Art N5.dvi


CUBO A Mathematical Journal
Vol.19, No¯ 02, (73–85). June 2017

A Trigonometrical Approach to Morley’s Observation
Ioannis Gasteratos1, Spiridon Kuruklis2 and Thedore Kuruklis3

1Department of Mathematics and Statistics, Boston University,

Boston, MA 02215, USA.

igaster@bu.edu
2Eurobank, Group Information and IT Security,

14234 Athens, Greece.

skuruklis@gmail.com
3Theoklitos,

17672 Kallithea, Greece.

tkuruklis@gmail.com

ABSTRACT

Simple trigonometrical arguments verify that in a triangle the trisectors, proximal to

sides respectively, meet at the vertices of an equilateral triangle by showing that the

length of each side is 8R times the sines of the angles between the sides of the triangle

and the trisectors that determine it, where R is the radius of the circumcircle of the

triangle. The 27 meeting points of the trisectors, proximal to a side, determine 18

such equilaterals, which in pairs share a vertex having two collinear sides and the third

parallel. Hence these points are located 6 by 6 on three triples of parallel lines.

RESUMEN

Argumentos trigonométricos simples verifican que en un triángulo los trisectores, próximos

a los lados respectivamente, se encuentran en los vértices de un triángulo equilátero

mostrando que la longitud de cada lado es 8R veces los senos de los ángulos entre los

lados del triángulo y los trisectores que lo determinan, donde R es el radio del cir-

cunćırculo del triángulo. Los 27 puntos de encuentro de los trisectores, próximos a un

lado, determinan 18 tales equiláteros, que a pares comparten un vértice teniendo dos

lados colineales y el tercero paralelo. Luego estos puntos están ubicados 6 por 6 en tres

triples de ĺıneas paralelas.

Keywords and Phrases: Angle trisection, proximal trisector, triangle trisectors, Morley’s theo-

rem, Morley triangle, Morley’s magic, Morley’s miracle, Morley’s mystery.

2010 AMS Mathematics Subject Classification: 00A05, 00A08



74 Ioannis Gasteratos, Spiridon Kuruklis & Thedore Kuruklis CUBO
19, 2 (2017)

1. Introduction

One of the infamous contemporary problems in mathematics refers to the angle trisection in a

triangle. The problem appeared suddenly around 1900, when Frank Morley made a shocking

observation, known since then as Morley’s theorem, usually expressed with the statement: In

a triangle, the trisectors of its angles, proximal to sides respectively, meet at the vertices of an

equilateral triangle.

The above theorem is considered among the most unexpected discoveries in mathematics,

as strangely went unnoticed during the ages, even though it expresses a property for trisectors

analog to bisectors. Ancient Greeks studied the triangle in depth and they could have discovered

it, but simply ignored it. More than one hundred years since its discovery and a very respectable

number of publications, several authored by distinguished mathematicians, we are still struggling

to fully comprehend it. Words like magic, miracle, mystery, or paradox have appeared in titles of

several articles. Notably, Morley’s theorem has been included in the list of The hundred greatest

theorems.[1]

Morley inferred it while studying the behavior of cardioids from an observation that plainly

asserts: In a triangle, the trisectors proximal to a side intersect on three sets of three parallel lines

forming equilateral triangles... Thus, if we take the interior trisectors of the angles of a triangle,

the points where those proximal to a side meet form an equilateral triangle. [13, p.469]

In Fig.1 appear 27 equilaterals. Their placement reveals a structure with startling symmetry,

where an impressive number of overlapping and interconnected equilaterals are arranged with

common vertices and parallel or collinear sides. Their existence, in fact with arrangement, is

interpreted as evidence of regularity in the behavior of angle trisectors in a triangle, like the

incenter and the excenters of a triangle express regularity in the behavior of its angle bisectors.[11]

Visual inspection easily verifies that only 18 from the 27 triangles determined by the meeting

points of trisectors of all three angles, proximal to sides respectively, are equilateral and they are

called Morley triangles.

Fig.2 illustrates the trisectors of ∠ABC = 3β, 0 < β < 60◦. The proximal to side BC trisector is

the one (BT) which coincides with side BA after two rotations around B towards BA by ∠CBT. Note

that the inclinations of the proximal trisectors to the corresponding proximal sides are β, 60◦ − β

and 60◦ + β.

