Morley's trisector theorem

There are many proofs of Morley's theorem, some of which are very technical.{{citation|url=http://www.cut-the-knot.org/triangle/Morley/index.shtml|title=Morley's Miracle|publisher=Cut-the-knot|last=Bogomolny|first=Alexander|authorlink= Alexander Bogomolny |accessdate=2010-01-02}}

Several early proofs were based on delicate trigonometric calculations. Recent proofs include an algebraic proof by {{harvs|first=Alain|last=Connes|authorlink=Alain Connes|txt|year=1998|year2=2004}} extending the theorem to general fields other than characteristic three, and John Conway's elementary geometry proof.{{Citation |last=Bogomolny |first=Alexander |title=J. Conway's proof |url=http://www.cut-the-knot.org/triangle/Morley/conway.shtml |publisher=Cut-the-knot |access-date=2021-12-03 |authorlink=Alexander Bogomolny}}{{citation|chapter-url=http://thewe.net/math/conway.pdf|title=Power|editor1-last=Blackwell|editor1-first=Alan|editor2-last=Mackay|editor2-first=David|editor2-link=David J. C. MacKay|year=2006|chapter=The Power of Mathematics|last=Conway|first=John|author-link=John Horton Conway|publisher=Cambridge University Press|accessdate=2010-10-08|pages=36–50|isbn=978-0-521-82377-7}} The latter starts with an equilateral triangle and shows that a triangle may be built around it which will be similar to any selected triangle. Morley's theorem does not hold in spherical[http://lienhard-wimmer.com/applets/dreieck/Morley.html Morley's Theorem in Spherical Geometry], Java applet. and hyperbolic geometry.

File:Morley Proof.svg

One proof uses the trigonometric identity

{{NumBlk|::|$\backslash sin(3\backslash theta)=4\backslash sin\backslash theta\backslash sin(60^\backslash circ+\backslash theta)\backslash sin(120^\backslash circ+\backslash theta)$|{{EquationRef|1}}}}

which, by using of the sum of two angles identity, can be shown to be equal to

::$\backslash sin(3\backslash theta)=-4\backslash sin^3\backslash theta+3\backslash sin\backslash theta.$

The last equation can be verified by applying the sum of two angles identity to the left side twice and eliminating the cosine.

Points $D,\; E,\; F$ are constructed on $\backslash overline\{BC\}$ as shown. We have $3\backslash alpha+3\backslash beta+3\backslash gamma=180^\backslash circ$, the sum of any triangle's angles, so $\backslash alpha+\backslash beta+\backslash gamma=60^\backslash circ.$ Therefore, the angles of triangle $XEF$ are $\backslash alpha,\; (60^\backslash circ+\backslash beta),$ and $(60^\backslash circ+\backslash gamma).$

From the figure

{{NumBlk|::|$\backslash sin(60^\backslash circ+\backslash beta)=\backslash frac\{\backslash overline\{DX\}\}\{\backslash overline\{XE\}\}$|{{EquationRef|2}}}}

and

{{NumBlk|::|$\backslash sin(60^\backslash circ+\backslash gamma)=\backslash frac\{\backslash overline\{DX\}\}\{\backslash overline\{XF\}\}.$|{{EquationRef|3}}}}

Also from the figure

::$\backslash angle\{AYC\}=180^\backslash circ-\backslash alpha-\backslash gamma=120^\backslash circ+\backslash beta$

and

{{NumBlk|::|$\backslash angle\{AZB\}=120^\backslash circ+\backslash gamma.$|{{EquationRef|4}}}}

The law of sines applied to triangles $AYC$ and $AZB$ yields

{{NumBlk|::|$\backslash sin(120^\backslash circ+\backslash beta)=\backslash frac\{\backslash overline\{AC\}\}\{\backslash overline\{AY\}\}\backslash sin\backslash gamma$|{{EquationRef|5}}}}

and

{{NumBlk|::|$\backslash sin(120^\backslash circ+\backslash gamma)=\backslash frac\{\backslash overline\{AB\}\}\{\backslash overline\{AZ\}\}\backslash sin\backslash beta.$|{{EquationRef|6}}}}

