proof of Morley’s theorem

proof of Morley’s theorem

The scheme of this proof, due to A. Letac,
is to use the

sines law

to get formulas for the

segments

A

⁢

R

,

A

⁢

Q

,

B

⁢

P

,

B

⁢

R

,

C

⁢

Q

, and

C

⁢

P

,
and then to apply the

cosines law

to the

triangles

A

⁢

R

⁢

Q

,

B

⁢

P

⁢

R

, and

C

⁢

Q

⁢

P

,
getting

R

⁢

Q

,

P

⁢

R

, and

Q

⁢

P

.

To simplify some formulas, let us denote the angle

π

/

3

, or 60 degrees,
by

σ

.
Denote the angles at

A

,

B

, and

C

by

3

⁢

a

,

3

⁢

b

, and

3

⁢

c

respectively,
and let

R

be the

circumradius

of

A

⁢

B

⁢

C

. We have

B

⁢

C

=

2

⁢

R

⁢

sin

⁡

(

3

⁢

a

)

. Applying the sines law to the triangle

B

⁢

P

⁢

C

,

B

⁢

P

/

sin

⁡

(

c

)

=

B

⁢

C

/

sin

⁡

(

π

-

b

-

c

)

=

2

⁢

R

⁢

sin

⁡

(

3

⁢

a

)

/

sin

⁡

(

b

+

c

)

(1)

=

2

⁢

R

⁢

sin

⁡

(

3

⁢

a

)

/

sin

⁡

(

σ

-

a

)

(2)

so

B

⁢

P

=

2

⁢

R

⁢

sin

⁡

(

3

⁢

a

)

⁢

sin

⁡

(

c

)

/

sin

⁡

(

σ

-

a

)

.

Combining that with the

identity

sin

⁡

(

3

⁢

a

)

=

4

⁢

sin

⁡

(

a

)

⁢

sin

⁡

(

σ

+

a

)

⁢

sin

⁡

(

σ

-

a

)

we get

B

⁢

P

=

8

⁢

R

⁢

sin

⁡

(

a

)

⁢

sin

⁡

(

c

)

⁢

sin

⁡

(

σ

+

a

)

.

Similarly,

B

⁢

R

=

8

⁢

R

⁢

sin

⁡

(

c

)

⁢

sin

⁡

(

a

)

⁢

sin

⁡

(

σ

+

c

)

.

Using the cosines law now,

P

⁢

R

2

=

B

⁢

P

2

+

B

⁢

R

2

-

2

⁢

B

⁢

P

⋅

B

⁢

R

⁢

cos

⁡

(

b

)

=

64

⁢

sin

2

⁡

(

a

)

⁢

sin

2

⁡

(

c

)

⁢

[

sin

2

⁡

(

σ

+

a

)

+

sin

2

⁡

(

σ

+

c

)

-

2

⁢

sin

⁡

(

σ

+

a

)

⁢

sin

⁡

(

σ

+

c

)

⁢

cos

⁡

(

b

)

]

.

But we have

(

σ

+

a

)

+

(

σ

+

c

)

+

b

=

π

.

whence the cosines law can be applied to those three angles, getting

sin

2

⁡

(

b

)

=

sin

2

⁡

(

σ

+

a

)

+

sin

2

⁡

(

σ

+

c

)

-

2

⁢

sin

⁡

(

σ

+

a

)

⁢

sin

⁡

(

σ

+

c

)

⁢

cos

⁡

(

b

)

whence

P

⁢

R

=

8

⁢

R

⁢

sin

⁡

(

a

)

⁢

sin

⁡

(

b

)

⁢

sin

⁡

(

c

)

.

Since this

expression

is

symmetric

in

a

,

b

, and

c

, we deduce

P

⁢

R

=

R

⁢

Q

=

Q

⁢

P

as claimed.

Remarks:

It is not hard to show that the triangles

R

⁢

Y

⁢

P

,

P

⁢

Z

⁢

Q

,
and

Q

⁢

X

⁢

R

are isoscoles.

By the sines law we have

A

⁢

R

sin

⁡

b

=

B

⁢

R

sin

⁡

a

B

⁢

P

sin

⁡

c

=

C

⁢

P

sin

⁡

b

C

⁢

Q

sin

⁡

a

=

A

⁢

Q

sin

⁡

c

whence

A

⁢

R

⋅

B

⁢

P

⋅

C

⁢

Q

=

A

⁢

Q

⋅

B

⁢

R

⋅

C

⁢

P

.

This implies that if we identify the various vertices
with

complex numbers

, then

(

P

-

C

)

⁢

(

Q

-

A

)

⁢

(

R

-

B

)

=

-

1

+

i

⁢

3

2

⁢

(

P

-

B

)

⁢

(

Q

-

C

)

⁢

(

R

-

A

)

provided that the triangle

A

⁢

B

⁢

C

has

positive

orientation, i.e.

ℜ

⁡

(

C

-

A

B

-

A

)

>

0

.

I found Letac’s proof at

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

cut-the-knot.org,
with the reference

Sphinx

,

9

(1939) 46.
Several shorter and prettier proofs of Morley’s

theorem

can also be
seen at cut-the-knot.

Title

proof of Morley’s theorem

Canonical

name

ProofOfMorleysTheorem

Date of creation

2013-03-22 13:45:44

Last modified on

2013-03-22 13:45:44

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Proof

Classification

msc 51M04

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