HYACINTHOS 25910

[Tran Quang Hung]¨

 

 

Let ABC be a triangle.

 

P is a point on Euler line.

 

A',B',C' are reflections of P in BC,CA,AB, reps.

 

da is Euler line of pedal triangle of P wrt A'BC.

 

db is Euler line of pedal triangle of P wrt B'CA.

 

dc is Euler line of pedal triangle of P wrt C'AB.

 

Then da,db,dc are concurrent.

 

It seems the intersection point lies on a parabola ?

 

I see that Pedal triangle of P wrt A'BC is isosceles triangle. So Euler line of this triangle is the perpendicular bisector of the segment connecting the projection of P on A'B,A'C.

 

[Angel Montesdeoca]:

****  When P moves on the line of Euler  the intersection point Q of Euler lines da, db, dc  lies on the hyperbola which passes through the five points:   X(3)=circumcenter,   X(5)=NPC center,  X(5663)= the point of intersection of the Euler line of  orthocentroidal triangle and the line at infinity,  X(10539) = homothetic center of  Johnson triangle  and  the triangle described in ADGEOM #2697 (8/26/2015, Tran Quang Hung), and  

((a^2-b^2-c^2) (4 a^4-(b^2-c^2)^2+a^2 (b^2+c^2)):...:..) = midpoint of centroid and X(6800) [= X(3)X(74)∩X(6)X(23)].


****  When P moves on the line of Euler the envelope od line PQ is the parabola determined by the four points:
 
   X(4)=orthocenter, 
   X(2777) = isogonal conjugate of  the  reflection of X(1304) in circumcenter
    [ Let A', B', C' be the intersections of the Euler line and lines BC, CA, AB, resp. The circumcircles of AB'C', BC'A', CA'B' concur in X(1304). (Randy Hutson, February 10, 2016)  ],
   X(6225) = intersection of  tangent lines to K007 (the Lucas cubic) at orthocenter  and  at  De Longchamps point.  (Randy Hutson, May 5, 2015),
     X(6759)  = inverse-in-circumcircle of X(6760)
[ Let A'B'C' be the reflection of the anticevian triangle of X(3) in the trilinear polar of X(3). The lines AA', BB', CC' concur in X(6760)].

The focus and the directirx of this parabola are X(1304) and the line X(107)X(110).

Angel Montesdeoca