Download Trisectors like Bisectors with equilaterals instead of Points

CUBO A Mathematical Journal
Vol.16, N¯o 01, (71–110). June 2014
Trisectors like Bisectors with equilaterals instead of Points
Spiridon A. Kuruklis
Eurobank,
Group IT Security and Risk Management Division
Athens, GREECE
[email protected]
ABSTRACT
It is established that among all Morley triangles of △ABC the only equilaterals are the
ones determined by the intersections of the proximal to each side of △ABC trisectors
of either interior, or exterior, or one interior and two exterior angles. It is showed
that these are in fact equilaterals, with uniform proofs. It is then observed that the
intersections of the interior trisectors with the sides of the interior Morley equilateral
form three equilaterals. These along with Pasch’s axiom are utilized in showing that
Morley’s theorem does not hold if the trisectors of one exterior and two interior angles
are used in its statement.
RESUMEN
Se establece que entre todos los triángulos de Morley de △ABC, los únicos equiláteros
son theones determinados por las intersecciones del proximal a cada lado de los trisectores △ABC de ángulos interior, o exterior, o uno interior y dos exteriores. Se muestra
que estos están en triángulos equiláteros de facto con demostraciones uniformes. Luego,
se observa que las intersecciones de trisectores interiores con los lados de un equilátero
Morley interior forman tres triángulos equiláteros. Junto con el axioma de Pasch, se
utilizan para probar que el Teorema de Morley no se satisface si se usan los trisectores
de un ángulo exterior y dos interiores.
Keywords and Phrases: Angle trisection, Morley’s theorem, Morley trisector theorem, Morley
triangle, Morley interior equilateral, Morley central equilateral, Morley exterior equilateral, Pasch’s
axiom, Morley’s magic, Morley’s miracle, Morley’s mystery.
2010 AMS Mathematics Subject Classification: 51M04
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Introduction
The systematic study of the angle trisectors in a triangle starts after 1899, when Frank Morley, a
Cambridge mathematician, who had just been recently appointed professor at Haverford College,
U.S.A. while investigating certain geometrical properties using abstract algebraic methods, made
the following astonishing observation, known since then as Morley’s theorem.
In any triangle the trisectors of its angles, proximal to the three sides respectively, meet at the
vertices of an equilateral.
A Morley triangle of △ABC is formed by the three
points of intersection of pairs of angle trisectors connected by each triangle side. Obviously for a particular
side there are four possibilities for pairing trisectors since
there are four of them that the side connects. Thus Morley’s theorem claims that a Morley triangle of △ABC is
equilateral, if it is formed by the intersections of trisectors
proximal to the three sides of △ABC respectively.
It should be noted that Morley’s theorem, as it is
stated, is subject to interpretation as the term angle
could mean either interior or exterior angle, or even a
combination of both for the different instances of the term
in the statement.
Fig.1
According to the angle meaning, Morley’s theorem
gives the following Morley equilaterals of △ABC. The
intersections of the proximal trisectors of the interior angles form the interior Morley equilateral
of △ABC. Also the intersections of the proximal trisectors of the exterior angles form the central
Morley equilateral of △ABC. In addition the intersections of the proximal trisectors of one interior
and two exterior angles form an exterior Morley equilateral of △ABC, and thus there are three
exterior Morley equilaterals of △ABC. Fig.1 depicts the above Morley equilaterals. Proofs that
the above Morley triangles are in fact equilaterals are given in Part 3 of this work.
But so an obvious question, that several authors have raised,
begs for an answer. In a △ABC are there other Morley equilaterals
besides the interior, the central and the three exterior Morley equilaterals?
Apparently the requirement of Morley’s theorem is satisfied by
three more Morley triangles formed by combinations of proximal trisectors of an exterior and two interior angles. One of them is portrayed in Fig.2. Some experimentation using computer generated
graphs for these triangles has tempted the belief that Morley’s theo-
Fig.2
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73
rem holds for them as well [14]. But in Part 5, it will be proved that these are not equilaterals.
After the examination of all Morley triangles it will be shown that the equilateral ones are
exactly the interior, the central and the three exterior Morley equilaterals.
This enables the establishment of an analogy between the structures of the angle bisectors
and the angle trisectors in a triangle. Namely, the structure of trisectors resembles the structure
of bisectors with the inner and the exterior Morley equilaterals of △ABC corresponding to the
incenter and the excenters of △ABC respectively, while the central Morley equilateral corresponds
to the triangle with vertices the excenters of △ABC.
Morley’s theorem is considered among the most surprising discoveries in mathematics as it
went curiously unnoticed across the ages. Ancient Greeks studied the triangle geometry in depth
and they could find it. But curiously they did not and it was overlooked during the following two
thousand years.
Angle trisectors exist regardless of how they can be constructed. If the structure of angle
trisectors maintains the regularity which characterizes the triangle geometry then theorems must
exist for expressing it.
The first observation about this regularity may have forgotten. Morley didn’t publish it until
25 years later by providing a sketchy proof, when the theorem had become already famous. But
Morley, excited by his discovery, travelled back to England to mention it to his expert friends. In
turn mathematical gossip spread it over the world and several journals proposed it for a proof.
Obviously, the simplicity of the theorem statement creates the expectation of an equally simple
proof. This simplicity challenges the mathematical talent.
The vast majority of publications on Morley’s theorem has treated only the trisectors of the
interior angles and gave proofs for the interior Morley equilateral. In the preface of the first
publication on the subject, by Taylor and Marr [12], it is recognized that the Morley’s work
on vector analysis, from which the above theorem follows, holds for both interior and exterior
trisectors. The paper’s treatment of the theorem with only the interior trisectors is explained
as ”Morley’s work never published and it was only the particular case of internal trisectors that
reached the authors”. The very respectable given effort has produced proofs of many kinds,
exploiting a variety of features. Trigonometric, analytic and algebraic proofs supplement the
proofs of a purely geometric kind. Site Cut the Knot [13] presents 27 different proofs of Morley’s
theorem from many more available. Notably, Roger Penrose [9] used a tiling technique, Edsger
Dijkstra applied the rule of sines three times and then the monotonicity of the function y = sin(x)
in the first quadrant [3], Alain Connes offered a proof in Algebraic Geometry [1], John Conway
showed it in plane geometry like a jigsaw puzzle solution [2], while Richard Guy proved that it is a
consequence of his Lighthouse theorem [5]. However, a geometric, concise and logically transparent
proof is still desirable.
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Richard Guy notes: “There are a few hints that there is more than one Morley triangle, but
Hosberger [p. 98] asks the reader to show that Morley’s theorem holds also in the case of the
trisection of the exterior angles of a triangle ” [5]. Rose [10] and Spickerman [11] have proposed
proofs, using different methods, for the central Morley equilateral. In Parts 3 and 4 proofs for the
exterior Morley equilaterals will be offered.
The most popular technique for proving Morley’s theorem is encountered as indirect, backwards
or reverse construction method and fits in the following scheme.
Given a triangle assume that its angles are trisected and equal to 3α, 3β and 3γ, respectively,
where α + β + γ = 60o . In order to show that one of its Morley triangles is equilateral, start with
an equilateral △A ′ B ′ C ′ and construct a △ABC with angles 3α, 3β and 3γ, so that △A ′ B ′ C ′ is
the appropriate (interior, central or exterior) Morley triangle of △ABC. Thus △ABC would be
similar to the given triangle and so would be their corresponding Morley triangles.
Proofs of the above method most often construct △ABC by erecting △B ′ AC ′ , △C ′ BA ′ and
△A CB ′ with proper choice of the angles formed on the sides of △A ′ B ′ C ′ . However repeated
requests have been recorded in geometry discussion forums for an explanation of the particular,
seemingly arbitrary, choice of angles made at the beginning of these proofs. Of course the reasoning of the choice is not necessary for their validity. But the readers unfulfilling understanding
may have encouraged the mathematical folklore the use of words “mystery”, “magic” or “miracle” for referring to Morley’s theorem. This is not justifiable as there is nothing mathematically
extraordinary related to the theorem.
′
The presented proofs for showing that the interior, the central and the exterior Morley triangles are equilaterals use the classical Analysis and Synthesis method. They exploit the inherent
symmetries of the problem and characterized by their uniform structure, logical transparency,
remarkable shortness and the distinct aesthetics of the Euclidean geometry. The Synthesis part
follows the previous method scheme. But it is empowered by two simple observations, supplying
necessary and sufficient conditions for a point to be the incenter or one of the excenters of a given
triangle. Even though they are almost trivial have a subtlety that enables to confront the messy
complexity of the triangle trisectors by enforcing clean simplicity and create proofs by harnessing
the power of the triangle angle bisector theorem. In addition these proofs reveal fundamental
properties of the Morley equilaterals stated as Corollaries. Besides their extensive use for showing
Morley triangles as not equilaterals, their fertility is demonstrated by proving the following: (1)
The two sides extensions of the inner Morley equilateral meet the corresponding inner trisectors at
two points which with the two sides common vertex form an equilateral. (2) The sides of Morley
equilaterals are collinear or parallel. (3) In any triangle the exterior trisectors of its angles, proximal to the three sides respectively, meet at the vertices of an equilateral, if and only if, the interior
trisectors of an angle and the exterior trisectors of the other two angles, proximal the three sides
respectively, meet at the vertices of an equilateral.
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In short, this work advocates that for Morley’s observation a natural theoretical setting is
Euclidean geometry.
2
Notation and Counting of all Morley triangles
In a Morley triangle of △ABC each vertex is the intersection of two trisectors, each of which is
either proximal or distal to a side of △ABC. Hence a vertex is called proximal, distal or mix with
respect to the triangle side it belongs in the case the trisectors are both proximal, both distal, or
one proximal and one distal to the side, respectively.
So we may denote a proximal, distal or mix vertex with respect to a side
by using as superscripts p, d or * to
the letter of the corresponding angle
of △ABC opposite to the side.
Thus Ap , Ad and A∗ denote the
proximal, distal and mix vertex of a
Morley triangle with respect to BC respectively. In Fig.3 the notations for
all intersections of the interior trisectors of △ABC are showed. Notice
that a Morley triangle may have either
proximal vertices, or distal vertices, or
exactly two mix vertices.
Specifically, △Ap Bp Cp denotes
Fig.3
the inner Morley triangle of proximal
vertices, which is the inner Morley triangle determined by the intersections
of proximal to each side trisectors. Also there is just one Morley triangle with distal vertices which
is denoted by △Ad Bd Cd . In addition there are three Morley triangles with one vertex proximal
and two vertices mix. They are denoted by △Ap B∗ C∗ , △Bp C∗ A∗ and △Cp A∗ B∗
Moreover there are three more Morley triangles with one vertex distal and two vertices mix.
They are denoted by △Ad B∗ C∗ , △Bd C∗ A∗ and △Cd A∗ B∗ . Notice that a proximal or a distal
vertex is uniquely determined but a mix vertex is not as there are two such denoted by the same
letter. However in a Morley triangle with a pair of mix vertices, given its proximal or distal vertex,
the mix vertices are uniquely specified due to the choice restrictions in pairing trisectors for the
second and then for the third vertex. Hence there are 8 interior Morley triangles formed by the
trisectors of the interior angles of △ABC.
Similarly the trisectors of the exterior angles of △ABC form Morley triangles. These are
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p p
d d d
denoted by △Ap
A BB CC , for the Morley triangle of proximal vertices, △AA BB CC , for the Morley
p ∗ ∗
p ∗ ∗
p ∗ ∗
triangle of distal vertices, △AA BB CC , △BB CC AA and △CC AA BB , for the Morley triangles with
∗ ∗
d ∗ ∗
d ∗ ∗
one proximal and two mix vertices, △Ad
A BB CC , △BB CC AA and △CC AA BB for the Morley triangles of one distal and two mix vertices. In this notation we use subscripts in order to distinguish
a vertex determined by the interior trisectors from the vertex of the same type determined by the
exterior trisectors. Hence, in general, there are 8 Morley triangles formed by the trisectors of the
exterior angles of △ABC. Their vertices are in the exterior of △ABC and due to their rather
central location with respect to △ABC are called central Morley triangles.
There are two more possibilities for the formation of a Morley triangle. One is by combining
the trisectors of an interior angle with the trisectors of the other two exterior angles of △ABC.
