perpendicular − bisectors of the original Triangle, showing again that δ 1/2 sends the orthocenter H to − the circumcenter C, and as in the Euler line theorem it follows that G lies on e = HC, dividing HC in the aﬃ ne ratio 2 : 1. We will see later that the central dilation also explains aspects of the various Nagel lines (there are four), since δ 1/2 takes any Incenter Ii to an incenter of the Medial − triangle, called a Spieker point Si. It follows that the four joins of Incenters and corresponding Spieker points all pass through G, and G divides each side IiSi in the aﬃ ne ratio 2 : 1.

The inverse of the central dilation δ 1/2 is δ 2, which takes the Points of A1A2A3 − − to the points of the Double triangle D1D2D3, which has A1A2A3 as its medial triangle.

Theorem 14 (Central dilation formula) The central dilation takes X = [x, y] to 1 δ 1/2 (X) = [1 x, 1 y] − 2 − − while the inverse central dilation δ 2 takes X to δ 2 (X) = [1 2x, 1 2y] . − − − − 1 2 Proof. If Y = δ 1/2 (X) then aﬃ nely 3 X + 3 Y = G so that − 3 1 1 Y = G X = [1 x, 1 y] . 2 − 2 2 − −

Inverting, we get the formula for δ 2 (X) . − Example 15 The central dilation of the Orthocenter is

1 b (c b) b (a b) 1 1 δ 1/2 (H) = 1 − , 1 − = [c (a b) , a (c b)] = [cc, aa] = C − 2 − ∆ − ∆ 2∆ − − 2∆ which is the Circumcenter. 2. Basic Universal Triangle Geometry 33

Example 16 The inverse central dilation of the Orthocenter is the De Longchamps point X20– the orthocenter of the Double triangle D1D2D3

2b (c b) 2b (a b) 1 2 2 δ 2 (H) = X20 1 − , 1 − = b 2cb + ac, b 2ab + ac . − ≡ − ∆ − ∆ ∆ − − 2.8.2 Reﬂections and Isogonal conjugates

Suppose that v is a non-null vector, so that v is not perpendicular to itself. It means that we can find a perpendicular vector w so that v and w are linearly independent. Now if u is an arbitrary vector, write u = rv + sw for some unique numbers r and s, and define the reflection of u in v to be

rv (u) rv sw. ≡ − If we replace v with a multiple, the reﬂection is unchanged. Now suppose that l and m are lines which meet at a point A, with respective direction vectors v and u. Then the reﬂection of m in l is the line through A with direction vector rv (u). It is important to note that if n is the perpendicular to l through A, then

rl (m) = rn (m) .

Our standard triangle A1A2A3 determines an important transformation of points.

Theorem 17 (Isogonal conjugate) If X is a point distinct from A1,A2,A3, then the reflections of the lines A1X,A2X,A3X in the bilines at A1,A2,A3 respectively meet in a point i (X), called the isogonal conjugate of X. If X = [x, y] then x + y 1 i (X) = − [cy, ax] . ax2 + 2bxy + cy2 ax cy − − Proof. First we reflect the vector a = (x, y) in the bilines v : w : 0 and v : w : 0 h i h − i through A1. We do this by writing (x, y) = r (w, v) + s (w, v) = (rw + sw, rv sv) 1 1 − − and solving to get r = (2vw)− (vx + wy) and s = (2vw)− (vx wy) . The reflection − is then 1 1 wy vx r (w, v) s (w, v) = (vx + wy)(w, v) (vx wy)(w, v) = , − − 2vw − 2vw − − v w which is, up to a multiple and using the quadratic relations,

w2y, v2x = (cdy, adx) = d (cy, ax) .

So reflectionin the biline at A1 takes the line A1X to the line A1+λ1 (cy, ax) . Similarly, by computing the reflectionsof (x 1, y) and (x, y 1) in the bilines at A2 and A3, we − − find that the lines A2X and A3X get sent to the lines A2+λ2 (ax + (a d) y a, ax ay + a) − − − − and A3 + λ3 ( cx cy + c, x (c d) + cy c) respectively. It is now a computation − − − − that these three reflected lines meet at the point i (X) as defined above. 2. Basic Universal Triangle Geometry 34

Example 18 The isogonal conjugate of the centroid G is the symmedian point

1 1 1 K i , = [c, a] = X6. ≡ 3 3 2 (a + c b) − Example 19 The isogonal conjugate of the Orthocenter H is the Circumcenter:

b (c b) b (a b) 1 i − , − = [c (a b) , a (c b)] = C = X3. ∆ ∆ 2∆ − −

2.8.3 Isotomic conjugates

Theorem 20 (Isotomic conjugates) If X is a point distinct from A1,A2,A3, then the lines joining the points A1,A2,A3 to the reﬂections in the midpoints M1,M2,M3 of the meets of A1X,A2X,A3X with the lines of the Triangle are themselves concurrent, meeting in the isotomic conjugate of X. If X = [x, y] then

y (x + y 1) x (x + y 1) t (X) = − , − . x2 + xy + y2 x y x2 + xy + y2 x y − − − −

Proof. The point X [x, y] has Cevian lines which meet the lines A2A3,A1A3,A1A2 ≡ respectively in the points

x y y x , 0, , 0 . x + y x + y 1 x 1 y − − These three points may be reﬂectedrespectively in the midpoints [1/2, 1/2] , [0, 1/2] , [1/2, 0] to get the points

y x 1 x y 1 x y , 0, − − − − , 0 . x + y x + y 1 x 1 y − − The lines x : y : 0 , 1 x y : 1 x : 1 + x + y and 1 y : 1 x y : 1 + x + y h − i h − − − − i h − − − − i joining these points to the original vertices meet at t (X) as deﬁned above.

Example 21 The isotomic conjugate of the Orthocenter H is

(c b) b (a b) b a b c b t − , − = − , − X69. ac b2 ac b2 a + c b a + c b ≡ − − − − 2.9 A Kimberling 6-9-13 triangle

We now introduce a Kimberling 6-9-13 triangle which has quadrances of sides (a, b, c) = (36, 81, 169) in Kimberling notations. The following example shows how to use a Kimberling triangle to check if the point X is in the Encyclopedia of Triangle Centers. 2. Basic Universal Triangle Geometry 35

Figure 2.5: The Euler line and X140 in blue geometry

Example 22 Our 6-9-13 triangle shown in Figure 2.5 has points A 4 √35, 13 , ≡ 3 − 3 B [0, 6] and C [0, 0] and lines a BC, b AC and c AB. ≡ ≡ ≡ ≡ ≡ The bilinear form we will consider is that of Euclidean (blue) geometry, determined

1 0 2 2 by the symmetric matrix Cb = and corresponding quadrance Q(x,y) = x + y . 0 1! The matrix N and its inverse M

13 3 0 6 1 840 √35 140 √35 N = 4 13 M = N − = 1 √35 ! 0 ! 3 − 3 6 send [1, 0] and [0, 1] to B and A, and B and A to [1, 0] and [0, 1] respectively. So the eﬀect of translation followed by multiplication by M is to send the original triangle to the standard triangle with points [0, 0] , [1, 0] and [0, 1]. T The bilinear form in these new standard coordinates is given by the matrix NCbN which is, up to a multiple,

36 26 a b D = − = . 26 81 ! b c! −

Now we check if the point X, which is the midpoint of the Circumcenter X3 and nine- point center X5, is in Encyclopedia of Triangle Centers. The point X in standard 1 coordinates is 8 ac b2 − 3ac bc 2b2 , 3ac ab 2b2 . In our example this −4751 2083 − − − − would be the point 8960 ,4480, and to convert that back into the original coordinates, 2. Basic Universal Triangle Geometry 36 we would multiply by N to get

4751 2083 , N = 2083 √35 7 = [3.667 617 317 915 63, 1.166 666 666666 67] . 8960 4480 3360 6 h i The x-coordinate of X tells us the distance from X to the side line a = BC. Now we check in the Encyclopedia of Triangle Centers and get the point X140 which has the same distance to the side line a as the point X. Therefore the midpoint of the

Circumcenter X3 and nine-point center X5 is X140. In general, if we have a point 4 X which has standard coordinates [x, y], we just look up the number 3 y√35 in the Encyclopedia of Triangle Centers. Chapter 3

Universal aﬃ ne triangle geometry and four-fold incenter symmetry

3.1 Introduction

In this chapter, we develop a generalized triangle geometry, using an arbitrary bilinear form in an affi ne plane over a general field. By using standardized coordinates we find canonical forms for some basic centers and lines. Strong concurrencies formed by quadruples of lines from the Incenter hierarchy are investigated, including joins of corresponding Incenters, Gergonne, Nagel, Spieker points, Mittenpunkts, Bevan points, Weill points, and so on. The diagrams are taken from relativistic (green) geometry. We summarize some main results of this chapter using Figure 3.1 from green geometry. As established in ([39]), the triangle A1A2A3 has a green Euler line CHG just as in the Euclidean setting, where C = X3 is the Circumcenter, G = X2 is the Centroid, and H = X4 is the Orthocenter, with the affi ne ratio −−→CG : −−→GH = 1 : 2, which we 2 1 may express as G = 3 C + 3 H. The reader might like to check that using the green notation of perpendicularity, the green altitudes really do meet at H, and the green midlines/perpendicular bisectors really do meet at C.

In the general situation there are four Incenters/Excenters I0,I1,I2 and I3 which algebraically are naturally viewed symmetrically. Associated to any one Incenter Ij is a Gergonne point Gj = X7, a Nagel point Nj = X8, a Mittenpunkt Dj = X9, a

Spieker point Sj = X10, a Bevan point Lj = X40, a Weill point Wj = X354 and so on. It is not at all obvious that these various points can be deﬁned for a general aﬃ ne geometry, but this is the case, as we shall show.

37 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 38

The four-fold symmetry between the four Incenters is maintained by all these points: so in fact there are four Gergonne, Nagel, Mittenpunkt, Spieker, Bevan points, Weill points and so on, each associated to a particular Incenter, as also pointed out in ([25]). Figure 3.1 shows just one Incenter and its related hierarchy: as we proceed in this chapter the reader will meet the other Incenters and hierarchies as well.

Figure 3.1: Aspects of the Incenter hierarchy in green geometry

The main aims of the chapter are to set-up a coordinate system for triangle geometry that incorporates the number-theoretical aspects of the Incenter hierarchy, and respects the four-fold symmetry inherent in it, and then to use this to catalogue ex- isting points and phenomena. Kimberling’s Triangle Center Encyclopedia ([19]) distinguishes the classical In- center X1 as the first and perhaps most important triangle center. Our embrace of the four-fold symmetry between incenters and excenters implies something of a re-evaluation of some aspects of classical triangle geometry; instead of certain distinguished centers we have rather distinguished quadruples of related points. Somewhat surprisingly, this point of view makes visible a number of remarkable strong concurrences– where four symmetrically-defined lines meet in a center. The proofs of these relations are reasonably straight-forward but not automatic, as in general certain important quadratic relations are needed to simplify expressions for incidence. Here is a summary of some of our main results which are reported in ([23]) . 3. Universal affi ne triangle geometry and four-fold incenter symmetry 39

Main Results i) The four lines IjGj, j = 0, 1, 2, 3, meet in the De Longchamps point X20 (orthocenter of the Double triangle)– these are the Soddy lines ([27]). 2 1 ii) The four lines IjNj meet in the Centroid G = X2, and in fact G = 3 Ij + 3 Nj – these are the Nagel lines. The Spieker points Sj also lie on the Nagel lines, and in 1 1 fact Sj = 2 Ij + 2 Nj. iii) The four lines IjDj meet in the Symmedian point K = X6 (isogonal conjugate of the Centroid G)– the standard such line is labelled L1,6 in [19]. 1 1 iv) The four lines IjLj meet in the Circumcenter C, and in fact C = 2 Ij + 2 Lj – the standard such line is labelled L1,3.

v) The four lines GjNj meet in the point X69 (isotomic conjugate of the Ortho- center H) – these lines are labelled L7,8. 2 1 vi) The four lines GjDj meet in the Centroid G = X2, and in fact G = 3 Dj + 3 Gj – the standard such line is labelled L2,7.

vii) The four lines DjSj meet in the Orthocenter H = X4 – the standard such line is labelled L4,7.

viii) The four lines NjLj meet in the point X20 (orthocenter of the Double triangle), 1 1 and in fact Lj = 2 X20 + 2 Nj – the standard such line is labelled L1,3. ix) The Bevan point Lj lies on the line DjSj which also passes through the Ortho- 1 1 center H, and in fact Sj = 2 H + 2 Lj. x) The four lines IjWj meet in the Circumcenter C – the standard such line is labelled L1,3.

xi) The four lines WjLj meet in the Circumcenter C – the standard such line is labelled L1,3. In particular the various points alluded to here have consistent definitions over general fields and with arbitrary bilinear forms! The Bevan points are the meets of the lines L1,3 and L4,7; they are the reflections of the Incenters Ij in the Circumcenter

C, and they are the reflections of the Orthocenter H in the Spieker points Sj. It is also worth pointing out a few additional relations between the triangle centers that appear here: the point X69, defined as the Isogonal conjugate of the Orthocenter H, is also the central dilation in the Centroid of the Symmedian point K; in our 2 1 notation X69 = δ 1/2 (K) . This implies that G = 3 K + 3 X69. In addition the De − Longchamps point X20, defined as the orthocenter of the Double (or anti-medial) triangle is also the reflection of the Orthocenter H in the Circumcenter C. These relations continue to hold in the general situation. 3. Universal affi ne triangle geometry and four-fold incenter symmetry 40

3.2 Bilines and Incenters

We now introduce the Incenter hierarchy. Unlike the Orthocenter hierarchy, this depends on number-theoretical conditions. Recall that d a + c 2b. ≡ −

Theorem 23 (Existence of Triangle bilines) The Triangle A1A2A3 has Bilines at each vertex precisely when we can ﬁnd numbers u, v, w in the ﬁeld satisfying

ac = u2 ad = v2 cd = w2. (3.1)

Proof. From the Vertex bilines theorem, bilines exist precisely when the spreads s1, s2, s3 of the Triangle have the property that 1 s1, 1 s2, 1 s3 are all squares. From − − − the Standard triangle quadrances and spreads theorem, this occurs for our standard triangle A1A2A3 precisely when we can ﬁnd u, v, w satisfying (3.1). There is an important ﬂexibility here: the three Incenter constants u, v, w are only determined up to a sign. The relations imply that

d2u2 = v2w2 c2v2 = u2w2 a2w2 = u2v2.

So we may choose the sign of u so that du = vw, and multiplying by u we get

acd = uvw.

From this we deduce that

du = vw cv = uw and aw = uv. (3.2)

The quadratic relations (3.1) and (3.2) will be very important for us, for they reveal that the existence of the Incenter hierarchy is a number-theoretical issue which depends not only on the given triangle and the bilinear form, but also on the nature of the ﬁeld over which we work, and they allow us to simplify many formulas involving u, v and w. Because only quadratic conditions are involved, we may always extend our ﬁeld by adjoining (algebraic!) square roots to ensure that a given triangle has bilines. The quadratic relations carry an important symmetry: we may replace any two of u, v and w with their negatives, and the relations remain unchanged. So if we have a formula F0 involving u, v, w, then we may obtain related formulas F1,F2,F3 by replacing v, w with their negatives, u, w with their negatives, and u, v with their negatives respectively. Adopting this convention allows us to exhibit the single formula

F0, since then F1,F2,F3 are determined– we refer to this as quadratic symmetry, and will make frequent use of it in the rest of this chapter.

From now on our working assumption is that: the standard triangle A1A2A3 has bilines at each vertex, implying that we have Incenter constants u, v and w satisfying 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 41

(3.1) and (3.2). So u, v and w now become ingredients in our formulas for various objects in the Incenter hierarchy, along with the numbers a, b and c (and d) from the a b bilinear form C = . b c!

Theorem 24 (Bilines) The Bilines of the Triangle are b1+ v : w : 0 and b1 ≡ h i − ≡ v : w : 0 through A1, b2+ u : u + w : u and b2 u : u w : u through h − i ≡ h − i − ≡ h − − i A2, and b3+ u v : u : u and b3 u + v : u : u through A3. ≡ h − − i − ≡ h − i

Proof. We use the Bilines theorem to ﬁnd bilines through A1 = [0, 0] . The lines meet- T ing at A1 have direction vectors v1 = (0, 1) and v2 = (1, 0) , with Qv1 = (0, 1) C (0, 1) = T v c and Qv2 = (1, 0) C (1, 0) = a. Now we renormalize and set u1 = w v1 to get v2 Qu1 = w2 c = a = Qv2 . So the bilines at A1 have direction vectors v v v v u1 + v2 = (0, 1) + (1, 0) = 1, and u1 v2 = (0, 1) (1, 0) = 1, w w − w − − w and the bilines are b1+ v : w : 0 and b1 v : w : 0 . Similarly we may check ≡ h i − ≡ h − i the other bilines through A2 and A3.

Theorem 25 (Incenters) The triples b1+, b2+, b3+ , b1+, b2 , b3 , b1 , b2+, b3 { } { − −} { − −} and b1 , b2 , b3+ of Bilines are concurrent, meeting respectively at the four Incen- { − − } ters uw uv 1 I0 = − , = [ w, v] uv uw + vw uv uw + vw (d + v w) − − − − uw uv 1 I1 = , − = [w, v] uv + uw + vw uv + uw + vw (d v + w) − − − − uw uv 1 I = , = [w, v] and 2 uv + uw + vw uv + uw + vw (d + v + w) uw uv 1 I3 = , = [ w, v] . uv + uw vw uv + uw vw (d v w) − − − − − − Proof. We may check concurrency of the various triples by computing

v w 0 v w 0 det u u + w u = det u u w u  −   − −  u v u u u + v u u  − −   −      v w 0 v w 0 − − = det u u w u = det u u + w u = 0.  − −   −  u v u u u + v u u  − −   −      3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 42

The corresponding meet of v : w : 0 , u : u + w : u and u v : u : u is h i h − i h − − i uw uv b b b = − , 1+ 2+ 3+ uv uw + vw uv uw + vw −u − = [ w, v] aw cv + du − −1 = [ w, v] I0. (d + v w) − ≡ − We have used the quadratic relations (3.1) and (3.2), and the last equality is valid since u (d + v w) (aw cv + du) = cv aw + uv uw = 0. − − − − − The computations are similar for the other Incenters.

The reader should check that the formulas for I1,I2,I3 may also be obtained from

I0 by the quadratic symmetry rule described above. From now on in such a situation we will only write down the formula corresponding to I0, and we will also often omit algebraic manipulations involving the quadratic relations.

Figure 3.2: Green bilines b, Incenters I, Contact points and Incircles

The Incenter altitude tij is the line through the Incenter Ij and perpendicular to the Line li of our Triangle. There are twelve Incenter altitudes; three associated to each Incenter. The Incenter altitudes associated to I0 are

t10 = c (d + v w): a (d + v w): av + cw h − − − i t20 = b (d + v w): c (d + v w): bw cv h − − − i t30 = a (d + v w): b (d + v w): aw bv . h − − − i Recall that a c b, b a c, c a b. ≡ − ≡ − ≡ − 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 43

The Contact points Cij are the meets of corresponding Incenter altitudes tij and Lines li. There are twelve Contact points; three associated to each Incenter. The

Contact points associated to the Incenter I0 are 1 C10 = [a w, c + v] (d + v w) − − 1 C20 = [0, cv bw] c (d + v w) − − 1 C30 = [bv aw, 0] . a (d + v w) − − In Figure 3.2 we see our standard example Triangle in the green geometry with Bilines b at each vertex, meeting in threes at the Incenters I. The Contact points are also shown, as are the Incircles, which are the circles with respect to the metrical structure centered at the Incenters and passing through the Contact points: they have equations in the variable point X of the form Q (X,I) = Q (C,I) where I is an incenter and C is one of its associated Contact points. In this green geometry such circles appear as rectangular hyperbolas, with axes parallel to the coordinate axes.

3.3 Strong concurrences

3.3.1 Bevan points

In the following section, we introduce a known center of the triangle, the Bevan point.

