Let I be the incenter of ABC. AI, BI, CI intersect the circumcircle of ABC again at A', B', C' respectively. Show that the area of A'B'C' >= the area of ABC
Is the following equivalent to our problem?
[sin(A+B)+sin(B+C)+sin(C+A)] >= [sin(2A)+sin(2B)+sin(2C)]
Edited on November 9, 2007, 5:03 pm
Posted by Bractals
on 2007-11-09 10:06:48