NOTES ON MORLEY’S THEOREM

MARKUS ROST

Contents

Introduction 1 1. Affine transformations and triangles 2 2. Affine transformations and quadrangles 3 3. Proof of Morley’s theorem 3 3.1. The Euclidean case 4 3.2. Trisectors 4 3.3. The geometric mean 4 3.4. Morley points 5 4. The incenter 6 4.1. The Euclidean case 6 4.2. Bisectors 6 4.3. Incenters 6 5. A group theoretic lemma 7 More sources 9 References 9

Introduction

Morley’s theorem states that The three points of intersection of the adjacent trisectors of the angles of any

triangle form an equilateral triangle.

This theorem is very curious. A standard source seems to be [4]. Among the many existing proofs we mention here D. J. Newman’s proof [10] (also in [6, Ch. 20, p. 163] and on the web) and the article [9]. For more sources see the end of the text.

These notes evolved from a study of the fairly recent proof of Connes ([2]; see also [7], [1]).

We briefly discuss the relation of Connes’ point of view of affine transformations with triangles and quadrangles. Then we give a proof of Morley’s theorem a la Connes [2]. Finally we consider a purely group theoretic lemma (Lemma 4) which implies Connes’ lemma on affine transformations.

In the context of Morley’s angle trisector theorem we found it useful to look also—as a toy model—at the fact that the angle bisectors of any triangle meet in one point. We call this for short the incenter theorem since the point of intersection is the center of the incircle of the triangle. We have complemented many considerations with the corresponding incenter variants.

Date: August 26, 2003.

Were we to give up, forever, understanding the Morley Miracle?

— D. J. Newman

1. Affine transformations and triangles

Let F be a field and let Aff(1, F) denote the group of affine transformations of the affine line over F. Elements f ∈ Aff(1, F) will be denoted by

f(t) = at + b  or  f = (\begin{matrix}a & b \\ 0 & 1 \end{matrix})

If a = 1, then f is a translation. Otherwise f has the unique fixed point

Fix(f) = \frac{b}{1 − a}

Note that if a ≠ 1, but a^{n} = 1 for some n > 1, then f^{n} = 1. Moreover, if f, g ∈ Aff(1, F) commute, then f and g are both translations or have a common fixed point.

Lemma 1. Let f_{0}, f_{1}, f_{2} ∈ Aff(1, F). Suppose that the f_{i} have no common fixed point and that none of them is a translation. Let x_{i} = Fix(f_{i}).

Then f_{0}f_{1}f_{2} = 1 if and only if there exists c ∈ F such that

(1)  f_{i}(t) = \frac{x_{i−1} − c}{x_{i+1} − c}(t − x_{i}) + x_{i}

(with the indices reduced mod 3). The element c is uniquely determined by f_{0}, f_{1}, f_{2}.

Proof. Let d_{i} = det(f_{i}). We may assume x_{0} = 0 and x_{1} = 1. The condition f_{0}f_{1}f_{2} = 1 is equivalent to the conditions

d_{0}d_{1}d_{2} = 1, x_{2} = \frac{d_{0}(1 − d_{1})}{(1 − d_{0}d_{1})} = \frac{1 − d_{0}d_{2}}{1 − d_{2}}

Consider the change of variables

c = \frac{1}{1 − d_{2}} , d_{2} = \frac{1 − c}{−c}

Then our conditions give indeed

d_{0} = \frac{1 − (1 − d_{2})x_{2}}{d_{2}} = \frac{x_{2} − c}{1 − c}, d_{1} = \frac{1}{d_{0}d_{2}} = \frac{−c}{x_{2} − c}

□

Example 1. Consider an Euclidean triangle with vertices x_{0}, x_{1}, x_{2} ∈ C and let c be its circumcenter. Then the affine transformation f_{i} given by (1) is the rotation with fixed point x_{i} and angle twice the angle at x_{i} (with appropriate orientation) of the triangle.

