14757ÈÝìá: Re: [EMHL] Construction
- Jan 14, 2007Dear Nikolaos
> > I tried to find for which points P the pedal triangle[JPE]
> > of P A'B'C' has vertices equidistant from A,B,C
> > i.e. AA' = BB' = CC'.
> > These points must be symmetric wrt H.
> > I found that these points must be on these lines.
> > From a sketch it seems to exist only two such points
> > on one of these lines.
> > Can we construct these points?
> Your points are the homothetic of the focii of the Steinerellipse.
> circumellipse in (L,3/2) where L = de Longchamps point.
> For both points, AA' = 3.T/2 where T = semimajor axis of the
A little explanation :
it is well known that the focii of the Steiner circumellipse have
cevian of equal lengths.
As they lie on the Lucas cubic, their cevian triangles are the pedal
triangles of the required points
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