Figure 3.3: Green Incenter altitudes, Bevan points L and In-New center C 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 44

Theorem 26 (Bevan points) The triples t11, t22,t33 , t10, t23, t32 , t20, t13, t31 { } { } { } and t30, t12, t21 of Incenter altitudes are concurrent. Each triple is associated to the { } Incenter which does not lie on any of the lines in that triple. The points where these triples meet are the Bevan points Li; for example t11, t22,t33 meet at { } 1 L0 = [au + cv + bw + cc, cu bv aw + aa] . 2∆ − −

Proof. We check that L0 as deﬁned is incident with

t11 = c (d v + w): a (d v + w): av cw h − − − − − i by computing

((c b) u + cv + bw + c (a b)) (a b)(a + c 2b v + w) − − − − − + ((a b) u bv aw + a (c b)) ( (c b)(a + c 2b v + w)) − − − − − − − − 2 ac b2 ((c b) v + (a b) w) − − − − 3 3 2 2 2 3 2 2 2 3 2 2 2 2 2 2 = ac + 2ab 2a cb a b ac + 2ac b + c b 2cb + b v b w acv + acw − − − − − − = ac b2 a2 c2 2ab + 2cb v2 + w2 = 0 − − − − using the quadratic relations (3.1). The computations for the other Incenter altitudes and B1,B2,B3 are similar. The In-Bevan lines are the joins of corresponding Incenter points and Bevan points. The In-Bevan line associated to I0 is

2 2 I0L0 = aad + ac + ab 2b v + aaw : ccd + ccv + ac + cb 2b w : aaw ccv . − − − − − Theorem 27 (In-Bevan center) The four In-Bevan lines IjLj are concurrent and meet at the circumcenter 1 C = [cc, aa] , 2∆ and in fact C is the midpoint of IjLj.

Proof. We check that C is the midpoint of IjLj by computing 1 1 1 1 Ij + Lj = [ w, v] 2 2 2 (d + v w) − − 1 1 + [(c b) u + cv + bw + c (a b) , (a b) u bv aw + a (c b)] 2 2∆ − − − − − − 1 1 = [2c (a b)(d + v w) , 2a (c b)(d + v w)] = [cc, aa] = C. 4∆ (d + v w) − − − − 2∆ − The In-Bevan center theorem shows that what we are calling the In-Bevan lines are also the In-Circumcenter lines, the standard one which is labelled L1,3 in [19]. The Incenter altitudes, Bevan points and In-Bevan lines are shown in Figure 3.3. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 45

The proofs in these two theorems are typical of the ones which appear in the rest of the chapter. Algebraic manipulations are combined with the quadratic relations to simplify expressions. Although sometimes long and involved, the veriﬁcations are in principle straightforward, and so from now on we omit the details for results such as these.

3.3.2 Sight Lines, Gergonne and Nagel points

We now adopt the principle that algebraic veriﬁcations of incidence, using the quadratic relations, will be omitted.

A Sight line sij is the join of a Contact point Cij with the Point Ai opposite to the Line that it lies on, and is naturally associated with the Incenter Ij. There are twelve Sight lines; three associated to each Incenter:

s10 = c + v : a + w : 0 s12 = c + v : a w : 0 h − i h − − i s20 = cv bw : c (d + v w): cv + bw s22 = cv + bw : c (d + v + w): cv bw h − − − i h − − i s30 = a (d + v w): aw + bv : aw bv s32 = a (d + v + w): aw + bv : aw bv h − − − i h − − i s11 = c v : a w : 0 s13 = c v : a + w : 0 h − − − i h − − i s21 = cv + bw : c (d v + w): cv bw s23 = cv bw : c (d v w): cv + bw h− − − i h− − − − i s31 = a (d v + w): aw bv : aw + bv s33 = a (d v w): aw bv : aw + bv . h − − − i h − − − − i

Figure 3.4: Green Sight lines, Gergonne points G, In-Gergonne lines and In-Gergonne center X20

Here we introduce a well-known center of the triangle, the Gergonne point (see for example [13], [27]). 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 46

Theorem 28 (Gergonne points) The triples s10, s20, s30 , s11, s21, s31 , s12, s22, s32 { } { } { } and s13, s23, s33 of Sight lines are concurrent. Each triple is associated to an Incen- { } ter, and the meets of these triples are the Gergonne points Gj. The Gergonne point associated to I0 is b u G0 = − [w a, v c] . 2 (du cv + aw) ∆ − − − − − The join of a corresponding Incenter Ij and Gergonne point Gj is an In-Gergonne line or Soddy line. There are four Soddy lines, and

2bcv + (∆ 2bc) w (∆ 2bc) d : I0G0 = − − − . * 2baw + (∆ 2ba) v + (∆ 2ba) d : (∆ 2ba) v (∆ 2bc) w + − − − − − −

Theorem 29 (In-Gergonne center) The four In-Gergonne/Soddy lines IjGj are concurrent, and meet at the De Longchamps point

1 2 2 X20 = b 2cb + ac, b 2ab + ac ∆ − − which is the orthocenter of the Double triangle. Furthermore the midpoint of HX20 is the Circumcenter C, so that X20 lies on the Euler line.

Proof. The concurrency of the In-Gergonne/Soddy lines IjGj is as usual. The equation 1 1 1 [b (c b) , b (a b)]+ b2 2cb + ac, b2 2ab + ac = [c (a b) , a (c b)] = C 2∆ − − 2∆ − − 2∆ − − 1 1 shows that C = 2 H + 2 X20. Since the Euler line is e = CH,X20 lies on e. Figure 3.4 shows the Gergonne points G and the In-Gergonne lines meeting at X20.

Theorem 30 (Nagel points) The triples s11, s22,s33 , s10, s32, s23 , s20, s31, s13 { } { } { } and s30, s21, s12 of Sight lines are concurrent. Each triple involves one Sight line { } associated to each of the Incenters, and so is associated to the Incenter with which it does not share a Sight line. The points where these triples meet are the Nagel points

Nj. For example, s11, s22,s33 meet at { } 1 N0 = [(b + u) a + cv + bw, (b + u) c bv aw] . ∆ − −

Proof. We check that N0 as deﬁned is incident with c v : a w : 0 by computing h − − − i (b + u) a + cv + bw (b + u) c bv aw (c v) + − − ( a w) ∆ − ∆ − − cduv2 + aduw2 + ccv2w + aavw2 acadv accdw = − − − = 0 ∆ 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 47

Figure 3.5: Green Sight lines, Nagel points N, In-Nagel lines and In-Nagel center G = X2 using the quadratic relations, (3.1) and (3.2). The computations for the other Sight lines and N1,N2,N3 are similar. The join of a corresponding Incenter and Nagel point is an In-Nagel line. There are four In-Nagel lines, and

I0N0 = 2v + w d : v + 2w + d : v w . h − − − i

In classical triangle geometry, the line I0N0 is called simply the Nagel line.

Theorem 31 (In-Nagel center) The four In-Nagel lines IjNj are concurrent, and 2 1 meet at the Centroid G = X2, and in fact G = 3 Ij + 3 Nj.

Proof. Using the formulas above for I0 and N0, we see that 2 1 2 1 1 1 I0 + N0 = [ w, v] + [(b + u) a + cv + bw, (b + u) c bv aw] 3 3 3 (d + v w) − 3 ∆ − − − 1 1 = [∆ (d + v w) , ∆ (d + v w)] = [1, 1] = G. 3∆ (d + v w) − − 3 − Note that a Nagel line is also an In-Centroid line which is the joint of an Incenter and the Centroid.

The join of a corresponding Gergonne point Gj and Nagel point Nj is a Gergonne- Nagel line. There are four Gergonne-Nagel lines, and

G0N0 = au + av + aw : cu + cv + cw : cw av . h− − − i 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 48

Figure 3.6: Green Gergonne-Nagel center X69 and Nagel-Bevan center X20

In Figure 3.6, we see the four green Bevan points L , the four green Nagel points N, the four green Gergonne points G, the green Gergonne-Nagel center X69 and Nagel-

Bevan center X20 of A1A2A3.

Theorem 32 (Gergonne-Nagel center) The four Gergonne-Nagel GjNj lines are concurrent, and meet at the isotomic conjugate of the Orthocenter, 1 X = [c, a] . 69 a + c b − The join of a corresponding Bevan point Lj and Nagel point Nj is a Nagel-Bevan line. There are four Nagel-Bevan lines, and

ac 3ab + 2b2 cu + bv + aw : L N = . 0 0 2 − − * 3cb ac 2b + au + cv + bw :(a c) b + (a c) u av + cw + − − − − − Theorem 33 (Nagel-Bevan center) The four Nagel-Bevan lines NjBj meet in the 1 1 De Longchamps point X20, and in fact Bj = 2 Nj + 2 X20.

Proof. We check that

1 1 1 1 2 2 X20 + N0 = b 2cb + ac, b 2ab + ac 2 2 2 ∆ − − 1 1 + [(b + u) a + cv + bw, (b + u) c bv aw] 2 ∆ − − 1 = [au + cv + bw + cc, cu bv aw + aa] = B0. 2∆ − − 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 49

3.3.3 InMid lines and Mittenpunkts

The join of an Incenter Ij with a Midpoint Mi is an InMid line. There are twelve InMid lines:

I0M1 = v + w d : v + w + d : v w I2M1 = v w d : v w + d : v + w h − − − i h − − − − i I0M2 = v + w d : 2w : w I2M2 = v w d : 2w : w h − − i h − − − i I0M3 = 2v : v + w + d : v I2M3 = 2v : v w + d : v h − i h − − i I1M1 = v + w + d : v + w d : v w I3M1 = v + w d : v + w + d : v w h − − − i h− − − − i I1M2 = v + w + d : 2w : w I3M2 = v + w d : 2w : w h − i h− − − i I1M3 = 2v : v + w d : v I3M3 = 2v : v + w + d : v . h − − i h− − i

Theorem 34 (InMid lines) The triples of InMid lines I1M1,I2M2,I3M3 , I0M1,I2M3,I3M2 , { } { } I0M2,I1M3,I3M1 and I0M3,I1M2,I2M1 are concurrent. Each triple involves one { } { } InMid line associated to each of three Incenters, and so is associated to the Incenter which does not appear. The points where these triples meet are the Mittenpunkts

Dj. For example, I1M1,I2M2,I3M3 meet at { } 1 D0 = [c + u + w, a + u v] . 2 (a + c b + u v + w) − − −

Figure 3.7: Green InMid lines and Mittenpunkts D

The Figure 3.7 shows the four Incenters I, three midpoints Mi, twelve InMid IjMi lines, and four Mittenpunkts D of A1A2A3. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 50

The join of a corresponding Incenter Ij and Mittenpunkt Dj is an In-Mitten line. There are four In-Mitten lines, and

I0D0 = (c + d) v + aw ad : cv + (a + d) w + cd : aw cv . h − − − i Theorem 35 (In-Mitten center) The four In-Mitten lines are concurrent and meet at the symmedian point (see Example 18) 1 K = X = [c, a] . 6 2 (a + c b) −

The join of a corresponding Gergonne point Gj and Mittenpunkt Dj is a Gergonne- Mitten line. There are four Gergonne-Mitten lines and

(∆ 4bd) u + (4cc ∆) v + 2 (4aa ∆) w + (a 2a) ∆ : − − − − D0G0 = (∆ 4bd) u + 2 (4cc ∆) v + (4aa ∆) w (c 2c) ∆ : . * − − − − − − + (∆ 4cc) v + (∆ 4aa) w b∆ − − −

Theorem 36 (Gergonne-Mitten center) The four Gergonne-Mitten lines GjDj 2 1 meet in the Centroid G = X2, and in fact G = 3 Dj + 3 Gj.

Proof. We use the formulas above for D0 and G0 to compute 2 1 2 1 D0 + G0 = [c + u + w, a + u v] 3 3 3 2 (a + c b + u v + w) − − − 1 b u 1 + − [w (c b) , v (a b)] = [1, 1] = G. 3 2 (du cv + aw) ∆ − − − − − 3 − − The join of a corresponding Mittenpunkt Dj and Bevan point Lj is a Mitten- Bevan line. There are four Mitten-Bevan lines and

D0L0 = av + bw cd : bv + cw + ad : b (v + w) . h − − i

Theorem 37 (Mitten-Bevan center) The four Mitten-Bevan lines DjLj are concurrent, and meet at the Orthocenter 1 H = [ba, bc] . ∆

Figure 3.8 shows the four In-Mitten lines meeting at K = X6, the four Gergonne-

Mitten lines meeting at G = X2 and the four Mitten-Bevan lines meeting at H = X4. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 51

Figure 3.8: Green Mitten-Bevan center H, Gergonne-Mitten center G and In-Mitten center K

3.3.4 Spieker points

The central dilation of an Incenter is a Spieker point. There are four Spieker points

S0, S1, S2, S3 which are central dilations of I0,I1,I2,I3 respectively.

Theorem 38 (Spieker points) The four Spieker points are 1 1 1 1 S0 = [v + d, w + d] S1 = [ v + d, w + d] 2 (d + v w) − 2 (d v + w) − − − 1 1 1 1 S2 = [v + d, w + d] S3 = [ v + d, w + d] . 2 (d + v + w) 2 (d v w) − − − − 1 Proof. We use the central dilation formula which takes I0 = (d + v w)− [ w, v] to − − the point 1 w v 1 S0 δ 1/2 (I0) = 1 − , 1 = [v + d, w + d] ≡ − 2 − d + v w − d + v w 2 (d + v w) − − − − and similarly for the other Spieker points.

Theorem 39 (Spieker-Nagel lines) The Spieker points lie on the corresponding In-

Nagel lines, and in particular S0, S1, S2, S3 are the midpoints of the sides I0N0, I1N1,

I2N2, I3N3 respectively.

Proof. We check that in fact S0 is the midpoint of I0N0 by computing 1 1 1 1 1 1 I0 + N0 = [ w, v] + [(b + u)(c b) + cv + bw, (b + u)(a b) bv aw] 2 2 2 (d + v w) − 2 ∆ − − − − − = S0. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 52

The computations for the other In-Nagel lines and S1, S2, S3 are similar.

Figure 3.9: Green Spieker points S and Mitten-Spieker center H

The joins of corresponding Mittenpunkts Dj and Spieker points Sj are the Mitten- Spieker lines. There are four Mitten-Spieker lines, and

D0S0 = av + bw cd : bv + cw + ad : b (v + w) . h − − i

Theorem 40 (Mitten-Spieker center) The four Mitten-Spieker lines DjSj are concurrent and meet at the Orthocenter H = X4.

Theorem 41 (Bevan Mitten-Spieker) The Spieker point Sj is the midpoint of

HBj, so that the corresponding Bevan point Bj also lies on the corresponding Mitten- Spieker line. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 53

Proof. The midpoint of HB0 is 1 1 1 H + B0 = [b (c b) , b (a b)] 2 2 2∆ − − 1 + [(c b) u + cv + bw + c (a b) , (a b) u bv aw + a (c b)] 4∆ − − − − − − 1 ac b2 + (c b)(u + b) + cv + bw, = 2 − 2 − 4 (ac b ) " ac b + (a b)(u + b) aw bv # − − − − − 1 ac b2 + (c b + w)(u + b) , = 2 − 2 − 4 (ac b ) " ac b + (a b v)(u + b) # − − − − 1 (c b + w) (a b v) = 1 + − , 1 + − − 4 u b u b − − 1 = [c 2b + u + w, a 2b + u v] . 4 (u b) − − − − Now a judicious use of the quadratic relations, which we leave to the reader, shows that this is S0. The computations for the other Spieker points are similar.

3.3.5 Contact triangle and Weill point

The Contact points C1i,C2i, and C3i form a new triangle called the Gergonne triangle or contact triangle. There are four contact triangles, each associated to an

Incenter, and the contact triangle associated to the Incenter I0 is C10C20C30. The side lines of C10C20C30 are

C10C20 = (cc + bw)(d + v w): c (w a)(d + v w):(bw cv)(w a) h − − − − − i C10C30 = a (c + v)(d + v w):(bv aa)(d + v w):(aw bv)(c + v) h − − − − i C20C30 = a (bw cv)(d + v w): c (aw bv)(d + v w):(aw bv)(bw cv) . h − − − − − − i

The Centroid of a contact triangle is called the Weill point W (X354). We have four Weill points, each associated to an Incenter. The Weill point associated to I0 ab + ac 2aw + bv ac bc bw + 2cv W = − − , − − 0 3a (a 2b + c + v w) 3c (a 2b + c + v w) − − − − 1 = [bw 2cv + u (c b) , 2aw bv + u (a b)] . 3 (aw cv + du) − − − − − For each Incenter Ij we deﬁne the In-Circum line CIj, which passes also through the circumcenter 1 C = [c (a b) , a (c b)] . 2 (ac b2) − − − 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 54

Figure 3.10: Green contact triangles and Weill points.

There are four In-Circum lines, and we may compute that

ab 2b2 + ac v + (c b) aw + a (b c)(a 2b + c): − 2 − − − CI0 = (a b) cv + cb 2b + ac w + c (a b)(a 2b + c): * − − − − + (b a) cv + (b c) aw − − = (bc + ∆) v + aaw aad : ccv + (ba + ∆) w + ccd : ccv aaw . h − − − i

Theorem 42 The In-Circum line CIj line contains the Weill point Wj.

Proof. We check the Weill point W0 lies on the In-Circum line CI0 by computing ab + ac 2aw + bv − − ab 2b2 + ac v + (c b) aw + a (b c)(a 2b + c) 3a (a 2b + c + v w) − − − − − − ac bc bw + 2cv + − − (a b) cv + cb 2b2 + ac w + c (a b)(a 2b + c) 3c (a 2b + c + v w) − − − − − − + (b a) cv + (b c) aw − − ac b2 = − − (ac 2bc) v2 + (2ab ac) w2 + 2ba2c a3c + ac3 2bac2 . 3ac (a 2b + c + v w) − − − − − − 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 55

Now we use v2 = a (a + c 2b) and w2 = c (a + c 2b) to deduce that − − (ac 2bc) v2 + (2ab ac) w2 + 2ba2c a3c + ac3 2bac2 − − − − 2 3 3 2 = (ac 2bc) a (a + c 2b) + (2ab ac) c (a + c 2b) + 2ba c a c + ac 2bac − − − − − − = 0.

The computations are similar for the other Weill points.

The Weill-Bevan lines are the joins of corresponding Weill points and Bevan points. The Weill-Bevan line associated to I0 is

7ab2 + 3a2b 3ac2 2a2c + b3 + 8abc u 2 : − 2 3 − − 2 3ac b + 4abc v + a 3ab + 2ac 5b w ! − − 2 − 2 2 2 3 − W0L0 = 2ac + 3a c 3bc + 7b c b 8abc u 2 : * − 2 2− −3 + +c 2ac + 3bc 5b v + 3a c + b 4abc w ! − − (a c) ac + b2 u + c 6ab 5ac b2 v a 5ac 6bc + b2 w − − − − − − Theorem 43 (Weill-Bevan center) The four Weill-Bevan lines WjLj are concurrent and meet at the circumcenter C.

Proof. It is obvious since the Bevan point Lj and the Weill point Wj lie on the line

CIj. 3. Universal aﬃ ne triangle geometry and four-fold incenter symmetry 56

3.3.6 Summary Table

Here is table which summarizes the various strong concurrences we have found. Note however that not all pairings yield concurrent quadruples: for example the lines joining corresponding Nagel points and Mittenpunkts are not in general concurrent.

Incenter Gergonne Nagel Mittenpunkt Spieker Bevan Weill

Incenter X20 G = X2 K = X6 G = X2 C = X3 C = X3 − Gergonne X20 X69 G = X2 − − − − Nagel G = X2 X69 G = X2 X20 − − − Mittenpunkt K = X6 G = X2 H = X4 H = X4 − − − Spieker G = X2 G = X2 H = X4 H = X4 − − − Bevan C = X3 X20 H = X4 H = X4 C = X3 − − Weill C = X3 C = X3 − − − − − In Chapter 4 we will extend this table to other interesting points associated to the Incenter hierarchy. Chapter 4

Midpoints, four-fold Incenters symmetry and the Euler line

In this chapter, we extend the strong concurrencies in Chapter 3. We investigate strong concurrencies formed by quadruples of lines from the Incenter hierarchy, including joins of corresponding these midpoints, Nagel, Bevan and Spieker points. In fact, these strong concurrencies lie on the Euler line of A1A2A3.

In Kimberling’s Triangle Center Encyclopedia, the point X551 is the midpoint between the X1 and X2, the point X1385 is the midpoint between the X1 and X3, the point X946 is the midpoint between the X1 and X4, and the point X4297 is the midpoint between the X1 and X20.

