# 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 $BPC$,

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

so

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

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,

 $PR^{2}=BP^{2}+BR^{2}-2BP\cdot BR\cos(b)$
 $=64\sin^{2}(a)\sin^{2}(c)[\sin^{2}(\sigma+a)+\sin^{2}(\sigma+c)-2\sin(\sigma+a% )\sin(\sigma+c)\cos(b)]\;.$

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

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

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.

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

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 ProofOfMorleysTheorem 2013-03-22 13:45:44 2013-03-22 13:45:44 mathcam (2727) mathcam (2727) 6 mathcam (2727) Proof msc 51M04