Incenter Circles, Chromogeometry, and the Omega Triangle

KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

Original scientiﬁc paper NGUYEN LE Accepted 4. 3. 2014. NORMAN JOHN WILDBERGER Incenter Circles, Chromogeometry, and the Omega Triangle

Incenter Circles, Chromogeometry, and the Upisane kruˇznice, kromogeometrija i Omega Omega Triangle trokut ABSTRACT SAZETAKˇ Chromogeometry brings together planar Euclidean geom- Kromogeometrija povezuje ravninsku euklidsku geo- etry, here called blue geometry, and two relativistic ge- metriju, ovdje zvanu plavom geometrijom, te dvije re- ometries, called red and green. We show that if a trian- lativistiˇcke geometrije, nazvane crvenom i zelenom geo- gle has four blue Incenters and four red Incenters, then metrijom. Pokazuje se da ukoliko trokut ima ˇcetiri these eight points lie on a green circle, whose center is plava i ˇcetiri crvena srediˇsta upisanih (odnosno pripisanih) the green Orthocenter of the triangle, and similarly for the kruˇznica, tada tih osam toˇcaka leˇzi na zelenoj kruˇznici ˇcije other colours. Tangents to the incenter circles yield inter- je srediˇste zeleni ortocentar trokuta. Vrijede i druge dvije esting additional standard quadrangles and concurrencies. analogne tvrdnje. Tangente na upisane kruˇznice stvaraju The proofs use the framework of rational trigonometry nove zanimljive ˇcetverokute i konkurentnosti. Dokazi se together with standard coordinates for triangle geometry, provode u okviru racionalne trigonometrije sa standar- while a dilation argument allows us to extend the results dnim koordinatama za geometriju trokuta. Transforma- also to Nagel and Speiker points. cija diletacije dozvoljava proˇsirenje rezultata na Nagelove i Speikerove toˇcke. Key words: triangle geometry, incenter circles, rational trigonometry, chromogeometry, four-fold symmetry, Nagel Kljuˇcne rijeˇci: geometrija trokuta, upisane kruˇznice, points, Spieker points, Omega triangle racionalna trigonometrija, kromogeometrija, ˇcetverostruka simetrija, Nagelove toˇcke, Spiekerove toˇcke, Omega trokut MSC 2000: 51M05, 51M10, 51N10

1 Introduction The first results of this paper concern the four Incenters of a planar triangle in one of the three geometries, and were announcedin [5]. As in that paper, we here refer to all four meets of the vertex bisectors, or bilines, as Incenters, so This paper investigates a surprising connection between do not distinguish between the classical incenter and the three closely related Incenter hierarchies of a fixed planar three excenters. If a triangle A A A has four blue Incen- triangle. The framework here is that of Rational Trigonom- 1 2 3 ters Ib,Ib,Ib and Ib, then all four points lie both on a red etry ([7], [8]) which allows a consistent universal trian- 0 1 2 3 incenter circle C b with center the red Orthocenter H , and gle geometry valid for any symmetric bilinear form, as de- r r on a green incenter circle C b with center the green Or- scribed in [5], together with the three-fold symmetry of g thocenter H ; this is illustrated in Figure 1. Similarly, if chromogeometry ([9], [10]), which connects the familiar g a triangle has red Incenters, then these lie both on a green Euclidean (blue) geometry based on the symmetric bilinear incenter circle C r with center H , and a blue incenter circle form x x + y y , and two relativistic geometries (red and g g 1 2 1 2 C r with center the blue Orthocenter H . If a triangle has green) based respectively on the bilinear forms x x y y b b 1 2 1 2 green Incenters, these lie both on a blue incenter circle C g and x y + y x . By working with the rational notions− of b 1 2 1 2 with center H , and on a red incenter circle C g with cen- quadrance and spread instead of the transcendental notions b r ter H . Furthermore, if both red and green Incenters exist, of distance and angle, the main laws of Rational Trigonom- r then they lie on the same blue incenter circle, and simi- etry allow metrical geometry, and so triangle geometry, to larly for the other colours. The proofs are algebraic, and be developed in each of these three geometries in a parallel rely on non-obvious simplifications found by the help of a fashion, with mostly identical formulas and theorems.

5 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle computer. So the Omega triangle formed by the three Or- of coordinates to place an arbitrary triangle into standard thocenters Ω HbHrHg, introduced in [9], has an intimate position, with vertices at [0,0], [1,0] and [0,1]. The var- connection with≡ the Incenter hierarchies. ious triangle centers and constructions are then expressed in terms of the coefﬁcients a,b and c of the matrix

b a b I2 C b b c Cr ≡ 10 of the resulting new bilinear form. This allows a system- b R23 Hr atic augmentation of Kimberling’s Encyclopedia of Trian- A b b 2 R13 G gle Centers ([2], [3], [4]) to arbitrary quadratic forms and Rb 02 b 02 G13 Rb Ib A general fields. 5 01 0 3 b b I1 Hb G12 Standard coordinates also have the advantage of yielding b b G01 R03 Hg b surprisingly simple equations for the three coloured Incen- Cg A1 ter Circles, which turn out to be, after pleasant simplifica- b tions, I3 0 5 10 15 Cb : Qb (X)= bb (2x + 2y 1) − Cr : Qr (X)= br (2x + 2y 1) Figure 1: The four blue Incenters of A1A2A3 and red and − Cg : Qg (X)= bg (2x + 2y 1). green Incenter Circles − These facts relate also to elegant classical properties of b However the formulas for the star lines s j become rather quadrangles. In [1] Haskell showed that if two quadrangles formidable, but seem to have interesting algebraic aspects. have the same diagonal triangle, then all eight points of Some intriguing number theoretical questions arise when these quadrangles lie on a single conic; and in [11] Woods we inquire into the existence of triangles, over a given field, found a synthetic derivation of the same result. Now it is which have simultaneously blue, red and green Incenters. obviousthat the four Incenters of a triangle, with respect to Studying concrete examples and using empirical computer any bilinear form, will form a standard quadrangle in this investigationsof Michael Reynolds [6], we make some ten- sense, meaning that the diagonal triangle coincides with tative conjectures on such triangles, both over the rational the original triangle. As a consequence, if blue and red In- numbers and over a finite prime field. Finally we extend centers exist, then they must lie on a conic. Our assertion the results to Spieker and Nagel points by suitable central C b C r C is that this conic is actually a green circle g = g g dilations. with center H ≡ g. In the rest of this introductionwe recall basic facts from [7] In the case of blue Incenters, the four tangent lines to the and [5] to formulate triangle geometry over a general bilin- C b red incenter circle r at the blue Incenters form a stan- ear form. We then specialize to the blue, red and green ge- dard quadrilateral, implying that they meet in six points b ometries, and use standard coordinates to develop formulas Ri j, which lie two at a time on the three lines of A1A2A3, for points and lines (always one of our key aims), and to where they are harmonic conjugates with respect to A1,A2 provide explicit computational proofs of the theorems. and A3; and similarly the four tangent lines to the green in- C b b center circle g at the blue Incenters meet in six points Gi j 1.1 Quadrilaterals and quadrangles on the three lines. This is also seen on the above Figure. Similarly there is a corresponding result when we look at We begin by reminding the reader of some basic facts from red Incenters, and at green Incenters. the projective geometry of a quadrangle (four points) or b quadrilateral (four lines), using a visual presentation to The six lines AkRi j, for i, j,k distinct, are the lines of a complete quadrangle, so they meet three at a time at avoid the need to introduce notation. b b In Figure 2 we see four blue lines forming a quadrilateral four quad points Qr j. Similarly, the six lines AkGi j meet [in this figure colours are not used in a metrical sense, but three at a time at points Qb . Somewhat remarkably, the g j only as an aide for explanation]. These four blue lines meet four star lines sb Qb Qb form a standard quadrilateral j ≡ r j g j in six points, also in blue. These six blue points determine b b b b s0s1s2s3. a further three green diagonal lines, forming the diagonal This paper also illustrates our novel approach to triangle triangle, in yellow, of the original quadrilateral, whose geometry initiated in [5]; using standard coordinates to es- vertices are three green points. Each green point may be tablish universal aspects of the subject which are valid over joined via a red line to the two blue points not on either a general bilinear form. This employs an affine change of the two green lines it lies on. This produces six red

