Difference between revisions of "2012 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
− | {{ | + | Examine the first term in the expression we want to evaluate, <math>\frac{\sin 2x}{\sin 2y}</math>, separately from the second term, <math>\frac{\cos 2x}{\cos 2y}</math>. |
+ | |||
+ | == The First Term == | ||
+ | |||
+ | Using the identity <math>\sin 2\theta = 2\sin\theta\cos\theta</math>, we have: | ||
+ | |||
+ | <math>\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}</math> | ||
+ | |||
+ | == The Second Term == | ||
+ | |||
+ | Let the equation <math>\frac{\sin x}{\sin y} = 3</math> be equation 1, and let the equation <math>\frac{\cos x}{\cos y} = \frac12</math> be equation 2. | ||
+ | Hungry for the widely-used identity <math>\sin^2\theta + \cos^2\theta = 1</math>, we cross multiply equation 1 by <math>\sin y</math> and multiply equation 2 by <math>\cos y</math>. | ||
+ | |||
+ | Equation 1 then becomes: | ||
+ | |||
+ | <math>\sin x = 3\sin y</math>. | ||
+ | |||
+ | Equation 2 then becomes: | ||
+ | |||
+ | <math>\cos x = \frac{1}{2} \cos y</math> | ||
+ | |||
+ | Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS: | ||
+ | |||
+ | <math>1 = 9\sin^2 y + \frac{1}{4} \cos^2 y</math> | ||
+ | |||
+ | Applying the identity <math>\cos^2 y = 1 - \sin^2 y</math> (which is similar to <math>\sin^2\theta + \cos^2\theta = 1</math> but a bit different), we can change <math>1 = 9\sin^2 y + \frac{1}{4} \cos^2 y</math> into: | ||
+ | |||
+ | <math>1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y</math> | ||
+ | |||
+ | Rearranging, we get <math>\frac{3}{4} = \frac{35}{4} \sin^2 y </math>. | ||
+ | |||
+ | So, <math>\sin^2 y = \frac{3}{35}</math>. | ||
+ | |||
+ | Squaring Equation 1 (leading to <math>\sin^2 x = 9\sin^2 y</math>), we can solve for <math>\sin^2 y</math>: | ||
+ | |||
+ | <math>\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}</math> | ||
+ | |||
+ | Using the identity <math>\cos 2\theta = 1 - 2\sin^2\theta</math>, we can solve for <math>\frac{\cos 2x}{\cos 2y}</math>. | ||
+ | |||
+ | <math>\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}</math> | ||
+ | |||
+ | <math>\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}</math> | ||
+ | |||
+ | Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>. | ||
+ | |||
+ | == Now Back to the Solution! == | ||
+ | |||
+ | Finally, <math>\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}</math>. | ||
+ | |||
+ | So, the answer is <math>49+58=\boxed{107}</math>. | ||
+ | |||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=8|num-a=10}} | {{AIME box|year=2012|n=II|num-b=8|num-a=10}} |
Revision as of 20:26, 3 April 2012
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 .
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |