HYACINTHOS 25046

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the cevian triangle of G (medial triangle).

Denote:

A", B", C" = the reflections of G in BC,CA, AB, resp.

Aa, Ab, Ac = the orthogonal projections of A" on AG, BG, CG, resp.
Ba, Bb, Bc = the orthogonal projections of B" on AG, BG, CG, resp.
Ca, Cb, Cc = the orthogonal projections of C" on AG, BG, CG, resp.

La, Lb, Lc = the Euler lines of AaAbAc, BaBbBc, CaCbCc, resp.

A*B*C* = the triangle bounded by La, Lb, Lc, resp.

1. ABC, A*B*C* are parallelogic.
The parallelogic center (ABC, A*B*C*) lies on the circumcircle.

2. A*B*C*, A'B'C' are parallelogic.
The parallelogic center (A'B'C', A*B*C*) lies on the NPC of ABC..

[Peter Moses]:

Hi Antreas,

1) (2 a^8+a^6 b^2-14 a^4 b^4+a^2 b^6+2 b^8-10 a^6 c^2+7 a^4 b^2 c^2+7 a^2 b^4 c^2-10 b^6 c^2+10 a^4 c^4-2 a^2 b^2 c^4+10 b^4 c^4-2 a^2 c^6-2 b^2 c^6) (2
a^8-10 a^6 b^2+10 a^4 b^4-2 a^2 b^6+a^6 c^2+7 a^4 b^2 c^2-2 a^2 b^4 c^2-2 b^6 c^2-14 a^4 c^4+7 a^2 b^2 c^4+10 b^4 c^4+a^2 c^6-10 b^2 c^6+2 c^8)::
oncircumcircle & {{110,11159},...}.

2) (2 a^6 b^2-10 a^4 b^4+10 a^2 b^6-2 b^8+2 a^6 c^2+2 a^4 b^2 c^2-7 a^2 b^4 c^2-b^6 c^2-10 a^4 c^4-7 a^2 b^2 c^4+14 b^4 c^4+10 a^2 c^6-b^2 c^6-2 c^8) (4
a^8-9 a^6 b^2-4 a^4 b^4-a^2 b^6+2 b^8-9 a^6 c^2+14 a^4 b^2 c^2+5 a^2 b^4 c^2-12 b^6 c^2-4 a^4 c^4+5 a^2 b^2 c^4+20 b^4 c^4-a^2 c^6-12 b^2 c^6+2 c^8)::
on NPC ... complement of 1).

Best regards,
Peter Moses.