This way Euclidean triangles appear as a special case of triples f_{0}, f_{1}, f_{2} ∈ Aff(1, F) with f_{0}f_{1}f_{2} = 1. This is the view point of Connes in his proof of Morley’s theorem. We state Connes’ generalization of Morley’s theorem [2]:

Lemma 2 (Connes). Let t_{0}, t_{1}, t_{2} ∈ Aff(1, F). Suppose none of t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0}, t_{0}t_{1}t_{2} is a translation and that t_{0}^{3}t_{1}^{3}t_{2}^{3} = 1. Let ζ = det(t_{0}t_{1}t_{2}).

Then 1 + ζ + ζ^{2} = 0 and

Fix(t_{0}t_{1}) + ζ Fix(t_{1}t_{2}) + ζ^{2} Fix(t_{2}t_{0}) = 0

The incenter theorem generalizes as follows:

Lemma 3. Let t_{0}, t_{1}, t_{2} ∈ Aff(1, F). Suppose none of t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0}, t_{0}t_{1}t_{2} is a translation and that t_{0}^{2}t_{1}^{2}t_{2}^{2} = 1.

Then the transformations t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0} commute. In particular, their fixed points coincide.

These lemmata will proved in Section 5.

2. Affine transformations and quadrangles

This section will not be used later on. We assume char F ≠ 2.

For (generic) points x_{0}, x_{1}, x_{2}, x_{3} ∈ F consider the affine transformations

(2)  f_{ijkℓ} = \frac{(x_{i} + x_{j}) − (x_{k} + x_{ℓ})}{(x_{i} + x_{k}) − (x_{j} + x_{ℓ})}(t − x_{i}) + x_{i}

where i, j, k, ℓ stand for any permutation of 0, 1, 2, 3.

If one takes in (1)

c = \frac{x_{0} + x_{1} + x_{2} − x_{3}}{2}

one finds

f_{i} = f_{i,i+1,i−1,3}

This way Lemma 1 shows that triples f_{0}, f_{1}, f_{2} ∈ Aff(1, F) with det(f_{i}) ≠ 1 and f_{0}f_{1}f_{2} = 1 (and no common fixed point) are in characteristic different from 2 essentially just quadruples of points in F. Thus the symmetric group S_{4} is a group of symmetries of such triples of affine transformations (this is true also in characteristic 2, and more generally over any commutative ring F).

Example 2. Let x_{0}, x_{1}, x_{2}, x_{3} ∈ C be an Euclidean orthocentric quadrangle. This means that all pairs x_{i} − x_{j}, x_{k} − x_{ℓ} are orthogonal, or, equivalently, that (at least) one of the x_{i} is the orthocenter of the triangle formed by the other points x_{j}, x_{k}, x_{ℓ}.

Let c be the circumcenter and let h = x_{3} be the orthocenter of the triangle x_{0}, x_{1}, x_{2}. Then

2c + h = x_{0} + x_{1} + x_{2}

(In fact, c, h and the center of mass (x_{0} + x_{1} + x_{2})/3 lie on the Euler line of the triangle.)

It follows that the affine transformation f_{ijkℓ} is the rotation with fixed point x_{i} and angle twice the angle at x_{i} (with appropriate orientation) of the triangle x_{i}, x_{j}, x_{k}.

3. Proof of Morley’s theorem

Let F be an algebraically closed field with char F ≠ 3 and let ζ ∈ F be a primitive cube root of 1.

Let x_{0}, x_{1}, x_{2} ∈ F. In the following the letters i, j, k stand for any permutation of 0, 1, 2. We assume that x_{i} ≠ 0 and x_{i} ≠ x_{j}.

Choose s_{ij} ∈ F^{∗} such that

s_{ij}^{3} = \frac{x_{j}}{x_{i}} ,  s_{ij}s_{ji} = 1,  s_{01}s_{12}s_{20} = ζ

It is easy to see that such families s_{ij} exist and that any such family is determined by s_{01}, s_{12}. Moreover, there are exactly 9 such families which one can get by multiplying s_{01}, s_{12} by powers of ζ.

We write

ζ_{ijk} = s_{ij}s_{jk}s_{ki}

Thus ζ_{ijk} = ζ_{jki}, ζ_{ijk} = ζ_{ikj}^{−1} and ζ_{012} = ζ.

3.1. The Euclidean case. As for the proof of Morley’s theorem we use the following setup.

One takes F = C, ζ = e^{2πi/3} and assumes that the circumcenter of the triangle x_{0}, x_{1}, x_{2} is the origin. In other words, |x_{0}| = |x_{1}| = |x_{2}| where | · | is the Euclidean norm. Moreover one assumes that the triangle is positively oriented.

