HYACINTHOS 29109

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the antipedal triangle of O (tangential triangle)

Denote:

Na, Nb, Nc = the NPC centers of OB'C', OC'A', OA'B', resp.

O' = the circumcenter of NaNbNc

(Oa), (Ob), (Oc) = the circumcircles of O'NbNc, O'NcNa, O'NaNb, resp.

(O1), (O2), (O3) = the reflections of (Oa), (Ob), (Oc) in OA, OB, OC, resp.

The radical center of (O1), (O2), (O3) lies on the Euler line of ABC.

 

 

[Peter Moses]:

Hi Antreas,

 = 36TH HATZIPOLAKIS-MOSES-EULER POINT

= a^2*(a^14 - 3*a^12*b^2 + a^10*b^4 + 5*a^8*b^6 - 5*a^6*b^8 - a^4*b^10 + 3*a^2*b^12 - b^14 - 3*a^12*c^2 + 9*a^10*b^2*c^2 - 7*a^8*b^4*c^2 - 4*a^6*b^6*c^2 + 9*a^4*b^8*c^2 - 5*a^2*b^10*c^2 + b^12*c^2 + a^10*c^4 - 7*a^8*b^2*c^4 + 16*a^6*b^4*c^4 - 9*a^4*b^6*c^4 - 4*a^2*b^8*c^4 + 3*b^10*c^4 + 5*a^8*c^6 - 4*a^6*b^2*c^6 - 9*a^4*b^4*c^6 + 12*a^2*b^6*c^6 - 3*b^8*c^6 - 5*a^6*c^8 + 9*a^4*b^2*c^8 - 4*a^2*b^4*c^8 - 3*b^6*c^8 - a^4*c^10 - 5*a^2*b^2*c^10 + 3*b^4*c^10 + 3*a^2*c^12 + b^2*c^12 - c^14) : :

= lies on these lines: {2,3}, {10627,25487}, {12901,23358}
= reflection of X(i) in X(j) for these {i,j}: {7488,15331}, {18377,1594}, {21950,7885}

Best regards,

Peter Moses.