Another is by combining the trisectors of two interior angles with the trisectors of the third exterior
angle of △ABC.
The Morley triangles formed by combining the trisectors of the interior ∠A with the trisectors
p p
of the exterior ∠B and ∠C are denoted by △Ap
A bA cA , for the Morley triangle of proximal vertices,
p ∗ ∗
p ∗ ∗
d d d
∗ ∗
△AA bA cA , for the Morley triangle of distal vertices, △Cp
C aC bC , △aC bC CC and △bC CC aC ,
∗ ∗
d ∗
∗
for the Morley triangles with one proximal and two mix vertices, and △Ad
A bA cA , △bA cA AA ,
d ∗ ∗
△cA AA bA for the Morley triangles with one distal and two mix vertices. The use of a small letter
is for denoting the intersection of an interior and an exterior trisector of △ABC. The vertices of
these 8 triangles formed by the trisectors of the interior ∠A with the trisectors of the exterior ∠B
and ∠C are in the exterior of △ABC and thus they are called exterior Morley triangles relative to
∠A.
Similarly are denoted the Morley triangles relative to ∠B, which are formed by combining the
trisectors of the interior angle ∠B with the trisectors of the exterior ∠C and ∠A, and also the
ones relative to ∠C formed by combining the trisectors of the interior ∠C with the trisectors of
the exterior ∠A and ∠B. Hence, in general, there are 24 exterior Morley triangles determined by
the intersections of trisectors of an interior and two exterior angles of △ABC.
The Morley triangles formed by combining the trisectors of the interior ∠B and ∠C with
p
the trisectors of the exterior ∠A are denoted by △Ap bp
A cA , for the Morley triangle of proximal
p ∗
p ∗ ∗
d
p ∗ ∗
∗
vertices, △Ad bd
A cA , for the Morley triangle of distal vertices, △A bA cA , △bA cA A , △cA A bA ,
∗
∗
d ∗ ∗
for the Morley triangles of one proximal vertex and two mix, △Ad b∗A c∗A , △bd
A cA A , △cA A bA ,
for the Morley triangles of one distal vertex and two mix. It should be remarked that in this
notation the same symbol for the intersection of an interior with an exterior trisector may refer to
two different points, an ambiguity which is clarified in a Morley triangle since one of its vertices
specifies its type and so the vertex that the symbolism refers. Hence, there are 8 Morley triangles
relative to the exterior ∠A, which obviously have one vertex inside and two outside △ABC. In
general, there are 24 Morley triangles of △ABC determined by the intersections of trisectors of
one exterior and two interior angles of △ABC.
Conclude that in total there are, in general, 64 Morley triangles of △ABC.
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Uniform Proofs for all Morley Equilaterals
In this part we will prove that five Morley triangles are
equilaterals. The proofs are uniform and utilize two basic
observations for determining the incenter or an excenter
of △ABC using only one of its bisectors.
Observe that the incenter I is lying on a unique arc
passing through two vertices and I. In Fig.4 the unique
arc passing through A, B and I is depicted. Obviously
∠AIB=1800 − 12 ∠ABC = 12 ∠BAC = 90o + 12 ∠ACB.
Thus I may be characterized as the intersection in the
interior of △ABC of a bisector with the arc of size 90o +
1
2 ∠ACB passing through A and B. Clearly an analogous
result holds for the other two pairs of vertices of △ABC.
We refer to this as the Incenter Lemma.
Fig.4
If IC is the excenter relative to ∠C then
∠AIC B = 90o − 21 ∠ACB, ∠BIC C = 12 ∠BAC and ∠CIC A = 12 ∠CBA.
Thus IC is determined by the intersection in the exterior of △ABC of a bisector, either of the
interior ∠C or the exterior ∠A or ∠B with the arc of size 90o − 21 ∠ACB passing through A and B,
or with the arc passing through B and C of size 12 ∠BAC, or with the arc of size 12 ∠CBA passing
through C and A. Evidently analogous results hold for the other two excenters IA and IB . We
refer to this as the Excenter Lemma.
Theorem 1. In any triangle the interior trisectors of its angles, proximal to the sides, meet at
the vertices of an equilateral.
Proof.
Analysis: Let △ABC be a triangle with ∠A =
3α, ∠B = 3β and ∠C = 3γ, where α+β+γ = 60o .
Suppose that △Ap Bp Cp is equilateral, where Ap ,
Bp and Cp are the intersections of the trisectors
proximal to the sides BC, CA and AB respectively. The aim of this step is to calculate the
angles formed by the sides of △Ap Bp Cp and the
trisectors of △ABC. See Fig.5.
Let Cd be the intersection of ABp and BAp . Since
ACp and BCp are angle bisectors in △ACd B , Cp
is the incenter.
Fig.5
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Spiridon A. Kuruklis
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Let P and Q be the orthogonal projections of Cp on ACd and BCd respectively. Thus Cp P =
C Q and Cd P = Cd Q. But so △Cp PBp = △Cp QBp as right triangles having two pairs of
sides equal. Hence Bp P = Ap Q. Then Cd Ap = Cd Bp and so △Ap Cd Bp is isosceles. Now from
△ACd B we have ∠ACd B = 180o − (2α + 2β) = 60o + 2γ.Therefore ∠Cd Bp Ap = ∠Cd Ap Bp =
1
o
o
o
2 [180 − (60 + 2γ)] = 60 − γ. Consequently
p p
∠C B A = ∠Cp Ap B = 180o − 60o − (60o − γ) = 60o + γ = γ+ .
p
Let Ad be the intersection of BCp and CBp . Also let Bd be the intersection of CAp and ACp .
Then from △BAd C and △CBd A find similarly
∠Ap Cp B = ∠Ap Bp C = α+ and ∠Bp Ap C = ∠Bp Cp A = β+ .
Synthesis: Suppose that a triangle is
given and assume that its angles are trisected
and equal to 3α, 3β and 3γ, respectively,
where α + β + γ = 60o . Then around an equilateral △Ap Bp Cp will construct △ABC with
angles 3α, 3β and 3γ so that Ap , Bp and Cp
will be the intersections of the proximal to the
sides interior trisectors.
On the side Bp Cp erect △Bp ACp with
adjacent angles γ+ = γ + 60o and β+ = β +
60o .
Similarly, erect △Cp BAp and △Ap CBp on
the sides Cp Ap and Ap Bp respectively with
corresponding angles as shown in Fig.6, which
were found in the Analysis step.
Fig.6
Let Cd be the intersection ABp and BAp .
Notice that △Ap Cd Bp is isosceles as two of
its angles are 180o − 60o − γ+ = 60o − γ.
Thus
Cd Ap = Cd Bp (1) and ∠Ap Cd Bp = 180o − 2(60o − γ) = 60o + 2γ (2)
Since △Ap Bp Cp has been taken equilateral, Cp Ap = Cp Bp . Combine this with (1) and infer
that Cp is on the Ap Bp bisector and so on the ∠ACd B bisector. Moreover from (2) ∠ACp B =
3600 −(α+ +60o +β+ ) = 180o −(α+β) = 90o + 21 (60o +2γ) = 90o + 12 ∠Ap Cd Bp = 90o + 12 ∠ACd B.
Hence, by the Incenter Lemma, Cp is the incenter of △ACd B.
Similarly it is shown that Ap and Bp are the incenters of △BAd C and △CBd A, respectively,
where Ad is the intersection of BCp and CBp , while Bd is the intersection of CAp and ACp . Thus
∠Cp AB = ∠Cp ABp = ∠CABp and so ABp , ACp are trisectors of ∠A.
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Also the choice of angles in the construction of △Bp ACp implies ∠Cp ABp = α. Hence
∠A = 3α. Likewise infer that BCp , BAp are trisectors of ∠B with ∠B = 3β and CAp , CBp are
trisectors of ∠C with ∠C = 3γ.
Corollary 1.
a) The angles between the trisectors of △ABC and the sides of its inner
Morley equilateral △Ap Bp Cp are: ∠Ap Bp C = ∠Ap Cp B = α+ , ∠Bp Cp A = ∠Bp Ap C = β+ ,
∠Cp Ap B = ∠Cp Bp A = γ+ .
b) The heights of the equilateral △Ap Bp Cp are: Ap Ad , Bp Bd and Cp Cd .
Theorem 2. In any triangle the exterior trisectors of its angles, proximal to the sides, meet at
the vertices of an equilateral.
Proof.
Analysis: Let △ABC be a triangle with ∠A = 3α, ∠B = 3β and ∠C = 3γ, where α + β + γ =
p
p
60o . Let Ap
A , BB and CC be the intersections of the exterior trisectors proximal to the sides BC,
p p
CA and AB respectively. Suppose △Ap
A BB CC is equilateral. The aim of this step is to calculate
p p
the angles formed by the sides of △Ap
A BB CC and the exterior trisectors of △ABC.
Fig.7a (γ < 30o )
Fig.7b (γ > 30o )
Let P and Q be the orthogonal projections of
Cp
C
on
ABp
B
Fig.7c (γ = 30o )
and
BAp
A
respectively.
p
Notice that ABp
B and BAA may intersect each other or be parallel since
∠PAB + ∠QBA = 2(β + γ) + 2(γ + α) = 120o + 2γ.
p
d
If ABp
B and BAA intersect each other let CC be their intersection. Next consider all possible
cases.
◃ If γ < 30o then
p
p
d
and Cp
C are at the same side of AB. In △ACC B, ACC and BCC are interior angle bisectors
p
o
and so CC is the incenter, while it is calculated ∠ACd
C B = 60 − 2γ. Fig.7a.
Cd
C
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Spiridon A. Kuruklis
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16, 2 (2014)
◃ If γ > 30o then
p
p
d
and Cp
C are on different sides of AB. In △ACC B, ACC and BCC are exterior angle bisectors
d
d
o
and so Cp
C is the excenter relative to ∠CC , while it is calculated ∠ACC B = 2γ − 60 . Fig.7b.
Cd
C
p
p
p
p
p
Hence in both the above cases (γ ̸= 30o ) it holds Cp
C P = CC Q. Thus △CC PBB = △CC QAA ,
p p
p p
as right triangles having two pairs of sides equal. Consequently ∠CC BB P = ∠CC AA Q and so
d p
△Ap
A CC BB is isosceles. Thus:
⋄ If γ > 30o then
p
p p
p d p
1
1
o
o
o
o
o
∠Cp
C BB P = ∠CC AA Q = 2 [180 − ∠AA CC BB ] − 60 = 2 [180 − (60 − 2γ)] − 60 = γ.
⋄ If γ < 30o then
p
p p
p d p
1
o
o
o
o
o1
o
o
∠Cp
C BB P = ∠CC AA Q = 180 − 60 − 2 [180 − ∠AA CC BB ] = 180 − 60 2 [180 − (2γ − 60 )] = γ.
p
p
p
p
Deduce that for γ ̸= 30o it holds ∠CC BB A = ∠CC AA B = γ.
p
◃ If γ = 30o then α + β = 30o and ABp
B //BAA . Fig.7c.
p
o
o
o
o
Notice ∠ACp
C B = 180 − (30 + β) − (30 + α) and so ∠ACC B = 90 .
p
p
Let M be the midpoint of AB. Since △ACC B is right triangle, CC M = MA = MB. Then Cp
CM =
p
p
p
p
MA gives ∠CC AM = ∠MCC A. But ACC is the ∠PAB bisector and thus ∠CC AM = ∠Cp
C AP.
p
p
p
p
Hence ∠MCp
A
=
∠C
AP
and
so
B
A//C
M//A
B.
C
C
B
C
A
p p
p
p
p p p
M
bisects
A
B
and
so
C
M
⊥ Ap
Since MA = MB, Cp
C
A B
C
A BB , as △AA BB CC is equilateral.
p
p
p p
p p
p p
o
o
Then ABB , BAA ⊥ AA BB . Therefore ∠ABB AA = 90 and given that ∠Cp
C BB AA = 60 infer
p
p p
o
o
∠Cp
C BB A = 30 . Similarly infer ∠CC AA B = 30 .
p p
p
o
Deduce that for γ = 30 it holds ∠CC BB A = ∠Cp
C AA B = γ.
p
p p
◦ Conclude that for any value of γ it holds ∠Cp
C BB A = ∠CC AA B = γ.
p
p
p
p p
p p
Then from △Cp
C ABB and △CC BAA deduce ∠BB CC A = β and ∠AA CC B = α respectively.
p
p p
p p
Also from △BAp
A C and △CBB A infer ∠BB AA C = β and ∠AA BB C = α.
