N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
NGUYEN LE
NORMAN JOHN WILDBERGER
Original scientific paper
Accepted 4. 3. 2014.
Incenter Circles, Chromogeometry,
and the Omega Triangle
Incenter Circles,
Omega Triangle
Chromogeometry,
and the
Upisane kružnice, kromogeometrija i Omega
trokut
ABSTRACT
SAŽETAK
Chromogeometry brings together planar Euclidean geometry, here called blue geometry, and two relativistic geometries, called red and green. We show that if a triangle has four blue Incenters and four red Incenters, then
these eight points lie on a green circle, whose center is
the green Orthocenter of the triangle, and similarly for the
other colours. Tangents to the incenter circles yield interesting additional standard quadrangles and concurrencies.
The proofs use the framework of rational trigonometry
together with standard coordinates for triangle geometry,
while a dilation argument allows us to extend the results
also to Nagel and Speiker points.
Kromogeometrija povezuje ravninsku euklidsku geometriju, ovdje zvanu plavom geometrijom, te dvije relativističke geometrije, nazvane crvenom i zelenom geometrijom. Pokazuje se da ukoliko trokut ima četiri
plava i četiri crvena središta upisanih (odnosno pripisanih)
kružnica, tada tih osam točaka leži na zelenoj kružnici čije
je središte zeleni ortocentar trokuta. Vrijede i druge dvije
analogne tvrdnje. Tangente na upisane kružnice stvaraju
nove zanimljive četverokute i konkurentnosti. Dokazi se
provode u okviru racionalne trigonometrije sa standardnim koordinatama za geometriju trokuta. Transformacija diletacije dozvoljava proširenje rezultata na Nagelove
i Speikerove točke.
Key words: triangle geometry, incenter circles, rational
trigonometry, chromogeometry, four-fold symmetry, Nagel
points, Spieker points, Omega triangle
MSC 2000: 51M05, 51M10, 51N10
1 Introduction
This paper investigates a surprising connection between
three closely related Incenter hierarchies of a fixed planar
triangle. The framework here is that of Rational Trigonometry ([7], [8]) which allows a consistent universal triangle geometry valid for any symmetric bilinear form, as described in [5], together with the three-fold symmetry of
chromogeometry ([9], [10]), which connects the familiar
Euclidean (blue) geometry based on the symmetric bilinear
form x1 x2 + y1y2 , and two relativistic geometries (red and
green) based respectively on the bilinear forms x1 x2 − y1 y2
and x1 y2 + y1 x2 . By working with the rational notions of
quadrance and spread instead of the transcendental notions
of distance and angle, the main laws of Rational Trigonometry allow metrical geometry, and so triangle geometry, to
be developed in each of these three geometries in a parallel
fashion, with mostly identical formulas and theorems.
Ključne riječi: geometrija trokuta, upisane kružnice,
racionalna trigonometrija, kromogeometrija, četverostruka
simetrija, Nagelove točke, Spiekerove točke, Omega trokut
The first results of this paper concern the four Incenters of
a planar triangle in one of the three geometries, and were
announced in [5]. As in that paper, we here refer to all four
meets of the vertex bisectors, or bilines, as Incenters, so
do not distinguish between the classical incenter and the
three excenters. If a triangle A1 A2 A3 has four blue Incenters I0b , I1b , I2b and I3b , then all four points lie both on a red
incenter circle Crb with center the red Orthocenter Hr , and
on a green incenter circle Cgb with center the green Orthocenter Hg ; this is illustrated in Figure 1. Similarly, if
a triangle has red Incenters, then these lie both on a green
incenter circle Cgr with center Hg , and a blue incenter circle
Cbr with center the blue Orthocenter Hb . If a triangle has
green Incenters, these lie both on a blue incenter circle Cbg
g
with center Hb , and on a red incenter circle Cr with center Hr . Furthermore, if both red and green Incenters exist,
then they lie on the same blue incenter circle, and similarly for the other colours. The proofs are algebraic, and
rely on non-obvious simplifications found by the help of a
5
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
computer. So the Omega triangle formed by the three Orthocenters Ω ≡ Hb Hr Hg , introduced in [9], has an intimate
connection with the Incenter hierarchies.
I 2b
b
Cr
10
b
R23
Hr
b
A2
R13
Gb02
Rb02
b
b
A3
I0
R01
b
I1b
G12
Hb
b
Hg
G01
Rb03
5
b
G13
b
Cg
A1
I 3b
0
5
10
15
Figure 1: The four blue Incenters of A1 A2 A3 and red and
green Incenter Circles
These facts relate also to elegant classical properties of
quadrangles. In [1] Haskell showed that if two quadrangles
have the same diagonal triangle, then all eight points of
these quadrangles lie on a single conic; and in [11] Woods
found a synthetic derivation of the same result. Now it is
obvious that the four Incenters of a triangle, with respect to
any bilinear form, will form a standard quadrangle in this
sense, meaning that the diagonal triangle coincides with
the original triangle. As a consequence, if blue and red Incenters exist, then they must lie on a conic. Our assertion
is that this conic is actually a green circle Cgb = Cgr ≡ Cg
with center Hg .
In the case of blue Incenters, the four tangent lines to the
red incenter circle Crb at the blue Incenters form a standard quadrilateral, implying that they meet in six points
Rbij , which lie two at a time on the three lines of A1 A2 A3 ,
where they are harmonic conjugates with respect to A1 , A2
and A3 ; and similarly the four tangent lines to the green incenter circle Cgb at the blue Incenters meet in six points Gbij
on the three lines. This is also seen on the above Figure.
Similarly there is a corresponding result when we look at
red Incenters, and at green Incenters.
The six lines Ak Rbij , for i, j, k distinct, are the lines of
a complete quadrangle, so they meet three at a time at
four quad points Qbr j . Similarly, the six lines Ak Gbij meet
three at a time at points Qbg j . Somewhat remarkably, the
four star lines sbj ≡ Qbr j Qbg j form a standard quadrilateral
sb0 sb1 sb2 sb3 .
This paper also illustrates our novel approach to triangle
geometry initiated in [5]; using standard coordinates to establish universal aspects of the subject which are valid over
a general bilinear form. This employs an affine change
6
of coordinates to place an arbitrary triangle into standard
position, with vertices at [0, 0] , [1, 0] and [0, 1]. The various triangle centers and constructions are then expressed
