Loading ...
Sorry, an error occurred while loading the content.

22493Re: A Circle

Expand Messages
  • Antreas Hatzipolakis
    Jul 3, 2014
    • 0 Attachment
      General Theorem:

      Let ABC be a triangle, P a point , Oa,Ob,Oc the circumcenters
      of PBC,PCA,PAB, resp. and Q an other point.

      The reflections of QOa in the sidelines of PBC bound a triangle,
      whose the incenter or an excenter Ja is lying on the circumcircle of PBC.
      Similarly Jb,Jc.
      The points P, Ja, Jb, Jc are concyclic.

      Applications:

      Let  P be a point on the Neuberg cubic.

      1. The Euler lines of PBC, PCA, PAB concur at a point Qe.
      The reflections of the Euler line of PBC in the sidelines PB,BC,CP
      concur at a point Epa on the circumcircle of PBC. Similarly Epb, Epc.

      Which point is the center of the circle (P, Epa, Epb, Epc)?
      For P = I, the center of the circle is the midpoint of OI.
      How about other simple points on the Neuberg cubic ? (Fermat's etc)

      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.
      Which is the center of the circle (P, Jpa, Jpb, Jpc)?

      For P = I:
      The circle has center X(4297), segment X(1)X(20) as diameter,
      and also passes through X(3109).
      (Randy Hutson, Anopolis #1738 and #1740)

      How about other simple points on the Neuberg cubic?

      PS: We have the locus problems:
      Which is the locus of the center of the circle:
      1. (P, Epa, Epb, Epc)
      2. (P, Jpa, Jpb, Jpc)
      as P moves on the Neuberg cubic?

      Antreas

      [APH]
      Sometime ago I posted this:

      Let ABC be a triangle and La, Lb, Lc the Euler lines of IBC,ICA,IAB, resp.

      Denote:

      Ea = the reflection of point of La wrt triangle IBC
      (ie the point of concurrence of La in the sidelines of IBC,
      lying on the circumcircle of IBC)
      Similarly Eb, Ec.
      The points Ea,Eb,Ec are lying on the circle with diameter OI.

      A natural question is this:
      How about if we replace the Euler lines with Brocard axes (OK-lines)
      (they are concurrent as well) and take the points by the generalized reflection theorem?

      That is:

      Let La, Lb, Lc be the Brocard axes of IBC,ICA,IAB resp.

      The reflections of La in the sidelines of IBC bound a triangle
      whose the incenter or an excenter, call it Ja, is lying on the
      circumcircle of IBC.
      Similarly, the reflections of Lb in the sidelines of ICA bound
      a triangle, whose the incenter or an excenter Jb is lying on
      the circumcircle of ICA and the reflections of Lc in the sidelines
      of IAB bound a triangle whose the incenter or an excenter Jc
      is lying on the circumcircle of IAB.

      I think the points Ja,Jb,Jc and I are concyclic.

      Which point is the center of the circle?

      Antreas



      --
      http://anopolis72000.blogspot.com/
    • Show all 3 messages in this topic