The used formulation of Morley’s theorem does not specify the type of trisectors, intersecting

in pairs at the vertices of an equilateral. This becomes crucial in the presence of many triangles

formed by the intersections of trisectors, proximal to sides respectively. However, several from these

triangles - but not all - are equilateral. As the theorem can be valid for other types of trisectors

this ambiguity may have been deliberately used, although most publications have focused only on

the interior trisectors. In contrast to the elementary, sharp and clean statement of the theorem,

approaches dealing with the observation remain abstract or in higher mathematics, where it was



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A Trigonometrical Approach to Morley’s Observation 75

Figure 1: Fig.1: The 27 intersections of trisectors, proximal to sides, determine 18 equilaterals

discovered. [8],[12], [16]

In this presentation, we prove several instances of Morley’s theorem yielding the 18 Morley

triangles. In fact, it is shown: The side length of a Morley triangle is 8R times the sin of the

angles between the sides of the triangle and the trisectors that determine it, where R is the radius

of the circumcircle of the triangle. Then, utilizing the arrangement of these triangles, we establish

Morley’s observation about the alignment of the intersections of trisectors proximal to a side. The

approach relies on the following trigonometrical property that combines the sine and cosine laws.

Proposition: In a ∆STV with ∠STV = φ, if ST = p sin θ and TV = p sin ω, where φ + θ + ω = 180◦,

then SV = p sin φ while ∠SVT = θ and ∠TSV = ω.

Proof. In ∆STV by applying the law of cosines we get (SV)2 = (TS)2+(TV)2−2(TS)(TV) cos φ = p2 sin2 φ

since sin2 θ + sin2 ω − 2 sin θ sin ω cos φ = sin2 φ by the law of cosines in the triangle with sides sin θ,

sin ω and sin φ. Thus SV = p sin φ.

Now from the law of sines ST/ sin(∠SVT) = TV/ sin(∠TSV) = SV/ sin φ. As SV = p sin φ, ST = p sin θ

and TV = p sin ω, we get sin θ = sin(∠SVT), sin ω = sin(∠TSV). Therefore, ∠SVT = θ or ∠SVT = 180◦−θ

and ∠TSV = ω or∠TSV = 180◦ − ω. Since φ + θ + ω = 180◦ and φ + ∠SVT + ∠TSV = 180◦, only the

case ∠SVT = θ and ∠TSV = ω may hold.



76 Ioannis Gasteratos, Spiridon Kuruklis & Thedore Kuruklis CUBO
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Figure 2: Fig.2: The three types of trisectors

Given a ∆ABC, with ∠A = 3α, ∠B = 3β and ∠C = 3γ, where α + β + γ = 60◦, we establish that

∆A′B′C′ is equilateral, where A′, B′ and C′ are the intersections of trisectors proximal to sides of

∆ABC respectively, by showing A′B′ = B′C′ = C′A′. To prove this we apply the above proposition to

the adjacent triangles ∆A′CB′, ∆B′AC′ and ∆C′BA′, formed by a side of ∆A′B′C′ and the trisectors

that determine it. The lengths of the trisectors are found from the surrounding triangles ∆BA′C,

∆CB′A and ∆AC′B after the sides are expressed by the formulas AB = 2R sin 3γ, BC = 2R sin 3α,

AC = 2R sin 3β, obtained from the law of sines.

In the appearing expressions, it is convenient to represent angles θ + 60◦ and 60◦ − θ as θ+

and θ− respectively. Hence θ+ = 60◦ + θ and θ− = 60◦ − θ, while θ++ = 120◦ + θ and θ−+ = 120◦ − θ.

Also, the formula

sin 3θ = 4 sin(60◦ − θ) sin θ sin(60◦ + θ) = 4 sin θ− sin θ sin θ+

will be used to simplify expressions. As sin 3θ = 3 sin θ−4 sin3 θ the above follows after factoring out

the right hand side. In the article Are These The Most Beautiful? Hofstadter is quoted that would

have given a very high score to Morley’s theorem as it follows from this trigonometrical identity.

[15]

In the sequel, Morley triangles are grouped according to the type of trisectors that determine

them and only representatives of groups are examined. The groups are: The primitive triangles

formed by the same type of trisectors, the mix triangles formed properly by one type of trisectors of

an angle and another type of trisectors of the other two angles, and the complete triangles formed

by trisectors of one distinct type for each angle. To avoid degenerate cases in which some of the

considered triangles are not formed, in the sequel we assume that the angles of the given triangle

are different and not multiples of 30◦.



