Universal Affine Triangle Geometry and Four-Fold Incenter Symmetry

Home , De Longchamps point, Incenter, Mittenpunkt, Spieker center, Triangle center

KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Original scientiﬁc paper NGUYEN LE Accepted 31. 12. 2012. NORMAN JOHN WILDBERGER Universal Aﬃne Triangle Geometry and Four-fold Incenter Symmetry

Universal Affine Triangle Geometry and Four-fold Univerzalna afina geometrija trokuta i Incenter Symmetry ˇcetverostruka simetrija srediˇsta upisane kruˇznice ABSTRACT SAZETAKˇ We develop a generalized triangle geometry, using an ar- Razvijamo opću geometriju trokuta koristeći proizvoljnu bitrary bilinear form in an affine plane over a general field. bilinearnu formu u afinoj ravnini nad općim poljem. By introducing standardized coordinates we find canonical Uvodeći standardizirane koordinate pronalazimo kanonske forms for some basic centers and lines. Strong concurren- oblike nekih osnovnih srediˇsta i pravaca. Prouˇcavamo cies formed by quadruples of lines from the Incenter hi- snaˇznu konkurentnost ˇcetvorki pravaca koji pripadaju “hierarchy are investigated, including joins of corresponding jerarhiji srediˇsta upisane kruˇznice”ukljuˇcujući i spojnice Incenters, Gergonne, Nagel, Spieker points, Mittenpunkts odgovarajućih sjeciˇsta simetrala kutova trokuta, Geor- and the New points we introduce. The diagrams are taken gonnovih toˇcaka, Nagelovih toˇcaka, Mittenpunkova (imen- from relativistic (green) geometry. ovano sa strane autora, op. ur.) te Novih toˇcaka koje se uvode u ˇclanku. Slike su prikazane u tzv. zelenoj ge- Key words: Triangle geometry, affine geometry, Rational ometriji. trigonometry, bilinear form, incenter hierarchy, Euler line, Gergonne, Nagel, Mittenpunkt, chromogeometry Kljuˇcne rijeˇci: geometrija trokuta, afina geometrija, racionalna trigonometrija, bilinearna forma, hijerarhija MSC 2000: 51M05, 51M10, 51N10 srediˇsta upisane kruˇznice, Eulerov pravac, Georgonnova toˇcka, Nagelova toˇcka, Mittenpunkt, kromogeometrija

1 Introduction formation on Triangle Centers in his Online Encyclopedia ([5], [6], [7]); and by the explorations and discussions of This paper repositions and extends triangle geometry the Hyacinthos Yahoo group ([4]). by developing it in the wider framework of Rational Trigonometry and Universal Geometry ([10], [11]), valid The increased interest in this rich and fascinating subject over arbitrary fields and with general quadratic forms. Our is to be applauded, but there are also mounting concerns main focus is on strong concurrency results for quadruples about the consistency and accessibility of proofs, which of lines associated to the Incenter hierarchy. have not kept up with the greater pace of discoveries. An- Triangle geometry has a long and cyclical history ([1], [3], other difficulty is that the current framework is modelled on the continuum as “real numbers”, which often leads [16], [17]). The centroid G = X2, circumcenter C = X3, synthetic treatments to finesse number-theoretical issues. orthocenter H = X4 and incenter I = X1 were known to the ancient Greeks. Prominent mathematicians like Euler One of our goals is to provide explicit algebraic formu- and Gauss contributed to the subject, but it took off mostly las for points, lines and transformations of triangle geom- in the latter part of the 19th century and the first part of etry which hold in great generality, over the rational num- the 20th century, when many new centers, lines, conics, bers, finite fields, and even the field of complex rational and cubics associated to a triangle were discovered and numbers, and with different bilinear forms determining the investigated. Then there was a period when the subject metrical structure without any recourse to transcendental languished; and now it flourishes once more—spurred by quantities or “real numbers”. Of course we proceed only the power of dynamic geometry packages like GSP, C.a.R., a very small way down this road, but far enough to es- Cabri, GeoGebra, and Cinderella; by the heroic efforts of tablish some analogs of results that have appeared first in Clark Kimberling in organizing the massive amount of in- Universal Hyperbolic Geometry ([14]); namely the con-

63 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

2 1 currency of some quadruples of lines associated to the G = 3C + 3 H. The reader might like to check that using classical Incenters, Gergonne points, Nagel points, Mitten- the green notation of perpendicularity, the green altitudes punkts, Spieker points as well as the New points which we really do meet at H, and the green midlines/perpendicular introduce here. We identify the resulting centers in Kim- bisectors really do meet at C. berling’s list. In the general situation there are four Incenters/Excenters Our basic technology is simple but powerful: we propose I0,I1,I2 and I3 which algebraically are naturally viewed to replace the affine study of a general triangle under a symmetrically. Associated to any one Incenter Ij is a Ger- particular bilinear form with the study of a particular tri- gonne point G j = X7 (not to be confused with the centroid angle under a general bilinear form—analogous to the also labelled G), a Nagel point Nj = X8, a Mittenpunkt projective situation as in ([14]), and using the framework D j = X9,a Spieker point S j = X10 and most notably a New of Rational Trigonometry ([10], [11]). By choosing a very point L j. It is not at all obvious that these various points elementary standard Triangle—with vertices the origin and can be defined for a general affine geometry, but this is the the two standard basis vectors—we get reasonably pleasant case, as we shall show. The New points L0,L1,L2,L3 are and simple formulas for various points, lines and construc- a particularly novel feature of this paper. They really do tions. An affine change of coordinateschanges any triangle appear to be new, and it seems remarkable that these im- under any bilinear form to the one we are studying, so our portant points have not been intensively studied, as they results are in fact very general. fit naturally and simply into the Incenter hierarchy, as we Our principle results center around the classical four shall see. points, but a big differencewith our treatment is that we ac- The four-fold symmetry between the four Incenters is knowledge from the start that the very existence of the In- maintained by all these points: so in fact there are four center hierarchy is dependent on number-theoretical con- Gergonne, Nagel, Mittenpunkt, Spieker and New points, ditions which end up playing an intimate and ultimately each associated to a particular Incenter, as also pointed out rather interesting role in the theory. Algebraically it be- in ([8]). Figure 1 shows just one Incenter and its related comes difficult to separate the classical incenter from the hierarchy: as we proceed in this paper the reader will meet three closely related excenters, and the quadratic relations the other Incenters and hierarchies as well. that govern the existence of these carry a natural four-fold symmetry between them. This symmetry becomes crucial X20 to simplifying formulas and establishing theorems. So in 12 our framework, there are four Incenters I0,I1,I2 and I3, not one. To showcase the generality of our results, we illustrate the- 8 Ij orems not over the Euclidean plane, but in the Minkowski C plane coming from Einstein’s special theory of relativity Lj A3 in null coordinates, where the metrical structure is deter- Gj 4 A2 G X69 mined by the bilinear form K Dj (x ,y ) · (x ,y ) ≡ x y + y x . Sj 1 1 2 2 1 2 1 2 A1 -4 4 8 12 In the language of Chromogeometry ([12] , [13]), this H is green geometry, with circles appearing as rectangu- N lar hyperbolas with asymptotes parallel to the coordinate j axes. Green perpendicularity amounts to vectors being Eu- clidean reflections in these axes, while null vectors are par- Figure 1: Aspects of the Incenter hierarchy in green ge- allel to the axes. It is eye-opening to see that triangle geometry ometry is just as rich in such a relativistic setting as it is in The main aims of the paper are to set-up a coordinate sys- the Euclidean one! tem for triangle geometry that incorporates the number- theoretical aspects of the Incenter hierarchy, and respects 1.1 Summary of results the four-foldsymmetryinherentin it, and then to use this to We summarize the main results of this paper using Figure catalogue existing as well as new points and phenomenon. 1 from green geometry. As established in ([13]), the trian- Kimberling’s Triangle Center Encyclopedia ([6]) distin- gle A1A2A3 has a green Euler line CHG just as in the Eu- guishes the classical Incenter X1 as the first and perhaps clidean setting, where C = X3 is the Circumcenter, G = X2 most important triangle center. Our embrace of the four- is the Centroid, and H = X is the Orthocenter, with the fold symmetry between incenters and excenters implies −→ −→ 4 affine ratio CG : GH = 1:2, which we may express as something of a re-evaluation of some aspects of classical

