HYACINTHOS 25105

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the cevian triangle of G.

Denote:

A"B"C" = the reflection trianle of A'B'C'

(ie A", B", C" = the reflections of A', B', C' in B'C', C'A', A'B', resp.)

Na, Nb, Nc = the NPC centers of AGA" BGB", CGC".

ABC, NaNbNc are orthologic.

The orthologic center (NaNbNc, ABC) lies on the Euler line of ABC: it is the midpoint of GN.

The other one (ABC, NaNbNc) ?


[César Lozada]:

Orthologic centers:
OC(NaNbNc, ABC) = X(547) = midpoint-of-GN

 

OC(ABC, NaNbNc)= a/(8*a^4-13*(b^2+c^2)*a^2+6*( b^2-c^2)^2) : : (trilinears)

= isogonal conjugate of {2,187}/\{39,632}

= [ 0.383358603501482, 0.60442116386660, 3.045284320691435 ]

 

César Lozada