HYACINTHOS 26953

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and P a point.

Denote:

(Oa), (Ob), (Oc) = the circumcircles of NBC, NCA, NAB, resp.

(O1), (O2), (O3) = the circumcircles of PBC, PCA, PAB, resp.

Ra, Rb, Rc = the radical axes of ((N),(Oa)), ((N),(Ob)), ((N),(Oc)), resp.

 

R1, R2, R3 = the radical axes of ((N),(O1)), ((N),(O2)), ((N),(O3)), resp.

AaBbCc = the triangle bounded by Ra, Rb, Rc

A1B2C3 = the triangle bounded by R1, R2, R3

 

Which is the locus of P such that:

1. AaBbCc, A1B2C3 are perspective.
The entire plane?

 

2. AaBbCc, A1B2C3 are orthologic.
The Euler line  + Something_else  ?


[Angel Montesdeoca]:

 

***AaBbCc, A1B2C3 are perspective for all P, with perspector the inverse-in-nine-point-circle of P.

*** The locus of P such that  AaBbCc, A1B2C3 are orthologic is a bicircular circum-quartic passing through X(4) and X(5).  
The fourth point of intersection of this quartic with the circumcircle is:

Z =  1 / (a^18 (b^2+c^2)   
    -a^16 (5 b^4+14 b^2 c^2+5 c^4)
    +a^14 (8 b^6+45 b^4 c^2+45 b^2 c^4+8 c^6)
    -a^12 (57 b^6 c^2+94 b^4 c^4+57 b^2 c^6)
    +a^10 (-14 b^10+23 b^8 c^2+74 b^6 c^4+74 b^4 c^6+23 b^2 c^8-14 c^10)
    +a^8 (14 b^12+19 b^10 c^2-29 b^8 c^4-32 b^6 c^6-29 b^4 c^8+19 b^2 c^10+14 c^12)   
    +a^6 b^2 c^2 (-41 b^10+30 b^8 c^2+2 b^6 c^4+2 b^4 c^6+30 b^2 c^8-41 c^10)
    -a^4 (b^2-c^2)^2 (8 b^12-29 b^10 c^2-15 b^8 c^4-17 b^6 c^6-15 b^4 c^8-29 b^2 c^10+8 c^12)   
    +a^2 (b^2-c^2)^4 (5 b^10-8 b^8 c^2-15 b^6 c^4-15 b^4 c^6-8 b^2 c^8+5 c^10)
    -(b^2-c^2)^6 (b^2+c^2)^2 (b^4-3 b^2 c^2+c^4) ) : ... : ....
   
with  (6 - 9 - 13) - search numbers:    0.369196524616796, -0.457212835608297, 3.78679804904391

Angel Montesdeoca