64 KoG•16–2012 N. Le, N. J. Wildberger: Universal Affine Triangle Geometry and Four-fold Incenter Symmetry triangle geometry; instead of certain distinguished centers In particular the various points alluded to here have consis- we have rather distinguished quadruples of related points. tent definitions over general fields and with arbitrary bilin- Somewhat surprisingly, this point of view makes visible a ear forms! The New points are the meets of the lines L1,3 number of remarkable strong concurrences—-where four and L4,7, they are the reflections of the Incenters Ij in the symmetrically-definedlines meet in a center. The proofs of Circumcenter C, and they are the reflections of the Ortho- these relations are reasonably straight-forward but not au- center H in the Spieker points S j. tomatic, as in general certain important quadratic relations It is also worth pointing out a few additional relations be- are needed to simplify expressions for incidence. Here is a tween the triangle centers that appear here: the point X69, summary of our main results. defined as the Isogonal conjugate of the Orthocenter H, is Main Results also the central dilation in the Centroid of the Symmedian δ point K; in our notation X69 = −1/2 (K). This implies that i) The four lines IjG j, j = 0,1,2,3, meet in the De 2 1 G = K + X69. In addition the De Longchampspoint X20, Longchamps point X20 (orthocenter of the Double 3 3 triangle) — these are the Soddy lines ([9]). defined as the orthocenter of the Double (or anti-medial) triangle is also the reflection of the Orthocenter H in the ii) The four lines IjNj meet in the Centroid G = X2, and Circumcenter C. These relations continue to hold in the 2 1 in fact G = 3 Ij + 3 Nj — these are the Nagel lines. general situation. The Spieker points S j also lie on the Nagel lines, and Table 1 summarizes the various strong concurrences we in fact S 1 I 1 N j = 2 j + 2 j. have found. Note however that not all pairings yield con- iii) The four lines IjD j meet in the Symmedian point current quadruples: for example the lines joining corre- K = X6 (isogonal conjugate of the Centroid G) — sponding Nagel points and Mittenpunkts are not in general the standard such line is labelled L1,6 in [6]. concurrent.

iv) The four lines IjL j meet in the Circumcenter C, and In the ﬁnal section of the paper, we give some further re- 1 1 in fact C = 2 Ij + 2 L j — the standard such line is sults and directions involving chromogeometry. labelled L1,3.

v) The four lines G jNj meet in the point X69 (isotomic 1.2 Affine structure and vectors conjugate of the Orthocenter H) — these lines are labelled L7,8. We begin with some terminology and concepts for elementary affine geometry in a linear algebra setting, follow- vi) The four lines G jD j meet in the Centroid G = X2, ing [10]. Fix a field F, of characteristic not two, whose 2 1 and in fact G = 3 D j + 3 G j — the standard such line elements will be called numbers. We work in a two- is labelled L2,7. dimensional affine space A2 over F, with V2 the associated two-dimensional vector space. A point is then an ordered vii) The four lines D jS j meet in the Orthocenter H = X4 pair A ≡ [x,y] of numbers enclosed in square brackets, typ- — the standard such line is labelled L4,7. ically denoted by capital letters, such as A,B,C etc. A vec- viii) The four lines NjL j meet in the point X20 (ortho- tor of V2 is an ordered pair v ≡ (x,y) of numbers enclosed center of the Double triangle), and in fact L j = in round brackets, typically u,v,w etc. Any pair of points 1 1 −→ 2 X20 + 2 Nj — the standard such line is labelled L1,3. A and B determines a vector v = AB; so for example if −→ ix) The New point L lies on the line D S which also A ≡ [2,−1] and B ≡ [5,1], then v = AB = (3,2), and this j j j −→ passes through the Orthocenter H, and in fact S j = is the same vector v = CD determined by C ≡ [4,1] and 1 1 2 H + 2 L j. D ≡ [7,3].

Incenter I Gergonne G Nagel N Mittenpunkt D Spieker S New L

Incenter I − X20 G = X2 K = X6 G = X2 C = X3 Gergonne G X20 − X69 G = X2 − − Nagel N G = X2 X69 − − G = X2 X20 Mittenpunkt D K = X6 G = X2 − − H = X4 H = X4 Spieker S G = X2 − G = X2 H = X4 − H = X4 New L C = X3 − X20 H = X4 H = X4 − Table 1

65 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

The non-zero vectors v1 ≡ (x1,y1) and v2 ≡ (x2,y2) are Three points A1 = [x1,y1],A2 = [x2,y2],A3 = [x3,y3] are parallel precisely when one is a non-zero multiple of the collinear precisely when they lie on a common line, which other, this happens precisely when amounts to the condition x1y2 − x2y1 = 0. x1y2 − x1y3 + x2y3 − x3y2 + x3y1 − x2y1 = 0.

Vectors may be scalar-multiplied and added component- Three lines ha1 : b1 : c1i, ha2 : b2 : c2i and ha3 : b3 : c3i are wise, so that if v and w are vectors and α,β are numbers, concurrent precisely when they pass through the same the linear combination αv + βw is defined. For points A point, which amounts to the condition and B and a number λ, we may define the affine combination C = (1 − λ)A + λB either by coordinates or by inter- a1b2c3 −a1b3c2 +a2b3c1 −a3b2c1 +a3b1c2 −a2b1c3 = 0. −→ preting it as the sum A+λAB. An important special case is 1.3 Metrical structure: quadrance and spread when λ = 1/2; in that case the point C ≡ A/2 + B/2 is the midpoint of AB, a purely affine notion independent of any We now introduce a metrical structure, which is deter- metrical framework. mined by a non-degeneratesymmetric 2×2 matrix C, with Once we fix an origin O ≡ [0,0], the affine space A2 and entries in the fixed field F over which we work. This ma- the associated vector space V2 are naturally identified: to trix defines a symmetric bilinear form on vectors, regarded every point A ≡ [x,y] there is an associated position vector as row matrices, by the formula −→ a = OA = (x,y). So points and vectors are almost the same v u vu vCuT thing, but not quite. The choice of distinguished point also · = = . allows us a useful notational shortcut: we agree that for a Here non-degenerate means detC 6= 0, and implies that if λ point A ≡ [x,y] and a number we write v · u = 0 for all vectors u then v = 0. Note our introduction of the simpler notation v · u = vu, λ[x,y] ≡ (1 − λ)O + λA = [λx,λy]. (1) so that also v · v = v2. There should be no confusion with A line is a proportion l ≡ ha : b : ci where a and b are matrix multiplication, even if v and u are viewed as 1 × 2 not both zero. The point A ≡ [x,y] lies on the line l ≡ matrices. Since C is symmetric, v · u = vu = uv = u · v. ha : b : ci, or equivalently the line l passes through the Two vectors v and u are perpendicular precisely when point A, precisely when v · u = 0. Since the matrix C is non-degenerate, for any vector v there is, up to a scalar, exactly one vector u which ax + by + c = 0. is perpendicular to v. The bilinear form determines the main metrical quantity: For any two distinct points A1 ≡ [x1,y1] and A2 ≡ [x2,y2], the quadrance of a vector v is the number there is a unique line l ≡ A1A2 which passes through them 2 both; namely the join Qv ≡ v · v = v .

2 A1A2 = hy1 − y2 : x2 − x1 : x1y2 − x2y1i. (2) A vector v is null precisely when Qv = v · v = v = 0, in other words precisely when v is perpendicular to itself. In vector form, this line has parametric equation l : A1 +λv, The quadrance between the points A and B is −−→ where v = A1A2 = (x2 − x1,y2 − y1) is a direction vector −→ for the line, and λ is a parameter. The direction vector Q(A,B) ≡ QAB. of a line is unique up to a non-zero multiple. The line In the Euclidean case, this is of course the square of the l ≡ ha : b : ci has a direction vector v = (−b,a). usual distance. But quadrance is a more elementary and Two lines are parallel precisely when they have parallel fundamental notion than distance, and its algebraic nature direction vectors. For every point P and line l, there is makes it ideal for metrical geometry using other bilinear then precisely one line m through P parallel to l, namely forms (as Einstein and Minkowski tried to teach us a cen- m : P + λv, where v is any direction vector for l. For any tury ago!) two lines l ≡ ha : b : c i and l ≡ ha : b : c i which are 1 1 1 1 2 2 2 2 Two lines l and m are perpendicular precisely when they not parallel, there is a unique point A ≡ l l which lies on 1 2 have perpendicular direction vectors. A line is null pre- them both; using (1) we can write this meet as cisely when it has a null direction vector (in which case all

b1c2 − b2c1 c1a2 − c2a1 direction vectors are null). A ≡ l1l2 = , a b − a b a b − a b We now make the important observation that the afﬁne no- 1 2 2 1 1 2 2 1 −1 tion of parallelism may also be recaptured via the bilinear = (a1b2 − a2b1) [b1c2 − b2c1,c1a2 − c2a1]. (3) form. (This result also appears with the same title in [15].)

