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Re: [EMHL] An Isocubic

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  • Antreas P. Hatzipolakis
    ... Let s improve the configuration: Let ABC be a triangle, P a point and A*B*C* the circumcevian triangle of P. Denote: A b := AB* / BC A c := AC* / BC Oab
    Message 1 of 4 , Aug 6, 2004
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      [APH]:

      >Let ABC be a triangle, and P a point.
      >
      >Denote:
      >
      >(Reflection of BP in BC) /\ AC := Ab
      >(Reflection of CP in BC) /\ AB := Ac
      >
      >Oab := The circumcenter of ABAb
      >Oac := The circumcenter of ACAc
      >
      >A' := BOab /\ COac. A" := AA' /\ BC
      >
      >Similarly B',B"; C',C"
      >
      >Which is the locus of P such that:
      >
      >(1) ABC, A'B'C' are perspective ?
      >
      >(2) A", B", C" are collinear?

      Let's improve the configuration:

      Let ABC be a triangle, P a point and A*B*C* the circumcevian
      triangle of P.

      Denote:

      A'b := AB* /\ BC
      A'c := AC* /\ BC

      Oab := The circumcenter of ABA'b
      Oac := The circumcenter of ACA'c

      A' := BOab /\ COac. A" := AA' /\ BC

      Similarly B',B"; C',C"

      Which is the locus of P such that:

      (1) ABC, A'B'C' are perspective ?

      (2) A", B", C" are collinear?

      The points ABAbA'b are concyclic, so
      the Oab is the same point in both configurations.

      Now in the new configuration, denote:

      Hab := The Orthocenter of ABA'b
      Hac := The Orthocenter of ACA'c

      A1 := BHab /\ CHac. A2 := AA1 /\ BC

      Similarly B1,B2; C1,C2

      Which is the locus of P such that:

      (1') ABC, A1B1C1 are perspective ?

      (2') A2, B2, C2 are collinear?

      We have Locus(1) = Locus(1'), and Locus(2) = Locus(2')


      Antreas




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