HYACINTHOS 26682

 [Antreas P. Hatzipolakis]:

 

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

Denote:

(Na), (Nb), (Nc) = the NPCs of IBC, ICA, IAB, resp.

The perpendicular bisector of BC intersects (Na) at A* other than A', the midpoint of BC 
The perpendicular bisector of CA intersects (Nb) at B* other than B', the midpoint of CA
The perpendicular bisector of AB intersects (Nc) at C* other than C', the midpoint of AB

Pa, Pb, Pc = same points on the Euler lines of  A*B'C', B*C'A', C*A'B', resp.

For (Pa, Pb, Pc) = orthocenters, circumcenters, NPC centers of  A*B'C', B*C'A', C*A'B', resp. the triangles ABC, A*B*C* are orthologic. Orthologic centers (PaPbPc, ABC) = O, N, X140, resp.

Conjecture:

ABC, PaPbPc are orthologic and the orthologic center (PaPbPc, ABC) lies on the Euler line of ABC.


[César Lozada]:

Conjecture proved.

 

If Pa=P-of-A*B’C’ and similarly Pb and Pc,  then, for P such that OP=t*OH, the orthologic centers are:

 

Qa=Q(A->Pa) = 1/(b+c)/((a^4-(b+c)*a^3+(b^2+ c^2)*a^2+(b^2-c^2)*(b-c)*a-2*( b^2-c^2)^2)*t+(b+c)*a^3-(b^2+ c^2)*a^2-(b^2-c^2)*(b-c)*a+(b^ 2-c^2)^2)*(a^2+2*(b+c)*a-(b+c) ^2) :: (trilinears)

= on a circumconic through X(21) and center with complicated coordinates and not related to ETCs.

 

Qp=Q(Pa->A) = ((2*a^4-(b^2+c^2)*a^2-(b^2-c^ 2)^2)*t-(b^2+c^2)*a^2+(b^2-c^ 2)^2)/a : : (trilinears)

= (1+t)*X(3) + (1-t)*X(4)

= complement of P

 

Example:

Qa(X (4) ) = X(21)X(90) ∩ X(28)X(662)

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

= On lines: {1, 1958}, {3, 1812}, {21, 90}, {28, 662}, {72, 1444}, {73, 7364}, {78, 1790}, {81, 386}, {86, 443}, {100, 3193}, {283, 4855}, {333, 631}, {572, 2287}, {1437, 1792}, {1959, 2939}, {2327, 10884}, {4197, 5333}, {4658, 8897}

= {X(1437), X(5440)}-Harmonic conjugate of X(1792)

= [ 3.824451550370398, -0.86460953696541, 2.474108830404862 ]

 

Qp(X(4)) = X(3)

 

César Lozada