2013 AIME II Problems/Problem 15
Problem 15
Let be angles of an acute triangle with There are positive integers , , , and for which where and are relatively prime and is not divisible by the square of any prime. Find .
Solutions
Solution 1
Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let .
By the Law of Sines, we must have and .
Now let us analyze the given:
Now we can use the Law of Cosines to simplify this:
Therefore: Similarly, Note that the desired value is equivalent to , which is . All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of . Thus, the answer is .
Solution 2
Let us use the identity .
Add to both sides of the first given equation.
Thus, as
we have so is and therefore is .
Similarily, we have and and the rest of the solution proceeds as above.
Solution 3
Let
Adding (1) and (3) we get: or or or
Similarly adding (2) and (3) we get: Similarly adding (1) and (2) we get:
And (4) - (5) gives:
Now (6) - (7) gives: or and so is and therefore is
Now can be computed first and then is easily found.
Thus and can be plugged into (4) above to give x = .
Hence the answer is = .
Kris17
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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