ADGEOM 1155

[Emmanuel José García]:

Dear César and all members,

Still in this configuration:

Let ABC be a triangle. Let A', B', C' be the touch points of its incircle.

Draw two internal semicircles with diameter BC', AC' & cyclic. Let semicircles

BC', AC' meet the incircle in X, X', respectively. Define YY', ZZ' cyclically.

Now the segments XX',YY', ZZ' form the triangle A_1, B_1, C_1. Also, let

B'' be the second intersection of semicircles BC', BA'. Define A'', C'' cyclically. (See image attached)

Then,

1) ABC is in perspective with A''B''C'' (in the incenter).

2) ABC is in perspective with A_1B_1C_1.

3) A''B''C'' is in perspective with A_1B_1C_1.

4) The three perspectors are collinear.

Best regards,

Emmanuel José García

 

[César Lozada]:

>>2) ABC is in perspective with A_1B_1C_1

  Perspector:

       X2 = 1/((s-a)*(s^2-SA)) :  :     (trilinears)

             = Isogonal conjugate of X(3601)

             = on Feuerbach hyperbola

             =  on lines (1, 1427), (4, 3671), (7, 950), (8, 226), (9, 65), (21, 57), (34, 1172), (40, 943), (72, 4866), (79, 3586), (84, 942), (85, 314), (104, 3333), (388, 3243), (405, 3339), (728, 2171), (946, 3427), (1000, 3487), (1420, 2320), (1490, 3577), (1697, 2346), (1728, 3467), (1896, 5342), (2099, 2900), (2263, 2298), (2335, 3247), (3296, 3488), (3419, 5290), (3600, 5558), (3612, 5424), (4332, 5269), (4355, 5557)

             = ( 0.826587435960589, 0.87840336478196, 2.651037181999674 )

 

>>3) A''B''C'' is in perspective with A_1B_1C_1.

  Perspector:

       X3 = (s*(s^2+4*r^2+2*b*c)-(2*r^2+s^2)*(b+c))/(s-a) : :   (trilinears)

             = on lines (1, 1427), (1054, 3339), (1707, 3361)

             =  ( 0.476312663037726, 0.54914210334871, 3.040652565879400 )

 

Both X2 and X3 are on line (1,1427).

 

Regards

César Lozada