HYACINTHOS 28722

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.

Denote:

Ha, Hb, Hc = the orthocenters of AB'C', BC'A', CA'B', resp.

Which is the locus of P such that 

 

1. A'B'C', HaHbHc are perspective?
I lies on the locus (Thanos Kalogerakis)

[César Lozada]:

 

 

Let P=u:v:w (trilinears)

 

1)      The entire plane.

Perspector: Q(P) = u+cos(C)*v+cos(B)*w : :

ETC pairs (P,Q(P)): (1,942), (2,5943), (3,5), (4,389), (5,5462), (6,6), (15,11542), (16,11543), (19,9119), (20,5907), (22,21243), (25,13567), (26,5449), (30,13754), (32,5305), (40,5777), (50,16310), (52,13292), (54,12242), (55,226), (56,1210), (64,4), (66,9969), (68,12235), (69,14913), (74,7687), (84,5908), (109,15252), (110,5972), (113,9826),….

 

[APH]:

 

Thanks!

The perspector Q(P) is the common midpoint of A'Ha, B'Hb, C'Hc.

So A', B', C', Ha, Hb, Hc are conconic (the center of the conic is the Q(P))

Which is the perspector of the conic in terms of P?

[César Lozada]:

 

 

For P=u:v:w (trilinears) the perspector of the conic is:

X(P) = (c^2*(3*b^2-c^2+a^2)*(b^2-c^2+3*a^2)*u^2*v^2+2*b^2*(c^2+a^2)*(a^2+b^2-c^2)*w^2*u^2+2*a^2*(b^2+c^2)*(a^2+b^2-c^2)*w^2*v^2+4*c*a^2*b*(a^2+b^2-c^2)*v^3*w+4*c^2*a*b*(a^2+b^2-c^2)*u*v^3+4*c^2*a*b*(a^2+b^2-c^2)*u^3*v+c*a*(3*a^4-2*c^2*a^2-6*b^2*c^2-c^4+14*a^2*b^2+7*b^4)*w*u*v^2+b*c*(-2*b^2*c^2-c^4+14*a^2*b^2+7*a^4-6*c^2*a^2+3*b^4)*w*u^2*v+4*c*a*b^2*(a^2+b^2-c^2)*w*u^3+a*b*(a^4+6*a^2*b^2+2*c^2*a^2+2*b^2*c^2+b^4-3*c^4)*w^2*u*v)*c*b*(2*c^2*(a^2+b^2)*(a^2-b^2+c^2)*v^2*u^2+b^2*(c^2-b^2+3*a^2)*(3*c^2-b^2+a^2)*u^2*w^2+2*a^2*(b^2+c^2)*(a^2-b^2+c^2)*v^2*w^2+4*b*c*a^2*(a^2-b^2+c^2)*v*w^3+4*b*c^2*a*(a^2-b^2+c^2)*u^3*v+c*a*(a^4+c^4+2*b^2*c^2+6*c^2*a^2+2*a^2*b^2-3*b^4)*w*u*v^2+b*c*(-b^4+14*c^2*a^2-2*b^2*c^2+3*c^4+7*a^4-6*a^2*b^2)*w*u^2*v+4*b^2*c*a*(a^2-b^2+c^2)*w^3*u+4*b^2*c*a*(a^2-b^2+c^2)*w*u^3+a*b*(3*a^4+14*c^2*a^2-2*a^2*b^2-6*b^2*c^2-b^4+7*c^4)*w^2*u*v) : :

 

This expression says:

-          If P lies on the circumcircle of ABC then X(P)=IsogonalConjugate( AntipodeInCIrcumcircle( P ) )

-          If P is in the infinity then X(P) is also in the infinity and

X(P) = IsogonalConjugate( SR( P*, InverseInPolarCircle(Complement(P*)))),

where P*=IsogonalConjugate(P) and SR(P,U) is as defined in the preamble just before X(2677)

 

 

ETC-pairs(P,X(P)):

-          (74,523), (98,512), (99,511), (100,517), (101,516), (102,522), (103,514), (104,513), (105,3309), (106,3667), (107,6000),

(108,6001), (109,515), (110,30), (111,1499), (112,1503), (691,542), ….

-           (512,2799), (513,2804), (523,9033), (1503,2799), (6000,9033), (6001,2804)

-          (3,3613), (6,21765), (64,253)

 

Some others:

 

X( X(1) ) = X(950)X(3664) ∩ X(3739)X(5745)

= ((b+3*c)*a^3+(2*b^2+3*b*c-c^2)*a^2+(b^2-c^2)*(b+3*c)*a+(b^2-c^2)*c*(3*b-c))*((3*b+c)*a^3-(b^2-3*b*c-2*c^2)*a^2-(b^2-c^2)*(3*b+c)*a+(b^2-c^2)*b*(b-3*c)) : : (barys)

= on lines: {950, 3664}, {3739, 5745}, {5088, 17175}

= [ -1.8602853858712190, -2.1290853747908670, 5.9732399194724650 ]

 

X( X(4) ) = X(140)X(6709) ∩ X(6748)X(11245)

= SB*SC*(S^2-4*R^2*(2*SW-SC)+SC*(2*SA+2*SB-SC)+2*SW^2)*(S^2-4*R^2*(2*SW-SB)+SB*(2*SA+2*SC-SB)+2*SW^2) : : (barys)

= on lines: {140, 6709}, {6748, 11245}

= [ 5.7101425149585270, 1.4462946474784700, 0.0039331805953494 ]

 

César Lozada