Forgot password?
 Register account
View 328|Reply 1

[几何] Napoleon – Morley triangles

[Copy link]

3152

Threads

7905

Posts

610K

Credits

Credits
64068
QQ

Show all posts

hbghlyj Posted 2022-3-24 08:55 |Read mode
这篇博文

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.

3152

Threads

7905

Posts

610K

Credits

Credits
64068
QQ

Show all posts

 Author| hbghlyj Posted 2022-3-24 08:59
Generalization:
Let ABC be a triangle, ANBNCN any Napoleon triangle of ABC (inner or outer) and AMBMCM any Morley triangle of ABC (1st, 2nd, 3rd, 1st adjunct, 2nd adjunct or 3rd adjunct). Let A’=(ANAM) ∩ (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 ∈ {-1,1} and m ∈ {-1,0,1}, i.e, for ANBNCN equal to the Napoleon inner or outer triangles and for AMBMCM 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 AMBMCM 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 INNERNAPOLEON OUTER
1st. MORLEY1/(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. MORLEY1/(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. MORLEY1/(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-ADJUNCTcos(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-ADJUNCTcos(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-ADJUNCTcos(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 )

Mobile version|Discuz Math Forum

2025-6-4 17:36 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit