HYACINTHOS 27151

[Tran Quang Hung]:

Let ABC be a triangle.

H is orthocenter.

Euler points Ea,Eb,Ec are midpoints of HA,HB,HC.

BEc meets CEb at A'.

Define similarly the points B' and C'.

HA meets B'C' at A''.

 

Define similarly the points B'' and C''.

 

Triangle A'B'C' and A''B''C'' are perspective with perspecto P.

 

Isogonal conjugate of P wrt A'B'C' is on Euler line of ABC. Which is this point ?

 

[César Lozada]:

 

P = X(14583)

 

isogonal P w/r to A’B’C’ =

Q = 4*X(5)-X(25)

= a^10-4*(b^4+c^4)*a^6+2*(b^2+c^ 2)^3*a^4+3*(b^2-c^2)^4*a^2-2*( b^4-c^4)*(b^2-c^2)^3 : : (barycentrics)

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

= X(4)+2*X(1368), 4*X(5)-X(25), 2*X(1352)+X(10602)

= on lines: {2, 3}, {6, 1568}, {1352, 10602}, {1853, 15030}, {2393, 10516}, {3167, 12022}, {5651, 13851}, {5654, 11402}, {5891, 14852}, {6761, 9308}, {7809, 14615}, {9140, 12825}, {9627, 11238}, {11178, 14913}, {11180, 13562}, {14644, 14984}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (5, 6816, 7395), (5, 12362, 3542), (235, 6643, 11414), (376, 3542, 10154), (376, 10154, 9715), (1370, 3091, 1596), (2043, 2044, 235), (2072, 9818, 5094), (3091, 6804, 7399), (3542, 12362, 9715), (3830, 7529, 13490), (10154, 12362, 376)

= [ 0.9192972345017282, 0.0473521822875346, 3.1835911705537500 ]

= As a point on the Euler line, this center has Shinagawa coefficients (E-F, -3*F)

 

César Lozada