66 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Theorem 1 (Parallel vectors) Vectors v and u are paral- Proof. We may that assume at least two of the vectors are lel precisely when linear independent, as otherwise all spreads are zero and 2 the relation is trivial. So suppose that v1 and v2 linearly in- QvQu = (vu) . dependent, and v3 = kv1 + lv2. Suppose the bilinear form a b is given by the matrix Proof. If C = , v = (x,y) and u = (z,w), then an b c a b explicit computation shows that C = b c 4 2 2 2 (xw − yz) ac − b QvQu −(vu) = − . with respect to the ordered basis v1,v2. Then in this basis 2 2 2 2 2 2 (ax + 2bxy + cy ) (az + 2bzw + cw ) v1 = (1,0),v2 = (0,1) and v3 = (k,l) and we may compute that Since the quadratic form is non-degenerate, ac − b2 6= 0, so we see that the left hand side is zero precisely when ac − b2 l2 ac − b2 s3 = s2 = xw − yz = 0, in other words precisely when v and u are ac a(ak2 + 2bkl + cl2) parallel. k2 ac − b2 This motivates the following measure of the non- s = . 1 b ak2 2bkl cl2 parallelism of two vectors; the spread between non-null ( + + ) vectors v and u is the number Then (4) is an identity, satisfied for all a,b,c,k and l. (vu)2 We now mention three consequences of the Triple spread s(v,u) ≡ 1 − . QvQu formula, taken from [10]. The Equal spreads theorem asserts that if s1 = s2 = s, then s3 = 0 or s3 = 4s(1 − s). This This is the replacement in rational trigonometry for the 2 2 2 2 θ follows from the identity (s + s + s3) − 2 s + s + s3 − transcendental notion of angle , and in the Euclidean case 2 2 2 4s s = −s s − 4s + 4s . The Complementary spreads it has the value sin θ. Spread is a more algebraic, log- 3 3 3 ical, general and powerful notion than that of angle, and theorem asserts that if s3 = 1 then s1 + s2 = 1. This follows by rewriting the Triple spread formula in the form together quadrance and spread provide the foundation for 2 Rational Trigonometry, a new approach to trigonometry (s3 − s1 − s2) = 4s1s2 (1 − s3). developed in [10]. The current pre-occupation with dis- And the Perpendicular spreads theorem asserts that if v and u are non-null planar vectors with perpendicular tance and angle as the basis for Euclidean geometry is a ⊥ ⊥ ⊥ ⊥ historical aberration contrary to the explicit orientation of vectors v and u , then s(v,u)= s v ,u . This follows from the Complementary spreads theorem, since if Euclid himself, and is a key obstacle to appreciating and ⊥ ⊥ ⊥ ⊥ ⊥ understanding the relativistic geometry introduced by Ein- s v,v = s u,u = 1, then s v ,u = 1 − s v ,u = 1 − (1 − s(v,u)) = s(v,u). stein and Minkowski. The spread s v u is unchanged if either v or u are multi- ( , ) 1.5 Altitudes and orthocenters plied by a non-zero number, and so we define the spread between any non-null lines l and m with direction vectors v Given a line l and a point P, there is a unique line n and u to be s(l,m) ≡ s(v,u). From the Parallel vectors the- through P which is perpendicular to the line l; it is the line orem, the spread between parallel lines is 0. Two non-null n : P + λw, where w is a perpendicular vector to the direc- lines l and m are perpendicular precisely when the spread tion vector v of l. We call n the altitude to l through P. between them is 1. 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 1.4 Triple spread formula P agrees with the parallel to l through P. We now derive one of the basic formulasin the subject: the We use the following conventions: a set {A,B} of two dis- relation between the three spreads made by three (copla- tinct points is a side and is denoted AB, and a set {l,m} nar) vectors, and give a linear algebra proof, following the of two distinct lines is a vertex and is denoted lm. A set same lines as the papers [11] and [15]. {A1,A2,A3} of three distinct non-collinear points is a triangle and is denoted A1A2A3. The triangle A1A2A3 has Theorem 2 (Triple spread formula) Suppose that lines l3 ≡ A1A2, l2 ≡ A1A3 and l1 ≡ A2A3 (by assumption v1,v2,v3 are (planar) non-null vectors with respective no two of these are parallel), sides A1A2,A1A3 and A2A3, spreads s1 ≡ s(v2,v3), s2 ≡ s(v1,v3) and s3 ≡ s(v1,v2). and vertices l1l2,l1l3 and l2l3. Then The triangle A1A2A3 also has three altitudes n1,n2,n3 passing through A1,A2,A3 and perpendicular to the oppo- (s + s + s )2 = 2 s2 + s2 + s2 + 4s s s . (4) 1 2 3 1 2 3 1 2 3 site lines A2A3, A1A3,A1A2 respectively. The following 67 KoG•16–2012 N. Le, N. J. Wildberger: Universal Affine Triangle Geometry and Four-fold Incenter Symmetry holds both for affine and projective geometries: we give by multiplication by M is to send the original triangle to a short and novel proof here for the general affine case. the standard triangle with points [0,0], [1,0] and [0,1]. The bilinear form in these new standard coordinates is Theorem 3 (Orthocenter) For any triangle A1A2A3 the T given by the matrix NCgN which is, up to a multiple, three altitudes n1,n2,n3 are concurrent at a point H. 1 1 a b Proof. Suppose that a1,a2,a3 are the associated position C = 4 = . 1 64 b c vectors to A1,A2,A3 respectively. Since no two of the 25 lines of the triangle A1A2A3 are parallel, the Perpendicular spreads corollary implies that no two of the three altitude We will shortly see that the Orthocenter in standard coor- lines are parallel. Define H to be the meet of n1 and n2, dinates is (ac − b2)−1 [b(c − b),b(a − b)]. In our example with h the associated position vector. In the identity 13 25 this would be the point − 3 , 12 , and to convert that back into the original coordinates, we would multiply by N to (h − a1)(a3 − a2)+ (h − a2)(a1 − a3) = (h − a3)(a1 − a2) get the left hand side equals 0 by assumption, so the right hand 13 25 is also equal to 0, implying that h − a3 is perpendicular to − 3 12 N = 9 −3 the line a1a2. Therefore, the three altitude lines n1,n2,n3 are concurrent at the point H. and translate by (3,1) to get the original orthocenter H = [12,−2]. This is shown in Figure 2, along with the We call H the orthocenter of the triangle A1A2A3. Centroid G = [82/15,18/5] and the Circumcenter C = 1.6 Change of coordinates and an explicit example [11/5,32/5]—we will meet these points shortly. If we change coordinates via either an affine transformation in the original affine space A2, or equivalently a linear 2 transformation in the associated vector space V , then the C matrix for the form changes in the familiar fashion. Sup- A3 pose φ : V → V is a linear transformation given by an in- A2 4 G vertible 2 × 2 matrix M, so that φ(v)= vM = w, with inverse matrix N, so that wN = v. Define a new bilinear form e ◦ by A1 T 4 8 w1 ◦ w2 ≡ (w1N) · (w2N) = (w1N)C (w2N) H T T = w1(NCN )w2 . (5) So the matrix C for the original bilinear form · becomes the T matrix D ≡ NCN for the new bilinear form ◦. Figure 2: Euler line in green geometry Example 1 We illustrate these abstractions in a concrete example that will be used throughout in our diagrams. 1.7 Bilines

