HYACINTHOS 29611

[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 NBC, NCA, NAB, resp.

A", B", C" = the midpoints of AN, BN, CN, 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 NPC center of M1M2M3 lies on the Euler line of ABC.

[Peter Moses]:

Hi Antreas,

A somewhat uninspiring:

2*a^16 + 3*a^14*b^2 - 69*a^12*b^4 + 231*a^10*b^6 - 365*a^8*b^8 + 313*a^6*b^10 - 143*a^4*b^12 + 29*a^2*b^14 - b^16 + 3*a^14*c^2 - 98*a^12*b^2*c^2 + 287*a^10*b^4*c^2 - 192*a^8*b^6*c^2 - 243*a^6*b^8*c^2 + 418*a^4*b^10*c^2 - 207*a^2*b^12*c^2 + 32*b^14*c^2 - 69*a^12*c^4 + 287*a^10*b^2*c^4 - 188*a^8*b^4*c^4 - 79*a^6*b^6*c^4 - 226*a^4*b^8*c^4 + 447*a^2*b^10*c^4 - 172*b^12*c^4 + 231*a^10*c^6 - 192*a^8*b^2*c^6 - 79*a^6*b^4*c^6 - 98*a^4*b^6*c^6 - 269*a^2*b^8*c^6 + 416*b^10*c^6 - 365*a^8*c^8 - 243*a^6*b^2*c^8 - 226*a^4*b^4*c^8 - 269*a^2*b^6*c^8 - 550*b^8*c^8 + 313*a^6*c^10 + 418*a^4*b^2*c^10 + 447*a^2*b^4*c^10 + 416*b^6*c^10 - 143*a^4*c^12 - 207*a^2*b^2*c^12 - 172*b^4*c^12 + 29*a^2*c^14 + 32*b^2*c^14 - c^16 : :

= lies on this line: {2,3}

Best regards,
Peter Moses.