HYACINTHOS 26835

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle, A'B'C' the cevian triangle of Ι and A"B"C" the pedal triangle of I.

Denote:

 Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

 

Mab, Mac = the orthogonal projections of Ma on BB', CC', resp.
Mbc, Mba = the orthogonal projections of Mb on CC', AA', resp.
Mca, Mcb = the orthogonal projections of Mc on AA', BB', resp.

R1, R2, R3 = the Euler lines of MaMabMac, MbMbaMbc, McMcaMcb, resp.

 

A*B*C* = the triangle bounded by R1, R2, R3

A"B"C", A*B*C* are parallelogic.
The parallelogic center (A"B"C", A*B*C*) is the antipode of the Feuerbach in the incircle.
The other one (A*B*C*, A"B"C") ?

[Peter Moses]:

Hi Antreas,

2 a^5 b-3 a^4 b^2-3 a^3 b^3+3 a^2 b^4+a b^5+2 a^5 c+2 a^4 b c+a^3 b^2 c-a^2 b^3 c-a b^4 c+b^5 c-3 a^4 c^2+a^3 b c^2-3 a^3 c^3-a^2 b c^3-2 b^3 c^3+3 a^2 c^4-a b c^4+a c^5+b c^5:: 

on lines {{500,950},{511,3911}}.
2 r X[500] - R X[950].

Best regards,
Peter Moses.