## Re: Cevian Trace Equal Area Points

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• Dear Nikos It was not an answer to me, but a comment to me as list-owner. [ND] ... [FR] ... ... and was [ the post] incomplete APH -- [Non-text portions of
Message 1 of 18 , Aug 4, 2007
Dear Nikos

It was not an answer to me, but a comment to me as list-owner.

[ND]
>
>Dear Francois,
>You answered to Antreas, I think by mistake.
>I am not Antreas.

[FR]

> > Dear Antreas
> > Sorry, I send my post too soon!

... and was [ the post] incomplete

APH

--

[Non-text portions of this message have been removed]
• 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
Message 2 of 18 , Aug 5, 2007
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)

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