SECOND TANGENT CIRCLES

Let ABC be a triangle, {I_A} 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 {I_A} 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 {I_A} 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 {I_A} 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 {I_A} 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