HYACINTHOS 25009

[Tran Quang Hung]:

Let ABC be a triangle, F the 1st Fermat point (X13) and A'B'C' the circumcevian
triangle of F.

The cevian triangles of F with respect to the triangles ABC, A'B'C' share
the same orthocenter lying on the OF line.

http://artofproblemsolving.com /community/c6t48f6h1319276

[Peter Moses]:


Hi Antreas,

(a^4+a^2 b^2-2 b^4+a^2 c^2+4 b^2 c^2-2 c^4+2 Sqrt[3] a^2 S) (2 a^12-a^10 b^2-16 a^8 b^4+34 a^6 b^6-26 a^4 b^8+7 a^2 b^10-a^10 c^2-18 a^8 b^2 c^2+21 a^6
b^4 c^2+22 a^4 b^6 c^2-30 a^2 b^8 c^2+6 b^10 c^2-16 a^8 c^4+21 a^6 b^2 c^4+20 a^4 b^4 c^4+23 a^2 b^6 c^4-24 b^8 c^4+34 a^6 c^6+22 a^4 b^2 c^6+23 a^2
b^4 c^6+36 b^6 c^6-26 a^4 c^8-30 a^2 b^2 c^8-24 b^4 c^8+7 a^2 c^10+6 b^2 c^10+2 Sqrt[3] (2 a^10-7 a^8 b^2+7 a^6 b^4+a^4 b^6-5 a^2 b^8+2 b^10-7 a^8 c^2+4
a^6 b^2 c^2+8 a^4 b^4 c^2+a^2 b^6 c^2-6 b^8 c^2+7 a^6 c^4+8 a^4 b^2 c^4+8 a^2 b^4 c^4+4 b^6 c^4+a^4 c^6+a^2 b^2 c^6+4 b^4 c^6-5 a^2 c^8-6 b^2 c^8+2
c^10) S)::

where S stands for twice area of ABC.

on lines {{3,13},{5,8014},...}
{X(13),X(8929)}-harmonic conjugate of X(17)

The point for X(14) is the same but with S changed to -S.
It is on lines {{3,14},{5,8015},...}


Best regards,
Peter Moses