More generally, we now introduce the notions of MIG point: the midpoint MIGj between the Incenter Ij and the Centroid X2, MIC point: the midpoint MICj between the Incenter Ij and the Circumcenter X3, MIH point: the midpoint MIHj between the Incenter Ij and the Orthocenter X4, MIN point: the midpoint MINj between the Incenter Ij and the Nine-point center X5, and MID point: the midpoint MIDj between the Incenter Ij and the De-Longchamp point X20.

4.1 Midpoints of Incenters I and the Centroid G

A MIG point is the midpoint of an Incenter and the Centroid. There are four MIG points each associated to an Incenter. The MIG point associated to I0 is (a 2b + c + v 4w) (a 2b + c + 4v w) MIG = − − , − − = X . 0 6 (a 2b + c + v w) 6 (a 2b + c + v w) 551 − − − −

We note that the MIG point MIGj, Spieker point Sj and Nagel point Nj lie on the Nagel line IjNj.

57 4. Midpoints, four-fold Incenters symmetry and the Euler line 58

Figure 4.1: Aﬃ ne ratio of Ij,MIGj, G, Sj, and Nj

Figure 4.2: Green Incenters, MIG, Centroid, Spiekers and Nagel points

The join of a corresponding MIGj and Bevan point Lj is a MIG-Bevan line. There are four MIG-Bevan lines, and

2 3ab + 2ac 5b2 v + b2 6ab + 5ac w + 6ab 5ac b2 (a 2b + c): 2 − − 2 − − − 2 MIG0L0 = b 6cb + 5ac v+ 2 2ac + 3bc 5b w + (a 2b + c) 5ac 6bc + b : * − − − − + (b + 4c)(a b) v + (b c) (4a + b) w b (a c)(a 2b + c) − − − − − − 2 (3bc + ∆) v + ( bc + 5aa) w ( bc + 5aa) d : − − − = ( ba + 5cc) v + 2 (3ba + ∆) w + ( ba + 5cc) d : . * − − + (b + 4c) cv a (4a + b) w bbd − − − Theorem 44 (MIG-Bevan center) The four MIG-Bevan lines are concurrent at the MIG-Bevan center, MIG-L, which lies on the Euler line. 4. Midpoints, four-fold Incenters symmetry and the Euler line 59

In fact the MIG-Bevan center is X3524 in the Encyclopedia of Triangle Centers and 4 1 4 1 1 1 b (c b) b (a b) MIG-L = X2 X4 = , − , − 3 − 3 3 3 3 − 3 (ac b2) (ac b2) − − 1 = [∆ + 3cc, ∆ + 3aa] . 9∆ Proof. We check that the MIG-Bevan center MIG-L as deﬁned lies on the MIG-Bevan line MIG0L0 by computing

4ac 3bc b2 − − 2 3ab + 2ac 5b2 v + b2 6ab + 5ac w + 6ab 5ac b2 (a 2b + c) 9 (ac b2) − − − − − − 4ac 3ab b2 + − − b2 6cb + 5ac v + 2 2ac + 3bc 5b2 w + (a 2b + c) 5ac 6bc + b2 9 (ac b2) − − − − − + ( (b + 4c)(a b) v + (b c) (4a + b) w b (a c)(a 2b + c)) − − − − − − b2 + 3ab 4ac b2 6cb + 5ac v 2 − −2 − 2  b + 3cb 4ac 6ab 10b + 4ac 9 ac b (a b)(b + 4c) !  − − − − − − 1 9 ac b2 (4a + b)(b c) b2 + 3ab 4ac 6cb 10b2 + 4ac =  + w  2  − 2 − − 2 − −  9 (ac b )  b 6ab + 5ac b + 3cb 4ac !  −  − − −   b2 6ab + 5ac b2 + 3cb 4ac   + (a 2b + c)   2 − 2 − 2   − b + 3ab 4ac b 6cb + 5ac 9b ac b (a c) !   − − − − − −    = 0.

Figure 4.3: Green MIG-Beven center 4. Midpoints, four-fold Incenters symmetry and the Euler line 60

4.2 Midpoints of Incenters I and Circumcenter

A MIC point is the midpoint of an Incenter and the Circumcenter. There are four

MIC points, each associated to an Incenter. The MIC point associated to I0 is

c(a b)v+(2b2+cb 3ac)w+c(a b)(a 2b+c) − 2 − − − , MIC = 4(ac b )(a 2b+c+v w) . 0 (3ac ab 2b2−)v+(b −c)aw a(−b c)(a 2b+c)  − − − − − −  4(ac b2)(a 2b+c+v w)  − − −  The join of a corresponding MIC point MICj and Nagel point Nj is a MIC-Nagel line. There are four MIC-Nagel lines and

6b2 + ab 7ac v 4b2 ab 3ac w (a 2b + c) ab + 3ac 4b2 : − 2 − − − − 2 − − 2 − MIC0N0 = cb 4b + 3ac v + 7ac cb 6b w + 3ac + bc 4b (a 2b + c): . * − − − − − + ab 4ac + bc + 2b2 v + ab 4ac + bc + 2b2 w + b (a c)(a 2b + c) − − − −

Figure 4.4: Green MIC-Nagel center

Theorem 45 (MIC-Nagel center) The four MIC-Nagel lines are concurrent and meet at the MIC-Nagel center, MIC-N which lies on the Euler line.

In fact, MIC-N is X631 which is the X2,X3 harmonic conjugate of X4 and 6 1 { } satisﬁes MIC-N = X2 X4, 5 − 5 2ac bc b2 b2 ab + 2ac (∆ + cc) (∆ + aa) MIC-N = − − , − − = , . 5 (ac b2) 5 (ac b2) 5∆ 5∆ " − − # 4. Midpoints, four-fold Incenters symmetry and the Euler line 61

Proof. We check that the MIC-Nagel center MIC-N as deﬁned lies on the MIC-Nagel line MIC0N0 by computing

2ac bc b2 − − 6b2 + ab 7ac v 4b2 ab 3ac w (a 2b + c) ab + 3ac 4b2 5 (ac b2) − − − − − − − − − b2 ab + 2ac + − − cb 4b2 + 3ac v + 7ac cb 6b2 w + 3ac + bc 4b2 (a 2b + c) 5 (ac b2) − − − − − − + ab 4ac + bc + 2b2 v + ab 4ac + bc+ 2b2 w + b (a c)(a 2b + c) − − − − 5 ac b2 ab 4ac + bc + 2b2 − − b2 + ab 2ac cb 4b2 + 3ac v   − − −   + b2 + cb 2ac 6b2 + ab 7ac   − −     2 2    5 ac b ab 4ac + bc + 2b  1  − −  =  + b2 + cb 2ac ab 4b2 + 3ac w  5 (ac b2)   − − −   −  2 2   + b + ab 2ac 6b + cb 7ac    − −     b2 + cb 2ac ab 4b2 + 3ac   − −   2 2   + (a 2b + c)  b + ab 2ac cb 4b + 3ac    − − − −   +5b ac b2 (a c)    − −       = 0.

We also have 6 1 6 1 1 1 b (c b) b (a b) X2 X4 = , + − , − 5 − 5 5 3 3 5 (ac b2) (ac b2) − − 2ac bc b2 b2 ab + 2ac = − − , − − = MIC-N. 5ac 5b2 5ac 5b2 " − − #

The join of a corresponding MIC point MICj and Spieker point Sj is a MIC- Spieker line. There are four MIC-Spieker lines and

2 ab + 3ac 2b2 v + 2 ab + ac 2b2 w 2 (a 2b + c) ab + ac 2b2 : − 2 − − 2 − − 2 − MIC0S0 = 2 cb 2b + ac v+ 2 3ac cb 2b w + 2 ac + bc 2b (a 2b + c): . * − − − − − + 2b2 + ab 3ac v + 2b2 + cb 3ac w + b (a c)(a 2b + c) − − − − Theorem 46 (MIC-Spieker center) The four MIC-Spieker lines are concurrent and meet at X140, the midpoint of the Circumcenter C and nine-point center N on the Euler line. 4. Midpoints, four-fold Incenters symmetry and the Euler line 62

Figure 4.5: Green MIC-Spieker center and MIC-Bevan center

Proof. The point 1 1 X = X + X 140 2 3 2 5 1 c (a b) a (c b) 1 2b2 + cb + ac 2b2 + ab + ac = − , − + − , − 2 2 (ac b2) 2 (ac b2) 2 4 (ac b2) 4 (ac b2) − − " − − # 3ac bc 2b2 3ac ab 2b2 = − − , − − 8 (ac b2) 8 (ac b2) " − − # lies on the MIC-Spieker line MIC0S0 since 3ac bc 2b2 − − 2 ab + 3ac 2b2 v + 2 ab + ac 2b2 w 2 (a 2b + c) ab + ac 2b2 8 (ac b2) − − − − − − − 3ac ab 2b2 + − − 2 cb 2b2 + ac v + 2 3ac cb 2b2 w + 2 ac + bc 2b2 (a 2b + c) 8 (ac b2) − − − − − − + 2b2 + ab 3ac v+ 2b2 + cb 3ac w +b (a c)(a 2b + c) − − − − 2b2 + cb 3ac 4b2 + 2ab 6ac v 2 − 2 − 2 2  2b + ab 3ac 2cb 4b + 2ac + 8 ac b 2b + ab 3ac !  − − − − − 1 2b2 + ab 3ac 4b2 + 2cb 6ac =  + w  2  2 2 − − 2 2  8 (ac b )  2ab 4b + 2ac 2b + cb 3ac + 8 ac b 2b + cb 3ac !  −  − − − − −   ab 2b2 + ac 2b2 + cb 3ac   +2 (a 2b + c)   2 − 2 − 2   − 2b + ab 3ac cb 2b + ac + 4b ac b (a c) !   − − − − −    = 0. 4. Midpoints, four-fold Incenters symmetry and the Euler line 63

We note that the four lines joining corresponding MIC points MICi and Bevan points Li, meet at X3 since In-Bevan lines are concurrent at X3.

4.3 Midpoints of Incenters I and Orthocenter H

A MIH point is the midpoint of an Incenter and the Orthocenter. There are four

MIH points, each associated to an Incenter. The MIH point associated to I0 is

b(c b)v+(2b2 cb ac)w b(b c)(a 2b+c) − −2 − − − − , MIH = 2(ac b )(a 2b+c+v w) . 0 (ab 2b2+ac)−v+(b −a)bw+b(a− b)(a 2b+c)  − − − −  2(ac b2)(a 2b+c+v w)  − − −  The join of a corresponding MIH point MIHj and Nagel point Nj is a MIH-Nagel line. There are four MIH-Nagel lines, and

4b2 ab 3ac v + b2 + ab 2ac w ab 2ac + b2 (a 2b + c): − − 2 2− − − 2 − MIH0N0 = 2ac cb b v cb 4b + 3ac w 2ac bc b (a 2b + c): . * − − − − − − − − − + ab + ac + bc 3b2 v + ab + ac + bc 3b2 w + b (a c)(a 2b + c) − − − −

Figure 4.6: Green MIH-Nagel center

Theorem 47 (MIH-Nagel center) The four MIH-Nagel lines are concurrent at the MIH-Nagel center, MIH-N, which lies on the Euler line. 4. Midpoints, four-fold Incenters symmetry and the Euler line 64

In fact, MIH-N is X3091 which is the midpoint of X4 and X631, and 3 2 3 1 1 2 b (c b) b (a b) MIH-N X2 + X4 = , + − , − ≡ 5 5 5 3 3 5 (ac b2) (ac b2) − − 1 = [∆ + 2ba, ∆ + 2bc] . 5∆ Proof. We check that the MIH-Nagel center MIH-N as deﬁned lies on the MIH-Nagel line MIH0N0 by computing

3b2 + 2cb + ac − 4b2 ab 3ac v + b2 + ab 2ac w ab 2ac + b2 (a 2b + c) 5 (ac b2) − − − − − − − 3b2 + 2ab + ac + − 2ac + cb + b2 v cb 4b2 + 3ac w 2ac bc b2 (a 2b + c) 5 (ac b2) − − − − − − − − + ab + ac + bc 3b2v + ab + ac +bc 3b2 w + b (a c)(a 2b + c) − − − − 3b2 + 2cb + ac 4b2 ab 3ac − − − + 3b2 + 2ab + ac 2ac + cb + b2 v   − −   +5 ac b2 ab + ac + bc 3b2   − −     2 2    3b + 2cb + ac b + ab 2ac  1  − −  =  + cb 4b2 + 3ac 3b2 + 2ab + ac w  5 (ac b2)   − − −   −  2 2   +5 ac b ab+ ac + bc 3b    − −     ab 2ac + b2 3b2 + 2cb +ac   − − −   2 2   + (a 2b + c)  2ac bc b 3b + 2ab + ac    − − − − −   +5 ac b2 b (a c)    − −       = 0.

The join of a corresponding MIH point MIHj and Bevan point Lj is a MIH- Bevan line. There are four MIH-Bevan lines, and

3ab 4b2 + ac v + b2 3ab + 2ac w + 3ab 2ac b2 (a 2b + c): 2 − − 2 − − − 2 MIH0L0 = b 3cb + 2ac v + 3cb 4b + acw + (a 2b + c) 2ac 3bc + b : * − − − − + (b + c)(a b) v + (b c)(a + b) w b (a c)(a 2b + c) − − − − − − (∆ + 3bc) v + (2aa bc) w + (bc 2aa) d : = − − . * (2cc ba) v + (∆ + 3ba) w + (2cc ba) d : (b + c) cv (a + b) aw bbd + − − − − − Theorem 48 (MIH-Bevan center) The four MIH-Bevan lines are concurrent and meet at the centroid G.

Proof. It is easy to check that the line MIHiLi passes through the centroid G by 4. Midpoints, four-fold Incenters symmetry and the Euler line 65 computing 1 3ab 4b2 + ac v + b2 3ab + 2ac w + 3ab 2ac b2 (a 2b + c) 3 − − − − − 1 + b2 3cb + 2ac v + 3cb 4b2 + ac w + (a 2b + c) 2ac 3bc + b2 3 − − − − (b+ c)(a b) v + (b c)(a + b) w b (a c)(a 2b + c) − − − − − − 3ab 4b2 + ac + b2 3cb + 2ac 3 (b + c)(a b) v a 2b + c − − − − = − + b2 3ab + 2ac + 3cb 4b2 + ac + 3 (b c)(a + b) w 3  − − −  + 3ab 2ac b2 + 2ac 3bc + b2 3b (a c)  − − − − −    = 0.

Figure 4.7: Green MIH-Spieker center and MIH-Bevan center

The join of a corresponding MIH point MIHj and Spieker point Sj is a MIH- Spieker line. There are four MIH-Spieker lines, and

2 ab 2b2 + ac v + 2 (c b) aw + 2a (b c)(a 2b + c): − 2 − − − MIH0S0 = 2 (a b) cv + 2 cb 2b + ac w + 2c (a b)(a 2b + c): * − − − − + 2b2 ab ac v + 2b2 cb ac w b (a c)(a 2b + c) − − − − − − − 2 (bc+ ∆) v + 2aaw 2aad : = − . * 2ccv + 2 (ba + ∆) w + 2ccd : (bc + ∆) v (ba + ∆) w bbd + − − − Theorem 49 (MIH-Spieker center) The four MIH-Spieker lines are concurrent at the nine-point center X5. 4. Midpoints, four-fold Incenters symmetry and the Euler line 66

Proof. We check that the line MIH0S0 passes through the nine-point center X5 by computing

2b2 + cb + ac − 2 ab 2b2 + ac v + 2 (c b) aw + 2a (b c)(a 2b + c) 4 (ac b2) − − − − − 2b2 + ab + ac + − 2 (a b) cv + 2 cb 2b2 + ac w + 2c (a b)(a 2b + c) 4 (ac b2) − − − − − + 2b2 ab ac v + 2b2 cb ac w b (a c)(a 2b + c) − − − − − − − 2 ab 2b2 + ac 2b2 + cb + ac − − +2 (a b) c 2b2 + ab + ac v   − −   +4 ac b2 2b2 ab ac 1   − − −   =   2   2  2 (c b) a 2b + cb + ac  4 (ac b )  − −  −  + +2 cb 2b2 + ac 2b2 + ab + ac w    − −    2 2   +4 ac b 2 b cb ac    − − −    + (a 2b+ c) 2a (b c) + 2c (a b) 4 ac b2b (a c)   − − − − − −    = 0.

4.4 Midpoints of Incenter I and nine-point center X5

A MIN point is a midpoint of an Incenter and nine-point center. There are four MIN points each associated to an Incenter. The MIN point associated to I0 is

(cb 2b2+ac)v+(6b2 cb 5ac)w+(ac+bc 2b2)(a 2b+c) − − 2 − − − , MIN = 8(ac b )(a 2b+c+v w) . 0 (ab 6b2+5ac)v+(2b−2 ab −ac)w+(ab−+ac 2b2)(a 2b+c)  − − − − −  8(ac b2)(a 2b+c+v w)  − − −  The join of a corresponding MIN point MINj and Nagel point Nj is a MIN-Nagel line. There are four MIN-Nagel lines, and

ab 14b2 + 13ac v + 7ac ab 6b2 w + (a 2b + c) ab 7ac + 6b2 : − 2 − 2 − − − 2 MIN0N0 = 7ac cb 6b v+ cb 14b + 13acw + (a 2b + c) 7ac bc 6b : . * − − − − − − + ab + 6ac + bc 8b2 v ab + 6ac + bc 8b2 w b (a c)(a 2b + c) − − − − − − − Theorem 50 (MIN-Nagel center)The four MIN-Nagel lines are concurrent and meet at the Min-Nagel center, MIN-N, which lies on the Euler line.

In fact, MIN-N is X1656 which is the reﬂection of X3 in X631, and 9 1 MIN-N X2 + X4 ≡ 10 10 (3∆ + ba) (3∆ + bc) = , . 10∆ 10∆ 4. Midpoints, four-fold Incenters symmetry and the Euler line 67

Proof. We have 9 1 9 1 1 1 b (c b) b (a b) MIN-N X2 + X4 = , + − , − ≡ 10 10 10 3 3 10 (ac b2) (ac b2) − − 4b2 + cb + 3ac 4b2 + ab + 3ac (3∆ + ba) (3∆ + bc) = − , − = , , 10 (ac b2) 10 (ac b2) 10∆ 10∆ " − − # and now we check the MIN-Nagel line MIN0N0 passes through the MIN-Nagel center, MIN-N as deﬁned by computing 4b2 + cb + 3ac − ab 14b2 + 13ac v + 7ac ab 6b2 w + (a 2b + c) ab 7ac + 6b2 10 (ac b2) − − − − − − 4b2 + ab + 3ac + − 7ac cb 6b2 v + cb 14b2 + 13ac w + (a 2b + c) 7ac bc 6b2 10 (ac b2) − − − − − − − + ab + 6ac + bc 8b2 v ab + 6ac +bc 8b2 w b (a c)(a 2b + c) − − − − − − − 4b2 + cb + 3ac ab 14b2 + 13ac − − + 4b2 + ab + 3ac 7ac cb 6b2 v   − − −   10 ac b2 ab + 6ac + bc 8b2   − − −     2 2    4b + cb + 3ac 7ac ab 6b  1  − − −  =  + + 4b2 + ab + 3ac cb 14b2 + 13ac w  10 (ac b2)   − −   −  2 2   10 ac b ab+ 6ac + bc 8b    − − −     4b2 + cb + 3ac ab 7ac + 6b2   − −   2 2   + (a 2b + c)  + 4b + ab + 3ac 7ac bc 6b    − − − −   10 ac b2 b (a c)    − − −       = 0.

The join of a corresponding MIN point MINj and Bevan point Lj is a MIN- Bevan line. There are four MIN-Bevan lines and 9ab 14b2 + 5ac v + 2b2 9ab + 7ac w + (a 2b + c) 9ab 7ac 2b2 : 2 − − 2 − 2 − − MIN0L0 = 2b 9cb + 7ac v + 9cb 14b + 5acw + 7ac 9bc + 2b (a 2b + c): . * − − − − + (2b + 5c)(a b) v + (b c) (5a + 2b) w 2b (a c)(a 2b + c) − − − − − − Theorem 51 (MIN-Bevan center) The four MIN-Bevan lines are concurrent and meet at X549, the midpoint of the centroid G and the circumcenter C on the Euler line.