Then

\frac{x_{2}}{x_{1}} ∈ S^{1} = \{ s ∈ C | |s| = 1 \}

is twice the angle of the triangle at x_{0}. We choose the unique family s_{ij} with

arg s_{ij} = \frac{1}{3} arg \frac{x_{j}}{x_{i}}

where 0 ≤ arg s < 2π is defined for s ∈ S^{1} by s = e^{i arg s}.

3.2. Trisectors. Consider the 6 elements

y_{ij} =^{def} s_{ij}x_{i} = s_{ji}^{2}x_{j} = ζ_{ijk}s_{kj}s_{ki}^{2}x_{k}

In the Euclidean case, the elements y_{ij} are points of the circumcircle. They trisect each of the arcs between the points x_{i}.

One has

y_{ij}^{3} = x_{i}^{2}x_{j}, y_{ij}y_{jk}y_{ki} = ζ_{ijk}x_{0}x_{1}x_{2}

3.3. The geometric mean. Consider the 3 elements

z_{i} = s_{ij}s_{ik}x_{i}

One has

z_{i}^{3} = z_{0}z_{1}z_{2} = x_{0}x_{1}x_{2}

Moreover

z_{i+1} = ζz_{i}

which can be seen for instance from

\frac{z_{1}}{z_{0}} = \frac{s_{12}s_{10}x_{1}}{s_{01}s_{02}x_{0}} = s_{12}s_{20}s_{01}s_{10}^{3} \frac{x_{1}}{x_{0}} = ζ

Hence

z_{0} + ζz_{1} + ζ^{−1}z_{2} = 0

In the Euclidean case, the elements z_{i} are points of the circumcenter and form an equilateral triangle.

3.4. Morley points. Let g_{jk} be the affine transformation with

g_{jk}(t) = s_{jk}(t − x_{i}) + x_{i}

We define the Morley points m_{i} by

g_{ij}(m_{i}) = g_{ik}(m_{i})

In the Euclidean case, the transformation g_{jk} is the rotation with center x_{i} and angle s_{jk}. Moreover m_{i} is the fixed point of g_{ik}^{−1} ∘ g_{ij} which can be easily seen as one of the “intersections of the adjacent trisectors” in Morley’s theorem. This description is due to Connes [2].

Let us compute m_{i}. The defining equation is

s_{ij}(m_{i} − x_{k}) + x_{k} = s_{ik}(m_{i} − x_{j}) + x_{j}

This gives

(s_{ij} − s_{ik})m_{i} = (x_{j} − x_{k}) − (s_{ik}x_{j} − s_{ij}x_{k})
= (s_{ij}^{3} − s_{ik}^{3})x_{i} − (s_{ij}^{2} − s_{ik}^{2})s_{ij}s_{ik}x_{i}

Hence

m_{i} = (s_{ij}s_{ik} + s_{ij}^{2} + s_{ik}^{2})x_{i} − (s_{ij} + s_{ik})s_{ij}s_{ik}x_{i}
= z_{i} + y_{ji} + y_{ki} − ζ_{ijk}^{−1}y_{jk} − ζ_{ikj}^{−1}y_{kj}
= z_{i} + v_{ji} + v_{ki}

where

v_{ij} = y_{ij} − ζ_{ijk}^{−1}y_{ki}

Next note that

v_{ij} + ζ_{ijk}v_{jk} + ζ_{ijk}^{−1}v_{ki} = 0

Indeed, one has

(y_{10} − ζy_{21}) + ζ(y_{21} − ζy_{02}) + ζ^{−1}(y_{02} − ζy_{10}) = 0

and

(y_{20} − ζ^{−1}y_{12}) + ζ(y_{01} − ζ^{−1}y_{20}) + ζ^{−1}(y_{12} − ζ^{−1}y_{01}) = 0

since all terms cancel out.

Hence

m_{0} + ζm_{1} + ζ^{−1}m_{2} = 0

which is Morley’s theorem.

