HYACINTHOS 23055

 

Antreas P. Hatzipolakis

Let ABC be a triangle and P a point.

Denote:

Pa = the reflection of P in BC
Pab, Pac = the reflections of Pa in PB, PC, resp.

N1 =  the NPC center of PaPabPac

Similarly N2, N3.

Which is the locus of P such that ABC, N1N2N3 are orthologic?

Points on the locus:

P = O
Orthologic center (PaPbPc, ABC) = O

P = N

Orthologic center (ABC, PaPbPc) on the circumcircle of ABC

P = H

Orthologic centers?

APH

[César Lozada]:

 

>Which is the locus of P such that ABC, N1N2N3 are orthologic?

Excentral-Circum-septic with trilinears:

-(a^2-c^2)*a*c*b^2*u^2*v^5*(c^2+a^2-b^2)-(a^2-c^2)*a*c*b^2*v^5*w^2*(c^2+a^2-b^2)-(b^2-c^2)*b*c*a^2*u^5*v^2*(-b^2-c^2+a^2)+(a^2-b^2)*a*b*c^2*u^2*w^5*(a^2+b^2-c^2)+(a^2-b^2)*a*b*c^2*v^2*w^5*(a^2+b^2-c^2)-(b^2-c^2)*b*c*a^2*u^5*w^2*(-b^2-c^2+a^2)-(a^2-c^2)*u*w*b^2*v^5*(c^2+a^2-b^2)^2+(5*(b^2-c^2))*b*c*a^2*u^3*v^2*w^2*(-b^2-c^2+a^2)-(5*(a^2-b^2))*a*b*c^2*u^2*v^2*w^3*(a^2+b^2-c^2)+(5*(a^2-c^2))*a*c*b^2*u^2*v^3*w^2*(c^2+a^2-b^2)-b*c*(-b^4*c^2+b^2*a^4+4*b^2*a^2*c^2-2*b^4*a^2+b^6+c^6-b^2*c^4-2*c^4*a^2+c^2*a^4)*v^2*u*w^4-c*a*(-c^4*a^2+c^6+a^6+b^4*c^2-2*b^2*a^4+4*b^2*a^2*c^2+b^4*a^2-2*b^2*c^4-c^2*a^4)*v*u^4*w^2+(b^2-c^2)*v*w*a^2*u^5*(-b^2-c^2+a^2)^2+(a^2-b^2)*u*v*c^2*w^5*(a^2+b^2-c^2)^2-b^2*c*a*(-c^4-b^4+2*b^2*c^2+2*a^4-a^2*b^2-3*c^2*a^2)*v^3*u^4-b*c*a^2*(3*b^2*c^2-2*b^4+a^4+a^2*b^2+c^4-2*c^2*a^2)*v^4*u^3-b^2*c*a*(b^4+3*c^2*a^2-2*c^4+b^2*c^2+a^4-2*a^2*b^2)*v^3*w^4-a*b*(a^6-2*b^4*c^2+4*b^2*a^2*c^2-b^2*a^4+b^2*c^4-b^4*a^2+c^4*a^2+b^6-2*c^2*a^4)*v^4*u^2*w+c*a*(-c^4*a^2+c^6+a^6+b^4*c^2-2*b^2*a^4+4*b^2*a^2*c^2+b^4*a^2-2*b^2*c^4-c^2*a^4)*v*u^2*w^4+a*b*(a^6-2*b^4*c^2+4*b^2*a^2*c^2-b^2*a^4+b^2*c^4-b^4*a^2+c^4*a^2+b^6-2*c^2*a^4)*v^2*u^4*w+b*c*a^2*(3*b^2*c^2-2*c^4+a^4+c^2*a^2+b^4-2*a^2*b^2)*u^3*w^4+b*c^2*a*(2*b^2*c^2-c^2*a^2-b^4-c^4+2*a^4-3*a^2*b^2)*u^4*w^3+b*c^2*a*(b^2*c^2+a^4-2*b^4-2*c^2*a^2+3*a^2*b^2+c^4)*v^4*w^3+b*c*(-b^4*c^2+b^2*a^4+4*b^2*a^2*c^2-2*b^4*a^2+b^6+c^6-b^2*c^4-2*c^4*a^2+c^2*a^4)*v^4*u*w^2-(4*(a^2-c^2))*b^2*a^2*c^2*v*u^3*w^3+(4*(b^2-c^2))*b^2*a^2*c^2*v^3*u*w^3+(4*(a^2-b^2))*b^2*a^2*c^2*v^3*u^3*w=0

 

Through:  I, O, H, N and ETC´s  X(1113) and X(1114)

 

>>Points on the locus:

>P = O
>Orthologic center (PaPbPc, ABC) = O

Orthologic center (ABC,Pa,Pb,Pc) = (4*cos(B)^2-1)*(4*cos(C)^2-1)*cos(B)*cos(C) : :

          = Polar conjugate of X(323)

          =  (4,94)/\(5,93)

          = ( 0.057823351506902, 0.10216734279162, 3.543245543894689 )

>>P = N

>Orthologic center (ABC, PaPbPc) on the circumcircle of ABC

Orthologic center (ABC,Pa,Pb,Pc) = F(A,B,C)  : :

       = On Circumcircle = ( 8.408292668302881, 0.33594393418884, -0.472662550209222 )

Orthologic center (PaPbPc, ABC) = G(A,B,C) ::

       = ( 0.681537831448283, 0.29986859608845, 3.118507223947013 )

 

Where

F(A,B,C) =

[3*cos(A)+cos(3*A)+cos(5*A)-cos(7*A)

+(3*cos(2*A)-cos(6*A)+1/2)*cos(B-C)

+(3*cos(A)+3*cos(3*A)-cos(5*A))*cos(2*(B-C))

+(-cos(2*A)+cos(4*A)+2)*cos(3*(B-C))

+(cos(A)-cos(3*A))*cos(4*(B-C))]^(-1)

 

G(A,B,C) =

5*cos(A)-4*cos(3*A)+3*cos(5*A)-cos(7*A)

+(-7*cos(2*A)-cos(4*A)+cos(6*A)+7)*cos(B-C)

+(5*cos(A)-3*cos(3*A)+cos(5*A))*cos(2*(B-C))

+(-3*cos(2*A)+5/2)*cos(3*(B-C))

+(cos(A)-2*cos(3*A))*cos(4*(B-C)) 

 

>>P = H

>>Orthologic centers?

Orthologic center (ABC,Pa,Pb,Pc) = X(265)

Orthologic center (PaPbPc, ABC) =X(6240)

 

César Lozada