RADICAL CENTERS - LOCI

[Antreas P. Hatzipolakis]:

(Anopolis #691)

Let ABC be a triangle, P a point and
Pa,Pb,Pc the reflections of P in BC,CA,AB, resp.

Let Rp be the radical center of the circles
(A,APa), (B, BPb), (C, CPc)
(ie the radical center of the circles centered at A,B,C,
with radii APa, BPb, CPc, resp.)


[Barry Wolk]:

(Anopolis/message/709)

Let Gp be the centroid of the pedal triangle of P. Then P,Gp,Rp are collinear,
with PGp / GpRp = 1/5.

If P=(x,y,z) barycentrics, then Rp = (- x/2 + y SC/b^2 + z SB/c^2,
x SC/a^2 - y/2 + z SA/c^2, x SB/a^2 + y SA/b^2 - z/2)

So P-->Rp is a linear map.



**** The linear map P-->Rp has three double points and three double lines because

W = a^6+b^6+3a^2b^2c^2+c^6 - (b^2+c^2)(a^4+b^2c^2) - a^2(b^4+c^4) >0.
W=4S^2 OH^2 (Francisco Javier García Capitan ADGEOM #410)

Double points:

K1=K=X(6),

K2= (a(a^5c + a^3c(b^2-2c^2) + a(-2b^4c+b^2c^3+c^5) + b(a^2-b^2+c^2)Sqrt[W]):
b(-2a^4b*c + b*c(b^2-c^2)^2 + a^2b*c(b^2+c^2) - a(a^2+b^2+c^2)Sqrt[W]):
-(c(a^6-a^4(b^2+2c^2) + (b^3-b*c^2)^2 + a^2(-b^4+2b^2c^2+c^4) +
2a*b*c*Sqrt[W])) )

K3= (a(a^5c + a^3c(b^2-2c^2) + a(-2b^4c+b^2c^3+c^5) - b(a^2-b^2+c^2)Sqrt[W]):
b(-2a^4b*c + b*c(b^2-c^2)^2 + a^2b*c(b^2+c^2) + a(a^2-b^2-c^2)Sqrt[W]):
-(c(a^6-a^4(b^2+2c^2) + (b^3-b*c^2)^2 + a^2(-b^4+2b^2c^2+c^4) -
2a*b*c*Sqrt[W])) )

Double lines:

K2K3= the line at infinity,
K1K2 and K1K3 (which are perpendicular)


**** The linear map P-->Rp transform ellipses into ellipses, hyperbolas into hyperbolas and parabolas into parabolas.

If P=O ---> R_p=X(382). Thus, the circumcircle is applied in an ellipse with its center at X(382)

Angel Montesdeoca, 30 Julio 2013, Creado con GeoGebra