NAPOLEON – MORLEY TRIANGLES – A Tran Quang Hung construction

NAPOLEON – MORLEY TRIANGLES – A Tran Quang Hung construction

Tran Quang Hung – Advanced Plane Geometry message #1740

https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/1740

Let DEF,XYZ be Morley and (outer) Napoleon triangle of ABC. DX,EY,FZ cuts BC,CA,AB at U,V,W, respectively. Then, AU,BV,CW are concurrent.

Generalization:

Let ABC be a triangle, A_NB_NC_N any Napoleon triangle of ABC (inner or outer) and A_MB_MC_M any Morley triangle of ABC (1st, 2nd, 3rd, 1st adjunct, 2nd adjunct or 3rd adjunct). Let A’=(A_NA_M) ∩ (BC) and B’, C’ built cyclically. Then lines (AA’), (BB’), (CC’) concur.

Considere the function:

N(A,B,C,n) = [1, -2*cos(C+n*Pi/3), -2*cos(B+n*Pi/3)]

Then N(A,B,C,1) is the trilinear A-vertex of the INNER Napoleon triangle and N(A,B,C,-1) is the trilinear A-vertex of the OUTER Napoleon triangle

Similarlly the function

M(A,B,C,m) = [1/(1-2*m^2), 2*cos(C/3+m*Pi/3), 2*cos(B/3+m*Pi/3)]

gives the trilinear coordinates of the A-vertex of 1st, 2nd and 3rd Morley triangles for m=0, m=1 and m=-1, respectively.

If n Î {-1,1} and m Î {-1,0,1}, i.e, for A_NB_NC_N equal to the Napoleon inner or outer triangles and for A_MB_MC_M equal to 1st, 2nd or 3rd Morley triangles of ABC, we find, in general:

A’ = [0, F(B,n,m), F(C,n,m)]

where

F(A,n,m)= 1/(cos(A+n*Pi/3)+cos(A/3+m*Pi/3)*(1-2*m^2))

therefore lines (AA’), (BB`), (CC’) concur at:

Z(n,m) = [ F(A,n,m) , F(B,n,m) , F(C,n,m) ]

Note that each vertex of 1st, 2nd or 3rd MORLEY-ADJUNCT triangle is the isogonal conjugate of the correspondant vertex of the 1st, 2nd and 3rd MORLEY triangle, respectively. Generalizating this property makes easier to find A’ when A_MB_MC_M is equal to 1st, 2nd or 3rd Morley-ADJUNCT triangle of ABC, i.e.:

A’ = [0, F’(B,n,m), F’(C,n,m)]

where

F’(A,n,m)= cos(A/3+m*Pi/3)/(4*cos(A+n*Pi/3)*cos(A/3+m*Pi/3)+1-2*m^2)

being the point of concurrence of  (AA’), (BB`), (CC’):

Z’(n,m) = [ F’(A,n,m) , F’(B,n,m) , F’(C,n,m) ]

The next table shows the central function and ETC-SEARCH values u,v for these 12 points of concurrence

	NAPOLEON INNER	NAPOLEON OUTER
1st. MORLEY	1/(cos(A+Pi/3)+cos(A/3)) ( -1.061175889876748, -1.37248299267917) Coordinates found by Chris Tienhoven, ADGEOM message #1760	1/(cos(A-Pi/3)+cos(A/3)) ( 1.449383712449321, 1.37198638995608 ) Coordinates found by Bernard Gibert, ADGEOM message #1753
2nd. MORLEY	1/(cos(A+Pi/3)-cos(A/3+Pi/3)) ( 2.993155354733740, 1.87035114072660 )	1/(cos(A-Pi/3)-cos(A/3+Pi/3)) ( 2.312842706106623, 1.60870667887718 )
3rd. MORLEY	1/(cos(A+Pi/3)-cos(A/3-Pi/3)) ( 3.133454149812978, 1.98256834592023 )	1/(cos(A-Pi/3)-cos(A/3-Pi/3)) ( 6.747193170763110, 5.24015654664033 )
1st. MORLEY-ADJUNCT	cos(A/3)/(4*cos(A+Pi/3)*cos(A/3)+1) ( 1.943228854328264, 5.45687972067145 )	cos(A/3)/(4*cos(A-Pi/3)*cos(A/3)+1) ( 1.554607572897988, 1.39778519618714 )
2nd. MORLEY-ADJUNCT	cos(A/3+Pi/3)/(4*cos(A+Pi/3)*cos(A/3+Pi/3)-1) ( -14.459440389293710, -8.44530315002158 )	cos(A/3+Pi/3)/(4*cos(A-Pi/3)*cos(A/3+Pi/3)-1) ( 2.308322612923908, 3.48600580865934 )
3rd. MORLEY-ADJUNCT	cos(A/3-Pi/3)/(4*cos(A+Pi/3)*cos(A/3-Pi/3)-1) ( 3.267771237439981, 1.96385627384925 )	cos(A/3-Pi/3)/(4*cos(A-Pi/3)*cos(A/3-Pi/3)-1) ( 1.422917983150731, 1.03292991115707 )

César Lozada

Sept. 19, 2014