Proof. We have the midpoint of X2 and X3 is 1 1 1 1 1 1 c (a b) a (c b) X549 X2 + X3 = , + − , − ≡ 2 2 2 3 3 2 2 (ac b2) 2 (ac b2) − − 2b2 + 3cb 5ac 2b2 + 3ab 5ac = − , − . 12 (b2 ac) 12 (b2 ac) − − 4. Midpoints, four-fold Incenters symmetry and the Euler line 68

Figure 4.8: Green MIN-Nagel center

Now we check X549 lies on MIN0L0 by computing

2b2 + 3cb 5ac 9ab 14b2 + 5ac v + 2b2 9ab + 7ac w 2 − − − 2 12 (b ac) + (a 2b + c) 9ab 7ac 2b ! − − − − 2b2 + 3ab 5ac 2b2 9cb + 7ac v + 9cb 14b2 + 5ac w + 2 − − 2 − 12 (b ac) + 7ac 9bc + 2b (a 2b + c) ! − − − (2b + 5c)(a b) v + (b c) (5a + 2b) w 2b(a c)(a 2b + c) − − − − − − = 0.

The computations for the other MIN-Bevan lines are similar.

The join of a corresponding MIN point MINj and Spieker point Sj is a MIN- Spieker line. There are four MIN-Spieker lines, and

2 6b2 ab 5ac v + 2 2b2 + ab 3ac w 2 ab 3ac + 2b2 (a 2b + c): − − 2 2− − − 2 − MIN0S0 = 2 3ac cb 2b v 2 cb 6b + 5ac w 2 3ac bc 2b (a 2b + c): . * − − − − − − − − − + ab 6b2 + 5ac v + cb 6b2 + 5ac w + b (a c)(a 2b + c) − − − − Theorem 52 (MIN-Spieker center) The four MIN-Spieker lines are concurrent and meet at the MIN-Spieker center, MIN-S, which lies on the Euler line.

In fact, MIN-S is X3628 which is the midpoint of X2 and X547 (the midpoint of

X2 and X5). 4. Midpoints, four-fold Incenters symmetry and the Euler line 69

Figure 4.9: Green MIN-Bevan center

Proof. We have the midpoint of X2 and X547 is

1 1 1 1 1 1 10b2 + 3cb + 7ac 10b2 + 3ab + 7ac X3628 X2 + X547 = , + − , − ≡ 2 2 2 3 3 2 24 (ac b2) 24 (ac b2) " − − # 6b2 + cb + 5ac 6b2 + ab + 5ac = − , − 16 (ac b2) 16 (ac b2) " − − # (5∆ + ba) (5∆ + bc) = , MIN-S. 16∆ 16∆ ≡

Now we check that the MIN-Spieker center as deﬁned lies on MIN0S0 by computing

6b2 + cb + 5ac 2 6b2 ab 5ac v + 2 2b2 + ab 3ac w − 2 − − 2 − 16 (ac b ) 2 ab 3ac + 2b (a 2b + c) ! − − − − 6b2 + ab + 5ac 2 3ac cb 2b2 v 2 cb 6b2 + 5ac w + − 2 − − − −2 − 16 (ac b ) 2 3ac bc 2b (a 2b + c) ! − − − − − + ab 6b2 + 5ac v + cb 6b2 + 5ac w + b (a c)(a 2b + c) − − − − = 0. 4. Midpoints, four-fold Incenters symmetry and the Euler line 70

Figure 4.10: Green MIN-Spieker center

4.5 Midpoints of Incenters I and De-Longchamp point

A MID point is a midpoint of an Incenter and the De-Longchamp point. There are four MID points each associated to an Incenter. The MID point associated to I0 is 1 1 MID = I + X 0 2 0 2 20 1 w v 1 b2 2cb + ac b2 2ab + ac = − , + − , − 2 (d + v w) (d + v w) 2 ∆ ∆ − − 1 = [(cc ba) v 2ccw + d (cc ba) , 2aav + (bc aa) w d (bc aa)] . 2∆ (d + v w) − − − − − − − The join of a corresponding MID point MIDj and Nagel point Nj is a MID-Nagel line. There are four MID-Nagel lines, and

2 ab 2ac + b2 v + 2ab 3b2 + ac w + (a 2b + c) 2ab ac + 3b2 : − − 2 − 2 − −2 − MID0N0 = 2cb 3b + acv + 2 2ac bc b w + ac + 2bc 3b (a 2b + c): . * − − − − − + 2ab 3ac + 2bc b2 v + 2ab 3ac + 2bc b2 w + 2b (a c)(a 2b + c) − − − − − − Theorem 53 (MID-Nagel center) The four MID-Nagel lines are concurrent and meet at the MID-Nagel center, MID-N, which lies on the Euler line.

In fact, MID-N is X3522 which is the midpoint of X20 and X3091, the reﬂection of 4. Midpoints, four-fold Incenters symmetry and the Euler line 71

X631 in the Circumcenter X3, and 9 4 MID-N X2 X4 ≡ 5 − 5 b2 4cb + 3ac b2 4ab + 3ac = − , − 5 (ac b2) 5 (ac b2) " − − # 1 = [3cc ba, 3aa bc] . 5∆ − −

Figure 4.11: Green MID-Nagel center

Proof. We check that MID-N is the reﬂection of X631 in the Circumcenter X3 by computing 1 1 MID-N + MIC-N 2 2 1 b2 4cb + 3ac b2 4ab + 3ac 1 2ac bc b2 b2 ab + 2ac = − , − + − − , − − 2 5 (ac b2) 5 (ac b2) 2 5ac 5b2 5ac 5b2 " − − # " − − # c (a b) a (c b) = − , − = C. 2 (ac b2) 2 (ac b2) − − 4. Midpoints, four-fold Incenters symmetry and the Euler line 72

The MID-Nagel center MID-N as deﬁned lies on MIN0S0 by computing

(b2 4cb+3ac) − 2 ab 2ac + b2 v + 2ab 3b2 + ac w + (a 2b + c) 2ab ac + 3b2 5(ac b2) − − − − − − (b2−4ab+3ac) − 2 2 2 + 5(ac b2) 2cb 3b + ac v + 2 2ac bc b w + ac + 2bc 3b (a 2b + c) − − − − − − + 2ab 3ac + 2bc b2 v + 2ab 3ac + 2bc b2 w + 2b (a c)(a 2b + c) − − − − − − 2 ab 2ac + b2 b2 4cb + 3ac − − − + 2cb 3b2 + ac b2 4ab + 3ac v   − −   + 2ab 3ac + 2bc b2 5 ac b2   − − −     2 2    2ab 3b + ac b 4cb + 3ac  1  − 2 2−  = 2  + +2 2ac bc b b 4ab + 3ac w  5(ac b )   − − −   −  2 2   + 2ab 3ac + 2bc b 5 ac b    − − −     2ab ac + 3b2 b2 4cb + 3ac   − − −   2 2   + (a 2b + c)  + ac + 2bc 3b b 4ab + 3ac    − − −   +2b (a c) 5 ac b2    − −       = 0.

The join of a corresponding MID point MIDj and Spieker point Sj is a MID- Spieker line. There are four MID-Spieker lines, and

2 (c b) av + 2 (a b) bw 2b (a b)(a 2b + c): − − − − − MID0S0 = 2 (c b) bv + 2 (a b) cw 2b (b c)(a 2b + c): . * − − − − − + (b c) av + (b a) cw + b (a c)(a 2b + c) − − − −

Theorem 54 (MID-Spieker center) The four MID-Spieker lines are concurrent and meet at the circumcenter C.

Proof. We check that the line MID0S0 passes through the circumcenter C by computing

c (a b) − (2 (c b) av + 2 (a b) bw 2b (a b)(a 2b + c)) 2 (ac b2) − − − − − − a (c b) + − (2 (c b) bv + 2 (a b) cw 2b (b c)(a 2b + c)) 2 (ac b2) − − − − − − + ((b c) av + (b a) cw + b (a c)(a 2b + c)) − − − − 2 (c b) ac (a b) + 2 (c b) ba (c b) + (b c) a2 ac b2 v 1 − − − − − − = + 2 (a b) bc (a b) + 2 (a b) ca (c b) + (b a) c2 ac b2 w 2 (ac b2)  − − − − − −  − b (a 2b + c) 2 (a b) c (a b) 2 (b c) a (c b) + (a c) 2 ac b2  − − − − − − − − −    = 0. 4. Midpoints, four-fold Incenters symmetry and the Euler line 73

Figure 4.12: Green MID-Bevan center and MID-Spieker center

The join of a corresponding MID point MIDj and Bevan point Lj is a MID- Bevan line. There are four MID-Bevan lines, and

ac b2 (2b a c + 2v + w): ac b2 (a 2b + c + v + 2w): MID0L0 = − − − − − . * (b 2c)(a b) v + (b c) (2a b) w + b (a c)(a 2b + c) + − − − − − − Theorem 55 (MID-Bevan center) The four MID-Bevan lines are concurrent and meet at the reﬂection of the Centroid G in the Circumcenter C, which is X376.

Proof. We get

b2 3cb + 2ac b2 3ab + 2ac X376 2C G = − , − . ≡ − 3 (ac b2) 3 (ac b2) " − − #

We check X376 lies on the MID-Bevan line MID0L0 by computing

b2 3cb + 2ac − ac b2 (2b a c + 2v + w) 3 (ac b2) − − − − b2 3ab + 2ac + − ac b2 (a 2b + c + v + 2w) 3 (ac b2) − − − + (b 2c)(a b) v + (b c) (2a b) w + b (a c)(a 2b + c) − − − − − − 1 b2 3cb + 2ac (2b a c + 2v + w) + b2 3ab + 2ac (a 2b + c + v + 2w) = − − − − − = 0. 3 +3 ((b 2c)(a b) v + (b c) (2a b) w + b (a c)(a 2b + c)) ! − − − − − − 4. Midpoints, four-fold Incenters symmetry and the Euler line 74

4.5.1 Summary Table

Here is the table which summarizes the various strong concurrences which lie on the Euler line we have found; for example the lines joining corresponding Bevan points

X40 and midpoints of Incenters X1 and Centroid X2 are concurrent at X3524.

Midpoints Nagel X8 Bevan X40 Spieker X10 4 1 (X1,X2) = X551 X2 X3524 = 3 X2 3 X4 X2 6 1 3 −1 9 1 (X1,X3) = X1385 X631 = 5 X2 5 X4 X3 = 2 X2 2 X4 X140 = 8 X2 8 X4 3 − 2 − 3 −1 (X1,X4) = X946 X3091 = 5 X2 + 5 X4 X2 X5 = 4 X2 + 4 X4 9 1 5 1 15 1 (X1,X5) X1656 = 10 X2 + 10 X4 X549 = 4 X2 4 X4 X3628 = 16 X2 + 16 X4 9 4 − 3 1 (X1,X20) = X4297 X3522 = X2 X4 X376 = 2X2 X4 X3 = X2 X4 5 − 5 − 2 − 2

Figure 4.13: Green Euler line. Chapter 5

Incenter circles, chromogeometry, and the Omega triangle

5.1 Introduction

In this chapter, we show that if a triangle has four blue Incenters and four red Incenters, then these eight points lie on a green circle, whose center is the green Orthocenter of the triangle, and similarly for the other colours. Tangents to the incenter circles yield interesting additional standard quadrangles and concurrencies. The proofs use the framework of rational trigonometry together with standard coordinates for triangle geometry, while a dilation argument allows us to extend the results also to Nagel and Spieker points. This chapter also illustrates our novel approach to triangle geometry initiated in [23]; using standard coordinates to establish universal aspects of the subject which are valid over a general bilinear form. Standard coordinates also have the advantage of yielding surprisingly simple equations for the three coloured Incenter Circles, which turn out to be, after pleasant simpliﬁcations,

b : Qb (X) = bb (2x + 2y 1) C − r : Qr (X) = br (2x + 2y 1) C − g : Qg (X) = bg (2x + 2y 1) . C − b However the formulas for the star lines sj become rather formidable, but seem to have interesting algebraic aspects.

75 5. Incenter circles, chromogeometry, and the Omega triangle 76

Some interesting number theoretical questions arise when we inquire into the existence of triangles, over a given field, which have simultaneously blue, red and green Incenters. Using empirical computer investigations of Michael Reynolds [28], we make some tentative conjectures on such triangles, both over the rational numbers and over a finite prime field. Finally we extend the results to Spieker and Nagel points by suitable central dilations.

5.2 The Incenter Circle theorem

Here is the main theorem of the thesis, illustrated for green Incenters of the triangle

A1A2A3 in Figure 5.1. The situation is completely symmetric between the three geometries blue, red and green.

Figure 5.1: Green Incenters and the blue and red Incenter Circles

b b b Theorem 56 (Incenter Circles) If a triangle A1A2A3 has four blue Incenters I0,I1,I2 b b and I3, then they all lie both on a red circle r with center the red Orthocenter Hr, b C and on a green circle with center the green Orthocenter Hg, and similarly for the Cg other colours. Furthermore, if both red and green Incenters exist, then they lie on the r g same blue circle, so that = = b, and similarly for the other colours. Cb Cb C Proof. To prove that the four blue Incenters Ib,Ib,Ib and Ib lie on a red circle b 0 1 2 3 Cr with center Hr, we need show that

b b b b Qr Hr,I0 = Qr Hr,I1 = Qr Hr,I2 = Qr Hr,I3 . 5. Incenter circles, chromogeometry, and the Omega triangle 77

First we ﬁnd the bilinear forms for the blue, red and green geometries. After translat- ing, and then applying a linear transformation with the matrix M, we send the original

1 α β triangle to the standard triangle A1A2A3. If M − = N = , then the bilinear γ δ! forms for the blue, red and green geometries become respectively the matrices

T α β 1 0 α β Db ≡ γ δ! 0 1! γ δ! α2 + β2 αγ + βδ a b = b b 2 2 αγ + βδ γ + δ ! ≡ bb cb!

T α β 1 0 α β Dr ≡ γ δ! 0 1! γ δ! − α2 β2 αγ βδ a b = r r − 2 − 2 αγ βδ γ δ ! ≡ br cr! − −

T α β 0 1 α β Dg ≡ γ δ! 1 0! γ δ! 2αβ αδ + βγ a b = g g . αδ + βγ 2γδ ! ≡ bg cg!

There are interesting relations between the introduced quantities; for example

2 2 2 2 2 ab = ag + ar abcb = bg + br 2 2 2 2 2 2 2 arcr = b b agcg = b b c = c + c b − g b − r b g r and

abcg 2bbbg + cbag = 0 abcr 2bbbr + cbar = 0 agcr 2bgbr + cgar = 0. − − −

The determinants of Db,Dr and Dg are respectively

2 2 ∆b = (αδ βγ) ∆r = ∆g = (αδ βγ) = ∆b − − − − and the mixed traces are

2 2 2 2 db = (α γ) + (β δ) dr = (α γ) (β δ) dg = 2 (α γ)(β δ) . − − − − − − − 2 2 2 Note also the relation db = dr + dg. 5. Incenter circles, chromogeometry, and the Omega triangle 78

If the original triangle has four blue Incenters, then the Existence of Triangle bilines theorem shows that we may choose numbers ub, vb, wb satisfying (3.1) and (3.2), so that

2 2 2 2 2 ub = α + β γ + δ v2 = α2 + β2 (α γ)2 + (β δ)2 b − − w2 = γ2 + δ2 (α γ)2 + (β δ)2 . b − − The blue Incenters are then

b 1 b 1 I0 = [ wb, vb] I1 = [wb, vb] db + vb wb − db vb + wb − − − b 1 b 1 I2 = [wb, vb] I3 = [ wb, vb] db + vb + wb db vb wb − − − − In exactly the same fashion

r 1 g 1 I0 = [ wr, vr] and I0 = [ wg, vg] . dr + vr wr − dg + vg wg − − − According to (2.9), the respective orthocenters are

bb br bg Hb = [cb bb, ab bb] Hr = [cr br, ar br] Hg = [cg bg, ag bg] . ∆b − − ∆r − − ∆g − −

If we set eb db + vb wb then ≡ − b br (cr br) wb br (ar br) vb H−−−→rI = − + , − 0 − ∆ e ∆ − e r b r b 1 = (br (cr br) eb + ∆rwb, br (ar br) eb ∆rvb) −∆reb − − − so that T b b b Qr Hr,I0 = H−−−→rI0 Dr H−−−→rI0 br(cr br) w − b br(cr br) w br(ar br) v ar br ∆ + e = − + b − b r b ∆r eb ∆r eb br(ar br) vb − br cr! − ! ∆r − eb 2 2 1 ar (br (cr br) eb + ∆rwb) + cr (br (ar br) eb ∆rvb) = 2 2 − − − ∆reb +2br (br (cr br) eb + ∆rwb)(br (ar br) eb ∆rvb) ! − − − 1 b2 (a 2b + c ) a c b2 e2 2∆ b (v w ) b2 + a c e = r r r r r r r b r r b b r r r b . 2 2 − 2 − 2 − 2 − − ∆reb +∆r arw + crv 2brvbwb ! b b − 2 Use the relation ∆r = arcr b to get − r 1 b2 (a 2b + c ) e2 Q H ,Ib = r r r r b (5.1) r r 0 2 − 2 2 ∆reb 2∆rbr (vb wb) eb + ∆r arw + crv 2brvbwb ! − − b b − 1 2b (b c )(a b )(v d v w w d ) = r r r r r b b b b b b (5.2) 2 − 2 2 − − 2 2 − 2 2 ∆reb +ar (br cr) v + cr (ar br) w + brdrd ! − b − b b 5. Incenter circles, chromogeometry, and the Omega triangle 79

2 2 2 where we have collected vb , wb and db of the numerator of (5.1), to rewrite it. 2 2 Replace vb = abdb, wb = cbdb and vbwb = ubdb and the values of ab, cb, db, ar, br, cr in terms of α, β, γ, δ to get the factorization

2 2br (br cr)(ar br)(vb ub wb) db + ar (br cr) abdb (5.3) − − − − − 2 2 2 + cr (ar br) cbdb + b (ar + cr 2br) d (5.4) − r − b 2b (b c )(a b )(v u w ) + a (b c )2 a = d r r r r r b b b r r r b b − − 2 − 2 − − +cr (ar br) cb + br (ar + cr 2br) db ! − − 2 2 2 2 = 2db (αγ βδ) α αγ + γ + β βδ + δ ub + vb wb − − − − − 2 2 2 2 α β αγ+ βδ γ + δ + αγ βδ (5.5) × − − − − and also note that

2 (db + vb wb) = db (ab + cb + db 2ub + 2vb 2wb) (5.6) − − − 2 2 2 2 = 2db α αγ + γ + β βδ + δ ub + vb wb . − − − − Combine (5.3) and (5.6), to get the surprisingly simple formula 2 2 2 2 b (αγ βδ) α β αγ + βδ γ + δ + αγ βδ Qr Hr,I0 = − − − − − ∆r br (ar br)(br cr) = − − Kr. ∆r ≡ b b b We may now repeat the calculation to see that Qr Hr,I1 = Qr Hr,I2 = Qr Hr,I3 = b Kr, showing that indeed the four blue Incenters lie on the red circle with quadrance Cr Kr and center Hr. Note that the expression for Kr depends only on the matrix Dr. Now a similar derivation shows that

b bg (ag bg)(bg cg) Qg Hg,Ii = − − Kg for i = 0, 1, 2, 3. ∆g ≡ b Hence the four blue Incenters also lie on a green circle with quadrance Kg and Cg center Hg. Similarly we ﬁnd that if a triangle has four red Incenters, then they lie on r a blue circle with center Hb and quadrance Cb g r bb (ab bb)(bb cb) Qb (Hb,Ii ) = Qb (Hb,Ii ) = − − Kb ∆b ≡ r as well as on a green circle with center Hg and quadrance Kg (the same one as Cg above!) Similarly if a triangle has four green Incenters, then they lie on a blue circle g g with center Hb and quadrance Kb, as well as on a red circle r with center Hr and Cb C quadrance Kr. The proof is complete. r g b g b r We now call b = = , r = = r and g = = the blue, red and green C Cb Cb C Cr C C Cg Cg Incenter Circles respectively. In Figure 5.2 we see a (small) triangle A1A2A3 with its Omega triangle HbHrHg and the three Incenter Circles, whose respective meets give the twelve blue, red and green Incenters. 5. Incenter circles, chromogeometry, and the Omega triangle 80

Figure 5.2: Three Incenter Circles Cb,Cr and Cg.