Synthesis: Let a triangle be given in which its angles are equal to 3α, 3β and 3γ respectively,
p p
where α + β + γ = 60o . Then around an equilateral, which is denoted by △Ap
A BB CC , will
p
p
p
construct a △ABC with angles 3α, 3β and 3γ so that AA , BB and CC will be the meeting points
p
p
p
of the exterior angle trisectors proximal to the sides of △ABC. On the side Bp
B CC erect △BB ACC
p
p
with adjacent angles γ and β which were calculated in the Analysis step. Similarly, erect △CC BAA
p
p p
p p
and △Ap
A CBB on the sides CC AA and AA BB with corresponding angles as they are depicted in
Fig.8. Hence △ABC has been determined. So it remains to be proved that the resulting △ABC
has angles 3α, 3β and 3γ respectively and the erected sides are the trisectors of its exterior angles.
p
p
Let P and Q be the orthogonal projections of Cp
C on the extensions of ABB and BAA respecp
tively. Next consider all cases regarding ABp
B and BAA .
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16, 2 (2014)
Trisectors like Bisectors with equilaterals instead of Points . . .
Fig.8a (γ < 30o )
Fig.8b (γ > 30o )
81
Fig.8c (γ = 30o )
p
◃ Assume γ ̸= 30o . Set s = (30o − γ)/|30o − γ| and let Cd
C be the meeting point of ABB and
p
p
p
p
Ap
A B. The choice of angles in the erection of △BB ACC and △CC BAA implies:
p
p d p
p
CC is lying on the ∠AA CC BB bisector (1) and ∠ACC B = 90o + 21 s∠ACd
C B (2)
d p
To verify (1) notice that △Ap
A CC BB is isosceles, as two of its angles are by construction
p p
either 60o + γ (γ < 30o ) or 120o − γ (γ > 30o ). But △Ap
A BB CC is assumed equilateral and so
p
p p
p p
p d p
p
d p
Cp
C AA = CC BB . Thus CC CC bisects side AA BB of the isosceles △AA CC BB and so CC is lying on
p d p
the ∠AA CC BB bisector.
d p
d
o
o
To verify (2) notice that in the isosceles △Ap
A CC BB either ∠ACC B = 180 − 2(60 + γ) −
o
o
o
o
o
d
60o −2γ (γ < 30o ) or ∠ACd
C B=180 −2(60 +γ) = 2γ−60 (γ > 30 ). Thus ∠ACC B = s(60 −2γ).
p
1
o
o
o
o
o
o
Hence
∠ACC B = β + 60 + β = 60 + (60 − γ) = 90 + 2 s(60 − 2γ) = 90 + s∠ACd
C B.
Therefore from (1) and (2), by the Incenter Lemma (γ < 30o , s = 1) or the Excenter Lemma
d
(γ > 30o , s = −1), Cp
C is the incenter or the excenter of △ACC B respectively.
p
p
p
d
Thus ACp
C and BCC are bisectors (interior or exterior) in △ACC B. So, using △CC ABB and
p
p
p
p
p
p
△CC BAA , deduce ∠CC AB = ∠CC AP = γ + β and ∠CC BA = ∠CC BQ = γ + β. Consequently
p
p
p
∠Cp
C AB = γ + β and ∠CC BA = γ + β while ACC and BCC bisect the angles formed by AB and
p
p
the extensions of ABB and BAA respectively.
p
p p
p
p
◃ Assume γ = 30o . Then α + β = 30o . Notice ABp
B , BAA ⊥ AA BB and so AA B//BB A. Also
p
o
o
∠ACp
C B = β+60 +α = 90 and so △ACC B is right triangle. Let M be the midpoint of AB. Hence
p
p
p
p
p
CC M=MA=MA. But CC M = MA implies ∠Cp
C AM = ∠MCC A. Since ∠CC AM = ∠CC AP,
p
p
p
p
p
∠Cp
C AM = ∠ACC M. Consequently CC M//AP. Thus AA B//CC M//BB A and since MA = MA,
p
p p
p
CC M passes through the midpoint of AA BB . As a result CC M is a height of the equilateral
p p
p p
p p
o
△Ap
A BB CC and so ∠AA CC M = ∠BB CC M = 30 . Therefore
p
p
p p
p
o
∠CC AB = ∠MCC A = ∠MCC BB = ∠Bp
B CC A = 30 + β = γ + β.
o
Similarly it is shown ∠Cp
C BA = 30 + α = γ + α.
p
p
p
p
p
p
Also Ap
A B//CC M//BB A implies ∠CC AP =∠ACC M and ∠QBCC = ∠BCC M. So
p
p
p
∠Cp
C AP = ∠CC AB and ∠CC BQ = ∠CC BA.
82
Spiridon A. Kuruklis
CUBO
16, 2 (2014)
p
p
p
◦ Conclude for any γ it holds ∠Cp
C AB = γ + β and ∠CC BA = γ + α, while ACC and BCC
p
bisect the angles formed by AB and the extensions of ABp
B and BAA respectively.
p
p
The rest cases are treated similarly. Considering BCp
C and CBB it shown that ∠AA BC = α +γ
p
p
p
and ∠AA CB = α+β while BAA and CAA bisect the angles formed by BC and the extensions of BCp
C
p
p
p
and CBp
respectively,
and
eventually
considering
CA
and
AC
it
is
proved
that
∠B
CA
=
β
+
α
B
A
C
B
p
p
p
and ∠BB AC = β + γ while CBB and ABB bisect the angles formed by AC and the extensions of
p
p
ACp
C and CAA respectively. Conclude that ABB bisects the angle between AC and the extension
p
p
p
p
of ACC , while ACC bisects the angle between AB and the extension of ABp
B . Thus ABB and ACC
are trisectors of the exterior ∠A.
p
1
o
o
Also ∠Cp
C AB = ∠BB AC = γ + β. Hence ∠A = 2 [360 − 6(β + γ)] = 180 − 3(β + γ) = 3α.
p
p
Similarly it is shown that BCC , BAA are trisectors of the exterior ∠B with ∠B = 3β and CAp
A,
p
CBB are trisectors of the exterior ∠C with ∠C = 3γ.
Corollary 2. a) The angles between the exterior trisectors of △ABC and the sides of its cenp p
p p
p p
p p
p p
tral Morley equilateral △Ap
A BB CC are: ∠AA BB C = ∠AA CC B = α, ∠BB CC A = ∠BB AA C = β,
p
p p
∠Cp
C AA B = ∠CC BB A = γ.
p p
p d
p d
p d
b) The heights of the equilateral △Ap
A BB CC are: AA AA , BB BB and CC CC .
Theorem 3. In any triangle the interior trisectors of an angle and the exterior trisectors of
the other two angles, proximal the three sides respectively, meet at the vertices of an equilateral.
Proof.
Analysis: Let △ABC be a triangle with ∠A = 3α, ∠B = 3β and ∠C = 3γ, where α + β + γ = 60o .
p
Let Cp
C be the intersection of the exterior trisectors of ∠B and ∠C, proximal to AB, while aC
p
and bC are the intersections of the interior with the exterior trisectors proximal to BC and CA
p p
respectively. Suppose that △ap
C CC bC is equilateral. The aim of this step is to calculate the angles
p p p
between the sides of △aC CC bC and the interior trisectors of ∠C and also the exterior trisectors
of ∠A and ∠B.
p
p
Let P and Q be the orthogonal projections of Cp
C on AbC and BaC , respectively. It was
p
observed in the course of the Analysis Step of Theorem 2 that the trisectors Abp
C and BaC interp
sect each other iff γ ̸= 30o . Recall that if γ ̸= 30o CC is the incenter (γ < 30o ) or the excenter
p
p
o
(γ > 30o ) of △BCd
C C while for γ = 30 AbC //BaC . But it was shown that in either case it holds
p
p
p
p
p
p
CC P = CC Q and hence △CC PbC = △CC QaC , as right triangles having two pairs of sides equal.
p
p p
This implies ∠Abp
C CC = ∠BaC CC (1). Consequently:
p p d
p p d
◃ If γ =
̸ 30o then Cd
C is determined. Hence (1) implies ∠bC aC CC = ∠aC bC CC . Thus
d
is isosceles. However from △ACC B it is calculated that
d p
△ap
C CC bC
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16, 2 (2014)
Trisectors like Bisectors with equilaterals instead of Points . . .
Fig.9a (γ < 30o )
Fig.9b (γ > 30o )
83
Fig.9c (γ = 30o )
o
o
d
o
o
∠ACd
C B = 60 − 2γ (γ < 30 ) or ∠ACC B = 2γ − 60 (γ > 30 ).
Then we have respectively.
p p
⋄ For γ < 30o , Cd
C is on the other side of aC bC from A, B and
p p
p p
∠AbC aC = ∠BaC bC = 180o − 12 [180o − (60o − 2γ)] = 120o − γ.
p p
⋄ For γ > 30o , Cd
C is on the same side of aC bC with A,B and
p
p p
1
o
o
o
∠Abp
C aC = ∠BaC bC = 2 [180 − (2γ − 60 )] = 120 − γ.
p
p p
o
In either case ∠Cp
C aC B = ∠CC bC A = 60 − γ = α + β.
p
p
p
o
◃ If γ = 30o then Abp
C //BaC . Also CC P, CC Q are collinear and α + β = 30 . Thus
p p p
p p
p p
o
o
o
o
∠aC CC bC = 180 − (30 + β) − (30 + α) = 60 and so by (1) ∠aC CC Q = ∠bC CC P =
p p p
p p
p p
1
1
o
o
o
o
o
2 (180 − ∠aC CC bC ) = 2 (180 − 60 ) = 60 . Hence ∠CC aC B = ∠CC bC A = 30 = α + β.
p
p p
o
◦ In conclusion for any γ it holds ∠Abp
C CC = ∠BaC CC = 60 − γ = α + β.
p
p
p
Finally from △Bap
C C and △CbC A we find ∠BaC C = α and ∠AbC C = β respectively,
p
p
p
p p
p p
and so ∠CC aC C = β and ∠CC bC C = α. Yet from △bC ACC and △Cp
C BaC calculate that
p
p p
+
+
∠bp
C CC A = (γ + α) and ∠aC CC B = (γ + β) .
Synthesis: Suppose that a triangle is given with angles equal to 3α, 3β and 3γ, respectively,
p p
where α + β + γ = 60o . Then from an equilateral, which we denote △ap
C CC bC , will construct
a △ABC with angles 3α, 3β and 3γ so that the sides of the erected triangles are the proper
p
p
p
angle trisectors of the resulting △ABC. On the side ap
C bC erect △aC CbC with adjacent angles
p
p
p
p
+
o
+
o
β = 60 + β and α = 60 + α so that CC is inside △aC CbC . Next on the side bp
C CC erect
p
p
p
p
p
△bC ACC with adjacent angles α + β and (γ + α)+ . Finally on the side CC aC erect △ap
C BCC
with adjacent angles α + β and (γ + β)+ . Thus △ABC has been determined. See Fig.10 for the
corresponding value of γ. So it remains to be proved that the resulting △ABC has angles 3α, 3β
p
p
p
and 3γ, respectively and the erected sides Cap
C , CbC are trisectors of ∠C, while AbC , ACC , and
84
CUBO
Spiridon A. Kuruklis
16, 2 (2014)
p
BCp
C , BaC are trisectors of the exterior angles ∠A and ∠B respectively.
Fig.10a (γ < 30o )
Fig.10b (γ > 30o )
Fig.10c (γ = 30o )
p
Notice that if either γ < 30o or γ > 30o then Abp
C and BaC intersect each other, while for
p
p
o
γ = 30 AbC //BaC .
p
First we deal with the erected sides Cap
C and CbC and prove that they are trisectors of ∠C .
We also show that ∠C = 3γ. See Fig.11.
p
p
Let bd
C be the intersection of ACC and CaC .
