Trisectors Like Bisectos With Equilaterals Intead of Points_Spiridon Kuruklis

7/24/2019 Trisectors Like Bisectos With Equilaterals Intead of Points_Spiridon Kuruklis

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CUBO A Mathematical JournalVol.16, No01, (71110). 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 ofABCthe only equilaterals are the

ones determined by the intersections of the proximal to each side ofABC 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 Paschs axiom are utilized in showing that

Morleys 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 triangulos de Morley de ABC, los unicos equilateros

son theones determinados por las intersecciones del proximal a cada lado de los trisec-

tores ABCde angulos interior, o exterior, o uno interior y dos exteriores. Se muestra

que estos estan en triangulos equilateros de facto con demostraciones uniformes. Luego,

se observa que las intersecciones de trisectores interiores con los lados de un equilatero

Morley interior forman tres triangulos equilateros. 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 angulo exterior y dos interiores.

Keywords and Phrases: Angle trisection, Morleys theorem, Morley trisector theorem, Morley

triangle, Morley interior equilateral, Morley central equilateral, Morley exterior equilateral, Paschs

axiom, Morleys magic, Morleys miracle, Morleys mystery.

2010 AMS Mathematics Subject Classification: 51M04


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72 Spiridon A. Kuruklis CUBO16, 2 (2014)

1 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 Morleys theorem.

In any triangle the trisectors of its angles, proximal to the three sides respectively, meet at the

vertices of an equilateral.

Fig.1

A Morley triangle ofABC is formed by the three

points of intersection of pairs of angle trisectors con-

nected 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 Mor-

leys theorem claims that a Morley triangle ofABC is

equilateral, if it is formed by the intersections of trisectors

proximal to the three sides ofABCrespectively.

It should be noted that Morleys 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.

According to the angle meaning, Morleys theorem

gives the following Morley equilaterals ofABC. The

intersections of the proximal trisectors of the interior angles form the interior Morley equilateral

ofABC. Also the intersections of the proximal trisectors of the exterior angles form the central

Morley equilateral ofABC. In addition the intersections of the proximal trisectors of one interiorand two exterior angles form an exterior Morley equilateral ofABC, and thus there are three

exterior Morley equilaterals ofABC. 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.

Fig.2

But so an obvious question, that several authors have raised,

begs for an answer. In aABC are there other Morley equilaterals

besides the interior, the central and the three exterior Morley equi-

laterals?

Apparently the requirement of Morleys theorem is satisfied by

three more Morley triangles formed by combinations of proximal tri-

sectors of an exterior and two interior angles. One of them is por-trayed in Fig.2. Some experimentation using computer generated

graphs for these triangles has tempted the belief that Morleys theo-


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Trisectors like Bisectors with equilaterals instead of Points . . . 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 bisectorsand the angle trisectors in a triangle. Namely, the structure of trisectors resembles the structure

of bisectors with the inner and the exterior Morley equilaterals ofABC corresponding to the

incenter and the excenters ofABCrespectively, while the central Morley equilateral corresponds

to the triangle with vertices the excenters ofABC.

Morleys 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 angletrisectors 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 didnt 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 Morleys theorem has treated only the trisectors of the

interior angles and gave proofs for the interior Morley equilateral. In the preface of the firstpublication on the subject, by Taylor and Marr [12], it is recognized that the Morleys work

on vector analysis, from which the above theorem follows, holds for both interior and exterior

trisectors. The papers treatment of the theorem with only the interior trisectors is explained

as Morleys 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 Morleys

theorem from many more available. Notably, Roger Penrose [9] used a tiling technique, Edsger

Dijkstraapplied the rule of sines three times and then the monotonicity of the functiony = 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], whileRichard Guyproved that it is aconsequence 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 Morleys 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 Morleys theorem is encountered asindirect, backwards

or reverse construction methodand fits in the following scheme.

Given a triangle assume that its angles are trisected and equal to 3, 3and3, respectively,

where+ += 60o. In order to show that one of its Morley triangles is equilateral, start with

an equilateralA B C and construct aABCwith angles 3, 3 and 3, so thatA B C is

the appropriate (interior, central or exterior) Morley triangle ofABC. ThusABC would be

similar to the given triangle and so would be their corresponding Morley triangles.

Proofs of the above method most often construct ABCby erecting B AC , C BA and

A CB with proper choice of the angles formed on the sides ofA 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 rea-

soning 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 mira-

cle for referring to Morleys 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 tri-

angles are equilaterals use the classicalAnalysis 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 partfollows 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, proxi-

mal to the three sides respectively, meet at the vertices of an equilateral, if and only if, the interiortrisectors 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 Morleys observation a natural theoretical setting is

Euclidean geometry.

2 Notation and Counting of all Morley triangles

In a Morley triangle ofABC each vertex is the intersection of two trisectors, each of which is

either proximal or distal to a side ofABC. 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.

Fig.3

So we may denote a proximal, dis-

tal or mix vertex with respect to a side

by using as superscripts p, d or * to

the letter of the corresponding angle

ofABCopposite to the side.Thus Ap, Ad and A denote the

proximal, distal and mix vertex of a

Morley triangle with respect toBC re-

spectively. In Fig.3 the notations for

all intersections of the interior trisec-

tors of ABC are showed. Notice

that a Morley triangle may have either

proximal vertices, or distal vertices, or

exactly two mix vertices.

Specifically, ApBpCp denotes

the inner Morley triangle of proximal

vertices, which is the inner Morley tri-

angle determined by the intersections

of proximal to each side trisectors. Also there is just one Morley triangle with distal vertices which

is denoted by AdBdCd. In addition there are three Morley triangles with one vertex proximal

and two vertices mix. They are denoted by ApBC, BpCA and CpAB

Moreover there are three more Morley triangles with one vertex distal and two vertices mix.

They are denoted by AdBC, BdCA and CdAB. 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 thesecond and then for the third vertex. Hence there are 8 interior Morley triangles formed by the

trisectors of the interior angles ofABC.

Similarly the trisectors of the exterior angles ofABC form Morley triangles. These are


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denoted by ApABpBC

pC , for the Morley triangle of proximal vertices, A

dAB

dBC

dC, for the Morley

triangle of distal vertices, ApAB

BC

C, BpBC

CA

A and CpCA

AB

B, for the Morley triangles with

one proximal and two mix vertices, AdAB

BC

C, BdBC

CA

A andCdCA

AB

B for the Morley trian-

gles 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 theexterior trisectors. Hence, in general, there are 8 Morley triangles formed by the trisectors of the

exterior angles ofABC. Their vertices are in the exterior ofABC and due to their rather

central location with respect to ABCare calledcentral 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 ofABC.

The Morley triangles formed by combining the trisectors of the interior Awith the trisectors

of the exterior Band Care denoted byApAbpAc

pA, for the Morley triangle of proximal vertices,

AdAb

dAc

dA, for the Morley triangle of distal vertices, C

p

Ca

Cb

C, ap

Cb

CC

C and bp

CC

Ca

C,for the Morley triangles with one proximal and two mix vertices, and AdAb

Ac

A, bdAc

AA

A,

cdAA

Ab

A 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 ofABC. The vertices of

these 8 triangles formed by the trisectors of the interior A with the trisectors of the exterior B

and Care in the exterior ofABCand 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 ofABC.

The Morley triangles formed by combining the trisectors of the interior B and C with

the trisectors of the exterior A are denoted by ApbpAcpA, for the Morley triangle of proximal

vertices,AdbdAcdA, for the Morley triangle of distal vertices, A

pbAc

A, bpAc

AA, cpAA

bA,

for the Morley triangles of one proximal vertex and two mix, AdbAc

A, bdAc

AA, cdAA

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 ofABC determined by the intersections of trisectors ofone exterior and two interior angles ofABC.

Conclude that in total there are, in general, 64 Morley triangles ofABC.


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3 Uniform Proofs for all Morley Equilaterals

Fig.4

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 excenterofABCusing 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 throughA, B and I is depicted. Obviously

AIB=1800 12ABC= 12BAC= 90o + 12ACB.

ThusI may be characterized as the intersection in the

interior ofABCof a bisector with the arc of size 90o +12ACB passing through A and B. Clearly an analogous

result holds for the other two pairs of vertices ofABC.

We refer to this as the Incenter Lemma.

IfIC is the excenter relative to Cthen

AICB= 90o

12ACB, BICC=

12BACand CICA=

12CBA.