Our basic Triangle shown in Figure 2 has points A1 ≡ A biline of the non-null vertex l1l2 is a line b which passes [3,1],A2 ≡ [4,4] and A3 ≡ [47/5,29/5], and lines A1A2 = through l1l2 and satisfies s(l1,b)= s(l2,b). The existence h−3:1:8i,A1A3 = h−3:4:5i and A2A3 = h1 : −3:8i. of bilines depends on number-theoretical considerations of The bilinear form we will consider is that of green ge- a particularly simple kind. ometry in the language of chromogeometry ([12], [13]), 0 1 determined by the symmetric matrix C = and Theorem 4 (Vertex bilines) If v and u are linearly inde- g 1 0 pendent non-null vectors, then there is a non-zero vector corresponding quadrance Q = 2xy. After translation (x,y) w with s(v,w)= s(u,w) precisely when 1 − s(v,u) is a by (−3,−1) we obtain A1 = [0,0], A2 = [1,3], A3 = square. In this case we may renormalize v and u so that [32/5,24/5]. The matrix N and its inverse M Qv = Qu, and then there are exactly two possibilities for e e e 1 3 − 1 5 w up to a multiple, namely v + u and v − u, and these are N = M = N−1 = 3 24 32 24 4 − 5 perpendicular. 5 5 9 72 send [1,0] and [0,1] to A2 and A3, and A2 and A3 to [1,0] Proof. Since v and u are linearly independent, any vector and [0,1] respectively. So the effect of translation followed can be written uniquely as w = kv + lu for some numbers e e e e 68 KoG•16–2012 N. Le, N. J. Wildberger: Universal Affine Triangle Geometry and Four-fold Incenter Symmetry k and l. The condition s(v,w)= s(u,w) amounts to they are mutually concurrent). Naturally this triangle has been chosen carefully to ensure that Incenters do exist. In 2 2 (vw) (uw) 2 2 = ⇐⇒ u2 kv2 + lvu = v2 kvu + lu2 green geometry, a vertex formed from a light-like line and QvQw QuQw a time-like line will not have bislines, not even approxi- 2 2 ⇐⇒ u2 k2 v2 + 2lkv2 (vu)+ l2 (vu) = mately over the rational numbers. 2 =v2 k2 (vu)2 + 2lku2 (vu)+ l2 u2 2 Standard coordinates and triangle 2 2 ⇐⇒ k2u2 v2 +l2u2 (vu)2 =k2v2 (vu)2+ l2v2u2 geometry ⇐⇒ v2u2 −(vu)2 k2v2 − l2u2 = 0. Our principle strategy to study triangle geometry is to ap- ply an affine transformation to move a general triangle to Since v and u are by assumption not parallel, the first term standard position: is non-zeroby the Parallel vectors theorem, and so the condition s(v,w)= s(u,w) is equivalent to k2v2 = l2u2. Since A1 = [0,0] A2 = [1,0] and A3 = [0,1]. (6) v,u are non-null, v2 and u2 are non-zero, so k and l are also, With this convention, A A A will be called the (standard) since by assumption w = kv + lu is non-zero. 1 2 3 Triangle, with Points A ,A ,A . The Lines of the Trian- So if s(v,w)= s(u,w) then we may renormalize v and u 1 2 3 gle are then 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)= l1 ≡ A2A3 = h1:1: −1i l2 ≡ A1A3 = h1:0:0i 2 2 (vu) / v2 is a square. There are thene two solutions:e w = v + u and w =eve− u, corresponding to l = ±k. Since l3 ≡ A2A1 = h0:1:0i. v u v u v2 u2 0 these vectors are perpendic- ( + )( − )= − = , All further objects that we define with capital letters ular. The converse is straightforward along the same lines. refer to this standard Triangle, and coordinates in this framework are called standard coordinates. In general

Example 2 In our example triangle of Figure 2, v1 = the standard coordinates of points and lines in the plane −−→ −−→ of the original triangle depend on the choice of affine A2A3 = (27/5,9/5), v2 = A1A3 = (32/5,24/5) and v3 = −−→ transformation—we are in principle free to permute the A A = (1,3), so 1 2 vertices—but triangle centers and central lines will have T 2 well-defined standard coordinates independent of such per- v2Cgv3 25 s(v2,v3)= 1 − = mutations. v C vT v C vT 16 2 g 2 3 g 3 Since we have performed an affine transformation, what- ever metrical structure we started with has changed as in is a square, so the vertex at A1 has bilines. Since Qv2 = T T (5). So we will assume that the new metrical structure, in v2Cgv2 = 1536/25 and Qv3 = v3Cgv3 = 6, we can renor- −−→ standard coordinates, is determined by a bilinear form with malize v2 by scaling it by 5/16 to get u2 = A1B = (2,3/2) −−→ generic symmetric matrix so that now Qu = Qv . This means that u2 + v3 = A1C1 −−→2 3 and u2 −v3 = A1C2 are the direction vectors for the bilines a b C ≡ . (7) of the vertex at A1. b c 8 b We assume that the form is non-degenerate, so that the de- I terminant C1 A3 b ∆ ≡ detC = ac − b2 4 A2 I v3 is non-zero. Another important number is the mixed trace I B u2 A1 d ≡ a + c − 2b. 4 8 C2 It will also be useful to introduce the closely related sec- ondary quantities

Figure3 : Green bilines b at A1 a ≡ c − b b ≡ a − c c ≡ a − b These are shown in Figure 3, along with three of the four Incenters I (the other two vertices also have bilines, and to simplify formulas. For example d = a + c.

69 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Theorem 5 (Standard triangle quadrances and spreads) The Circumlines are the lines of the Medial triangle The quadrances and spreads of A1A2A3 are M1M2M3, these are

Q1 ≡ Q(A2,A3)= d Q2 ≡ Q(A1,A3)= c b1 ≡ M2M3 = h2:2: −1i b2 ≡ M3M1 = h2:0: −1i

Q3 ≡ Q(A1,A2)= a b3 ≡ M1M2 = h0:2: −1i. and The Double triangle of A1A2A3 (usually called the anti- ∆ ∆ medial triangle) is formed from lines through the Points s ≡ s(A A ,A A )= s ≡ s(A A ,A A )= 1 1 2 1 3 ac 2 2 3 2 1 ad parallel to the opposite Lines. This is D1D2D3 where ∆ s ≡ s(A A ,A A )= . D = [1,1] D = [−1,1] D = [1,−1]. 3 3 1 3 2 cd 1 2 3

Furthermore The lines of D1D2D3 are

2 2 2 b (c) (a) D2D3 = h1:1:0i D1D3 = h1:0: −1i 1 − s1 = 1 − s2 = 1 − s3 = . ac ad cd D1D2 = h0:1: −1i. Proof. Using the deﬁnition of quadrance, Figure 4 shows these objects for our example Triangle. T Q1 ≡ Q(A2,A3)= Q−−−→ = (−1,1)C (−1,1) A2A3 D1 = a + c − 2b = d

M1 and similarly for Q2 and Q3. Using the definition of spread, A3 A2 4 G s ≡ s(A A ,A A )= s((1,0),(0,1)) 1 1 2 1 3 M2 2 M3 D 2 (1,0)C (0,1)T = 1 − A1 (1,0)C(1,0)T (0,1)C (0,1)T D 3 4 8 1 ∆ = 1 − b2 = ac ac Figure 4: The Medial triangle M1M2M3 and Double triangle D D D and similarly for s2 and s3. 1 2 3 2.2 The Orthocenter hierarchy 2.1 Basic affine objects in triangle geometry We now introduce some objects involving the metrical We now write down some basic central objects which fig- structure, and so the entries a b c of C from (7). Recall ure prominently in triangle geometry, all with reference to , , that a c b and c a b the standard triangle A A A in the form (6). The deriva- ≡ − = − . 1 2 3 The Altitudes of A A A are the lines tions of these formulas are mostly immediate using the 1 2 3 two basic operations of joins (2) and meets (3). We be- n = hc : −a :0i n = hb : c : −bi n = ha : b : −bi. gin with some purely affine notions, independent of the 1 2 3 bilinear form. Theorem 6 (Orthocenter formula) The three Altitudes The Midpoints of the Triangle are meet at the Orthocenter 1 1 1 1 b M1 = , M2 = 0, M3 = ,0 . H = [a,c]. 2 2 2 2 ∆ The Medians are Proof. We know that the altitudes meet from the Ortho- center theorem. We check that n1 passes through H by d1 ≡ A1M1 = h1 : −1:0i d2 ≡ A2M2 = h1:2: −1i computing b∆−1 (ac − ac)= 0. d3 ≡ A3M3 = h2:1: −1i. Also n2 passes through H since

The Centroid is the common meet of the Medians b b 2 ∆ (ba + cc) − b = ∆ b(c − b)+ c(a − b)− ac + b = 0 1 1 G = , . 3 3 and similarly for n3. 70 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

The Midlines m1,m2 and m3 are the lines through the and similarly for t2 and t3. midpoints M1,M2 and M3 perpendicular to the respective sides— these are usually called perpendicular bisectors. The existence of an Euler line in relativistic geometries was established in [13], here we extend this to the general case. They are also the altitudes of M1M2M3:

m1 = −2c :2a : b m2 = h2b :2c : −ci Theorem 9 (Euler line) The points H,C and G are concurrent, and satisfy G = 1 H + 2C. The Euler line e ≡ CH m3 = h 2a :2b : −ai . 3 3 is Theorem 7 (Circumcenter) The Midlines m1,m2,m3 meet at the Circumcenter e = ∆ − 3bc : −∆ + 3ba : bb . 1 C = [cc,aa]. 2∆ Proof. Using the formulas above for H and C, we see that Proof. We check that m1 passes through C by computing 1 2 1 1 2 1 H + C = [ba,bc]+ [cc,aa] 1 3 3 3 ∆ 3 2∆ −2c2c + 2a2a + b 2∆ 1 1 = ac − b2,ac − b2 = [1,1]= G. 1 2 2 ∆ = −2(a − b) c + 2(c − b) a +(a − c)= 0 3 3 2(ac − b2) Computing the equation for the Euler line CH is straight- and similarly for m2 and m3. forward.