Remark 1. The only thing which might be new in this deduction is that the Morley triangle appears as a superposition of three terms, the triple z_{0}, z_{1}, z_{2}, the triple v_{10}, v_{21}, v_{02}, and triple v_{20}, v_{01}, v_{12}, each of which is subject by itself to the equilaterality relation X_{0} + ζX_{1} + ζ^{−1}X_{2} = 0:

m_{0} = z_{0} + (y_{10} − ζy_{21}) + (y_{20} − ζ^{−1}y_{12})
m_{1} = z_{1} + (y_{21} − ζy_{02}) + (y_{01} − ζ^{−1}y_{20})
m_{2} = z_{2} + (y_{02} − ζy_{10}) + (y_{12} − ζ^{−1}y_{01})

I don’t know a geometric or algebraic interpretation of this observation.

4. The incenter

Let F be an algebraically closed field with char F ≠ 2.

Let x_{0}, x_{1}, x_{2} ∈ F. In the following the letters i, j, k stand for any permutation of 0, 1, 2. We assume that x_{i} ≠ 0 and x_{i} ≠ x_{j}.

Choose s_{ij} ∈ F^{∗} such that

s_{ij}^{2} = \frac{x_{j}}{x_{i}} ,  s_{ij}s_{ji} = 1,  s_{01}s_{12}s_{20} = −1

It is easy to see that such families s_{ij} exist and that any such family is determined by s_{01}, s_{12}. Moreover, there are exactly 4 such families which one can get by multiplying s_{01}, s_{12} by powers of −1.

4.1. The Euclidean case. As for the classical fact that the angle bisectors of an Euclidean triangle meet in one point, the incenter, we use the following setup.

One takes F = C, and assumes that the circumcenter of the triangle x_{0}, x_{1}, x_{2} is the origin. In other words, |x_{0}| = |x_{1}| = |x_{2}| where | · | is the Euclidean norm. Moreover one assumes that the triangle is positively oriented.

Then

\frac{x_{2}}{x_{1}} ∈ S^{1} = \{ s ∈ C | |s| = 1 \}

is twice the angle of the triangle at x_{0}. We choose the unique family s_{ij} with

arg s_{ij} = \frac{1}{2} arg \frac{x_{j}}{x_{i}}

where 0 ≤ arg s < 2π is defined for s ∈ S^{1} by s = e^{i arg s}.

4.2. Bisectors. Consider the 3 elements

y_{ij} =^{def} s_{ij}x_{i} = s_{ji}x_{j} = −s_{kj}s_{ki}x_{k}

One has y_{ij} = y_{ji}.

In the Euclidean case, the elements y_{ij} are points of the circumcircle. They bisect each of the arcs between the points x_{i}.

We also write

z_{i} = y_{jk} = −s_{ij}s_{ik}x_{i}

One has

z_{i}^{2} = x_{j}x_{k}, z_{0}z_{1}z_{2} = −x_{0}x_{1}x_{2}

4.3. Incenters. We write

z = z_{0} + z_{1} + z_{2}

Let g_{jk} be the affine transformation with

g_{jk}(t) = s_{jk}(t − x_{i}) + x_{i}

and let m_{i} be the fixed point of g_{ik}^{−1} ∘ g_{ij}.

In the Euclidean case, the transformation g_{jk} is the rotation with center x_{i} and angle s_{jk}. The fixed point m_{i} is therefore the intersection of the bisectors of the angles at x_{j} and x_{k}. Thus m_{1} = m_{2} = m_{3} is the incenter of the triangle x_{0}, x_{1}, x_{2}.

In general we have

(3)  g_{ij}(z) = g_{ik}(z)

so that m_{1} = m_{2} = m_{3} = z.

Proof of (3). Let us compute m_{i}. The defining equation is

s_{ij}(m_{i} − x_{k}) + x_{k} = s_{ik}(m_{i} − x_{j}) + x_{j}

This gives

(s_{ij} − s_{ik})m_{i} = (x_{j} − x_{k}) − (s_{ik}x_{j} − s_{ij}x_{k})
= (s_{ij}^{2} − s_{ik}^{2})x_{i} − (s_{ij} − s_{ik})s_{ij}s_{ik}x_{i}

Hence

m_{i} = (s_{ij} + s_{ik} − s_{ij}s_{ik})x_{i} = z_{k} + z_{j} + z_{i}
□

One can set up things also this way: Choose a, u_{1}, u_{2}, u_{3} with

x_{i} = au_{i}^{2}

Then one can take

s_{ij} = −\frac{u_{j}}{u_{i}} , z_{i} = −au_{i}u_{j}

and for the incenter one has

z = −a(u_{0}u_{1} + u_{1}u_{2} + u_{2}u_{0})

5. A group theoretic lemma

Lemma 4. Let t_{0}, t_{1}, t_{2} be elements of a group G.