Thus IC is determined by the intersection in the exterior ofABCof a bisector, either of the

interiorCor the exterior Aor Bwith the arc of size90o 12ACBpassing through A and B,

or with the arc passing through B and C of size 12BAC, or with the arc of size 12CBApassing

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 atthe vertices of an equilateral.

Proof.

Analysis: Let ABC be a triangle with A =3, B= 3 and C= 3, where ++= 60o.Suppose that ApBpCp is equilateral, whereAp,Bp and Cp are the intersections of the trisectorsproximal to the sides BC, CA and AB respec-tively. The aim of this step is to calculate theangles formed by the sides ofApBpCp and the

trisectors ofABC. See Fig.5.Let Cd be the intersection ofABp and BAp. SinceACp andBCp are angle bisectors inACdB, Cp

is the incenter.

Fig.5


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LetP and Q be the orthogonal projections ofCp on ACd andBCd respectively. ThusCpP=

CpQ and CdP = CdQ. But so CpPBp = CpQBp as right triangles having two pairs of

sides equal. HenceBpP = ApQ. ThenCdAp = CdBp and so ApCdBpis isosceles. Now from

ACdB we have ACdB = 180o (2+2) = 60o +2.Therefore CdBpAp = CdApBp =

12 [180o

(60o + 2)] = 60o. Consequently

CpBpA= CpApB= 180o 60o (60o ) =60o += +.

LetAd be the intersection ofBCp andCBp. Also letBd be the intersection ofCAp andACp.

Then from BAdC and CBdA find similarly

ApCpB= ApBpC= + and BpApC= BpCpA= +.

Fig.6

Synthesis: Suppose that a triangle is

given and assume that its angles are trisected

and equal to 3, 3 and 3, respectively,

where ++= 60o. Then around an equi-

lateral ApBpCp will constructABCwithangles 3, 3 and 3 so that Ap, Bp and Cp

will be the intersections of the proximal to the

sides interior trisectors.

On the side BpCp erect BpACp with

adjacent angles + = +60o and + = +

60o.

Similarly, erect CpBAp andApCBp on

the sides CpAp and ApBp respectively with

corresponding angles as shown in Fig.6, which

were found in the Analysis step.Let Cd be the intersection ABp and BAp.

Notice that ApCdBp is isosceles as two of

its angles are180o 60o + =60o .

Thus

CdAp =CdBp (1) and ApCdBp =180o 2(60o ) =60o + 2 (2)

Since ApBpCp has been taken equilateral, CpAp =CpBp. Combine this with (1) and infer

that Cp is on the ApBp bisector and so on the ACdB bisector. Moreover from (2) ACpB =

3600(++60o++)=180o(+) = 90o+ 12

(60o+2)=90o+ 12ApCdBp =90o+ 1

2ACdB.

Hence, by the Incenter Lemma, Cp is the incenter ofACdB.

Similarly it is shown that Ap andBp are the incenters ofBAdCand CBdA, respectively,whereAd is the intersection ofBCp andCBp, whileBd is the intersection ofCAp andACp. Thus

CpAB= CpABp = CABp and so ABp, ACp are trisectors ofA.


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Also the choice of angles in the construction ofBpACp implies CpABp = . Hence

A = 3. Likewise infer thatBCp, BAp are trisectors ofB with B = 3 and CAp, CBp are

trisectors ofC with C= 3.

Corollary 1. a) The angles between the trisectors ofABC and the sides of its inner

Morley equilateralApBpCp are: ApBpC = ApCpB = +, BpCpA = BpApC = +,

CpApB= CpBpA= +.

b) The heights of the equilateralApBpCp are: ApAd, BpBd andCpCd .

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 ABCbe a triangle with A= 3, B= 3and C= 3, where ++=60o. Let ApA, B

pB and C

pC be the intersections of the exterior trisectors proximal to the sides BC,

CA and AB respectively. Suppose ApABpBC

pC is equilateral. The aim of this step is to calculate

the angles formed by the sides ofApABpBC

pC and the exterior trisectors ofABC.

Fig.7a (< 30o) Fig.7b (> 30o) Fig.7c (= 30o)

Let P andQ be the orthogonal projections ofCpC on ABpB and BA

pA respectively.

Notice that ABpB and BApA may intersect each other or be parallel since

PAB + QBA= 2( +) +2( + ) =120o + 2.

IfABpB and BApA intersect each other let C

dC be their intersection. Next consider all possible

cases.

If< 30o thenCdC and C

pC are at the same side ofAB. In AC

dCB, AC

pC and BC

pC are interior angle bisectors

and soCpC is the incenter, while it is calculated ACdCB= 60

o 2. Fig.7a.


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If> 30o then

CdC and CpC are on different sides ofAB. In AC

dCB, AC

pC and BC

pC are exterior angle bisectors

and so CpC is the excenter relative to CdC, while it is calculated AC

dCB= 2 60

o. Fig.7b.

Hence in both the above cases ( =30o)it holds CpCP = CpCQ. Thus C

pCPB

pB = C

pCQA

pA,

as right triangles having two pairs of sides equal. Consequently CpCBpBP = C

pCA

pAQ and so

ApACdCB

pB is isosceles. Thus:

If> 30o then

CpCBpBP= C

pCA

pAQ=

12 [180

oA

pAC

dCB

pB] 60

o = 12 [180o (60o 2)] 60o =.

If< 30o then

CpCBpBP = C

pCA

pAQ= 180

o60o 1

2[180oA

pAC

dCB

pB] = 180

o60o 1

2[180o (260o)] = .

Deduce that for =30o it holds CpCBpBA= C

pCA

pAB= .

If =30o then + = 30o andABpB//BApA. Fig.7c.

Notice ACp

C

B= 180o (30o + ) (30o + ) and so ACp

C

B= 90o.

LetM be the midpoint ofAB. Since ACpCBis right triangle, CpCM= MA = MB. ThenC

pCM=

MA gives CpCAM = MCpCA. But AC

pC is the PAB bisector and thus C

pCAM = C

pCAP.

Hence MCpCA= CpCAP and so B

pBA//C

pCM//A

pAB.

SinceMA= MB, CpCM bisects ApAB

pB and so C

pCM A

pAB

pB, as A

pAB

pBC

pC is equilateral.

Then ABpB, BApA A

pAB

pB. Therefore AB

pBA

pA = 90

o and given that CpCBpBA

pA = 60

o infer

CpCBpBA= 30

o. Similarly infer CpCApAB= 30

o.

Deduce that for = 30o it holds CpCBpBA= C

pCA

pAB= .

Conclude that for any value ofit holds CpCBpBA= C

pCA

pAB= .

Then from CpCAB

pB and C

pCBA

pA deduce B

pBC

pCA = and A

pAC

pCB = respectively.

Also from BApAC and CBpBA infer B

pBA

pAC= and A

pAB

pBC= .

Synthesis: Let a triangle be given in which its angles are equal to 3, 3and 3 respectively,

where + + = 60o. Then around an equilateral, which is denoted by ApABpBC

pC, will

construct a ABCwith angles 3, 3 and3 so that ApA, BpB and C

pC will be the meeting points

of the exterior angle trisectors proximal to the sides ofABC. On the side BpBCpC erect B

pBAC

pC

with adjacent angles and which were calculated in the Analysis step. Similarly, erect CpCBApA

and ApACBpB on the sides C

pCA

pA and A

pAB

pB with corresponding angles as they are depicted in

Fig.8. Hence ABChas been determined. So it remains to be proved that the resulting ABC

has angles3,3and 3respectively and the erected sides are the trisectors of its exterior angles.Let P andQ be the orthogonal projections ofCpC on the extensions ofAB

pB and BA

pA respec-

tively. Next consider all cases regarding ABpB andBApA.


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Assume =30o. Set s = (30o )/|30o |and let CdC be the meeting point ofABpB and

ApAB. The choice of angles in the erection ofBpBAC

pC and C

pCBA

pA implies:

CpC is lying on the ApAC

dCB

pB bisector (1) and AC

pCB= 90

o + 12sACdCB (2)

To verify (1) notice that ApACdCB

pB is isosceles, as two of its angles are by construction

either 60o + ( < 30o) or 120o ( > 30o). But ApABpBC

pC is assumed equilateral and so

CpCApA = C

pCB

pB. ThusC

dCC

pC bisects side A

pAB

pB of the isosceles A

pAC

dCB

pB and so C

pC is lying on

the ApACdCB

pB bisector.