5.2.1 Equations of Incenter Circles

Theorem 57 (Incenter Circles equations) In standard coordinates with X = [x, y], the blue, red and green Incenter circles, when they exist, have respective equations

b : Qb (X) = bb (2x + 2y 1) C − r : Qr (X) = br (2x + 2y 1) C − g : Qg (X) = bg (2x + 2y 1) . C − Proof. The derivation of these equations, using the formulas established above for the orthocenters Hr and coloured Incenters, is somewhat involved algebraically, although the basic idea is simple. We show how to ﬁnd the equation of the red Incenter Circle

r, with center Hr, which four blue Incenters and four green Incenters lie on if they C exist. From the deﬁnition of a red circle, we get the equation Qr (Hr,X) = Kr, and then substitute the values of Hr and Kr to get

br(cr br) br(cr br) br(ar br) ar br ∆− x − x − y r − (5.7) ∆r ∆r br(ar br) − − br cr! − y ! ∆r − br (ar br)(br cr) = − − (5.8) ∆r or after expansion

2 2 1 ar (br (cr br) ∆rx) + cr (br (ar br) ∆ry) 2 − − − − ∆r +2br (br (cr br) ∆rx)(br (ar br) ∆ry) ! − − − − br (ar br)(br cr) = − − . ∆r 5. Incenter circles, chromogeometry, and the Omega triangle 81

2 This may be rewritten, using ∆r = arcr b , in the form − r 1 2 2 2 2 2 2 2 2 ∆rarx + 2∆rbrxy + ∆rcry + ∆rbr (ar 2br + cr) 2∆rbrx 2∆rbry ∆r − − − = br (ar br)(br cr) . − −

Now cancel ∆r, and rearrange to get

2 2 2 ∆rarx + 2∆rbrxy + ∆rcry 2∆rbrx 2∆rbry + br arcr b = 0 − − − r or more simply 2 2 arx + 2brxy + cry 2brx 2bry + br = 0 − − which has the form stated in the theorem. The same kind of calculation establishes the formulas for b and g. C C Note that the equations for the Incenter Circles b, r and g allow them to be C C C deﬁned whether or not the corresponding Incenters exist! Incenters then exist precisely b b b b as meets of these Incenter Circles: for example the blue Incenters I0,I1,I2,I3 are just the meets of r and g, if these exist in the ﬁeld in which we work. C C

5.2.2 Tangent lines of Incenter Circles

Now we consider tangent lines to Incenter circles. Figure 5.3 shows the four blue

Incenters of A1A2A3, together with the red and green Incenter Circles passing through b them, namely Cr and Cg. At each of the four Incenters Ii , i = 1, 2, 3, 4 we have the b b tangent lines t and t to the red and green Incenter Circles r and g respectively. ri gi C C b b b b Theorem 58 (Incenter tangent meets) The tangent lines tr0tr1tr2tr3 to the red Incenter circle r at the blue Incenters form a standard quadrilateral, as do the tangent b b b bC lines t t t t at the green Incenter circle g. The same holds for the red and green g0 g1 g2 g3 C Incenters, if they exist.

b b b b b b This implies that the meets R01 tr0tr1 and R23 tr2tr3 lie on l1 = A2A3, ≡ ≡ b b b and are harmonic conjugates with respect to A2 and A3. Similarly R02 tr0tr2 and b b b ≡ R13 tr1tr3 lie on l2 = A1A3, and are harmonic conjugates with respect to A1 and ≡ b b b b b b A3; and R03 tr0tr3 and R12 tr1tr2 lie on l3 = A1A2, and are harmonic conjugates ≡ ≡ b b b b b b with respect to A1 and A2. The points G01 tg0tg1 and G23 tg2tg3 lie on l1, ≡ ≡ b b b and are harmonic conjugates with respect to A2 and A3. Similarly G02 tg0tg2 and b b b ≡ G13 tg1tg3 lie on l2, and are harmonic conjugates with respect to A1 and A3, and b ≡ b b b b b G t t and G t t lie on l3, and are harmonic conjugates with respect to 03 ≡ g0 g3 12 ≡ g1 g2 A1 and A2. 5. Incenter circles, chromogeometry, and the Omega triangle 82

Figure 5.3: Incenter tangent meets

b b Proof. We prove the result for the meets Gij of the green tangent lines tgi associated to the blue Incenters; the other cases are similar. We ﬁnd the joins of a blue Incenter b Ii and the green Orthocenter Hg to be

b (bg ag) bgdb + (cg bg) agvb + (ag bg) bgwb : HgI0 = − − − * (cg bg) bgdb + (cg bg) bgvb + (ag bg) cgwb :(bg cg) bgvb + (bg ag) bgwb + − − − − − b (bg ag) bgdb (cg bg) agvb (ag bg) bgwb : HgI1 = − − − − − * (cg bg) bgdb (cg bg) bgvb (ag bg) cgwb : (bg cg) bgvb (bg ag) bgwb + − − − − − − − − − b (bg ag) bgdb + (cg bg) agvb (ag bg) bgwb : HgI2 = − − − − * (cg bg) bgdb + (cg bg) bgvb (ag bg) cgwb :(bg cg) bgvb (bg ag) bgwb + − − − − − − − b (bg ag) bgdb (cg bg) agvb + (ag bg) bgwb : HgI3 = − − − − . * (cg bg) bgdb (cg bg) bgvb + (ag bg) cgwb : (bg cg) bgvb + (bg ag) bgwb + − − − − − − − 5. Incenter circles, chromogeometry, and the Omega triangle 83

b b b The tangent line tgi is the line green perpendicular to HgIi passing through Ii . These we may calculate to be

(ag bg) ub + bgvb + (ag 2bg) wb + (dbbg cb (ag bg)) : b − − − − tg0 = (cg bg) ub + (2bg cg) vb bgwb + (dbbg + ab (bg cg)) : * − − − − + bg ( vb + wb db) − − (ag bg) ub bgvb (ag 2bg) wb + (bgdb cb (ag bg)) : b − − − − − − tg1 = (cg bg) ub (2bg cg) vb + bgwb + (dbbg + ab (bg cg)) : * − − − − + bg (vb wb db) − − (ag bg) ub + bgvb (ag 2bg) wb + (bgdb cb (ag bg)) : b − − − − − − tg2 = (cg bg) ub + (2bg cg) vb + bgwb + (dbbg + ab (bg cg)) : * − − − − + bg ( vb wb db) − − − (ag bg) ub bgvb + (ag 2bg) wb + (bgdb cb (ag bg)) : b − − − − − − tg3 = (cg bg) ub (2bg cg) vb bgwb + (dbbg + ab (bg cg)) : . * − − − − − − + bg (vb + wb db) − We could verify directly that these four lines form a standard quadrilateral. But we prefer to verify that the meets of these four tangent lines agree with the following meets with the side lines of A1A2A3 :

b b b b 1 G01 tg0tg1 = tg0l1 = [(bg cg)( ub + vb + ab) , (ag bg)( ub wb + cb)] ≡ λ01 − − − − − b b b b 1 G23 tg2tg3 = tg2l1 = [(bg cg)(ub + vb + ab) , (ag bg)(ub + wb + cb)] ≡ λ23 − − b b b b 1 G02 tg0tg2 = tg0l2 = [0, bg ( vb + wb db)] ≡ λ02 − − b b b b 1 G13 tg1tg3 = tg1l2 = [0, bg ( vb + wb + db)] ≡ λ13 − b b b b 1 G03 tg0tg3 = tg0l3 = [bg (vb wb + db) , 0] ≡ λ03 − b b b b 1 G12 tg1tg2 = tg1l3 = [bg (vb wb db) , 0] ≡ λ12 − − where

λ01 = (cg ag) ub + (bg cg) vb + (bg ag) wb + (abbg abcg + cbag cbbg) − − − − − λ23 = (ag cg) ub + (bg cg) vb + (ag bg) wb + (abbg abcg + cbag cbbg) − − − − − λ02 = (bg cg) ub + (cg 2bg) vb + bgwb + (abcg 2abbg + 2bbbg cbbg) − − − − λ13 = (cg bg) ub + (cg 2bg) vb + bgwb + (2abbg abcg 2bbbg + cbbg) − − − − λ03 = (ag bg) ub + bgvb + (ag 2bg) wb + (abbg 2bbbg cbag + 2cbbg) − − − − λ12 = (bg ag) ub + bgvb + (ag 2bg) wb + (2bbbg abbg + cbag 2cbbg) . − − − − 5. Incenter circles, chromogeometry, and the Omega triangle 84

b b The fact that A2,A3,G01,G23 form a harmonic range etc. is an immediate consequence of a well known fact about standard quadrilaterals in projective geometry, since we have shown that the points A1,A2 and A3 are diagonal points of the quadrilateral formed by the four green tangent lines. Following the construction of the red lines in the introductory section on Quad- b rangles and quadrilaterals, we join a point Gij with the triangle point Ak opposite to b the triangle line that it lies on; giving six lines AkGij:

b A1G = (ag bg)(ub + wb cb):(bg cg)( ub + vb + ab) : 0 01 h − − − − i b A1G = (bg ag)(ub + wb + cb):(bg cg)(ub + vb + ab) : 0 23 h − − i b bg ( vb + wb db): A2G02 = − − * (bg cg) ub + (cg 2bg) vb + bgwb bgdb + ab (cg bg): bg (vb wb + db) + − − − − − b bg ( vb + wb + db): A2G13 = − * (cg bg) ub + (cg 2bg) vb + bgwb + bgdb + ab (bg cg): bg (vb wb db) + − − − − − b (ag bg) ub + bgvb + (ag 2bg) wb + bgdb + cb (bg ag): A3G03 = − − − * bg (vb wb + db): bg ( vb + wb db) + − − − b (bg ag) ub + bgvb + (ag 2bg) wb bgdb + cb (ag bg): A3G12 = − − − − . * bg (vb wb db): bg ( vb + wb + db) + − − −

Figure 5.4: Quad points and star lines

b b b b b b Theorem 59 (Quad points) The triples A1G23,A2G13,A3G12 , A1G23,A2G02,A3G03 , b b b b b b A1G01,A2G13,A3G03 and A1G01,A2G02 ,A3G12 of lines are concurrent in the 5. Incenter circles, chromogeometry, and the Omega triangle 85

b b b b respective points Qg0,Qg1,Qg2 and Qg3, called the blue/green quad points. The b b b b b b b b b triples A1R23,A2R13,A3R12 , A1R23,A2R02,A3R03 , A1R01,A2R13,A3R03 and b b b b b b A1R01,A2R02,A3R12 are also concurrent in the respective points Qr0,Qr1,Qr 2 and b Qr3, called the blue/red quad points. Similar results hold for the red and green Incenters, if they exist.

Proof. We verify this for the blue/green quad points: this is a consequence of the projective geometry of the complete quadrilateral we mentioned in the first section– if the original four tangent lines are regarded as the blue lines in Figure 5.4, then b the quad points Qgj correspond to the red points. However we want to find explicit b formulas and check things directly. The quad point Qgj is naturally associated to the b Incenter Ij . After some calculation, we find that these are

b bg Qg0 = [(bg cg)(dbub (bb cb) vb) , (ag bg) ((cb bb) wb + cbdb)] λ0 − − − − − b bg Qg1 = [(bg cg) ((ab bb) vb + abdb) , (ag bg)(dbub + (ab bb) wb)] λ1 − − − − b bg Qg2 = [(cg bg)(abwb bbvb) , (bg ag)(bbwb cbvb)] λ2 − − − −

b bg (bg cg)( dbub + (db + bb) vb abwb + abdb) , Qg3 = − − − λ3 " (bg ag)(dbub cbvb + (db + bb) wb cbdb) # − − − where

(bg cg)(bgdb + (bb cb)(ag bg)) ub (bg cg)(bbbg + cbag 2cbbg) vb λ0 = − − − − − − bg (ag bg)(bb cb) wb + cbbg (ag bg) db ! − − − − (ag bg)(bgdb + (ab bb)(bg cg)) ub + bg (bg cg)(ab bb) vb λ1 = − − − − − + (ag bg) (2abbg abcg bbbg) wb + abbg (bg cg) db ! − − − − bb (bg cg)(ag bg) ub + bg (bbbg + cbag bbcg cbbg) vb λ2 = − − − − bg (abbg + bbag abcg bbbg) wb abcb (bg cg)(ag bg) ! − − − − − − 2 (db + bb) b + agcg agbg (2db + bb) bbbgcg ub g − −  + (bg ((bg cg)(ab bb) + cb (2ag bg)) cbagcg) vb  λ3 = − − − − . + ((d + b )(b a ) b a a (b c )) w  b b g g g b g g g b   2 − − −   +bg (ab (db cb) cbdb) + bg (cbag (db + ab) abcg (db cb)) agcgabcb   − − − − −    5. Incenter circles, chromogeometry, and the Omega triangle 86

b b We may then check directly that for example Qg0 is incident with A3G12 by computing

bg (bg cg)(dbub (bb cb) vb) ((bg ag) ub + bgvb + (ag 2bg) wb bgdb + cb (ag bg)) − − − − − − − λ 0 bg (ag bg) ((cb bb) wb + cbdb) + bg (vb wb db) − − + bg ( vb + wb + db) − − λ − 0 2 2 bg (ag bg)(bg cg) dbu + (bb ab) ubwb bbdbub cbv + bbvbwb + cb (ab bb) vb = − − b − − − b − = 0 (bg cg)(bgdb + (bb cb)(ag bg)) ub + (bg cg)(bbbg + cbag 2cbbg) vb − − − − − − +bg (ag bg)(bb cb) wb cbbg (ag bg) db ! − − − − 2 2 since dbu + (bb ab) ubwb bbdbub cbv + bbvbwb + cb (ab bb) vb = 0 by using (3.1) b − − − b − and (3.2), and similarly for the other indices. In a parallel fashion, we ﬁnd that the b b four blue/red quad points Qrj have exactly the same formulas as the Qgj, except for the replacements ag ar, bg br and cg cr, and similarly for the other colours −→ −→ −→ red and green. b Now introduce the blue star line sj to be the join of the corresponding blue/red b b quad point Qrj and the blue/green quad point Qgj, and similarly for the other colours. b b b b There are then four blue star lines s0, s1, s2 and s3. b b The blue star point Bij is the meet of the two blue star lines si and sj, that is b b Bij s s , and similarly for the other colours. ≡ i j Note that following the introductory section on Quadrangles and quadrilaterals, we b b use the correspondence between the Qgj; and the tangent lines tgj; and the Incenters b Ij to match up the indices.

Theorem 60 (Star quadrangle) The four blue star lines form a standard quadran- b b b b gle s0s1s2s3. This holds also for the other colours.

b Proof. The proof we have is surprisingly complicated. The star lines sj have quite involved formulas; for example we ﬁnd that

E0dbub + F0cbvb : 2 bgbr (bb cb) + cbdb (agbr bgar agcr + cgar + bgcr cgbr) dbub − − − −  2cbbgbr (agbr bgar agcr + cgar + bgcr cgbr)(bb cb) dbvb  − − − − − : b b0 b0 2 s = Q Q =  ab (br cr)(bg cg)(agbr bgar) (bb cb) + cbdb wb  0 g r  − − − − −  *   +  +2abcb (br cr)(bg cg)(agbr bgar)(bb cb) db   − − − −   bgbrdb (agbr bgar agcr + cgar + bgcr cgbr)  − − 2 − − × (bb cb) + cbdb ub 2cb (bb cb) vb − − − where E0 and F0 are both homogeneous polynomials of degree 6 in the variables ai, bi and ci, with the former having 32 terms and the latter 46 terms. After some trial and 5. Incenter circles, chromogeometry, and the Omega triangle 87 error we can present these in the somewhat pleasant, but still mysterious, forms:

2 2 E0 = bgbr (agcr cgar bgcr + cgbr) b + 4c + abcb 6bbcb − − − b b − 2 2 + bgbr (agbr bgar) b + 2c + abcb 4bbcb − b b − 2 2 2cb agcgb arcrb + agarcrbg agcgarbr (bb cb) − r − g − − and

2 2 2 2 F0 = agcgb arcrb + agarcrbg agcgarbr b 4bbcb + 2c + abcb r − g − b − b 2 2 + bgbr (agcr cgar bgcr + cgbr) 5b 4 c + 2abbb 3abcb + 10bbcb − − − b − b − 2bgbr (agbr bgar)(bb cb)(ab 2bb + cb) . − − − − We can then calculate the blue star points, for example

1 bgbrdb (agbr bgar agcr + cgar + bgcr cgbr) B03 = −2 − − × E0dbub + F0cbvb (bb cb) + cbdb ub 2cb (bb cb) vb , 0 " − − − # from which clearly B03 lies on l3. The computations are similar for the other indices, and the other colours.

5.3 Explicit examples

5.3.1 An example over Q √30, √217, √741, √2470, √82 297 We will now explore in detail a particular triangle which has both blue, red and green Incenters; for us this is not only an important tool for checking the consistency of our formulas, but also a way to get a sense of the level of complexity of various constructions. In fact this kind of explicit calculation of examples is much to be encouraged in this subject: especially as working over concrete fields, including finite fields and explicit extension fields of the rationals, allows us to appreciate the number theoretic aspects of our geometrical investigations. For example, finding a triangle with blue, red and green Incenters approximately is easy with a geometry package: finding a concrete example and working out all the points precisely is more challenging. In particular we were unable, despite a reasonable computer search, to find any triangles with purely rational points that have blue, red and green Incenters! We would like to thank Michael Reynolds for his contributions to this search. So we tentatively conjecture that there are no such triangles. In any case, to get an explicit example we use an algebraic extension field of the rationals; so by √30 we mean an appropriate symbol in the extension field Q √30 etc.. Note that although our use of square roots is entirely algebraic, our representation 5. Incenter circles, chromogeometry, and the Omega triangle 88 of these square roots as approximate rational numbers (we prefer to avoid discussion of "real numbers"), necessarily brings an approximate aspect into our diagrams.

Example 61 One may check that the basic Triangle with points

X1 [ 21/59, 58/59] X2 [ 13/3, 2] and X3 [35/3, 8/5] ≡ − − ≡ − ≡ − in Q √30, √217, √741, √2470, √82 297 has both blue, red and green Incenters. After

Figure 5.5: An example triangle X1X2X3 translation by (21/59, 58/59) we obtain X1 = [0, 0] , X2 = [ 704/177, 176/59] and − X3 = [2128/177, 182/295] . The matrix N and its inverse M, where − e e 704 176 13 5 e 177 59 α β 1 704 56 N = −2128 182 = and M = N − = 95 5 ! γ δ! ! 177 − 295 264 42 respectively send [1, 0] and [0, 1] to X2 and X3, and X2 and X3 to [1, 0] and [0, 1].