Notice the choice of angles in the construction of
p
p
p
p p
△ap
C CbC and △aC BCC yields ∠CaC CC = β and
p d
o
+
o
= β. Hence
∠ap
C CC bC = 180 − (γ + α) − 60
p d p
d
△aC bC CC is isosceles. Thus ∠AbC C = 2β and
1
d
∠Abp
C C = 2 ∠AbC C.
d p
Since △ap
C bC CC is isosceles and from the asp p p
p
sumption △aC CC bC is equilateral, infer that bd
C bC
p
p d p
d p
bisects ap
C CC and so bC bC is the ∠aC bC CC bisector.
d
Hence bC is lying on the exterior bisector of △Abd
C C.
p
Thus, by the Excenter Lemma, bC is the excenter of
p
p
△Abd
C C relative to ∠C. But so CbC is the ∠ACaC bisector.
Fig.11
p
Similarly show that Cap
C is the ∠BCbC bisector.
p
Therefore Cbp
C and CaC are trisectors of ∠C. Also
p
p
p
p p
o
+
+
∠aC CbC = 180o − ∠Cbp
C aC − ∠CaC bC = 180 − α − β = γ.
Conclude ∠C = 3γ.
p
p
p
Next we deal with the erected sides ACp
C , AbC and BCC , BaC and prove that are trisectors
of ∠A and ∠B. We also show that ∠A = 3α and ∠B = 3β.
CUBO
16, 2 (2014)
Trisectors like Bisectors with equilaterals instead of Points . . .
85
p
◃ Assume γ ̸= 30o . Set s = (30o − γ)/|30o − γ| and let Cd
C be the meeting point of AbC and
p
p
p
p
The choice of angles in the erection of △CC AbC and △CC BaC implies:
p
1
d
d
o
Cp
C is lying on the ∠ACC B bisector (1) and ∠ACC B = 90 + 2 s∠ACC B (2)
Bap
C.
d p
o
To verify (1) notice that △ap
A CC bC is isosceles, as two of its angles are either 120 − (α + β)
p p p
o
o
o
(γ < 30 ) or 60 + (α + β) (γ > 30 ). Using the fact that △aC CC bC is equilateral, infer that
p
p p
p
p d p
Cd
C CC bisects aC bC and so CC is lying on the ∠aC CC bC bisector.
d p
To verify (2) notice that in the isosceles △ap
C CC bC
o
o
o
o
⋄ if γ < 30o then ∠ACd
C B = 180 − 2[120 − (α + β)] = 60 − 2γ = s(60 − 2γ) ,
o
d
o
o
o
o
⋄ if γ > 30 then ∠ACC B = 180 − 2[60 + (α + β)] = 2γ − 60 − s(60 − 2γ).
So for the cases γ < 30o and γ > 30o have respectively:
o
⋄ ∠ACd
C B = s(60 − 2γ) and
1
1
o
+
+
o
o
o
o
d
⋄ ∠ACp
C B = 360 − (α + γ) − (β + γ) = 120 − γ = 90 + 2 (60 − 2γ) = 90 + 2 s∠ACC B.
Therefore from (1) and (2), by the Incenter Lemma (γ < 30o , s = 1) or the Excenter
d
Lemma (γ > 30o , s = −1), Cp
C is the incenter or the excenter of △ACC B respectively. Thus
p
p
p
p
p
p
∠CC AbC = ∠CC AB and ∠CC BaC = ∠CC BA.
p
p
p
p
p
p
p
Moreover from △CC AbC and △Cp
C BaC infer ∠CC AbC = γ + β and ∠CC BaC = γ + α.
p
p
p
p
p
Deduce for γ ̸= 30o it holds ∠CC AbC = ∠CC AB = γ + β and ∠CC BaC = ∠Cp
C BA = γ + α.
p
◃ Assume γ = 30o and so α + β = 30o . Then the choice of angles in construction of △Cp
C BaC
p
p p
p p
p
p
p p
o
and △Cp
C AbC implies ∠CC bC A = ∠CC aC B = α + β = 30 . Hence AbC , BaC ⊥ aC bC and thus
p
p
p
p p p
AbC //BaC . Draw from CC the height of the equilateral △aC bC CC meeting AB at M. Hence
p
p
p
p p
Cp
C M//AbC //BaC , and also MCC bisects aC bC . But so M is the midpoint of AB. Also notice
p
p
that △ACC B is right triangle as ∠ACC B = 360o − 60o − (β + 30o )+ − (α + 30o )+ = 90o . Then
p
p
p
p
p
Cp
C M = MA = MB. Now CC M = MA implies ∠CC AM = ∠MCC A. Yet CC M//AbC implies
p
p
p
p
p
p
p
p
∠CC AbC = ∠ACC M. Thus ∠CC AbC = ∠CC AM and so ACC is the ∠bC AB bisector.
p
Similarly it is shown that BCp
C is the ∠aC BA bisector.
p
Also the choice of angles in the construction of △Cp
C AbC gives
p
p
o
o
o +
∠CC AbC = 180 − 30 − (α + 30 ) = 30o + β.
p
p
o
Deduce for γ = 30o it holds ∠Cp
C AbC = ∠CC AB = 30 + β = γ + β.
p
p
p
Similarly it is shown that ∠CC BaC = ∠CC BA = γ + α.
◦ Conclude for all γ it holds
p
p
p
p
p
o
∠Cp
C AbC = ∠CC AB = 30 + β = γ + β and ∠CC BaC = ∠CC BA = γ + α.
p
o
From △Abp
C C it follows ∠bC AC = 180 − β − γ, since from the construction choice of angles
p
p
p
p
∠AbC C = β and ∠ACbC = γ, as found in the first step. But clearly ∠bp
C AC = ∠bC ACC +
∠Cp
C AB + ∠CAB = 2(γ + β) + ∠A. Then ∠A = 3α and similarly ∠B = 3β. Therefore the angles
p
p
p
p
of △ABC are 3α, 3β and 3γ. Since ∠Cp
C AbC = ∠CC AB = γ + β it follows that AbC and ACC
86
CUBO
Spiridon A. Kuruklis
16, 2 (2014)
are trisectors of ∠A in △ABC.
p
Similarly it shown that Bap
C and BCC are trisectors of ∠B in △ABC.
Corollary 3.
a) In any △ABC, the angles formed by the side’s of the exterior Morley
p p
p p
C
b
equilateral △ap
C C C relative to the ∠C and the exterior trisectors of ∠A and ∠B are: ∠aC CC B =
p
p
p
p
p
p
(γ + β)+ , ∠bA CC A = (γ + α)+ , ∠CC aC B = ∠CC bA A = α + β, while with the interior trisectors
p
p p
of ∠C are: ∠Cp
C aC C = β and ∠CC bC C = α.
p p
p d
p d
p d
b) The heights of the equilateral △ap
C CC bC are: aC aC , bC bC and CC CC .
4
4.1
Implications
Companion Equilaterals of the inner Morley equilateral
Theorem 4. The two sides’ extensions of the inner Morley equilateral meet the corresponding
inner trisectors at two points which with the two sides’ common vertex form an equilateral.
Proof. As usually △Ap Bp Cp denotes the interior Morley equilateral of △ABC. Let SA be the intersection of the
extension of side Ap Cp with the trisector CBp and let KA be
the intersection of the extension of side Ap Bp with the trisector BCp . Moreover let Ad be the intersection of the trisectors
BCp and CBp . By Corollary 1a, ∠Ap Bp C = ∠Cp Ap B = α+
and so △Bp Ad Cp is isosceles. Thus Ap Ad is bisector of both
∠Bp Ap Cp and ∠Bp Ad Cp .
Hence ∠Bp Ap Ad = ∠Cp Ap Ad and ∠Bp Ad Ap = ∠Cp Ad Ap .
Also ∠SA Ad Ap = ∠KA Ad Ap as obviously ∠SA Ad Cp =
∠KA Ad Bp and ∠SA Ad Ap = ∠SA Ad Cp + ∠Cp Ad Ap while
∠KA Ad Ap = ∠KA Ad Ap + ∠Bp Ad Ap .
Therefore △Ap SA Ad and △Ap KA Ad are equal because in
addition they have side Ap Ad in common. Consequently
△SA Ap KA is equilateral.
Fig.12
The previous equilateral is named companion equilateral relative to vertex Ap and it will be
denoted by △SA Ap KA . Obviously there are two more companion equilaterals relative to vertices
Bp and Cp , denoted by △SB Bp KB and △SC Cp KC , respectively.
Corollary 4. For the companion equilateral relative to vertex Ap , △SA Ap KA , it holds
∠BASA = ∠CAKA = |β + γ − α|.
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Trisectors like Bisectors with equilaterals instead of Points . . .
87
In fact, the points SA and KA , for α < 30o are outside △ABC, for α = 30o are on AB and AC,
respectively, and for α > 30o are inside △ABC.
Proof. Corollary 1a asserts ∠Cp Ap B = γ+ and
∠B Ap C = β+ . Thus from △KA Ap B and △KA Ap B it
is calculated ∠Cp SA Bp = α and ∠Cp KA Bp = α, respectively. Then the points Bp , Cp , SA and KA are cyclic as
Bp Cp is seen from SA and KA with the same angle. Thus
∠ASA KA = ∠ABp KA and ∠AKA SA = ∠ACp SA . From
△BCp A and △CBp A infer that ∠ACp KA = β + γ and
∠ABp SA = α + γ, respectively. Hence
∠ASA KA = β + γ and ∠AKA SA = α + γ.
p
Next notice that ∠SA ACp = 180o − ∠ASA Cp −
∠SA Cp A = 180o − (60o + β + γ) − (α + γ) = β + γ
and similarly ∠KA AC = β + γ.
However
Fig.13
∠SA ACp = ∠SA AB ± ∠BACp and
∠KA AC = ∠KA AC ± ∠CABp ,
where ± may be either + or − depending on the location of SA and KA with respect to AB and
AC respectively. Therefore ∠SA AB = ∠KA AC = |β + γ − α|.
Because α + β + γ = 60o for α < 30o the points SA and KA are outside △ABC, for α = 30o
the points SA and KA are on AB and AC, respectively and for α > 30o the points SA and KA are
inside △ABC.
4.2
Relation of the Morley equilaterals
Theorem 5. In any triangle the sides of Morley equilaterals are either collinear or parallel.
p
Proof. Corollary 2a claims ∠BCp
C AA = α. Also
p
+
Corollary 3a confirms ∠BCp
C aC = (β + γ) . Hence
p p p
p p p
p p
p
∠BB CC aC = ∠BB CC AA + ∠AA CC B + ∠BCp
C aC =
= 60o + α + (β + γ)+ = 180o .
p
p p
Thus ap
C CC is extension of BB CC .
p
p p
Similarly it is shown that bp
C CC is extension of AA CC .
p p
p p p
o
As ∠bp
C aC CC = ∠AA BB CC = 60 , it follows
p
p p
ap
C bC //AA BB .
p
p p p
p p
o
Since ∠bp
C aC C = ∠bC aC CC + ∠CC aC C = 60 + β
Fig.14
88
CUBO
Spiridon A. Kuruklis
16, 2 (2014)
and ∠Bp Ap C = β+ then
p
p p
ap
C bC //A B .
The previous result is mentioned as a fact in [8] and it might be in print elsewhere. At any
event, it inspires the next theorem.
4.3
Interrelationship between central and exterior Morley equilaterals
Theorem 6. In any triangle, the exterior trisectors of its angles, proximal to the three sides respectively, meet at the vertices of an equilateral, if and only if, the interior trisectors of an angle
and the exterior trisectors of the other two angles, proximal the three sides respectively, meet at
the vertices of an equilateral.
Proof. Let △ABC be given with ∠A = 3α, ∠B = 3β
and ∠C = 3γ, where α + β + γ = 600 .
p p
(=⇒) Assume that △Ap
A BB CC is the central Morley equilateral formed by the intersections of the exterior
trisectors, proximal to the sides of △ABC. See Fig.15a.
p
p p
Extend Ap
A CC and BB CC to meet the extensions of
p
′′
′′
′′
Ap
A B and BB A at A and B respectively. Then AB and
′′
BA are trisectors of the exterior ∠A and ∠B respectively,
as extensions of the corresponding trisectors. Thus it suf′′
fices to show that △A ′′ Cp
C B is equilateral and also that
CA ′′ and CB ′′ are trisectors of the interior ∠C.
