HYACINTHOS 24922

 
[Tran Quang Hung]:

I have found another similar problem Mr APH

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

Denote:

Ab = HB intersection circle (H, HA') near to B
Ac = HC intersection circle (H, HA') near to C.

A" = BC intersection AbAc.

Similarly B", C".

The points A", B", C" are collinear.

 

Which is the trilinear pole of this line?

 

 

[César Lozada]:

 

-Line {A”,B”,C”} passes through X(3064). Its trilinear pole is:

 

P = 1/((-a+b+c)*(-a^2+b^2+c^2)*(a^ 2+2*(b+c)*a+(b-c)^2)) : : (barycentrics)

= On lines: {4,1435}, {29,1396}, {34,281}, {278,318}, {2376,6574}

= [ -0.509972831898809, -0.62851107680563, 4.311159611341575 ]

 

-If Ab,Ac are further from B and C and similarly the other ones, then A”,B”,C” are also collinear on the polar trilinear of X(278), so they are collinear with these ETC’s: 513, 1835, 1874, 1875, 1876, 1877, 1878, 1884, 7178, 7649.

 

If circle (A’, A’H) is used instead, then A”,B”,C” are collinear, together with X(924),  on the trilinear polar of:

Q= (SA^2-S^2)/SA^2 : : (barycentrics)

= polar conjugate of X(68)

= On lines: {2,216}, {4,54}, {25,3425}, {107,3563}, {158,10198}, {254,1093}, {297,315}, {317,467}, {343,9308}, {371,1585}, {372,1586}, {406,1896}, {427,9744}, {458,7803}, {648,6515}, {1217,7400}, {2165,8794}, {3618,6819}, {4232,6525}, {6995,10002}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2,393,2052), (184,6747,4), (467,1993,317), (578,6750,4), (6353,6524,107)

= [ -5.810327047777666, -2.00948785923712, 7.713537791122844 ]

 

If circle (A, AH) is used instead, then A”,B”,C” are collinear, together with X(186) and X(523),  on the trilinear polar of X(275)

 

 

Regards,

César Lozada