- Jul 3 11:18 AMAntreas1. (P, Epa, Epb, Epc)PS: We have the locus problems:How about other simple points on the Neuberg cubic?(Randy Hutson, Anopolis #1738 and #1740)For P = I:Which is the center of the circle (P, Jpa, Jpb, Jpc)?How about other simple points on the Neuberg cubic ? (Fermat's etc)For P = I, the center of the circle is the midpoint of OI.Which point is the center of the circle (P, Epa, Epb, Epc)?The reflections of the Euler line of PBC in the sidelines PB,BC,CPApplications:The points P, Ja, Jb, Jc are concyclic.Similarly Jb,Jc.whose the incenter or an excenter Ja is lying on the circumcircle of PBC.The reflections of QOa in the sidelines of PBC bound a triangle,of PBC,PCA,PAB, resp. and Q an other point.General Theorem:Let ABC be a triangle, P a point , Oa,Ob,Oc the circumcenters

LetÂ P be a point on the Neuberg cubic.

1. The Euler lines of PBC, PCA, PAB concur at a point Qe.

concur at a point Epa on the circumcircle of PBC. Similarly Epb, Epc.

2. The Brocard axes of PBC, PCA, PAB concur at a point Qb.

The reflections of the Euler line of PBC in the sidelines PB,BC,CP

bound a triangle whose the incenter or an excenter Jpa is lying on the

circumcircle of PBC. Similarly Jpb, Jpc.

The circle has center X(4297), segment X(1)X(20) as diameter,

and also passes through X(3109).

Which is the locus of the center of the circle:

2. (P, Jpa, Jpb, Jpc)

as P moves on the Neuberg cubic?[APH]AntreasWhich point is the center of the circle?I think the points Ja,Jb,Jc and I are concyclic.is lying on the circumcircle of IAB.of IAB bound a triangle whose the incenter or an excenter Jcthe circumcircle of ICA and the reflections of Lc in the sidelinesSimilarly, the reflections of Lb in the sidelines of ICA boundcircumcircle of IBC.whose the incenter or an excenter, call it Ja, is lying on theThe reflections of La in the sidelines of IBC bound a triangleLet La, Lb, Lc be the Brocard axes of IBC,ICA,IAB resp.That is:(they are concurrent as well) and take the points by the generalized reflection theorem?How about if we replace the Euler lines with Brocard axes (OK-lines)A natural question is this:The points Ea,Eb,Ec are lying on the circle with diameter OI.Similarly Eb, Ec.lying on the circumcircle of IBC)(ie the point of concurrence of La in the sidelines of IBC,Ea = the reflection of point of La wrt triangle IBCDenote:Sometime ago I posted this:Let ABC be a triangle and La, Lb, Lc the Euler lines of IBC,ICA,IAB, resp.

a triangle, whose the incenter or an excenter Jb is lying on

--

http://anopolis72000.blogspot.com/ - << Previous post in topic