HYACINTHOS 28999

Tran Quang Hung]:

 

Let ABC be a triangle with circumcircle (O).

Excircles (Ia), (Ib), (Ic).

Circle (wa) is tangent to (Ib), (Ic) externally and is tangent to (O) internally.

Define similarly circle (wb) and (wc).

Circle (w) is tangent to (wa), (wb) and (wc) internally. P is center of (w).

The P lies on line connecting incenter and center of Taylor circle i.e line X(1)X(389). Please see the figure.

Which is the point P?

 

[César Lozada]

 

 

Circle (w) has center:

P = X(1)X(389) ∩ X(3)X(2262)

= a*(2*S^4+a*((SA+SB)*b+(SA+SC)*c)*S^2-(SB+SC)*(S^2+2*(4*R^2-SW)*SA)*b*c) : : (barys)

=  lies on these lines: {1, 389}, {3, 2262}

= [ -12.4539048245845900, -12.5826216468643700, 18.0996663103141400 ]

 

César Lozada