From now on we discuss only the standard triangle A1A2A3 associated to X1X2X3; e e e e to convert back into the original coordinates, we would multiply by N and translate by ( 21/59, 58/59). The bilinear forms in these new standard coordinates, for the blue, − − 5. Incenter circles, chromogeometry, and the Omega triangle 89 red and green geometries respectively, are given by matrices

774 400 7778 848 1 0 T 31 329 156 645 ab bb Db = N N = 7778 848 −113 507 716 = 0 1! ! bb cb! − 156 645 783 225 216 832 7202 272 1 0 T 31 329 156 645 ar br Dr = N N = 7202 272 −112 911 484 = 0 1! ! br cr! − − 156 645 783 225 247 808 2000 768 0 1 T 10 443 52 215 ag bg Dg = N N = −2000 768 774 592 = . 1 0! ! bg cg! 52 215 − 52 215 97 140 736 97 140 736 The determinants of Db,Dr and Dg are ∆b = 87 025 and ∆r = ∆g = 87 025 , 6724 6076 576 − while the mixed traces are db = , dr = and dg = . The orthocenters of 25 25 − 5 A1A2A3 are 8825 537 84 337 87 833 227 55 537 7105 377 Hb = , Hr = , Hg = , . −1019 520 −25 488 11 214 720 25 488 3894 177 Blue, red and green Incenters exist over F = Q √30, √217, √741, √2470, √82 297 and we may choose 1875 104 14 432 873 628 u = v = w = b 31 329 b 177 b 4425 17 248 2464 196 u = √82 297 v = √217 w = √217√82297 r 156 645 r 885 r 4425 19 712 2816 448 u = √2470 v = √30 w = √741. g 52 215 g 295 g 295 Then the four blue Incenters, the four red Incenters and the four green Incenters of

A1A2A3 respectively are 761 220 5327 25 761 25 5327 220 Ib = , Ib = , Ib = , Ib = , 0 −590 413 1 10 384 −118 2 2112 168 3 270 27

1 r 22 429 440 4032 553√217 20 461√217√82297 + 210 343√82 297 76 618 507 , I0 = 1 − − " 2923√217 7√217√82297 + 133√82 297 30 049 # 50 976 − − 1 r 22 429 440 20 461√ 217√82297 4032 553√217 + 210 343√82 297 76 618 507 , I1 = 1 − − " 7√217√82297 2923√217 + 133√82 297 30 049 # 50 976 − − 1 r 22 429 440 4032 553 √217 + 20 461√217√82297 210 343√82 297 76 618 507 , I2 = 1 − − " 7√217√82297 + 2923√217 133√82 297 30 049 # 50 976 − − 1 r 22 429− 440 4032 553 √217 + 20 461√217√82297 + 210 343√82 297 + 76 618 507 , I3 = 1 " − 7√217√82297 2923√217 133√82 297 30 049 # 50 976 − − − 5. Incenter circles, chromogeometry, and the Omega triangle 90

203 247 35 3211 g 7788 √741 3894 √30 + 3894 √2470 7788 , I0 = 13 − 29 20 − 100 " √2470 √30 + √741 # 1239 − 177 1239 − 177 247 203 35 3211 g 3894 √30 7788 √741 + 3894 √2470 7788 , I1 = 29 − 13 20 − 100 " √30 + √2470 √741 # 177 1239 − 1239 − 177 247 203 35 3211 g 3894 √30 7788 √741 3894 √2470 7788 , I2 = − 29 − 13 − 20 − 100 " √30 √2470 √741 # − 177 − 1239 − 1239 − 177 247 203 35 3211 g 3894 √30 + 7788 √741 3894 √2470 7788 , I3 = 29 13 − 20 − 100 . " √30 √2470 + √741 # 177 − 1239 1239 − 177 The Incenter circle quadrances are 18 154 129 609 11 681 819 191 1182 272 Kb = Kr = Kg = . 28 196 100 − 28 196 100 10 443 The blue, red and green Incenter Circles themselves have respective equations

4840 000x2 19 447 120xy + 28 376 929y2 + 19 447 120x + 19 447 120y 9723 560 = 0 − − 19 360x2 62 524xy + 12 103y2 + 62 524x + 62 524y 31 262 = 0 − − 193 600x2 2572 240xy + 4032 553y2 + 2572 240x + 2572 240y 1286 120 = 0. − − b The four tangent lines tgj are

tb = 1570 : 11823 : 8323 tb = 127 512 : 33 761 : 58 261 g0 h − i g1 h− − i tb = 18 216 : 11 823 : 8323 tb = 1570 : 4823 : 8323 . g2 h− − i g3 h− i The meets of these four tangent lines agree with the following meets with the side lines of A1A2A3 :

b b b b 3500 9893 b b b b 3500 9893 G t t = t l1 = , G t t = t l1 = , 01 ≡ g0 g1 g0 13 393 13 393 23 ≡ g2 g3 g2 −6393 6393 b b b b 1189 b b b b 1189 G t t = t l2 = 0, G t t = t l2 = 0, 02 ≡ g0 g2 g0 1689 13 ≡ g1 g3 g1 689 b b b b 8323 b b b b 8323 G t t = t l3 = , 0 G t t = t l3 = , 0 . 03 ≡ g0 g3 g0 −1570 12 ≡ g1 g2 g1 18 216 b b b b b The blue/red quad points Qrj associated to I0,I1,I2,I3 respectively are

18 005 811 535 12 129 669 559 18 005 811 535 12 129 669 559 Qb = , Qb = , r0 21 082 889 161 −21 082 889 161 r1 − 9330 605 209 9330 605 209 18 005 811 535 12 129 669 559 18 005 811 535 12 129 669 559 Qb = , Qb = , . r2 14 928 733 909 14 928 733 909 r3 45 342 228 279 45 342 228 279 5. Incenter circles, chromogeometry, and the Omega triangle 91

b The respective blue/green quad points Qgj are 4161 500 11 762 777 4161 500 11 762 777 Qb = , Qb = , g0 −2654 777 2654 777 g1 −12 547 777 12 547 777 4161 500 11 762 777 4161 500 11 762 777 Qb = , Qb = , . g2 10 977 777 10 977 777 g3 20 870 777 20 870 777 The blue star lines are then

sb = Qb Qb = 1796 063 533 088 : 868 804 574 039 : 1034 074 074 039 0 r0 g0 h − i sb = 272 084 614 990 : 1199 343 574 039 : 1034 074 074 039 1 h − i sb = 272 084 614 990 : 868 804 574 039 : 1034 074 074 039 2 h − i sb = 1796 063 533 088 : 1199 343 574 039 : 1034 074 074 039 3 h − i and they meet at the blue star points 165 269 500 000 761 989 459 049 165 269 500 000 761 989 459 049 B01 = , B23 = , 927 258 959 049 927 258 959 049 −596 719 959 049 596 719 959 049 1034 074 074 039 1034 074 074 039 B = 0, B = 0, 02 868 804 574 039 13 1199 343 574 039 1034 074 074 039 1034 074 074 039 B03 = , 0 B12 = , 0 . 1796 063 533 088 272 084 614 990

5.3.2 An example over F13

Theorem 62 (Null quadrances incenters) Suppose that the ﬁeld F contains an element i, where i2 = 1, and the characteristic of F is not 2. If − bb (ab bb)(bb cb) br (ar br)(br cr) bg (ag bg)(bg cg) Kb − − = Kr − − = Kg − − = 0 ≡ ∆b ≡ ∆r ≡ ∆g then the standard Triangle A1A2A3 has four distinct blue, red and green Incenters.

Proof. If Kb = 0 then from the deﬁnition of the blue incenter circle b, which is C Qb (Hb,X) = Kb, b is a null circle, so it is a product of lines. Similarly, if Kr = 0 C then r is a null circle, and if Kg = 0 then g is a null circle. These null lines have C C distinct direction vectors (1, i) , (1, 1) and (1, 0) , (0, 1) respectively, and they are ± ± never parallel since char(F) = 2, so i = 1. Therefore, any two null circles meet in 6 6 ± exactly four points. Here is an example found by Michael Reynolds [28] which illustrates explicitly the above theorem.

Example 63 The triangle X1X2X3 with points X1 [3, 4] ,X2 [1, 9] and X3 ≡ ≡ ≡ [12, 3] in F13 has four blue, red and green Incenters. In F13 the squares are 0, 1, 3, 4, 9, 10 5. Incenter circles, chromogeometry, and the Omega triangle 92 and 12, and in particular 1 = 12 = 52 is a square. After translation by (3, 4) we − obtain X1 = [0, 0] , X2 = [11, 5] and X3 = [9, 12] . The matrix N and its inverse M

e e 11 5 e 1 10 11 N = M = N − = 9 12! 12 7 ! send [1, 0] and [0, 1] to X2 and X3, and X2 and X3 to [1, 0] and [0, 1] respectively. The bilinear form in these new standard coordinates for the blue, red and green geometries e e e e respectively are

1 0 T 3 3 Db = N N = 0 1! 3 4!

1 0 T 5 0 Dr = N N = 0 1! 0 2! − 0 1 T 6 8 Dg = N N = . 1 0! 8 8!

We can see immediately that Kb = Kr = Kg = 0 from the deﬁnitions

bb (ab bb)(bb cb) br (ar br)(br cr) Qb (Hb,Iir) = − − Kb Qr (Hr,Iib) = − − Kr ∆b ≡ ∆r ≡ bg (ag bg)(bg cg) Qg (Hg,Iir) = − − Kg for i = 0, 1, 2, 3. ∆g ≡ The four blue, red and green Incenters respectively are

b b b b I0 = [4, 8] I1 = [3, 6] I2 = [8, 10] I3 = [11, 4] r r r r I0 = [10, 9] I1 = [8, 2] I2 = [6, 5] I3 = [4, 12] g g g g I0 = [9, 8] I1 = [5, 3] I2 = [12, 11] I3 = [2, 4] and the blue, red and green Incenter Circles respectively have equations

b :(y x + 1) (x + 3y 1) = 0 C − − r :(x 6y)(x + 6y) = 0 C − g :(x + 2y 2) (x + 5y 5) = 0. C − − From Michael Reynolds’ computer investigations, we tentatively conjecture that for finite fields Fp where p 3 mod 4, there are no triangles which have both blue, red ≡ and green Incenters, and for finite fields Fp where p 1 mod 4, blue, red and green ≡ Incenters exists precisely when Kb = Kr = Kg = 0, as in the above example. 5. Incenter circles, chromogeometry, and the Omega triangle 93

5.4 Spieker circles and Nagel circles

Now we recall from ([23]) that the central dilation δ 1/2 about the centroid takes the − Orthocenter to the Circumcenter, and the Incenters to the Spieker centers. In standard coordinates

δ 1/2 ([x, y]) = (1/2) [1 x, 1 y] . − − − The inverse central dilation δ 2 takes the Orthocenter to the De Longchamps point − X20, and takes the Incenters to the Nagel points. In standard coordinates

δ 2 ([x, y]) = [1 2x, 1 2y] . − − −

Figure 5.6: Blue, red and green Spieker circles

b b b Theorem 64 (Spieker circles) If a triangle has four blue Incenters I0,I1,I2 and b I3, then the four blue Spieker centers all lie both on a red Spieker circle with center the red Circumcenter Cr, and on a green Spieker circle with center the green Circumcenter

Cg. If both say blue and red Incenters exist, then all 8 blue and red Spieker points lie on the same green circle. The same holds for the other colours.

Proof. We see that if we use the central dilation formula to transform Incenter circles centred at the Orthocenters, we get the Spieker circles centred at Circumcenters, so this theorem is a direct consequence of the Incenter circles theorem and the fact that a dilation preserves circles of any colour. 5. Incenter circles, chromogeometry, and the Omega triangle 94

Here are the formulas for the coloured Circumcenters in standard coordinates: 1 Cb = [cb (ab bb) , ab (cb bb)] 2∆b − − 1 Cr = [cr (ar br) , ar (cr br)] 2∆r − − 1 Cg = [cg (ag bg) , ag (cg bg)] . 2∆g − −

b b b b Theorem 65 (Nagel circles) If a triangle has four blue Incenters I0,I1,I2 and I3, then the four blue Nagel centers all lie both on a red Nagel circle with center the red

De Longchamps point X20r, and on a green Nagel circle with center the green De

Longchamps point X20g. If both say blue and red Incenters exist, then all 8 blue and red Nagel points lie on the same green circle. The same holds for the other colours.

Figure 5.7: Blue, red and green Nagel circles

Proof. In the same fashion as in the previous theorem, if we use the inverse central dilation δ 2 to transform Incenter circles centred at the Orthocenters, we get the Nagel − circles centred at De Longchamps points. Here are the formulas for the blue, red and green De Longchamps points:

1 2 2 X20b = bb 2bbcb + abcb, bb 2abbb + abcb ∆b − − 1 2 2 X20r = br 2brcr + arcr, br 2arbr + arcr ∆r − − 1 2 2 X20g = bg 2bgcg + agcg, bg 2agbg + agcg . ∆g − − 5. Incenter circles, chromogeometry, and the Omega triangle 95

Figure 5.8: Blue, red, green Orthocenters, Circumcenters and De Longchamps points

In Figure 5.8 we see the relations between the three coloured Orthocenters, Cir- cumcenters and De Longchamps points. The lines joining these are the three coloured Euler lines. Note that the centroids of the triangles of Orthocenters, Circumcenters and

De Longchamps points all agree with the centroid G of the original triangle A1A2A3. We conclude with a simple observation about De Longchamps points.

Theorem 66 (Orthocenters as midpoints) For any triangle, a coloured orthocenter H is the midpoint of the two De Longchamps points X20 of the other two colours.

Proof. This follows by considering the action of the central dilation δ 2 which − takes the circumcenter Ci to the orthocenter Hi, and the orthocenter Hi to the De

Longchamps point X20i. Chapter 6

Euler lines and Schiﬄ er points

In this chapter, we investigate another aspect of four-fold symmetry arising from the Incenter hierarchy: the rich relations between the twelve Incenter Euler lines of triangles formed from two of the original points of the triangle A1A2A3 and one of the

Incenters I0,I1,I2 or I3. These lines meet three at a time at four points which we call the Euler points P0,P1,P2 and P3 and also form concurrencies with bilines at six

Bi-Euler points E01,E02,E03,E12,E13, and E23. We show that these ten points all lie on the Circumcircle.

The classical Schiffl er point S = X21 is also a meet of three such Incenter Euler lines. We extend this classical approach to a four-fold symmetry, to find four Schiffl er points S = S0,S1,S2 and S3, all of which lie on the classical Euler line e of A1A2A3. The alternate approach of L. Emelyanov and T. Emelyanov to S also generalizes, again in a four-fold manner. This chapter also shows some associated results, for example that the four In-Schiffl er lines sj IjSj form a standard quadrilateral s0s1s2s3, ≡ which is the tangent quadrilateral at I0, I1, I2, I3 to the In-Circum conic which passes through the Circumcenter and four Incenters. As with almost all of the other theorems in this thesis, these results, together with their proofs, hold in considerable generality. We give illustrations over both the Euclidean blue geometry and the green relativistic geometry.

6.1 Incenter Euler lines

In this section, we focus on the Incenter Euler line eji of AkAlIj, where k, l = i, 6 for i, k, l = 1, 2, 3. There are twelve Incenter Euler lines eji, three associated to each

Incenter Ij. In standard coordinates, the three Incenter Euler lines associated to the

96 6. Euler lines and Schiﬄ er points 97

Incenter I0 are

(2 (aa + 3ac 2bc) du 2a (aa + 3ac 2bc) v + 6acdw 6aacd): − − − − e01 = ( 6acdu + 2c (cc 2ba + 3ca) v 6aadw + 2ad (cc 2ba + 3ca)) : , * − − − − + 2b (cdu acv + adw aad) − − − 2 ac + 2b2 du 2a ac + 2b2 v + 6abaw 6abcd : − − 2 e02 = 2 (∆ + 3bc) du 2c (∆ + 3ba) v + 2a (∆ + 3ca) w 6ac d : and * − − + ( 2c (a + b) du + 2ca (a + b) v 2aa (b + c) w + 2ac (b + c) d) − − 2 (3ab + ∆) du 2c (∆ + 3ac) v + 2a (∆ + 3bc) w 6a2cd : 2 − 2 − e03 = 2 ac + 2b du 6bc (a b) v + 2c ac + 2b w 6abcd : . * − − − + ( 2a (b + c) du + 2cc (a + b) v 2a (b + c) cw + 2ac (a + b) d) − −

6.1.1 Euler points P0,P1,P2 and P3

Theorem 67 (Euler points) The triples P0 e11e22e33, P1 e01e23e32,, P2 ≡ ≡ ≡ e02e13e31, and P3 e03e12e21 of Euler lines are concurrent. ≡

We call the points P0,P1,P2 and P3 the Euler points. It depends on how we choose the side of u, v, w then one of the Euler points P is X100 which is the anticomplement of Feuerbach point (X11) in the Encyclopedia of Triangle Centers. For example, if we choose u and w are negative values and v is positive value then P0 is X100.

Proof. Each triple of Incenter Euler lines is associated to the Incenter Ij not associated to any of the lines in that triple. The point P0 e22e33 may be computed, using the ≡ formulas for the Incenter Euler lines to be

ab2 + ac2 + a2c + 4bc2 3b2c 2b3 2abc u  2 − −2 2− 2 2   + 2ab + ac + 4bc + b cv + 2abc ab a c + 2ac + 2b c w   − − −   2 2 2 2 3   +2 2ab + a b + ac + b c b c    − −  ,    2 2 2 2 3    13ab + 5ac + 5a c + 13b c 18b 18abc u   − −    5ac2 13b2c + 4b3 + 14abc v + 13ab2 5a2c + 4b3 + 14abc w     − − − − −     3 3 2 2 4 2 2 2     +2 2ab + 2b c 2a c 2b + 7abc 14ab c + 7a bc   P0 =   − − −   .   3ab2 + 4a2b + ac2 + a2c + b2c 2b3 2abc u       2 − 2 2 2 − − 2    2ab ac + 2a c b c + 2abc v 4ab + ac 2bc + b aw    − − − − −    2 2 2 2 3    +2a ab + a c + bc 2b c b         − −    13ab2 + 5ac2 + 5a2c + 13b2c 18b3 18abc u   − −    2 2 3 2 2 3    5ac 13b c + 4b + 14abc v + 13ab 5a c + 4b + 14abc w   − − − − −    3 3 2 2 4 2 2 2     +2 2ab + 2b c 2a c 2b + 7abc 14ab c + 7a bc     − − −       6. Euler lines and Schiﬄ er points 98

We check that P0 lies on e11 by computing

ab2 + ac2 + a2c + 4bc2 3b2c 2b3 2abc u − − − + 2ab + ac + 4bc + b2 cv + 2abc ab2 a2c + 2ac2 + 2b2c w  − − −  +2 2ab2 + a2b + ac2 + b2c b3 c  − −    13ab2 + 5ac2 + 5a2c + 13b2c 18b3 18 abc u × − − 5ac2 13b2c + 4b3 + 14abc v + 13ab2 5a2c + 4b3 + 14abc w  − − − − −  +2 2ab3 + 2b3c 2a2c2 2b4 + 7abc2 14ab2c + 7a2bc  − − −    2 6ab +ac + 3a2 + 2b2 (a 2b + c) u 2 (b c) 6ab + ac+ 3a2 + 2b2 v − − − − − 6a (a b)(a 2b + c) w + 6a (b c)(a b)(a 2b + c) ! − − − − − − 3ab2 + 4a2b + ac2 + a2c + b2c 2b3 2abc u − − − 2ab2 ac2 + 2a2c b2c + 2abc v 4ab + ac 2bc + b2 aw  − − − − −  +2a ab2 + a2c + bc2 2b2c b3  − −  +   13ab2 + 5ac 2 + 5a2c + 13b2c 18b3 18 abc u × − − 5ac2 13b2c + 4b3 + 14abc v + 13ab2 5a2c + 4b3 + 14abc w  − − − − −  +2 2ab3 + 2b3c 2a2c2 2b4 + 7abc2 14ab2c + 7a2bc  − − −    6 (b c)(a b)(a 2b + c) u 2 (a b) ac 6bc + 2b2 + 3c2 v − − − − − − 2 2 6a (b c)(a 2b + c) w + 2a (a 2b + c) ac 6bc + 2b + 3c ! − − − − − 2 (a c) ((a b)(a 2b + c) u (b c)(a b) v a (a 2b + c)w + a (b c)(a 2b + c)) − − − − − − − − − − −

2 ac b2 = − 13ab2 + 5ac2 + 5a2c + 13b2c 18b3 18abc u × − − 5ac2 13b2c + 4b3 + 14abc v + 13ab2 5a2c + 4b3 + 14abc w  − − − − −  +2 2ab3 + 2b3c 2a2c2 2b4 + 7abc2 14ab2c + 7a2bc  − − −    (a 2b + c) ab2 2a2b 3ac2 a2c 2bc2 + 3b2c + 2a3 + 2b3 + 2abc u2 − − − − − − − + (a 2b + c) ab2 + 6ac2 + a2c + 4bc2 4b2c 2abc uv  − − − −  +a (a 2b + c) 2ab + 3ac + 4a2 5b2 uw    − 2 2 2 2 − 2 3   +a (a 2b + c) ab + 3ac + a c + 2bc 5b c 4b + 2abc u   − − −   +c 3ab2 2a2b + 3ac2 + 3a2c + 2bc2 5b2c + 2a3 + 4b3 4abc v2    .  − − 2 2 2 2 − 2 3 −   +a ab 3ac a c 2bc + 5b c + 4b 2abc vw   − − − − −   2 2   ac (a 2b + c) 2ab + 3ac + 4a 5b v   − − − 2 2   2a (a 2b + c) 2ab + 2ac + bc + a w   − −   2 2 2 2 2   +a (a 2b + c) ab 6ac a c 4bc + 4bc + 2abc w   − 2 − 2 − 2 − 2 2 3 3   +2ac (a 2b + c) 3ab ac + a c bc + 2b c + a b + 2abc   − − − − −    6. Euler lines and Schiﬄ er points 99

Using the quadratic relations (3.1) and (3.2), the second factor is

ab2 2a2b 3ac2 a2c 2bc2 + 3b2c + 2a3 + 2b3 + 2abc c − − − − − − + ab2 + 6ac2 + a2c + 4bc2 4b2c 2abc w  − − −  + 2ab + 3ac + 4a2 5b2 cv    2 2 2 2 −2 3   + ab + 3ac + a c + 2bc 5b c 4b + 2abc u   − −   +c 3ab2 2a2b + 3ac2 + 3a2c + 2bc2 5b2c + 2a3 + 4b3 4abc  a (a 2b + c)    − − 2 2 2 2 −2 3 −  −  + ab 3ac a c 2bc + 5b c + 4b 2abc u   − − − − −   2 2   c 2ab + 3ac + 4a 5b v   − 2 −   2 2ab + 2ac + bc + a (a 2b + c) c   − −   2 2 2 2 2   + ab 6ac a c 4bc + 4b c + 2abc w   2 − 2 − 2 − 2 2 3 3   +2c 3ab ac + a c bc + 2b c + a b + 2abc   − − − −    = 0.

