HYACINTHOS 24351

[Antreas P. Hatzipolakis]:

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

Denote:

Na, Nb, Nc = the the NPC centers of PBC, PCA, PAB, resp.

N1, N2, N3 = the reflections of Na, Nb, Nc in NA', NB', NC', resp.

Which is the the locus of P such that the circumcenter of N1N2N3 lies on
the Euler line ?

I, O lie on the locus

 

[César Lozada]:

Locus = The excentral circum-quintic with barycentrics equation:

a^2*y*z*(((a^8-4*a^6*b^2+(6*b^ 4-2*b^2*c^2-c^4)*a^4-(b^2-c^2) *a^2*((2*b^2+c^2)^2-4*b^2*c^2) +(b^2+c^2)*(b^2-c^2)^3)*b^2*y- (a^8-4*a^6*c^2+(6*c^4-b^4-2*b^ 2*c^2)*a^4+(b^2-c^2)*a^2*((b^ 2+2*c^2)^2-4*b^2*c^2)-(b^2+c^ 2)*(b^2-c^2)^3)*c^2*z)*a^2*y* z+3*(b^2-c^2)*b^4*c^4*(a^2-b^ 2-c^2)*x^3+(b^2-c^2)*(a^8+b^8+ 2*b^6*c^2+2*b^2*c^6+c^8-4*(b^ 2+c^2)*a^6+2*(3*b^4+5*b^2*c^2+ 3*c^4)*a^4-4*(-b^2*c^2+(b^2+c^ 2)^2)*(b^2+c^2)*a^2)*a^2*x*y* z)+… = 0

 

This curve passes through ETC’s  I, O, H, X(54) and X(110)

 

Pairs (P,Points of concurrence): (I, N), (H, N), ( X(54), X(5576) ),  ( X(110), X(403) )

 

For P=O the point of concurrence is:

Z(O)= (cos(2*A)+1/2)*cos(B-C)-cos(A) *cos(2*(B-C)) : :  (trilinears)

= (9*R^2-2*SW)*X(3)+(7*R^2-2*SW) *X(4)

= [E-8*F, -3*E-8*F]

= anticomplement of X(10125)

= complement of X(1658)

= reflection of X(i) in X(j) for these (i,j): (3,5498), (1658,10125), (10020,3628)

= On lines: (2,3), (125,6102), (569,8254), (1154,5449), (1568,5876), (3574,5946), (5448,5663), (7741,8144)

= [ 2.777656473803460, 1.90080902299303, 1.042724478079915 ]

 

César Lozada