HYACINTHOS 27235

[Antreas P. Hatzipolakis]:

 

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

Denote:

Ab, Ac = the reflections of A' in BO, CO, resp.
Bc, Ba = the reflections of B' in CO, AO, resp.
Ca, Cb = the reflections of C' in AO, BO, resp.

P, Pa, Pb, Pc = same points on the Euler lines of ABC, AAbAc, BBcBa, CCaCb, resp. such that PO/PH = t (number) etc.

ABC, PaPbPc are orthologic.
The orthologic center (PaPbPc, ABC) is the P.
Which is the locus of the other one (ABC, PaPbPc) as P moves on the Euler line?


[César Lozada]:

 

When P is an OH-absolutely positioned center on the Euler line(*),   triangles ABC and PaPbPc are orthologic and the orthologic center A->Pa = Qa(P) lies on the circumconic:

 

q = ∑ [(SB-SC)*(SA^2+SB*SC)*(SA-8*R^ 2+SW)*y*z ] =0 (barys)

This conic passes through ETC’s 264, 1975, 10621, 10622 and has perspector:

 

q_pers = X(297)X(525) ∩ X(2451)X(17215)

= (b^2-c^2)*(a^4-(b^2+c^2)*a^2+ 2*b^2*c^2)*((b^2+c^2)*a^4-2*( b^2-c^2)^2*a^2+(b^4-c^4)*(b^2- c^2)) : : (barys)

= on lines: {297, 525}, {2451, 17215}

= [ 5.4619738886654760, -0.7348553712368464, 1.6284994826104380 ]

 

The center of q is not related to any ETC center X(i) for 1<=i<=17772

 

IF OP/OH=t, then

Qa(P) = f(SB,SC,SA)*f(SC,SA,SB) : : (barys)

where

f(SA,SB,SC)=(SB+SC)* ((35*R^2+4*SA-8*SW)*S^2+(3*R^ 2-4*SW)*SA^2)*t-(S^2+SA^2)*R^2

 

ETC pairs (P,Qa): (2,264), (3,1975), (4,10621), (5,10622)

 

No other interesting Qa(P) found.

 

César Lozada

 

 

 

(*) Centers P for which OP/OH=t=real constant not depending on the shape of the triangle.