To verify (2) notice that in the isosceles ApACdCB

pB either AC

dCB = 180

o 2(60o +)

60o2 (< 30o)or ACdCB=180o2(60o+) =260o (> 30o). Thus ACdCB= s(60

o2).

Hence ACpCB= + 60o + = 60o + (60o ) = 90o + 1

2s(60o 2) =90o +sACdCB.

Therefore from (1) and (2), by the Incenter Lemma (< 30o, s= 1) or the Excenter Lemma

(> 30o, s= 1), CpC is the incenter or the excenter ofACdCB respectively.

Thus ACpC and BCpC are bisectors (interior or exterior) in AC

dCB. So, using C

pCAB

pB and

CpCBApA, deduce C

pCAB = C

pCAP = + and C

pCBA = C

pCBQ = + . Consequently

CpCAB= + and CpCBA= + while AC

pC and BC

pC bisect the angles formed by AB and

the extensions ofABpB andBApA respectively.

Assume = 30o . Then+= 30o. NoticeABpB , BApA A

pAB

pB and soA

pAB//B

pBA. Also

ACpCB= +60o+= 90o and so ACpCBis right triangle. LetMbe the midpoint ofAB. Hence

CpCM=MA=MA. But C

pCM = MA implies C

pCAM = MC

pCA. Since C

pCAM = C

pCAP,

CpCAM= ACpCM. Consequently C

pCM//AP. Thus A

pAB//C

pCM//B

pBAand since MA= MA,

CpCM passes through the midpoint of ApAB

pB. As a result C

pCM is a height of the equilateral

ApABpBC

pC and so A

pAC

pCM= B

pBC

pCM= 30

o. Therefore

Cp

C

AB= MCp

C

A= MCp

C

Bp

B

= Bp

B

Cp

C

A= 30o + = + .

Similarly it is shown CpCBA= 30o + = + .

Also ApAB//CpCM//B

pBA implies C

pCAP=AC

pCMand QBC

pC = BC

pCM. So

CpCAP= CpCABand C

pCBQ= C

pCBA.


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Conclude for any it holds CpCAB = + and CpCBA = + , while AC

pC and BC

pC

bisect the angles formed by AB and the extensions ofABpB and BApA respectively.

The rest cases are treated similarly. Considering BCpC and CBpB it shown that A

pABC= +

and ApACB= + while BApAand CA

pAbisect the angles formed by BC and the extensions ofBC

pC

andCBpB respectively, and eventually consideringCApA and AC

pCit is proved that B

pBCA= +

and BpBAC= + while CBpB and AB

pB bisect the angles formed by ACand the extensions of

ACpC and CApA respectively. Conclude that AB

pB bisects the angle between AC and the extension

ofACpC, whileACpC bisects the angle between AB and the extension ofAB

pB. ThusAB

pB andAC

pC

are trisectors of the exterior A.

Also CpCAB= BpBAC= + . Hence A=

12 [360

o 6( +)] =180o 3( +) =3.

Similarly it is shown that BCpC, BApA are trisectors of the exterior B with B = 3 and CA

pA,

CBpB are trisectors of the exterior Cwith C= 3.

Corollary 2. a) The angles between the exterior trisectors ofABCand the sides of its cen-

tral Morley equilateralApABpBC

pC are: A

pAB

pBC = A

pAC

pCB = , B

pBC

pCA = B

pBA

pAC = ,

CpCApAB= C

pCB

pBA= .

b) The heights of the equilateralApABpBC

pC are: A

pAA

dA, B

pBB

dB andC

pCC

dC.

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: LetABCbe a triangle with A= 3, B= 3and C= 3, where ++= 60o.

Let CpC be the intersection of the exterior trisectors of B and C, proximal to AB, while apCand bpC are the intersections of the interior with the exterior trisectors proximal to BC and CA

respectively. Suppose that apCCpCb

pC is equilateral. The aim of this step is to calculate the angles

between the sides ofapCCpCb

pC and the interior trisectors ofC and also the exterior trisectors

ofAand B.

Let P and Q be the orthogonal projections of CpC on AbpC and Ba

pC, respectively. It was

observed in the course of the Analysis Step of Theorem 2 that the trisectors AbpC and BapC inter-

sect each other iff = 30o. Recall that if = 30o CpC is the incenter ( < 30o) or the excenter

(> 30o)ofBCdCCwhile for = 30o AbpC//Ba

pC. But it was shown that in either case it holds

CpCP = CpCQ and hence C

pCPb

pC =C

pCQa

pC, as right triangles having two pairs of sides equal.

This implies

Ab

p

CC

p

C =

Ba

p

CC

p

C (1). Consequently:

If = 30o then CdC is determined. Hence (1) implies bpCa

pCC

dC = a

pCb

pCC

dC. Thus

apCC

dCb

pC is isosceles. However from AC

dCB it is calculated that


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CUBO16, 2 (2014)



ACdCB= 60o 2 (< 30o)or ACdCB= 2 60

o (> 30o).

Then we have respectively.

For < 30o, CdC is on the other side ofapCbpC from A, B andAbpCa

pC = Ba

pCb

pC = 180

o

12 [180

o (60o 2)] = 120o .

For > 30o, CdC is on the same side ofapCb

pC with A,B and

AbpCapC = Ba

pCb

pC =

12

[180o (2 60o)] = 120o .

In either case CpCapCB= C

pCb

pCA= 60

o= + .

If = 30o then AbpC//BapC. Also C

pCP, C

pCQ are collinear and + = 30

o. Thus

apCCpCb

pC = 180

o (30o + ) (30o + ) = 60o and so by (1) apCC

pCQ = b

pCC

pCP =

12

(180o apCC

pCb

pC) =

12

(180o 60o) =60o. Hence CpCapCB= C

pCb

pCA= 30

o =+ .

In conclusion for any it holds AbpCCpC = Ba

pCC

pC = 60

o= + .

Finally from BapCC and CbpCA we find Ba

pCC = and Ab

pCC = respectively,

and so CpCapCC = and C

pCb

pCC = . Yet from b

pCAC

pC and C

pCBa

pC calculate that

bpCCpCA= ( + )

+ and apCCpCB= ( + )

+.

Synthesis: Suppose that a triangle is given with angles equal to 3, 3 and 3, respectively,

where + + = 60o. Then from an equilateral, which we denote apCCpCb

pC, will construct

a ABC with angles 3, 3 and 3 so that the sides of the erected triangles are the proper

angle trisectors of the resulting ABC. On the side apCbpC erect a

pCCb

pC with adjacent angles

+ = 60o + and + = 60o + so that CpC is inside apCCb

pC. Next on the sideb

pCC

pC erect

bpCACpC with adjacent angles + and (+ )+. Finally on the side CpCapC erect apCBCpCwith adjacent angles + and ( + )+. ThusABChas been determined. See Fig.10 for the

corresponding value of. So it remains to be proved that the resulting ABChas angles 3, 3

and 3, respectively and the erected sides CapC, CbpC are trisectors ofC, while Ab

pC, AC

pC, and


14/40


BCpC, BapC are trisectors of the exterior angles Aand B respectively.


Notice that if either < 30o or > 30o then AbpC and BapC intersect each other, while for

= 30o Abp

C//Bap

C.

First we deal with the erected sides CapC andCbpC and prove that they are trisectors ofC.

We also show that C= 3. See Fig.11.

Fig.11

Let bdC be the intersection of ACpC and Ca

pC.

Notice the choice of angles in the construction of

apCCbpC and a

pCBC

pC yields Ca

pCC

pC = and

apCCpCb

dC = 180

o ( + )+ 60o = . Hence

apCb

dCC

pC is isosceles. Thus Ab

dCC = 2 and

AbpCC= 12AbdCC.

Since a

p

Cbd

CC

p

C is isosceles and from the as-sumption apCC

pCb

pC is equilateral, infer that b

dCb

pC

bisectsapCCpC and sob

dCb

pCis the a

pCb

dCC

pCbisector.

HencebdC is lying on the exterior bisector ofAbdCC.

Thus, by the Excenter Lemma, bpC is the excenter of

AbdCCrelative to C. But so CbpC is the ACa

pC bisector.

Similarly show that CapC is the BCbpC bisector.

