Re: (A,B,F) construction problem

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• ... 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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