HYACINTHOS 19710

Hi Jean-Louis,

The Radical triangle of ABC is perspective to
ABC at X(19)
Medial triangle at X(10) .. Spieker point.
Incentral triangle at X(612)
The cevian triangle of points on the cubic cyclic[a y z (b (a^2+b^2-c^2) y-c (a^2-b^2+c^2) z)] = 0, ETC points on this cubic are {1, 2, 19, 75, 279, 304, 346, 2184}

The cevian triangle of X(346) gives rise to a simple non-ETC point (b + c) (a^2 - b^2 - c^2) (a^2 + b^2 + c^2 + 2 b c)::

Barycentric of the A-vertex of the radical triangle is
{-(b + c) (a^2 + b^2 + c^2 + 2 b c), b (a^2 + b^2 - c^2), c (a^2 - b^2 + c^2)}

Best regards,
Peter Moses

 

----- Original Message -----
From: Jean-Louis Ayme



Dear Hyacinthists,
this is a conjecture:
the Spieker's point of ABC is the center of perspective of the medial triangle of ABC with the radical triangle of ABC.
The radical triangle of ABC is determined by the intersection of the radical axes of the circumcircle of ABC with the resp. excircles of ABC.
Please, can some one confirms or not this result.
Sincerely
Jean-Louis Ayme