Fig.15a
′′
First show that △A ′′ Cp
C B is equilateral.
p
′′ p p
Notice that △A ′′ Ap
A BB = △B AA BB because they
p
′′ p p
′′ p p
o
′′ p p
′′ p p
o
have Ap
A BB common, ∠A BB AA = ∠B AA BB = 60 and ∠A AA BB = ∠B BB AA = 60 + γ by
p ′′
p ′′
p p
p p
p p
′′ p
′′ p
Corollary 2a. Thus A BB = B AA . But so CC A = CC B , as CC AA = CC BB since △Ap
A BB CC
p p p
′′
o
′′ p p
is equilateral. Also ∠A ′′ Cp
C B = ∠AA CC BB = 60 and hence △A BB CC is equilateral.
Next show that CA ′′ and CB ′′ are trisectors of the interior ∠C.
p ′′ p
p
′′ p
′′ p
′′ p
From △Ap
A A BB and △AA B BB it is easily calculated that ∠AA A BB = α + β and ∠AB BB =
p
p
p
p
p
p
α + β. Hence AA , BB , A ′′ , B ′′ are cyclic. But ∠BB AA C = β and ∠AA BB C = α, by Corollary
p
p
p
o
2a. So from △Ap
A CBB , ∠AA CBB = 180 − (α + β). But so C is also on this circle. Thus
p
p
p
p
p
′′
∠CA ′′ AA = ∠AA BB C = α and ∠CB ′′ BB = ∠Bp
B AA C = β. Finally note that in △BA C and
′′
′′
′′
△CB A it holds respectively ∠A CB = (α + γ) − α = γ and ∠B CA = (β + γ) − β = γ.
Conclude CA ′′ and CB ′′ are trisectors of the interior ∠C, as ∠C = 3γ.
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89
p p
(⇐=) Assume that the exterior triangle △ap
C CC bC is
equilateral formed by the intersections of the trisectors of
the interior ∠C and the exterior ∠A and ∠B. See Fig.15b.
p
p p
Extend ap
C CC and bC CC to meet the extensions of
p
′′
′′
′′
ap
C B and bC A at A and B respectively. Then AB and
BA ′′ are trisectors of the exterior ∠A and ∠B respectively,
as extensions of the corresponding trisectors.
Fig.15b
′′
Thus it suffices to show that △A ′′ Cp
C B is equilateral
′′
′′
and that CA and CB are trisectors of the interior ∠C.
p
′′ p p
Note △A ′′ ap
C bC = △B aC bC , because they have
p p
p
p
p
o
aC bC common, ∠A ′′ BB AA = ∠B ′′ Ap
A BB = 60 and
0
′′ p p
′′ p p
∠A AA BB = ∠B BB AA = 60 + α + β by Corollary 3a. Thus A ′′ bp
C
p ′′
′′ p ′′
′′
is equilateral then Cp
C A = CC B and so △A CC B is equilateral.
Cbp
C
5
p p p
= B ′′ ap
C . Since △aC CC bC
p
p
p
p
Also from △ap
C BC and △bC CA it is calculated that ∠aC CB = ∠bC CA = γ and so CaC and
are trisectors of ∠C.
The non Equilateral Morley Triangles
For a given △ABC there are in general 64 Morley triangles, as the
trisectors of its three angles meet at many points. Among them are
the inner, the central and the exterior Morley equilaterals.
A number of authors (see for example [4] or [5]) have wondered:
Are there more Morley equilaterals for △ABC ?
This part examines all the remaining Morley triangles of △ABC
systematically and shows that none of them is equilateral.
In
used.
the
sequel
the
following
easily
proved
lemma
is
Fig.16
The Equilateral Center Lemma. The incenter of an equilateral is the unique interior point from which its sides are seen with
120o . Similarly the excenter relative to an angle is the unique exterior point from which the side
opposite to the angle is seen with 120o while the other two sides are seen with 60o .
5.1
Morley triangles by trisectors of interior angles
This section treats the non equilateral Morley triangles formed by the trisectors of the interior
angles of △ABC . The proximal to the sides trisectors meet at Ap , Bp and Cp and △Ap Bp Cp
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Spiridon A. Kuruklis
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denotes the inner Morley equilateral.
5.1.1
The Interior Morley triangle of distal vertices
The interior Morley triangle of distal vertices is
denoted by △Ad Bd Cd where Ad , Bd and Cd are
the meeting points of the distal trisectors with respect to the sides BC, CA and AB, respectively,
as shown in Fig.17. If △ABC is equilateral then
△Ad Bd Cd is equilateral as well. Thus in the following we assume that △ABC is not equilateral.
From Corollary 1b, we have that Ap Ad ,
Bp Bd and Cp Cd are the heights of the inner Morley equilateral △Ap Bp Cp .
Let M be the center of △Ap Bp Cp . Thus
∠A MBd =∠Bd MCp = ∠Cp MAd =∠Ad MBp
p
= ∠Bp MCd =∠Cd MAp = 60o
So ∠Ad MBd = ∠Bd MCd = ∠Cd MAd = 120o .
Hence the sides of △Ad Bd Cd are seen from M
with 120o .
Fig.17
Assume towards a contradiction that △Ad Bd Cd is equilateral. Then, by the Equilateral
Center Lemma, Ap Ad is a height of △Ad Bd Cd . Thus Ap Ad bisects Bd Cd . Hence Ap Ad bisects
∠Bd Ap Cd and so ∠BAp C. Since Ap is the incenter of △BAd C, Ap Ad bisects also ∠BAd C. But
so the exterior angles of △Ad Ap B and △Ad Ap C at vertex Ap are 21 ∠BAp C = 12 ∠BAd C + β and
1
1
p
d
2 ∠BA C = 2 ∠BA C + γ. Hence β = γ. Similarly it is shown that α = β. Thus △ABC is
equilateral contrary to the assumption.
Conclude that △Ad Bd Cd cannot be equilateral (if △ABC is not equilateral).
5.1.2
Interior Morley triangles with one proximal and two mix vertices
There are three interior Morley triangles with one proximal and two mix vertices denoted by
△Ap B∗ C∗ , △Bp C∗ A∗ and △Cp A∗ B∗ . We will study only △Ap B∗ C∗ as the other two are similar.
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91
Since Ap is the intersection of the proximal trisectors, B∗
must be the intersection of the remaining trisector CBp
(proximal to CA) with ACp as distal. Then C∗ is the
intersection of the left trisectors BCp (distal to AB) and
ABp (proximal). So B∗ is on ACp and C∗ is on ABp .
See Fig.18. Corollary 1a asserts ∠ACp Bp = β+ and
∠ABp Cp = γ+ . Hence ∠ACp Ap = 60o + β+ < 180o
and ∠ABp Ap = 60o + γ+ < 180o . Therefore the quadrangle ABp Ap Cp is convex and so ∠B∗ Ap C∗ is inside
∠Bp Ap Cp .
Fig.18
∗
p
∗
o
p
∗
∗
Conclude that ∠B A C < 60 and thus △A B C cannot be equilateral.
5.1.3
Interior Morley triangles with one distal and two mix vertices
There are three interior Morley triangles with
one distal and two mix vertices which are denoted by △Ad B∗ C∗ , △Bd C∗ A∗ and △Cd A∗ B∗ .
We will study only △Ad B∗ C∗ as the other two
are similar. Since Ad is the intersection of the
distal trisectors, B∗ must be the intersection of
the remaining trisectors CAp (distal to CA) and
ABp , as proximal. So C∗ is the intersection of
the left trisectors ACp and BAp (mix to AB).
First notice that if β = γ then △Ad B∗ C∗
is isosceles, because Corollary 1a with β = γ
yields △Bp Ap B∗ = △Cp Ap C∗ and so Ap B∗ =
Ap C∗ which implies △Ad Ap B∗ = △Ad Ap C∗ .
Thus if α = β = γ then △Ad B∗ C∗ is equilateral.
Fig.19
Assume △ABC is not equilateral. Then it
has two sides not equal and thus in the following we may assume γ < β. Fig.19.
Suppose, towards a contradiction, that Ad B∗ = Ad C∗ . Let Z be the symmetric point of C∗
with respect to Ap Ad . We will fist show that Z is inside △Ad Ap B∗ .
From Corollary 1b, Ap Ad is height of the equilateral △Ap Bp Cp and so Bp and CP are symmetric
with respect to Ap Ad . Consequently ∠Bp Ap Z = ∠Cp Ap C∗ and since ∠Cp Ap C∗ = γ+ infer
∠Bp Ap Z = γ+ . Since ∠Bp Ap B∗ = β+ and by assumption γ < β, deduce ∠Bp Ap Z < ∠Bp Ap B∗ .
Moreover ∠Ap Bp Z = ∠Ap Cp C∗ = α + (α + β) = α + γ while ∠Ap Bp B∗ = α + (α + γ) = α + β.
So ∠Ap Bp Z < ∠Ap Bp B∗ .
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Since Z is inside △Ad Ap B∗ then ∠Ad ZB∗ > ∠Ad Ap B∗ = ∠Ad Ap Bp + ∠Bp Ap B∗ = 30o +
β = 90o + β. However, the assumption Ad B∗ = Ad C∗ implies Ad Z = Ad B∗ and so ∠Ad ZB∗ =
∠Ad B∗ Z. Thus ∠Ad ZB∗ + ∠Ad B∗ Z > 2(90o + β) > 180o . Hence two angles of △Ad ZB∗ have
sum greater than 180o , which is a contradiction.
+
Conclude that △Ad B∗ C∗ cannot be equilateral (if △ABC is not equilateral).
5.2
Morley triangles by trisectors of exterior angles
This section treats the non equilateral Morley triangles formed by the trisectors of the exterior
angles. So throughout this section trisectors mean trisectors of the exterior angles of △ABC.
p
p
p p p
The proximal trisectors meet at the points Ap
A , BB and CC and so △AA BB CC denotes the
p
p
central Morley equilateral. Notice that the trisectors BCC and CBB are parallel iff
p
o
o
∠Bp
B CB + ∠CC BC = 180 ⇔ 2(α + β) + 2(α + γ) = 180 .
p
p
o
So for α = 30 BCC //CBB . In this case the distal trisectors with respect to BC do not intersect
p
o
d
and hence the distal to BC vertex Ad
A is not determined. Also if 30 > α then AA and AA are on
p
d
o
o
d
the same side of BC while ∠BAA C = 60 − 2α. If 30 < α then AA and AA are on different sides
o
of BC while ∠BAd
A C = 2α − 60 . See Fig.20.
5.2.1
The Central Morley triangle of distal vertices
Fig.20a (γ < 30o )
Fig.20b (γ > 30o )
d d
d
d
d
The Morley triangle of distal vertices is denoted by △Ad
A BB CC , where AA , BB and CC are
the meeting points of the distal trisectors of the exterior angles with respect to the sides BC, CA
d d
and AB, respectively. In Fig.20a and Fig.20b the different locations of △Ad
A BB CC with respect to
p p p
p d
p d
p d
△AA BB CC are illustrated. Note that AA AA , BB BB and CC CC , by Corollary 1b, are the heights
p p
of the central Morley equilateral △Ap
A BB CC and let N be their intersection.
d d
Notice that if △ABC is equilateral then △Ad
A BB CC is equilateral as well. Thus in the following
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93
d d
we assume that △ABC is not equilateral. Also suppose towards a contradiction that △Ad
A BB CC
is equilateral.
p p
d d d
◃ If △ABC is an acute triangle the equilateral △Ap
A BB CC is inside △AA BB CC . Hence
d
d
d
d
d
o
∠Ad
A NBB = ∠B NCC = ∠CC NAA = 120 .
p
◃ If △ABC is an obtuse triangle (assume α > 30o ) Ad
A and AA are on different sides of BC.
Hence
d
d
d
o
d
d
o
∠Ad
A NBB = ∠AA NCC = 60 and ∠BB NCC = 120 .
