Re: (A,B,F) construction problem
> Now, back to (B,C,F).orthocenter.
> 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-
> As N travels on r, A and H travel on a rectangular hyperbolaBeta.
> 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
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 ?