Let a, b, and c be the lengths of sides BC,
CA, and AB respectively. Let s be the
semiperimeter (a+b+c)/2. Let O be the
circumcircle of triangle ABC and R the
circumradius.
Let point D be a point on side BC such that
AC+CD = AB+BD = s. Let M be the
midpoint of line segment AD. Let N be the
circle with center M and radius s/2.
Triangle ABC lies inside circle N since
AM = DM = AD/2 < (AB+BD)/2 = s/2
and points B and C lie on an elipse with
center M and major axis s/2.
If more than 180 degrees of arc of circle O
lies outside of circle N, then angle A would
be greater than 90 degrees. Therefore, R < s/2.
From the rule of sines,
a b c
 =  =  = 2R
sin(A) sin(B) sin(C)
Therefore,
a+b+c s
sin(A) + sin(B) + sin(C) =  =  > 2
2R R

Posted by Bractals
on 20071015 05:41:03 