ThereforeCbpC and CapC are trisectors ofC. Also

apCCbpC = 180

o CbpCa

pC Ca

pCb

pC = 180

o + + =.

Conclude C= 3.

Next we deal with the erected sides ACpC, AbpC and BC

pC, Ba

pC and prove that are trisectors

ofAand B. We also show that A= 3and B= 3.


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CUBO16, 2 (2014)


Assume =30o. Set s = (30o )/|30o |and let CdC be the meeting point ofAbpC and

BapC. The choice of angles in the erection ofCpCAb

pCandC

pCBa

pC implies:

CpC is lying on the ACdCB bisector (1) and AC

pCB= 90

o + 12

sACdCB (2)

To verify (1) notice that ap

A

CdCbp

C

is isosceles, as two of its angles are either 120o (+)

( < 30o) or 60o + (+ ) ( > 30o). Using the fact thatapCCpCb

pC is equilateral, infer that

CdCCpC bisects a

pCb

pC and so C

pC is lying on the a

pCC

dCb

pC bisector.

To verify (2) notice that in the isosceles apCCdCb

pC

if< 30o then ACdCB= 180o 2[120o (+ )] = 60o 2= s(60o 2) ,

if> 30o then ACdCB= 180o 2[60o + (+ )] = 2 60o s(60o 2).

So for the cases < 30o and > 30o have respectively:

ACdCB= s(60o 2)and

ACpCB= 360o (+)+ (+)+ =120o = 90o + 12 (60

o 2) =90o + 12sAC

dCB.

Therefore from (1) and (2), by the Incenter Lemma ( < 30o, s = 1) or the Excenter

Lemma ( > 30o, s = 1), CpC is the incenter or the excenter ofACdCB respectively. Thus

CpCAbpC = CpCAB and CpCBapC = CpCBA.Moreover from CpCAb

pC and C

pCBa

pC infer C

pCAb

pC = + and C

pCBa

pC = + .

Deduce for =30o it holds CpCAbpC = C

pCAB= + and C

pCBa

pC = C

pCBA= + .

Assume = 30o and so += 30o. Then the choice of angles in construction ofCpCBapC

and CpCAbpC implies C

pCb

pCA= C

pCa

pCB= + = 30

o. HenceAbpC, BapC a

pCb

pC and thus

AbpC//BapC. Draw from C

pC the height of the equilateral a

pCb

pCC

pC meeting AB at M. Hence

CpCM//AbpC//Ba

pC, and also MC

pC bisects a

pCb

pC. But soM is the midpoint ofAB. Also notice

that ACpCB is right triangle as ACpCB = 360

o 60o ( +30o)+ (+30o)+ = 90o. Then

CpCM = MA = MB. Now CpCM = MA implies C

pCAM = MC

pCA. Yet C

pCM//Ab

pC implies

CpCAbpC = AC

pCM. Thus C

pCAb

pC = C

pCAMand so AC

pC is the b

pCAB bisector.

Similarly it is shown that BCpC is the apCBAbisector.

Also the choice of angles in the construction ofCpCAbpC gives

CpCAbpC = 180

o 30o (+ 30o)+ =30o + .

Deduce for = 30o it holds CpCAbpC = C

pCAB= 30

o + = + .

Similarly it is shown that CpCBapC = C

pCBA= + .

Conclude for all it holds

CpCAbpC = C

pCAB= 30

o + = + and CpCBapC = C

pCBA= + .

From Abp

CCit follows

b

p

CAC= 180o

, since from the construction choice of anglesAbpCC = and ACb

pC = , as found in the first step. But clearly b

pCAC = b

pCAC

pC +

CpCAB+ CAB= 2( + ) + A. Then A= 3and similarly B= 3. Therefore the angles

ofABCare 3, 3 and 3. Since CpCAbpC = C

pCAB = + it follows that Ab

pC and AC

pC


16/40


are trisectors ofAin ABC.

Similarly it shown that BapC and BCpC are trisectors ofB in ABC.

Corollary 3. a) In anyABC, the angles formed by the sides of the exterior Morley

equilateralapCCpCb

pC relative to theCand the exterior trisectors ofAandBare: a

pCC

pCB=

( + )+, bpACpCA= ( + )

+, CpCapCB= C

pCb

pAA= + , while with the interior trisectors

ofC are: CpCapCC= and C

pCb

pCC= .

b) The heights of the equilateralapCCpCb

pC are: a

pCa

dC, b

pCb

dC andC

pCC

dC.

4 Implications

4.1 Companion Equilaterals of the inner Morley equilateral

Theorem 4. The two sides extensions of the inner Morley equilateral meet the correspondinginner trisectors at two points which with the two sides common vertex form an equilateral.

Fig.12

Proof. As usually ApBpCp denotes the interior Mor-

ley equilateral ofABC. Let SA be the intersection of the

extension of sideApCp with the trisector CBp and letKA be

the intersection of the extension of side ApBp with the trisec-

torBCp. Moreover letAd be the intersection of the trisectors

BCp and CBp. By Corollary 1a, ApBpC= CpApB= +

and soBpAdCp is isosceles. ThusApAd is bisector of both

BpApCp and BpAdCp.

Hence BpApAd = CpApAd and BpAdAp = CpAdAp.

Also SAAdAp = KAA

dAp as obviously SAAdCp =

KAAdBp and SAA

dAp = SAAdCp + CpAdAp while

KAAdAp = KAA

dAp + BpAdAp.

Therefore ApSAAd and ApKAA

d are equal because in

addition they have side ApAd in common. Consequently

SAApKA is equilateral.

The previous equilateral is named companion equilateral relative to vertex Ap and it will be

denoted by SAApKA. Obviously there are two more companion equilaterals relative to vertices

Bp

andCp

, denoted by SBBp

KB and SCCp

KC, respectively.

Corollary 4. For the companion equilateral relative to vertex Ap, SAApKA, it holds

BASA = CAKA = |+ |.


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CUBO16, 2 (2014)


In fact, the pointsSA andKA , for< 30o are outsideABC, for= 30o are onAB andAC,

respectively, and for> 30o are insideABC.

Fig.13

Proof. Corollary 1a asserts CpApB = + and

BpApC = +. Thus from KAApB and KAA

pB it

is calculated CpSABp =and CpKAB

p =, respec-

tively. Then the points Bp, Cp, SA and KA are cyclic as

BpCp is seen fromSA andKA with the same angle. Thus

ASAKA = ABpKA and AKASA = AC

pSA. From

BCpA and CBpA infer that ACpKA = + and

ABpSA = +, respectively. Hence

ASAKA = + and AKASA = +.

Next notice that SAACp = 180o ASAC

p

SACpA= 180o (60o + +) (+) = +

and similarly KAAC= +.However

SAACp = SAAB BAC

p and

KAAC= KAAC CABp,

where may be either + or depending on the location ofSA and KA with respect to AB and

ACrespectively. Therefore SAAB= KAAC= | + |.

Because + += 60o for < 30o the points SA andKA are outside ABC, for = 30o

the points SA andKA are onAB and AC, respectively and for > 30o the points SA andKA 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.

Fig.14

Proof. Corollary 2a claims BCpCApA = . Also

Corollary 3a confirms BCpCapC = ( +)

+. Hence

BpBCpCa

pC = B

pBC

pCA

pA+ A

pAC

pCB + BC

pCa

pC =

=60o + + ( +)+ =180o.

Thus apCCpC is extension ofB

pBC

pC.

Similarly it is shown that bpCCpC is extension ofA

pAC

pC.

As bpCapCC

pC = A

pAB

pBC

pC = 60

o, it follows

apCbpC//A

pAB

pB.

Since bpCapCC= b

pCa

pCC

pC+ C

pCa

pCC= 60

o +


18/40


and BpApC= + then

apCbpC//A

pBp.

The previous result is mentioned as a fact in [8] and it might be in print elsewhere. At anyevent, 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 re-

spectively, 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.

Fig.15a

Proof. LetABCbe given with A= 3, B= 3

and C= 3, where + += 600.

(=) Assume that ApABpBC

pC is the central Mor-

ley equilateral formed by the intersections of the exterior

trisectors, proximal to the sides ofABC. See Fig.15a.

Extend ApACpC and B

pBC

pC to meet the extensions of

ApABand B

pBAat A

andB respectively. ThenAB and

BA are trisectors of the exterior Aand Brespectively,

as extensions of the corresponding trisectors. Thus it suf-

fices to show that A CpCB is equilateral and also that

CA

andCB

are trisectors of the interior

C.First show that A CpCB

is equilateral.

