HYACINTHOS 28756

[César Lozada]:

 

 

Let  P=X(20)-of-ABC ( De Longchamps point), A’B’C’ the pedal triangle of P, P’ = X(20)-of-A’B’C’ and P”=the isogonal conjugate of P w/r to A’B’C’. Then P, P’, P” are collinear.. (https://artofproblemsolving.com/community/c374081t48f374081h1757574)

 

The preceding collinearity remains not only for P=X(20)=De Longchamps point, but for any “independent point” on the Euler line.

 

I say P is a “independent point” on the Euler line of ABC if OP/OH = t = real number not depending of the shape of ABC.

 

If OP/OH = t then P, P’, P” are collinear on the trilinear polar of Q whose isogonal conjugate Q^-1 is given by

Q^-1(t) = (SB^2-SC^2)*(2*(SA-SB)*(SA-SC)*S^2*t^2+(S^2-3*SB*SC)*b^2*c^2*t-SA*a^2*b^2*c^2) : : (barys)

 

ETC pairs (P, Q^-1(P)): (2,3288), (3,647), (5,2623)

 

For P=X(4):

P,P’,P” are collinear on the line {4, 52, 68, 324, 847, 3060, 5392, 5446, 5889, 5962, 6515, 9927, 11411, 11439, 11442, 12002, 12111, 12162, 12235, 13428, 13439, 13754, 14593, 15305, 16194, 16269, 18392, 18474, 18555, 18934, 21268, 27367}, whose tripole is:

 

Q(X(4)) = X(4)X(6754) ∩ X(107)X(925)

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

= on lines: {4, 6754}, {107, 925}, {687, 4558}, {847, 17983}, {2165, 16081}, {5392, 15466}, {6330, 20563}, {6528, 16813}, {15352, 16237}, {18817, 18883}

= [ 0.9642597519627971, -1.8730591442014600, 4.4923547731410220 ]

 

whose isogonal conjugate is:

 

Q^-1(X(4)) = X(6)X(2501) ∩ X(184)X(669)

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

= 2*X(6753)-3*X(14397)

= on lines: {6, 2501}, {184, 669}, {520, 647}, {523, 2623}, {526, 16040}, {826, 3288}, {924, 6753}, {1181, 1499}, {1409, 7180}, {1899, 23301}, {1993, 6563}, {2451, 12077}, {5926, 19357}, {9009, 19459}, {10601, 14341}, {11422, 11450}, {13366, 21646}

= [ 11.3849113402472100, -5.8610064818564230, 2.4437098892324250 ]

 

César Lozada