HYACINTHOS 25689

 

[Antreas P Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

Oa, Ob, Oc = the circumcenters of  BCI, CAI, ABI, resp.

Oab, Oac = the circumcenters of OaIB, OaIC, resp.
Obc, Oba = the circumcenters of ObIC, ObIA, resp.
Oca, Ocb = the circumcenters of OcIA, OcIB, resp.

Na, Nb, Nc = the NPCs of OOabOac, OOcaOcb, OOcaOcb, resp.

I think the following are true.

1 The radical center of (Na), (Nb), (Nc) lies on the OI line.

2. The centroid of NaNbNc lies on the OI line.

 

 

[César Lozada]:

1)      Radical center =

= 2*a^6-3*(b+c)*a^5-(3*b^2+2*b*c +3*c^2)*a^4+6*(b^3+c^3)*a^3+4* b*c*(b^2+b*c+c^2)*a^2-3*(b^4- c^4)*(b-c)*a+(b^2-c^2)^2*(b-c) ^2 : :  (trilinears)

= 3*X(1006)-X(3219) = 6*X(1385)-X(3748)

= On lines: {1,3}, {30,5249}, {214,5745}, {443,3897}, {500,1104}, {515,6881}, {549,5440}, {912,1006}, {944,6989}, {968,7986}, {971,7489}, {1125,6841}, {2320,9776}, {2771,3683}, {2772,11709}, {3305,6883}, {3534,6173}, {3560,10884}, {3616,6851}, {3740,12738}, {3916,5428}, {5226,6827}, {5731,6826}, {5787,6861}, {6914,10167}

= midpoint of X(1) and X(7688)

= {X(5709), X(7987)}-Harmonic conjugate of X(3)

= [ 4.702996560584783, 4.17796378818908, -1.422385784031804 ]

 

2)      Centroid = X(3576) = X(1)+2*X(3)

 

César Lozada