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A Trigonometrical Approach to Morley’s Observation 77

2. The 18 Morley triangles

In this section, we will prove that the 18 Morley triangles are indeed equilaterals by producing

formulas for the lengths of their sides.

I. Primitive Morley triangles (3)

The primitive triangles are formed exclusively by the intersections of same type trisectors. Hence

the two pairs of trisectors determining one of its sides form the same three angles with the corre-

sponding sides of ∆ABC. In the sequel, we find an expression of the length for one side. As this is

independent of the side then all sides have the same length and so the specific Morley triangle is

equilateral.

Figure 3: Fig.3: The primitive Morley triangles

Theorem 0.1. In a triangle, the same type trisectors of its angles, proximal to sides respectively,

meet at the vertices of a corresponding equilateral.

Proof. Assume that the intersections of the interior, exterior and explementary trisectors, proximal

to the sides, BC, CA and AB, meet at A′, B′ and C′, A′′, B′′ and C′′, A′′′, B′′′ and C′′′, defining

∆A′B′C′, ∆A′′B′′C′′ and ∆A′′′B′′′C′′′, called inner triangle, central triangle and peripheral triangle

respectively.

The proximal trisectors that determine one of their sides form angles with the corresponding



78 Ioannis Gasteratos, Spiridon Kuruklis & Thedore Kuruklis CUBO
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sides of ∆ABC equal to α, β and γ, for the inner, or α−, β− and γ−, for the central, or α+, β+ and

γ+, for the peripheral triangle, regardless of the side.

a. The lengths of the interior trisectors determining side B′C′ are found from the law of

sines in ∆AB′C and ∆BC′A :

AB′/ sin γ = AC/ sin(180◦ − α − γ) = 2R sin 3β/ sin β− = 8R sin β sin β+;

so AB′ = 8R sin β sin β+ sin γ. Similarly, AC′ = 8R sin γ sin γ+ sin β.

Then in ∆B′AC′, ∠B′AC′ = α, AB′ = p sin β+, AC′ = p sin γ+, where p = 8R sin β sin γ. As α+β++γ+ =

180◦, B′C′ = p sin α = 8R sin β sin γ sin α while ∠AB′C′ = γ+ and ∠AC′B′ = β+. We conclude that

∆A′B′C′ is equilateral.

b. The lengths of the exterior trisectors determining side B′′C′′ are found from the law of

sines in ∆AB′′C and ∆BC′′A:

AB′′/ sin γ− = AC/ sin(180◦ − α− − γ−) = 2R sin 3β/ sin β+ = 8R sin β sin β−

and so AB′′ = 8R sin β sin β− sin γ−. Similarly, AC′′ = 8R sin γ sin γ− sin β−.

Then in ∆B′′AC′′, ∠B′′AC′′ = 2α−+3α = α++, AB′′ = p sin β, AC′′ = p sin γ, where p = 8R sin β− sin γ−.

As α++ +β+γ = 180◦, B′′C′′ = p sin α++ = 8R sin α− sin β− sin γ− while ∠AB′′C′′ = γ and ∠AC′′B′′ = β.

We conclude that ∆A′′B′′C′′ is equilateral.

c. The lengths of the explementary trisectors determining side B′′′C′′′ are found from the law

of sines in ∆AB′′′C and ∆BC′′′A:

AB′′′/ sin γ+ = AC/ sin(180◦ − α+ − γ+) = 2R sin 3β/ sin β = 8R sin β+ sin β−;

so AB′′′ = 8R sin β+ sin β− sin γ+. Similarly, AC′′′ = 8R sin γ+ sin γ− sin β+.

Then in ∆B′′′AC′′′, ∠B′′′AC′′′ = 2α+ − 3α = α−+, AB′′′ = p sin β−, AC′′′ = p sin γ−, where p =

8R sin β+ sin γ+. As α−+ + β− + γ− = 180◦,

B′′′C′′′ = p sin α−+ = 8R sin α+ sin β+ sin γ+ while ∠AB′′′C′′′ = γ− and ∠AC′′′B′′′ = β−. We conclude

that ∆A′′′B′′′C′′′ is equilateral.

II. Mix Morley triangles (9)

The mix triangles are formed by the intersections of same type trisectors of an angle combined with

a different type trisectors of the other two. Hence, they share a vertex with a primitive triangle.

