## 15388Re: Cevian Trace Equal Area Points

Expand Messages
• Jul 31, 2007
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

> 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
> 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.

___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών