357. Morley theorem.
by Virgil Nicula, Sep 22, 2012, 8:05 AM
PP1 - Morley's trisector theorem. Prove that in any 
the three points of intersection of the adjacent angle trisectors form an equilateral triangle
called the first Morley triangle or simply the Morley triangle (this theorem was discovered in 1899 by Anglo-American mathematician Frank Morley)
Proof. Let circumradius
of , 
where 
and Morley triangle 
, where 
.
The theorem of SINUS in
and 
![$\sin (60^{\circ}-z)\left[3-4\sin^2\left(60^{\circ}-z\right)\right]=$](//latex.artofproblemsolving.com/8/5/f/85fb7300d503d771914a0f019ba430780777d9c5.png)
![$\sin (60^{\circ}-z)\left[1+2\cos \left(120^{\circ}-2z\right)\right]=$](//latex.artofproblemsolving.com/c/e/1/ce182d8bf64449421da3d437ba8846d472da0f56.png)
![$\sin (60^{\circ}-z)\left[1+2\sin\left(2z-30^{\circ}\right)\right]$](//latex.artofproblemsolving.com/8/e/4/8e4c7e9b18b4f7ad98f8c3847735701f7fc65426.png)
. Thus, 
. Obtain analogously 
. In conclusion,
, where 
. The generalized Pythagoras' theorem to the side ![$[NP]$](//latex.artofproblemsolving.com/4/b/f/4bf4a2a9ab593d9ee3bd6f3316f38ad187f4cbb9.png)
in the triangle 
![$k^2\left[\cos^2(30^{\circ}-y)+\cos^2(30^{\circ} -z)-2\cos (30^{\circ}-y)\cos (30^{\circ}-z)\cos x\right]\implies$](//latex.artofproblemsolving.com/2/e/b/2ebc833835b21f8ea93672d459d41ef4adc7d8ed.png)
![$\frac {2}{k^2}\cdot NP^2=2+\cos(60^{\circ}-2y)+\cos (60^{\circ} -2z)-2\cos x\left[\cos x+\cos (y-z)\right]=$](//latex.artofproblemsolving.com/5/7/e/57e5e31a20dabd681d31d77f3a3e3bf75c7020c3.png)
. In conclusion, 
, i.e. 
,
what is a symmetrical expression of
, i.e. 
is an equilateral triangle.
PP2. Let , 
so that 
and 
. Prove that 
and 
, where 
.
Proof 1.![$[ABM]+[AMC]=[ABC]\iff$](//latex.artofproblemsolving.com/0/1/4/0142f2b47b6a0d471dfd840dec78b616c4b95ba3.png)
. Obtain similarly that 
. Observe that 
.
Proof 2.
. Apply the theorem
of Sinus in
.
Remark. Apply the theorem of Sinus's in the triangle
. I'll prove directly the relation 
.
Indeed,
![$\sin C[\sin (B+2\theta )+\sin B]+$](//latex.artofproblemsolving.com/7/8/2/78245defaa3df4defb09db3e2ac182b3fc8dac8e.png)
![$\frac 12\cdot [\cos \theta -\cos (2B+\theta )]=$](//latex.artofproblemsolving.com/c/e/a/ceaaad75ce9759414d16ca301349af142c96d5f5.png)
![$\sin B[\sin (C+2\theta )+\sin C]+$](//latex.artofproblemsolving.com/b/3/a/b3a58120430cdcb9748aa12188191ab3a96f0566.png)
![$\frac 12[\cos\theta -\cos (2C+\theta )]\iff$](//latex.artofproblemsolving.com/1/2/9/12960b1ae96ae573fe29540e21bad9e346c3573d.png)
, what is truly. Otherwise, 
![$\frac {[\cos\theta -\cos (2B+\theta )]-[\cos\theta -\cos (2C+\theta )]}{[\cos (C+\theta -B)-\cos (B+C+\theta )]-[\cos (B+\theta -C )-\cos (B+C+\theta )]}=$](//latex.artofproblemsolving.com/b/a/c/bac7f7f8b51d3c6dc07fef4b02a53c62845818a1.png)
, i.e. 
