Difference between revisions of "2013 AIME II Problems/Problem 15"
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Now let us analyze the given: | Now let us analyze the given: | ||
− | <cmath>\begin{align} | + | <cmath>\begin{align*} |
− | \cos^2A + \cos^2B + 2\ | + | \cos^2A + \cos^2B + 2\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\ |
− | &= 2-(\sin^2A + \sin^2B - 2\ | + | &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) |
− | \end{align}</cmath> | + | \end{align*}</cmath> |
Now we can use the Law of Cosines to simplify this: | Now we can use the Law of Cosines to simplify this: |
Revision as of 20:30, 13 March 2015
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.