Thus, by the Equilateral Center Lemma, N is the incenter (acute) or the excenter (obtuse) of
p d
p d
d d
d d d
the assumed equilateral △Ad
A BB CC . Hence AA AA is a height of △AA BB CC and so AA AA bisects
d
d p d
Bd
B CC and ∠BB AA CC .
p
◃ In the case of the acute triangle, notice in △BAd
A C that AA is the incenter while the bisector
p
d
d
bisects ∠BAA C. As a result ∠AA BC = ∠AA CB ⇔ 2(α + γ) = 2(α + β) ⇔ γ = β.
d
Ap
A AA
p
d
d d
◃ In the case of the obtuse triangle, note that since Ap
A AA bisects BB CC it bisects ∠BAA C.
p
p
p
p
p
p
d
d
Also it bisects ∠BAd
A C as a height of △AA BB CC . Then △AA BAA = △AA CAA and so AA B =
p
AA C.
Therefore α + γ = α + β and so γ = β. Similarly we show that α = β.
Deduce that △ABC is equilateral contrary to the assumption that it is not.
d d
Conclude that △Ad
A BB CC cannot be equilateral (if △ABC is not equilateral).
5.2.2
The Central Morley triangles with one proximal and two mix vertices
There are three Morley triangles of △ABC formed
by exterior trisectors with one proximal and two
p ∗ ∗
∗ ∗
mix vertices denoted by △Ap
A BB CC , △BB CC AA and
p ∗ ∗
∗ ∗
△Cp
C AA BB . We will study only △AA BB CC as the other
two are similar.
As vertex Ap
A is the intersection of the proximal to
BC trisectors, vertex B∗B is the intersection of the remaining trisector CBPB (proximal to CA) with ACPC , as
distal. Then vertex C∗C is the intersection of the left trisectors ABPB (distal to AB) and BCPC (proximal). Thus
B∗B is on CBPB while C∗C is on BCPC . Using the angle
p p
values between the sides of △Ap
A BB CC and the trisectors of △ABC given by Corollary 2a it is easily deduced
p p
that all the angles of the quadrangles B∗B Bp
B AA CC and
Fig.21
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Spiridon A. Kuruklis
16, 2 (2014)
p p
o
C∗C Cp
C AA BB are less than 180 and so they are convex.
For instance
p
p p
o
o
∠C∗C Cp
C AA = 180 − ∠BCC AA = 180 − α,
while
p
o
∠ABp
B AA = 60 + α.
p p
p ∗
p p p
∗ p p p
Since B∗B Bp
B AA CC is convex infer AA CC is inside ∠BB AA CC . Also since CC CC AA BB is
p ∗
p p p
convex infer AA BB is inside ∠BB AA CC .
p p p
p ∗ ∗
∗
o
Conclude that ∠B∗C Ap
A CC < LBB AA CC = 60 and so △AA BB CC cannot be equilateral.
5.2.3
The Central Morley triangles with one distal and two mix vertices
There are three Morley triangles formed by exterior trisectors with one distal and two mix vertices
d ∗ ∗
d ∗ ∗
d ∗ ∗
∗ ∗
which are denoted by △Ad
A BB CC , △BB CC AA and △CC AA BB . We will study only △AA BB CC as
the other two are similar.
p
p
∗
Since Ad
A is the intersection of the distal to BC trisectors BCC and CBB , BB is the intersection
p
of the remaining trisector CAA (proximal to CA) with ABPB as distal. Then C∗C is the intersection
p
of the left trisectors ACp
C (proximal to AB) and BAA (distal).
If α = 30o then Ad
A is not determined.
is
on the same side of
If 30o > α then Ad
A
p
d
BC with AA and ∠BAA C = 60o − 2α. If
p
30o < α then Ad
A and AA are on different
o
sides of BC while ∠BAd
A C = 2α − 60 .
Consider the case β = γ.
Then
p p ∗
= △CC AA CC by Corolp ∗
∗
lary 2a.
So Ap
A BB = AA CC and in
p ∗
d p ∗
Thus
turn △Ad
A AA BB = △AA AA CC .
d ∗ ∗
△AA BB CC is isosceles. So if △ABC is equi∗ ∗
lateral then △Ad
A BB CC is equilateral.
p ∗
△Bp
B AA BB
Next consider that △ABC is not equilateral and let β > γ.
Since the location of Ad
A depends on the
value of α, assume that 30o > α. Suppose,
∗ ∗
towards a contradiction that △Ad
A BB CC is
equilateral. Let W be the symmetric point
d
of C∗C with respect to Ap
A AA and let V be
d
the intersection of AA W and B∗B Bp
B . In the
Fig.22
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95
sequel we find the location of W.
p p p
p
p
d
From Corollary 2b, Ap
A AA is height of the equilateral △AA BB CC and so AA and BB are
d
symmetric with respect to Ap
A AA .
Also by Corollary 2a the angles between the trisectors of △ABC and the central Morley
equilateral are as depicted in Fig.22. Consequently,
p
d p ∗
o
o
o
d p ∗
o
o
∠Ad
A CC W = ∠AA CC CC = 180 − 60 − α − β = 60 + γ and ∠AA BB BB = 180 − 60 − α − β =
p
p ∗
o
∗ p
d p ∗
60 + β. Therefore ∠BB BB W < ∠AA BB BB and the line BB W is inside ∠CBB BB .
p
p p ∗
p
p p ∗
Furthermore ∠Bp
B AA W = ∠CC AA CC = γ and since β > γ, AA W is inside ∠BB AA BB . Thus
p ∗
W is inside △CBB BB .
p d ∗
p d
Clearly C∗C is outside △ABCp
C . Thus we may set ∠CC AA CC = ϕ. Then by symmetry ∠BB AA W =
∗
d ∗ ∗
∗ d ∗
o
ϕ. Also set ∠Ad
A WBB = ω. Since △AA BB CC is assumed equilateral then ∠CC AA BB = 60 and
d ∗
so ∠WAA BB = 2(α − ϕ).
∗
d
d ∗
d ∗
d
d ∗
∗ d
However Ad
A BB = AA W and so AA CC = AA BB implies AA W = AA BB . Thus ∠WBB AA = ω
d
∗
o
o
∗
and from △AA WBB , 2ω + 2(α − ϕ) = 180 ⇔ ω = 90 + ϕ − α. Given that W is inside △CBp
B BB
p ∗
∗
infer △CBB BB , ω > ∠Ad
A VBB .
p d
o
∗
d p ∗
∗
d
But from △Ad
A VBB we deduce ∠AA VBB = ∠AA BB BB + ∠BB AA V = 60 + β + ϕ.
Thus ω > 60o + β + ϕ =⇒ ω > 60o + β + (ω + α − 90o ) =⇒ α + β < 30o which contradicts
the assumption 30o > α.
The case α > 30o is similar and it is omitted.
∗ ∗
Conclude that △Ad
A BB CC cannot be equilateral (if △ABC is not equilateral).
5.3
Morley triangles by trisectors of one interior and two exterior angles
This section deals with the non equilateral Morley triangles formed by the trisectors of one interior
and two exterior angles. Even crude figures of these triangles indicate clearly that they are too
asymmetric to be equilaterals. Nevertheless it must be shown rigorously that they are not. We will
consider only those formed by the interior trisectors of ∠C and the exterior trisectors of ∠A and
p p
∠B, as the other two cases are similar. △ap
c CC bc denotes the exterior Morley equilateral relative
to ∠C.
p
Notice that the trisectors Abp
C and BaC are parallel iff
p
p
o
∠aC BA + ∠bC AB = 180 ⇔ 2(α + γ) + 2(β + γ) = 180o ⇔ γ = 30o .
In this case the distal trisectors with respect to AB do not intersect and hence the distal vertex
p
d
is
not determined. Also if 30o > γ then Cd
Cd
C
C and CC are on the same side of AB with ∠ACC B =
p
o
o
d
d
o
60 − 2γ. If 30 < γ then CC and CC are on different sides of AB with ∠ACC B = 2γ − 60 . See
Fig.23.
p p
p
Futhermore note that from Corollary 3a the trisectror Cap
C is inside ∠CC aC B and so CaC
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Spiridon A. Kuruklis
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p
intersects the trisector BCp
C between B and CC . Moreover Corollary 3a implies that the extension
p
p p
p
p
p
p
of ACC is inside ∠aC CC B and so CaC intersects ACp
C inside △CC BaC . In addition ACC intesects
p
p
p
Bap
C between B and aC . Similarly the trisectror CbC intersects the trisector ACC between A and
p
p
p
p
p
p
CC and the trisector BCC inside △CC AbC . Also BCC intersects AbC between A and bp
C.
5.3.1
The Morley triangle of distal vertices
p
d d
d
This is denoted by △ad
C CC bC . Vertex CC is the intersection of the distal to AB trisectors AbC
p
and BaC and is determined iff γ ̸= 30o . Vertex ad
C is the intersection of the distal to BC trisectors
p
p
p
p
BCC and CbC and hence it is inside △CC AbC . Vertex bd
C is the intersection of the distal to AC
p
p
p
trisectors Cap
and
AC
and
hence
it
is
inside
△C
Ba
.
C
C
C
C See Fig.23.
d
d
d
d d d
Thus, for ∠C ̸= 90o Cd
C is determined while aC and bC are inside ∠ACC B. But so ∠aC CC bC <
d
o
d
o
d d d
o
∠ACd
C B. However ∠ACC B = |60 − 2γ|. Hence ∠ACC B < 60 . Therefore ∠aC CC bC < 60 .
d d
Conclude △ad
C CC bC is not equilateral.
Fig.23a (γ < 30o )
5.3.2
Fig.23b (γ > 30o )
The Morley triangles with one proximal and two mix vertices
p ∗ ∗
p ∗ ∗
∗ ∗
There are three such triangles denoted by △ap
C bC CC , △bC aC CC and △CC aC bC .
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97
p
∗ ∗
a. △ap
C bC CC : Vertex aC is the intersection of
p
the proximal to BC trisectors Bap
C and CaC . Vertex
b∗C must be the intersection of the remaining interior
trisector Cbp
C (proximal to ) with the exterior trisecp
tor AbC , as distal. Hence C∗C is the intersection of
p
the left trisectors, BCp
C (proximal to AB) and AbC
(distal). See Fig.24.
p
So b∗C is on ACp
C and it is between A and CC .
Fig.24
p
Also C∗C is on Abp
C and it is between A and bC . Nop p p
p p
o
tice ∠ap
C bC CC = ∠aC CC bC = 60 while, by Corolp
p p
p p
p p
o
o
lary 3a, ∠CC bC A = α+β and ∠bC CC A = (γ+α)+ . Thus ∠ap
C CC A < 180 and ∠aC bC A < 180 .
p p p
p
p
p
p
Hence the quadrangle AbC aC CC is convex. Therefore ∠b∗C aC C∗C is inside ∠bC aC CC and so
∗
o
∠b∗C ap
C CC < 60 .
∗ ∗
Conclude that △ap
C bC CC is not equilateral.
∗ ∗
b. △bp
C aC CC : It is shown as above that it is not equilateral.
p
∗ ∗
c. △Cp
C aC bC : Vertex CC is the intersection of the proximal to AB exterior trisectors. Thus
a∗C is the intersection of the remaining exterior trisector Bap
C (proximal to BC) with the interior
∗
,
as
distal.
Then
b
is
the
intersection
of
the
left
trisectors Cap
trisector Cbp
C
C
C (distal to AC) and
p
AbC (proximal).
So a∗C and Cp
C are on the same side of BC iff
p
o
o
∠BCbp
C + ∠CBaC < 180 ⇔ 2γ + 2(α + γ) + 3β < 180 ⇔ γ < α.
p
If γ = α then a∗C is not determined as Bap
C //CbC .
Also b∗C and Cp
C are on different sides of AC iff
p
o
o
∠ACap
C + ∠CAbC < 180 ⇔ 2γ + 2(β + γ) + 3α < 180 ⇔ γ < β.
p
If γ = β then b∗C is not determined as Abp
C //CaC .
p
p
Since a∗C , b∗C and CC are outside △ABC while a∗C and b∗C are on Abp
C and BaC respectively we
deduce
a∗C and Cp
C are on the same side of AB iff γ < α
while
b∗C and Cp
C are on the same side of AB iff γ < β.
∗
The above conditions correlate the ranges of α, βCp
C and γ with the different locations of aC
∗
and bC and vise versa.
p
p
d
o
d
o
Recall that Abp
C and BaC intersect at CC iff γ ̸= 30 , with ∠ACC B = |60 − 2γ|, while CC and
o
Cd
C are on the same side of AB iff γ < 30 .
Next all the different locations of a∗C and b∗C are considered.
