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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