in terms of the coefficients a, b and c of the matrix
a b
C≡
b c
of the resulting new bilinear form. This allows a systematic augmentation of Kimberling’s Encyclopedia of Triangle Centers ([2], [3], [4]) to arbitrary quadratic forms and
general fields.
Standard coordinates also have the advantage of yielding
surprisingly simple equations for the three coloured Incenter Circles, which turn out to be, after pleasant simplifications,
Cb : Qb (X) = bb (2x + 2y − 1)
Cr : Qr (X) = br (2x + 2y − 1)
Cg : Qg (X) = bg (2x + 2y − 1).
However the formulas for the star lines sbj become rather
formidable, but seem to have interesting algebraic aspects.
Some intriguing number theoretical questions arise when
we inquire into the existence of triangles, over a given field,
which have simultaneously blue, red and green Incenters.
Studying concrete examples and using empirical computer
investigations of Michael Reynolds [6], we make some tentative conjectures on such triangles, both over the rational
numbers and over a finite prime field. Finally we extend
the results to Spieker and Nagel points by suitable central
dilations.
In the rest of this introduction we recall basic facts from [7]
and [5] to formulate triangle geometry over a general bilinear form. We then specialize to the blue, red and green geometries, and use standard coordinates to develop formulas
for points and lines (always one of our key aims), and to
provide explicit computational proofs of the theorems.
1.1 Quadrilaterals and quadrangles
We begin by reminding the reader of some basic facts from
the projective geometry of a quadrangle (four points) or
quadrilateral (four lines), using a visual presentation to
avoid the need to introduce notation.
In Figure 2 we see four blue lines forming a quadrilateral
[in this figure colours are not used in a metrical sense, but
only as an aide for explanation]. These four blue lines meet
in six points, also in blue. These six blue points determine
a further three green diagonal lines, forming the diagonal
triangle, in yellow, of the original quadrilateral, whose
vertices are three green points. Each green point may be
joined via a red line to the two blue points not on either
of the two green lines it lies on. This produces six red
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
four red points, giving the opposite quadrangle from the
original blue quadrilateral. Note that there is a natural correspondence between the four original blue lines and the
four red points.
metric bilinear form on vectors:
v · u ≡ vCuT .
Non-degenerate means detC 6= 0, and implies that if v · u =
0 for all vectors u, then v = 0.
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.
Figure 2: A quadrilateral and its opposite quadrangle
The situation is completely symmetric with regard to
points and lines. If we had started out with a quadrilateral
of four red points, we would join them to form six red lines.
These six red lines determine a further three green diagonal
points, forming the diagonal triangle of the original quadrilateral, whose sides form three green lines. Each green line
meets two of the red lines in two new blue points. These six
new blue points lie three at a time on four blue lines, giving
the opposite quadrilateral from the original red quadrangle.
The diagonal green points on a green line are harmonic
conjugates with respect to the two blue points on the same
line. The diagonal green lines through a green point are
harmonic conjugates with respect to the two red lines
through the same point.
There is another more subtle remark to be made here concerning symmetry: each of the three diagonal points is
canonically associated to a subdivision of the four original
blue lines into two subsets of two: namely those subsets
whose joins meet at that diagonal point.
If we start with a triangle, say the yellow triangle in the
Figure formed by three green points and three green lines,
then any quadrilateral or quadrangle which has that triangle as its diagonal triangle is called standard.
1.2 Quadrance, spread and standard coordinates
In this section we briefly summarize the main facts needed
from rational trigonometry in the general affine setting (see
[7], [8]). We work in the standard two-dimensional vector
space V , consisting of row vectors v = [x, y] , over a field F.
A line l is given by an equation of the form ax + by+ c = 0,
or equivalently the proportion l ≡ ha : b : ci.
We assume a metrical structure determined by a nondegenerate symmetric 2 × 2 matrix C: this gives a sym-
The quadrance between the points A and B is Q(A, B) ≡
→ . A vector v is null precisely when Qv = v · v = 0, in
Q−
AB
other words precisely when v is perpendicular to itself. A
line is null precisely when it has a null direction vector.
The following basic fact appears in [5].
Theorem 1 (Parallel vectors) Vectors v and u are parallel precisely when
Qv Qu = (v · u)2 .
This motivates the following measure of the nonparallelism of two vectors; the spread between non-null
vectors v and u is the number
s (v, u) ≡ 1 −
(v · u)2
(v · u)2
= 1−
.
Qv Qu
(v · v)(u · u)
The spread s (v, u) is unchanged if either v or u are multiplied by a non-zero number. We define the spread between
any non-null lines l and m with direction vectors v and u
to be s (l, m) ≡ s (v, u). From Theorem 1, the spread between parallel lines is 0. Two non-null lines l and m are
perpendicular precisely when the spread between them is
1.
A circle is given by an equation of the form Q (A, X) = K
for some fixed point A called the center, and a number K
called the quadrance. Note that it is not required that a
circle have any points X lying on it: in this case by enlarging the field to a quadratic extension we can guarantee that
it does.
The three particular planar geometries we are most interested in come from the blue, red and green bilinear forms
given by the respective matrices
1 0
1 0
0 1
Cb ≡
, Cr ≡
and Cg ≡
.
0 1
0 −1
1 0
The corresponding formulas for the blue, red and green
quadrances between points A1 ≡ [x1 , y1 ] and A2 ≡ [x2 , y2 ]
7
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
and
are
2
Qb (A1 , A2 ) = (x2 − x1) + (y2 − y1)
2
Qr (A1 , A2 ) = (x2 − x1)2 − (y2 − y1)2
Qg (A1 , A2 ) = 2 (x2 − x1 ) (y2 − y1 ) .
It will be useful to discuss triangle geometry then in a general setting: suppose v1 ◦ v2 ≡ v1 BvT2 is a symmetric bilinear form, with B a symmetric 2 × 2 matrix. Suppose
