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9170Re: [EMHL] Orthologic Triangles

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  • jpehrmfr
    Feb 1, 2004
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      Dear Antreas, Fred and Darij
      > Let ABC be a triangle, P a point, and (Oa, Ob, Oc)
      > the Circumcenters of the triangles PBC, PCA, PAB, resp.
      > The locus of P such that the perpendiculars
      > from A,B,C to POa, POb, POc are concurrent is:
      > infinity line*circumcircle*KjP cubic.

      Consider three lines La, Lb, Lc going respectively through A, B, C
      and such as (La,POa) = (Lb,POb) = (Lc,POc) = phi where (L1,L2) is
      the measure modulo Pi of the oriented angle of lines L1, L2.
      Then La, Lb, Lc concur if and only if
      (AP,BC)+(BP,CA)+(CP,AB) = Pi/2-phi
      Thus, we get the MacCay cubic for phi = 0, the Kjp cubic for phi =
      Pi/2 and, for any theta, a member of the pencil generated by Kjp and
      Friendly. Jean-Pierre
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