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Re: (A,B,F) construction problem

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  • Giovanni Artico
    ... orthocenter. ... Beta. ... Dear Jack, playing with C.a.R. around this interesting figure, it seems that one can notice two facts (I have no proof, only
    Message 1 of 3 , Oct 31, 2008
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      > Now, back to (B,C,F).
      > The incenter I must lie, for sure, on the parabola Alpha
      > for which d(I,F) = d(I,BC).
      > If we call M the midpoint of BC,
      > N (center of the nine-point circle) must lie on r, the
      > perpendicular bisector of FM, that is tangent to Alpha.
      > So, pick a candidate N on r, draw the nine-point circle
      > Gamma (having center N and radius NM) then intersect
      > - the image of Gamma under a homothety with center B and factor 2
      > - the image of Gamma under a homothety with center C and factor 2
      > in order to find the candidate A-vertex and the candidate H-
      orthocenter.
      > As N travels on r, A and H travel on a rectangular hyperbola
      > having center F, that is, obviously, the Feuerbach hyperbola
      > of the real-ABC triangle. Moreover, the incenter of the candidate
      > triangle ABC (built in that way, from N) travels on a hyperbola
      Beta.
      >

      Dear Jack,
      playing with C.a.R. around this interesting figure, it seems that one
      can notice two facts (I have no proof, only visual evidence):
      - Beta seems not to be an hyperbola (but if you have a proof ...)
      - the positions of I solving the problem are at the intersections of
      the parabola Alpha and the Feuerbach hyperbola (2 solutions of the
      problem). So the question would be: to find the intersections of a
      parabola with an hyperbola having the center at the focus of the
      parabola: is this solvable by ruler and compass ?

      Giovanni
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