Suppose that G is metabelian (i. e., [G, G] is abelian) and

(t_{0}t_{1}t_{2})^{3} = (t_{0}^{2}t_{1}^{2}t_{2}^{2})^{3} = t_{0}^{3}t_{1}^{3}t_{2}^{3} = 1

Then

(4)  [[t_{0}t_{1}, t_{1}t_{2}], t_{2}t_{0}] = (t_{0}t_{1}t_{2})[[t_{2}t_{0}, t_{0}t_{1}], t_{1}t_{2}](t_{0}t_{1}t_{2})^{−1}

Proof. We have to show

[[t_{0}t_{1}, t_{1}t_{2}], t_{2}t_{0}](t_{0}t_{1}t_{2}) =^{?} (t_{0}t_{1}t_{2})[[t_{2}t_{0}, t_{0}t_{1}], t_{1}t_{2}]

One multiplies out and collects appropriate terms.

(t_{0}t_{1}^{2}t_{2}t_{1}^{−1})[(t_{0}^{−1}t_{2}^{−1}t_{1}^{−1})(t_{2}t_{0}t_{1})(t_{2}t_{0}t_{1})(t_{2}^{−1}t_{1}^{−2}t_{0}^{−2}t_{2}^{−1})(t_{0}t_{1}t_{2})]

=^{?} (t_{0}t_{1}t_{2})(t_{2}t_{0}^{2}t_{1}t_{0}^{−1})(t_{2}^{−1}t_{1}^{−1}t_{0}^{−1})[(t_{1}t_{2}t_{0})(t_{1}t_{2}t_{0})(t_{1}^{−1}t_{0}^{−2}t_{2}^{−2}t_{1}^{−1})]

The terms in square brackets are commutators and therefore commute. Moreover (t_{2}t_{0}t_{1})^{3} = (t_{1}t_{2}t_{0})^{3} = 1. This yields

(t_{0}t_{1}^{2}t_{2}t_{1}^{−1})[(t_{1}t_{2}t_{0})^{−1}(t_{1}^{−1}t_{0}^{−2}t_{2}^{−2}t_{1}^{−1})]^{−1}

=^{?} (t_{0}t_{1}t_{2})(t_{2}t_{0}^{2}t_{1}t_{0}^{−1})(t_{2}^{−1}t_{1}^{−1}t_{0}^{−1})[(t_{0}^{−1}t_{2}^{−1}t_{1}^{−1})(t_{2}t_{0}t_{1})^{−1}(t_{2}^{−1}t_{1}^{−2}t_{0}^{−2}t_{2}^{−1})(t_{0}t_{1}t_{2})]^{−1}

Then, using (t_{0}t_{1}t_{2})^{3} = 1,

t_{0}t_{1}^{2}t_{2}^{3}t_{0}^{2}t_{1}(t_{1}t_{2}t_{0})

=^{?} (t_{0}t_{1}t_{2})(t_{2}t_{0}^{2}t_{1}t_{0}^{−1})(t_{0}t_{1}t_{2})(t_{2}^{−1}t_{1}^{−2}t_{0}^{−2}t_{2}^{−1})^{−1}(t_{2}t_{0}t_{1})(t_{0}^{−1}t_{2}^{−1}t_{1}^{−1})^{−1}

Finally, using (t_{0}^{2}t_{1}^{2}t_{2}^{2})^{3} = 1,

t_{0}t_{1}^{2}t_{2}^{3}t_{0}^{2}t_{1} =^{?} (t_{0}t_{1}t_{2})(t_{2}t_{0}^{2}t_{1}t_{0}^{−1})(t_{0}t_{1}t_{2})(t_{2}t_{0}^{2}t_{1}^{2}t_{2})(t_{2}t_{0}t_{1}) =