φ : V → V is a linear transformation given by an invertible 2 × 2 matrix M, so that φ (v) = vM = w, with inverse matrix N, so that wN = v. The new bilinear form
w1 · w2 ≡ (w1 N) ◦ (w2 N) then has matrix D = NBN T .
Suppose that X1 X2 X3 is a triangle in the vector space V
which has a distinguished symmetric bilinear form ◦. We
may move this triangle by a combination of a translation
(which does not effect the bilinear form), and a linear transformation φ, so that the triangle is in what we call standard
form, with points
A1 ≡ [0, 0],
A2 ≡ [1, 0]
and A3 ≡ [0, 1]
and lines
l1 ≡ A2 A3 = h1 : 1 : −1i
l2 ≡ A1 A3 = h1 : 0 : 0i
∆
ac
∆
s2 ≡ s (A2 A3 , A2 A1 ) =
ad
∆
s3 ≡ s (A3 A1 , A3 A2 ) = .
cd
Furthermore
s1 ≡ s (A1 A2 , A1 A3 ) =
1 − s1 =
b2
,
ac
1 − s2 =
(a − b)2
,
ad
1 − s3 =
(c − b)2
.
cd
Note that the centroid of A1 A2 A3 is
1 1
G= , .
3 3
1.3 Bilines, Incenters and some other triangle centers
A biline of the non-null vertex l1 l2 is a line b which passes
through l1 l2 and satisfies s(l1 , b) = s(b, l2 ). The existence
of bilines depends on number theoretical considerations of
a particularly simple kind.
Theorem 3 (Existence of Triangle bilines) The Triangle
A1 A2 A3 has bilines at each vertex precisely when we can
find numbers u, v, w in the field satisfying
l3 ≡ A2 A1 = h0 : 1 : 0i .
ac = u2 ,
Whatever the initial matrix B, the new bilinear form · is
given by
a b
T
T
v · u ≡ vDu
where D ≡ NBN =
(1)
b c
In this case we can choose u, v, w so that acd = uvw and
for some numbers a, b, and c. We may then replace arguments involving the general triangle X1 X2 X3 and the
bilinear form ◦ with ones involving the simpler triangle
A1 A2 A3 . What we prove for the standard triangle A1 A2 A3
with bilinear form given by the matrix D will be true for
the original triangle with bilinear form given by the original matrix B.
We will assume that D is invertible, so that
∆ ≡ det D = ac − b2
is non-zero. Another important quantity is the mixed trace
d ≡ a + c − 2b
that appears in many formulas. With these notations, we
have the following result from [5].
Theorem 2 (Standard quadrances and spreads) The
quadrances and spreads of A1 A2 A3 are
Q1 ≡ Q(A2 , A3 ) = d
Q2 ≡ Q(A1 , A3 ) = c
Q3 ≡ Q(A1 , A2 ) = a
8
du = vw,
ad = v2 ,
cd = w2 .
cv = uw and aw = uv.
(2)
(3)
We now summarize some basic triangle centers of the standard triangle A1 A2 A3 , assuming the existence of bilines.
These formulas involve the entries a, b, c of D from (1), as
well as the secondary quantities u, v and w from (2), satisfying (3). The formulas and proofs are found in [5].
The four Incenters are
1
1
[−w, v] ,
I1 =
[w, −v] ,
I0 =
d+v−w
d −v+w
1
1
[w, v] ,
I3 =
[−w, −v] .
I2 =
d+v+w
d−v−w
Notice that I1 , I2 and I3 may be obtained from I0 by changing signs of: both v and w, just w, and just v respectively.
This four-fold symmetry will hold more generally and note
that it means that we can generally just record the values of I0 . The Orthocenter H, Circumcenter C and De
Longchamps point X20 (the orthocenter of the double triangle) are
b
[c − b, a − b]
∆
1
[c (a − b), a (c − b)]
C=
2∆
1 2
b − 2bc + ac, b2 − 2ab + ac .
X20 =
∆
H=
(4)
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
2 The Incenter Circle theorem
Here is the main theorem of the paper, illustrated for green
Incenters of the triangle A1 A2 A3 in Figure 3. The situation is completely symmetric between the three geometries
blue, red and green.
g
Cb
10
Hb
I 3g
g
I2
A2
g
I1
5
g
Cr
5
A3
bg
.
cg
There are interesting relations between the introduced
quantities; for example
a2b = a2g + a2r ,
ab cb = b2g + b2r ,
ar cr = b2b − b2g ,
ag cg = b2b − b2r ,
c2b = c2g + c2r
and
g
I0
Hr
ab cg − 2bbbg + cb ag = 0,
A1
Hg
ab cr − 2bbbr + cb ar = 0,
ag cr − 2bgbr + cgar = 0.
The determinants of Db , Dr and Dg are respectively
10
15
Figure 3: Green Incenters and the blue and red Incenter
Circles
Theorem 4 (Incenter Circles) If a triangle A1 A2 A3 has
four blue Incenters I0b , I1b , I2b and I3b , then they all lie both
on a red circle Crb with center the red Orthocenter Hr , and
on a green circle Cgb with center the green Orthocenter Hg ,
and similarly for the other colours. Furthermore, if both
red and green Incenters exist, then they lie on the same
blue circle, so that Cbr = Cbg = Cb , and similarly for the
other colours.
Proof. To prove that the four blue Incenters I0b , I1b , I2b and
I3b lie on a red circle Crb with center Hr , we need show that
Qr Hr , I0b = Qr Hr , I1b = Qr Hr , I2b = Qr Hr , I3b .
First we find the bilinear forms for the blue, red and green
geometries. After translating, and then applying a linear
transformation with the matrix M, we send the original
−1
triangle
to the standard triangle A1 A2 A3 . If M = N =
α β
, then the bilinear forms for the blue, red and
γ δ
green geometries become respectively the matrices
T
α β
1 0
α β
Db ≡
γ δ
0 1
γ δ
2
ab bb
α + β2 αγ + βδ
≡
=
b b cb
αγ + βδ γ2 + δ2
T
α β
1 0
α β
γ δ
0 −1
γ δ
2
2
ar br
α − β αγ − βδ
≡
=
b r cr
αγ − βδ γ2 − δ2
Dr ≡
α
γ
T
β
0 1
α β
δ
1 0
γ δ
2αβ
αδ + βγ
a
=
≡ g
αδ + βγ
2γδ
bg
Dg ≡
∆b = (αδ − βγ)2 ,
∆r = ∆g = − (αδ − βγ)2 = −∆b
and the mixed traces are
db = (α − γ)2 + (β − δ)2 ,
dr = (α − γ)2 − (β − δ)2 ,
dg = 2 (α − γ)(β − δ).
Note also the relation db2 = dr2 + dg2 .
If the original triangle has four blue Incenters, then the
Existence of Triangle bilines theorem shows that we may
choose numbers ub , vb , wb satisfying (2) and (3), so that
u2b = α2 + β2 γ2 + δ2
v2b = α2 + β2 (α − γ)2 + (β − δ)2
w2b = γ2 + δ2 (α − γ)2 + (β − δ)2 .
The blue Incenters are then
1
1
[−wb , vb ] , I1b =
[wb , −vb ] ,
db + vb − wb
db − vb + wb
1
1
I2b =
[wb , vb ] , I3b =
[−wb , −vb ] .
db + vb + wb
db − vb − wb
I0b =
In exactly the same fashion
I0r =
1
1
g
[−wr ,vr ] and I0 =
[−wg ,vg ] .
dr + vr −wr
dg +vg −wg
According to (4), the respective orthocenters are
Hb =
bb
[cb − bb , ab − bb] ,
∆b
Hg =
bg
[cg − bg , ag − bg] .
∆g
Hr =
br
[cr − br , ar − br ] ,
∆r
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N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
If we set eb ≡ db + vb − wb then
−−→
br (cr − br ) wb br (ar − br ) vb
b
Hr I0 = −
+ ,
−
∆r
eb
∆r
eb
1
(br (cr − br ) eb + ∆r wb , br (ar − br ) eb − ∆r vb )
=−
∆ r eb
so that
−−→ −−→T
Qr Hr , I0b = Hr I0b Dr Hr I0b
=
br (cr −br ) wb
+ eb
∆r
br (ar −br ) vb
− eb
∆r
a
r
br
br
cr
br (cr −br )
+ webb
∆r
br (ar −br )
− vebb
∆r
1 ar (br (cr −br ) eb + ∆r wb )2 +cr (br (ar −br ) eb −∆r vb )2
2
∆r e2b +2br (br (cr − br ) eb + ∆r wb ) (br (ar −br ) eb −∆r vb )
 2

