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Re: Envelope problem

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  • jpehrmfr
    Dear Antreas ... The comnon points between the NPC and the pedal circle of P are the centers of the rectangular circumhyperbolae through P and through P*
    Message 1 of 2 , Oct 19, 2010
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      Dear Antreas
      > Le ABC be a triangle, (c) a curve on the plane
      > of ABC, and P a point on (c).
      >
      > General Problem:
      >
      > Which is the envelope of the radical axis of the NPC of ABC
      > and the pedal circle of P, as P moves on (c)?
      >
      > If (c) is the McCay cubic, then the two circles are tangent,
      > and therefore the envelope is the NPC.
      >
      > How about simple (c)'s?
      >
      > For example, (c) := the Euler line of ABC.

      The comnon points between the NPC and the pedal circle of P are the centers of the rectangular circumhyperbolae through P and through P* (isogonal conjugate of P)
      Now, if P moves on a line L going through O, the rectangular circumhyperbola going through P* is the isogonal conjugate of L (and doesn't depend on the position of P upon L); hence the radical axis goes through a fixed point of the NPC : the center of the isogonal conjugate of L
      In the case of the Euler line, the fixed point is the center of the Jerabek hyperbola
      Friendly. Jean-Pierre
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