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 $AR$, $AQ$, $BP$, $BR$, $CQ$, 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 $BPC$,

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

$=$ $2R\sin (3a)/\sin (\sigma -a)$ (2)

so

$$BP=2R\sin (3a)\sin (c)/\sin (\sigma -a).$$

Combining that with the identity

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

we get

$$BP=8R\sin (a)\sin (c)\sin (\sigma +a).$$

Similarly,

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

Using the cosines law now,

$$P{R}^2=B{P}^2+B{R}^2-2BP\cdot BR\cos (b)$$

$$=64{\sin }^2(a){\sin }^2(c)[{\sin }^2(\sigma +a)+{\sin }^2(\sigma +c)-2\sin (\sigma +a)\sin (\sigma +c)\cos (b)].$$

But we have

$$(\sigma +a)+(\sigma +c)+b=\pi .$$

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

$${\sin }^2(b)={\sin }^2(\sigma +a)+{\sin }^2(\sigma +c)-2\sin (\sigma +a)\sin (\sigma +c)\cos (b)$$

whence

$$PR=8R\sin (a)\sin (b)\sin (c).$$

Since this expression is symmetric in $a$, $b$, and $c$, we deduce

$$PR=RQ=QP$$

as claimed.

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

By the sines law we have

$$\frac{AR}{\sin b}=\frac{BR}{\sin a}\mathit{\hspace{1em}\hspace{1em}}\frac{BP}{\sin c}=\frac{CP}{\sin b}\mathit{\hspace{1em}\hspace{1em}}\frac{CQ}{\sin a}=\frac{AQ}{\sin c}$$

whence

$$AR\cdot BP\cdot CQ=AQ\cdot BR\cdot CP.$$

This implies that if we identify the various vertices with complex numbers, then

$$(P-C)(Q-A)(R-B)=\frac{-1+i\sqrt{3}}{2}(P-B)(Q-C)(R-A)$$

provided that the triangle $ABC$ has positive orientation, i.e.

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

I found Letac’s proof at http://www.cut-the-knot.org/triangle/Morley/index.shtmlcut-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