HYACINTHOS 28065

[César Lozada]:

 

Let A'B'C' be the intouch triangle of ABC.

Denote: A" = the intersection other than A' of the incircle of ABC and the line AA', and cyclically B" and C".

Then the radical center of  (AB"C") , (BC"A") and (CA"B") lie on the Soddy line of ABC.

(Rewritten from: AoPS - Nice geometry)

 

The radical center of the three circles is:

 

O* = X(1)X(7) ∩ X(57)X(1358)

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

= on lines: {1, 7}, {57, 1358}, {85, 1698}, {241, 5219}, {738, 3338}, {948, 3911}, {1088, 1111}, {1212, 20195}, {1418, 4859}, {1565, 1699}, {1743, 7960}, {3337, 7177}, {3361, 7195}, {3632, 9312}, {3633, 6604}, {3665, 5290}, {3679, 9436}, {5526, 6180}, {10491, 18886}

= X(1384)-of-intouch triangle

= barycentric product X(i)*X(j) for these {i,j}: {7, 6173}, {85, 4860}, {279, 5231}

= trilinear product X(i)*X(j) for these {i,j}: {7, 4860}, {57, 6173}, {269, 5231}

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 20121, 7), (7, 279, 1323), (7, 1323, 1), (7, 10481, 20121), (279, 10481, 1), (1323, 10481, 7), (3638, 3639, 390), (3668, 7271, 4862), (5542, 11200, 1)

= [ 0.3405226216783257, 0.4112979116742127, 3.1987554868968500 ]

 

César Lozada