Dear Francois

thank you very much for explaining

what is equicenter E and areal point Q.

Another simple construction for Q comes from the

following property:

If A'B'C', A"B"C" are two inscribed triangles

in ABC we construct the line La that passes

through A and is parallel to the line that connects

the mid points of B'C", B"C' and similarly

construct the lines Lb, Lc. These three lines

are concurrent at Q the areal point such that

the signed area (QA'A") = (QB'B") = (QC'C").

If in barycentrics the points are

A' = (0 : p : 1-p)

B' = (1-q : 0 : q)

C' = (r : 1-r : 0)

A" = (0 : p' : 1-p')

B" = (1-q' : 0 : q')

C" = (r' : 1-r' : 0)

then the point Q is the isotomic conjugate

of (p-p' : q-q' : r-r') and the signed area is

(ABC).(p-p').(q-q').(r-r')/K where

K = pq+qr+rp+p'q'+q'r'+r'p'+pp'+qq'+rr'-

-(p+q+r)(p'+q'+r').

Best regards

Nikos Dergiades

> Given any 2 inscribed triangles PaPbPc and UaUbUc of

> the reference triangle

> ABC, (cevian triangles are not needed!), there

> exists (in general), a unique

> point Q in the plane, such that:

>

> Area(QPaUa) = Area(QPbUb) = Area(QPcUc)

>

> In this formula , area means signed area of course!

>

> Now I give you a way (without proof for lack of

> room) to construct this

> point.

> 1° Let E be the fixed point of the affine map f

> sending PaPbPc on UaUbUc.

> E is the equicenter, following Neuberg notation. Why

> equicenter? Simply

> because E has same barycentrics in PaPbPc and

> UaUbUc.

> There is a very known construction of E, due to

> Sollertinski. I have already

> talked about it in some previous posts.

>

> 2° Let g be the affine map sending PaPbPc on ABC,

> then Q = g(E).

> It is also very easy to have a construction of Q.

>

> This area equality is the reason why I have called Q

> the areal center in

> choo-choo theory.

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