HYACINTHOS 26383

[Le Viet An]:

Let ABC be a triangle, MaMbMc the cevian triangle of G and Fe the Feuerbach point.
The line IG intersects MbMc, McMa, MaMb at Ka, Kb, Kc, resp.
The perpendiculars to IA, IB, IC from Ka, Kb, Kc, resp. bound a triangle A'B'C'.

Then
The circumcircle of A'B'C' touches the incircle at a point T.
1. Which point is the center of the circle?
2. Which point is T?

Let K be the orthocenter of A'B'C'.
The points I, T, K, Fe are concyclic
3. Which point is the center of the circle (lying on the line IG) ?

 

 

[César Lozada]:

 

1)       

O’ = (b+c)*a^6-2*(2*b^2+b*c+2*c^2)* a^5-(b+c)*(2*b^2-13*b*c+2*c^2) *a^4+(7*b^4+7*c^4-b*c*(5*b^2+ 18*b*c+5*c^2))*a^3-b*c*(b+c)*( 11*b^2-24*b*c+11*c^2)*a^2-(b^ 2-c^2)^2*(3*b^2-7*b*c+3*c^2)* a+(b^2-c^2)^3*(b-c) : : (barycentrics)

= (5*r^2+s^2-16*R*r)*X(3)+2*(2* r^2-R*r)*X(8)

= On lines: {3,8}, {10,3667}, {40,2957}, {517,1647}, {5697,6018}

= [ -1.117652475040984, -8.05440541667940, 9.732630912704421 ]

 

The 1^st circle passes through X(1145)

 

2)

T = (b-c)^2*(2*a-b-c)^2*(a-b+c)*( a+b-c) : : (barycentrics)

= On the incircle and lines: {8,6079}, {11,3667}, {55,2743}, {56,2222}, {65,13756}, {214,519}, {244,1365}, {513,1357}, {1155,3021}, {1358,3676}, {1366,1447}, {1397,2720}, {1647,3259}, {4014,5577}, {5061,5211}, {5433,6789}, {6790,7288}

= midpoint of X(2718) and X(6788)

= {X(244), X(6075)}-Harmonic conjugate of X(7336)

= [ 1.214676066748715, 0.03967941818046, 3.052574392360040 ]

 

3)

O” = (a^3-(b+c)*a^2-(3*b^2-7*b*c+3* c^2)*a-(b^2-c^2)*(b-c))*(2*a- b-c) : : (barycentrics)

= (3*r^2+s^2-15*R*r)*X(1)+3*(2* r^2-R*r)*X(2)

= on lines: {1,2}, {11,11717}, {900,1387}

= midpoint of X(1) and X(1647)

= {X(1), X(6788)}-Harmonic conjugate of X(3244)

= [ 6.581958892885720, 2.01641853501895, -0.793144762129839 ]

 

The 2^nd circle passes through ETC’s: 1, 11, 1647

 

César Lozada