Notice that A ApABpB = B

ApABpB because they

haveApABpB common, A

BpBApA = B

ApABpB =60

o and A ApABpB = B

BpBApA = 60

o + by

Corollary 2a. Thus A BpB =BApA. But so C

pCA

=CpCB

, as CpCApA = C

pCB

pB since A

pAB

pBC

pC

is equilateral. Also A CpCB = ApAC

pCB

pB = 60

o and hence A BpBCpC is equilateral.

Next show that CA and CB are trisectors of the interiorC.

From ApAABpB and A

pAB

BpB it is easily calculated that ApAA

BpB = + and ABBpB =

+ . Hence ApA, BpB, A

, B are cyclic. But BpBApAC = and A

pAB

pBC = , by Corollary

2a. So from ApACBpB, A

pACB

pB = 180

o(+ ). But so C is also on this circle. Thus

CA

Ap

A = Ap

ABp

BC = and CB

Bp

B = Bp

BAp

AC = . Finally note that inBA

C andCB Ait holds respectively A CB= (+) = and B CA= ( +) = .

ConcludeCA and CB are trisectors of the interiorC, as C= 3.


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CUBO16, 2 (2014)


Fig.15b

(=) Assume that the exterior triangle apCCpCb

pCis

equilateral formed by the intersections of the trisectors of

the interior Cand the exterior Aand B. See Fig.15b.

Extend ap

C

Cp

C

and bp

C

Cp

C

to meet the extensions of

apCBandbpC A atA

andB respectively. ThenAB and

BA are trisectors of the exterior Aand Brespectively,

as extensions of the corresponding trisectors.

Thus it suffices to show that A CpCB is equilateral

and that CA and CB are trisectors of the interior C.

Note A apCbpC = B

apCbpC, because they have

apCbpC common, A

BpBApA = B

ApABpB = 60

o and

A ApABpB = B

BpBApA = 60

0 + + by Corollary 3a. Thus A bpC =BapC. Since a

pCC

pCb

pC

is equilateral then CpCA =CpCB

and so A CpCB is equilateral.

Also from apC

BCand bpC

CAit is calculated that apC

CB= bp

CCA= and so Cap

Cand

CbpC are trisectors ofC.

5 The non Equilateral Morley Triangles

Fig.16

For a given ABCthere 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 forABC?

This part examines all the remaining Morley triangles ofABCsystematically and shows that none of them is equilateral.

In the sequel the following easily proved lemma is

used.

The Equilateral Center Lemma. The incenter of an equi-

lateral 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 with120o while the other two sides are seen with60o.

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 ofABC. The proximal to the sides trisectors meet at Ap, Bp and Cp and ApBpCp


20/40


denotes the inner Morley equilateral.

5.1.1 The Interior Morley triangle of distal vertices

Fig.17

The interior Morley triangle of distal vertices is

denoted byAdBdCd whereAd,Bd andCd are

the meeting points of the distal trisectors with re-

spect to the sides BC, CA and AB, respectively,

as shown in Fig.17. IfABCis equilateral then

AdBdCd is equilateral as well. Thus in the fol-

lowing we assume that ABCis not equilateral.

From Corollary 1b, we have that ApAd,

BpBd andCpCd are the heights of the inner Mor-

ley equilateral ApBpCp.

Let M be the center ofApBpCp. Thus

ApMBd=BdMCp= CpMAd=AdMBp

= BpMCd=CdMAp= 60o

So AdMBd = BdMCd = CdMAd = 120o.

Hence the sides ofAdBdCd are seen from M

with 120o.

Assume towards a contradiction that AdBdCd is equilateral. Then, by the Equilateral

Center Lemma, Ap

Ad

is a height ofAd

Bd

Cd

. Thus Ap

Ad

bisectsBd

Cd

. Hence Ap

Ad

bisectsBdApCd and so BApC. Since Ap is the incenter ofBAdC, ApAd bisects also BAdC. But

so the exterior angles ofAdApB and AdApCat vertexAp are 12BApC= 1

2BAdC + and

12BApC = 1

2BAdC+ . Hence = . Similarly it is shown that = . ThusABC is

equilateral contrary to the assumption.

Conclude that AdBdCd cannot be equilateral (ifABC 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

ApBC, BpCA andCpAB. We will study onlyApBC as the other two are similar.


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CUBO16, 2 (2014)


SinceAp 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 trisectorsBCp

(distal toAB) andABp (proximal). So B is on ACp and C is on ABp.See Fig.18. Corollary 1a asserts ACpBp = + andABpCp = +. Hence ACpAp = 60o ++ < 180o

and ABpAp =60o ++ < 180o. Therefore the quad-rangle ABpApCp is convex and so BApC is insideBpApCp.

Fig.18

Conclude that BApC < 60o and thus ApBC cannot be equilateral.

5.1.3 Interior Morley triangles with one distal and two mix vertices

Fig.19

There are three interior Morley triangles with

one distal and two mix vertices which are de-

noted byAdBC,BdCA andCdAB.

We will study only AdBC as the other two

are similar. SinceAd is the intersection of the

distal trisectors, B must be the intersection of

the remaining trisectorsCAp (distal toCA) and

ABp, as proximal. So C is the intersection of

the left trisectors ACp and BAp (mix to AB).

First notice that if = then AdBC

is isosceles, because Corollary 1a with = yields BpApB = CpApC and so ApB =

ApC which implies AdApB =AdApC.

Thus if= = then AdBC is equi-

lateral.

Assume ABC is not equilateral. Then it

has two sides not equal and thus in the following we may assume < . Fig.19.

Suppose, towards a contradiction, that AdB = AdC. Let Z be the symmetric point ofC

with respect to ApAd. We will fist show that Z is inside AdApB.

From Corollary 1b,ApAd is height of the equilateralApBpCp and soBp andCP are symmetric

with respect to ApAd. Consequently BpApZ = CpApC and since CpApC = + inferBpApZ= +. Since BpApB = + and by assumption< , deduce BpApZ < BpApB.

Moreover ApBpZ= ApCpC =+ (+ ) =+while ApBpB =+ (+) = + .

So ApBpZ < ApBpB.


22/40


Since Z is inside AdApB then AdZB > AdApB = AdApBp + BpApB = 30o +

+ =90o + . However, the assumptionAdB =AdC implies AdZ= AdB and so AdZB =

AdBZ. Thus AdZB + AdBZ > 2(90o +) > 180o. Hence two angles ofAdZB have

sum greater than 180o, which is a contradiction.

Conclude that AdBC cannot be equilateral (ifABC 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 ofABC.

The proximal trisectors meet at the points ApA, BpB and C

pC and so A

pAB

pBC

pC denotes the

central Morley equilateral. Notice that the trisectorsBCpC and CBpB are parallel iff

BpBCB + CpCBC= 180

o 2(+ ) + 2(+) =180o.

So for = 30o BCpC//CBpB. In this case the distal trisectors with respect to BCdo not intersect

and hence the distal to BC vertexAdA is not determined. Also if30o >then AdA andApA are onthe same side ofBC while BAdAC= 60

o 2. If30o 30o)

The Morley triangle of distal vertices is denoted by AdABdBC

dC, where A

dA, B

dB and C

dC are

the meeting points of the distal trisectors of the exterior angles with respect to the sides BC, CA

andAB, respectively. In Fig.20a and Fig.20b the different locations ofAdAB

dBC

dC with respect to

ApAB

pBC

pC are illustrated. Note thatA

pAA

dA, B

pBB

dB and C

pCC

dC, by Corollary 1b, are the heights

of the central Morley equilateral ApABpBC

pC and let N be their intersection.

Notice that ifABCis equilateral then AdABdBC

dCis equilateral as well. Thus in the following


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CUBO16, 2 (2014)


we assume that ABCis not equilateral. Also suppose towards a contradiction that AdABdBC

dC

is equilateral.

IfABCis an acute triangle the equilateral ApABpBC

pC is inside A

dAB

dBC

dC. Hence

AdANBdB = BdNCdC = CdCNAdA = 120o.

IfABCis an obtuse triangle (assume > 30o) AdA and ApA are on different sides ofBC.

Hence

AdANBdB = A

dANC

dC = 60

o and BdBNCdC = 120

o.