For a short statement of the next theorem we define the mixable type of explementary, interior and

exterior trisectors to be the interior, exterior and explementary trisectors respectively.

Theorem 0.2. In a triangle, the same type trisectors of an angle and the corresponding mixable

type trisectors of the other two, proximal to sides respectively, meet at the vertices of an equilateral.

Proof. a. The intersections of the explementary trisectors of an angle and the interior trisectors of

the other two proximal to sides respectively, form three triangles, referred as mix inner triangles

since each of them shares a vertex with the inner triangle. We will study only one representative

from them. Consider the explementary trisectors of ∠B and the interior trisectors of the other two.



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A Trigonometrical Approach to Morley’s Observation 79

Assume the interior trisectors proximal to AC meet at B′ while the explementary and the interior

proximal to BC and AB meet at BC and BA defining ∆B
′BABC, called B-inner triangle as it shares

vertex B′ with the inner triangle. The angles between the trisectors that determine it and the

corresponding sides are equal to β+ (explementary) and α, γ (interior).

Find the lengths of the trisectors defining ∆B′BABC:

In ∆BBCA, as ∠BBCA = 180
◦ − α − β+ = γ+,

BBC/ sin α = ABC/ sin β
+
= AB/ sin γ+ = 2R sin 3γ/ sin γ+ = 8R sin γ sin γ−

and so BBC = 8R sin γ sin γ
− sin α and ABC = 8R sin γ sin γ

− sin β+.

Similarly, in ∆CBAB, BBA = 8R sin α sin α
− sin γ and CBA = 8R sin α sin α

− sin β+.

In ∆AB′C, as ∠AB′C = 180◦ − α − γ = β++,

AB′/ sin γ = CB′/ sin α = AC/ sin β− = 2R sin 3β/ sin β− = 8R sin β sin β+

and so AB′ = 8R sin β sin β+ sin γ and CB′ = 8R sin β sin β+ sin α.

Now in ∆BCAB
′, AB′ = 8R sin β sin β+ sin γ = p sin β and

ABC = 8R sin γ sin γ
++ sin β+ = p sin γ++ where p = 8R sin β+ sin γ. As α + β + γ++ = 180◦, B′BC =

8R sin β+ sin γ sin α while ∠BCB
′A = γ++. Similarly, in ∆BACB

′, B′BC = 8R sin β
+ sin γ sin α while

∠BAB
′C = α++. Also, in ∆BBABC, ∠BCBBA = 2β

+ − 3β = β−+, BBA = p sin α
−, BBC = p sin γ

−,

where p = 8R sin α sin γ. As β−+ + α− + γ− = 180◦, BABC = p sin β
−+

= 8R sin α sin γ sin β−+ while

∠BBCBA = α
−, ∠BBABC = γ

−. Since sin β−+ = sin β+, we conclude that ∆B′BABC is equilateral.

Corollary 2a. A mix inner and the inner triangle have two collinear sides and the third parallel.

Proof. Consider for instance the B-inner ∆B′BABC. It was seen ∠AB
′BC = γ

++. As it is equilateral

∠AB′BA = γ
+. Also, it was shown in the inner ∆A′B′C′ that ∠AB′C′ = γ+. Thus, B′BA and B

′C′

are collinear.

b. The intersections of the interior trisectors of an angle and the exterior trisectors of the other

two, proximal to sides respectively, form three triangles, referred as mix central triangles since each

of them shares a vertex with the central triangle. We will study only a representative from them.

Consider the interior trisectors of ∠C and the exterior trisectors of the other two. Assume the

exterior trisectors proximal to AB meet at C′′ while the interior and the exterior proximal to BC

and AC meet at C′′A and C
′′

B defining ∆C
′′C′′AC

′′

B, called C-central triangle as it shares vertex C
′′ with

the central triangle. The angles between the trisectors that determine it and the corresponding

sides are equal to γ (interior) and α−, β− (exterior).

Find the lengths of the trisectors defining ∆C′′C′′AC
′′

B:

In ∆CAC′′B, as ∠C
′′

BAC = 2α
− + 3α = α++ and so ∠CC′′BA = β. Hence

AC′′B/ sin γ = CB
′′

A/ sin α
++

= AC/ sin β = 2R sin 3β/ sin β = 8R sin β− sin β+

and so AC′′B = 8R sin β
− sin β+ sin γ and CC′′B = 8R sin β

− sin β+ sin α++.

