HYACINTHOS 23032

 

 

 

 

Antreas Hatzipolakis wrote:

Let ABC be a triangle and A'B'C' the orthic triangle.

The perpendicular bisector of the altitude AA' intersects

B'C' at A*.Similarly B*, C*

The circumcircles of A*BC, B*CA, C*AB are concurrent.

Which point is the point of concurrence?

APH

[CL]

 

The point of concurrence is (trilinears):

    Z = (b^2-c^2)^2/(a*(a^6-(b^2+c^2)*a^4+(b^4-b^2*c^2+c^4)*a^2-(b^2+c^2)*(b^2-c^2)^2)) : :

       = sin(B-C)^2/(cos(A)*(-7+4*cos(B-C)^2+4*cos(A)^2)+cos(B-C)) : :

       = Antigonal conjugate of X(115)

       = Inverse of X(3447) in Circumcircle

       = On loci:  K072, K474, K567

       = On lines:  (3,3447)

       = ( 3.705235940218338, 3.23560265347782, -0.309477019831803 )

 

César Lozada

[Randy Hutson]:

Dear Antreas and César,

Another construction of this point:  Let A'B'C' be the tangential triangle of the Kiepert hyperbola.  This point is the radical center of circumcircles of A'BC, B'CA, C'AB.

Best regards,
Randy Hutson