HYACINTHOS 26643

[Antreas P. Hatzipolakis]:

 

Orthic version:

Let ABC be a triangle and A'B'C', A"B"C" the cevian triangles of H,O, resp.

Denote:

Ma, Mb, Mc = the midpoints of HA", HB", HC", resp.

The centroid G* of MaMbMc lies on the Euler line of A'B'C'.

 

Excentral version:

Let ABC be a triangle and IaIbIc the antipedal triangle of I

Denote:

O' = the circumcenter of IaIbIc

A"B"C" = the cevian triangle of O' wrt triangle IaIbIc

Ma, Mb, Mc = the midpoints of IA", IB", IC", resp.

The centroid G* of MaMbMc lies on the Euler line of ABC.

[Peter Moses]:

Hi Antreas,

Orthic version:

(a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^12-10 a^8 b^4+20 a^6 b^6-15 a^4 b^8+4 a^2 b^10-5 a^8 b^2 c^2+4 a^6 b^4 c^2+10 a^4 b^6 c^2-12 a^2 b^8 c^2+3 b^10 c^2-10 a^8 c^4+4 a^6 b^2 c^4+10 a^4 b^4 c^4+8 a^2 b^6 c^4-12 b^8 c^4+20 a^6 c^6+10 a^4 b^2 c^6+8 a^2 b^4 c^6+18 b^6 c^6-15 a^4 c^8-12 a^2 b^2 c^8-12 b^4 c^8+4 a^2 c^10+3 b^2 c^10)::
on lines {{5,51},{30,12012},...}.
Reflection of X(10184) in X(5).

Excentral version:

 
a (3 a^5 b+3 a^4 b^2-3 a^3 b^3-3 a^2 b^4+3 a^5 c+4 a^4 b c-5 a^2 b^3 c-3 a b^4 c+b^5 c+3 a^4 c^2-6 a^2 b^2 c^2-3 a b^3 c^2-3 a^3 c^3-5 a^2 b c^3-3 a b^2 c^3-2 b^3 c^3-3 a^2 c^4-3 a b c^4+b c^5)::
on the line {2,3}.


Best regards,
Peter Moses.