HYACINTHOS 26522

[Le Viet An]:

Let ABC be a triangle.

Denote:

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

The perpendicular to AN at N intersects again (Ob), (Oc) at Ab, Ac, resp.

Similarly Bc, Ba and Ca, Cb.

Let O1, O2, O3 be the circumcenters of NBcCb, NCaAc, NAbBa, resp.

1. The Euler lines of the triangles NBaCa, NCbAb, NAcBc are concurrent.
Point of concurrence?

2. The six point Oa, Ob, Oc, O1, O2, O3 are concyclic.
Center of the circle?

 

 

[César Lozada]:

1)      X(14072)

 

2)      Squared-radius: S^8*(5*R^2-2*SW)^2* CyclicProduct[ ((2*SA+R^2)/(S^2+SB*SC)^2 ]/4

 

Center:

OO = X(3)X(2888) ∩ X(5)X(252)

= (cos(2*A)-cos(4*A)+7/2)*cos(B- C)+(2*cos(A)-2*cos(3*A))*cos( 2*(B-C))+(cos(2*A)+1)*cos(3*( B-C))-cos(5*A)+cos(A)-cos(3*A) : :  (trilinears)

= on lines: {3, 2888}, {4, 14051}, {5, 252}, {1209, 6150}

= {X(3), X(2888)}-Harmonic conjugate of X(14072)

= [ 5.428024100252400, 8.11965160778560, -4.485874677445219 ]

 

No ETC centers lie on this circle.

X(5) and X(1157) are inverse points w/r to this circle.

 

César Lozada