b (a − 2br + cr ) ar cr − b2r e2b
1  r r
= 2 2
−2∆r br (vb − wb ) −b2r + ar cr eb  .
∆r eb
+∆2r ar w2b + cr v2b − 2br vb wb
=
!
Use the relation ∆r = ar cr − b2r to get
(5)
Qr Hr , I0b
1
b2r (ar − 2br + cr ) e2b
= 2
∆r eb −2∆r br (vb − wb ) eb + ∆r ar w2b + cr v2b − 2br vb wb
2br (br − cr ) (ar − br ) (vb db − vb wb − wb db )
+ ar (br − cr )2 v2b + cr (ar − br )2 w2b + b2r dr db2
=
∆2r eb
where we have collected v2b , w2b and db2 of the numerator of
(5), to rewrite it.
Replace v2b = ab db , w2b = cb db and vb wb = ub db and the
values of ab , cb , db , ar , br , cr in terms of α, β, γ, δ to get the
factorization
2
2br (br − cr ) (ar − br ) (vb − ub − wb ) db + ar (br − cr ) ab db
+ cr (ar − br )2 cb db + b2r (ar + cr − 2br ) db2
2br (br −cr )(ar −br )(vb −ub −wb ) + ar (br −cr )2 ab
= db
+cr (ar − br )2 cb + b2r (ar + cr − 2br ) db
= 2db (αγ − βδ) α2 −αγ + γ2 + β2 −βδ + δ2 −ub + vb −wb
× α2 − β2 − αγ + βδ −γ2 + δ2 + αγ − βδ
(6)
and also note that
(db + vb − wb )2 = db (ab + cb + db − 2ub + 2vb − 2wb)
= 2db α2 − αγ + γ2 + β2 − βδ + δ2 − ub + vb − wb . (7)
Combine (6) and (7), to get the surprisingly simple formula
Qr Hr , I0b
(αγ − βδ) α2 − β2 − αγ + βδ −γ2 + δ2 + αγ − βδ
=
∆r
br (ar − br ) (br − cr )
≡ Kr .
=
∆r
10
We may now repeat the calculation
to see that
Qr Hr , I1b = Qr Hr , I2b = Qr Hr , I3b = Kr , showing that
indeed the four blue Incenters lie on the red circle Crb with
quadrance Kr and center Hr . Note that the expression for
Kr depends only on the matrix Dr . Now a similar derivation
shows that
b (a − b ) (b − c )
g g
g
g
g
Qg Hg , Iib =
≡ Kg , i = 0, 1, 2, 3.
∆g
Hence the four blue Incenters also lie on a green circle Cgb
with quadrance Kg and center Hg . Similarly we find that
if a triangle has four red Incenters, then they lie on a blue
circle Cbr with center Hb and quadrance
bb (ab − bb) (bb − cb )
≡ Kb
Qb Hb , Iig = Qb (Hb , Iir ) =
∆b
as well as on a green circle Cgr with center Hg and quadrance Kg (the same one as above!) Similarly if a triangle
g
has four green Incenters, then they lie on a blue circle Cb
with center Hb and quadrance Kb , as well as on a red circle
Crg with center Hr and quadrance Kr . The proof is complete.
8
Hr
A2
4
Cg
A3
Hg
A1
4
Cb
Hb
8
Cr
Figure 4: Three Incenter Circles Cb , Cr and Cg .
g
g
We now call Cb = Cbr = Cb , Cr = Crb = Cr and Cg = Cgb =
Cgr the blue, red and green Incenter Circles respectively. In
Figure 4 we see a (small) triangle A1 A2 A3 with its Omega
triangle Hb Hr Hg and the three Incenter Circles, whose respective meets give the twelve blue, red and green Incenters.
2.1 Equations of Incenter Circles
Theorem 5 (Incenter Circles equations) In standard coordinates with X = [x, y], the blue, red and green Incenter
circles, when they exist, have respective equations
Cb : Qb (X) = bb (2x + 2y − 1)
Cr : Qr (X) = br (2x + 2y − 1)
Cg : Qg (X) = bg (2x + 2y − 1).
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 Hr and coloured
Incenters, is somewhat involved algebraically, although the
basic idea is simple. We show how to find the equation
of the red Incenter Circle Cr , with center Hr , which four
blue Incenters and four green Incenters lie on if they exist. From the definition of a red circle, we get the equation
Qr (Hr , X) = Kr , and then substitute the values of Hr and
Kr to get
!
a b br (cr −br ) − x
r
r
br (ar −br )
br (cr −br )
∆r
−x
−y
br (ar −br )
∆r
∆r
b r cr
−y
b
R 12
trb1
=
b
G 02
I b0
b
R 03
Cg
b
G 12
ar (br (cr − br ) − ∆r x)2 + cr (br (ar − br ) − ∆r y)2
+2br (br (cr − br ) − ∆r x) (br (ar − br ) − ∆r y)
br (ar − br ) (br − cr )
.
∆r
A2
Rb
b
R 01
A3
02
b
G 01
Rb
b
G 23
t gb2
b
G 03
23
b
R 13
trb0
trb3
Cr
br (ar − br ) (br − cr )
∆r
I 1b
or after expansion
1
∆2r
13
A1
tgb0
I 2b
∆r
=
I 3b
tgb1 G b
tgb3
trb 2
Figure 5: Incenter tangent meets
This may be rewritten, using ∆r = ar cr − b2r , in the form
1 2 2
(∆ ar x + 2∆2r br xy + ∆2r cr y2 + ∆r b2r (ar − 2br + cr )
∆r r
− 2∆2r br x − 2∆2r br y) = br (ar − br ) (br − cr ) .
Now cancel ∆r , and rearrange to get
∆r ar x2+2∆r br xy+∆r cr y2−2∆r br x−2∆r br y+br ar cr −b2r =0
or more simply
ar x2 + 2br xy + cr y2 − 2br x − 2br y + br = 0
which has the form stated in the theorem. The same kind
of calculation establishes the formulas for Cb and Cg . Note that the equations for the Incenter Circles Cb , Cr and
Cg allow them to be defined whether or not the corresponding Incenters exist! Incenters then exist precisely as meets
of these Incenter Circles: for example the blue Incenters
I0b , I1b , I2b , I3b are just the meets of Cr and Cg , if these exist in
the field in which we work.
Theorem 6 (Incenter tangent meets) The tangent lines
b ,t b ,t b ,t b to the red Incenter circle C at the blue Intr0
r
r1 r2 r3
centers form a standard quadrilateral, as do the tangent
b ,t b ,t b ,t b at the green Incenter circle C . The
lines tg0
g
g1 g2 g3
same holds for the red and green Incenters, if they exist.
b t b and Rb ≡ t b t b lie
This implies that the meets Rb01 ≡ tr0
r1
23
r2 r3
on l1 = A2 A3 , and are harmonic conjugates with respect to
b t b and Rb ≡ t b t b lie on
A2 and A3 . Similarly Rb02 ≡ tr0
r2
13
r1 r3
l2 = A1 A3 , and are harmonic conjugates with respect to A1
b t b and Rb ≡ t b t b lie on l = A A ,
and A3 ; and Rb03 ≡ tr0
3
1 2
r3
12
r1 r2
and are harmonic conjugates with respect to A1 and A2 .
b t b and Gb ≡ t b t b lie on l , and are
The points Gb01 ≡ tg0
1
g1
23
g2 g3
harmonic conjugates with respect to A2 and A3 . Similarly
b t b and Gb ≡ t b t b lie on l , and are harmonic
Gb02 ≡ tg0
2
g2
13
g1 g3
b t b and
conjugates with respect to A1 and A3 , and Gb03 ≡ tg0
g3
b
b b
G12 ≡ tg1tg2 lie on l3 , and are harmonic conjugates with
respect to A1 and A2 .
Proof. We prove the result for the meets Gbij of the green
b associated to the blue Incenters; the other
tangent lines tgi
cases are similar. We find the joins of a blue Incenter Iib
and the green Orthocenter Hg to be
* (b −a ) b d + (c −b ) a v + (a −b ) b w : +
Hg I0b =
Hg I1b =
2.2 Tangent lines of Incenter Circles
Now we consider tangent lines to Incenter circles. Figure 5 shows the four blue Incenters of A1 A2 A3 , together
with the red and green Incenter Circles passing through
them, namely Cr and Cg . At each of the four Incenters Iib ,
b to the red
i = 1, 2, 3, 4 we have the tangent lines trib and tgi
and green Incenter Circles Cr and Cg respectively.
