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a problem of II OMCC 2000

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  • Ricardo Barroso
    Dear Hyacinthos: In II Olimpiada matematica de centroamerica y del Caribe San Salvador 10 julio 2000, the problem 5 is: Let ABC a triangle with acutes
    Message 1 of 2 , Aug 4, 2002
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      Dear Hyacinthos:
      In "II Olimpiada matematica de centroamerica y del Caribe"
      San Salvador 10 julio 2000, the problem 5 is:

      Let ABC a triangle with acutes angles.
      C1 and C2 circles with diameters AB and CA.
      C2 meet to AB in point F (F other of A).
      C1 meet to CA in point E (E other of A).
      BE meet to C2 in P.
      CF meet to C1 in Q.
      Proof with length AP and lengt AQ they are equals.

      Reference:
      http://www.geocities.com/CollegePark/7174

      Friendly
      Ricardo

      Home Page: http://www.pdipas.us.es/r/rbarroso






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    • jpehrmfr
      Dear Ricardo ... The triangle APC is rectangular in P and E is the projection of P on AC; hence AP^2 = AE*AC; same way AQ^2 = AF*AB. As B,C,E,F are concyclic,
      Message 2 of 2 , Aug 4, 2002
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        Dear Ricardo

        > In "II Olimpiada matematica de centroamerica y del Caribe"
        > San Salvador 10 julio 2000, the problem 5 is:
        >
        > Let ABC a triangle with acutes angles.
        > C1 and C2 circles with diameters AB and CA.
        > C2 meet to AB in point F (F other of A).
        > C1 meet to CA in point E (E other of A).
        > BE meet to C2 in P.
        > CF meet to C1 in Q.
        > Proof with length AP and lengt AQ they are equals.

        The triangle APC is rectangular in P and E is the projection of P on
        AC; hence AP^2 = AE*AC; same way AQ^2 = AF*AB. As B,C,E,F are
        concyclic, we have AE*AC = AF*AB.
        Friendly. Jean-Pierre
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