HYACINTHOS 27030

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the cevian triangle of H.

Denote:

MaMbMc = the midheight triangle.
(ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)

Ab, Ac = the orthogonal projections of Ma on AB, AC, resp.
A2, A3 = the orthogonal projections of Ma on BB', CC', resp.

Na, N1 = the NPC centers of MaAbAc, MaA2A3, resp.

Similarly Nb, N2 and Nc, N3.

A*B*C* = the triangle bounded by NaN1, NbN2, NcN3.

ABC, A*B*C* are perallelogic.
The parallelogic center (ABC, A*B*C*) lies on the Euler line.

 

 

[Peter Moses]:


Hi Antreas,

>ABC, A*B*C* are perallelogic

(ABC,  A*B*C*): X(24).

(A*B*C*,  ABC): 

a^2 (3 a^12 b^2-12 a^10 b^4+15 a^8 b^6-15 a^4 b^10+12 a^2 b^12-3 b^14+3 a^12 c^2-6 a^10 b^2 c^2+5 a^8 b^4 c^2-20 a^6 b^6 c^2+45 a^4 b^8 c^2-38 a^2 b^10 c^2+11 b^12 c^2-12 a^10 c^4+5 a^8 b^2 c^4+8 a^6 b^4 c^4-30 a^4 b^6 c^4+44 a^2 b^8 c^4-15 b^10 c^4+15 a^8 c^6-20 a^6 b^2 c^6-30 a^4 b^4 c^6-36 a^2 b^6 c^6+7 b^8 c^6+45 a^4 b^2 c^8+44 a^2 b^4 c^8+7 b^6 c^8-15 a^4 c^10-38 a^2 b^2 c^10-15 b^4 c^10+12 a^2 c^12+11 b^2 c^12-3 c^14):: 

= (1 + 3 J^2) X[389] + (3 + J^2) X[1596].

on lines {{51,12173},{389,1596},{3567,6776},{3853,10110},{5462,13371}}.

 

 

Best regards,
Peter Moses.