8 In Figure 5 we illustrate the situation with our basic example triangle with the Altitudes, Medians and Midlines C meeting to form the Orthocenter H, Centroid G and Cir- A3 cumcenter C respectively on the Euler line e. A2 4 G The bases of altitudes of M1M2M3 are: e 1 1 1 A1 E = [c,a] E = [c,a] E = [c,a]. 1 2d 2 2c 3 2a 4 8 H The joins of Points and corresponding bases of altitudes of M1M2M3 are

Figure 5: The Euler line of a triangle A1E1 = ha : −c :0i A2E2 = ha : c : −ai As Gauss realized, this is also a consequence of the Ortho- A3E3 = ha : c : −ci. center theorem applied to the Medial triangle M1M2M3, since the altitudes of the Medial triangle are the Midlines Theorem 10 (Medial base perspectivity) The three lines of the original Triangle. A1E1,A2E2,A3E3 meet at the point The three altitudes of the Double triangle D1D2D3 are 1 X69 = [c,a]. t1 = c : −a : −b t2 = hb : c : −ai t3 = ha : b : −ci. a + c − b Theorem 8 (Double orthocenter formula) The three al- Proof. Straightforward. titudes of the Double triangle meet in the De Longchamps point 2.3 Bilines and Incenters 1 2 2 We now introduce the Incenter hierarchy. Unlike the Or- X20 ≡ b − 2bc + ac,b − 2ab + ac . ∆ thocenter hierarchy, this depends on number-theoretical Proof. We check that t1 passes through X20 by computing conditions. Recall that d ≡ a + c − 2b. 1 2 2 Theorem 11 (Existence of Triangle bilines) The Trian- ∆ c b − 2bc + ac − a b − 2ab + ac − b gle A A A has Bilines at each vertex precisely when we 1 1 2 3 2 2 can find numbers u,v,w in the field satisfying = ∆((a − b) b − 2bc + ac − (c − b) b − 2ab + ac 2 − (a − c)ac − b )= 0 ac = u2 ad = v2 cd = w2. (8) 71 KoG•16–2012 N. Le, N. J. Wildberger: Universal Affine Triangle Geometry and Four-fold Incenter Symmetry

Proof. From the Vertex bilines theorem, bilines exist pre- Proof. We use the Bilines theorem to find bilines through cisely when the spreads s1, s2, s3 of the Triangle have the A1 = [0,0]. The lines meeting at A1 have direction vectors T property that 1 − s1,1 − s2,1 − s3 are all squares. From v1 = (0,1) and v2 = (1,0), with Qv1 = (0,1)C (0,1) = c the Standard triangle quadrances and spreads theorem, this T and Qv2 = (1,0)C (1,0) = a. Now we renormalize and set 2 occurs for our standard triangle A1A2A3 precisely when we u = v v to get Q = v c = a = Q . So the biliness at A 1 w 1 u1 w2 v2 1 can find u, v, w satisfying (8). have direction vectors There is an important flexibility here: the three Incenter v v u + v = (0,1)+ (1,0)= 1, and constants u,v,w are only determined up to a sign. The 1 2 w w relations imply that v v u − v = (0,1) − (1,0)= −1, 1 2 w w d2u2 = v2w2 c2v2 = u2w2 a2w2 = u2v2. and the bilines are b1+ ≡ hv : w :0i and b1− ≡ hv : −w :0i. So we may choose the sign of u so that du = vw, and mul- Similarly you may check the other bilines through A2 and tiplying by u we get A3. Theorem 13 (Incenters) The triples {b ,b ,b }, acd = uvw. 1+ 2+ 3+ {b1+,b2−,b3−}, {b1−,b2+,b3−} and {b1−,b2−,b3+} of From this we deduce that Bilines are concurrent, meeting respectively at the four Incenters du = vw cv = uw and aw = uv. (9) −uw uv 1 I0 = , = [−w,v] uv − uw + vw uv − uw + vw (d + v − w) The quadratic relations (8) and (9) will be very important for us, for they reveal that the existence of the Incenter uw −uv 1 I1 = , = [w,−v] −uv + uw + vw −uv + uw + vw (d − v + w) hierarchy is a number-theoretical issue which depends not only on the given triangle and the bilinear form, but also uw uv 1 I2 = , = [w,v] on the nature of the field over which we work, and they uv + uw + vw uv + uw + vw (d + v + w) allow us to simplify many formulas involving u,v and w. uw uv 1 Because only quadratic conditions are involved, we may I3 = , = [−w,−v]. uv + uw − vw uv + uw − vw (d − v − w) always extend our field by adjoining (algebraic!) square roots to ensure that a given triangle has bilines. Proof. We may check concurrency of the various triples The quadratic relations carry an important symmetry: we by computing may replace any two of u,v and w with their negatives, and the relations remain unchanged. So if we have a formula v w 0 v w 0 F involving u v w, then we may obtain related formulas det u u + w −u = det u u − w −u 0 , ,     F1,F2,F3 by replacing v,w with their negatives, u,w with u − v u −u u + v u −u their negatives, and u,v with their negatives respectively.  v −w 0  v −w 0 Adopting this convention allows us to exhibit the single = det u u − w −u = det u u + w −u = 0.     formula F0, since then F1,F2,F3 are determined—we refer u − v u −u u + v u −u to this as quadratic symmetry, and will make frequent     use of it in the rest of this paper. The corresponding meet of hv : w :0i, hu : u + w : −ui and From now on our working assumption is that: the stan- hu − v : u : −ui is dard triangle A A A has bilines at each vertex, implying 1 2 3 −uw uv that we have Incenter constants u,v and w satisfying (8) b1+b2+b3+ = , and (9). So u,v and w now become ingredients in our for- uv − uw + vw uv − uw + vw u mulas for various objects in the Incenter hierarchy, along = [−w,v] with the numbers a,b and c (and d) from the bilinear form aw − cv + du a b 1 C = . = [−w,v] ≡ I0. b c (d + v − w) We have used the quadratic relations, and the last equality Theorem 12 (Bilines) The Bilines of the Triangle are is valid since b1+ ≡ hv : w :0i and b1− ≡ hv : −w :0i through A1, b2+ ≡ hu : u + w : −ui and b2− ≡ hu : u − w : −ui through u(d + v − w)− (aw − cv + du)= cv − aw + uv − uw = 0. A2, and b3+ ≡ hu − v : u : −ui and b3− ≡ hu + v : u : −ui through A3. The computations are similar for the other Incenters.

72 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

The reader should check that the formulas for I1,I2,I3 may 2.4 New points also be obtained from I by the quadratic symmetry rule 0 One of the main novelties of this paper is the introduc- described above. From now on in such a situation we will tion of the four New points L associated to each Incenter only write down the formula corresponding to I , and we j 0 I . It is surprising that these points have seemingly slipped will also often omit algebraic manipulations involving the j through the radar: they deserve to be among the top twenty quadratic relations. in Kimberling’s list, in our opinion. The Incenter altitude ti j is the line through the Incenter Ij and perpendicular to the Line li of our Triangle. There are I twelve Incenter altitudes; three associated to each Incenter. L The Incenter altitudes associated to I are 0 L

