HYACINTHOS 26767

[Antreas P. Hatzipolakis]:

Let ABC be a triangle, A'B'C', HaHbHc the cevian triangles of G, H, resp. and A"B"C" the midheight triangle (ie A",B",C' = the midpoints of AHa, BHb, CHc)

The circumcircles of GA'A", GB'B", GC'C" are coaxial.

 

[Peter Moses]:

Hi Antreas,

2nd Intersection:

(5 a^2-b^2-c^2) (2 a^8-9 a^6 b^2+21 a^4 b^4+41 a^2 b^6+9 b^8-9 a^6 c^2+6 a^4 b^2 c^2-57 a^2 b^4 c^2-108 b^6 c^2+21 a^4 c^4-57 a^2 b^2 c^4+198 b^4 c^4+41 a^2 c^6-108 b^2 c^6+9 c^8):: 

on line {2,2418}.

Best regards,
Peter Moses.