(In reply to
re: Solution by Charlie)
After the step:
BE + EC < BZ + ZY + YC = (1/2)*(AB + BC + AC).......(*)
But, BE = EC = R, where R is the length of circumradius of the triangle ABC, while E is the circumcenter of the triangle.
Hence, BE+ EC = 2R........(i)
In terms of the law of sines,
(Reference: http://en.wikipedia.org/wiki/Law_of_sines), we have:
BC/sin A = CA/sin B = AB/sin C = 2R, giving:
(1/2)*(AB + BC + AC) = R(sin C + sin A + sin B).......(ii)
Substituting (i) and (ii) in (*), we have:
2R< R(sin A + sin B + sin C)
Or, sin A + sin B + sin C > 2
Edited on October 14, 2007, 3:13 pm
re(2): Solution