, what is truly.
the three points of intersection of the adjacent angle trisectors form an equilateral trianglecalled the first Morley triangle or simply the Morley triangle (this theorem was discovered in 1899 by Anglo-American mathematician Frank Morley)
Proof. Let circumradius
of , 
where 
and Morley triangle 
, where 
.The theorem of SINUS in
and 
![$\sin (60^{\circ}-z)\left[3-4\sin^2\left(60^{\circ}-z\right)\right]=$](http://latex.artofproblemsolving.com/8/5/f/85fb7300d503d771914a0f019ba430780777d9c5.png)
![$\sin (60^{\circ}-z)\left[1+2\cos \left(120^{\circ}-2z\right)\right]=$](http://latex.artofproblemsolving.com/c/e/1/ce182d8bf64449421da3d437ba8846d472da0f56.png)
![$\sin (60^{\circ}-z)\left[1+2\sin\left(2z-30^{\circ}\right)\right]$](http://latex.artofproblemsolving.com/8/e/4/8e4c7e9b18b4f7ad98f8c3847735701f7fc65426.png)
. Thus, 
. Obtain analogously 
. In conclusion,
, where 
. The generalized Pythagoras' theorem to the side ![$[NP]$](http://latex.artofproblemsolving.com/4/b/f/4bf4a2a9ab593d9ee3bd6f3316f38ad187f4cbb9.png)
in the triangle 
![$k^2\left[\cos^2(30^{\circ}-y)+\cos^2(30^{\circ} -z)-2\cos (30^{\circ}-y)\cos (30^{\circ}-z)\cos x\right]\implies$](http://latex.artofproblemsolving.com/2/e/b/2ebc833835b21f8ea93672d459d41ef4adc7d8ed.png)
![$\frac {2}{k^2}\cdot NP^2=2+\cos(60^{\circ}-2y)+\cos (60^{\circ} -2z)-2\cos x\left[\cos x+\cos (y-z)\right]=$](http://latex.artofproblemsolving.com/5/7/e/57e5e31a20dabd681d31d77f3a3e3bf75c7020c3.png)
. In conclusion, 
, i.e. 
,what is a symmetrical expression of
, i.e. 
is an equilateral triangle.PP2. Let
, 
so that 
and 
. Prove that 
and 
, where 
.Proof 1.
![$[ABM]+[AMC]=[ABC]\iff$](http://latex.artofproblemsolving.com/0/1/4/0142f2b47b6a0d471dfd840dec78b616c4b95ba3.png)
. Obtain similarly that 
. Observe that 
.Proof 2.
. Apply the theoremof Sinus in
.Remark. Apply the theorem of Sinus's in the triangle
. I'll prove directly the relation 
.Indeed,
![$\sin C[\sin (B+2\theta )+\sin B]+$](http://latex.artofproblemsolving.com/7/8/2/78245defaa3df4defb09db3e2ac182b3fc8dac8e.png)
![$\frac 12\cdot [\cos \theta -\cos (2B+\theta )]=$](http://latex.artofproblemsolving.com/c/e/a/ceaaad75ce9759414d16ca301349af142c96d5f5.png)
![$\sin B[\sin (C+2\theta )+\sin C]+$](http://latex.artofproblemsolving.com/b/3/a/b3a58120430cdcb9748aa12188191ab3a96f0566.png)
![$\frac 12[\cos\theta -\cos (2C+\theta )]\iff$](http://latex.artofproblemsolving.com/1/2/9/12960b1ae96ae573fe29540e21bad9e346c3573d.png)
, what is truly. Otherwise, 
![$\frac {[\cos\theta -\cos (2B+\theta )]-[\cos\theta -\cos (2C+\theta )]}{[\cos (C+\theta -B)-\cos (B+C+\theta )]-[\cos (B+\theta -C )-\cos (B+C+\theta )]}=$](http://latex.artofproblemsolving.com/b/a/c/bac7f7f8b51d3c6dc07fef4b02a53c62845818a1.png)
, i.e. 
, what is truly.This post has been edited 38 times. Last edited by Virgil Nicula, Nov 16, 2015, 5:21 PM
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