HYACINTHOS 26750

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

 

Naa, Nab, Nac = the orthogonal projections of Na on AN, AB, AC, resp.

Nba, Nbb, Nbc = the orthogonal projections of Nb on BA, BN, BC, resp.

Nca, Ncb, Ncc = the orthogonal projections of Nc on CA, CB, CN, resp.

 

O1, O2, O3 = the circumcenters of NaaNabNac, NbaNbbNbc, NcaNcbNcc, resp.

N1, N2, N3 = the NPC centers of NaaNabNac, NbaNbbNbc, NcaNcbNcc, resp.

 

Which is the locus of P such that:

 

1. ABC, O1O2O3 are orthologic?
The entire plane?

2. ABC, N1N2N3 are orthologic?
I lies on the locus. Orthologic centers ?


[César Lozada]:


The entire plane. Orthologic centers are:

Qao = ABC->O1O2O3 = antigonal conjugate of P

Qoa = O1O2O3->ABC = complement-of-complement of P

 
Locus = { L∞} ∪ {circumcircle} ∪ { K= circum-rectangular hyperbola through ETCs X(4) and X(10121)}

K: ∑[ (SB-SC)*((R^4-4*R^2*SW+12*S^2)*SB*SC-(27*R^4-12*R^2*SW+4*S^2)*S^2)*y*z ] = 0 (barys)

 
I=X(1) does not lie on the locus.

 
Center of Κ:

K * = X(3)X(10120) ∩ X(4)X(10121)

(3*SA^2-2*SW*SA+(2*S-SW)*(2*S+SW))*((R^4-4*R^2*SW+12*S^2)*SA^2+(53*R^4-20*R^2*SW-4*S^2)*S^2)*((-R^4+4*R^2*SW-12*S^2)*SA^2+(R^4-4*R^2*SW+12*S^2)*SW*SA+2*(13*R^4-4*R^2*SW-4*S^2)*S^2) : : (barys)

= On the nine-points circle and these lines: {3, 10120}, {4, 10121}

= midpoint of X(4) and X(10121)

= reflection of X(3) in X(10120)

= [ 2.998722050393190, 3.30229644335113, -0.029566309824797 ]


ETC orthologic centers:

For P=X(4), N1, N2, N3, O1, O2, O3 coincide with N


For P=X(10121), the orthologic centers are:

Qna = N1N2N3->ABC = X(10120)

Qan = ANC->N1N2N3 = a complicated center not related to ETC’s


Properties for P on K:

The locus of Qan is K.  
The locus of Qao is K.
Qna = Qoa
The locus of Qna is another hard to calculate conic

 
César Lozada