Similarly, in ∆CBC′′A, BC
′′

A = 8R sin α
− sin α+ sin γ and CC′′A = 8R sin α

− sin α+ sin β++. Also in ∆AC′′B,



80 Ioannis Gasteratos, Spiridon Kuruklis & Thedore Kuruklis CUBO
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Figure 4: Fig.4: Mix Morley triangles

as ∠AC′′B = 180◦ − β− − α− = γ−+,

AC′′/ sin β− = BC′′/ sin α− = AB/ sin γ−+ = 8R sin 3γ/ sin γ = 8R sin γ sin γ−

and so AC′′ = 8R sin γ sin γ− sin β− and BC′′ = 8R sin γ sin γ− sin α−.

Then in ∆C′′BAC
′′, ∠C′′BAC

′′
= α−, AC′′B = p sin β

−+, AC′′ = p sin γ− where p = 8R sin γ sin β−. As

α− + β−+ + γ− = 180◦, C′′BC
′′ = p sin α− = 8R sin γ sin β− sin α−, ∠AC′′BC

′′ = γ−, ∠AC′′C′′B = β
−+. Simi-

larly in ∆C′′ABC
′′, C′′AC

′′
= 8R sin γ sin β− sin α− and ∠BC′′AC

′′
= γ−, ∠BC′′C′′A = α

−+. Also in ∆C′′ACC
′′

B,

∠C′′ACC
′′

B = γ, CC
′′

B = p sin β
+, CC′′A = p sin α

+, where p = 8R sin α− sin β−. As γ + β+ + α+ = 180◦,

C′′AC
′′

B = p sin γ = 8R sin α
− sin β− sin γ and ∠CC′′AC

′′

B = β
+, ∠CC′′BC

′′

A = α
+. We conclude that ∆CC′′AC

′′

B

is equilateral. �

Corollary 2b. A mix central and the central triangle have two collinear sides and the third parallel.

Proof. Consider for instance the mix central ∆C′′C′′AC
′′

B. It was seen ∠AC
′′C′′B = β

−+. Also, it was

shown that in the central ∆A′′B′′C′′, ∠AC′′B′′ = β. As it is equilateral, ∠AC′′C′′B + ∠AC
′′B′′ +

∠B′′C′′A′′ = 180◦. So C′′BC
′′ and C′′A′′ are collinear.

c. The intersections of the exterior trisectors of an angle and the explementary trisectors of the

other two, proximal to sides respectively, define three triangles, referred as mix peripherals since

each of them shares a vertex with the peripheral triangle. We will study only one representative

from them. Consider the exterior trisectors of ∠B and the explementary trisectors of ∠C and ∠A.



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A Trigonometrical Approach to Morley’s Observation 81

Assume that the explementary trisectors proximal to AC meet at B′′′ while the explementary and

the exterior proximal to AB and BC meet at C′A and A
′

C defining ∆B
′′′C′AA

′

C, called B-peripheral tri-

angle as it shares vertex B′′′ with the peripheral. The angles between the trisectors that determine

it and the corresponding sides are equal to β− (exterior) and α+, γ+ (explementary).

The lengths of trisectors that determine ∆B′′′C′AA
′

C are:

In ∆BC′AA, ∠ABC
′

A = β
−, ∠BAC′ = 180◦ − α+ = α−+ and so ∠BC′AA = γ

−,

AC′A/ sin β
−
= BC′A/ sin α

−+
= AB/ sin γ− = 2R sin 3γ/ sin γ− = 8R sin γ sin γ+;

thus AC′A = 8R sin γ sin γ
+ sin β− and BC′A = 8R sin γ sin γ

+ sin α−+.

Similarly, in ∆CA′CB, CA
′

C = 8R sin α sin α
+ sin β− and BA′C = 8R sin α sin α

+ sin γ−+.

Also in ∆AB′′′C, as ∠AB′′′C = 180◦ − α+ − γ+ = β,

AB′′′/ sin γ+ = CB′′′/ sin α+ = AC/ sin β = 2R sin 3β/ sin β = 8R sin β+ sin β−;

thus AB′′′ = 8R sin β+ sin β− sin γ+ and CB′′′ = 8R sin β+ sin β− sin α+.

