HYACINTHOS 27193

 

[Antreas P. Hatzipolakis]:

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

Denote:

 

Aa, Pa = the reflections of A, P in B'C', resp.
Bb. Pb = the reflections of B, P in C'A', resp.
Cc, Pc = the reflections of C,P in A'B', resp.

Which is the locus of P such that the circumcircles of AAaPa, BBbPb, CCcPc are coaxial.?
And which is the locus of the other than P point of concurrence?

O, H lie on the locus.

 

[César Lozada]:

 

Locus: {circumcircle} ∪ {orthocubic K006}

 

K006 passes through ETCs 1, 3, 4, 46, 90, 155, 254, 371, 372, 485, 486, 487, 488, 6212, 6213, 8946, 8947, 8948, 8949 and vertices of triangles: ABC, EXCENTRAL, ORTHIC, VECTEN INNER, VECTEN OUTER

 

General expression for P’ (2^nd point of intersection) is very, very long.

 

Some values for P’:

P’( X(3) ) = X(403)

 

P’(X(4)) = isogonal conjugate of X(15478)

= SB*SC*(2*R^2-SB)*(2*R^2-SC)* (6*R^2-SA-SW) : : (barys)

= on cubics K025, K339 and lines: {4, 155}, {30, 13398}, {131, 403}

= antigonal conjugate of X(403)

= isogonal conjugate of X(15478)

= anticomplementary circle-inverse-of X(12318)

= polar circle-inverse-of X(155)

= [ 8.5110075414246540, 0.9565281816629016, -0.9497046629782389 ]

 

César Lozad