Note that this solution requires that one circle be inscribed into the triangle BCP and the other be inscribed into the quadrilateral bounded by lines AB, AC, BP and CP. Please change the order of the vertices in the triangle of the figure (counter-clockwise), with A as the top one.
Let s=(a+b+c)/2, sa=(-a+b+c)/2, sb=(a-b+c)/2 and sc=(a+b-c)/2
ResponderEliminarPoint P ( =Pa) must have trilinear coordinates:
Pa = 1 : (sqrt(a*s)-sb)/b : (sqrt(a*s)-sc)/c
The radii of both circles is:
ra =| (s-sqrt(a*s))*tan(A/2)|
The incircle of BCP has center:
Ia = a-sqrt(a*s) : (sqrt(a*s)*(a-c)-SC)/b : (sqrt(a*s)*(a-b)-SB)/c
The other circle has center:
Ja = -(b+c-sqrt(a*s))/(a-sqrt(a*s)) : 1 : 1
Best regards.
César Lozada
Note that this solution requires that one circle be inscribed into the triangle BCP and the other be inscribed into the quadrilateral bounded by lines AB, AC, BP and CP. Please change the order of the vertices in the triangle of the figure (counter-clockwise), with A as the top one.
ResponderEliminar