HYACINTHOS 26826

[Antreas P. Hatzipolakis]:

 

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

The Euler line of AB'C' intersects B'C' at A"

The Euler line of BC'A' intersects C'A' at B"

The Euler line of CA'B' intersects A'B' at C"

For P = H, P = G (cevian case) or O (pedal case): the A", B", C" are collinear.

1. Which is the line A"B"C" wrt the triangles ABC, A'B'C'?
2. Which point is the centroid of the degenerated triangle A"B"C" ?
3. Which are the loci (cevian, pedal cases) of P such that A", B", C" are collinear?

[César Lozada]:

 

 

a)    Cevian case

 

Locus = {sidelines of ABC} ∪ {sidelines of anticomplementary triangle} ∪ {q12=a degree-12-circumcurve through ETC’s 2, 4}

 

For P=G:

1)    Line A”B”C” = line through ETC’s {402, 5972, 6130, 6716, 9033} of ABC which is the polar trilinear of:

T = cyclocevian conjugate of X(110)

= (2*SB^2-4*(6*R^2-SW)*SB+S^2- SW^2+4*R^2*SW)*(2*SC^2-4*(6*R^ 2-SW)*SC+S^2-SW^2+4*R^2*SW) : : (barys)

= on lines: {20, 9218}, {401, 11064}, {3566, 14721}

= cyclocevian conjugate of X(110)

= trilinear pole of the line {402, 5972}

= [ 5.4913439946392200, 6.3162488430382320, -3.2665896377217290]

 

2)    Centroid G” = X(402) = ZEEMAN-GOSSARD PERSPECT

 

 

For P=H:

1)    Line A”B”C” = line through ETC’s {4, 51, 185, 389,…} of ABC which is the polar trilinear of:

T = polar conjugate of X(520)

= SB^3*SC^3*(SA^2-SB^2) *(SA^2-SC^2) : : (barys)

= on lines: {29, 8764}, {53, 1987}, {107, 1624}, {235, 1942}, {264, 6330}, {338, 2052}, {648, 1625}, {653, 823}, {1989, 8794}, {6331, 14570}, {14249, 15274}

= polar conjugate of X(520)

= trilinear pole of the line {4, 51}

= [ -0.2669748631857882, 0.2690319696382736, 3.5776307474744000 ]

 

2)    Centroid =

G” = X(4)X(51) ∩ X(526)X(1637)

= (S^4+(432*R^4+12*(SA-16*SW)*R^ 2-(SA+SW)^2+22*SW^2)*S^2+3*(4* R^2-SW)*(6*(3*SA-SW)*R^2-(5* SA-2*SW)*SW)*SA)*(SB+SC) : : (barys)

= on lines: {4, 51}, {511, 11050}, {526, 1637}, {5502, 11402}

= [ 1.6717260284025810, 1.5495244386602840, 1.7963509343414530 ]

 

b)    Pedal case

 

Locus= {sidelines} ∪ {infinity} ∪ {circumcircle} ∪ {q3: circum-cubic through ETC’s 3, 4, 110, 523, 7471, 14264}

 

q3 = ∑ [ (c^2*y^2+b^2*z^2)*(6*R^2-SA- SW)*x] + 4*(3*SW*R^2-SW^2+S^2)*x*y*z = 0 (barys)

 

For P=O: Same result than in previous results for P=G.

 

For P=H: Same result than in previous results for P=H

 

For P=X(110)

1)    Line A”B”C” = line through ETC’s { 30, 113, 1495, 1511, 1514, 1524, 1525,…} of ABC which is the trilinear polar of X(2407)

 

2)    Centroid =

G” = X(110)X(11050) ∩ X(125)X(11049)

= ((18*R^2+3*SA-5*SW)*S^2+3*(4* R^2-SW)*(54*R^2*SA-9*SA*SW-SW^ 2))*(2*S^2+3*(SA-SW)*SA) : : (barys)

= on lines: {2, 9033}, {30, 113}, {110, 11050}, {125, 11049}, {542, 1650}, {1651, 5972}

= midpoint of X(110) and X(11050)

= reflection of X(i) in X(j) for these (i,j): (125, 11049), (1651, 5972)

= [ 2.8796335170939170, 1.5343230135162770, 1.2493792338913030 ]

 

César Lozada