6 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle lines, which somewhat remarkably meet three at a time at metric bilinear form on vectors: four red points, giving the opposite quadrangle from the v u vCuT . original blue quadrilateral. Note that there is a natural cor- · ≡ respondence between the four original blue lines and the Non-degenerate means detC = 0, and implies that if v u = four red points. 0 for all vectors u, then v = 06 . · Two vectors v and u are then perpendicular 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 is perpendicular to v. Two lines l and m are perpendicular precisely when they have perpendicular direction vectors. The bilinear form determines the main metrical quantity: the quadrance of a vector v is the number

Qv v v. ≡ · The quadrance between the points A and B is Q(A,B) Q . A vector v is null precisely when Q = v v = 0, ≡in −→AB v Figure 2: A quadrilateral and its opposite quadrangle other words precisely when v is perpendicular to· itself. A line is null precisely when it has a null direction vector. The situation is completely symmetric with regard to The following basic fact appears in [5]. points and lines. If we had started out with a quadrilateral of four red points, we wouldjoin them to formsix red lines. Theorem 1 (Parallel vectors) Vectors v and u are paral- These six red lines determinea further three green diagonal lel precisely when points, forming the diagonal triangle of the original quadri- 2 lateral, whose sides form three green lines. Each green line QvQu = (v u) . 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 This motivates the following measure of the non- the opposite quadrilateral from the original red quadran- parallelism of two vectors; the spread between non-null gle. vectors v and u is the number The diagonal green points on a green line are harmonic (v u)2 (v u)2 conjugates with respect to the two blue points on the same s(v,u) 1 · = 1 · . ≡ − QvQu − (v v)(u u) line. The diagonal green lines through a green point are · · harmonic conjugates with respect to the two red lines The spread s(v,u) is unchanged if either v or u are multi- through the same point. plied by a non-zeronumber. We define the spread between There is another more subtle remark to be made here con- any non-null lines l and m with direction vectors v and u cerning symmetry: each of the three diagonal points is to be s(l,m) s(v,u). From Theorem 1, the spread be- canonically associated to a subdivision of the four original tween parallel≡ lines is 0. Two non-null lines l and m are blue lines into two subsets of two: namely those subsets perpendicular precisely when the spread between them is whose joins meet at that diagonal point. 1. If we start with a triangle, say the yellow triangle in the A circle is given by an equation of the form Q(A,X)= K Figure formed by three green points and three green lines, for some fixed point A called the center, and a number K then any quadrilateral or quadrangle which has that trian- called the quadrance. Note that it is not required that a gle as its diagonal triangle is called standard. circle have any points X lying on it: in this case by enlarg- ing the field to a quadratic extension we can guarantee that 1.2 Quadrance, spread and standard coordinates it does. In this section we briefly summarize the main facts needed The three particular planar geometries we are most inter- from rational trigonometryin the general affine setting (see ested in come from the blue, red and green bilinear forms [7], [8]). We work in the standard two-dimensional vector given by the respective matrices space V, consisting of row vectors v = [x,y], over a field F. 1 0 1 0 0 1 A line l is givenby an equationof the form ax+by+c = 0, Cb , Cr and Cg . ≡ 0 1 ≡ 0 1 ≡ 1 0 or equivalently the proportion l a : b : c . − We assume a metrical structure≡ h determinedi by a non- The corresponding formulas for the blue, red and green degenerate symmetric 2 2 matrix C: this gives a sym- quadrances between points A1 [x1,y1] and A2 [x2,y2] × ≡ ≡

7 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle are and 2 2 ∆ Qb (A1,A2) = (x2 x1) + (y2 y1) s s(A A ,A A )= − − 1 1 2 1 3 2 2 ≡ ac Qr (A1,A2) = (x2 x1) (y2 y1) ∆ − − − s2 s(A2A3,A2A1)= Qg (A1,A2)= 2(x2 x1)(y2 y1). ≡ ad − − ∆ It will be useful to discuss triangle geometry then in a gen- s3 s(A3A1,A3A2)= . T ≡ cd eral setting: suppose v1 v2 v1Bv2 is a symmetric bilinear form, with B a symmetric◦ ≡ 2 2 matrix. Suppose Furthermore × φ : V V is a linear transformation given by an invert- b2 (a b)2 (c b)2 → φ 1 s1 = , 1 s2 = − , 1 s3 = − . ible 2 2 matrix M, so that (v) = vM = w, with in- − ac − ad − cd verse matrix× N, so that wN = v. The new bilinear form T Note that the centroid of A1A2A3 is w1 w2 (w1N) (w2N) then has matrix D = NBN . Suppose· ≡ that X X◦ X is a triangle in the vector space V 1 1 1 2 3 G = , . which has a distinguished symmetric bilinear form . We 3 3 may move this triangle by a combination of a translation◦ (which does not effect the bilinear form), and a linear trans- 1.3 Bilines, Incenters and some other triangle centers φ formation , so thatthetriangleis in whatwe call standard A biline of the non-null vertex l1l2 is a line b which passes form, with points through l1l2 and satisfies s(l1,b)= s(b,l2). The existence of bilines depends on number theoretical considerations of A1 [0,0], A2 [1,0] and A3 [0,1] ≡ ≡ ≡ a particularly simple kind. and lines Theorem 3 (Existence of Triangle bilines) The Triangle l1 A2A3 = 1:1: 1 A A A has bilines at each vertex precisely when we can ≡ h − i 1 2 3 l2 A1A3 = 1:0:0 find numbers u,v,w in the field satisfying ≡ h i 2 2 2 l3 A2A1 = 0:1:0 . ac = u , ad = v , cd = w . (2) ≡ h i Whatever the initial matrix B, the new bilinear form is In this case we can choose u,v,w so that acd = uvw and given by · du = vw, cv = uw and aw = uv. (3) a b v u vDuT where D NBNT = (1) We now summarize some basic triangle centers of the stan- · ≡ ≡ b c dard triangle A1A2A3, assuming the existence of bilines. for some numbers a,b, and c. We may then replace ar- These formulas involve the entries a,b,c of D from (1), as guments involving the general triangle X1X2X3 and the well as the secondary quantities u,v and w from (2), satis- bilinear form with ones involving the simpler triangle fying (3). The formulas and proofs are found in [5]. ◦ A1A2A3. What we prove for the standard triangle A1A2A3 The four Incenters are with bilinear form given by the matrix D will be true for 1 1 I0 = [ w,v], I1 = [w, v], the original triangle with bilinear form given by the origi- d + v w − d v + w − nal matrix B. 1− −1 I2 = [w,v], I3 = [ w, v]. We will assume that D is invertible, so that d + v + w d v w − − 2 − − ∆ detD = ac b Notice that I1,I2 and I3 may be obtained from I0 by chang- ≡ − ing signs of: both v and w, just w, and just v respectively. is non-zero. Another important quantity is the mixed trace This four-fold symmetry will hold more generally and note d a + c 2b that it means that we can generally just record the val- ≡ − ues of I0. The Orthocenter H, Circumcenter C and De that appears in many formulas. With these notations, we Longchamps point X20 (the orthocenter of the double tri- have the following result from [5]. angle) are Theorem 2 (Standard quadrances and spreads) The b H = [c b,a b] (4) quadrances and spreads of A1A2A3 are ∆ − − 1 Q1 Q(A2,A3)= d C = [c(a b),a(c b)] ≡ 2∆ − − Q Q A A c 2 ( 1, 3)= 1 ≡ X = b2 2bc + ac,b2 2ab + ac . Q3 Q(A1,A2)= a 20 ∆ ≡ − − 8 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

