15397Re: [EMHL] Re:Cevian Trace Equal Area Points
- Aug 2, 2007Dear Nikos
I give you my own construction of the equal area axis based on vectors.
So you could compare it with yours.
Given 4 points A, B, A', B' such that line AB and A'B' are through point O,
let L+ be the line locus of points M such that:
Area(MAB) = Area(MA'B')
and L- be the line locus of points M such that:
Area(MAB) = -Area(MA'B')
where Area means signed area.
Then L+ is the line through O directed by the vector V(AB) - V(A'B')
and L- is the line through O directed by the vector V(AB) + V(A'B').
That's why the four lines AB, A'B', L+, L- are in an harmonic range.
On 8/2/07, Francois Rideau <francois.rideau@...> wrote:
> Dear Nikos and Tuan
> > > Two segments XY, X'Y' are on given two lines. To
> > > find the locus of Z such that Area(ZXY) =
> > > Area(ZX'Y').
> > > If areas are not signed then the locus is two
> > > lines passing intersection of two given lines.
> > I note this fact:
> > These 2 lines of this locus are harmonic conjugate wrt lines XY and
> > X'Y'.
> So if abc and a'b'c' are 2 inscribed triangles in ABC and if we use
> ordinary area ( i.e not signed), the Tuan-Nikos six lines cut themselves
> in 4 points forming an harmonic quad wrt ABC, each of these four points
> beeing an equal area center wrt abc and a'b'c'. I will have to think about
> it in choo-choo theory!
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