The computations for P1,P2,P3 are similar.

Figure 6.1: Green Euler lines and Euler points on the Circumcircle

In Figure 6.1 we see the twelve green Incenter Euler lines (in red), and their meets the Euler points Pi. Note that there are other meets in this diagram, these will be discussed shortly. The most obvious fact about the Euler points is clearly visible in this diagram.

Theorem 68 (Circumcircle Euler points) The four Euler points P0,P1,P2 and

P3 lie on the Circumcircle of A1A2A3.

Proof. The Circumcircle of A1A2A3 has equation

ax2 + 2bxy ax + cy2 cy = 0. (6.1) − − 6. Euler lines and Schiﬄ er points 100

We check that P0 lies on the Circumcircle of A1A2A3 by replacing [x, y] in (6.1) with the coordinates of P0, so that the left hand side becomes 2 ac b2 − 2 13ab2 + 5ac2 + 5a2c+ 13b2c 18b3 18abc u × 2 2 3− −  5ac 13b c + 4b + 14abc v  − − − + 13ab2 5a2c + 4b3 + 14abc w    3 3− 2−2 4 2 2 2   +2 2ab + 2b c 2a c 2b + 7abc 14ab c + 7a bc   − − −    5ab3 2ac3 2a3c 5b3c + 2a2b2 4a2c2 (a 2b + c) u2 − −2 2 −4 − 2 2 −2 − +2b c + 2b + 3abc 2ab c + 3a bc !  −  +c (a 2b + c)(a 4b + 4c) ac + b2 uv    − 2 − −   +a ac b (a 2b + c) (4a 4b + c) uw   − − −   ac (a 2b + c) 3ab2 + ac2 + a2c + 3b2c + 2b3 + 6abc u     − − 3 3 3 3 2 2 2 2   4ab 2ac + 2a c 3b c 2a b + a c   +c − − − v2   +2b2c2 + 5abc2 + 3ab2c 2a2bc   !  .  2 2 2 2 − 3   +ac 3ab + ac + a c + 3b c + 2b + 6abc vw     2   +ac (4a 4b + c) ac + b (a 2b + c)v   3 − 3 3 − 3 −2 2 2 2   3ab 2ac + 2a c 4b c 2a b a c   a − − − − w2   − 2 2 2 2 2   +2b c + 2abc 3ab c 5a bc !   − 2 −   +ac (a 2b + c) ac b (a 4b + 4c) w   − − −   2ab3 + ac3 + a3c + 2b3c a2b2 + a2c2   +2ac (a 2b + c)   2 2 4 2 − 2 2   − b c b 3abc 2ab c 3a bc !   − − − − −    After simpliﬁcation using the quadratic relations (3.1) and (3.2), the second factor is

5ab3 2ac3 2a3c 5b3c + 2a2b2 4a2c2 − −2 2 −4 − 2 2 −2 +2b c + 2b + 3abc 2ab c + 3a bc !  −  + (a 4b + 4c) ac + b2 w + ac b2 (4a 4b + c) v  − − − −   2 2 2 2 3   3ab +ac + a c+ 3bc + 2b + 6abc u   −   4ab3 2ac3 + 2a3c 3b3c 2a2b2 + a2c2   + − − −   2 2 2 2 2   +2b c + 5abc + 3ab c 2a bc !   −  (a 2b + c) ac  + 3ab2 + ac2 + a2c + 3b2c + 2b3 + 6abc u  = 0 −    2   + (4a 4b + c) ac + b v   − −   3ab3 2ac3 + 2a3c 4b3c 2a2b2 a2c2   − − − −   2 2 2 2 2   − +2b c + 2abc 3ab c 5a bc !   − −   + ac b2 (a 4b + 4c) w   − −   2ab3 + ac3 + a3c + 2b3c a2b2 + a2c2   +2   2 2 4 2 − 2 2   b c b 3abc 2ab c 3a bc !   − − − − −    and similarly for the other Euler points. 6. Euler lines and Schiﬄ er points 101

6.1.2 Bi-Euler points

We remind the reader that the six Bilines of A1A2A3 are b1+ v : w : 0 and b1 ≡ h i − ≡ v : w : 0 through A1, b2+ u : u + w : u and b2 u : u w : u through h − i ≡ h − i − ≡ h − − i A2, and b3+ u v : u : u and b3 u + v : u : u through A3. Also eji is the ≡ h − − i − ≡ h − i Euler line of AkAlIj.

Figure 6.2: Blue Bi-Euler points

Theorem 69 (Bi-Euler points) The triples E01 e01e11b1+, E02 e02e22b2+, ≡ ≡ E03 e03e33b3+, E12 e13e23b3 , E13 e12e32b2 and E23 e21e31b1 are con- ≡ ≡ − ≡ − ≡ − current.

We call the points Eij the Bi-Euler points. Note that there is exactly one Bi-Euler point on each Biline. In Figure 6.2 we see the blue Bi-Euler points Eij. We used a Kimberling 6-9-13 triangle to check and found that Bi-Euler points are not in the Encyclopedia of Triangle Centers. 6. Euler lines and Schiﬄ er points 102

Proof. We show that the triple b3+, e03, e33 is concurrent by computing { } u v u u − − 2 (3ab + ∆) du 2 ac + 2b2 du 2a (b + c) du  −   2c (∆ + 3ac) v   6bc (a b) v  +2cc (a + b) v   − − −   +2a (∆ + 3bc) w +2c ac + 2b2 w 2a (b + c) cw         2    −  det  6a cd   6abcd  +2ac (a + b) d  −   −     2   2 (3ab + ∆) du  2 ac + 2b du  2a (b + c) du   − −  +2c (∆ + 3ac) v   +6bc (a b) v   2cc (a + b) v   − −   +2a (∆ + 3bc) w +2c ac + 2b2 w 2a (b + c) cw         2    −   6a cd   6abcd  +2ac (a + b) d  − −          b2 (a 2b + c) u2w c (a b)2 v2w  = 8 ac b2 (a 2b) − − − . − − − +ac (a 2b)(a 2b + c) uv ! − − Now we use the quadratic relations to replace u2 = ac, uv = aw and v2 = a (a + c 2b) to get − b2 (a 2b + c) u2w c (a b)2 v2w + ac (a 2b)(a 2b + c) uv − − − − − = b2c (a 2b + c) aw c (a b)2 (a + c 2b) aw + ac (a 2b)(a 2b + c) aw = 0. − − − − − − The computations are similar for the other triples.

Explicit formulas for the six Bi-Euler points Eij are

(a c)(a 2b+c)((b c)(a b)u+(a b)cv+(b c)aw+ac(a 2b+c)) − 2(−a2c −ab2 2abc+−ac2+2− b3 b2c−)((b c)u+(− c a)v+(b −a)a) , E01 = a(a 2−b+c)(−a c)((a 2b+c)−u+(b c)−v+(a b)−w+(b c)(−a b)) " 2(a−2c ab2 −2abc+ac−2+2b3 b2c−)((b c)u−+(c a)v+(− b a−)a) # − − − − − − c((ac bc+b2)(a 2b+c)u+c(2b c)(a b)v a(ac 3bc+b2+c2)w ac(2b c)(a 2b+c)) − − − − − − − − − 2(ac b2)((c b)(a 2b+c)u+(2b c)cv abw) , E02 =  a(c 2b)( −c(a 2b−+c)u+(− c b)cv+(b −c)bw+−bc(a 2b+c))  − − 2(ac− b2)((− c b)(a 2b−+c)u+(2−b c)cv abw−) − − − − −  c(a 2b)(a(a 2b+c)u (a b)bv (b a)aw ab(a 2b+c))  − 2(ac b−2)((a b)(−a −2b+c)−u+bcv− +(a−2b)aw−) , E03 = a((a 2b+c)( ab+ac+b2)u−+c( 3ab− +ac−+a2+b2)v a(b c−)(a 2b)w+ac(a 2b)(a 2b+c))  − − − − − − − −  2(ac b2)((a b)(a 2b+c)u+bcv+(a 2b)aw) − − − −  c(a 2b)( a(a 2b+c)u+(b a)bv+(a b)aw+ab(a 2b+c))  − − 2(ac− b2−)((a b)(a 2−b+c)u bcv−+(2b a)aw)− , E12 = a(( ab+ac+b2)(a 2b+c)u− c( 3ab− +ac−+a2+b2)−v+a(b c)(−a 2b)w+ac(a 2b)(a 2b+c))  − − − − − − − −  2(ac b2)((a b)(a 2b+c)u bcv+(2b a)aw) − − − − −  c((ac bc+b2)(a 2b+c)u c(2b c)(a b)v+a(ac 3bc+b2+c2)w ac(2b c)(a 2b+c))  − − − − − − − − − 2(ac b2)((c b)(a 2b+c)u+(c 2b)cv+abw) , E13 =  a(c 2b)( −c(a 2b−+c)u+(− b c)cv+(c− b)bw+bc(a 2b+c))  − − 2(ac− b2)((− c b)(a 2b−+c)u+(c−2b)cv+abw−) − − − −  (a c)(a 2b+c)((c b)(a b)u+(a b)cv+(c b)aw+ac(a 2b+c))  − 2(−a2c −ab2 2abc+−ac2+2− b3 b2c−)((c b)u+(− c a)v+(b −a)a) , E23 = a(a −c)(a −2b+c)((a 2b+c−)u+(c b−)v+(a b−)w (b c−)(a b)) . " −2(a2−c ab−2 2abc+ac−2+2b3 b2c)((−c b)u+(−c a−)v+(−b a)−a) # − − − − − − 6. Euler lines and Schiﬄ er points 103

Figure 6.3: Green Bi-Euler points

Theorem 70 (Bi-Euler concurrencies) The lines E01E23, E02E13, and E03E12 are concurrent at the Circumcenter C and C is the midpoint of each of these sides

Proof. Rather remarkably, the formulas for the lines in question are very simple:

E01E23 = 2 (a b) : 2 (b c):(c a) h − − − i E02E13 = 2b : 2c : c h − i E03E12 = 2a : 2b : a . h − i

We may easily check that the line E03E12 passes through the Circumcenter 1 C = [c (a b) , a (c b)] 2 (ac b2) − − − by calculating

c (a b) a (c b) 2a − + 2b − a 2 (ac b2) 2 (ac b2) − − − 1 = 2ac (a b) + 2ab (c b) 2a ac b2 = 0. 2 (ac b2) − − − − − The computations for the other lines are similar. 6. Euler lines and Schiﬄ er points 104

The Circumcenter is the midpoint of E03E12 since 1 1 E + E 2 03 2 12 1 = (a b)2 (a 2b + c)2 u2 b2c2v2 × 2 (ac b2) − − 2 − 2 2 − 2abc (a 2b) vw a (a 2b) w ! − − − − a (a 2b + c)2 u2 − 2 2 2  ab (a 2b + c) u + b cv   c (a 2b)(a b) − − ,  − − +ab (a 2b c) vw   − −     2 2     a (a 2b) w     − −    (a b) ab + ac + b2 (a 2b + c)2 u2    − − −   2    +ac (a 2b)(a b)(a 2b + c) u    − − −   a bc2 3ab + ac + a2 + b2 v2    − −     2     ac (a 2b) 3ab + ac + bc +a vw     − − −     +a2 (a 2b)2 (b c) w2     − −       After using the quadratic relations (3.1) and (3.2), the right hand side can be reduced to c(a 2b)(a b)ac(b2 2ub+ac)(a 2b+c) − − − − 2(ac b2)ac(a 2b)(b2 2ub+ac)(a 2b+c) , ac−(a 2b)(b− c)(b2 −2ub+ac)(a−2b+c)  − − − − −  2(ac b2)ac(a 2b)(b2 2ub+ac)(a 2b+c) − − − −  c (a b) a (c b)  = − , − = C. 2 (ac b2) 2 (ac b2) − −

Figure 6.4: Green Circumcircle, Euler points and Bi-Euler points

Theorem 71 (Bi-Euler cyclic) The six Bi-Euler points Eij lie on the Circumcircle of A1A2A3. 6. Euler lines and Schiﬄ er points 105

Figure 6.5: Blue Circumcircle, Euler points and Bi-Euler points

Proof. We check that E01 lies on the Circumcircle of A1A2A3 by replacing [x, y] in

(6.1) with the coordinates of E01, so that the left hand side becomes

a (a c)(a 2b + c) − − − 4 (ac b2)(ab av + bu cu + cv a2)2 × − − − − 2ab a2 ac b2 + 2bc c2 u2 2c (b c) uv 2a (a b) uw − − − − − − − − + (a c) cv2 + 2ac (a b) v a (a c) w2 + 2ac (b c) w .  − − − − −  +ac 2ab + ac 2bc + a2 + b2 + c2  − −    After simpliﬁcation using the quadratic relations (3.1) and (3.2), the second factor is

2ab a2 ac b2 + 2bc c2 ac 2c (b c) aw 2a (a b) cv − − − − − − − − + (a c) c (a 2b + c) a + 2ac (a b) v a (a c)(a 2b + c) c + 2ac (b c) w = 0,  − − − − − − −  +ac 2ab + ac 2bc + a2 + b2 + c2  − −    and similarly for the other Bi-Euler points. 6. Euler lines and Schiﬄ er points 106

6.2 The classical Schiﬄ er point X21, and the Emelyanov’s point of view

An accomplished geometer, Kurt Schiffl er (1896-1986) discovered one of the most attractive of the "twentieth-century" triangle centers, now known as the Schiffl er point, which we will in this chapter denote by S = X21 (not to be confused with the Spieker point also discussed in a previous chapter). The original definition of the Schiffl er point of the triangle A1A2A3 was as the intersection of the Euler lines of the three triangles A1A2I, A1A3I, A2A3I, where I = I0 = X1 is the classical incenter of the triangle A1A2A3, see [18] and [19]. In our notation, this is

S = e10e20e30.

In addition the Schiﬄ er point S lies on the Euler line e of the original triangle A1A2A3.

Figure 6.6: The classical Schiﬄ er point S = X21

Figure 6.6 shows the classical Euclidean Schiﬄ er point S (in blue): the meet of the four Euler lines (in red), where for example G1 and C1 are the centroid and circumcenter of the triangle A2A3I, and similarly for the other indices. In [9], L. Emelyanov and T. Emelyanova give an interesting alternate characterization of the Schiﬄ er point involving also the excenters I1,I2 and I3, as well as the

Circumcenter C. Let D1 be the meet of CI1 and A2A3,and similarly D2 (CI2)(A1A3) ≡ and D3 (CI3)(A1A2) . Then the (classical) Schiﬄ er point S is also determined as ≡ the point of concurrence of A1D1,A2D2,A3D3. Note that this construction assumes a correspondence between the incenters I1,I2 and I3 and the lines l1, l2 and l3.

In our approach, we treat the classical incenter I0 and the excenters I1,I2 and I3 symmetrically. We will see that then there are in fact four Schiffl er points Si, i = 1, 2, 3 and 4, each associated to a particular Incenter Ii, and in fact these four Schiffl er points 6. Euler lines and Schiffl er points 107

Figure 6.7: The Emelyanov characterization of S

lie on the Euler line of the triangle A1A2A3. Both the classical situation as well as the approach of L. Emelyanov and T. Emelyanov generalize to our more general set-up with an arbitrary quadratic form.

6.3 Four-fold views of Schiﬄ er points

6.3.1 Generalization of the classical Schiﬄ er point

We now begin by generalizing the classical Schiﬄ er point.

Theorem 72 The Incenter Euler lines ej1, ej2, ej3 of the respective triangles IjA2A3, A1IjA3, and A1A2Ij are concurrent. The points where these Euler lines meet is the Schiﬄ er point Sj, associated to the Incenter Ij. The four Schiﬄ er points S0,S1,S2 and S3 all lie on the Euler line e.

Proof. If we deﬁne S0 to be the meet of e01 and e02 then we may obtain

c( b(a b)(a 2b+c)u+c(a b)2v+a(b c)(a b)w ab(b c)(a 2b+c)) − − 2 − − − − − − − , S = (ac b )( 2b(a 2b+c)u+2c(a b)v+2a(b c)w+ac(a 2b+c)) 0 a( b(−b c)(−a 2b+−c)u+c(b c)(a −b)v+a(b −c)2w bc(a −b)(a 2b+c))  − − − − − − − − − −  (ac b2)( 2b(a 2b+c)u+2c(a b)v+2a(b c)w+ac(a 2b+c)) − − − − − − c bcdu + cc2v aacw + abad a badu cacv + aa2w bccd = − − , − − − . ∆ ( 2bdu + 2ccv 2aaw + acd) ∆ ( 2bdu + 2ccv 2aaw + acd) " − − − − # 6. Euler lines and Schiﬄ er points 108

Figure 6.8: Blue Schiﬄ er points

We check that S0 as deﬁned is incident with the Euler line e03 by computing

c bcdu + cc2v aacw + abad 2 (3ab + ∆) du 2c (∆ + 3ac) v − − − 2 ∆ ( 2bdu + 2ccv 2aaw + acd) +2a (∆ + 3bc) w 6a cd ! − − − a badu cacv + aa2w bccd 2 ac + 2b2 du 6bc (a b) v + − − − 2− − ∆ ( 2bdu + 2ccv 2aaw + acd) +2c ac + 2b w 6abcd ! − − − + ( 2a (b + c) du + 2cc (a + b) v 2a (b + c) cw + 2ac (a + b) d) − − = 0.

To check that S0 lies on the Euler line e = ∆ 3bc : ∆ + 3ba : bb of A1A2A3 − − we compute that c bcdu + cc2v aacw + abad − − (∆ 3bc) ∆ ( 2bdu + 2ccv 2aaw + acd) − − − a badu cacv + aa2w bccd − − ( ∆ + 3ba) + bb − ∆ ( 2bdu + 2ccv 2aaw + acd) − − − c bcdu + cc2v aacw + abad (∆ 3bc) 1 − − − = a badu cacv + aa2w bccd ( ∆ + 3ba) . ∆ ( 2bdu + 2ccv 2aaw + acd)  − − − −  − − +bb∆ ( 2bdu + 2ccv 2aaw + acd)  − −    6. Euler lines and Schiﬄ er points 109

Figure 6.9: Green Schiﬄ er points

Now replace ∆ = ac b2, d = a + c 2b, a = c b, b = a c and c = a b to get − − − − − c bcdu + cc2v aacw + abad (∆ 3bc) − − − a badu cacv + aa2w bccd ( ∆ + 3ba)  − − − −  +bb∆ ( 2bdu + 2ccv 2aaw + acd)  − −    b (a b)(a 2b + c) u + c (a b)2 v c − − − − 2b2 3ab + ac +a (b c)(a b) w ab (b c)(a 2b + c) ! −  − − − − −  = b (b c)(a 2b + c) u + c (b c)(a b) v  a − − − − − 2b2 + 3cb ac   2   − +a (b c) w bc (a b)(a 2b + c) ! − −   − − − −   + ac b2 ( 2b (a 2b + c) u + 2c (a b) v + 2a (b c) w +ac (a 2b + c)) b (a c)   − − − − − − −    c (a b) 2b2 + 3ab ac b (a 2b + c) u 2 − − − 2  − 2b ac b (a c) a (b c) 2b 3cb + ac !  − − − − − − 2b ac b2 (a c)  +c (a b) v   2 − − 2   − c (a b) 2b + 3ab ac + a (b c) 2b 3cb + ac !  =  − − − − − −   a (b c) 2b2 3cb + ac   +a (b c) w   2 − − 2   − c (a b) 2b + 3ab ac + 2b ac b (a c) !   − − − − − −   (b c) 2b2 + 3ab ac   +abc (a 2b + c)   2 − − − 2   − + ac b (a c) (a b) 2b 3cb + ac !   − − − − −    = 0.

The computations for the other Euler lines and Schiﬄ er points are similar. 6. Euler lines and Schiﬄ er points 110

6.3.2 Generalization of L. Emelyanov and T. Emelyanova theorem

We now generalize the approach of L. Emelyanov and T. Emelyanova to the Schiﬄ er point.

