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Floor's Monthly problem and points of intersection of circumcircles

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  • Angel
    Dear Hyacinthists, Let ABC be a triangle and G its centroid. The midpoints of sides BC, CA, AB are labeled D, E, F. The circumcenters Ab, Ac, Bc, Ba, Ca, Cb
    Message 1 of 1 , Apr 2, 2013
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      Dear Hyacinthists,

      Let ABC be a triangle and G its centroid.
      The midpoints of sides BC, CA, AB are labeled D, E, F. The circumcenters Ab, Ac, Bc, Ba, Ca, Cb of the 6 triangles AGE, AGF, BGF, BGD, CGD ,CGE lie on the Van Lamoen Circle. Its center is X(1153).

      Figure:
      http://amontes.webs.ull.es/pdf/SeisTriangulosX2.pdf

      Points of intersection of the circumcircles of these triangles, other than G:

      Ka=(Bc)/\(Cb), Kb=(Ca)/\(Ac), Kc=(Ab)/\(Ba);
      Gab=(Ab)/\(Ca), the symmetrical point de G w/r to the line AbCa,
      Gac=(Ac)/\(Ba), the symmetrical point de G w/r to the line AcBa,
      Gbc=(Bc)/\(Ab),
      Gba=(Ba)/\(Cb),
      Gca=(Ca)/\(Bc),
      Gcb=(Cb)/\(Ac).

      1. The lines AKa, BKb, CKc concur in the symmedian, X(6).

      2. If A', B', C' are the points of intersection of the lines (AbCa, AcBa), (BcAb,BaCb), (CaBc,CbAc), respectively, then AA', BB', CC' concur in the triangle center with (6-9-13)-search number 1.88868384192281470263 and first barycentric coordinate:

      1/(SA^2-SB SC-2 S^2).


      Note:
      The triangle center of the first barycentric coordinate
      1/(SA^2-SB SC+2 S^2)
      is X(262)= isogonal conjugate of midpoint of Brocard diameter (X(182))

      3. The lines GabGac, GbcGba, GcaGcb determine a triangle A''B''C'' perspective with ABC, and the first coordinate of the perspector is:

      1/((2a^2 - b^2 - c^2)^2 - 9b^2c^2).

      Best regards
      Angel Montesdeoca
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