Hg I2b =
g
g
g b
g
g
g b
g
g
g b
(cg −bg ) bg db + (cg −bg ) bg vb + (ag −bg ) cg wb :
(bg −cg ) bg vb + (bg −ag ) bg wb
* (b −a ) b d − (c −b ) a v − (a −b ) b w : +
g
g g b
g
g g b
g
g g b
(cg −bg ) bg db − (cg −bg ) bg vb − (ag −bg ) cg wb :
− (bg −cg ) bg vb − (bg −ag ) bg wb
* (b −a ) b d + (c −b ) a v − (a −b ) b w : +
g
g
g b
g
g
g b
g
g
g b
(cg −bg ) bg db + (cg −bg ) bg vb − (ag −bg ) cg wb :
(bg −cg ) bg vb − (bg −ag ) bg wb
* (b −a ) b d − (c −b ) a v + (a −b ) b w : +
g
g g b
g
g g b
g
g g b
Hg I3b = (cg −bg ) bg db − (cg −bg ) bg vb + (ag −bg ) cg wb : .
− (bg −cg ) bg vb + (bg −ag ) bg wb
11
KoG•18–2014
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
b is the line green perpendicular to H I b
The tangent line tgi
g i
b
passing through Ii . These we may calculate to be
* a −b u + b v + a −2b w + d b −c a −b : +
g
g b
g b
g
g
b
b g
b g
g
b
tg0
= cg −bg ub + 2bg −cg vb −bg wb + db bg + ab bg −cg :
bg (−vb + wb − db )
* a −b u −b v − a −2b w + b d −c a −b : +
g
g b
g b
g
g b
g b
b g
g
b
tg1
= cg −bg ub − 2bg − cg vb + bg wb + db bg + ab bg −cg :
bg (vb − wb − db )
*− a −b u + b v − a −2b w + b d −c a −b : +
g
g b
g b
g
g b
g
b
b g
g
b
tg2
= − cg −bg ub + 2bg −cg vb + bg wb + db bg + ab bg −cg :
bg (−vb − wb − db )
*− a −b u −b v + a −2b w + b d −c a −b : +
g
g b
g b
g
g b
g
b
b g
g
b
tg3
= − cg −bg ub − 2bg −cg vb −bg wb + db bg + ab bg −cg : .
bg (vb + wb − db )
We could verify directly that these four lines form a standard quadrilateral. But we prefer to verify that the meets
of these four tangent lines agree with the following meets
with the side lines of A1 A2 A3 :
b b
b
Gb01 ≡ tg0
tg1 = tg0
l1
1
[(bg −cg ) (−ub + vb + ab ), (ag −bg )(−ub −wb + cb )]
=
λ01
b b
b
Gb23 ≡ tg2
tg3 = tg2
l1
1
[(bg − cg) (ub + vb + ab ) , (ag − bg) (ub + wb + cb )]
=
λ23
1
b b
b
[0, bg (−vb + wb − db)]
Gb02 ≡ tg0
tg2 = tg0
l2 =
λ02
1
b b
b
Gb13 ≡ tg1
tg3 = tg1
l2 =
[0, bg (−vb + wb + db)]
λ13
1
b b
b
Gb03 ≡ tg0
tg3 = tg0
l3 =
[bg (vb − wb + db) , 0]
λ03
1
b b
b
[bg (vb − wb − db) , 0]
Gb12 ≡ tg1
tg2 = tg1
l3 =
λ12
where
λ01 = (cg − ag) ub + (bg − cg) vb + (bg − ag ) wb
+ (ab bg − ab cg + cb ag − cb bg )
λ23 = (ag − cg ) ub + (bg − cg) vb
+ (ag − bg) wb + (ab bg − abcg + cbag − cb bg )
λ02 = (bg − cg ) ub + (cg − 2bg) vb + bgwb
+ (ab cg − 2abbg + 2bbbg − cb bg )
λ13 = (cg − bg) ub + (cg − 2bg) vb + bgwb
+ (2ab bg − abcg − 2bbbg + cb bg )
λ03 = (ag − bg ) ub + bg vb + (ag − 2bg) wb
+ (ab bg − 2bbbg − cb ag + 2cbbg )
λ12 = (bg − ag ) ub + bg vb + (ag − 2bg) wb
+ (2bb bg − abbg + cb ag − 2cbbg ) .
12
The fact that A2 , A3 , Gb01, Gb23 form a harmonic range etc.
is an immediate consequence of a well known fact about
standard quadrilaterals in projective geometry, since we
have shown that the points A1 , A2 and A3 are diagonal
points of the quadrilateral formed by the four green tangent lines.
Following the construction of the red lines in the introductory section on Quadrangles and quadrilaterals, we join a
point Gbij with the triangle point Ak opposite to the triangle
line that it lies on; giving six lines Ak Gbij :
* a − b (u + w − c ) : +
g
g
b
b
b
bg − cg (−ub + vb + ab ) :
0
* b − a (u + w + c ) : +
g
g b
b
b
A1 Gb23 =
bg − cg (ub + vb + ab ) :
0
+
*
bg (−v
b + wb − db ) :
b
A2 G02 = bg −cg ub + cg −2bg vb + bg wb −bg db + ab cg −bg :
bg (vb − wb + db )
+
*
b
g (−v
b + wb + db ) :
b
A2 G13 = cg −bg ub + cg −2bg vb + bg wb + bg db + ab bg −cg :
bg (vb − wb − db )
* a −b u +b v + a −2b w +b d +c b −a : +
g
g b
g b
g
g
g b
g
b
b g
bg (vb − wb + db ) :
A3 Gb03 =
bg (−vb + wb − db )
* b −a u +b v + a −2b w −b d +c a −b : +
A1 Gb01 =
A3 Gb12 =
g
g
b
g b
g
g
b
g b
b
g
g
bg (vb − wb − db ) :
bg (−vb + wb + db )
Theorem7 (Quad points) The triples
{A1 Gb23 , A2 Gb13,
b
b
b
b
b
A3 G12 }, A1 G23 , A2 G02 , A3 G03 , A1 G01 , A2 Gb13 , A3 Gb03
and
A1 Gb01 , A2 Gb02 , A3 Gb12
of lines are concurrent in the respective points Qbg0 , Qbg1 , Qbg2 and
Qbg3 , called the blue/green quad points.
The
b
b
b
b
b
b
triples A1 R23 , A2 R13 , A3 R12 ,
A1 R23 , A2 R02 , A3 R03 ,
A1 Rb01 , A2 Rb13 , A3 Rb03 and A1 Rb01 , A2 Rb02 , A3 Rb12 are
also concurrent in the respective points Qbr0 , Qbr1 , Qbr2 and
Qbr3 , called the blue/red quad points. Similar results hold
for the red and green Incenters, if they exist.
Proof. We verify this for the blue/green quad points: this
is a consequence of the projective geometry of the complete quadrilateral we mentioned in the first section—if the
original four tangent lines are regarded as the blue lines
in Figure 6, then the quad points Qbg j correspond to the
red points. However we want to find explicit formulas and
check things directly. The quad point Qbg j is naturally associated to the Incenter I bj . After some calculation, we find
.
KoG•18–2014
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
that these are
bg
[(bg − cg ) (db ub − (bb − cb ) vb ) ,
λ0
(ag − bg) ((cb − bb ) wb + cb db )]
bg
Qbg1 = [(bg − cg ) ((ab − bb) vb + abdb ) ,
λ1
(ag − bg) (db ub + (ab − bb ) wb )]
bg
Qbg2 = [(cg − bg) (ab wb − bbvb ) ,
λ2
(bg − ag) (bb wb − cb vb )]
bg
Qbg3 = [(bg −cg ) (−db ub + (db + bb) vb −ab wb + abdb ),
λ3
(bg− ag) (db ub −cb vb + (db + bb) wb −cb db )]
Qbg0 =
where
λ0 = (bg − cg ) (bg db + (bb − cb ) (ag − bg)) ub
since db u2b + (bb − ab ) ub wb − bb db ub − cb v2b + bb vb wb +
cb (ab − bb) vb = 0 by using (2), and similarly for the other
indices. In a parallel fashion, we find that the four blue/red
quad points Qbr j have exactly the same formulas as the
Qbg j , except for the replacements ag −→ ar , bg −→ br and
cg −→ cr , and similarly for the other colours red and green.
sb1
I0
Qrb3 B13 Qrb1
trb1
A3
B23
Qrb2
trb0
−bg (ab bg +bb ag− abcg−bb bg)wb −ab cb (bg −cg)(ag −bg)
λ3 = (db + bb) b2g + ag cg −ag bg (2db + bb )−bb bg cg ub
+ (bg ((bg − cg ) (ab − bb ) + cb (2ag − bg )) − cbag cg ) vb
+ ((db + bb ) (bg − ag ) bg − abag (bg − cg )) wb
+b2g (ab (db −cb)−cb db)+bg (cb ag (db + ab)− abcg (db−cb))
− a g cg a b cb .
We may then check directly that for example Qbg0 is incident with A3 Gb12 by computing
bg − ag ub + bg vb + ag − 2bg wb − bg db + cb ag − bg ·
!
bg bg − cg (db ub − (bb − cb ) vb )
·
λ0
!
bg ag − bg ((cb − bb ) wb + cb db )
+ bg (vb − wb − db )
λ0
+ bg (−vb + wb + db )
db u2 +(bb − ab ) ub wb − bb db ub
b
bg ag − bg bg − cg
2
−cb vb + bb vb wb + cb (ab − bb ) vb