t c d v w : a d v w : av cw 8 10 = h ( + − ) − ( + − ) + i I C t20 = hb(d + v − w) : c(d + v − w) : bw − cvi L A3 t30 = ha(d + v − w) : b(d + v − w) : aw − bvi. 2 4 A I The Contact points Ci j are the meets of corresponding I Incenter altitudes ti j and Lines li. There are twelve Con- tact points; three associated to each Incenter. The Contact A1 L 4 8 12 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) Figure 7: Green Incenter altitudes, New points L and In- 1 New center C C30 = [bv − aw,0]. a(d + v − w) Theorem 14 (New 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 I 12 Incenterwhich does not lie on any of the lines in that triple. b b The points where these triples meet are the New points Li; b b for example {t11,t22,t33} meet at b 8 I 1 L = [au + cv + bw + cc,cu − bv − aw + aa]. 0 2∆ A3 A2 Proof. We check that L0 as deﬁned is incident with t11 = 4 I hc(d − v + w) : −a(d − v + w) : −av − cwi by computing I ((c − b)u + cv + bw + c(a − b))(a−b)(a + c − 2b−v + w) b A1 + ((a−b)u−bv−aw+a(c−b))(−(c−b)(a+c−2b−v+w)) 4 8 12 − 2 ac − b2 ((c − b)v + (a − b)w) 3 3 2 2 2 3 2 2 2 Figure 6: Green bilines b, Incenters I, Contact points and = ac + 2ab − 2a cb − a b − ac + 2ac b + c b Incircles − 2cb3 + b2v2 − b2w2 − acv2 + acw2 In Figure 6 we see our standard example Triangle in the 2 2 2 2 2 green geometry with Bilines b at each vertex, meeting = ac − b a − c − 2ab + 2cb − v + w = 0 in threes at the Incenters I. The Contact points are also using the quadratic relations (8). The computations for the shown, as are the Incircles, which are the circles with re- other Incenter altitudes and L1,L2,L3 are similar. spect to the metrical structure centered at the Incenters and passing through the Contact points: they have equations in The In-New lines are the joins of corresponding Incenter the variable point X of the form Q(X,I)= Q(C,I) where I points and New points. The In-New line associated to I0 is is an incenter and C is one of its associated Contact points. 2 I0L0 = h−aad + ac + ab − 2b v + aaw : In this green geometry such circles appear as rectangular 2 hyperbolas, with axes parallel to the coordinate axes. ccd +ccv + ac + cb− 2b w : −aaw − ccvi. 73 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Theorem 15 (In-New center) The four In-New lines IjL j Since the central dilation preserves spread, the three alti- δ are concurrent and meet at the circumcenter tudes of A1A2A3 are sent by −1/2 to the three altitudes of the medial triangle, which are the midlines/perpendicular 1 bisectors of the original Triangle, showing again that δ C = ∆ [cc,aa], −1/2 2 sends the orthocenter H to the circumcenter C, and as in the Euler line theorem it follows that G lies on e = HC, and in factC is the midpoint of IjL j. dividing HC in the affine ratio 2 : 1. We will see later that the central dilation also explains Proof. We check that C is the midpoint of IjL j by computing aspects of the various Nagel lines (there are four), since δ −1/2 takes any Incenter Ii to an incenter of the Medial 1 1 1 1 triangle, called a Spieker point Si. It follows that the I + L = [−w,v] 2 j 2 j 2 (d + v − w) four joins of Incenters and corresponding Spieker points 1 1 all pass through G, and G divides each side IiSi in the affine + [(c−b)u+cv+bw+c(a−b),(a−b)u−bv−aw+a(c−b)] 2 2∆ ratio 2:1. δ δ 1 The inverse of the central dilation −1/2 is −2, which takes = [2c(a − b) (d + v − w),2a(c − b) (d + v − w)] 4∆(d + v − w) the Points of A1A2A3 to the points of the Double triangle 1 D1D2D3, which has A1A2A3 as its medial triangle. = ∆ [cc,aa] = C. 2 Theorem 16 (Central dilation formula) The central dilation takes X = [x,y] to The In-New center theorem shows that what we are calling 1 δ (X)= [1 − x,1 − y] the In-New lines are also the In-Circumcenter lines, the −1/2 2 standard one which is labelled L in [6]. The Incenter al- 1,3 while the inverse central dilation δ takes X to δ (X)= titudes, New points and In-New lines are shown in Figure −2 −2 [1 − 2x,1 − 2y]. 7. δ 1 2 The proofs in these two theorems are typical of the ones Proof. If Y = −1/2 (X) then affinely 3 X + 3Y = G so that which appear in the rest of the paper. Algebraic manipula- 3 1 1 tions are combined with the quadratic relations to simplify Y = G − X = [1 − x,1 − y]. 2 2 2 expressions. Although sometimes long and involved, the δ verifications are in principle straightforward, and so from Inverting, we get the formula for −2 (X). now on we omit the details for results such as these. Example 3 The central dilation of the Orthocenter is 1 b(c − b) b(a − b) δ (H)= 1 − ,1 − 3 Transformations −1/2 2 ∆ ∆ Important classical transformations of points associated to 1 1 = ∆ [c(a − b),a(c − b)] = ∆ [cc,aa]= C a triangle include dilations in the centroid, and the isogo- 2 2 nal and isotomic conjugates. It is useful to have general which is the Circumcenter. formulae for these in our standard coordinates. Example 4 The inverse central dilation of the Orthocen- ter is the De Longchamps point X —the orthocenter of 3.1 Dilations about the Centroid 20 the Double triangle D1D2D3 δ λ The dilation of factor centered at the origin takes [x,y] 2b(c − b) 2b(a − b) λ δ to [x,y]. This also acts on vectors by scalar multiplying, −2 (H)= X20 ≡ 1 − ∆ ,1 − ∆ and in particular it leaves spreads unchanged and multi- 1 plies any quadrance by a factor of λ2. Similarly the dila- b2 2cb ac b2 2ab ac −→ = ∆ − + , − + . tion centered at a point A takes a point B to A + λAB. Any dilation preserves directions of lines, so preserves spreads, 3.2 Reflections and Isogonal conjugates and changes quadrances between points proportionally. Suppose that v is a non-null vector, so that v is not perpen- Given our Triangle A A A with centroid G, define the 1 2 3 dicular to itself. It means that we can find a perpendicular central dilation δ to be the dilation by the factor −1/2 −1/2 vector w so that v and w are linearly independent. Now if centered at G. It takes the three Points of the Triangle to the u is an arbitrary vector, write u = rv + sw for some unique midpoints M ,M ,M of the opposite sides. This medial 1 2 3 numbers r and s, and define the reflection of u in v to be triangle M1M2M3 then clearly has lines which are parallel to the original triangle. rv (u) ≡ rv − sw.

74 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

If we replace v with a multiple, the reflection is unchanged. 3.3 Isotomic conjugates Now suppose that l and m are lines which meet at a point Theorem 18 (Isotomic conjugates) If X is a point dis- A, with respective direction vectors v and u. Then the re- tinct from A ,A ,A , then the lines joining the points flection of m in l is the line through A with direction vector 1 2 3 A ,A ,A to the reflections in the midpoints M ,M ,M of r (u). It is important to note that if n is the perpendicular 1 2 3 1 2 3 v the meets of A X,A X,A X with the lines of the Triangle to l through A, then 1 2 3 are themselves concurrent, meeting in the isotomic conjugate ofX. IfX = [x,y] then rl (m)= rn (m). y(x + y − 1) x(x + y − 1) Our standard triangle A A A determines an important t (X)= , . 1 2 3 x2 + xy + y2 − x − y x2 + xy + y2 − x − y transformation of points. Proof. The point X ≡ [x,y] has Cevian lines which meet Theorem 17 (Isogonal conjugate) If X is a point dis- the lines A2A3,A1A3,A1A2 respectively in the points tinct from A ,A ,A , then the reflections of the lines 1 2 3 x y y x A1X,A2X,A3X in thebilinesatA1,A2,A3 respectively meet , 0, ,0 . x + y x + y 1 − x 1 − y in a point i(X), called the isogonal conjugate of X. If X = [x,y] then These three points may be reflected respectively in the midpoints [1/2,1/2],[0,1/2],[1/2,0] to get the points x + y − 1 i(X)= [cy,ax]. y x 1 − x − y 1 − x − y ax2 + 2bxy + cy2 − ax − cy , 0, ,0 . x + y x + y 1 − x 1 − y Proof. First we reflect the vector a = (x,y) in the bi- The lines hx : −y :0i, h1 − x − y :1 − x : −1 + x + yi and lines hv : w :0i and hv : −w :0i through A . We do this 1 h1 − y :1 − x − y : −1 + x + yi joining these points to the by writing (x,y)= r (w,v)+ s(w,−v) = (rw + sw,rv − sv) original vertices meet at t (X) as defined above. and solving to get r = (2vw)−1 (vx + wy) and s = (2vw)−1 (vx − wy). The reflection is then Example 7 The isotomic conjugate of the Orthocenter H is 1 (c − b)b (a − b)b a − b c − b r (w,v) − s(w,−v)= (vx + wy)(w,v) t , = , ≡ X . 2vw ac − b2 ac − b2 a + c − b a + c − b 69 1 wy vx − (vx − wy)(w,−v)= , 2vw v w 4 Strong concurrences which is, up to a multiple and using the quadratic relations, 4.1 Sight Lines, Gergonne and Nagel points w2y,v2x = (cdy,adx)= d (cy,ax). We now adopt the principle that algebraic verifications of incidence, using the quadratic relations, will be omitted. So reflection in the biline at A1 takes the line A1X to A Sight line s is the join of a Contact point C with the the line A1 + λ1 (cy,ax). Similarly, by computing the re- i j i j Point A opposite to the Line that it lies on, and is natu- flections of (x − 1,y) and (x,y − 1) in the bilines at A2 i and A3, we find that the lines A2X and A3X get sent rally associated with the Incenter Ij. There are twelve Sight lines; three associated to each Incenter: to the lines A2 + λ2 (ax + (a − d)y − a,−ax − ay + a) and λ A3 + 3 (−cx − cy + c,x(c − d)+ cy − c) respectively. It is s10 = hc + v : −a+ w :0i now a computation that these three reflected lines meet at s = hcv − bw : c(d + v − w) : −cv + bwi the point i(X) as defined above. 20 s30 = ha(d + v − w) : −aw + bv : aw − bvi