Thus, by the Equilateral Center Lemma, N is the incenter (acute) or the excenter (obtuse) of

the assumed equilateralAdABdBC

dC. HenceA

pAA

dA is a height ofA

dAB

dBC

dCand soA

pAA

dAbisects

BdBCdC and B

dBA

pAC

dC.

In the case of the acute triangle, notice in BAdACthatApAis the incenter while the bisector

Ap

AAdA bisects BA

p

AC. As a result AdABC= A

dACB 2(+) =2(+ ) = .

In the case of the obtuse triangle, note that since ApAAdA bisects B

dBC

dC it bisects BA

pAC.

Also it bisects BAdAC as a height ofApAB

pBC

pC. ThenA

pABA

dA = A

pACA

dA and so A

pAB =

ApAC.

Therefore += + and so = . Similarly we show that = .

Deduce that ABCis equilateral contrary to the assumption that it is not.

Conclude that AdABdBC

dC cannot be equilateral (ifABCis not equilateral).

5.2.2 The Central Morley triangles with one proximal and two mix vertices

Fig.21

There are three Morley triangles of ABC formed

by exterior trisectors with one proximal and two

mix vertices denoted by ApAB

BC

C, BpBC

CA

A and

CpCA

AB

B. We will study only ApAB

BC

Cas the other

two are similar.

As vertexApA is the intersection of the proximal to

BC trisectors, vertex BB is the intersection of the re-

maining trisector CBPB (proximal to CA) with ACPC, as

distal. Then vertexCC is the intersection of the left tri-

sectors ABPB (distal to AB) and BCPC (proximal). Thus

BB is on CBPB while CC is on BCPC. Using the anglevalues between the sides ofApAB

pBC

pC and the trisec-

tors ofABCgiven by Corollary 2a it is easily deduced

that all the angles of the quadrangles BBBpBA

pAC

pC and


24/40


CCCpCA

pAB

pB are less than180

o and so they are convex.

For instance

CCCpCA

pA = 180

o BCpCA

pA =180

o ,

whileABpBA

pA = 60

o + .

Since BBBpBA

pAC

pC is convex infer A

pAC

C is inside BpBA

pAC

pC. Also since C

CCpCA

pAB

pB is

convex infer ApAB

B is inside BpBA

pAC

pC.

Conclude that BCApAC

C < LBpBA

pAC

pC = 60

o and so ApAB

BC

C 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

which are denoted by AdAB

BC

C, BdBC

CA

A andCdCA

AB

B. We will study onlyAdAB

BC

C as

the other two are similar.

SinceAdA is the intersection of the distal to BCtrisectorsBCpCand CB

pB,B

B is the intersection

of the remaining trisectorCApA (proximal to CA) withABPB as distal. Then C

C is the intersection

of the left trisectors ACpC (proximal to AB) and BApA (distal).

Fig.22

If= 30o then AdA is not determined.

If 30o > then AdA is on the same side of

BC with ApA and BAdAC = 60

o 2. If

30o < then AdA and ApA are on different

sides ofBC while BAdAC= 2 60o.

Consider the case = . Then

BpBApAB

B = CpCA

pAC

C by Corol-

lary 2a. So ApAB

B = ApAC

C and in

turn AdAApAB

B = AdAA

pAC

C. Thus

AdAB

BC

Cis isosceles. So ifABCis equi-

lateral then AdAB

BC

C is equilateral.

Next consider that ABC is not equi-

lateral and let > .

Since the location ofAdAdepends on the

value of, assume that 30o > . Suppose,

towards a contradiction that AdABBCC isequilateral. Let Wbe the symmetric point

ofCC with respect to ApAA

dA and let V be

the intersection ofAdAWand B

BBpB. In the


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CUBO16, 2 (2014)


sequel we find the location ofW.

From Corollary 2b, ApAAdA is height of the equilateral A

pAB

pBC

pC and so A

pA and B

pB are

symmetric with respect to ApAAdA.

Also by Corollary 2a the angles between the trisectors ofABC and the central Morleyequilateral are as depicted in Fig.22. Consequently,

AdACpCW= A

dAC

pCC

C = 180o 60o = 60o +and AdAB

pBB

B = 180o 60o =

60o + . Therefore BBBpBW < A

dAB

pBB

B and the line BpBWis inside CB

pBB

B.

Furthermore BpBApAW= C

pCA

pAC

C = and since > , ApAWis inside B

pBA

pAB

B. Thus

Wis inside CBpBB

B.

ClearlyCCis outside ABCpC. Thus we may set C

pCA

dAC

C = . Then by symmetry BpBA

dAW=

. Also set AdAWB

B = . Since AdAB

BC

C is assumed equilateral then C

CAdAB

B =60o and

so WAdAB

B = 2( ).

HoweverAdAB

B = AdAWand soA

dAC

C = AdAB

BimpliesAdAW= A

dAB

B. Thus WB

BAdA =

and fromAdA

WBB

, 2+ 2() =180o = 90o +. Given thatWis insideCBpB

BB

inferCBpBB

B, > AdAVB

B.

But from AdAVB

B we deduce AdAVB

B = AdAB

pBB

B+ BpBA

dAV=60

o + + .

Thus > 60o + += > 60o + + ( + 90o)= + < 30o which contradicts

the assumption 30o >.

The case > 30o is similar and it is omitted.

Conclude that AdAB

BC

C cannot be equilateral (ifABCis 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 ofC and the exterior trisectors ofA and

B, as the other two cases are similar. apcCpCb

pc denotes the exterior Morley equilateral relative

to C.

Notice that the trisectors AbpC and BapC are parallel iff

apCBA + bpCAB= 180

o 2(+) + 2( +) = 180o = 30o.

In this case the distal trisectors with respect to AB do not intersect and hence the distal vertex

CdC is not determined. Also if30

o

> thenCdC and C

p

Care on the same side ofAB with ACdCB=

60o 2. If30o


26/40


intersects the trisector BCpC betweenB and CpC. Moreover Corollary 3a implies that the extension

ofACpC is inside apCC

pCBand soCa

pC intersectsAC

pC insideC

pCBa

pC. In additionAC

pC intesects

BapC between B and apC. Similarly the trisectror Cb

pC intersects the trisector AC

pC between A and

CpC and the trisector BCpC inside C

pCAb

pC. AlsoBC

pC intersects Ab

pC between A and b

pC.

5.3.1 The Morley triangle of distal vertices

This is denoted by adCCdCb

dC. VertexC

dC is the intersection of the distal to AB trisectors Ab

pC

andBapC and is determined iff =30o. VertexadC is the intersection of the distal to BC trisectors

BCpC and CbpC and hence it is inside C

pCAb

pC. Vertexb

dC is the intersection of the distal to AC

trisectorsCapC andACpC and hence it is inside C

pCBa

pC. See Fig.23.

Thus, for C =90o CdCis determined whileadCand b

dCare inside AC

dCB. But so a

dCC

dCb

dC 30o)

5.3.2 The Morley triangles with one proximal and two mix vertices

There are three such triangles denoted byapCb

CC

C, bpCa

CC

C and CpCa

Cb

C.


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CUBO16, 2 (2014)


Fig.24

a. apCb

CC

C : Vertex apC is the intersection of

the proximal to BCtrisectors BapC andCapC. Vertex

bC must be the intersection of the remaining interior

trisectorCbpC (proximal to ) with the exterior trisec-

tor AbpC, as distal. Hence CC is the intersection ofthe left trisectors, BCpC (proximal to AB) and Ab

pC

(distal). See Fig.24.

So bC is on ACpC and it is between A and C

pC.

Also CC is onAbpC and it is between A and b

pC. No-

tice apCbpCC

pC = a

pCC

pCb

pC = 60

o while, by Corol-

lary 3a, CpCbpCA= + and b

pCC

pCA= (+)

+. Thus apCCpCA < 180

o and apCbpCA < 180

o.

Hence the quadrangle AbpCapCC

pC is convex. Therefore b

CapCC

C is inside bpCa

pCC

pC and so

bCapCC

C < 60o.

Conclude that apCb

CC

C is not equilateral.

b. bpCa

CC

C : It is shown as above that it is not equilateral.

c. CpCa

Cb

C : VertexCpC is the intersection of the proximal to AB exterior trisectors. Thus

aC is the intersection of the remaining exterior trisector BapC (proximal to BC) with the interior

trisectorCbpC, as distal. Then b

C is the intersection of the left trisectors CapC (distal to AC) and

AbpC (proximal).

