Dear Cesar and Antreas,

[CL]: Let A'B'C' be the medial triangle of ABC and G its centroid.

Let A" be the intersection of the BC with the angle bisector of B'GC'. Build B" and C" cyclically.

The triangles A" B" C" and ABC are perspective with perspector

X = (a*sqrt( -a^2 + 2*b^2+2*c^2) : : )

...

[APH]: This is true for any point P instead of G.

*** Yes, Antreas. But Cesar, I do not get your coordinates. In homogeneous barycentric coordinates, the perspector is

(1/m_a : 1/m_b : 1/m_c) where m_a etc are the lengths of the medians. With m_a = sqrt(2b^2+2c^2-a^2)/2 etc, this is

(1/sqrt{2b^2+2c^2-a^2) : ... : ...).

The trilinear coordinates are

(1/(a*sqrt(2b^2+2c^2-a^2)) : ... : ...).

Best regards

Sincerely

Paul