2 The Incenter Circle theorem α β 0 1 α β T Dg ≡ γ δ 1 0 γ δ Here is the main theorem of the paper, illustrated for green 2αβ αδ + βγ ag bg Incenters of the triangle A1A2A3 in Figure 3. The situa- = . αδ + βγ 2γδ ≡ bg cg tion is completely symmetric between the three geometries blue, red and green. There are interesting relations between the introduced quantities; for example g Cb 2 2 2 2 2 ab = ag + ar , abcb = bg + br , H 2 2 2 2 2 2 2 10 g b arcr = b b , agcg = b b , c = c + c I3 b g b r b g r g − − I2 A3 and A2 g I 0 g 5 I Hr 1 abcg 2bbbg + cbag = 0, abcr 2bbbr + cbar = 0, − −

g A1 Hg agcr 2bgbr + cgar = 0. Cr − The determinants of Db,Dr and Dg are respectively 5 10 15 2 2 ∆b = (αδ βγ) , ∆r = ∆g = (αδ βγ) = ∆b − − − − Figure 3: Green Incenters and the blue and red Incenter Circles and the mixed traces are 2 2 2 2 Theorem 4 (Incenter Circles) If a triangle A1A2A3 has db = (α γ) + (β δ) , dr = (α γ) (β δ) , b b b b − − − − − four blue Incenters I0 ,I1 ,I2 and I3 , then they all lie both b on a red circle C with center the red Orthocenter H , and dg = 2(α γ)(β δ). r r − − on a green circle C b with center the green Orthocenter H , g g Note also the relation d2 = d2 + d2. and similarly for the other colours. Furthermore, if both b r g If the original triangle has four blue Incenters, then the red and green Incenters exist, then they lie on the same Existence of Triangle bilines theorem shows that we may blue circle, so that C r = C g = C , and similarly for the b b b choose numbers u ,v ,w satisfying (2) and (3), so that other colours. b b b b b b u2 = α2 + β2 γ2 + δ2 Proof. To prove that the four blue Incenters I0 ,I1 ,I2 and b b C b 2 2 I3 lie on a red circle r with center Hr, we need show that v2 = α2 + β2 (α γ) + (β δ) b − − b b b b 2 2 Qr Hr,I0 = Qr Hr,I1 = Qr Hr,I2 = Qr Hr,I3 . w2 = γ2 + δ2 (α γ) + (β δ) . b − − First we ﬁnd the bilinear forms for the blue, red and green The blue Incenters are then geometries. After translating, and then applying a linear transformation with the matrix M, we send the original b 1 b 1 1 I0 = [ wb,vb], I1 = [wb, vb], triangle to the standard triangle A1A2A3. If M− = N = db + vb wb − db vb + wb − α β − − , then the bilinear forms for the blue, red and b 1 b 1 γ δ I2 = [wb,vb], I3 = [ wb, vb]. db + vb + wb db vb wb − − green geometries become respectively the matrices − − In exactly the same fashion α β 1 0 α β T Db γ δ γ δ ≡ 0 1 r 1 g 1 I = [ wr,vr] and I = [ wg,vg]. 2 2 0 0 α + β αγ + βδ ab bb dr + vr wr − dg+vg wg − = 2 2 − − αγ + βδ γ + δ ≡ bb cb According to (4), the respective orthocenters are

T α β 1 0 α β bb br Dr Hb = [cb bb,ab bb], Hr = [cr br,ar br], ≡ γ δ 0 1 γ δ ∆b − − ∆r − − − α2 β2 αγ βδ ar br b = − 2 − 2 g αγ βδ γ δ ≡ br cr Hg = [cg bg,ag bg]. − − ∆g − −

