Dear 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.

Friendly

Francois

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!

> friendly

> Francois

>

> ___________________________________________________________

> > �������������� Yahoo!;

> > ���������� �� ���������� �������� (spam); �� Yahoo! Mail

> > �������� ��� �������� ������ ��������� ���� ��� �����������

> > ��������� http://login.yahoo.com/config/mail?.intl=gr

> >

> >

> >

> >

> > Yahoo! Groups Links

> >

> >

> >

> >

>

[Non-text portions of this message have been removed]