HYACINTHOS 25612

[Antreas P. Hatzipolaks]:

 

Let ABC be a triangle and P a point.

Denote:

A', B', C' = the reflections of P in BC, CA, AB, resp.

Ab, Ac = the orthogonal projections of A' on AC, AB, resp.

Bc, Ba = the orthogonal projections of B' on BA, BC, resp.

Ca, Cb = the orthogonal projections of C' on CB, CA, resp.

 

Na, Nb, Nc = the NPC centers of A'AbAc, B'BcBa, C'CaCb, resp.

For P = G, the triangles ABC, NaNbNc are perspective (homothetic). Perspector?

Which is the locus of P such that ABC, NaNbNc are perspective?

[César Lozada]:

Locus = This cubic through ETCs 2,4,6 and vertices of REFLECTION triangle:

 

CyclicSum[ 3*(b^2-c^2)*b^2*c^2*(-a^2+b^2+ c^2)*x^3+((2*a^6+(-5*b^2+c^2)* a^4+(b^2-c^2)*(4*b^2+5*c^2)*a^ 2+(b^2-c^2)*(-b^4+6*b^2*c^2-2* c^4))*y+(-2*a^6+(-b^2+5*c^2)* a^4+(b^2-c^2)*(5*b^2+4*c^2)*a^ 2+(b^2-c^2)*(-2*b^4+6*b^2*c^2- c^4))*z)*a^2*y*z ] =0

 

Perspectors Z(P):

 

Z(H) = X(1173)

Z(X(6)) = X(6)

 

Z(G) = (a^4-3*(b^2+c^2)*a^2-7*b^2*c^ 2+2*c^4+2*b^4)*a : : (trilinears)

= On lines: {2,576}, {6,11451}, {22,10541}, {25,5012}, {51,5092}, {110,5943}, {140,1173}, {182,5645}, {184,10545}, {186,5462}, {323,6688}, {373,1994}, {589,8956}, {597,11416}, {694,3108}, {1350,3060}, {1597,10574}, {2979,5644}, {3567,7514}, {3580,11548}, {5020,11422}, {5133,9140}, {5899,10095}, {9781,12083}, {9815,20009}, {10546,11402}

= {X(5422), X(5640)}-Harmonic conjugate of X(5012)

= [ 1.462931516903858, 1.11801595090958, 2.191454277322120 ]

 

César Lozada