Tuesday, August 12, 2014

SECOND TANGENT CIRCLES


Let ABC be a triangle, {IA} its A-excircle, {I} its incircle and A’B’C’, A”B”C” its medial and orthic triangle, respectively.

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Let {a’} be the circle (not the NPC) through B’, C’ and tangent to {IA} at A1, as done in X(5213). Build B1, C1 cyclically.

A1 has trilinears:
(2*a+b+c)^2/(a*(a+b+c)) : -(a+c)^2/(b*(a+b-c)) : -(a+b)^2/(c*(a-b+c))
 A1B1C1 and ABC are perspective with perspector:
Z1 = (b+c)^2*(-a+b+c)/a : :
    =  Isotomic conjugate of X(552)
    = on lines (8,3058), (10,3175), (11,312), (12,1089), (42,3943), (55,346) and others
    = (-5*R*r-2*r^2+2*s^2)*X(1)+(3*(4*R*r-s^2))*X(2)+3*r^2*X(3)
    = ( 5.174982699686370, 1.60827199947209, 0.138791851648454 )
 The triangle T’aT’bT’c bounded by the common tangents of {a’} and {IA} is perspective to:
ABC, at:
(b+c)*(2*a+b+c)/(a*(a*(a+b-c)*(a-b+c)-(b+c)*(b^2+c^2-a^2))) : :
 =  ( 4.466295198982971, 6.20844398653125, -2.718856062144783 )
MEDIAL triangle of ABC, at:
                (2,3953) (594,3697)
=  ( 3.144749911827034, -0.31563697048064, 2.407759348320035 )

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Let {a’’} be the circle (not the NPC) through B’’, C’’ and tangent to {IA} at A2. Build B2, C2 cyclically.

A2 has trilinears:
(2*a^3+b*a^2+b^3+c*a^2-c*b^2-b*c^2+c^3)^2/(a*(a+b+c)) : -(a+c)^2*(a+b-c)^3/b : -(a+b)^2*(a-b+c)^3/c
 A2B2C2 and ABC are perspective with perspector:
Z2 = (b+c)^2/(a*(-a+b+c)^3) : :
     = on lines (7,3486), (11,273), (55,347), (56,1119), (65,1439) and others
     = ( 0.032585839353912, 0.06636810864421, 3.579677711605812 )
A2B2C2 is also perspective to the EXTANGENTS triangle of ABC at
                Q2 = on lines (10,12), (40,1723), (44,1842), (55,387), (71,1834) and others
                 =  ( -3.938480038745505, -1.09801096466883, 6.218585936868418 )
The triangle T’’aT’’bT’’c bounded by the common tangents of {a’’} and {IA} is perspective to:
                ABC, at: X(440) = Complement of X(27).
                ORTHIC triangle of ABC, at:
(1,5757) (4,1736)
= ( -0.860892635710262, -0.58571714842946, 4.443496031917200 )

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Let {a’} be the circle (not the NPC) through B’, C’ and tangent to the incircle {I} at A1. Build B1, C1 cyclically.

A1 has trilinears:
                (2*a-b-c)^2/(a*(b+c-a)) : (a-c)^2/(b*(a-b+c)) : (a-b)^2/(c*(a+b-c))
A1B1C1 and ABC are perspective with perspector X(1358) = BRISSE TRANSFORM OF X(101)
The triangle T’aT’bT’c bounded by the common tangents of {a’} and {I} is perspective to:
INTOUCH triangle of ABC, at X(3676) = X(650)com(INTOUCH TRIANGLE)
MEDIAL triangle of ABC, at X(3669) = X(663)com(INTOUCH TRIANGLE)

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Let {a’’} be the circle (not the NPC) through B’’, C’’ and tangent to the incircle {I} at A2. Build B2, C2 cyclically.

A2 has trilinears:
(2*a^3-(b+c)*(a^2+(b-c)^2))^2/(a*(-a+b+c)) : (a-c)^2*(a-b+c)^3/b : (a-b)^2*(a+b-c)^3/c
A2B2C2 and ABC are perspective with perspector X(4081) = X(651)com(EXTOUCH TRIANGLE)
A2B2C2 is also perspective with the INTANGENTS triangle of ABC at:
       Q2 = (11,116) (3900,5514)
                = ( 3.043142280414568, 3.26195719549706, -0.022140782858769 )
 The triangle T’’aT’’bT’’c bounded by the common tangents of {a’’} and {I} is perspective to:
                INTOUCH triangle of ABC, at: X(522) = ISOGONAL CONJUGATE OF X(109)
                ORTHIC triangle of ABC, at: X(3900) = X(514)com[INVERSE(n(3rd EULER TRIANGLE))]
                INCENTRAL triangle of ABC, at:
(522,905) (3064,3700)
= ( -24.297514066099830, 20.17023232928474, 0.890894745987175 )


César E. Lozada
August 12, 2014