2008 AIME I Problems/Problem 8
Find the positive integer such that
Since we are dealing with acute angles, .
Note that , by tangent addition. Thus, .
Applying this to the first two terms, we get .
We now have . Thus, ; and simplifying, .
Solution 2 (generalization)
From the expansion of , we can see that and If we divide both of these by , then we have which makes for more direct, less error-prone computations. Substitution gives the desired answer.
Solution 3: Complex Numbers
Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, , is the argument of . The sum of these angles is then just the argument of the product
and expansion give us . Since the argument of this complex number is , its real and imaginary parts must be equal. So we set them equal and expand the product to get Therefore, equals .
Solution 4 Sketch
You could always just bash out (where a,b,c,n are the angles of the triangles respectively) using the sum identities again and again until you get a pretty ugly radical and use a triangle to get and from there you use a sum identity again to get and using what we found earlier you can find by division that gets us
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