HYACINTHOS 28750

[Tran Quang Hung]:

 

Let ABC be a triangle with de Longchamps point L. 

A'B'C' is the pedal triangle of L. 

L'  is de Longchamps point of triangle A'B'C'. 

L* is the isogonal conjugate of L wrt triangle A'B'C'. 

Prove that L, L' and L* are collinear.  
AoPs

 

Which are s the points L', L* and the line (L, L', L*)?  

 

 

[César Lozada]:

 

The line through ETC’s {20, 2979, 5562, 6000, 9833, 12058, 12111, 12250, 12279, 15318, 15606, 15644, 20213, 20427, 30263}, tripolar of

 

Q = BARYCENTRIC PRODUCT X(99)*X(15318)

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

= on line {2404, 14570}

= barycentric product X(99)*X(15318)

= barycentric quotient X(i)/X(j) for these (i,j): (99, 20477), (110, 6759)

= trilinear product X(i)*X(j) for these {i,j}: {662, 15318}, {811, 18890}

= trilinear quotient X(i)/X(j) for these (i,j): (662, 6759), (799, 20477)

= [ -4.4379558311419880, 2.1326964595016270, 4.2124696243180160 ]

 

whose isogonal conjugate is:

Q^-1 = X(6)X(2430) ∩ X(421)X(2501)

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

= on lines: {6, 2430}, {421, 2501}, {647, 657}, {2485, 17434}, {3265, 10601}, {3288, 7927}, {6753, 14398}, {11792, 15140}, {15609, 23438}

= [ -1.7911174157574130, 3.7271595515182220, 1.8869928304367240 ]

 

César Lozada