= (t_{0}t_{1}t_{2}^{2})(t_{0}^{2}t_{1}^{2}t_{2}^{2})^{2}(t_{0}t_{1}) = (t_{0}t_{1}t_{2}^{2})(t_{0}^{2}t_{1}^{2}t_{2}^{2})^{−1}(t_{0}t_{1}) = t_{0}t_{1}^{−1}t_{0}^{−1}t_{1} This amounts to t_{0}^{3}t_{1}^{3}t_{2}^{3} = 1. □

Corollary 1. Let t_{0}, t_{1}, t_{2} be elements of Aff(1, F) with t_{0}^{3}t_{1}^{3}t_{2}^{3} = 1. Suppose d_{0}d_{1}d_{2} ≠ 1 where d_{i} = det(t_{i}).

Then (4) holds.

Proof. One uses the fact that any affine transformation whose determinant is a primitive n-th root of unity has order n itself (n > 1).

Let t = t_{0}t_{1}t_{2} and d = det(t) = d_{0}d_{1}d_{2}. Then d^{3} = d_{0}^{3}d_{1}^{3}d_{2}^{3} = 1 and d ≠ 1. Therefore d^{2} + d + 1 = 0. Thus t^{3} = 1. Similarly one finds (t_{0}^{2}t_{1}^{2}t_{2}^{2})^{3} = 1. By Lemma 4 the claim is clear. □

Formula (4) translates apparently the cyclic permutation t_{i} ↦ t_{i+1} into multiplication with a cube root of unity.

Proof of Lemma 2. One finds that

[[t_{0}t_{1}, t_{1}t_{2}], t_{2}t_{0}]

is the translation with vector

(∏_{0}^{2}(1 − d_{i}d_{i+1})) (Fix(t_{0}t_{1}) − Fix(t_{1}t_{2}))

where d_{i} = det(t_{i}). By (4) one gets

(Fix(t_{0}t_{1}) − Fix(t_{1}t_{2})) = d(Fix(t_{2}t_{0}) − Fix(t_{0}t_{1}))

with d^{2} + d + 1 = 0. The claim is now clear. □

Corollary 2. Morley’s theorem.

Proof. [Connes, [2]] For an Euclidean triangle with vertices x_{0}, x_{1}, x_{2} ∈ C one takes for t_{i} the rotation with fixed point x_{i} and angle 2/3 the angle at x_{i} (with appropriate orientation) of the triangle. Then indeed t_{0}^{3}t_{1}^{3}t_{2}^{3} = 1 and the fixed points Fix(t_{i}t_{i+1}) are the intersections of the trisectors in Morley’s theorem. □

Remark 2. Lemma 4 suggests to consider the group Ĝ generated by elements t and σ with relations

σ^{3} = 1, (tσ)^{9} = (t^{2}σ)^{9} = (t^{3}σ)^{3} = 1

and some commutation relations. Indeed if we put t_{i} = σ^{i}tσ^{−i}, then (tσ)^{3} = t_{0}t_{1}t_{2} etc.

However I don’t know whether this really helps. Anyway, let us note the following general formulas for elements x and σ in a group with relation σ^{3} = 1:

[σxσ^{−1}, σ^{2}xσ^{−2}] = (σx)^{3}(x^{−1}σ)^{3}

and

[x, [σxσ^{−1}, σ^{2}xσ^{−2}]] = (xσ)^{3}(σx^{−1})^{3}(σ^{−1}x)^{3}(x^{−1}σ^{−1})^{3}

Remark 3. Let s_{ij} = t_{i}t_{j}. In the situation of Lemma 4 the elements t_{i} are in the subgroup generated by the s_{12}, s_{20}, s_{01}. Maybe one can simplify things by using the s_{ij} as generators. Similarly for Lemma 5.

We conclude with similar (and much simpler) considerations for the incenter theorem.

Lemma 5. Let G be a group and let t_{0}, t_{1}, t_{2} be elements of G with t_{0}^{2}t_{1}^{2}t_{2}^{2} = 1 and (t_{0}t_{1}t_{2})^{2} = 1. Then the elements t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0} commute.

