# Theorem of Morley

The theorem of Morley , discovered by Frank Morley in 1898, is a Théorème of Géométrie. That is to say ABC a unspecified Triangle. One traces the Trissectrice S of his angles. Their intersections are cut to form a equilateral Triangle PQR .

## First Demonstration

As a Theorem, this discovery can be proven. Moreover, there are several ways of showing it, we will use here a simple method which uses the trigonometrical laws.

One can indeed determine, according to the Loi of the sines, the length of the majority of the segments starting from the sides of the triangle. In addition, the Théorème of Al-Kashi enables us to determine and compare the others, in particular QR, PR, and PQ - the three sides of the red triangle, that which is supposed equilateral being.

One defines the angles has, B and C such as:

• $\ widehat \left\{VAT\right\} = 3 \ times a$
• $\ widehat \left\{ABC\right\} = 3 \ times b$
• $\ widehat \left\{ACB\right\} = 3 \ times c$

Since in any triangle one a:

$\ widehat \left\{VAT\right\} + \ widehat \left\{ABC\right\} + \ widehat \left\{ACB\right\} = 180^ \ circ$
our change of variable above gives:
$a + B + C = 60^ \ circ$.

Moreover, to simplify calculations one adopts a unit such as the radius of the circle circumscribed with the triangle is 1. One has then:

• AB = 2 sin (3c)

• BC = 2 sin (3a)
• AC = 2 sin (3b).

In triangle BPC, according to the Law of the Sines, one a:

$\ frac \left\{BP\right\} \left\{\ sin \left(c\right)\right\} = \ frac \left\{BC\right\} \left\{\ sin \left(180^ \ circ - B - c\right)\right\} = \ frac \left\{2 \ sin \left(3a\right)\right\}\left\{\ sin \left(B + c\right)\right\} = \ frac \left\{2 \ sin \left(3a\right)\right\}\left\{\ sin \left(60^ \ circ - a\right)\right\}$
$BP = \ frac \left\{2 \ sin \left(3a\right) \ sin \left(c\right)\right\} \left\{\ sin \left(60^ \ circ - a\right)\right\}$

One can develop sin (3a):

• sin (3a) = 3sin (a) - 4sin3 (a)

• 3sin (a) - 4sin3 (a) = 4sin (a) - sin ² (A)
• 4sin (a) - sin ² (A) = 4sin (a) - sin ² (A)
• 4sin (a) - sin ² (A) = 4sin (a) + sin (a) - sin (a)
• 4sin (a) + sin (a) - sin (a) = 4sin (a) 2sin + a)/2cos - a)/2 × 2sin - a)/2cos + a)/2
• sin (a) 2sin + a)/2cos - a)/2 × 2sin - a)/2cos + a)/2 = 4sin (A) sin(60° + A) sin (60° - a)

What enables us to simplify the expression of BP:

$BP = 8 \ sin \left(a\right) \ sin \left(c\right) \ sin \left(60^ \ circ + a\right)$
But that one can also apply to Br:
$BR = 8 \ sin \left(a\right) \ sin \left(c\right) \ sin \left(60^ \ circ + c\right)$

Al-Kashi gives us: PR ² = BP ² + Br ² - 2BPBR cos (b). If one develops, one obtains:

$PR^2 = 64 \ sin^2 \left(A\right) \ sin^2 \left(c\right) \ sin^2 \left(60^ \ circ + a\right) + \ sin^2 \left(60^ \ circ + c\right) - 2 \ sin \left(60^ \ circ + a\right) \ sin \left(60^ \ circ + c\right) \ cos \left(b\right)$

However, (60° + a) + (60° + c) + B = 120° + (has + B + c) = 120° + 60° = 180 °. Among the triangles having for angle 60° + has, 60° + C and B whose ray of the circumscribed circle is 1, if Al-Kashi is applied, one a:

sin ² (b) = sin ² (60° + a) + sin ² (60° + c) - 2 sin (60° + a) sin (60° + c) cos (b)
PR = 8 sin (a) sin (b) sin (c)
PQ = 8 sin (b) sin (a) sin (c)
QR = 8 sin (a) sin (c) sin (b)
PR = PQ = QR

Triangle PQR is thus quite equilateral.

## Second Demonstration

This demonstration is based on an article of Alain Connes. It uses the complex numbers and gives a fast calculation of the Affixe of the tops of the equilateral triangle.

