2012 AIME II Problems/Problem 9
Contents
Problem 9
Let and be real numbers such that and . The value of can be expressed in the form , where and are relatively prime positive integers. Find .
Solution
Examine the first term in the expression we want to evaluate, , separately from the second term, .
The First Term
Using the identity , we have:
The Second Term
Let the equation be equation 1, and let the equation be equation 2. Hungry for the widely-used identity , we cross multiply equation 1 by and multiply equation 2 by .
Equation 1 then becomes:
.
Equation 2 then becomes:
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
Applying the identity (which is similar to but a bit different), we can change into:
Rearranging, we get .
So, .
Squaring Equation 1 (leading to ), we can solve for :
Using the identity , we can solve for .
Thus, .
Now Back to the Solution!
Finally, .
So, the answer is .
Solution 2
As mentioned above, the first term is clearly For the second term, we first wish to find Now we first square the first equation getting Squaring the second equation yields Let and We have the system of equations Multiplying the first equation by yields and so We then find Therefore the second fraction ends up being so that means our desired sum is so the desired sum is
Solution 3
We draw 2 right triangles with angles x and y that have the same hypotenuse.
We get . Then, we find .
Now, we can scale the triangle such that , . We find all the side lengths, and we find the hypotenuse of both these triangles to equal This allows us to find sin and cos easily.
The first term is , refer to solution 1 for how to find it.
The second term is . Using the diagram, we can easily compute this as
Summing these you get
-Alexlikemath
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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