9 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

If we set eb db + vb wb then We may now repeat the calculation to see that ≡ − b b b Qr Hr,I1 = Qr Hr,I2 = Qr Hr,I3 = Kr, showing that b br (cr br) wb br (ar br) vb b H−−→rI = − + , − indeed the four blue Incenters lie on the red circle C with 0 − ∆ e ∆ − e r r b r b quadrance K and center H . Note that the expression for 1 r r ∆ ∆ K dependsonly on the matrix D . Now a similar derivation = ∆ (br (cr br)eb + rwb,br (ar br)eb rvb) r r − reb − − − shows that so that b bg (ag bg)(bg cg) T Qg Hg,Ii = − ∆ − Kg, i = 0,1,2,3. b −−→b −−→b g ≡ Qr Hr,I0 = HrI0 Dr HrI0 b br(cr br ) w Hence the four blue Incenters also lie on a green circle C − + b g br(cr br) wb br (ar br) vb ar br ∆r eb = ∆− + ∆− with quadrance Kg and center Hg. Similarly we ﬁnd that r eb r eb b c br (ar br) vb − r r ∆− ! r − eb if a triangle has four red Incenters, then they lie on a blue 2 2 r 1 ar (br (cr br)e + ∆rw ) +cr (br (ar br)e ∆rv ) circle C with center Hb and quadrance = − b b − b− b b ∆2e2 +2br (br (cr br)e + ∆rw )(br (ar br)e ∆rv ) r b − b b − b− b 2 2 2 g r bb (ab bb)(bb cb) b (ar 2br + cr) arcr b e Qb Hb,I = Qb (Hb,I )= − − Kb 1 r − − r b i i ∆ = 2∆ b (v w ) b2 + a c e . b ≡ ∆2 2 r r b b r r r b r e  − 2 −2 2−  b +∆ arw + crv 2brvbwb C r r b b − as well as on a green circle g with center Hg and quad-   ∆ 2 rance Kg (the same one as above!) Similarly if a triangle Use the relation r = arcr br to get C g − has four green Incenters, then they lie on a blue circle b b Qr Hr,I0 (5) with center Hb and quadrance Kb, as well as on a red circle C g with center H and quadrance K . The proof is com- 1 2 2 r r r br (ar 2br + cr)eb = − 2 2 plete. ∆2e 2∆rbr (v w )e + ∆r arw + crv 2brv w r b − b − b b b b − b b 2br (br cr)(ar br)(vbdb vbwb wbdb) − 2 2− − 2 2 − 2 2 + ar (br cr) v + cr (ar br) w + b drd = − b − b r b ∆2 r eb 8 Hr where we have collected v2,w2 and d2 of the numerator of b b b C (5), to rewrite it. g A2 Hg 2 2 4 A3 Replace v = abdb, w = cbdb and vbwb = ubdb and the b b A1 values of ab,cb,db,ar,br,cr in terms of α,β,γ,δ to get the factorization 4 8 Cr Hb 2 Cb 2br (br cr)(ar br)(vb ub wb)db + ar (br cr) abdb − − − − − 2 2 2 + cr (ar br) cbdb + b (ar + cr 2br)d − r − b 2 2br (br cr)(ar br)(vb ub wb)+ ar (br cr) ab = db − − 2 − 2 − − Figure 4: Three Incenter Circles Cb, Cr and Cg. +cr (ar br) cb + br (ar + cr 2br)db − − C C r C g C C b C g C C b αγ βδ α2 αγ γ2 β2 βδ δ2 We now call b = b = b , r = r = r and g = g = = 2db ( ) + + + ub + vb wb r − − − − − Cg the blue, red andgreen Incenter Circles respectively. In α2 β2 αγ + βδ γ2 + δ2 + αγ βδ (6) × − − − − Figure 4 we see a (small) triangle A1A2A3 with its Omega and also note that triangle HbHrHg and the three Incenter Circles, whose respective meets give the twelve blue, red and green Incen- 2 (db + vb wb) = db (ab + cb + db 2ub + 2vb 2wb) ters. − − − 2 2 2 2 = 2db α αγ + γ + β βδ + δ ub + vb wb . (7) − − − − 2.1 Equations of Incenter Circles Combine (6) and (7), to get the surprisingly simple formula Theorem 5 (Incenter Circles equations) In standard co- b ordinates with X = [x,y], the blue, red and green Incenter Qr Hr,I 0 circles, when they exist, have respective equations (αγ βδ ) α2 β2 αγ + βδ γ2 + δ2 + αγ βδ = − − − − − Cb : Qb (X)= bb (2x + 2y 1) ∆r − C : Q (X)= b (2x + 2y 1) br (ar br)(br cr) r r r = − − Kr. − ∆ Cg : Qg (X)= bg (2x + 2y 1). r ≡ −

10 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

Proof. The derivation of these equations, using the formulas established above for the orthocenters H and coloured Ib r b 3 tg1 b Incenters, is somewhat involved algebraically, although the G13 basic idea is simple. We show how to find the equation b A1 Cg tg0 b C R b R 03 of the red Incenter Circle r, with center Hr, which four 12 Ib b 2 G02 b G b I0 12 blue Incenters and four green Incenters lie on if they ex- b A tr1 b 2 ist. From the definition of a red circle, we get the equation R 02 b b G23 tb R 01 A b g2 Qr (Hr,X)= Kr, and then substitute the values of Hr and 3 b G01 R 23 b Kr to get G03 b R 13 br(cr br) b − x tr0 br(cr br) br(ar br) ar br ∆r ∆− x ∆− y − r r b c br(ar br) tb C − − r r ∆− y ! r3 r b r − tg3 b br (ar br)(br cr) b tr 2 = − − I1 ∆r or after expansion Figure 5: Incenter tangent meets 2 2 1 ar (br (cr br) ∆rx) + cr (br (ar br) ∆ry) Theorem 6 (Incenter tangent meets) The tangent lines 2 − − − − b b b b ∆ +2br (br (cr br) ∆rx)(br (ar br) ∆ry) t ,t ,t ,t to the red Incenter circle Cr at the blue In- r − − − − r0 r1 r2 r3 b (a b )(b c ) centers form a standard quadrilateral, as do the tangent = r r r r r . b b b b C − ∆ − lines tg0,tg1,tg2,tg3 at the green Incenter circle g. The r same holds for the red and green Incenters, if they exist. ∆ 2 b b b b b b This may be rewritten, using r = arcr br , in the form This implies that the meets R t t and R t t lie − 01 ≡ r0 r1 23 ≡ r2 r3 on l1 = A2A3, and are harmonic conjugates with respect to 1 b b b b b b (∆2a x2 + 2∆2b xy + ∆2c y2 + ∆ b2 (a 2b + c ) A2 and A3. Similarly R t t and R t t lie on ∆ r r r r r r r r r r r 02 ≡ r0 r2 13 ≡ r1 r3 r − l2 = A1A3, and are harmonic conjugates with respect to A1 ∆2 ∆2 b b b b b b 2 brx 2 bry)= br (ar br)(br cr). and A3; and R t t and R t t lie on l3 = A1A2, − r − r − − 03 ≡ r0 r3 12 ≡ r1 r2 and are harmonic conjugates with respect to A1 and A2. ∆ b b b b b b Now cancel r, and rearrange to get The points G t t and G t t lie on l1, and are 01 ≡ g0 g1 23 ≡ g2 g3 harmonic conjugates with respect to A2 and A3. Similarly ∆ 2 ∆ ∆ 2 ∆ ∆ 2 rarx +2 rbrxy+ rcry 2 rbrx 2 rbry+br arcr br =0 Gb tb tb and Gb tb tb lie on l and are harmonic − − − 02 g0 g2 13 g1 g3 2, ≡ ≡ b b b conjugates with respect to A1 and A3, and G03 tg0tg3 and or more simply b b b ≡ G t t lie on l3, and are harmonic conjugates with 12 ≡ g1 g2 2 2 respect to A and A . arx + 2brxy + cry 2brx 2bry + br = 0 1 2 − − b Proof. We prove the result for the meets Gi j of the green which has the form stated in the theorem. The same kind tangent lines tb associated to the blue Incenters; the other gi of calculation establishes the formulas for Cb and Cg. b cases are similar. We find the joins of a blue Incenter Ii Note that the equations for the Incenter Circles Cb,Cr and and the green Orthocenter Hg to be Cg allow them to be defined whether or not the correspond- (bg ag)bgdb + (cg bg)agvb + (ag bg)bgwb : ing Incenters exist! Incenters then exist precisely as meets b − − − H I = (cg bg)bgd + (cg bg)bgv + (ag bg)cgw : of these Incenter Circles: for example the blue Incenters g 0 b b b * − (b c )b −v + (b a )b −w + b b b b C C g g g b g g g b I0 ,I1 ,I2 ,I3 are just the meets of r and g, if these exist in − − the field in which we work. (bg ag)bgdb (cg bg)agvb (ag bg)bgwb : b − − − − − HgI = (cg bg)bgdb (cg bg)bgvb (ag bg)cgwb : 1 − − − − − 2.2 Tangent lines of Incenter Circles * (bg cg)bgvb (bg ag)bgwb + − − − − (bg ag)bgdb + (cg bg)agvb (ag bg)bgwb : Now we consider tangent lines to Incenter circles. Fig- b − − − − HgI = (cg bg)bgdb + (cg bg)bgvb (ag bg)cgwb : ure 5 shows the four blue Incenters of A1A2A3, together 2 − − − − * (bg cg)bgvb (bg ag)bgwb + with the red and green Incenter Circles passing through − − − b them, namely Cr and Cg. At each of the four Incenters Ii , (bg ag)bgdb (cg bg)agvb + (ag bg)bgwb : b b b − − − − i = 1,2,3,4 we have the tangent lines t and t to the red HgI = (cg bg)bgdb (cg bg)bgvb + (ag bg)cgwb : . ri gi 3 − − − − and green Incenter Circles Cr and Cg respectively. * (bg cg)bgvb + (bg ag)bgwb + − − −