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Case 1: a∗C and b∗C are on the other side of AB from Cp
C.
This happens iff γ ≤ 30o (and so α < γ and β < γ ) or γ < 30o with α < γ and β < γ. Fig.25a,b.
p
p
p
o
d
If γ = 30o then Abp
C //BaC , while for γ ̸= 30 AbC and BaC meet at CC .
p
p
But so, a∗C and b∗C are on the extensions (to the other side of AB from Cp
C ) of CbC , BaC and
p
Cap
C , bC A respectively.
p p p
∗ p ∗
∗
∗
Since Cp
C is inside △aC CC bC , then ∠aC CC bC encompasses ∠aC CbC . Therefore
∗
∗
∗
∗
∗ p ∗
o
< ∠aC CbC . However ∠aC CbC = γ. Deduce ∠aC CC bC < 60 .
∗
∠a∗C Cp
C bC
Fig.25a (γ > 30o )
Fig.25b (γ < 30o , β < γ, α < γ)
Case 2 : a∗C and b∗C are on the same side of AB with Cp
C.
o
This happens iff γ < 30 , β > γ and α > γ . Fig.25c.
p
p
d
Then Cd
C and CC are on the same side of AB. Note that CaC intersects sides AB and BCC of
d
∗
△ACd
C B internally and so, by Pasch’s axiom, it intersects the third side ACC externally. Thus bC
p
d
∗
d
d
is on the extension of ACC . Similarly aC is on the extension of BCC . Since CC is inside △ACC B
∗
∗ d ∗
∗ p ∗
∗ d ∗
∗ d ∗
then ∠a∗C Cp
C bC encompasses ∠aC CC bC . Therefore ∠aC CC bC < ∠aC CC bC . But ∠aC CC bC =
o
d
o
∗ p ∗
∠ACC B = 60 − 2γ. Deduce ∠aC CC bC < 60 .
Case 3: a∗C and b∗C are on different sides of AB. Fig.25d.
This happens iff γ < 30o with β > γ and α < γ or with β < γ and α > γ.
Next consider the case γ < 30o with β > γ and α < γ.
p
p
p
p
d
d
Then Cp
C is on the same side with CC . Hence CC is inside △ACC B and also CC is inside △aC CbC .
p
d
By Pasch’s axiom on △ACd
C B , since CaC intersects sides AB and BCC at interior points, infer
p
p
∗
d
CaC intersects the third side ACd
C at an exterior point. Thus bC is on extensions of ACC and BaC
p
p
p p
∗
d
on the same side of AB with Cp
C . So CC is inside △bC CbC on the other side of aC bC from CC and
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Trisectors like Bisectors with equilaterals instead of Points . . .
Fig.25c (γ < 30o , β > γ, α > γ)
99
Fig.25d (γ < 30o , β < γ, α < γ)
p
d
∗ p
∗
∗ p
∗ p ∗
b∗C . Consequently ap
C CC intersects bC CC between bC and CC . Hence aC aC is inside △aC CC bC .
p p
∗ p ∗
∗ p ∗
Consequently ap
C and B are inside ∠aC CC bC . Therefore ∠aC CC bC encompasses ∠aC CC B and so
p
p
p
∠a∗C CC b∗C > ∠aC CC B.
p
+
∗ p ∗
o
However Corollary 3a asserts ∠ap
C CC B = (β + γ) . Deduce ∠bC CC aC > 60 .
The case γ < 30o with β > γ and α < γ is similar and it is omitted.
∗ ∗
Conclude that △Cp
C aC bC is not equilateral.
5.3.3
The Morley triangles with one distal and two mix vertices
∗ ∗
d ∗ ∗
d ∗ ∗
These Morley triangles are denoted by △Cd
C aC bC , △bC CC aC and △aC cC BC .
∗ ∗
d
a. △Cd
C aC bC : Vertex CC is the intersection of the distal to AB trisectors
p
∗
Abp
C and BaC . Thus vertex aC is determined by the intersection of the remaining trisector CPC , distal to BC, with Cap
C,
∗
as proximal. Vertex bC , is determined by
the left trisectors ACPC and Cbp
C which are
distal and proximal to CA, respectively.
o
Vertex Cd
C is determined iff γ ̸= 30
P
d
with CC and CC to be on the same side
of AB iff γ > 30o .
Fig.26
Moreover a∗C is always located between B and CPC while b∗C is always located between A and CPC . Fig.39 depicts the case for CPC and Cd
C to be on the same side of AB.
d
∗
∗
∗ d ∗
d
Regardless the location of CC , vertices aC and bC are inside ∠ACd
C B. Thus ∠aC CC bC < ∠ACC B.
d
o
o
∗ d ∗
o
Since ∠ACC B = |60 − 2γ| < 60 then ∠aC CC bC < 60 .
100
Spiridon A. Kuruklis
CUBO
16, 2 (2014)
∗ ∗
Conclude that △Cd
C aC bC is not equilateral.
p
p
∗ ∗
d
b. △bd
C CC aC : Vertex bC is the intersection of the distal to CA trisectors CaC and ACC .
p
Hence vertex C∗C is determined by the intersection of the remaining trisector bC , distal to AB,
with BCp
C , as proximal.
p
Vertex a∗C is determined by the intersection of the left trisectors Cbp
C and BaC which are
distal and proximal to BC, respectively.
p
d
Trisectors Cap
C and ACC always intersect each other and so bC is located on the same side of AB
p
p
p
with CC . Also AbC and BCC always intersect each other and so C∗C is located on the same side of
p
p
p
p
p
∗
o
AB with Cp
C . However aC is not always determined as CbC //BaC iff ∠bC CB + ∠bC CaC = 180
∗
d
∗
⇔ γ = α. In fact aC is on the same side with bC and CC iff α = γ . It should also be noted that
30o < γ implies α < γ. See Fig.27a.
∗ ∗
d ∗ ∗
o
For establishing that △bd
C aC CC is not equilateral we will show that ∠bC aC CC < 60 . Recall
p
p
d
that bC is inside △aC BCC .
∗ ∗
d ∗ ∗
o
Thus ∠Ba∗C C∗C encompasses ∠bd
C aC CC . Hence for proving ∠bC aC CC < 60 it suffices to
∗ ∗
o
show ∠BaC CC < 60 .
d ∗ ∗
∗ d
∗
∗
Notice that ∠Ba∗C C∗C = ∠Ba∗C ad
C + ∠aC aC CC and ∠BaC aC = ∠BaC C. From △BaC C it is
∗
∗
∗ ∗
d ∗ ∗
calculated ∠BaC C = |γ − α| regardless the location of aC . Hence ∠BaC CC = |γ − α| + ∠aC aC CC .
Moreover:
Fig.27a (30o < γ, α < γ)
Fig.27b (α < γ < 30o )
∗
◃ If α < γ then a∗C is on the other side of AB from ad
C and CC . See Fig.27a,b. Thus
d
d ∗ ∗
d ∗ ∗
d
∠BaC C is exterior angle in △aC aC CC . Hence ∠aC aC CC < ∠BaC C. In △Bad
C C it is calculated
d
d ∗ ∗
∗ ∗
a
C
=
2α
and
so
∠a
a
C
<
2α.
Infer
∠Ba
C
<
(γ
−
α)
+
2α
=
γ
+
α
< 60o .
∠Cp
C C C
C C
C C
∗
◃ If γ < α then a∗C is on the same side of AB, with ad
C and CC . See Fig.27c.
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101
Fig.27c (γ < 30o , γ < α)
p ∗
p ∗ ∗
p ∗ ∗
d p ∗
Thus ∠ad
C bC CC is exterior angle in △bC aC CC . Hence ∠bC aC CC < ∠aC bC CC . But
p ∗
p
p
p
p ∗ ∗
∠ad
C bC CC = ∠CbC A. In △CbC A it is calculated ∠CbC A = β and so ∠bC aC CC < β.
Infer ∠Ba∗C C∗C < (α − γ) + β < 60o .
∗ ∗
Conclude that △bd
C aC CC is not equilateral.
∗ ∗
c. △ad
C BC cC : It is shown as above that it is not equilateral.
5.4
Morley triangles by trisectors of one exterior and two interior angles
Eventually the non equilateral Morley triangles formed by trisectors of one exterior and two interior
angles are treated. Obviously these Morley triangles have one vertex in the interior and two in
the exterior of △ABC . As previously we will consider only those formed by the trisectors of the
exterior ∠A combined with the interior trisectors of ∠B and ∠C as the other two cases are similar.
5.4.1
The Morley triangle of proximal vertices
p
p
This is denoted by △Ap bp
A cA . Vertex A is
the intersection of the proximal to BC interior
p
trisectors, vertex bp
A is the intersection of CB
with the exterior trisector of ∠A proximal to AC,
p
while vertex cp
A is the intersection of BC with
the exterior trisector of ∠A proximal to AB.
Fig.28
102
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Spiridon A. Kuruklis
16, 2 (2014)
Consider the companion equilateral △SA Ap KA relative to vertex A. In Fig.28 the case α < 30o
is depicted for which Corollary 4 asserts that ∠BASA = ∠CAKA = β + γ − α while SA and KA
d p
are outside of △ABC . Consider the intersections of line Ap Cp with the sides of △bp
A A cA . Then
p p
d p
p
d p
A C intersects side A cA at C and so externally, while it intersects side A bA at SA internally
p p
since ∠BAbp
A = 2(β + γ) and ∠BASA = β + γ − α. Thus, by Pasch’s axiom, A C intersects
p p
p p
p p
p p
the third side bA cA internally. Similarly line A B intersects bA cA internally. Thus ∠bp
A A cA
p p
encompasses ∠SA Ap KA and so ∠SA Ap KA < Lbp
A A cA .
p p
o
o
p p p
But ∠SA Ap KA = 60o and so ∠bp
A A cA > 60 . Therefore for α < 30 the △A bA cA is not
equilateral. The cases α > 30o and α = 30o are similar.
p
Conclude that △Ap bp
A cA cannot be equilateral.
p
◃ Note that the non equilateral △Ap bp
A cA fails the original statement of Morley’s theorem.
5.4.2
The Morley triangle of distal vertices
d
d
This is denoted by △Ad bd
A cA . Vertex bA is the
p
intersection of CA , the distal to CA trisector of the
interior ∠C, and the distal to CA trisector of the
p
exterior ∠A. Vertex cd
A is the intersection of BA ,
the remaining trisector of ∠B (distal to AB) with the
distal to AB trisector of the exterior ∠A. Also it is
d
easily seen that bd
A and cA are determined iff β ̸= γ.
d
d
Hence, for β ̸= γ, A is inside ∠Acd
A B and ∠CbA A.
d
From △CbA A it is calculated that
o
∠Cbd
A A = 180 −3α−(β+γ)−2γ = 2β and similarly
d
o
from △AcA B, ∠Acd
A B = 2γ. Since α + β + γ = 60 ,
o
at least one of β and γ is less than 30 . Thus either
d d
d d d
o
∠bd
A cA A or ∠cA bA A is less than 60 .
d d d
Conclude that △A bA cA is not equilateral.
Fig.29
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5.4.3
Trisectors like Bisectors with equilaterals instead of Points . . .
103
The Morley triangles with a proximal and two mix vertices
p ∗ ∗
∗ ∗
These triangles are denoted by △Ap b∗A c∗A , △bp
A A cA and △cA A bA .
a. △Ap b∗A c∗A : Vertex Ap is the intersection of the proximal to BC interior trisectors. Hence b∗A and c∗A are the intersections
of the two remaining interior trisectors, CBp
and BCp , with the trisectors of the exterior
∠A. Since each of these interior trisectors is
proximal to the side it belongs, it must be
paired with the distal to the corresponding
side exterior trisector.
Consider the companion equilateral
△SA Ap KA relative to vertex Ap . In Fig.30
the case α > 30o is depicted for which Corollary 4 asserts that vertices SA and KA are
inside △ABC.
Fig.30
d p
p p
Consider the intersections of line Ap Cp with the sides of △bp
A A cA . A C intersects side
p
p p
d p
at C and so externally, while A C intersects side A bA at SA and so internally. Thus,
by Pasch’s axiom, Ap Cp intersects the third side Ab∗A internally. Similarly Ap Bp intersects
p p p
p p
p
p
Ac∗A internally. Hence ∠bp
A A cA encompasses ∠SA A KA and so ∠SA A KA < ∠bA A cA . But
∠SA Ap KA = 60o and so ∠b∗A Ap c∗A > 60o . Therefore for α > 30o △Ap b∗A c∗A is not equilateral.