=
− bg − cg bg db + (bb − cb ) ag − bg ub


+ bg − cg bb bg + cb ag − 2cb bg vb +bg ag − bg (bb − cb ) wb − cb bg ag − bg db
=0
B02
sb3
B03
Qgb3
A2
B 01
Qrb0
sb0
Qgb0
B12
t gb2
Qgb1
Qgb2
sb2
Crb
trb3
trb 2
λ1 = (ag − bg) (bg db + (ab − bb ) (bg − cg )) ub
λ2 = bb (bg − cg ) (ag − bg) ub
+ bg (bb bg + cb ag − bbcg − cb bg ) vb
Cbg
A1
tgb0
I2
− (bg − cg ) (bb bg + cb ag − 2cb bg ) vb
− bg (ag − bg) (bb − cb) wb + cbbg (ag − bg ) db
+ bg (bg − cg) (ab − bb) vb
+ (ag −bg ) (2ab bg −ab cg −bb bg ) wb + ab bg (bg −cg ) db
I3
tgb1
tgb3
I1
Figure 6: Quad points and star lines
Now introduce the blue star line sbj to be the join of the
corresponding blue/red quad point Qbr j and the blue/green
quad point Qbg j , and similarly for the other colours. There
are then four blue star lines sb0 , sb1 , sb2 and sb3 .
The blue star point Bi j is the meet of the two blue star
lines sbi and sbj , that is Bi j ≡ sbi sbj , and similarly for the other
colours.
Note that following the introductory section on Quadrangles and quadrilaterals, we use the correspondence between the Qbg j ; and the tangent lines tgb j ; and the Incenters
I bj to match up the indices.
Theorem 8 (Star quadrilateral) The four blue star lines
form a standard quadrilateral sb0 sb1 sb2 sb3 . This holds also for
the other colours.
Proof. The proof we have is surprisingly complicated. The
star lines sbj have quite involved formulas; for example we
find that
b0
sb0 = Qb0
g Qr =
E0 db ub + F0 cb vb : 
bg br (bb − cb )2 + cb db ·


·(ag br − bg ar − ag cr + cg ar + bg cr − cg br ) db ub
+
*


bg cr −cg br ) (bb −cb) db vb :
 −2cb bg br (ag br −bg ar − ag cr + cg ar + 

 −ab (br − cr ) (bg − cg ) (ag br − bg ar ) (bb − cb )2 + cb db wb 
+2ab cb (br − cr ) (bg − cg ) (ag br − bg ar ) (bb − cb ) db !
−bg b
r db (ag br − bg ar − ag cr + cg ar + bg cr − cgbr ) ·

·
(bb − cb )2 + cb db 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
degree 6 in the variables ai , bi and ci , with the former having 32 terms and the latter 46 terms. After some trial and
error we can present these in the somewhat pleasant, but
still mysterious, forms:
E0 = −bg br (ag cr − cg ar − bgcr + cg br ) ·
· b2b + 4c2b + ab cb − 6bbcb
+ bgbr (ag br − bgar ) b2b + 2c2b + ab cb − 4bbcb
− 2cb ag cg b2r − ar cr b2g + agar cr bg − ag cg ar br (bb −cb )
search. So we tentatively conjecture that there are no such
triangles.
In any case, to get an explicit example we√use an algebraic extension field of the rationals; so by 30√we mean
an appropriate symbol in the extension field Q 30 etc..
Note that although our use of square roots is entirely algebraic, our representation of these square roots as approximate rational numbers (we prefer to avoid discussion of
“real numbers”), necessarily brings an approximate aspect
into our diagrams.
50
and
45
Cr
F0 = ag cg b2r − ar cr b2g + agar cr bg − agcg ar br ·
· b2b − 4bbcb + 2c2b + ab cb
+ bgbr (ag cr − cg ar − bgcr + cg br ) ·
· −5b2b − 4c2b + 2abbb − 3abcb + 10bbcb
40
35
30
25
Hr
Cg
20
15
10
X2
5
50
30
40
20
10
− 2bgbr (ag br − bgar ) (bb − cb ) (ab − 2bb + cb) .
X3
10
X1
Hg
20
30
40
50
60
70
80
5
10
15
We can then calculate the blue star points, for example
! 

bg
br db (ag br −bg ar −agcr +cg ar +bg cr −cg br)·

 · (b − c )2 + c d u − 2c (b − c ) v
b
b b
b b
b
b
b b



, 0
B03 =

E0 db ub + F0 cb vb


20
Hb
25
30
Cb
35
40
Figure 7: An example triangle X1 X2 X3
from which clearly B03 lies on l3 . The computations are
similar for the other indices, and the other colours.
Example 1 One may check that the basic Triangle with
points
3 Explicit examples and some conjectures
X1 ≡ [−21/59, −58/59],
3.1 An example over Q
√
√
√ √
√
30, 217, 741, 2470, 82 297
We will now explore in detail a particular triangle which
has both blue, red and green Incenters; for us this is not
only an important tool for checking the consistency of our
formulas, but also a way to get a sense of the level of complexity of various constructions. In fact this kind of explicit
calculation of examples is much to be encouraged in this
subject: especially as working over concrete fields, including finite fields and explicit extension fields of the rationals, allows us to appreciate the number theoretic aspects
of our geometrical investigations. For example, finding a
triangle with blue, red and green Incenters approximately
is easy with a geometry package: finding a concrete example and working out all the points precisely is more challenging.
In particular we were unable, despite a reasonable computer search, to find any triangles with purely rational
points that have blue, red and green Incenters! We would
like to thank Michael Reynolds for his contributions to this
14
X2 ≡ [−13/3, 2]
and
X3 ≡ [35/3, −8/5]
√
√ √
√
√
has both
in Q 30, 217, 741, 2470, 82 297
blue, red and green Incenters.
After translation
by (21/59, 58/59) we obtain Xe1 = [0, 0], Xe2 =
[−704/177, 176/59] and Xe3 = [2128/177, −182/295].
The matrix N and its inverse M, where
704
176
− 177
α β
59
N = 2128
=
and
γ δ
− 182
177
295
M=N
−1
=
13
704
95
264
5
56
5
42
respectively send [1, 0] and [0, 1] to Xe2 and Xe3 , and Xe2 and
Xe3 to [1, 0] and [0, 1]. From now on we discuss only the
standard triangle A1 A2 A3 associated to X1 X2 X3 ; to convert
back into the original coordinates, we would multiply by N
and translate by (−21/59, −58/59). The bilinear forms in
these new standard coordinates, for the blue, red and green
KoG•18–2014
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
geometries respectively, are given by matrices
774 400
848
− 7778
ab bb
1 0 T
31
329
156
645
Db = N
N =
113 507 716 = b
848
0 1
− 7778
b cb
156 645
783 225
216 832
7202 272
1 0
ar br
− 156 645
T
31
329
Dr = N
N =
112 911 484 = b
272
0 −1
− 7202
r cr
156 645
783 225
247 808 2000 768 0 1 T
− 10 443
ag bg
52 215
Dg = N
N = 2000
.
=
768
592
1 0
b g cg
− 774
52 215
52 215
140 736
The determinants of Db , Dr and Dg are ∆b = 9787
025 and
97 140 736
∆r = ∆g = − 87 025 , while the mixed traces are db = 6724
25 ,
576
A
A
A
dr = 6076
and
d
=
−
.
The
orthocenters
of
1 2 3 are
g
25
5
8825 537 84 337
87 833 227 55 537
Hb = −
, Hr =
,
,−
,
1019 520 25 488
11 214 720 25 488
7105 377
.
,
Hg =
3894 177
Blue,
√ and
√ green
√ red
√ Incenters
√
exist over F =
Q 30, 217, 741, 2470, 82 297 and we may
choose
1875 104
14 432
873 628
, vb =
, wb =
31 329
177
4425
2464 √
17 248 √
82 297, vr =
217
ur =
156 645
885
√
√
196
217 82297
wr =
4425
2816 √
19 712 √
448 √
ug =
2470, vg =
30, wg =
741.
52 215
295
295
ub =
Then the four blue Incenters, the four red Incenters and the
four green Incenters of A1 A2 A3 respectively are
5327
25
761 220
b
b
I1 =
,
,−
I0 = −
590 413
10 384 118
5327 220
761 25
I3b =
,
,
I2b =
2112 168
270 27


