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
