Difference between revisions of "2012 AIME II Problems/Problem 9"
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Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>. | Thus, <math>\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}</math>. | ||
− | + | Plugging in the numbers we got back into the original equation : | |
− | + | We get <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>. | So, the answer is <math>49+58=\boxed{107}</math>. | ||
+ | == Solution 2== | ||
− | + | As mentioned above, the first term is clearly <math>\frac{3}{2}.</math> For the second term, we first wish to find <math>\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.</math> Now we first square the first equation getting <math>\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.</math> Squaring the second equation yields <math>\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.</math> Let <math>\cos^2x = a</math> and <math>\cos^2y = b.</math> We have the system of equations | |
− | + | <cmath> | |
− | As mentioned above, the first term is clearly <math> | ||
− | |||
\begin{align*} | \begin{align*} | ||
1-a &= 9-9b \\ | 1-a &= 9-9b \\ | ||
4a &= b \\ | 4a &= b \\ | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
+ | Multiplying the first equation by <math>4</math> yields <math>4-4a = 36 - 36b</math> and so <math>4-b =36 - 36b \implies b =\frac{32}{35}.</math> We then find <math>a =\frac{8}{35}.</math> Therefore the second fraction ends up being <math>\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}</math> so that means our desired sum is <math>\frac{49}{58}</math> so the desired sum is <math>\boxed{107}.</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[Image:Dgxje.png|400px]] | ||
+ | |||
+ | We draw 2 right triangles with angles x and y that have the same hypotenuse. | ||
+ | |||
+ | We get <math>b^2 + 9a^2 = 4b^2 + a^2</math>. Then, we find <math>8a^2 = 3b^2</math>. | ||
+ | |||
+ | Now, we can scale the triangle such that <math>a = \sqrt{3}</math>, <math>b = \sqrt{8}</math>. We find all the side lengths, and we find the hypotenuse of both these triangles to equal <math>\sqrt{35}</math> This allows us to find sin and cos easily. | ||
+ | |||
+ | The first term is <math>\frac{3}{2}</math>, refer to solution 1 for how to find it. | ||
+ | |||
+ | The second term is <math>\frac{\cos^2(x) - \sin^2(x)}{\cos^2(y) - \sin^2(y)}</math>. Using the diagram, we can easily compute this as <math>\frac{\frac{8}{35} - \frac{27}{35}}{\frac{32}{35} - \frac{3}{35}} = \frac{-19}{29}</math> | ||
+ | |||
+ | Summing these you get <math>\frac{3}{2} + \frac{-19}{29} = \frac{49}{58} \implies \boxed{107}</math> | ||
+ | |||
+ | -Alexlikemath | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let <math>a = \sin(x), b = \sin(y)</math> | ||
+ | The first equation yields <math>\frac{a}{b} = 3.</math> Using <math>sin^2(x) + cos^2(x) = 1</math> the second equation yields | ||
+ | <cmath>\frac{\sqrt{1-a^2}}{\sqrt{1-b^2}} = \frac{1}{2} \rightarrow \frac{1-a^2}{1-b^2} = \frac{1}{4}</cmath> | ||
− | + | Solving this yields <math>\left(a, b\right) = \left(3\sqrt{\frac{3}{35}},\sqrt{\frac{3}{35}}\right).</math> | |
+ | Finding the first via double angle for sin yields | ||
+ | <cmath>\frac{\sin(2x)}{\sin(2y)} = \frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}} = 3 \cdot \frac{1}{2} = \frac{3}{2}</cmath> | ||
+ | Double angle for cosine is | ||
+ | <cmath>\cos(2x) = 1-2\sin^2{x}</cmath> | ||
+ | so <math>\frac{\cos(2x)}{\sin(2x)} = \frac{1-2a^2}{1-2b^2} = -\frac{19}{29}.</math> | ||
+ | Adding yields <math>\frac{49}{58} \rightarrow 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}} | ||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:42, 13 January 2024
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, .
Plugging in the numbers we got back into the original equation :
We get .
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
Solution 4
Let The first equation yields Using the second equation yields
Solving this yields Finding the first via double angle for sin yields Double angle for cosine is so Adding yields
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.