11 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

b b b b The tangent line tgi is the line green perpendicular to HgIi The fact that A2,A3,G01,G23 form a harmonic range etc. b is an immediate consequence of a well known fact about passing through Ii . These we may calculate to be standard quadrilaterals in projective geometry, since we ag bg ub + bgvb + ag 2bg wb + dbbg cb ag bg : have shown that the points A ,A and A are diagonal b − − − − 1 2 3 t = cg bg u + 2bg cg v bgw + d bg + a bg cg : g0 − b − b− b b b − points of the quadrilateral formed by the four green tan- * bg ( v + w d ) + − b b − b gent lines. ag bg ub bgvb ag 2bg wb + bgdb cb ag bg : b − − − − − − t = cg bg u 2bg cg v + bgw + d bg + a bg cg : Following the construction of the red lines in the introduc- g1 − b− − b b b b − * bg (vb wb db) + tory section on Quadrangles and quadrilaterals, we join a − − b point G with the triangle point Ak opposite to the triangle ag bg ub + bgvb ag 2bg wb + bgdb cb ag bg : i j b − − − − − − b t = cg bg u + 2bg cg v + bgw + d bg + a bg cg : line that it lies on; giving six lines AkG : g2 − − b − b b b b − i j * bg ( v w d ) + − b − b − b ag bg ub bgvb + ag 2bg wb + bgdb cb ag bg : ag bg (ub + wb cb) : b − − − − − − b − − t = cg bg u 2bg cg v bgw + d bg + a bg cg : . A1G = bg cg ( ub + vb + ab) : g3 − − b− − b− b b b − 01 − − * bg (v + w d ) + * 0 + b b − b bg ag (u + w + c ) : − b b b We could verify directly that these four lines form a stan- A Gb = b c (u + v + a ) : 1 23 g g b b b dard quadrilateral. But we prefer to verify that the meets * − 0 + of these four tangent lines agree with the following meets bg ( vb+ wb db) : with the side lines of A1A2A3 : b − − A2G02 = bg cg ub+ cg 2bg vb+ bgwb bgdb+ ab cg bg : * − − − − + b b b b bg (vb wb + db) G01 tg0tg1 = tg0l1 − ≡ bg ( vb + wb + db) : 1 b − A G = cg bg u + cg 2bg v + bgw + bgd + a bg cg : = λ [(bg cg)( ub + vb + ab),(ag bg)( ub wb + cb)] 2 13 − b − b b b b − 01 − − − − − * bg (vb wb db) + b b b b − − G23 tg2tg3 = tg2l1 ag bg ub+bgvb+ ag 2bg wb+bgdb+cb bg ag : ≡ b − − − A G = bg (v w + d ) : 1 3 03 b − b b = [(bg cg)(ub + vb + ab),(ag bg)(ub + wb + cb)] * bg ( vb + wb db) + λ23 − − − − 1 bg ag ub+bgvb+ ag 2bg wb bgdb+cb ag bg : b b b b b − − − − G02 tg0tg2 = tg0l2 = [0,bg ( vb + wb db)] A3G = bg (vb wb db) : . ≡ λ02 − − 12 − − * bg ( vb + wb + db) + b b b b 1 − G13 tg1tg3 = tg1l2 = [0,bg ( vb + wb + db)] ≡ λ13 − b b b b 1 G t t = t l3 = [bg (vb wb + db),0] 03 ≡ g0 g3 g0 λ − b b 03 Theorem 7 (Quad points) The triples A1G23,A2G13, 1 A Gb , A Gb A Gb A Gb , A Gb {A Gb A Gb Gb tb tb tb l b v w d 0 3 12 1 23, 2 02, 3 03 1 01, 2 13, 3 03 12 g1 g2 = g1 3 = [ g ( b b b), ] } b b b ≡ λ12 − − and A G ,A G ,A G of lines are concur- 1 01 2 02 3 12 rent in the respective points Qb ,Qb ,Qb and where g0 g1 g2 b Qg3, called the blue/green quad points. The λ01 =(cg ag)ub + (bg cg)vb + (bg ag)wb b b b b b b − − − triples A1R23,A2R13,A3R12 , A1R23,A2R02,A3R03 , + (abbg abcg + cbag cbbg) A Rb ,A Rb ,A Rb and A Rb ,A Rb ,A Rb are − − 1 01 2 13 3 03 1 01 2 02 3 12 λ =(a c )u + (b c )v also concurrent in the respective points Qb ,Qb ,Qb and 23 g g b g g b r0 r1 r2 − − Qb called the blue/red quad points. Similar results hold + (ag bg)wb + (abbg abcg + cbag cbbg) r3, − − − for the red and green Incenters, if they exist. λ02 =(bg cg)ub + (cg 2bg)vb + bgwb − − + (abcg 2abbg + 2bbbg cbbg) − − Proof. We verify this for the blue/green quad points: this λ13 =(cg bg)ub + (cg 2bg)vb + bgwb − − is a consequence of the projective geometry of the com- + (2abbg abcg 2bbbg + cbbg) plete quadrilateral we mentioned in the first section—if the − − λ03 =(ag bg)ub + bgvb + (ag 2bg)wb original four tangent lines are regarded as the blue lines − − in Figure 6, then the quad points Qb correspond to the + (abbg 2bbbg cbag + 2cbbg) g j − − red points. However we want to find explicit formulas and λ12 =(bg ag)ub + bgvb + (ag 2bg)wb b − − check things directly. The quad point Qg j is naturally as- + (2bbbg abbg + cbag 2cbbg). b − − sociated to the Incenter Ij . After some calculation, we find