We in the directed Euclidean plan place which we will be able to later on identify with the body of the complexes. Let us indicate by P, Q and R the 3 intersections the trisecting ones which one wants to show that they form an equilateral triangle. Moreover let us place the P' points, Q' and R' symmetrical of P, Q and R respectively compared to BC, CA, AB (see figure opposite). Let us indicate finally respectively by $\ quad \ alpha, \ beta, \ gamma$ the principal determination (including between $\ quad - \ pi$ and $\ quad \ pi$) of the angles $\ widehat \left\{\left(\ overrightarrow \left\{AB\right\}, \ overrightarrow \left\{AC\right\}\right)\right\}, \; \ widehat \left\{\left(\ overrightarrow \left\{BC\right\}, \ overrightarrow \left\{BA\right\}\right)\right\}, \; \ widehat \left\{\left(\ overrightarrow \left\{CA\right\}, \ overrightarrow \left\{CB\right\}\right)\right\}$.

Are now $\; F, G, H \ quad$ rotations of respective centers has, B, C and of respective angles $2 \ alpha/3, \; 2 \ beta/3, \; 2 \ gamma/3$.

• (I)   P (resp. Q, R) is the fixed point of $G \ circ h$ (resp. $H \ circ F, \; F \ circ g$) .
Indeed $h$ transforms P into P' and $g$ transforms P' into P (immediate: to see figure). It is in a similar way for Q and R.

• (II) $f^3 \ circ g^3 \ circ h^3 = Id$ (identical application).
Indeed the sum of the angles of component rotations is $2 \ pi$ and one thus obtains a translation. But has is invariant since $h^3$ (rotation of center C and angle $2 \ gamma$) transforms has in symmetrical A' of has compared to BC, $g^3$ transforms A' into has and finally $f^3$ leaves has invariant. Consequently this translation is the identical application.

It is completely remarkable that the only proposals (I) and (II) above are sufficient to deduce the equilateral character from it from triangle PQR. It is even need to suppose that $f, \, G \, \, H$ are rotations but only that they are applications closely connected of the body of the complexes (identified in the plan) with the only restriction however that neither $f \ circ g$ neither $g \ circ h$ neither $h \ circ f$ nor $f \ circ G \ circ H$ne is translations .

Thus, we from now on will work in the body of the complexes by preserving the notations which we introduced.

We define simply $F, \, G, \, H$ (applications closely connected) by

$\ quad F \left(X\right) =a_1.x+b_1$
$\ quad G \left(X\right) =a_2.x+b_2$
$\ quad H \left(X\right) =a_3.x+b_3 \ qquad \left(a_1a_2, \; a_2a_3, \; a_3a_1, \; a_1a_2a_3$ different from 1).

A fast calculation shows that (I) is equivalent to

$\ quad P= \left(a_2b_3+b_2\right)/\left(1-a_2a_3\right)$
$\ quad Q= \left(a_3b_1+b_3\right)/\left(1-a_3a_1\right)$
$\ quad P= \left(a_1b_2+b_1\right)/\left(1-a_1a_2\right)$

As for (II) one shows easily equivalence with

$\ quad \left(a_1a_2a_3\right) ^3=1$
$\ quad \left(a_1^2+a_1+1\right) b_1+a_1^3 \left(a_2^2+a_2+1\right) b_2+a_1^3a_2^3 \left(a_3^2+a_3+1\right) b_3=0$

Like $a_1a_2a_3 \ 1$, one sees that $\ quad a_1a_2a_3 = J$ or $\ quad j^2$. Let us suppose to fix the ideas that $\ quad a_1a_2a_3 = j$ (that will correspond in the application to a triangle ABC of positive direction).

Now, after 2 lines of calculation, one obtains:

$\ quad P+jQ+j^ 2R =$

$\ quad =$
$\ quad =-a_3/a_1/$

And thus $\ quad P+jQ+j^ 2R =0.$ This shows well that triangle PQR is equilateral (positive direction) (traditional characterization).

Naturally if the triangle ABC is negative direction, one must take $\ quad a_1a_2a_3=j^2$ and one obtains equilateral PQR negative direction .

If ABC is positive direction, there is in fact

$\ quad F \left(Z\right) =e^ \left\{2i \ alpha /3\right\} \left(z-A\right) +A \ qquad G \left(Z\right) =e^ \left\{2i \ beta /3\right\} \left(z-B\right) +B \ qquad H \left(Z\right) =e^ \left\{2i \ gamma /3\right\} \left(z-C\right) +C$

and thus $\ quad a_1=e^ \left\{2i \ alpha /3\right\} \ quad b_1= \left(1-e^ \left\{2i \ alpha /3\right\}\right) has similar$ and expressions for $\ quad a_2, b_2, a_3, b_3$. It is thus very easy to determine P, Q and R.

## External bonds

• See the 27 equilateral triangles formed by trisecting the

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