Then in ∆C′AAB
′′′, ∠C′AAB

′′′
= 180◦ − 2α+ + 3α = α+, AC′A = p sin γ,

AB′′′ = p sin β+ where p = 8R sin β− sin γ+. As α+ + γ + β+ = 180◦,

B′′′C′A = p sin α
+

= 8R sin β− sin γ+ sin α+,∠AB′′′C′A = γ, ∠AC
′

AB
′′′

= β+. Similarly, in ∆CB′′′A′C,

B′′′A′C = 8R sin β
− sin α+ sin γ+, ∠CB′′′A′C = α, ∠CA

′

CB
′′′

= β+.

Also, in ∆A′CBC
′

A, as ∠A
′

CBC
′

A = 2β
−
+ 3β = β++, BC′A = p sin γ, BA

′

C = p sin α, where p =

8R sin γ+ sin α+. As β++ + α + γ = 180◦,

C′AA
′

C = p sin β
++ = 8R sin γ+ sin α+ sin β++ while ∠BA′CC

′

A = γ, ∠BC
′

AA
′

C = α. Since sin β
++ = sin β−,

we conclude that ∆B′′′C′AA
′

C is equilateral.

Corollary 2c. A mix peripheral and the peripheral triangle have two collinear sides and the third

parallel.

Proof. Consider for instance the mix peripheral ∆B′′′C′AA
′

C. It was seen ∠AB
′′′C′A = γ. Also in

the peripheral equilateral, it was shown ∠AB′′′C′′′ = γ−. Hence ∠AB′′′A′′′ = γ. Thus, B′′′C′A and

B′′′C′′′ are collinear.

III. Complete Morley triangles (6)

The complete Morley triangles are formed by the intersections of the interior, exterior and exple-

mentary trisectors from each angle, proximal to sides respectively. Apparently, there are 3x2 such

triangles. For example, the interior trisectors of ∠C combined with the explementary trisectors

proximal to CB or CA form two different complete triangles.

Theorem 0.3. In a triangle, the trisectors of a distinct type from each angle, proximal to sides

respectively, meet at the vertices of an equilateral.

Proof. From the 6 triangles formed by the intersections of the interior, exterior and explementary

trisectors from each angle, proximal to sides, we will study only a representative as the rest are



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Figure 5: Fig.5: The BA-complete Morley triangle

similar. Consider for instance the interior trisectors of ∠C, the exterior trisectors of ∠A and the

explementary trisectors of ∠B, proximal to sides respectively. Assume that the interior and the

explementary proximal to CB meet at BA, the interior with the exterior proximal to CA meet at C
′′

B

and the exterior with the explementary meet at C′B defining ∆BAC
′′

BC
′

B called BA-complete triangle

as it shares vertex BA with the B-inner triangle. It also shares vertex C
′′

B with the C-central, and

vertex C′B with the A-peripheral triangle. The angles between the trisectors that determine it and

the corresponding sides are equal to α−(exterior), β+ (explementary) and γ (interior).

The lengths of the trisectors that determine it are:

In ∆AC′BB, ∠BAC
′

B = α
−, ∠C′BBA = 180

◦
− β+ = β−+ and so ∠AC′BB = γ

−. Hence

AC′B/ sin β
−+ = C′B/ sin α

− = AB/ sin γ− = 2R sin 3γ/ sin γ− = 8R sin γ sin γ+;

thus AC′B = 8R sin γ sin γ
+ sin β−+ and BC′B = 8R sin γ sin γ

+ sin α−.

Similarly, in ∆BBAC, as ∠BABC = β
+, ∠BBAC = α

+,

CBA/ sin β
+ = BBA/ sin γ = BC/ sin α

+ = 2R sin 3α/ sin α+ = 8R sin α sin α−;

thus CBA = 8R sin α sin α
− sin β+ and BBA = 8R sin α sin α

− sin γ.

Also in ∆CC′′BA, as ∠ACC
′′

B = γ and ∠C
′′

BAC = 2α
− + 3α = α++, ∠AC′′BC = β,

CC′′B/ sin α
++

= AC′′B/ sin γ = AC/ sin β = 2R sin 3β/ sin β = 8R sin β
− sin β+;

thus CC′′B = 8R sin β
− sin β+ sin α++ and AC′′B = 8R sin β

− sin β+ sin γ.