So aC and CpC are on the same side ofBC iff

BCbpC+ CBapC < 180

o 2 +2(+) + 3< 180o < .

If = then aC is not determined as BapC//Cb

pC.

Also b

C and CpC are on different sides ofAC iffACapC+ CAb

pC < 180

o 2 + 2( +) + 3< 180o < .

If = then bC is not determined as AbpC//Ca

pC.

Since aC, b

C and CpC are outside ABC while a

C and b

C are on AbpC and Ba

pC respectively we

deduce

aC and CpC are on the same side ofAB iff<

while

bC and CpC are on the same side ofAB iff< .

The above conditions correlate the ranges of,CpC and with the different locations ofa

C

and bC and vise versa.

Recall that AbpC and BapC intersect at CdC iff =30o, with ACdCB = |60o

2|, while CpC andCdC are on the same side ofABiff< 30

o.

Next all the different locations ofaC and b

C are considered.


28/40


Case 1: aC andb

C are on the other side ofAB from CpC.

This happens iff 30o (and so < and < ) or < 30o with< and < . Fig.25a,b.

If= 30o

then Abp

C//Bap

C, while for =30o

Abp

C andBap

C meet at CdC.

But so, aC and b

C are on the extensions (to the other side ofAB fromCpC) ofCb

pC,Ba

pC and

CapC, bpCA respectively.

SinceCpC is inside apCC

pCb

pC, then a

CCpCb

C encompasses a

CCb

C . Therefore

aCCpCb

C < a

CCb

C. However a

CCb

C = . Deduce a

CCpCb

C< 60o.

Fig.25a (> 30o) Fig.25b (< 30o,< ,< )

Case 2 : aC and b

C are on the same side ofAB with CpC.

This happens iff< 30o, > and > . Fig.25c.

Then CdC and CpC are on the same side ofAB. Note that Ca

pC intersects sides AB and BC

dC of

ACdCBinternally and so, by Paschs axiom, it intersects the third side ACdC externally. Thus b

C

is on the extension ofACdC. Similarlya

C is on the extension ofBCdC. Since C

pC is inside AC

dCB

then aCCpCb

C encompasses a

CCdCb

C. Therefore a

CCpCb

C < a

CCdCb

C. But a

CCdCb

C =

ACdCB= 60o 2. Deduce aCC

pCb

C< 60o.

Case 3: aC andb

C are on different sides ofAB. Fig.25d.

This happens iff < 30o with > and < or with < and > .

Next consider the case < 30o with > and < .

ThenCpC is on the same side with CdC. HenceCpC is inside ACdCBand alsoCpCis insideapCCbpC.By Paschs axiom on ACdCB , since Ca

pC intersects sides AB and BC

dC at interior points, infer

CapC intersects the third sideACdC at an exterior point. Thusb

C is on extensions ofACdC and Ba

pC

on the same side ofAB withCpC. SoCpC is inside b

pCCb

C on the other side ofapCb

pCfromC

dC and


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CUBO16, 2 (2014)


Fig.25c (< 30o,> ,> ) Fig.25d (< 30o, < , < )

bC. ConsequentlyapCC

dC intersects b

CCpC between b

C and CpC. Hence a

CapC is inside a

CCpCb

C.

ConsequentlyapC and B are inside a

CCpCb

C. Therefore a

CCpCb

C encompasses apCC

pCBand so

aCCpCb

C> apCC

pCB.

However Corollary 3a asserts apCC

pCB= ( +)

+

. Deduce b

CCpCa

C > 60o

.

The case < 30o with > and < is similar and it is omitted.

Conclude that CpCa

Cb


5.3.3 The Morley triangles with one distal and two mix vertices

These Morley triangles are denoted byCdCa

Cb

C, bdCC

Ca

C and adCc

CB

C.

Fig.26

a. CdCa

Cb

C: Vertex CdC is the in-

tersection of the distal to AB trisectors

AbpC and BapC. Thus vertexa

C is deter-

mined by the intersection of the remain-

ing trisectorCPC , distal toBC, withCapC,

as proximal. VertexbC, is determined by

the left trisectors ACPCand CbpCwhich are

distal and proximal to CA, respectively.

VertexCdC is determined iff =30o

with CPC and CdC to be on the same side

ofAB iff> 30o.

Moreover aC is always located be-

tween B and CPC while bC is always lo-cated between A and CPC. Fig.39 depicts the case for C

PC and C

dC to be on the same side ofAB.

Regardless the location ofCdC, verticesa

Cand b

C are inside ACdCB. Thus a

CCdCb

C< ACdCB.

Since ACdCB= |60o 2|< 60o then aCC

dCb

C< 60o.


30/40


Conclude that CdCa

Cb


b. bdCC

Ca

C: Vertex bdC is the intersection of the distal to CA trisectors Ca

pC and AC

pC.

Hence vertex CC is determined by the intersection of the remaining trisector bp

C

, distal to AB,

with BCpC, as proximal.

Vertex aC is determined by the intersection of the left trisectors CbpC and Ba

pC which are

distal and proximal toBC, respectively.

Trisectors CapC andACpC always intersect each other and sob

dC is located on the same side ofAB

withCpC. AlsoAbpC andBC

pC always intersect each other and so C

C is located on the same side of

AB with CpC. However a

C is not always determined as CbpC//Ba

pC iffb

pCCB+ b

pCCa

pC =180

o

= . In fact aC is on the same side with bdC and C

C iff= . It should also be noted that

30o


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CUBO16, 2 (2014)


Fig.27c (< 30o, < )

Thus adCbpCC

C is exterior angle in bpCa

CC

C. Hence bpCa

CC

C < adCb

pCC

C. But

adCbpCC

C = CbpCA. InCb

pCAit is calculated Cb

pCA= and so b

pCa

CC

C < .

Infer BaCC

C < () + < 60o.

Conclude that bdCa

CC


c. adCB

Cc

C: 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 ofABC. As previously we will consider only those formed by the trisectors of the

exterior Acombined with the interior trisectors ofBand Cas the other two cases are similar.

5.4.1 The Morley triangle of proximal vertices

This is denoted by ApbpAcpA . Vertex A

p isthe intersection of the proximal to BC interiortrisectors, vertex bpA is the intersection of CB

p

with the exterior trisector ofAproximal toAC,while vertex cpA is the intersection of BC

p withthe exterior trisector ofAproximal to AB.

Fig.28


32/40


Consider the companion equilateral SAApKArelative to vertex A. In Fig.28 the case< 30

o

is depicted for which Corollary 4 asserts that BASA = CAKA = + while SA and KA

are outside ofABC. Consider the intersections of lineApCp with the sides ofbpAAdcpA. Then

ApCp intersects sideAdcpA at Cp and so externally, while it intersects side AdbpA atSA internally

since BAbpA = 2(+ ) and BASA = + . Thus, by Paschs axiom, ApCp intersects

the third side bpAcpA internally. Similarly line A

pBp intersectsbpAcpA internally. Thus b

pAA

pcpAencompasses SAA

pKA and so SAApKA < Lb

pAA

pcpA.

But SAApKA = 60

o and so bpAApcpA > 60

o. Therefore for< 30o the ApbpAcpA is not

equilateral. The cases > 30o and = 30o are similar.

Conclude that ApbpAcpA cannot be equilateral.

Note that the non equilateralApbpAcpA fails the original statement of Morleys theorem.

5.4.2 The Morley triangle of distal vertices

This is denoted by AdbdAcdA. Vertex b

dA is the

intersection ofCAp, the distal to CA trisector of the

interior

C, and the distal to CA trisector of theexterior A. Vertex cdA is the intersection of BAp,

the remaining trisector ofB(distal toAB) with thedistal to AB trisector of the exterior A. Also it iseasily seen that bdA and c

dA are determined iff =.

Hence, for =, Ad is inside AcdABand CbdAA.

From CbdAAit is calculated thatCbdAA= 180

o3(+)2 =2 and similarly

fromAcdAB, AcdAB= 2. Since ++= 60

o,at least one of and is less than30o. Thus eitherbdAc

dAA

d or cdAbdAA

d is less than 60o.Conclude that AdbdAc

dA is not equilateral.

Fig.29


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CUBO16, 2 (2014)


5.4.3 The Morley triangles with a proximal and two mix vertices

These triangles are denoted by ApbAc

A, bpAA

cA and cpAA

bA.