Express the height of triangle $ABC$ in two ways

::$h=\backslash overline\{AB\}\; \backslash sin(3\backslash beta)=\backslash overline\{AB\}\backslash cdot\; 4\backslash sin\backslash beta\backslash sin(60^\backslash circ+\backslash beta)\backslash sin(120^\backslash circ+\backslash beta)$

and

::$h=\backslash overline\{AC\}\; \backslash sin(3\backslash gamma)=\backslash overline\{AC\}\backslash cdot\; 4\backslash sin\backslash gamma\backslash sin(60^\backslash circ+\backslash gamma)\backslash sin(120^\backslash circ+\backslash gamma).$

where equation (1) was used to replace $\backslash sin(3\backslash beta)$ and $\backslash sin(3\backslash gamma)$ in these two equations. Substituting equations (2) and (5) in the $\backslash beta$ equation and equations (3) and (6) in the $\backslash gamma$ equation gives

::$h=4\backslash overline\{AB\}\backslash sin\backslash beta\backslash cdot\backslash frac\{\backslash overline\{DX\}\}\{\backslash overline\{XE\}\}\backslash cdot\backslash frac\{\backslash overline\{AC\}\}\{\backslash overline\{AY\}\}\backslash sin\backslash gamma$

and

::$h=4\backslash overline\{AC\}\backslash sin\backslash gamma\backslash cdot\backslash frac\{\backslash overline\{DX\}\}\{\backslash overline\{XF\}\}\backslash cdot\backslash frac\{\backslash overline\{AB\}\}\{\backslash overline\{AZ\}\}\backslash sin\backslash beta$

Since the numerators are equal

::$\backslash overline\{XE\}\backslash cdot\backslash overline\{AY\}=\backslash overline\{XF\}\backslash cdot\backslash overline\{AZ\}$

or

::$\backslash frac\{\backslash overline\{XE\}\}\{\backslash overline\{XF\}\}=\backslash frac\{\backslash overline\{AZ\}\}\{\backslash overline\{AY\}\}.$

Since angle $EXF$ and angle $ZAY$ are equal and the sides forming these angles are in the same ratio, triangles $XEF$ and $AZY$ are similar.

Similar angles $AYZ$ and $XFE$ equal $(60^\backslash circ+\backslash gamma)$, and similar angles $AZY$ and $XEF$ equal $(60^\backslash circ+\backslash beta).$ Similar arguments yield the base angles of triangles $BXZ$ and $CYX.$

In particular angle $BZX$ is found to be $(60^\backslash circ+\backslash alpha)$ and from the figure we see that

::$\backslash angle\{AZY\}+\backslash angle\{AZB\}+\backslash angle\{BZX\}+\backslash angle\{XZY\}=360^\backslash circ.$

Substituting yields

::$(60^\backslash circ+\backslash beta)+(120^\backslash circ+\backslash gamma)+(60^\backslash circ+\backslash alpha)+\backslash angle\{XZY\}=360^\backslash circ$

where equation (4) was used for angle $AZB$ and therefore

::$\backslash angle\{XZY\}=60^\backslash circ.$

Similarly the other angles of triangle $XYZ$ are found to be $60^\backslash circ.$

The first Morley triangle has side lengths{{MathWorld |id= FirstMorleyTriangle|title=First Morley Triangle |access-date=2021-12-03}}

a^\prime=b^\prime=c^\prime=8R\,\sin\tfrac13A\,\sin\tfrac13B\,\sin\tfrac13C,

where R is the circumradius of the original triangle and A, B, and C are the angles of the original triangle. Since the area of an equilateral triangle is $\backslash tfrac\{\backslash sqrt\{3\}\}\{4\}a\text{'}^2,$ the area of Morley's triangle can be expressed as

\text{Area} = 16 \sqrt{3}R^2\, \sin^2\!\tfrac13A\, \sin^2\!\tfrac13B\, \sin^2\!\tfrac13C.

Morley's theorem entails 18 equilateral triangles. The triangle described in the trisector theorem above, called the first Morley triangle, has vertices given in trilinear coordinates relative to a triangle ABC as follows:

$$\backslash begin\{array\}\{ccccccc\}$$

A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13 C &:& 2 \cos\tfrac13 B \\[5mu]

B \text{-vertex} &=& 2 \cos\tfrac13 C &:& 1 &:& 2 \cos\tfrac13 A \\[5mu]

C \text{-vertex} &=& 2 \cos\tfrac13 B &:& 2 \cos\tfrac13 A &:& 1

\end{array}

Another of Morley's equilateral triangles that is also a central triangle is called the second Morley triangle and is given by these vertices:

$$\backslash begin\{array\}\{ccccccc\}$$

A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13(C - 2\pi) &:& 2 \cos\tfrac13(B - 2\pi) \\[5mu]

B \text{-vertex} &=& 2 \cos\tfrac13(C - 2\pi) &:& 1 &:& 2 \cos\tfrac13(A - 2\pi) \\[5mu]