12 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

2 2 that these are since dbu + (bb ab)ubwb bbdbub cbv + bbvbwb + b − − − b cb (ab bb)vb = 0 by using (2), and similarly for the other b bg − Qg0 = λ [(bg cg)(dbub (bb cb)vb), indices. In a parallel fashion, we ﬁnd that the four blue/red 0 − − − b quad points Qr j have exactly the same formulas as the (ag bg)((cb bb)wb + cbdb)] b − − Qg j, except for the replacements ag ar, bg br and b bg c c , and similarly for the other−→ colours red−→ and green. Qg1 = [(bg cg)((ab bb)vb + abdb), g r λ1 − − −→ (ag bg)(dbub + (ab bb)wb)] − − b bg Q = [(cg bg)(abwb bbvb), I3 sb g2 λ b 0 2 − − tg1 b Cg (bg ag)(bbwb cbvb)] b A1 sb − − tg0 3 b b b g I Qg0 Q = [(bg cg)( dbub + (db + bb)vb abwb + abdb), 2 g3 λ b I0 B 3 − − − tr1 03 b b B b b s1 Qr3 13 Qr1 Qg3 b (bg ag)(dbub cbvb + (db + bb)wb cbdb)] A2 tg2 − − − B01 B12 b A3 Qg1 where b B23 Q b b r0 s2 Qg2 b B02 Qr2 λ0 = (bg cg)(bgdb + (bb cb)(ag bg))ub b − − − tr0 (bg cg)(bbbg + cbag 2cbbg)vb b Cb tr3 r b − − − tg3 bg (ag bg)(bb cb)wb + cbbg (ag bg)db b − − − − tr 2 λ1 = (ag bg)(bgdb + (ab bb)(bg cg))ub I1 − − − + bg (bg cg)(ab bb)vb − − + (ag bg)(2abbg abcg bbbg)wb + abbg (bg cg)db Figure 6: Quad points and star lines − − − − λ2 = bb (bg cg)(ag bg)ub b − − Now introduce the blue star line s j to be the join of the + bg (bbbg + cbag bbcg cbbg)vb b − − corresponding blue/red quad point Qr j and the blue/green bg(abbg+bbag abcg bbbg)wb abcb(bg cg)(ag bg) b − − − − − − quad point Qg j, and similarly for the other colours. There 2 b b b b λ3 = (db + bb) b + agcg agbg (2db + bb) bbbgcg ub are then four blue star lines s ,s ,s and s . g − − 0 1 2 3 The blue star point Bi j is the meet of the two blue star + (bg ((bg cg)(ab bb)+ cb (2ag bg)) cbagcg)vb b b b b − − − − lines si and s j , that is Bi j si s j , and similarly for the other + ((db + bb)(bg ag)bg abag (bg cg))wb ≡ − − − colours. 2 +bg(ab(db cb) cbdb)+bg(cbag(db+ ab) abcg(db cb)) Note that following the introductory section on Quadran- − − − − gles and quadrilaterals, we use the correspondence be- agcgabcb. b b − tween the Qg j; and the tangent lines tg j; and the Incenters b b We may then check directly that for example Qg0 is inci- Ij to match up the indices. b dent with A3G by computing 12 Theorem 8 (Star quadrilateral) The four blue star lines b b b b form a standard quadrilateral s0s1s2s3. This holds also for bg ag u + bgv + ag 2bg w bgd + c ag bg − b b − b − b b − · the other colours. bg bg cg (dbub (bb cb)vb) − λ − − Proof. The proof we have is surprisingly complicated. The · 0 ! b star lines s j have quite involved formulas; for example we bg ag bg ((cb bb)wb + cbdb) ﬁnd that + bg (vb wb db) − − − − λ b b0 b0 0 ! s0 = Qg Qr = + bg ( vb + wb + db) E0dbub + F0cbvb : − 2 2 bgbr (bb cb) + cbdb dbub +(bb ab)ubwb bbdbub − · bg ag bg bg cg 2 − − (agbr bgar agcr + cgar + bgcr cgbr)dbub − − cbvb + bbvbwb + cb (ab bb)vb  · − − −  = − − 2cbbgbr (agbr bgar agcr + cgar + bgcr cgbr)(bb cb)dbvb : bg cg bgdb +(bb cb) ag bg ub − − − − 2 −   ab (br cr)(bg cg)(agbr bgar) (bb cb) + cbdb wb  − − − − *  − − − − −  + + bg cg bbbg + cbag 2cbbg vb    − −   +2abcb (br cr)(bg cg)(agbr bgar)(bb cb)db  +bg ag bg (bb cb)wb cbbg ag bg db − − − − − − − −  bgbrdb (agbr bgar agcr + cgar + bgcr cgbr)    − −2 − − · = 0 (bb cb) + cbdb ub 2cb (bb cb)vb · − − − ! 13 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

where E0 and F0 are both homogeneous polynomials of search. So we tentatively conjecture that there are no such degree 6 in the variables ai,bi and ci, with the former hav- triangles. ing 32 terms and the latter 46 terms. After some trial and In any case, to get an explicit example we use an alge- error we can present these in the somewhat pleasant, but braic extension ﬁeld of the rationals; so by √30 we mean still mysterious, forms: an appropriate symbol in the extension ﬁeld Q √30 etc.. Note that although our use of square roots is entirely alge- E0 = bgbr (agcr cgar bgcr + cgbr) braic, our representation of these square roots as approx- − − − · 2 2 b + 4c + abcb 6bbcb imate rational numbers (we prefer to avoid discussion of · b b − 2 2 “real numbers”), necessarily brings an approximate aspect + bgbr (agbr bgar) b + 2c + abcb 4bbcb − b b − into our diagrams. 2 2 2cb agcgb arcrb + agarcrbg agcgarbr (bb cb) − r − g − − 50 and 45 Cr 40 F a c b2 a c b2 a a c b a c a b 35 0 = g g r r + g r r g g g r r 30 r − g − · 25 2 2 C bb 4bbcb + 2cb + abcb H 20 g · − r 15 + bgbr (agcr cgar bgcr + cgbr) − − · 10 2 2 X25 Hg 5bb 4cb + 2abbb 3abcb + 10bbcb X3 · − − − 50 40 30 20 10 10 20 30 40 50 60 70 80 X15 2bgbr (agbr bgar)(bb cb)(ab 2bb + cb). − − − − 10 15 We can then calculate the blue star points, for example 20 Hb 25

30 Cb bgbrdb(agbr bgar agcr +cgar +bgcr cgbr) 35 −2 − − · 40 (bb cb) + cbdb ub 2cb (bb cb)vb  · − − − !  B03 = ,0 E d u + F c v  0 b b 0 b b  Figure 7: An example triangle X1X2X3         Example 1 One may check that the basic Triangle with from which clearly B03 lies on l3. The computations are similar for the other indices, and the other colours. points

X1 [ 21/59, 58/59], X2 [ 13/3,2] and 3 Explicit examples and some conjectures ≡ − − ≡ −

X3 [35/3, 8/5] 3.1 Anexampleover Q √30,√217,√741,√2470,√82297 ≡ − We will now explore in detail a particular triangle which in Q √30,√217,√741,√2470,√82297 has both has both blue, red and green Incenters; for us this is not blue, red and green Incenters. After translation only an important tool for checking the consistency of our by (21/59,58/59) we obtain X1 = [0,0], X2 = formulas, but also a way to get a sense of the level of com- [ 704/177,176/59] and X3 = [2128/177, 182/295]. − − plexity of various constructions. In fact this kind of explicit The matrix N and its inverse M, wheree e calculation of examples is much to be encouraged in this e 704 176 subject: especially as working over concrete fields, includ- α β N = − 177 59 = and ing finite fields and explicit extension fields of the ratio- 2128 182 γ δ 177 − 295 nals, allows us to appreciate the number theoretic aspects of our geometrical investigations. For example, finding a 13 5 1 704 56 triangle with blue, red and green Incenters approximately M = N− = 95 5 264 42 is easy with a geometry package: finding a concrete example and working out all the points precisely is more chal- respectively send [1,0] and [0,1] to X2 and X3, and X2 and lenging. X3 to [1,0] and [0,1]. From now on we discuss only the In particular we were unable, despite a reasonable com- standard triangle A1A2A3 associatede to X1Xe2X3; to converte puter search, to find any triangles with purely rational backe into the original coordinates, we would multiply by N points that have blue, red and green Incenters! We would and translate by ( 21/59, 58/59). The bilinear forms in like to thank Michael Reynolds for his contributions to this these new standard− coordinates,− for the blue, red and green

