HYACINTHOS 29613

[Antreas P. Hatzipolakis]:


Let ABC be a triangle and A'B'C' the pedal triangle of O

Denote:

Na, Nb, Nc = the NPC centers of OBC, OCA, OAB, resp.

A", B", C" = the midpoints of AO, BO, CO, resp,

Ma, Mb, Mc = the midpoints of A'Na, B'Nb, C'Nc, resp

M1, M2, M3 = the midpoints of A"Ma, B"Mb, C"Mc, resp.

Then the circumcenter of M1M2M3 lies on the Euler line of ABC.

[Peter Moses]:


Hi Antreas,

4*a^10 - 10*a^8*b^2 + 4*a^6*b^4 + 8*a^4*b^6 - 8*a^2*b^8 + 2*b^10 - 10*a^8*c^2 + 18*a^6*b^2*c^2 - 9*a^4*b^4*c^2 + 7*a^2*b^6*c^2 - 6*b^8*c^2 + 4*a^6*c^4 - 9*a^4*b^2*c^4 + 2*a^2*b^4*c^4 + 4*b^6*c^4 + 8*a^4*c^6 + 7*a^2*b^2*c^6 + 4*b^4*c^6 - 8*a^2*c^8 - 6*b^2*c^8 + 2*c^10 : : 

= 3 (2 J^2 - 3) X[2] + (2 J^2 + 1) X[3]

= lies on these lines: {2,3}, {15311,32415}
= midpoint of X(i) and X(j) for these {i,j}: {140,5498}, {10125,23336}
= reflection of X(3628) in X(12043)

Best regards,
Peter Moses.