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Acute triangle, Trigonometric function! (Posted on 2007-10-13) Difficulty: 3 of 5
Prove that in an acute triangle sin(A) + sin(B) + sin(C) > 2

See The Solution Submitted by Chesca Ciprian    
Rating: 4.0000 (1 votes)

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Solution Solution Comment 9 of 9 |

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 2007-10-15 05:41:03
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