C \text{-vertex} &=& 2 \cos\tfrac13(B - 2\pi) &:& 2 \cos\tfrac13(A - 2\pi) &:& 1

\end{array}

The third of Morley's 18 equilateral triangles that is also a central triangle is called the third Morley triangle and is given by these vertices:

$$\backslash begin\{array\}\{ccccccc\}$$

A \text{-vertex} &=& 1 &:& 2 \cos\tfrac13(C + 2\pi) &:& 2 \cos\tfrac13(B + 2\pi) \\[5mu]

B \text{-vertex} &=& 2 \cos\tfrac13(C + 2\pi) &:& 1 &:& 2 \cos\tfrac13(A + 2\pi) \\[5mu]

C \text{-vertex} &=& 2 \cos\tfrac13(B + 2\pi) &:& 2 \cos\tfrac13(A + 2\pi) &:& 1

\end{array}

The first, second, and third Morley triangles are pairwise homothetic. Another homothetic triangle is formed by the three points X on the circumcircle of triangle ABC at which the line XX ⁻¹ is tangent to the circumcircle, where X ⁻¹ denotes the isogonal conjugate of X. This equilateral triangle, called the circumtangential triangle, has these vertices:

$$\backslash begin\{array\}\{lllllll\}$$

A \text{-vertex} &=& \phantom{-}\csc\tfrac13(C - B) &:& \phantom{-}\csc\tfrac13(2C + B) &:& -\csc\tfrac13(C + 2B) \\[5mu]

B \text{-vertex} &=& -\csc\tfrac13(A + 2C) &:& \phantom{-}\csc\tfrac13(A - C) &:& \phantom{-}\csc\tfrac13(2A + C) \\[5mu]

C \text{-vertex} &=& \phantom{-}\csc\tfrac13(2B + A) &:& -\csc\tfrac13(B + 2A) &:& \phantom{-}\csc\tfrac13(B - A)

\end{array}

A fifth equilateral triangle, also homothetic to the others, is obtained by rotating the circumtangential triangle {{pi}}/6 about its center. Called the circumnormal triangle, its vertices are as follows:

$$\backslash begin\{array\}\{lllllll\}$$

A \text{-vertex} &=& \phantom{-}\sec\tfrac13(C - B) &:& -\sec\tfrac13(2C + B) &:& -\sec\tfrac13(C + 2B) \\[5mu]

B \text{-vertex} &=& -\sec\tfrac13(A + 2C) &:& \phantom{-}\sec\tfrac13(A - C) &:& -\sec\tfrac13(2A + C) \\[5mu]

C \text{-vertex} &=& -\sec\tfrac13(2B + A) &:& -\sec\tfrac13(B + 2A) &:& \phantom{-}\sec\tfrac13(B - A)

\end{array}

An operation called "extraversion" can be used to obtain one of the 18 Morley triangles from another. Each triangle can be extraverted in three different ways; the 18 Morley triangles and 27 extravert pairs of triangles form the 18 vertices and 27 edges of the Pappus graph.{{harvtxt|Guy|2007}}.

The Morley center, X(356), centroid of the first Morley triangle, is given in trilinear coordinates by

\cos\tfrac13A + 2\cos\tfrac13B\,\cos\tfrac13C \,:\, \cos\tfrac13B + 2\cos\tfrac13C\,\cos\tfrac13A \,:\, \cos\tfrac13C + 2\cos\tfrac13A\,\cos\tfrac13B

1st Morley–Taylor–Marr center, X(357): The first Morley triangle is perspective to triangle {{nobr|$\backslash triangle\; ABC${{sfn|Taylor Marr|1913}} :Fox, M. D.; and Goggins, J. R. "Morley's diagram generalised", Mathematical Gazette 87, November 2003, 453–467.}}<< the lines each connecting a vertex of the original triangle with the opposite vertex of the Morley triangle concur at the point

\sec\tfrac13A \,:\, \sec\tfrac13B \,:\, \sec\tfrac13C

• Angle trisection
• Hofstadter points
• Morley centers

{{reflist|2}}

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• [http://mathworld.wolfram.com/MorleysTheorem.html Morleys Theorem] at MathWorld
• [http://www.mathpages.com/home/kmath376/kmath376.htm Morley's Trisection Theorem] at MathPages
• [http://demonstrations.wolfram.com/MorleysTheorem/ Morley's Theorem] by Oleksandr Pavlyk, The Wolfram Demonstrations Project.

Category:Theorems about triangles