HYACINTHOS 26212

[Antreas P. Hatzipolakis]:

 

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

Denote:

Ab, Ac = the orthogonal projections of A' on BB', CC', resp.

 

A* = the orthogonal projection of P on AbAc. Similarly B*,C*.

Which is the locus of P such that ABC, A*B*C* are: 

 

1. perspective

 

2. orthologic?
G lies on the locus.
The orthologic center (A*B*C*, ABC) lies on the Euler line.

[César Lozada]:

 

1)      Very complicated: a degree-21 circumcurve through X(13) (at least). The perspector for P=X(13) is X(13)

2)      {sidelines} \/ { a circum-sixtic through no ETCs} \/ {an excentral circum-nonic through I,G}

Orthologic centers:

 

For P=I
O(A->A*)=X(65); O(A*->A)=X(1)

 

For P=G

O(A->A*) = X(3)X(524)∩X(4)X(111)

= (5*a^2-b^2-c^2)/(a^4-b^4+4*b^ 2*c^2-c^4)  : :  (barycentrics)

=  on cubics K009, K043 and these lines: {2,12505}, {3,524}, {4,111}

= [ -0.124762628233785, 4.78256741562842, 0.387239022580294 ]

 

O(A*->A) = 18*SW*S^4+SA*(4*SW^3-9*(12*R^ 2-SW)*S^2)*(SB+SC) : : (barycentrics)

= (108*R^2*S^2-27*S^2*SW-4*SW^3) *X(3)-9*S^2*SW*X(4)

= On the Euler line {2,3}

= [ 3.162585203199529, 2.28472230068163, 0.599278949189623 ]

 

César Lozada