Then, in ∆C′′BAC
′

B, ∠C
′′

BAC
′

B = α
−, AC′′B = p sin β

−, AC′B = p sin γ
+

= p sin γ−+ where p =

8R sin γ sin β−+. As α− + β− + γ−+ = 180◦,

C′′BC
′

B = p sin α
−
= 8R sin γ sin β−+ sin α− while ∠AC′BC

′′

B = β
−, ∠AC′′BC

′

B = γ
−+.



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A Trigonometrical Approach to Morley’s Observation 83

Also in ∆C′BBBA, ∠C
′

BBBA = 180
◦ − 2β+ + 3β = β+, BC′B = p sin γ

+, BBA = p sin α where p =

8R sin γ sin α−. As β+ + γ+ + α = 180◦,

C′BBA = p sin β
+
= 8R sin γ sin α− sin β+ while ∠BBAC

′

B = γ
+ and ∠BC′BBA = α.

Furthermore, in ∆BACC
′′

B, ∠BACC
′′

B = γ, CC
′′

B = p sin β
−

= p sin β++, CBA = p sin α, where p =

8R sin α++ sin β+. As γ + β++ + α = 180◦,

C′′BAC = p sin γ = 8R sin α
++ sin β+ sin γ while ∠CACC

′′

B = 120
◦ + β = β++ and

∠CC′′BBA = α. We conclude that ∆BAC
′′

BC
′

B is equilateral.

Corollary 3. A complete and a mix inner, or a mix central or a mix peripheral triangle, sharing

a vertex, have two collinear sides and the third parallel.

Proof. Consider for instance the complete ∆BAC
′′

BC
′

B. This shares a vertex with the mix inner

∆B′BABC, the mix central ∆C
′′C′′AC

′′

B, and the mix peripheral ∆A
′′′B′CC

′

B.

It was shown ∠BBAC
′

B = γ
+. However, in the B-inner ∆B′BABC, ∠BBABC = γ

−. Since it is

equilateral, ∠BCBAB
′
= 60◦. Then ∠C′BBAB

′
= ∠C′BBAB+∠BBABC +∠BCBAB

′
= γ+ +γ− +60◦ = 180◦

and so C′BBA and BAB
′ are collinear. Hence, ∆BAC

′′

BC
′

B and ∆B
′BABC have two collinear sides with

the third ones parallel.

It was shown ∠AC′′BC
′

B = γ
−+. As it is equilateral ∠AC′′BBA = γ

−. However, in the C-central,

∆C′′C′′AC
′′

B ∠AC
′′

BC
′′
= γ−. Thus, C′′BC

′′ and C′′BBA are collinear. Hence, ∆BAC
′′

BC
′

B and ∆C
′′C′′AC

′′

B

have two collinear sides with the third parallel.

It was shown ∠AC′BC
′′

B = β
−. Since is ∆BAC

′′

BC
′

B equilateral, ∠AC
′

BBA = β. However, in the

B-peripheral, ∠BC′AA
′

C = α. Symmetrically, in the A-peripheral ∆A
′′′B′CC

′

B, ∠AC
′

BB
′

C = β. Hence

∠AC′BBA = ∠AC
′

BB
′

C and so C
′

BBA and C
′

BB
′

C are collinear. Thus ∆BAC
′′

BC
′

B and ∆A
′′′B′CC

′

B have two

collinear sides with the third parallel.

From the above we infer that a precise formulation of the general Morley’s theorem yielding 18

equilaterals is: In a triangle, the same type trisectors of the three angles, the same type trisectors

of an angle with its mixable type trisectors of the other two, and the trisectors of a distinct type

from each angle, proximal to sides respectively, meet at the vertices of an equilateral.

3. Arrangement of Morley triangles and alignment of inter-

sections of proximal trisectors

The trisectors of a triangle, proximal to one of its sides, meet at 27 points. Each of them is a

common vertex of two Morley triangles which are arranged with two sides collinear and the third

parallel.

From corollary 3, sides C′′BBA and C
′′

BC
′

B of the complete ∆BAC
′′

BC
′

B and the mix central

∆C′′C′′BC
′′

A are collinear. Hence, C
′′

B and BA lie on the line determined by the side C
′′A′′ of the



84 Ioannis Gasteratos, Spiridon Kuruklis & Thedore Kuruklis CUBO
19, 2 (2017)

central triangle ∆A′′B′′C′′. Thus, on the line determined by a side of the central triangle lie 6

intersections of trisectors proximal to a side, two of the interior with the exterior trisectors, two of

the interior with the explementary trisectors and two between the exterior trisectors of the central

triangle.