Figure 6.10: Green generalization of L. Emelyanov and T. Emelyanova theorem

The In-Circum points Kij are the meets of corresponding In-Circum lines CIj and Lines li. Here we avoid any assumption about a correspondence between incenters and lines. There are twelve In-Circum points; three associated to each Incenter Ij, and four associated to each Ai point. The In-Circum points associated to the Incenter

I0 are c (b a) bw a (b c) + bv K = − − , − 10 bv bw + (ab 2ac + bc) bv bw + (ab 2ac + bc) − − − − (a b) cv + (c b) aw K = 0, − − 20 (a b) cv + (cb 2b2 + ac) w + c (a b)(a 2b + c) − − − − (a b) cv + (c b) aw K = − − , 0 . 30 (ab 2b2 + ac) v + (c b) aw + a (b c)(a 2b + c) − − − − The InCircum-Vertex line kij is the line passing through the In-Circum point

Kij and the corresponding point Ai. There are twelve InCircum-Vertex lines; three associated to each Incenter. The InCircum-Vertex lines associated to the Incenter I0 6. Euler lines and Schiﬄ er points 111 are

k10 = bv + a (b c): bw + c (a b) : 0 h − − i = aa + bv : cc + bw : 0 h− i (a b) cv + (c b) aw : − 2 − k20 = (ac bc) v + cb 2b + ac w + c (a b)(a 2b + c) * − − − − + :(b a) cv + (b c) aw − − = ccv + aaw : ccv + (ba + ∆) w + ccd : ccv aaw h − − i ab 2b2 + ac v + (c b) aw + a (b c)(a 2b + c): k30 = − − − − * (a b) cv + (c b) aw :(b a) cv + (b c) aw + − − − − = (bc + ∆) v + aaw aad : ccv + aaw : ccv aaw . h − − − i Here is our generalization of the Emelyanov-Emelyanova characterization of the classical Schiﬄ er point.

Figure 6.11: Blue generalization of L. Emelyanov and T. Emelyanova theorem

Theorem 73 The triples S0 k11k22k33, S1 k10k23k32, S2 k20k13k31 and S3 ≡ ≡ ≡ ≡ k30k12k21 of InCircum-Vertex lines are concurrent. Each triple is associated to the

Incenter Ij which does not lie on any of the lines in that triple, and the common meet is the Schiﬄ er point Sj. 6. Euler lines and Schiﬄ er points 112

We recall that we obtained that the classical Schiﬄ er point associated to the In- center I0 is

c( b(a b)(a 2b+c)u+c(a b)2v+a(b c)(a b)w ab(b c)(a 2b+c)) − − 2 − − − − − − − , S = (ac b )( 2b(a 2b+c)u+2c(a b)v+2a(b c)w+ac(a 2b+c)) . 0 a( b(−b c)(−a 2b+−c)u+c(b c)(a −b)v+a(b −c)2w bc(a −b)(a 2b+c))  − − − − − − − − − −  (ac b2)( 2b(a 2b+c)u+2c(a b)v+2a(b c)w+ac(a 2b+c))  − − − − − −  Proof. We check that S0 as deﬁned earlier is incident with k11 by computing

b (a b)(a 2b + c) u + c (a b)2 v c − − − − +a (b c)(a b) w ab (b c)(a 2b + c) ! − − − − − (ab ac bv) 2b (a 2b + c) u + 2c (a b) v × − − (ac b2) − − − − +2a (b c) w + ac (a 2b + c) ! − − b (b c)(a 2b + c) u + c (b c)(a b) v a − − −2 − − − +a (b c) w bc (a b)(a 2b + c) ! + − − − − (ac bc bw) 2b (a 2b + c) u + 2c (a b) v × − − (ac b2) − − − − +2a (b c) w + ac (a 2b + c) ! − − 1 = (ac b2)( 2b (a 2b + c) u + 2c (a b) v + 2a (b c) w + ac (a 2b + c))× − − − − − − b2c (a b)(a 2b + c) uv ab2 (b c)(a 2b + c) uw − − − − −  + (ab (b c)(ac bc)(a 2b + c) bc (a b)(ab ac)(a 2b + c)) u  − − − − − − − bc2 (a b)2 v2    2 − −   c c (a b) (ab ac) + ab2 (b c)(a 2b + c)   + − − − − v     ac (a b)(b c)(ac bc) !  .  − − − −   2 2 2   + a b (b c) w   −     ac (a b)(b c)(ab ac)   + 2 − − − w   a a (b c) (ac bc) + b2c (a b)(a 2b + c)   − − − − − !     + (abc (a b)(ac bc)(a 2b + c) abc (b c)(ab ac)(a 2b + c))   − − − − − − −    After simpliﬁcation using quadratic relations (3.1) and (3.2), the second factor becomes

b (a 2b + c)(a (b c)(ac bc) c (a b)(ab ac)) u − − − − − − + c2 (a b)2 (ab ac) ac (a b)(b c)(ac bc) v  − − − − − −  2 2  + ac (a b)(b c)(ab ac) a (b c) (ac bc)w  = 0.  − − − − − −   2 2   a (b c) c (a b)   +abc (a 2b + c) − − −   − + (a b)(ac bc) (b c)(ab ac) !   − − − − −    The computations for the other InCircum-Vertex lines k22, k33, and then the other

Schiﬄ er points S1,S2,S3 are similar. 6. Euler lines and Schiﬄ er points 113

6.3.3 In-Schiﬄ er lines and points, and standard quadrilaterals and quadrangles

Now we introduce the In-Schiffl er line sj which is the join of an Incenter Ij with the corresponding Schiffl er point Sj. There are four In-Schiffl er lines s0, s1, s2 and s3.

The In-Schiffl er point Rij is the meet of the two In-Schiffl er lines si and sj, that is Rij sisj. ≡ Theorem 74 (In-Schiffl er quadrilateral) The four In-Schiffl er lines form a standard quadrilateral s0s1s2s3. In particular the In-Schiffl er points R01 and R23 lie on l1 = A2A3, and are harmonic conjugates with respect to A2 and A3. Similarly R02 and R13 lie on l2 = A1A3, and are harmonic conjugates with respect to A1 and A3; and R03 and R12 lie on l3 = A1A2, and are harmonic conjugates with respect to A1 and A2.

Figure 6.12: Green In-Schiﬄ er quadrilateral

Proof. The In-Schiﬄ er line s0 may be calculated to be

s0 I0S0 = (a c) u + (c 2b) v :(a c) u + (a 2b) w : (a c) u . ≡ h − − − − − − i 6. Euler lines and Schiﬄ er points 114

We can then calculate the In-Schiﬄ er points, and they are

(a 2b) w (2b c) v R = − , − 01 (2b c) v + (a 2b) w (2b c) v + (a 2b) w − − − − (a c) u R = 0, − 02 (a c) u + (a 2b) w − − (a c) u R = − , 0 03 (a c) u + (c 2b) v − − (a c) u R = − , 0 12 (a c) u + (2b c) v − − (a c) u R = 0, − 13 (a c) u + (2b a) w − − (a 2b) w (c 2b) v R = − , − 23 (c 2b) v + (a 2b) w (c 2b) v + (a 2b) w − − − − from which clearly R01 and R23 lie on l1 since (a 2b) w (2b c) v − + − 1 = 0 and (2b c) v + (a 2b) w (2b c) v + (a 2b) w − − − − − (a 2b) w (c 2b) v − + − 1 = 0. (c 2b) v + (a 2b) w (c 2b) v + (a 2b) w − − − − − Also it is clear that R02 and R13 lie on l2, while R03 and R12 lie on l3.

Figure 6.13: Blue In-Schiﬄ er quadrilateral 6. Euler lines and Schiﬄ er points 115

k The r-Schiffl er line rij is the join of an Incenter Ik with an In-Schiffl er point Rij sisj for which si and sj don’tpass through Ik. There are twelve r-Schiffl er lines; ≡ three associated to each Incenter Ik.

Theorem 75 (r-Schiﬄ er quadrilaterals) The twelve r-Schiﬄ er lines form three 0 1 2 3 0 1 2 3 0 1 2 3 standard quadrilaterals which are r23r23r01r01, r13r02r13r02 and r12r03r03r12.

Figure 6.14: Green r-Schiﬄ er standard quadrilaterals

Proof. The twelve r-Schiﬄ er lines are

r0 = (a c) u + (2b c) v :(a c) u + (a + 2b 2c) w :(c a) u 12 h − − − − − i r0 = (a c) u + (2b 2a + c) v :(a c) u + (2b a) w :(c a) u 13 h − − − − − i r0 = (a 4b + c) u + (2b c) v :(a 4b + c) u + (a 2b) w : (4b a c) u 23 h − − − − − − i r1 = (c a) u + (2b 2a + c) v :(c a) u + (2b a) w :(a c) u 02 h − − − − − i r1 = (c a) u + (2b c) v :(c a) u + (a + 2b 2c) w :(a c) u 03 h − − − − − i r1 = (a 4b + c) u (2b c) v :(a 4b + c) u (a 2b) w : (4b a c) u 23 h − − − − − − − − i r2 = (a 4b + c) u (2b c) v :(a 4b + c) u + (a 2b) w : (4b a c) u 01 h − − − − − − − i r2 = (a c) u (2b c) v :(a c) u + (a + 2b 2c) w :(c a) u 03 h − − − − − − i r2 = (a c) u (2b 2a + c) v :(a c) u + (2b a) w :(c a) u 13 h − − − − − − i r3 = (a 4b + c) u + (2b c) v :(a 4b + c) u (a 2b) w : (4b a c) u 01 h − − − − − − − i r3 = (a c) u + (2b 2a + c) v :(a c) u (2b a) w :(c a) u 02 h − − − − − − i r3 = (a c) u + (2b c) v :(a c) u (a + 2b 2c) w :(c a) u . 12 h − − − − − − i 6. Euler lines and Schiﬄ er points 116

Figure 6.15: Blue r-Schiﬄ er standard quadrilaterals

We now calculate the meets of two r-Schiﬄ er lines, for example

(a 2b) w (c 2b) v r0 r1 = − , − 23 23 (c 2b) v + (a 2b) w (c 2b) v + (a 2b) w − − − − and clearly this point lies on the side line l1 since (a 2b) w (c 2b) v − + − 1 = 0. (c 2b) v + (a 2b) w (c 2b) v + (a 2b) w − − − − − The computations are similar for the other meets of the other two r-Schiﬄ er lines.

1 2 3 0 2 3 Theorem 76 (r-Schiffl er Quadrangle) The triples T0 r23r13r12, T1 r23r03r02, 0 1 3 0 1 2 ≡ ≡ T2 r r r and T3 r r r of r-Schiffl er lines are concurrent. Each triple is ≡ 13 03 01 ≡ 12 02 01 associated to an Incenter Ij which does not lie on any of the lines in that triple. The meet of these triples form a standard quadrangle T0T1T2T3. 6. Euler lines and Schiffl er points 117

Figure 6.16: Green r-Schiﬄ er Quadrangle

Proof. We may check concurrency of the various triples by computing

(a 4b + c) u (2b c) v (a 4b + c) u (a 2b) w (4b a c) u − − − − − − − − det (a c) u (2b 2a + c) v (a c) u + (2b a) w (c a) u  − − − − − −  (a c) u + (2b c) v (a c) u (a + 2b 2c) w (c a) u  − − − − − −    (a 4b + c) u + (2b c) v (a 4b + c) u + (a 2b) w (4b a c) u − − − − − − = det (a c) u (2b c) v (a c) u + (a + 2b 2c) w (c a) u  − − − − − −  (a c) u + (2b 2a + c) v (a c) u (2b a) w (c a) u  − − − − − −    (a c) u + (2b 2a + c) v (a c) u + (2b a) w (c a) u − − − − − = det (c a) u + (2b c) v (c a) u + (a + 2b 2c) w (a c) u  − − − − −  (a 4b + c) u + (2b c) v (a 4b + c) u (a 2b) w (4b a c) u  − − − − − − −    (a c) u + (2b c) v (a c) u + (a + 2b 2c) w (c a) u − − − − − = det (c a) u + (2b 2a + c) v (c a) u + (2b a) w (a c) u = 0  − − − − −  (a 4b + c) u (2b c) v (a 4b + c) u + (a 2b) w (4b a c) u  − − − − − − −    6. Euler lines and Schiﬄ er points 118

Figure 6.17: Blue r-Schiﬄ er Quadrangle and the meet of these triples respectively are

(2b c)w 2 3 (a c)(a 2b+c)+(2− b c)w (a 2b)v , T0 r13r12 = − − (2b a)−v − − ≡ " (a c)(a 2b+c)+(2− b c)w (a 2b)v # − − − − − (2b c)w 0 2 (a c)(a 2b+−c) (2−b c)w+(a 2b)v , T1 r23r03 = − − (2−b a−)v − ≡ " (a c)(a 2b+−c) (2−b c)w+(a 2b)v # − − − − − (2b c)w 0 3 (a c)(a 2b+−c) (2−b c)w (a 2b)v , T2 r13r01 = − − (2−b a)−v − − ≡ " (a c)(a 2b+c) −(2b c)w (a 2b)v # − − − − − − (2b c)w 0 2 (a c)(a 2b+c)+(2− b c)w+(a 2b)v , T3 r12r01 = − − (2b a−)v − . ≡ " (a c)(a 2b+−c)+(2−b c)w+(a 2b)v # − − − −

The line passing through T0 and T1 is

(a 2b) v : (2b c) w : 0 , h − − i and it is obvious that A1 = [0, 0] lies on this line. The computations are similar for the other indices. 6. Euler lines and Schiﬄ er points 119

6.4 The In-Circum conic and its tangent lines

Here we introduce the In-Circum conic which passes through the Circumcenter 1 C = [c (a b) , a (c b)] 2 (ac b2) − − − and the four Incenters I0,I1,I2,I3. We remind the readers that if [x0, y0] is a point lying on the conic Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, then the tangent line to the conic at the point [x0, y0] is

2 2 2Ax0 + By0 + D : Bx0 + 2Cy0 + E : Ax Bx0y0 Cy + F . − 0 − − 0

Theorem 77 (In-Circum conic) The In-Circum conic has the equation

2a (b c)(a 2b) x2 2ac (a c) xy 2c (2b c)(a b) y2 − − − − − − − + 2ac (a c) x + 2ac (a c) y ac (a c) − − − − = 0.

Figure 6.18: Green In-Circum conic and tangent lines 6. Euler lines and Schiﬄ er points 120

Proof. We check that the Circumcenter C lies on the In-Circum conic as deﬁned, by computing

c (a b) 2 c (a b) a (c b) 2a (b c)(a 2b) − 2ac (a c) − − − − 2 (ac b2) − − 2 (ac b2) 2 (ac b2) − − − c (a b) a (c b) 2 + 2ac (a c) − 2c (2b c)(a b) − − 2 (ac b2) − − − 2 (ac b2) − − a (c b) + 2ac (a c) − ac (a c) − 2 (ac b2) − − − 2ac2 (b c)(a 2b)(a b)2 2a2c2 (a c)(a b)(c b) 1 − − − − − − − = +4ac2 (a c)(a b) ac b2 2a2c (2b c)(a b)(c b)2 2 2   4 (ac b ) − − − − − − −2 − +4a2c (a c)(c b) ac b2 4ac (a c) ac b2  − − − − − −    = 0.

The Incenter I0 lies on the In-Circum conic since

w 2 2a (b c)(a 2b) − − − (a + c 2b) + v w − − w v 2ac (a c) − − − (a + c 2b) + v w a + c 2b + v w − − − − w v 2 + 2ac (a c) − 2c (2b c)(a b) − (a + c 2b) + v w − − − a + c 2b + v w − − − − v + 2ac (a c) ac (a c) − a + c 2b + v w − − − − (a 2b + c) (ac 2bc) v2 + (2ab ac) w2 + 2ba2c a3c + ac3 2bac2 = − − − − − . (a 2b + c + v w)2 − − Now we use v2 = a (a 2b + c) and w2 = c (a 2b + c) to obtain − − (ac 2bc) v2 + (2ab ac) w2 + 2ba2c a3c + ac3 2bac2 − − − − 2 3 3 2 = (ac 2bc) a (a 2b + c) + (2ab ac) c (a 2b + c) + 2ba c a c + ac 2bac − − − − − − = 0.

The same kind of calculation holds for I1, I2 and I3. Now we consider tangent lines to the In-Circum conic. Figure 6.18 shows the four green Incenters and Circumcenter C of A1A2A3, together with the four In-Schiﬄ er lines and the Euler line.

Theorem 78 (Euler line tangent) The Euler line of A1A2A3 is the tangent at the Circumcenter C to the In-Circum conic. 6. Euler lines and Schiﬄ er points 121

Figure 6.19: Blue In-Circum conic and tangent lines

Proof. From the general formula of the tangent line to the In-Circum conic, we get the tangent line to the In-Circum conic at the point [x0, y0] lying on it is

4a (b c)(a 2b) x0 2ac (a c) y0 + 2ac (a c): − − − − − 2ac (a c) x0 4c (2b c)(a b) y0 + 2ac (a c): . * − −2 − − − − 2 + 2a (b c)(a 2b) x + 2ac (a c) x0y0 + 2c (2b c)(a b) y ac (a c) − − − 0 − − − 0 − − c(a b) a(c b) Now we substitute C = 2(ac−b2) , 2(ac−b2) for [x0, y0] to get the tangent at C to − − the In-Circum conic, andh it is i ac b2 3b (a b): ac b2 + 3b (c b): b (a c) = e. − − − − − − −

Theorem 79 (In-Schiﬄ er tangents) The four In-Schiﬄ er lines s0, s1, s2, and s3 are respectively the tangents at I0, I1, I2, I3 to the In-Circum conic.

w v Proof. We substitute I0 = a+c −2b+v w , a+c 2b+v w for [x0, y0] to get the tangent − − − − at I0 to the In-Circum conic,h and it is i ac (a c) + (c 2b) aw : ac (a c) + (a 2b) cv : ac (a c) . h − − − − − − i Using the quadratic relations (3.1) and (3.2), we get that

(a c) ac + (c 2b) aw :(a c) ac + (a 2b) cv : (a c) ac h − − − − − − i = (a c) u2 + (c 2b) uv :(a c) u2 + (a 2b) uw : (a c) u2 − − − − − − = (a c) u + (c 2b) v :(a c) u + (a 2b) w : (a c) u = s0 . h − − − − − − i The computations are similar for the other Incenters. 6. Euler lines and Schiﬄ er points 122

6.5 Conclusion and future directions

Although we have shown the power of standard coordinates and a large number of new results of four-fold symmetry in this thesis, it seems clear that there are many more interesting results to be investigated and explored. Using standard coordinates and four-fold symmetry, we could study many more triangle centers from Kimberling’s Encyclopedia of Triangle Centers. For example, the following table summarizes the various strong concurrences which lie on the Euler line that we have observed from the Geometer’sSketchpad.

Midpoints (X10,X4) (X10,X3) (X10,X2)(X3828) 2 1 (X1,X4) X381 = 1/2X2 + 1/2X4 X2 X5055 = 3 X5 + 3 X2 2 1 (X1,X2) X3545 = 2/3X5 + 1/3X381 X5054 = 3 X2 + 3 X3 X2 1 1 5 1 (X1,X3) X5 X549 = 2 X2 + 2 X3 6 X2 + 6 X3 1 1 5 1 (X1,X5) X5066 = 1/2X5 + 1/2X381 X140 = 2 X5 + 2 X3 6 X2 + 6 X5 2 1 (X1,X20) X2 X3 X3524 = 3 X3 + 3 X2 The results of Chapter six on the Schiﬄ er points opens the door for investigating a corresponding ﬁve-fold symmetry in the projective plane. This will follow from a con- siderably more general theory, which extends the point of view of L. Emelyanov and

T. Emelyanova to a projective setting involving a general projective triangle A1A2A3, an associated standard quadrangle P0P1P2P3, and an arbitrary point X. If we introduce an aﬃ ne metrical structure on the projective plane then the quadrangle can be replaced by the quadrangle of Incenters, and the point X by the Circumcenter C. There is the posibility of generalizing our results on the Incenter hierachy by replacing the quadrangle of Incenters by a more general standard quadrangle. Finally, it seems clear that in the context of universal triangle geometry we should be able to similarly investigate other triangle centers. Much work can be done in this direction, which reveals further the richness and depth of standard coordinates to embrace the four-fold symmetry inherent in triangle geometry. Bibliography

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