Problem:
In equilateral △ABC let points D and E trisect BC. Then sin(∠DAE) can be expressed in the form cab, where a and c are relatively prime positive integers, and b is an integer that is not divisible by the square of any prime. Find a+b+c.
Solution:
Let θ=∠DAE. Suppose, without loss of generality, that the triangle has sides of length 6 and that E is between D and C. Applying the Law of Cosines to △AEC gives AE=62+22−2⋅6⋅2⋅cos60∘=27. The area [DAE]=31[ABC]=3⋅4623 is also 2AD⋅AEsinθ=2(27)2sinθ. Therefore sinθ=3⋅4623⋅4⋅72=1433. The requested sum is 3+3+14=20.
OR
Let θ=∠DAE. Suppose, without loss of generality, that the triangle has sides of length 6 and that E is between D and C. Let M be the midpoint of DE. Then DM=1,AM=33, and tan2θ=331. Because tan2θ=1+cosθsinθ,