Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P.
The line B'C' intersects the circumcircle at Ba, Ca
The line C'A' intersects the circumcircle at Cb, Ab
The line A'B' intersects the circumcircle at Ac, Bc
Denote:
A1, B1, C1 = the midpoints of AA', BB', CC', resp.
A2, B2, C2 = the midpoints of BaCa, CbAb, AcBc, resp.
A*, B*, C* = the reflections of A2, B2, C2 in B1C1, C1A1, A1B1, resp.
Which is the locus of P such that ABC, A*B*C* are:
1. perspective
2. orthologic
?
Note: The A*B*C* is inscribed in the NPC. See
APH
[César Lozada]:
Algebraically simpler:
Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P.The line B'C' intersects the circumcircle at Ba, Ca
The line C'A' intersects the circumcircle at Cb, Ab
The line A'B' intersects the circumcircle at Ac, Bc
These intersections lead generally to 2nd degree equations with squares roots.
Denote:
A1, B1, C1 = the midpoints of AA', BB', CC', resp.A2, B2, C2 = the midpoints of BaCa, CbAb, AcBc, resp.
A2B2C2 = the pedal triangle of O w/r to A’B’C’
A*, B*, C* = the reflections of A2, B2, C2 in B1C1, C1A1, A1B1, resp.
1) Perspective:
Locus = {Gibert’s Q066=Stammler quartic through ETC’s 1, 2, 4, 254, 1113, 1114, 1138, 2184, 3223, 3346, 3459, 8049, 9510, 13483, 13484, 13574, 13575)}
ETC-pairs (P, Z1(P)=perspector): (1,12), (2,2), (4,4), (254,68), (1113,1313), (1114,1312), (1138,5627), (3346,6526), (3459,252), (13574,10415)
Z1( X(2184) ) = (b+c)^2*(a-b+c)^2*(a+b-c)^2//( a^3+(b+c)*a^2-(b+c)^2*a-(b^2- c^2)*(b-c)) : : (barycentrics)
= On lines: {2,7367}, {4,6611}, {11,1435}, {84,5715}, {225,1427}, {226,1439}, {1422,2006}, {1436,7490}, {1440,6612}, {2184,5514}, {3772,7129}, {6356,6358}
= {X(226), X(8808)}-Harmonic conjugate of X(1903)
= [ -0.214603731338937, -0.32118572451519, 3.962071705651325 ]
Z1( X(13575) ) = (4)X(251) ∩ X(6)X(66)
= (S^2-2*SA*SC+SB^2)*(S^2-2*SA* SB+SC^2)*SB*SC : : (barycentrics)
= On the cubic K701 and these lines: {2,1235}, {4,251}, {6,66}, {22,5523}, {25,2353}, {111,1289}, {112,7391}, {127,13575}, {232,2165}, {468,8770}, {1383,6995}, {1400,2156}, {2395,6753}, {2987,6515}, {3172,5064}, {5133,8743}, {7735,8882}
= polar conjugate of X(315)
= [ 0.771108835007703, 1.28820741641366, 2.392932192848292 ]
2) Orthologic:
Locus = {sidelines} \/ { circum-nonic q9 through ETC’s 2,4,7, vertices of triangles anticomplementary and cevian-of-X(264)}
q9: CyclicSum[ y*z*((SB*b^2*y-SC*c^2*z)*a^4* y^3*z^3+2*(-b^2*z^2+c^2*y^2)* SA^2*a^2*x^5-(b^2-c^2)*(3*S^2- SB*SC)*a^2*x*y^3*z^3-2*(-(SC* S^2+(2*SA*SB+SA*SC-2*SC^2)*SA) *c^2*y+(SB*S^2+(SA*SB+2*SA*SC- 2*SB^2)*SA)*b^2*z)*x^4*y*z) ] = 0 (barys)
ETC triads: (2,4,3), (4,3,4), ( 7,1,5)
César Lozada
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