Proof. By symmetry, it suffices to show that t_{0}t_{1} and t_{2}t_{0} commute. Indeed,

(t_{0}t_{1})(t_{2}t_{0})(t_{0}t_{1})^{−1}(t_{2}t_{0})^{−1} = (t_{0}t_{1}t_{2})t_{0}(t_{1}^{−1}t_{0}^{−1})(t_{0}^{−1}t_{2}^{−1})
= (t_{0}t_{1}t_{2})^{−1}t_{0}t_{1}^{−1}t_{0}^{−2}t_{2}^{−1}
= t_{2}^{−1}t_{1}^{−2}t_{0}^{−2}t_{2}^{−1} = t_{2}(t_{0}^{2}t_{1}^{2}t_{2}^{2})^{−1}t_{2}^{−1} = 1
□

Corollary 3. Let t_{0}, t_{1}, t_{2} be elements of Aff(1, F) with t_{0}^{2}t_{1}^{2}t_{2}^{2} = 1. Suppose d_{0}d_{1}d_{2} ≠ 1 where d_{i} = det(t_{i}). Then t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0} have the same fixed point.

Proof. Let t = t_{0}t_{1}t_{2} and d = det(t) = d_{0}d_{1}d_{2}. Then d^{2} = d_{0}^{2}d_{1}^{2}d_{2}^{2} = 1 and d ≠ 1. Therefore d + 1 = 0. Thus t^{2} = 1. By Lemma 5 the elements t_{0}t_{1}, t_{1}t_{2}, t_{2}t_{0} commute. Hence their fixed points coincide. □

Corollary 4. The bisectors of the angles of a triangle meet in one point.

Proof. For an Euclidean triangle with vertices x_{0}, x_{1}, x_{2} ∈ C one takes for t_{i} the rotation with fixed point x_{i} and angle the angle at x_{i} (with appropriate orientation) of the triangle. Then indeed t_{0}^{2}t_{1}^{2}t_{2}^{2} = 1 and the fixed points Fix(t_{i}t_{i+1}) are the intersections of the bisectors. □

More sources

Here is a list of other possible sources for Morley’s theorem: [3, 5, 8, 11] and, of course, the web:

           http://www.google.com/search?q=morley+triangle
http://www-cabri.imag.fr/abracadabri/GeoPlane/Classiques/Morley/Morley1.htm
       http://www.cut-the-knot.org/triangle/Morley/index.shtml
         http://mathforum.org/library/drmath/view/51789.html

Under the last address one finds a proof of Conway.

References

[1] A. Connes, Symétries, Pour la Science, N^{∘} 292, 2002.

[2] , A new proof of Morley’s theorem, Les relations entre les mathématiques et la physique théorique, Inst. Hautes Études Sci., Bures, 1998, pp. 43–46, MR 99m:51027.

[3] J. H. Conway and R. K. Guy, The book of numbers, Copernicus, New York, 1996, MR 98g:00004.

[4] H. S. M. Coxeter, Introduction to geometry, Wiley Classics Library, John Wiley & Sons Inc., New York, 1989, Reprint of the 1969 edition, MR 90a:51001.

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[6] D. Gale, Tracking the automatic ant, Springer-Verlag, New York, 1998, And other mathematical explorations, A collection of Mathematical Entertainments columns from The Mathematical Intelligencer, MR 1 661 863.

[7] H. Geiges, Beweis des Satzes von Morley nach A. Connes, Elem. Math. 56 (2001), no. 4, 137–142, MR 2003b:51024.

[8] H. Lebesgue, Leçons sur les Constructions Géométriques, Gauthier-Villars, Paris, 1950, MR 11,678d.

[9] C. Lubin, A proof of Morley’s theorem, Amer. Math. Monthly 62 (1955), 110–112, MR 16,848c.

[10] D. J. Newman, The Morley Miracle, Math. Intelligencer 18 (1996), no. 1, 31–32.

[11] C. O. Oakley and J. C. Baker, The Morley trisector theorem, Amer. Math. Monthly 85 (1978), no. 9, 737–745, MR 80i:01012.

Department of Mathematics, The Ohio State University, 231 W 18th Avenue, Columbus, OH 43210, USA

   E-mail address: rost@math.ohio-state.edu
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