14 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

1 √ √ geometries respectively, are given by matrices 22429440− (4032553 217+ 20461 217√82297 +210343√82297+ 76618507), 774400 7778848 Ir =   1 0 T ab bb 3 1 √ √ √ D =N N = 31329 156645 = 50976− (7 217 82297 2923 217 b 0 1 7778848 −113507716 b c − 156645 783225 b b  133√82297 30049)  −   216832 7202272  − −  1 0 T 31329 156645 ar br Dr =N N = 7202272 −112911484 = 0 1 br cr − − 156645 783225 247808 2000768 0 1 T 10443 52215 ag bg Dg=N N = −2000768 774592 = . 1 0 52215 52215 bg cg − 203 247 35 3211 g √741 √30 + √2470 , The determinants of D D and D are ∆ 97140736 and I = 7788 − 3894 3894 − 7788 b, r g b = 87025 0 13 √2470 29 √30 + 20 √741 100 ∆ ∆ 97140736 6724 1239 − 177 1239 − 177 r = g = 87025 , while the mixed traces are db = 25 , 247 203 35 3211 6076 − 576 g √30 √741+ √2470 , dr = and dg = . The orthocenters of A1A2A3 are I = 3894 − 7788 3894 − 7788 25 − 5 1 29 √30+ 13 √2470 20 √741 100 177 1239 − 1239 − 177 247 203 35 3211 8825537 84337 87833227 55537 g √30 √741 √2470 , Hb = , ,Hr = , , I = − 3894 − 7788 − 3894 − 7788 −1019520 −25488 11214720 25488 2 29 √30 13 √2470 20 √741 100 − 177 − 1239 − 1239 − 177 247 √30 + 203 √741 35 √2470 3211 , 7105 377 Ig = 3894 7788 3894 7788 . Hg = , . 3 29 √ 13 √ − 20 √ − 100 3894 177 177 30 1239 2470+ 1239 741 177 − − Blue, red and green Incenters exist over F = Q √30,√217,√741,√2470,√82297 and we may choose The Incenter circle quadrances are 1875104 14432 873628 ub = , vb = , wb = 31329 177 4425 18154129609 11681819191 17248 2464 Kb = , Kr = ur = √82297, vr = √217 28196100 − 28196100 156645 885 1182272 196 Kg = . w = √217√82297 10443 r 4425 19712 2816 448 ug = √2470, vg = √30, wg = √741. 52215 295 295 The blue, red and green Incenter Circles themselves have Then the four blue Incenters, the four red Incenters and the respective equations four green Incenters of A1A2A3 respectively are 761 220 5327 25 Ib = , Ib = , 4840000x2 19447120xy + 28376929y2 0 −590 413 1 10384 −118 − +19447120x + 19447120y 9723560 = 0 761 25 5327 220 − Ib = , Ib = , 2 2 2 2112 168 3 270 27 19360x 62524xy + 12103y − +62524x + 62524y 31262 = 0 − 2 2 1 (4032553√217 20461√217√82297 193600x 2572240xy + 4032553y 22429440 − − +210343√82297 76618507), +2572240x + 2572240y 1286120 = 0. Ir =  −  − 0 1 (2923√217 7√217√82297 50976 −  +133√82297 30049)   −   1  b (20461√217√82297 4032553√217 The four tangent lines tg j are 22429440 − +210343√82297 76618507), Ir =  −  1 1 (7√217√82297 2923√217 50976 −  +133√82297 30049)  b   tg0 = 1570 : 11823 : 8323  −  h − i 1 (4032553√217+ 20461√217√82297 b 22429440 tg1 = 127512 : 33761: 58261 210343√82297 76618507), h− − i Ir = b 2  −1 √ − √  tg2 = 18216 : 11823 : 8323 50976 (7 217√82297+ 2923 217 h− − i  133√82297 30049)  tb = 1570: 4823: 8323 .   g3 h− i  − −  15 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

The meets of these four tangent lines agree with the follow- and they meet at the blue star points ing meets with the side lines of A1A2A3 : 165269500000 761989459049 B01 = , b b b b 3500 9893 927258959049 927258959049 G t t = t l1 = , 01 ≡ g0 g1 g0 13393 13393 165269500000 761989459049 B23 = , b b b b 3500 9893 596719959049 596719959049 G t t = t l1 = , − 23 ≡ g2 g3 g2 −6393 6393 1034074074039 1034074074039 B02 = 0, ,B13 = 0, b b b b 1189 868804574039 1199343574039 G t t = t l2 = 0, 02 ≡ g0 g2 g0 1689 1034074074039 1034074074039 B03 = ,0 ,B12 = ,0 . b b b b 1189 1796063533088 272084614990 G t t = t l2 = 0, 13 ≡ g1 g3 g1 689 Note the pleasant rationality of the previous objects. b b b b 8323 G t t = t l3 = ,0 03 ≡ g0 g3 g0 −1570 3.2 An exampleover F13 b b b b 8323 G t t = t l3 = ,0 . 12 ≡ g1 g2 g1 18216 Now we look at an example over a finite field. The blue/red quad points Qb associated to Ib,Ib,Ib,Ib re- Theorem 9 (Null quadrances incenters) Suppose that r j 0 1 2 3 2 spectively are the field F contains an element i, where i = 1, and the characteristic of F is not 2. If − 18005811535 12129669559 Qb = , r0 21082889161 −21082889161 bb (ab bb)(bb cb) br (ar br)(br cr) Kb − − = Kr − − ≡ ∆b ≡ ∆r b 18005811535 12129669559 Qr1 = , b a b b c − 9330605209 9330605209 g ( g g)( g g) = Kg − ∆ − = 0 18005811535 12129669559 ≡ g Qb = , r2 14928733909 14928733909 then the standard Triangle A1A2A3 has four distinct blue, 18005811535 12129669559 Qb = , . red and green Incenters. r3 45342228279 45342228279 Proof. If Kb = 0 then from the definition of the blue incen- b The respective blue/green quad points Qg j are ter circle Cb, which is Qb (Hb,X)= Kb, Cb is a null circle, so it is a product of lines. Similarly, if Kr = 0 then Cr is a b 4161500 11762777 null circle, and if Kg = 0 then Cg is a null circle. These Qg0 = , −2654777 2654777 null lines have distinct direction vectors (1, i),(1, 1) ± ± 4161500 11762777 and (1,0),(0,1) respectively, and they are never parallel Qb = , g1 −12547777 12547777 since char(F) = 2, so i = 1. Therefore, any two null circles meet in exactly6 four6 points.± 4161500 11762777 Qb = , g2 10977777 10977777 Here is an example found by Michael Reynolds [6] which 4161500 11762777 illustrates explicitly the above theorem. Qb = , . g3 20870777 20870777 Example 2 The triangle X1X2X3 with points X1 ≡ The blue star lines are then [3,4],X2 [1,9] and X3 [12,3] in F13 has four blue, red ≡ ≡ b b b and green Incenters. In F13 the squares are 0,1,3,4,9,10 s0 = Qr0Qg0 = and 12, and in particular 1 = 12 = 52 is a square. After − 1796063533088: 868804574039: 1034074074039 translation by (3,4) we obtain X1 = [0,0],X2 = [11,5] and h − i X = [9,12]. The matrix N and its inverse M sb = 3 1 e e 272084614990: 1199343574039: 1034074074039 e 11 5 1 10 11 h − i N = , M = N− = sb 9 12 12 7 2 = 272084614990: 868804574039: 1034074074039 h − i send [1,0] and [0,1] to X and X , and X and X to [1,0] b 2 3 2 3 s3 = and [0,1] respectively. The bilinear form in these new stan- 1796063533088:1199343574039: 1034074074039 dard coordinates for thee blue,e red ande green geometriese h − i

