HYACINTHOS 29577

 

[Antreas P. Hatzipolakis]:

Let ABC be a triangle.

Denote:

La, Lb, Lc = the Euler lines of IBC, ICA, IAB, resp. concurrent at X(21)

Aa, Ab, Ac = the orthogonal projections of A on La, Lb, Lc, resp.
Ba, Bb, Bc = the orthogonal projections of B on La, Lb, Lc, resp.
Ca, Cb, Cc = the orthogonal projections of C on La, Lb, Lc, resp.

A*, B*, C* = the X(110) of AaAbAc, BaBbBc, CaCbCc, resp.

1. A*, B*, C*, X(21) are collinear,
Which is the line? (whose point trilinear polar?)

 

2. The orthogonal projections of A, B, C on the line A*B*C*X(21) are the A*, B*, C*, resp.
Therefore the perpendiculars from A*, B*, C* to BC, CA, AB, resp. are concurrent (orthopole of the line A*B*C*X(21))
Which point is it?

[César Lozada]:

 

1)

Q1 = ISOGONAL CONJUGATE OF Q1*

= (a^5-(b-c)*a^4-(2*b^2-b*c+2*c^2)*a^3+2*(b-c)*(b^2+c^2)*a^2+(b^2-c^2)*(b^2-b*c-c^2)*a-(b^2-c^2)^2*(b-c))*(a^2-c^2)*(a^5+(b-c)*a^4-(2*b^2-b*c+2*c^2)*a^3-2*(b-c)*(b^2+c^2)*a^2+(b^2-c^2)*(b^2+b*c-c^2)*a+(b^2-c^2)^2*(b-c))*(a^2-b^2) : : (barys)

= on lines {}

= trilinear pole of the line {21, 33668}

= barycentric quotient X(110)/X(11849)

= trilinear quotient X(662)/X(11849)

= [ 3.1130562462572170, 3.6955267996360850, -0.3545723391900861 ]

 

Q1* = X(37)X(12077) ∩ X(647)X(661)

= (a^5-(b+c)*a^4-(2*b^2-b*c+2*c^2)*a^3+2*(b+c)*(b^2+c^2)*a^2+(b^4+c^4-b*c*(b+c)^2)*a-(b^2-c^2)^2*(b+c))*(b^2-c^2)*a^2 : : (barys)

= lies on these lines: {37, 12077}, {647, 661}, {2081, 2245}

= [ -0.4706728487351667, -0.3964877353461426, 4.1323896134864010 ]

 

2) Unrelated, not interesting, point:

Q2 =

= (b-c)^2*(2*a^4-(b+c)*a^3-(3*b^2+4*b*c+3*c^2)*a^2+(b+c)*(b^2+b*c+c^2)*a+(b^2-c^2)^2)*((b+c)*a^5-(b-c)^2*a^4-(b+c)*(2*b^2-b*c+2*c^2)*a^3+2*(b^4+c^4)*a^2+(b^3+c^3)*(b^2+c^2)*a-(b^2-c^2)^2*(b+c)^2) : : (barys)

= [ -6.1310520920996980, -7.1415563670136630, 11.4145352400398500 ]

 

César Lozada