HYACINTHOS 25233

[Antreas P. Hatzipolakis]:

 

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

Denote:

Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

The circumcircles of NA'Na, NB'Nb, NC'Nc are coaxial.

Second point of intersection?

Locus ? (Locus of P, instead of N, such that PA'Na, PB'Nb, PC'Nc are coaxial)

I lies on the locus (Hyacinthos 25224)

[César Lozada]:

Locus = {sidelines }\/ { perpendicular to sidelines through O } \/ { circumcircle } \/ { Napoleon-Feuerbach cubic K005 }

If P lies on the circumcircle then the 2^nd point of intersection Q(P)=midpoint(O,P)

If P lies on the line through O perpendicular to BC then A’, P, Na are collinear

 

Some Q(P) for P on K005:

 

Q( I ):  See Hyacinthos 25224

 

Q( H ) = N

 

Q(N) = (2*a^18-8*(b^2+c^2)*a^16+2*(5* b^4+8*b^2*c^2+5*c^4)*a^14-(b^2 +c^2)*(b^4+6*b^2*c^2+c^4)*a^12 -2*(3*b^8+3*c^8-(b^4+b^2*c^2+ c^4)*b^2*c^2)*a^10+(b^4-c^4)*( b^2-c^2)*(3*b^4+b^2*c^2+3*c^4) *a^8-(b^2-c^2)^2*(b^2+2*c^2)*( 2*b^2+c^2)*(b^4+c^4)*a^6+(b^4- c^4)*(b^2-c^2)^3*(5*b^4+3*b^2* c^2+5*c^4)*a^4-(b^2-c^2)^6*(4* b^4+5*b^2*c^2+4*c^4)*a^2+(b^2+ c^2)*(b^2-c^2)^8)*((b^2+c^2-a^ 2)^2-b^2*c^2)  : : (barycentrics)

= (1-4*cos(A)^2)*((3*cos(2*A)-2* cos(4*A)+1/2)*cos(B-C)-2*cos(A )*cos(2*(B-C))+(cos(2*A)+1/2)* cos(3*(B-C))-cos(A)*cos(4*(B-C ))-cos(3*A)+cos(5*A))  : :  (trilinears)

= On lines: {2,8157}, {128,11597}, {140,10628}, {186,3258}, {252,933}, {468,10214}

= [ 2.437940635868892, 2.66207582274628, 0.672485541912850 ]

 

Q( X(54) ) = (-1+4*cos(A)^2)*((cos(2*A)-3/2 )*cos(B-C)+cos(A)*cos(2*(B-C)) +cos(A)-cos(3*A)) : : (trilinears)

= On lines: {54,5663}, {110,143}, {113,137}, {125,8254}, {186,323}, {5944,10274}, {5965,6593}, {10610,10628}

= midpoint of X(i) and X(j) for these {i,j}: {110,195}, {2914,11597}

= reflection of X(i) in X(j) for these (i,j): (125,8254), (1511,11597), (10113,3574)

= {X(3043), X(11561)}-Harmonic conjugate of X(1511)

= [ -9.125159012516049, 7.35652583977264, 2.759296906303036 ]

 

César Lozada