16 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle respectively are Theorem 10 (Spieker circles) If a triangle has four blue b b b b Incenters I0 ,I1 ,I2 and I3 , then the four blue Spieker cen- 1 0 3 3 D = N NT = ters all lie both on a red Spieker circle with center the red b 0 1 3 4 Circumcenter Cr, and on a green Spieker circle with center 1 0 5 0 the green Circumcenter Cg. If both say blue and red Incen- D = N NT = r 0 1 0 2 ters exist, then all 8 blue and red Spieker points lie on the − same green circle. The same holds for the other colours. 0 1 6 8 D = N NT = . g 1 0 8 8 Proof. We see that if we use the central dilation formula We can see immediately that Kb = Kr = Kg = 0 from the to transform Incenter circles centred at the Orthocenters, deﬁnitions we get the Spieker circles centred at Circumcenters, so this theorem is a direct consequence of the Incenter circles the- bb (ab bb)(bb cb) orem and the fact that a dilation preserves circles of any Qb (Hb,Iir)= − − Kb ∆b ≡ colour. br (ar br)(br cr) Here are the formulas for the coloured Circumcenters in Qr (Hr,Iib)= − − Kr ∆r ≡ standard coordinates: bg (ag bg)(bg cg) Qg (Hg,Iir)= − ∆ − Kg, i = 0,1,2,3. 1 g ≡ Cb = [cb (ab bb),ab (cb bb)] 2∆b − − The four blue, red and green Incenters respectively are 1 Cr = [cr (ar br),ar (cr br)] 2∆ − − b b b b r I0 = [4,8], I1 = [3,6], I2 = [8,10], I3 = [11,4] 1 r r r r Cg = [cg (ag bg),ag (cg bg)]. I0 = [10,9], I1 = [8,2], I2 = [6,5], I3 = [4,12] 2∆g − − 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 8 Cb

Cb : (y x + 1)(x + 3y 1)= 0 − − 4 A1 Cr : (x 6y)(x + 6y)= 0 − A Cg : (x + 2y 2)(x + 5y 5)= 0. Cg 2 A3 − − Cr 4 8 From Michael Reynolds’ computer investigations, we tentatively conjecture that for finite fields Fp where p 3mod4, there are no triangles which have both blue, red≡ and green Incenters, and for finite fields Fp where p 1mod4, blue, red and green Incenters exists precisely≡ when Kb = Kr = Kg = 0, as in the above example. Figure 8: Blue, red and green Speiker circles

4 Spieker circles and Nagel circles Theorem 11 (Nagel circles) If a triangle hasfour blueIn- centers Ib,Ib,Ib and Ib, then the four blue Nagel centers Now we recall from [5] that the central dilation δ about 0 1 2 3 1/2 all lie both on a red Nagel circle with center the red De the centroid takes the Orthocenter to the Circumcenter,− and Longchamps point X , and on a green Nagel circle with the Incenters to the Spieker centers. In standard coordi- 20r center the green De Longchamps point X . If both say nates 20g blue and red Incenters exist, then all 8 blue and red Nagel δ points lie on the same green circle. The same holds for the 1/2 ([x,y])=(1/2)[1 x,1 y]. − − − other colours.

The inverse central dilation δ 2 takes the Orthocenter to − the De Longchamps point X20, and takes the Incenters to Proof. In the same fashion as in the previous theorem, if the Nagel points. In standard coordinates we use the inverse central dilation δ 2 to transform Incen- ter circles centred at the Orthocenters,− we get the Nagel δ 2 ([x,y]) = [1 2x,1 2y]. circles centred at De Longchamps points. − − −

17 KoG•18–2014 N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle

X20b Proof. This follows by considering the action of the central 8 dilation δ 2 which takes the circumcenter Ci to the ortho- − center Hi, and the orthocenter Hi to the De Longchamps 4 point X20i. A1 A2 A3 X20g 4 8 12 References X20r -4 [1] M. W. HASKELL, The Construction of Conics under given conditions, Bulletin of the Amer. Math. Soc. 11 (5) (1905), 268–273. -8 [2] C. KIMBERLING, Triangle Centers and Central Tri- angles, vol. 129 Congressus Numerantium, Utilitas Figure 9: Blue, red and green Nagel circles Mathematica Publishing, Winnepeg, MA, 1998. Here are the formulas for the blue, red and green De Longchamps points: [3] C. KIMBERLING, Encyclopedia of Triangle Centers, http://faculty.evansville.edu/ck6/ 1 2 2 encyclopedia/ETC.html. X20b = bb 2bbcb + abcb,bb 2abbb + abcb ∆b − − 1 2 2 [4] C. KIMBERLING, Major Centers of Triangles, Amer. X20r = br 2brcr + arcr,br 2arbr + arcr ∆r − − Math. Monthly 104 (1997), 431–438. 1 2 2 X20g = ∆ bg 2bgcg + agcg,bg 2agbg + agcg . [5] N. LE,N.J.WILDBERGER, Universal Afﬁne Trian- g − − gle Geometry and Four-fold Incenter Symmetry, KoG In Figure 10 we see the relations between the 16 (2012), 63–80. three coloured Orthocenters, Circumcenters and De Longchamps points. The lines joining these are the three [6] M.REYNOLDS, Personal Communication, 2013. coloured Euler lines. Note that the centroids of the trian- [7] N. J. WILDBERGER, Divine Proportions: Ratio- gles of Orthocenters, Circumcenters and De Longchamps nal Trigonometry to Universal Geometry, Wild Egg points all agree with the centroid G of the original triangle Books, Sydney, 2005. A1A2A3. We conclude with a simple observation about De Longchamps points. [8] N. J. WILDBERGER, Afﬁne and projective metrical geometry, arXiv: math/0612499v1.

12 X20b [9] N. J. WILDBERGER, Chromogeometry, Mathemati- cal Intelligencer 32 (1) (2010), 26–32. H r [10] N. J. WILDBERGER, Chromogeometry and Rela- 8 C b tivistic Conics, KoG 13 (2009), 43–50. A1 Hg [11] B. M. WOODS, The Construction of Conics under Cg A2 A3 Cr given conditions, Amer. Math. Monthly 21 (6) (1914), 4 X20g 173–180. Hb

X20r Nguyen Le 4 8 12 e-mail: [email protected] Norman John Wildberger Figure 10: Blue, red, green Orthocenters, Circumcenters e-mail: [email protected] and De Longchamps points School of Mathematics and Statistics UNSW Theorem 12 (Orthocenters as midpoints) For any tri- Sydney 2052 Australia angle, a coloured orthocenter H is the midpoint of the two De Longchamps points X20 of the other two colours.