Ad cp
A
The cases α < 30o and α = 30o are similar and they are omitted.
Conclude that △Ap b∗A c∗A is not equilateral.
p
∗ ∗
p
b. △bp
A A cA : Vertex bA is the intersection of the proximal to AC trisectors, which are CB
∗
and the corresponding trisector of the exterior ∠A. Thus A is the intersection of the remaining
interior trisector of ∠C, CAp , which is proximal to BC, with BCp as distal to BC. Then c∗A is the
intersection of the left trisectors BAp , which is distal to AB, with the proximal to AB trisector of
the exterior ∠A.
Notice that the last two trisectors are parallel iff 2β = β + γ ⇔ β = γ. Thus c∗A exists iff
β ̸= γ.
p
∗
∗
Also if β > γ then bp
A and cA are on the same side of AC, while for β < γ, bA and cA are on
different sides of AC.
∗
∗
o
Case β > γ: We will show ∠bp
A cA A < 60 .
104
Spiridon A. Kuruklis
CUBO
16, 2 (2014)
Notice that A∗ is inside △ABAp and so c∗A Ap is a right bound for the right side c∗A A∗ of
∗
∗
∠bp
A cA A .
∗
∗
In following we will find a left bound for the left side c∗A b∗A of ∠bp
A cA A .
p ∗
∗
Let b∗A be the intersection of CAp with Abp
A and note that the points A, A , cA , and bA are
p
∗
p ∗
∗ ∗
cyclic, because from △BA C it follows that bA A cA = β + γ and so bA cA is seen from A and Ap
with angle β + γ.
The extension of Ap Cp meets the exterior trisector Ab∗A between A and b∗A . Then it crosses
∗
∗
the circle, say at T . We will show that a left bound for side c∗A b∗A of ∠bp
A cA A is the bisector of
∗ ∗
∠bA cA A.
Note that in a triangle the bisector of an angle crosses its opposite side at a point which is
between the side’s middle point and the side’s common vertex with the shortest of the other two
sides.
Let G be the intersection of the ∠b∗A c∗A A bisector with Ab∗A . Also let M be the middle of
Ab∗A .
◃ Cbp
A
is angle bisector in △ACb∗A . It is easily calculated from △ACb∗A that ∠Ab∗A C = β + γ and
so ∠Ab∗A C < ∠b∗A AC. Thus CA < Cb∗A . Hence bp
A is between A and M.
◃ c∗A G is angle bisector in △Ab∗A c∗A . Obviously ∠b∗A Ac∗A = β+γ. Also ∠c∗A b∗A A = ∠c∗A b∗A C+
∠Cb∗A A while ∠c∗A b∗A A = ∠c∗A b∗A C + ∠Cb∗A A. But ∠c∗A b∗A C = ∠c∗A AAp = ∠c∗A AB + ∠BAAp =
(β+γ)+∠BAAp . So ∠b∗A Ac∗A < ∠c∗A b∗A A and thus b∗A c∗A < b∗A A. Hence G is between b∗A and M.
p ∗
∗
∗
p
∗
Therefore bp
A is on AbA and it is between A and G. So ∠GcA A encompasses ∠bA cA A and thus
∗
∗
∗
p
∠bp
.
A cA A < ∠GcA A
In the sequel we calculate ∠Gc∗A Ap . Notice that ∠Gc∗A Ap = ∠Gc∗A A + ∠Ac∗A Ap whereas
= β − γ and ∠Gc∗A A = 21 ∠b∗A c∗A A = 12 ∠b∗A Ap A = 12 [∠b∗A Ap T + ∠TAp A].
∠Ac∗A Ap
But
∠b∗A Ap T = ∠BAp Cp − ∠BAp A∗ = γ+ − (β + γ) = α + γ
.
Also
∠TAb∗A = ∠TAp b∗A = α + γ and ∠Ap TA = ∠Ac∗A Ap = β − γ.
Then from △TAAp it is calculated
.
∠TAp A = α + γ − ∠Cp AAp
Thus
∠Gc∗A A = α + γ − 12 ∠TAp A
and so ∠Gc∗A Ap = α +β− 12 ∠Cp AAp where 0 < ∠Cp AAp < α. Hence ∠Gc∗A Ap < 60o . Therefore
∗
∗
o
∠bp
A cA A < 60 .
∗ ∗
Conclude that for β > γ △bp
A A cA is not equilateral.
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16, 2 (2014)
Trisectors like Bisectors with equilaterals instead of Points . . .
Fig.31a (β > γ)
Case β < γ: We will show that ∠bPA A∗ c∗A >
60o .
Consider the intersections of Ac∗A with BCp and
∗
CBp denoted by cp
A and bA respectively. Notice that
P ∗ ∗
∠bA A cA encompasses ∠b∗A A∗ cp
A . So it suffices to
∗
∗ p
o
show that ∠bA A cA > 60 .
Observe that b∗A , A∗ , C and cp
A are cyclic, because side b∗A A∗ is seen from cp
and
C with angle γ
A
p
p
as from △AcA B it is calculated ∠AcA B = γ. Thus
p
∗
∠b∗A A∗ cp
A = ∠bA CcA .
p
d
∗
Moreover ∠b∗A Ccp
A = ∠A CcA , since CbA
passes through Ad . But
p
p
d
∠Ad Ccp
A = ∠A CA + ∠ACcA = γ + ∠ACcA and so
p
∠b∗A Ccp
A = γ + ∠ACcA .
In adition A, Ad , C and cp
A are also cyclic since
p
d
AA is seen from C and cA with angle γ.
d p
Consequently ∠ACcp
A = ∠AA cA and so
∠b∗A Ccp
A
=γ+
Yet from △AAd B infer
∠AAd cp
A.
Fig.31b
105
106
CUBO
Spiridon A. Kuruklis
16, 2 (2014)
d
d
∠AAd cp
A = ∠ABA + ∠A AB.
p
However ∠A AB > ∠C AB and thus
d
p
∠AAd cp
A > ∠ABA + ∠C AB = β + α.
∗
∗ p
Therefore ∠bA A cA > γ + β + α = 60o .
d
∗ ∗
Conclude that for β < γ, △bp
A A cA is not equilateral.
p ∗ ∗
∗ ∗
c. △cp
A A bA : It is showed that it is not equilateral similarly as △bA A cA .
5.4.4
The Morley triangles with one distal and two mix vertices
∗ ∗
d ∗ ∗
These triangles are denoted by △Ad b∗A c∗A , △cd
A A bA and △bA A cA .
a. △Ad b∗A c∗A : Obviously Ad is the intersection
of CBp and BCp . Then b∗A is the intersection of the
remaining interior trisectors CAp (distal to AC) with
the proximal to AC exterior trisector of ∠A. Notice
these two lines are parallel iff β + γ = 2γ ⇔ β = γ.
Moreover c∗A is the intersection of the left trisectors,
the interior BCp (proximal) with the distal to AB
trisector of the exterior ∠A. Notice these lines are
parallel iff β = γ. Therefore △Ad b∗A c∗A is determined iff β ̸= γ. From △Ac∗A B and △Ab∗A C it follows ∠Ap b∗A A = ∠Ap b∗A A = |β − γ| and so b∗A and
c∗A are on the same side of AC.
We will consider only the case γ > β as the other
one is similar.
Notice that Ap is inside ∠BAd C.
∠Ad c∗A b∗A encompasses ∠Ap c∗A b∗A and so
Fig.32
Thus
∠Ad c∗A b∗A > ∠Ap c∗A b∗A .
Also notice that A, Ap , b∗A and c∗A are cyclic as AAp is seen from b∗A and c∗A with angle γ − β.
Thus ∠Ap c∗A b∗A = ∠Ap Ab∗A . But ∠Ap Ab∗A = ∠Ap AC + ∠CAb∗A = ∠Ap AC + (β + γ). Moreover
∠Ap AC > ∠Bp AC = α and so ∠Ap Ab∗A > α + (β + γ) = 60o . Therefore ∠Ad c∗A b∗A > 60o .
Conclude that △Ad b∗A c∗A is not equilateral.
∗ ∗
d
p
b. △bd
A A cA : Obviously bA is the intersection of CA with the distal to AC exterior
∗
trisector of ∠A. Then A is the intersection of the remaining trisector CBp (proximal to AB)
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107
Fig.33
with the BAp (distal). Thus c∗A is the intersection of the left trisectors, BCp and the distal to AB
exterior trisector of ∠A.
∗
d ∗ ∗
o
Notice that bd
A and cA exist iff β ̸= γ. We will show that ∠bA A cA > 60 .
p ∗ ∗
∗ ∗
d ∗ ∗
Note that Bp is inside △bd
A A cA and so ∠bA A cA encompasses ∠bA A cA . Thus it suffices
p ∗ ∗
o
to prove ∠bA A cA > 60 .
o
For this we use a symmetric argument to the proof of ∠b∗A A∗ cp
A > 60 (5.4.4.a case β < γ).
p
∗
∗
∗
∗
Let bp
A be the intersection of AcA with CA . Notice that cA , A , B, bA are cyclic as
p
∗
∗ p
∗
∗
∗ p
c∗A A∗ is seen from B and bp
A with angle β. Thus ∠cA A bA = ∠cA BbA . Moreover ∠cA A bA =
p
p
p
d
∗
d
d
d
∠A BbA , since BcA passes through A . But ∠A BbA = ∠A BA + ∠ABbA = β + ∠ABbp
A and
p
∗
∗ p
so ∠cA A bA = β + ∠ABbA .
p
d
However A, Ad , B, bp
A are cyclic as AA is seen from B and bA with angle β. Consequently
d p
∗
∗ p
d p
∠ABbp
A = ∠AA bA and so ∠cA A bA = β + ∠AA bA .
d
d
d
p
Yet from △AAd C infer ∠AAd bp
A = ∠ACA + ∠A AC. However ∠A AC > ∠B AB and
d p
d
p
∗
∗ p
o
thus ∠AA cA > ∠ACA + ∠B AB = γ + α. Therefore ∠cA A bA > β + γ + α = 60 .
∗ ∗
Conclude that △bd
A A cA is not equilateral.
∗ ∗
c. △cd
A A bA : This case is similar to the above and it is omitted.
108
6
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Spiridon A. Kuruklis
16, 2 (2014)
Analogy between Bisectors and Trisectors in a triangle
The essence of the previous work is portrayed in the following two figures illustrating the analogy
between the (well understood) structure of angle bisectors and the (under study) structure of angle
trisectors in a triangle.
The structure of angle bisectors
Fig.34a
The structrure of angle trisectors
Fig.34b
The interior angle bisectors pass through a
unique point (incenter).
The interior angle trisectors proximal to the triangle sides pass through the vertices of a unique
equilateral (inner Morley equilateral).
The bisector of an interior angle and the bisectors of the other two exterior angles pass
through a unique point (excenter).
The trisectors of an interior angle and the trisectors of the other two exterior angles proximal to
the triangle sides pass through the vertices of a
unique equilateral (exterior Morley equilateral).
The exterior bisectors pass through the vertices The exterior trisectors proximal to the triangle
of a unique triangle with orthocenter the interior sides pass through the vertices of a unique equilateral (central Morley equilateral). +
angle bisectors common point (incenter). +
+
This fact follows from the previous one
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109
This structural similarity suggests that the triangle trisectors with the proper pairing meet
at equilaterals which correspond to the triangle bisectors common points. The perception that
trisectors behave like bisectors with equilaterals instead of points invites further exploration. New
results could be inspired from the vast variety of the angle bisectors’ point-line-circle theorems
revealing more exciting analogies between the two structures.
This work owes gratitude to John Conway, Gerry Ladas, George Metakides, Stanley Tennenbaum, Thanasis Fokas, Fotis Fragos and foremost to Frank Morley. Its title is the instantaneous Ladas response to the showing of the last figure, when the project was trying to take off.
If it revealed any of the Morley triangles’ mystery hopefully it has left their charm untouched.
Received:
January 2014.
Revised:
April 2014.
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(Is it a proof?)
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16, 2 (2014)
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