Also, sides BAC
′

B and BAB of the complete ∆BAC
′′

BC
′

B and the mix inner are collinear. Hence,

C′′B and BA lie on the line which is determined by a side of the inner triangle. Thus, on the line

determined by a side of the inner triangle lie 6 intersections of trisectors proximal to a side, two of

the exterior with the explementary trisectors, two of interior with the explementary trisectors and

two intersections between interior trisectors of the inner triangle.

Finally, sides C′BC
′′

B and C
′

BA
′′′ of the complete ∆BAC

′′

BC
′

B and the mix peripheral triangle

∆A′′′B′CC
′

B are collinear. As the A-peripheral and the peripheral have sides A
′′′C′B and A

′′′B′′′

collinear, C′′B and C
′

B lie on side A
′′′B′′′. Symmetrically C′′B and C

′

B lie on side A
′′′B′′′ as well. Thus,

on a side of the peripheral triangle lie 6 intersections of trisectors proximal to a side, two of the

interior with the exterior trisectors, two of the explementary with the exterior trisectors and two

between the explementary trisectors of the peripheral triangle.

Conclude that the intersections of trisectors proximal to a side lie 6 by 6 on three triples of

parallel lines intersecting with 60◦ angles.

4. Open Problems

Next there are three basic questions stemming out from this work inviting further exploration.

1. How many equilaterals do the intersections of trisectors in a triangle determine?

2. Are there lines or circles, beyond Morley’s 3 triples of parallel lines, on which the intersections

of trisectors lie?

3. Do theorems exist, analog to Morley’s theorem regarding angle trisectors, for the side or per-

pendicular trisectors?

Very special thanks to Gerry Ladas and Thanasis Fokas for their unceasing encouragement to our

long efforts in demystifying Morley’s mystery.

References

[1] Paul and Jack Abad, The Hundred Greatest Theorems, in Actes de la confrence JFLA2008.

[2] B. Bollobás, The Art of Mathematics, Cambridge University Press, (2006), 126-127.

[3] Alain Connes, A new proof of Morley’s theorem, Publications Mathmatiques de l’ IHÉS, (1998),

43-46.



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19, 2 (2017)

A Trigonometrical Approach to Morley’s Observation 85

[4] John Conway, The power of mathematics, In Power, Cambridge University Press, (2006), 36-50.

[5] John Conway, On Morley’s Trisector Theorem, The Mathematical Intelligencer, 36, no. 3(2014),

3-3

[6] Edsger W. Dijkstra, A Collection of Beautiful Proofs, Selected Writings on Computing: A

personal Perspective, Springer, (1982), 174-183.

[7] Edsger W. Dijkstra, On the design of a simple proof for Morley’s Theorem, Programming and

Mathematical Method, Volume 88 of the NATO ASI Series, Springer (1992) 3-9.

[8] W. J. Dobbs, Morley’s triangle, The Mathematical Gazette, 1938 JSTOR.

[9] Richard K. Guy, The Lighthouse Theorem A Budget of Paradoxes, Amer. Math. Monthly, 114

(2007), 97-141

[10] Clark Kimberling, Central points and central lines in the plane of a triangle, Mathematics

Magazine, 67 (1994), 163-187.

[11] Spiridon A. Kuruklis, Trisectors like Bisectors with Equilaterals instead of Points, CUBO 16,

2 (2014), 71-110.

[12] Henri Lebesgue, Sur les n-sectrices d’un triangle [En memoire de Frank Morley (1860-1930)],

Enseingnement Math., 38 (1940), 39-58.

[13] Frank Morley, Extensions of Clifford’s Chain-Theorem, American J. of Math., 51 (1929), 465-

472.

[14] Roger Penrose, Morley’s trisector theorem, Eureka The Archimedeans Journal, Cambridge,

16 (1953), 6-7.

[15] David Wells, Are These The Most Beautiful? The Mathematical Intelligencer, 12 (1990),

37-41.

[16] Glanville F. Taylor and L.W. Marr, The six trisectors of each of the angles of a triangle,

Proceedings of the Edinburgh Mathematical Society 33 (1913-14), 119-131.

[17] Alexander Bogomolny, Morley’s Miracle, Interactive Mathematics Miscellany and Puzzles,

http://www.cut-the-knot.org/triangle/Morley/