HYACINTHOS 27240

 

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, P a point, A'B'C' the pedal triangle of P and (N) = the common pedal circle of H and O [ = the NPC].

Denote:

(Npa), (Npb), (Npc) = the reflections of (N) in PA', PB', PC', resp.

Ra = the radical axis of (O), (Npa)
Rb = the radical axis of (O), (Npb)
Rc = the radical axis of (O), (Npc)

A*B*C* = the triangle bounded by Ra, Rb, Rc.

1. P  = H:
1.1. A'B'C', A*B*C* are perspective. Perspector?

1.2. The Euler lines of ABC, A*B*C* are parallels.
Which is the Euler line of A*B*C* ? (trilinear polar of which point?)

 

[César Lozada]:

 

1.1.

 

Perspector = X(4)X(74) ∩ X(132)X(468)

= (2*a^4-(b^2+c^2)*a^2-(b^2-c^2) ^2)^2*(a^2-b^2+c^2)*(a^2+b^2- c^2) : : (barys)

= on the orthic inconic, the cubic K496 and these lines: {4, 74}, {25, 1989}, {51, 6749}, {132, 468}, {184, 6525}, {1495, 1990}, {1637, 9409}, {1842, 2969}, {1859, 3270}, {3081, 3163}, {3517, 13558}, {4232, 9752}, {4240, 5642}, {5095, 9003}, {6618, 15004}, {13857, 15144}

= {X(13202), X(14847)}-Harmonic conjugate of X(125)

= [ -0.3796522173282774, -0.9776308567336585, 4.4927099445668870 ]

 

1.2

Trilinear pole: Not interesting

(-a+b+c)*((4*a^6+10*(b+c)*a^5- 2*(9*b^2-4*b*c+9*c^2)*a^4-2*( b+c)*(4*b^2+19*b*c+4*c^2)*a^3+ 12*(b^2+4*b*c+c^2)*(b-c)^2*a^ 2-2*(b^2-c^2)*(b-c)*(b^2-3*b* c+c^2)*a+2*(b^2-c^2)^2*(b-c)^ 2)*S+(a+b+c)*(2*a^7-5*(b+c)*a^ 6-(b^2+12*b*c+c^2)*a^5+(b+c)*( 13*b^2+b*c+13*c^2)*a^4-12*(b^ 2+c^2)*(b-c)^2*a^3+(b^2-c^2)*( b-c)*(b^2-16*b*c+c^2)*a^2+(3* b^2+8*b*c+3*c^2)*(b-c)^4*a-(b^ 2-c^2)*(b-c)^3*(b^2+3*b*c+c^2) ))*((2*a^2+6*c*a-2*b^2+2*c^2)* S+(a^2-(b+2*c)*a-c*(b-c))*(a- b+c)*(a+b+c))*((2*a^2+6*b*a+2* b^2-2*c^2)*S+(a^2-(2*b+c)*a+b* (b-c))*(a+b-c)*(a+b+c)) : : (barys)

= [ -0.9665269622312740, 1.4078004559710970, 3.1121227641880540 ]

 

César Lozada