I0r = 



I1r = 



I2r = 

√
√ √
217 − 20 461 217 82297
+210 343 √82 297 −√76 618
√507),
1
(2923
217
−
7
217
82297
50 976
√
+133 82 297 − 30 049)
√ √
√
1
217 82297 − 4032 553 217
22 429 440 (20 461 √
+210 343
√ 82
√
√ 297 − 76 618 507),
1
(7
217
50 976
√ 82297 − 2923 217
+133 82 297 − 30 049)
√
√ √
1
22 429 440 (4032 553
√ 217 + 20 461 217 82297
−210 343
√
√ 82
√ 297 − 76 618 507),
1
50 976 (7 217
√ 82297 + 2923 217
−133 82 297 − 30 049)
1
22 429 440 (4032 553
√














I3r = 

I0g
=
I1g =
I2g
I3g
=
=
√ √
√
217 + 20 461 217 82297
+210 343
√
√ 82
√ 297 + 76 618 507),
−1
217
(7
50 976
√ 82297 − 2923 217
−133 82 297 − 30 049)
−1
22 429 440 (4032 553
√
√
√
√




247
35
3211
203
741 − 3894
30 + 3894
2470 − 7788
,
7788 √
√
√
29
20
100
13
1239 2470 − 177 30 + 1239 741 − 177
√
√
√
203
35
3211
247
2470 − 7788
,
3894 √30 − 7788 √741 + 3894 √
13
20
100
29
177 30 + 1239 2470 − 1239 741 − 177
√
√
√
203
35
3211
247
30 − 7788
741 − 3894
2470 − 7788
,
− 3894
√
√
√
13
20
29
30 − 1239
2470 − 1239
741 − 100
− 177
177
√
√
√
203
35
3211
247
30
+
741
−
2470
−
3894 √
7788 √
3894 √
7788 ,
.
13
20
100
29
177 30 − 1239 2470 + 1239 741 − 177
The Incenter circle quadrances are
18 154 129 609
,
28 196 100
1182 272
.
Kg =
10 443
Kb =
Kr = −
11 681 819 191
28 196 100
The blue, red and green Incenter Circles themselves have
respective equations
4840 000x2 − 19 447 120xy + 28 376 929y2
+19 447 120x + 19 447 120y − 9723 560 = 0
19 360x2 − 62 524xy + 12 103y2
+62 524x + 62 524y − 31 262 = 0
193 600x2 − 2572 240xy + 4032 553y2
+2572 240x + 2572 240y − 1286 120 = 0.
The four tangent lines tgb j are
b
tg0
= h1570 : −11823 : 8323i
b
tg1
= h−127 512 : −33 761 : 58 261i
b
tg2
= h−18 216 : −11 823 : 8323i
b
tg3
= h−1570 : 4823 : 8323i.
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 following meets with the side lines of A1 A2 A3 :
3500 9893
b b
b
,
Gb01 ≡ tg0
tg1 = tg0
l1 =
13 393 13 393
3500 9893
b b
b
Gb23 ≡ tg2
tg3 = tg2
l1 = −
,
6393 6393
1189
b b
b
Gb02 ≡ tg0
tg2 = tg0
l2 = 0,
1689
1189
b b
b
Gb13 ≡ tg1
tg3 = tg1
l2 = 0,
689
8323
b b
b
Gb03 ≡ tg0
tg3 = tg0
l3 = −
,0
1570
8323
b b
b
,0 .
Gb12 ≡ tg1
tg2 = tg1
l3 =
18 216
The blue/red quad points Qbr j associated to I0b , I1b , I2b , I3b respectively are
18 005 811 535 12 129 669 559
b
Qr0 =
,−
21 082 889 161 21 082 889 161
18 005 811 535 12 129 669 559
b
Qr1 = −
,
9330 605 209 9330 605 209
18 005 811 535 12 129 669 559
b
,
Qr2 =
14 928 733 909 14 928 733 909
18 005 811 535 12 129 669 559
b
Qr3 =
,
.
45 342 228 279 45 342 228 279
The respective blue/green quad points Qbg j are
4161 500 11 762 777
Qbg0 = −
,
2654 777 2654 777
4161 500 11 762 777
,
Qbg1 = −
12 547 777 12 547 777
4161
500 11 762 777
Qbg2 =
,
10 977 777 10 977 777
4161 500 11 762 777
b
.
,
Qg3 =
20 870 777 20 870 777
The blue star lines are then
sb0 = Qbr0 Qbg0 =
h1796 063 533 088 : 868 804 574 039 : −1034 074 074 039i
sb1 =
h272 084 614 990 : 1199 343 574 039 : −1034 074 074 039i
sb2 =
h272 084 614 990 : 868 804 574 039 : −1034 074 074 039i
sb3 =
h1796 063 533 088 : 1199 343 574 039 : −1034 074 074 039i
16
and they meet at the blue star points
165 269 500 000 761 989 459 049
B01 =
,
927 258 959 049 927 258 959 049
165 269 500 000 761 989 459 049
B23 = −
,
596 719 959 049 596 719 959 049
1034 074 074 039
1034 074 074 039
B02 = 0,
, B13 = 0,
868 804 574 039
1199 343 574 039
1034 074 074 039
1034 074 074 039
, 0 , B12 =
,0 .
B03 =
1796 063 533 088
272 084 614 990
Note the pleasant rationality of the previous objects.
3.2 An example over F13
Now we look at an example over a finite field.
Theorem 9 (Null quadrances incenters) Suppose that
the field F contains an element i, where i2 = −1, and
the characteristic of F is not 2. If
bb (ab − bb ) (bb − cb )
br (ar − br ) (br − cr )
= Kr ≡
∆b
∆r
bg (ag − bg) (bg − cg)
=0
= Kg ≡
∆g
Kb ≡
then the standard Triangle A1 A2 A3 has four distinct blue,
red and green Incenters.
Proof. If Kb = 0 then from the definition of the blue incenter 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
null circle, and if Kg = 0 then Cg is a null circle. These
null lines have distinct direction vectors (1, ±i), (1, ±1)
and (1, 0), (0, 1) respectively, and they are never parallel
since char(F) 6= 2, so i 6= ±1. Therefore, any two null circles meet in exactly four points.
Here is an example found by Michael Reynolds [6] which
illustrates explicitly the above theorem.
Example 2 The triangle X1 X2 X3 with points X1 ≡
[3, 4], X2 ≡ [1, 9] and X3 ≡ [12, 3] in F13 has four blue, red
and green Incenters. In F13 the squares are 0, 1, 3, 4, 9, 10
and 12, and in particular −1 = 12 = 52 is a square. After
translation by (3, 4) we obtain Xe1 = [0, 0] , Xe2 = [11, 5] and
Xe3 = [9, 12]. The matrix N and its inverse M
11 5
10 11
N=
,
M = N −1 =
9 12
12 7
send [1, 0] and [0, 1] to Xe2 and Xe3 , and Xe2 and Xe3 to [1, 0]
and [0, 1] respectively. The bilinear form in these new standard coordinates for the blue, red and green geometries
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
respectively are
1 0
3 3
T
Db = N
N =
0 1
3 4
1 0
5 0
T
Dr = N
N =
0 −1
0 2
0 1
6 8
T
Dg = N
N =
.
1 0
8 8
Theorem 10 (Spieker circles) If a triangle has four blue
Incenters I0b , I1b , I2b and I3b , then the four blue Spieker centers all lie both on a red Spieker circle with center the red
Circumcenter Cr , and on a green Spieker circle with center