Fig.30

a. ApbAc

A : VertexAp is the inter-

section of the proximal to BC interior trisec-

tors. Hence bA 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

SAApKA relative to vertex A

p. In Fig.30

the case > 30o is depicted for which Corol-

lary 4 asserts that vertices SA and KA are

inside ABC.

Consider the intersections of line ApCp with the sides ofbpAAdcpA. A

pCp intersects side

AdcpA at Cp and so externally, while ApCp intersects side AdbpA at SA and so internally. Thus,

by Paschs axiom, ApCp intersects the third side AbA internally. Similarly ApBp intersects

AcA internally. Hence bpAA

pcpA encompasses SAApKA and so SAA

pKA < bpAA

pcpA. But

SAApKA = 60

o and so bAApcA > 60

o. Therefore for > 30o ApbAc

A is not equilateral.

The cases < 30o and = 30o are similar and they are omitted.

Conclude that ApbAc


b. bpAAcA: Vertex b

pA is the intersection of the proximal to ACtrisectors, which are CB

p

and the corresponding trisector of the exterior A. Thus A is the intersection of the remaining

interior trisector ofC, CAp, which is proximal to BC, with BCp as distal to BC. Then cA 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 iff2 = + = . ThuscA exists iff

=.

Also if > thenbp

A andc

A are on the same side ofAC, while for < , bp

A and c

A are ondifferent sides ofAC.

Case > : We will show bpAc

AA < 60o.


34/40


Notice that A is inside ABAp and so cAAp is a right bound for the right side cAA

of

bpAc

AA.

In following we will find a left bound for the left side cAb

A ofbpAc

AA.

Let bA be the intersection ofCAp withAbp

A

and note that the pointsA, Ap, cA, andb

A are

cyclic, because fromBApCit follows thatbAApcA =+and sob

Ac

A is seen from A and Ap

with angle +.

The extension ofApCp meets the exterior trisector AbA between Aand b

A. Then it crosses

the circle, say at T. We will show that a left bound for side cAb

A ofbpAc

AA is the bisector of

bAc

AA.

Note that in a triangle the bisector of an angle crosses its opposite side at a point which is

between the sides middle point and the sides common vertex with the shortest of the other two

sides.

Let G be the intersection of the bAc

A A bisector with Ab

A. Also letM be the middle of

AbA

.

CbpA is angle bisector in ACb

A. It is easily calculated from ACb

Athat Ab

AC= +and

so AbAC < b

AAC. Thus CA < Cb

A. Hence bpA is between A and M.

cAGis angle bisector in Ab

Ac

A. Obviously b

AAc

A = +. Also c

Ab

AA= c

Ab

AC+

CbAAwhile c

Ab

AA= c

Ab

AC + Cb

AA. But c

Ab

AC= c

AAAp = cAAB + BAA

p =

(+)+BAAp. So bAAc

A < c

Ab

AAand thusb

Ac

A < b

AA. HenceGis betweenb

Aand M.

ThereforebpA is onAb

A and it is betweenA and G. So Gc

AAp encompasses bpAc

AA and thus

bpAc

AA < GcAA

p .

In the sequel we calculate GcAAp. Notice that GcAA

p = GcAA+ Ac

AAp whereas

AcAAp = and GcAA=

12bAc

AA= 12bAA

pA= 12

[bAApT+ TApA].

ButbAA

pT = BApCp BApA =+ ( +) = + .

Also

TAbA = TApbA = +and A

pTA= AcAAp = .

Then from TAAp it is calculated

TApA= + CpAAp .

Thus

GcAA= + 12TApA

and so GcAAp =+ 1

2CpAAp where0 < CpAAp bpAAcA is not equilateral.


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CUBO16, 2 (2014)


Fig.31a (> )

Fig.31b

Case < : We will show that bPAAcA >

60o.

Consider the intersections ofAcA withBCp and

CBp denoted bycpAand b

A respectively. Notice that

bPAAcA encompasses b

AAcpA. So it suffices to

show that bAAcpA > 60

o.

Observe that b

A, A

, C and c

p

A are cyclic, be-cause sidebAA is seen from cpA andC with angle

as from AcpAB it is calculated AcpAB= . Thus

bAAcpA = b

ACcpA.

Moreover bACcpA = A

dCcpA, since Cb

A

passes through Ad. But

AdCcpA = AdCA+ACc

pA = +ACc

pA and so

bACcpA = + ACc

pA.

In aditionA, Ad, C and cpA are also cyclic since

AAd is seen from C and cpA with angle .

ConsequentlyACcpA = AAdcpA and so

bACcpA = + AA

dcpA.

Yet from AAdB infer


36/40


AAdcpA = ABAd + AdAB.

However AdAB > CpABand thus

AAdcpA > ABAd + CpAB= + .

Therefore bAAcpA > + + = 60

o.

Conclude that for < , bpAAcA is not equilateral.

c. cpAA

bA : It is showed that it is not equilateral similarly as bpAA

cA.

5.4.4 The Morley triangles with one distal and two mix vertices

These triangles are denoted byAdbAc

A, cdAA

bA andbdAA

cA.

Fig.32

a. AdbAc

A: ObviouslyAd is the intersection

ofCBp and BCp. Then b

A is the intersection of theremaining interior trisectorsCAp (distal toAC) with

the proximal to AC exterior trisector ofA. Notice

these two lines are parallel iff += 2 = .

Moreover cA 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 = . Therefore AdbAc

A is deter-

mined iff = . FromAcAB and Ab

AC it fol-

lows ApbAA= ApbAA= || and so b

A and

cA are on the same side ofAC.

We will consider only the case > as the other

one is similar.

Notice that Ap is inside BAdC. Thus

AdcAb

A encompasses ApcAb

A and so

AdcAb

A > ApcAb

A.

Also notice thatA,Ap, bAand c

A are cyclic asAAp is seen frombAand c

A with angle.

Thus ApcAb

A = ApAbA. But A

pAbA = ApAC + CAbA = A

pAC + (+). Moreover

ApAC > BpAC= and so ApAbA > + ( +) =60o. Therefore AdcAb

A > 60o.

Conclude that Ad

b

Ac


b. bdAAcA : Obviously b

dA is the intersection of CA

p 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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CUBO16, 2 (2014)


Fig.33

with the BAp (distal). Thus cA is the intersection of the left trisectors,BCp and the distal to AB

exterior trisector ofA.

Notice that bdA and c

A exist iff =. We will show that bdAA

cA > 60o.

Note that Bp is inside bdAAcA and so b

dAA

cA encompasses bpAA

cA. Thus it suffices

to prove bpAAcA > 60

o.

For this we use a symmetric argument to the proof ofbAAcpA > 60

o (5.4.4.a case < ).

Let bpA be the intersection of Ac

A with CA. Notice that cA, A

, B, bpA are cyclic as

cAA is seen from B and bpA with angle . Thus c

AAbpA = c

ABbpA. Moreover c

AAbpA =

AdBbpA, since Bc

A passes through Ad. But AdBbpA = A

dBA+ ABbpA = + ABb

pA and

so cAAbpA= + ABb

pA.

HoweverA, Ad, B, bpA are cyclic as AAd is seen from B andbpA with angle . Consequently

ABbpA = AAdbpA and so c

AAbpA = + AA

dbpA.

Yet from AAdC infer AAdbpA = ACAd + AdAC. However AdAC > BpAB and

thus AAdcp

A

> ACAd + BpAB= + . Therefore cAAbp

A

> + + = 60o.

Conclude that bdAAcA is not equilateral.

c. cdAAbA: This case is similar to the above and it is omitted.


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6 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 The structrure of angle trisectors

Fig.34a Fig.34b

The interior angle bisectors pass through a

unique point (incenter).

The interior angle trisectors proximal to the tri-

angle sides pass through the vertices of a unique

equilateral (inner Morley equilateral).

The bisector of an interior angle and the bi-

sectors of the other two exterior angles pass

through a unique point (excenter).

The trisectors of an interior angle and the trisec-

tors 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

of a unique triangle with orthocenter the interior

angle bisectors common point (incenter). +

The exterior trisectors proximal to the triangle

sides pass through the vertices of a unique equi-

lateral (central Morley equilateral). +

+ This fact follows from the previous one


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CUBO16, 2 (2014)


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

Documents

Trisectors Like Bisectos With Equilaterals Intead of Points_Spiridon Kuruklis