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[parent] proof of Morley's theorem (Proof)
\includegraphics{morley}
The scheme of this proof, due to A. Letac, is to use the sines law to get formulas for the segments $ AR$, $ AQ$, $ BP$, $ BR$, $ CQ$, and $ CP$, and then to apply the cosines law to the triangles $ ARQ$, $ BPR$, and $ CQP$, getting $ RQ$, $ PR$, and $ QP$.

To simplify some formulas, let us denote the angle $ \pi/3$, or 60 degrees, by $ \sigma$. Denote the angles at $ A$, $ B$, and $ C$ by $ 3a$, $ 3b$, and $ 3c$ respectively, and let $ R$ be the circumradius of $ ABC$. We have $ BC = 2R \sin(3a)$. Applying the sines law to the triangle $ BPC$,

$\displaystyle BP/\sin(c)$ $\displaystyle =$ $\displaystyle BC/\sin(\pi - b - c) = 2R \sin(3a)/\sin(b + c)$ (1)
  $\displaystyle =$ $\displaystyle 2R\sin(3a)/\sin(\sigma - a)$ (2)

so

$\displaystyle BP = 2R\sin(3a)\sin(c)/\sin(\sigma - a)\;.$
Combining that with the identity

$\displaystyle \sin(3a) = 4\sin(a)\sin(\sigma + a)\sin(\sigma -a)$
we get

$\displaystyle BP = 8R\sin(a)\sin(c)\sin(\sigma + a)\;.$
Similarly,

$\displaystyle BR = 8R\sin(c)\sin(a)\sin(\sigma + c)\;.$

Using the cosines law now,

$\displaystyle PR^2 = BP^2 + BR^2 - 2 BP\cdot BR\cos(b)$

$\displaystyle = 64 \sin^2(a)\sin^2(c)[ \sin^2(\sigma + a) + \sin^2(\sigma+ c) - 2\sin(\sigma + a)\sin(\sigma + c)\cos(b)]\;.$
But we have

$\displaystyle (\sigma + a) + (\sigma + c) + b = \pi\;.$
whence the cosines law can be applied to those three angles, getting

$\displaystyle \sin^2(b) = \sin^2(\sigma + a) + \sin^2(\sigma + c) - 2\sin(\sigma + a)\sin(\sigma + c)\cos(b)$
whence

$\displaystyle PR = 8R\sin(a)\sin(b)\sin(c)\;.$
Since this expression is symmetric in $ a$, $ b$, and $ c$, we deduce

$\displaystyle PR = RQ = QP$
as claimed.

Remarks:It is not hard to show that the triangles $ RYP$, $ PZQ$, and $ QXR$ are isoscoles.

By the sines law we have

$\displaystyle \frac{AR}{\sin b}=\frac{BR}{\sin a}\qquad \frac{BP}{\sin c}=\frac{CP}{\sin b}\qquad \frac{CQ}{\sin a}=\frac{AQ}{\sin c}$
whence

$\displaystyle AR\cdot BP\cdot CQ = AQ\cdot BR\cdot CP\;.$
This implies that if we identify the various vertices with complex numbers, then

$\displaystyle (P-C)(Q-A)(R-B)=\frac{-1+i\sqrt{3}}{2}(P-B)(Q-C)(R-A)$
provided that the triangle $ ABC$ has positive orientation, i.e.

$\displaystyle \Re\left(\frac{C-A}{B-A}\right)>0\;.$

I found Letac's proof at 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.



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Cross-references: reference, orientation, positive, complex numbers, implies, expression, circumradius, degrees, angle, triangles, cosines law, formulas, sines law, proof

This is version 3 of proof of Morley's theorem, born on 2003-07-17, modified 2003-08-04.
Object id is 4465, canonical name is ProofOfMorleysTheorem.
Accessed 4078 times total.

Classification:
AMS MSC: 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)
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