the green Circumcenter Cg . If both say blue and red Incenters exist, then all 8 blue and red Spieker points lie on the
same green circle. The same holds for the other colours.
We can see immediately that Kb = Kr = Kg = 0 from the
definitions
bb (ab − bb) (bb − cb )
≡ Kb
∆b
br (ar − br ) (br − cr )
≡ Kr
Qr (Hr , Iib ) =
∆r
bg (ag − bg) (bg − cg )
Qg (Hg , Iir ) =
≡ Kg ,
∆g
Qb (Hb , Iir ) =
i = 0, 1, 2, 3.
The four blue, red and green Incenters respectively are
I0b = [4, 8] ,
I1b = [3, 6],
I2b = [8, 10],
I3b = [11, 4]
I0r = [10, 9], I1r = [8, 2], I2r = [6, 5] , I3r = [4, 12]
I0g = [9, 8] , I1g = [5, 3], I2g = [12, 11], I3g = [2, 4]
and the blue, red and green Incenter Circles respectively
have equations
Cb : (y − x + 1)(x + 3y − 1) = 0
Cr : (x − 6y)(x + 6y) = 0
Cg : (x + 2y − 2)(x + 5y − 5) = 0.
From Michael Reynolds’ computer investigations, we tentatively conjecture that for finite fields F p where p ≡
3 mod4, there are no triangles which have both blue, red
and green Incenters, and for finite fields F p where p ≡
1 mod4, blue, red and green Incenters exists precisely
when Kb = Kr = Kg = 0, as in the above example.
4 Spieker circles and Nagel circles
Now we recall from [5] that the central dilation δ−1/2 about
the centroid takes the Orthocenter to the Circumcenter, and
the Incenters to the Spieker centers. In standard coordinates
δ−1/2 ([x, y]) = (1/2)[1 − x, 1 − y].
The inverse central dilation δ−2 takes the Orthocenter to
the De Longchamps point X20 , and takes the Incenters to
the Nagel points. In standard coordinates
δ−2 ([x, y]) = [1 − 2x, 1 − 2y].
Proof. We see that if we use the central dilation formula
to transform Incenter circles centred at the Orthocenters,
we get the Spieker circles centred at Circumcenters, so this
theorem is a direct consequence of the Incenter circles theorem and the fact that a dilation preserves circles of any
colour.
Here are the formulas for the coloured Circumcenters in
standard coordinates:
1
[cb (ab − bb) , ab (cb − bb)]
2∆b
1
Cr =
[cr (ar − br ) , ar (cr − br )]
2∆r
1
[cg (ag − bg) , ag (cg − bg)] .
Cg =
2∆g
Cb =
8
4
Cb
A1
Cg
A2
4
A3
Cr
8
Figure 8: Blue, red and green Speiker circles
Theorem 11 (Nagel circles) If a triangle has four blue Incenters I0b , I1b , I2b and I3b , then the four blue Nagel centers
all lie both on a red Nagel circle with center the red De
Longchamps point X20r , and on a green Nagel circle with
center the green De Longchamps point X20g . If both say
blue and red Incenters exist, then all 8 blue and red Nagel
points lie on the same green circle. The same holds for the
other colours.
Proof. In the same fashion as in the previous theorem, if
we use the inverse central dilation δ−2 to transform Incenter circles centred at the Orthocenters, we get the Nagel
circles centred at De Longchamps points.
17
N. Le, N. J. Wildberger: Incenter Circles, Chromogeometry, and the Omega Triangle
KoG•18–2014
Proof. This follows by considering the action of the central
dilation δ−2 which takes the circumcenter Ci to the orthocenter Hi , and the orthocenter Hi to the De Longchamps
point X20i .
X20b
8
4
X20g
A1
A2 A3
4
12
8
References
X20r
-4
[1] M. W. H ASKELL, The Construction of Conics under
given conditions, Bulletin of the Amer. Math. Soc. 11
(5) (1905), 268–273.
-8
[2] C. K IMBERLING, Triangle Centers and Central Triangles, vol. 129 Congressus Numerantium, Utilitas
Mathematica Publishing, Winnepeg, MA, 1998.
Figure 9: Blue, red and green Nagel circles
Here are the formulas for the blue, red and green De
Longchamps points:
1 2
X20b =
bb − 2bbcb + abcb , b2b − 2abbb + abcb
∆b
1 2
X20r =
br − 2br cr + ar cr , b2r − 2ar br + ar cr
∆r
1 2
bg − 2bgcg + agcg , b2g − 2agbg + agcg .
X20g =
∆g
In Figure 10 we see the relations between the
three coloured Orthocenters, Circumcenters and De
Longchamps points. The lines joining these are the three
coloured Euler lines. Note that the centroids of the triangles of Orthocenters, Circumcenters and De Longchamps
points all agree with the centroid G of the original triangle
A1 A2 A3 . We conclude with a simple observation about De
Longchamps points.
12
X20b
Hg
Cg A2
4
[6] M. R EYNOLDS, Personal Communication, 2013.
[7] N. J. W ILDBERGER, Divine Proportions: Rational Trigonometry to Universal Geometry, Wild Egg
Books, Sydney, 2005.
[8] N. J. W ILDBERGER, Affine and projective metrical
geometry, arXiv: math/0612499v1.
[11] B. M. W OODS, The Construction of Conics under
given conditions, Amer. Math. Monthly 21 (6) (1914),
173–180.
A3
Cr
X20g
[5] N. L E , N. J. W ILDBERGER, Universal Affine Triangle Geometry and Four-fold Incenter Symmetry, KoG
16 (2012), 63–80.
[10] N. J. W ILDBERGER, Chromogeometry and Relativistic Conics, KoG 13 (2009), 43–50.
Cb
A1
[4] C. K IMBERLING, Major Centers of Triangles, Amer.
Math. Monthly 104 (1997), 431–438.
[9] N. J. W ILDBERGER, Chromogeometry, Mathematical Intelligencer 32 (1) (2010), 26–32.
Hr
8
[3] C. K IMBERLING, Encyclopedia of Triangle Centers,
http://faculty.evansville.edu/ck6/
encyclopedia/ETC.html.
Hb
X20r
4
8
12
Figure 10: Blue, red, green Orthocenters, Circumcenters
and De Longchamps points
Theorem 12 (Orthocenters as midpoints) For any triangle, a coloured orthocenter H is the midpoint of the two
De Longchamps points X20 of the other two colours.
18
Nguyen Le
e-mail: [email protected]
Norman John Wildberger
e-mail: [email protected]
School of Mathematics and Statistics UNSW
Sydney 2052 Australia