2013 AIME II Problems/Problem 15
Contents
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 .
Note that the problem has a flaw because which contradicts with the statement that it's an acute triangle. Would be more accurate to state that and are smaller than 90. -Mathdummy
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
Solution 4
Let's take the first equation . Substituting for C, given A, B, and C form a triangle, and that , gives us:
Expanding out gives us .
Using the double angle formula , we can substitute for each of the squares and . Next we can use the Pythagorean identity on the and terms. Lastly we can use the sine double angle to simplify.
.
Expanding and canceling yields, and again using double angle substitution,
.
Further simplifying yields:
.
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation yields:
and .
Substituting the identity , we get:
and .
Since the third expression simplifies to the expression , taking inverse cosine and using the angles in angle addition formula yields the answer, , giving us the answer .
Solution 5
We will use the sum to product formula to simply these equations. Recall Using this, let's rewrite the first equation: Now, note that . We apply the sum to product formula again. Now, recall that . We apply this and simplify our expression to get: Analogously, We can find this value easily by angle sum formula. After a few calculations, we get , giving us the answer . ~superagh
Solution 6
According to LOC , we can write it into . We can simplify to . Similarly, we can generalize . After solving, we can get that Assume the value we are looking for is , we get , while which is , so , giving us the answer .~bluesoul
Video Solution by The Power Of Logic
~Hayabusa1
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 | ||
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