HYACINTHOS 27147

[Antreas P. Hatzipolakis]:

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

 

Denote:

A", B", C" = the orthogonal projections of A', B', C' on AO, BO, CO, resp.

ABC, A"B"C" are orthologic.

Generalization:

Let A*, B*, C* be points on A'A", B'B", C'C" such that:

A'A*/A'A" = B'B*/B'B" = C'C*/C'C" = t

ABC, A*B*C* are orthologic.

Which are the loci of the orthologic centers as t varies?

[César Lozada]:

 

ABC->A*B*C*: Jerabek hyperbola

A*B*C*->ABC: Line {3,161}

 

Related centers:

Q1 = Polar trilinear of line X(3)X(161)

= (SA-SB)*(SA-SC)*(3*SB^2-8*R^2* SB+3*S^2-4*SC*SA)*(3*SC^2-8*R^ 2*SC+3*S^2-4*SA*SB) : : (barys)

= on the MacBeath circumconic and line {895, 6145}

= isogonal conjugate of Q2

= trilinear pole of the line {3, 161}

= barycentric product X(99)*X(6145)

= barycentric quotient X(110)/X(7488)

= trilinear product X(662)*X(6145)

= trilinear quotient X(662)/X(7488)

= [ 10.8180831988006100, -0.9876979585244879, -0.6685061000989516 ]

 

Q2 = isogonal conjugate of Q1

= (SB^2-SC^2)*(3*SA^2-8*R^2*SA+ 3*S^2-4*SB*SC) : : (barys)

= on lines: {6, 14346}, {230, 231}, {1510, 2623}, {2165, 10412}

= midpoint of X(647) and X(6753)

= isogonal conjugate of Q1

= barycentric product X(523)*X(7488)

= barycentric quotient X(512)/X(6145)

= trilinear product X(661)*X(7488)

= trilinear quotient X(661)/X(6145)

= [ -0.1562271428824485, 1.7111286046779220, 2.5281418215377160 ]

 

César Lozada