Example 5 The isogonal conjugate of the centroid G is s11 = hc − v : −a− w :0i the symmedian point s21 = h−cv + bw : c(d − v + w) : cv − bwi 1 1 1 s = ha(d − v + w) : aw − bv : −aw + bvi K ≡ i , = [c,a]= X . 31 3 3 2(a + c − b) 6 s12 = hc + v : −a− w :0i Example 6 The isogonal conjugate of the Orthocenter H s22 = hcv + bw : c(d + v + w) : −cv − bwi is the Circumcenter: s32 = ha(d + v + w) : aw + bv : −aw − bvi b(c − b) b(a − b) 1 s13 = hc − v : −a+ w :0i i , = [c(a − b),a(c − b)] ∆ ∆ 2∆ s = h−cv − bw : c(d − v − w) : cv + bwi 23 = C = X3. s33 = ha(d − v − w) : −aw − bv : aw + bvi.

75 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

X20 1 1 shows that C = 2 H + 2 X20. Since the Euler line is e = CH, I X20 lies on e. 12 Figure 8 shows the Gergonne points G and the In- Gergonne lines meeting at X20.

Theorem 21 (Nagel points) The triples {s11,s22,s33}, 8 {s ,s ,s }, {s ,s ,s } and {s ,s ,s } of Sight I 10 32 23 20 31 13 30 21 12 A3 lines are concurrent. Each triple involves one Sight line G associated to each of the Incenters, and so is associated to G A2 4 the Incenter with which it does not share a Sight line. The I G points where these triples meet are the Nagel points Nj. I For example, {s11,s22 s33} meet at A1 , G -4 4 8 12 1 N0 = ∆ [(b + u)a + cv + bw,(b + u)c − bv − aw].

Figure 8: Green Sight lines, Gergonne points G,In- Proof. We check that N0 as deﬁned is incident with Gergonne lines and In-Gergonne center X20 hc − v : −a − w :0i by computing Here we introduce a well-known center of the triangle, the (b + u)a + cv + bw (b + u)c − bv − aw Gergonne point (see for example [2], [9]). ∆ (c − v)+ ∆ (−a− w) 2 2 2 2 Theorem 19 (Gergonne points) The triples {s ,s ,s }, −cduv +aduw +ccv w+aavw −acadv−accdw 10 20 30 = ∆ = 0 {s11,s21,s31}, {s12,s22,s32} and {s13,s23,s33} of Sight lines are concurrent. Each triple is associated to an Incen- using the quadratic relations, (8) and (9). ter, and the meets of these triples are the Gergonne points The computations for the other Sight lines and N1,N2,N3 G j. The Gergonne point associated to I0 is are similar.

b − u G0 = [w − a,−v − c]. I 2(du − cv + aw)− ∆ 12

The join of a correspondingIncenter Ij and Gergonne point

G j is an In-Gergonne line or Soddy line. There are four 8 Soddy lines, and I A3 N I0G0 =h2bcv + (∆ − 2bc)w − (∆ − 2bc)d : 4 A2 G 2baw + (∆ − 2ba)v + (∆ − 2ba)d : N I − (∆ − 2ba)v − (∆ − 2bc)wi. I A1

Theorem 20 (In-Gergonne center) The four In- -4 4 8 12 Gergonne/Soddy lines IjG j are concurrent, and meet at the De Longchamps point N 1 2 2 X20 = ∆ b − 2cb + ac,b − 2ab + ac Figure 9: Green Sight lines, Nagel points N,In-Nagel lines and In-Nagel center G = X2 which is the orthocenter of the Double triangle. Further- The join of a corresponding Incenter and Nagel point is an more the midpoint of HX20 is the Circumcenter C, so that In-Nagel line. There are four In-Nagel lines, and X20 lies on the Euler line. I0N0 = h2v + w − d : v + 2w + d : −v − wi. Proof. The concurrency of the In-Gergonne/Soddy lines IjG j is as usual. The equation In classical triangle geometry, the line I0N0 is called simply the Nagel line. 1 1 [b(c − b),b(a − b)] + b2 − 2cb + ac,b2 − 2ab + ac 2∆ 2∆ Theorem 22 (In-Nagel center) The four In-Nagel lines 1 IjNj are concurrent, and meet at the Centroid G = X2, and = [c(a − b),a(c − b)] = C 2 1 2∆ in fact G = 3 Ij + 3 Nj.

76 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Proof. Using the formulas above for I0 and N0, we see that Proof. We check that 2 1 2 1 I0 + N0 = [−w,v] 1 1 1 1 3 3 3 (d + v − w) X + N = b2 − 2cb + ac,b2 − 2ab + ac 2 20 2 0 2 ∆ 1 1 + [(b + u)a + cv + bw,(b + u)c − bv − aw] 3 ∆ 1 1 + ∆ [(b + u)a + cv + bw,(b + u)c − bv − aw] 1 2 = [∆(d + v − w),∆(d + v − w)] 3∆(d + v − w) 1 = ∆ [au + cv + bw + cc,cu − bv − aw + aa]= L0. 1 2 = [1,1]= G. 3

The join of a corresponding Gergonne point G j and Nagel 4.2 InMid lines and Mittenpunkts point Nj is a Gergonne-Nagel line. There are four Gergonne-Nagel lines, and The join of an Incenter Ij with a Midpoint Mi is an InMid G0N0 = h−au + av + aw : cu + cv + cw : −cw − avi. line. There are twelve InMid lines:

X20 I0M1 = hv + w − d : v + w + d : −v − wi

12 I0M2 = hv + w − d :2w : −wi

L I0M3 = h2v : v + w + d : −vi L I1M1 = hv + w + d : v + w − d : −v − wi 8 I1M2 = hv + w + d :2w : −wi

L A3 I1M3 = h2v : v + w − d : −vi G N G I2M1 = hv − w − d : v − w + d : −v + wi 4 G A2 N X69 I2M2 = hv − w − d : −2w : wi

A1 I2M3 = h2v : v − w + d : −vi G I3M1 = h−v + w − d : −v + w + d : v − wi 4 8 12 L I3M2 = h−v + w − d :2w : −wi I M 2v : v w d : v N 3 3 = h− − + + i.

Figure 10: Green Gergonne-Nagel center X69 and Nagel- New center X20 Theorem 25 (InMid lines) The triples of In- Mid lines {I1M1,I2M2,I3M3}, {I0M1,I2M3,I3M2}, Theorem 23 (Gergonne-Nagel center) The four {I0M2,I1M3,I3M1} and {I0M3,I1M2,I2M1} are concur- Gergonne-Nagel G jNj lines are concurrent, and meet at rent. Each triple involves one InMid line associated to the isotomic conjugate of the Orthocenter, each of three Incenters, and so is associated to the Incenter which does not appear. The points where these triples meet 1 X69 = [c,a]. are the Mittenpunkts D j. For example, {I1M1,I2M2,I3M3} a + c − b meet at The join of a corresponding New point L j and Nagel point Nj is a Nagel-New line. There are four Nagel-New lines, 1 D = [c + u + w,a + u − v]. and the one associated to I0 is 0 2(a + c − b + u − v + w) 2 L0N0 =hac − 3ab + 2b − cu + bv + aw : 2 3cb − ac − 2b + au + cv + bw : The join of a corresponding Incenter Ij and Mittenpunkt (a − c)b + (a − c)u − av + cwi. D j is an In-Mitten line. There are four In-Mitten lines, and Theorem 24 (Nagel-New center) The four Nagel-New lines NjL j meet in the De Longchamps point X20, and in 1 1 I0D0 =h(c + d)v + aw−ad : cv+(a + d)w + cd : −aw−cvi. fact L j = 2 N0 + 2 X20.

