- 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.

Χρησιμοποιείτε Yahoo!;

Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail

διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών

μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - Dear Francois and Nikos,

We denote S(T1, T2) = equal area point of two inscribed triangles T1, T2. Of course S(T1, T2) = S(T2, T1).

Suppose T1, T2, T3 are three inscribed triangles then three equal area points S(T1, T2), S(T2, T3), S(T3, T1) are always on one circumconic. Denote this circumconic as C(T1, T2, T3).

Suppose now T1, T2, T3 are three Cevian triangles of P=X(2), U=X(4) and X respectively.

There is one special locus:

The locus of X such that circumconic C(T1, T2, T3) is passing also through X is one quintic:

CyclicSum[ (b^2 - c^2)*y*z*(a^2*SA*(x^3 - y*z*(y + z - x)) + b^2*c^2*(x^3 - y*z*x)) ] = 0

This circumquintic passes:

- Vetices of antimedial triangle.

- X(2), X(4), X(69), X(110), X(2574), X(2575)

Best regards,

Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: Given triangle ABC and two points P, U with barycentrics P = (p : q : r), U = (u : v : w). PaPbPc and UaUbUc are Cevian triangles of P, U respectively.

We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))

here:

c(P) = complement of P

c(U) = complement of U

t(cd(P, U)) = isotomic conjugate of cross difference of P, U

* = barycentric product

The results:

a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :

b). Area(QPaUa) = Area(QPbUb) = Area(QPcUc)

---------------------------------

Luggage? GPS? Comic books?

Check out fitting gifts for grads at Yahoo! Search.

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