[Antreas P. Hatzipolakis]:
Let ABC be a triangle.
Denote:
Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.
N*a, N*b, N*c = the isogonal conjugates of Na, Nb, Nc, resp. (wrt triangle ABC)
N'a, N'b, N'c = the isogonal conjugates of Na, Nb, Nc, wrt triangles IBC, ICA, IAB, resp.
1. ABC, N*aN*bN*c are cyclologic.
2. ABC, N'aN'bN'c are circumcyclologic.
ie the circumcircles of AN'bN'c, BN'cN'a, CN'aN'b and ABC are concurrent
the circumcircles of N'aBC, N'bCA, N'cAB and N'aN'bN'c [ = the line OI] are concurrent.
PS Probably the loci (for P instead of I) are complicated.
[César Lozada]:
· For P=I
1) Q11(N*aàA) = X(501)
Q12(AàN*a) = Not interesting
= a^2*(a^12+2*(b+c)*a^11-4*(b^2-b*c+c^2)*a^10-2*(b+c)*(5*b^2-3*b*c+5*c^2)*a^9+(5*b^4+5*c^4-4*(2*b^2-b*c+2*c^2)*b*c)*a^8+2*(b+c)*(10*b^4+10*c^4-(5*b^2-7*b*c+5*c^2)*b*c)*a^7+2*(b^4+12*b^2*c^2+c^4)*b*c*a^6-2*(b+c)*(10*b^6+10*c^6+(3*b^4+3*c^4+(4*b-c)*(b-4*c)*b*c)*b*c)*a^5-(5*b^8+5*c^8-(2*b^6+2*c^6+(10*b^4+10*c^4-(14*b^2+25*b*c+14*c^2)*b*c)*b*c)*b*c)*a^4+2*(b+c)*(5*b^8+5*c^8+(9*b^6+9*c^6+(7*b^4+7*c^4-(11*b^2+14*b*c+11*c^2)*b*c)*b*c)*b*c)*a^3+2*(2*b^8+2*c^8-(3*b^6+3*c^6+4*(b^4-3*b^2*c^2+c^4)*b*c)*b*c)*(b+c)^2*a^2-2*(b^4-c^4)*(b^2-c^2)*(b+c)^5*a+(b^2-c^2)^2*(b+c)^4*(-b^4-c^4+2*(b^2+b*c+c^2)*b*c))*(a^3-(b-3*c)*a^2-(b^2+b*c-3*c^2)*a+(b^2-c^2)*(b-c))*(a^2-c^2)*(a^3+(3*b-c)*a^2+(3*b^2-b*c-c^2)*a+(b^2-c^2)*(b-c))*(a^2-b^2) : :
2) Q21(N’aàA) = X(484)
Q22(AàN’a) = X(6584)
· Other P
1) Locus = degree(11) through ETCS 1, 4
2) Locus = degree(11). Same than last.
For P=X(4):
Q21(N’aàA) = X(11584)
Q21(AàN’a) = X(4)X(30248) ∩ X(5)X(930)
= (SB+SC)*((4*SB+R^2-2*SW)*S^2-(3*R^2-2*SW)*SA*SC)*((4*SC+R^2-2*SW)*S^2-(3*R^2-2*SW)*SA*SB) : : (barys)
= lies on the circumcircle, cubics K067, K464 and these lines: {4, 30248}, {5, 930}, {110, 143}, {112, 14577}, {933, 10214}, {1291, 6150}
= reflection of X(74) in the line X(3)X(15532)
= (ABC-X3 reflections)-isogonal conjugate of-X(13152)
= [ 0.3713216250973606, -0.4593946653215054, 3.7873277309312550 ]
César Lozada
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