77 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Theorem 28 (Mitten-New center) The four Mitten-New lines D jL j are concurrent, and meet at the Orthocenter I 12 1 H = ∆ [ba,bc].

8 Figure 12 shows the four In-Mitten lines meeting at K = I X6, the four Gergonne-Mitten lines meeting at G = X2 and D A3 the four Mitten-New lines meeting at H = X . A2 M1 4 4 D M2 D I I M3 D

A1 I 12 4 8 12 L L

8 Figure 11: Green InMid lines and Mittenpunkts D I A3 D L G G A2 Theorem 26 (In-Mitten center) The four In-Mitten lines 4 G G are concurrent and meet at the symmedian point (see Ex- K D DI D ample 3) I A1 G 1 4 8 12 K = X6 = [c,a]. L 2(a + c − b) H

The join of a corresponding Gergonne point G j and Mit- tenpunkt D j is a Gergonne-Mitten line. There are four Figure 12: Green Mitten-New center H, Gergonne-Mitten Gergonne-Mitten lines and center G and In-Mitten center K ∆ ∆ ∆ ∆ ( −4bd) u + (4cc− )v + 2(4aa− )w + (a−2a) : 4.3 Spieker points D0G0 = −(∆ − 4bd) u + 2(4cc−∆)v + (4aa−∆)w−(c−2c)∆ : . * (∆−4cc)v + (∆−4aa)w − b∆ + 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 27 (Gergonne-Mitten center) The four Gergonne-Mitten lines G jD j meet in the Centroid G = X2, Theorem 29 (Spieker points) The four Spieker points are 2 1 and in fact G = 3 D j + 3 G j. 1 1 S = [v + d,−w + d] Proof. We use the formulas above for D0 and G0 to com- 0 2 (d + v − w) pute 1 1 S1 = [−v + d,w + d] 2 1 2 (d − v + w) D + G 3 0 3 0 1 1 S2 = [v + d,w + d] 2 1 2 (d + v + w) = [c + u + w,a + u − v] 3 2(a + c − b + u − v + w) 1 1 S = [−v + d,−w + d]. 1 b − u 3 2 (d − v − w) + [w − (c − b),−v − (a − b)] 3 2(du − cv + aw)− ∆ Proof. We use the central dilation formula which takes 1 = [1,1]= G. I = (d + v − w)−1 [−w,v] to the point 3 0 1 −w v S0 ≡ δ (I0)= 1 − ,1 − The join of a correspondingMittenpunkt D and New point −1/2 2 d + v − w d + v − w j L j is a Mitten-New line. There are four Mitten-New lines 1 = [v + d,−w + d] and 2(d + v − w)

D0L0 = hav + bw − cd : bv + cw + ad : −b(v + w)i. and similarly for the other Spieker points.

78 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

Theorem 30 (Spieker-Nagel lines) The Spieker points lie Proof. The midpoint of HL0 is on the corresponding In-Nagel lines, and in particular S0, 1 1 1 S1,S2,S3 are the midpoints of the sides I0N0, I1N1, I2N2, H + L0 = ∆ [b(c − b),b(a − b)] I3N3 respectively. 2 2 2 1 + [(c − b)u + cv + bw + c(a − b), 4∆ Proof. We check that in fact S0 is the midpoint of I0N0 by (a − b)u − bv − aw + a(c − b)] computing 1 = [ac−b2 +(c−b)(u+b)+cv+bw, 4(ac−b2) 1 1 1 1 2 I0 + N0 = [−w,v] ac − b + (a − b)(u + b)− aw − bv] 2 2 2 (d + v − w) 1 1 1 = [ac − b2 + (c − b + w)(u + b), + [(b + u)(c − b)+ cv + bw,(b + u)(a − b)− bv − aw] 4(ac − b2) 2 ∆ ac b2 a b v u b = S0. − + ( − − )( + )] 1 (c − b + w) (a − b − v) = 1 + ,1 + 4 u − b u − b The computations for the other In-Nagel lines and S1, S2, S are similar. 1 3 = [c − 2b + u + w,a − 2b + u − v]. 4(u − b)

Now a judicious use of the quadratic relations, which we I 12 leave to the reader, shows that this is S0. The computations L for the other Spieker points are similar. L The proof shows in fact that there is quite some variety 8 possible in the formulas for the various points and lines in I A3 N this paper. L D A2 S 4 G S I N 5 Future Directions D D D I S This paper might easily be the starting point for many more A1 investigations, as there are lots of additional points in the 8 4 L 12 H Incenter hierarchy that might lead to similar phenomenon. S In a related but slightly different direction, the basic idea N of Chromogeometry ([12], [13]) is that we can expect wonderful relations between the corresponding geometri- cal facts in the blue (Euclidean bilinear form x1x2 + y1y2), Figure 13: Green Spieker points S and Mitten-Spieker cen- red (bilinear form x1x2 − y1y2) and green (bilinear form ter H x1y2 + y1x2) geometries.

The joins of corresponding Mittenpunkts D j and Spieker points S j are the Mitten-Spieker lines. There are four Mitten-Spieker lines, and Hg 4 A3

D0S0 = hav + bw − cd : bv + cw + ad : −b(v + w)i. A2 Hb A1 Hr

4 8 Theorem 31 (Mitten-Spieker center) The four Mitten- Spieker lines D jS j are concurrent and meet at the Ortho- center H = X4.

Figure 14: Blue, red and green Incenter circles Theorem 32 (New Mitten-Spieker) The Spieker point S j is the midpoint of HL j, so that the corresponding New A spectacular illustration of this is the following, which we point L j also lies on the corresponding Mitten-Spieker line. will describein detailin afuturework: if we haveatriangle

79 KoG•16–2012 N. Le, N. J. Wildberger: Universal Afﬁne Triangle Geometry and Four-fold Incenter Symmetry

A1A2A3 that has both blue, red and green Incenters(a rather [8] B. ODEHNAL, Generalized Gergonne and Nagel delicate issue, as it turns out), then remarkably the four red points, Beitrage¨ Algebra Geom. 51 (2) (2010), 477– Incenters and four green Incenters lie on a conic, in fact a 491. blue circle, as in Figure 14. Similarly, the four red Incen- ters and four blue Incenters lie on a green circle, and the [9] A. OLDKNOW, The Euler-Gergonne-Soddy Triangle four green Incentersand four blue Incenterslie on a red cir- of a Triangle, Amer. Math. Monthly 103 (4) (1996), cle. The centers of these three coloured Incenter circles are 319–329. exactly the respective orthocenters H ,H ,H which form b r g [10] N. J. WILDBERGER, Divine Proportions: Ratio- the Omega triangle of the given triangle A1A2A3, intro- nal Trigonometry to Universal Geometry, Wild Egg duced in [12]. Books, Sydney, 2005. In particular the four green Incenters I that have appeared http://wildegg.com. in our diagrams are in fact concyclic in a Euclidean sense, as well as in a red geometry sense. By applying central di- [11] N. J. WILDBERGER, Afﬁne and projective metrical lations, we may conclude similar facts about circles pass- geometry, arXiv: math/0612499v1,(2006), to appear, ing through Nagel points and Spieker points. Many more J. of Geometry. interesting facts wait to be discovered. [12] N. J. WILDBERGER, Chromogeometry, Mathemati- cal Intelligencer 32 (1) (2010), 26–32.

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[4] Hyacinthos Yahoo Group [17] P. Yiu, A Tour of Triangle Geometry, Florida Atlantic http://tech.groups.yahoo.com/group/ University lecture notes, 2004. hyacinthos.

[5] C. KIMBERLING, Triangle Centers and Central Tri- Nguyen Le angles, vol. 129 Congressus Numerantium, Utilitas e-mail: [email protected] Mathematica Publishing, Winnepeg, MA, 1998. Norman John Wildberger [6] C. KIMBERLING, Encyclopedia of Triangle Centers, e-mail: [email protected] http://faculty.evansville.edu/ck6/ School of Mathematics and Statistics UNSW encyclopedia/ETC.html. Sydney 2052 Australia [7] C. KIMBERLING, Major Centers of Triangles, Amer. Math. Monthly 104 (1997), 431–438.