Let ABC be a triangle and A'B'C' the pedal triangle of H (orthic triangle).
Denote:
La = the reflection of HA' in B'C'
Lb = the reflection of HB' in C'A'
Lc = the reflection of HC' in A'B'
Ma = the reflection of La in OA.
Mb = the reflection of Lb in OB.
Mc = the reflection of Lb in OC.
The lines Ma,Mb,Mc concur at a point Q. Denote:
A" = BC /\ Ma
B" = CA /\ Mb
C" = AB /\ Mc
The triangles ABC, A"B"C" are orthologic, with orthologic centers Q,S.
Locus:
Let ABC be a triangle P,P* two isogonal conjugate points and A'B'C' the pedal triangle of P.
Denote:
La = the reflection of PA' in B'C'
Lb = the reflection of PB' in C'A'
Lc = the reflection of PC' in A'B'
Ma = the reflection of La in AP*.
Mb = the reflection of Lb in BP*.
Mc = the reflection of Lb in CP*.
Denote:
A" = BC /\ Ma
B" = CA /\ Mb
C" = AB /\ Mc
(Ma,Mb,Mc are perpendiculars to BC,CA,AB, resp.)
Which is the locus of P such that:
1. ABC, triangle bounded by (Ma,Mb,Mc) are perspective? Special case: Ma,Mb,Mc be concurrent.
2. ABC, A"B"C" are orthologic?
Antreas P. Hatzipolakis, 14 March 2013
ADDENDUM (9/9/19)
Q = X(34224) = X(3)X(70)∩X(4)X(6)
S = X(34225) = ISOGONAL CONJUGATE OF X(34224)
Q=-4*b^4*a^6-2*c^2*a^6*b^2-4*c^4*a^6-2*a^10+2*b^6*c^4-3*c^2*b^8+c^10-3*c^8*b^2-2*c^8*a^2+2*c^6*a^4+5*a^8*c^2+b^10+2*c^6*b^4-2*c^4*b^2*a^4+2*c^6*b^2*a^2+2*c^2*b^6*a^2-2*c^2*b^4*a^4-2*b^8*a^2+2*b^6*a^4+5*a^8*b^2 : :
ΑπάντησηΔιαγραφή(Barycentrics)
Cesar Lozada
More simple form (barycentrics)
ΑπάντησηΔιαγραφή-2*a^10+(5*(c^2+b^2))*a^8-(c^2*(b^2+4*c^2)+b^2*(4*b^2+c^2))*a^6+(2*(c^2+b^2))*(b^2-c^2)^2*a^4+(b^2-c^2)^2*((c^2+b^2)^2-b^2*c^2)*a^2+(c^2+b^2)*(b^2-c^2)^4 : :