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: | ||
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<math>\begin{align*} | <math>\begin{align*} | ||
\cos^2A + \cos^2B + 2\sinA\sinB\cosC &= 1-\sin^2A + 1-\sin^2B + 2\sinA\sinB\cosC \\ | \cos^2A + \cos^2B + 2\sinA\sinB\cosC &= 1-\sin^2A + 1-\sin^2B + 2\sinA\sinB\cosC \\ | ||
&= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC) | &= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC) | ||
\end{align*}</math> | \end{align*}</math> | ||
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Now we can use the Law of Cosines to simplify this: | Now we can use the Law of Cosines to simplify this: | ||
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<cmath>= 2-\sin^2C</cmath> | <cmath>= 2-\sin^2C</cmath> | ||
Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}}\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>. | Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}}\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>. | ||
Revision as of 13:42, 4 April 2013
\[ \begin{align*}
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}
\end{align*} \]
There are positive integers 
, 
, 
, and 
for which ![\[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\]](http://latex.artofproblemsolving.com/9/0/0/9002465ecd454b8602947063c4eff3ea7634af70.png)
where 
and are relatively prime and is not divisible by the square of any prime. Find 
.
Solution
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:
$\begin{align*} \cos^2A + \cos^2B + 2\sinA\sinB\cosC &= 1-\sin^2A + 1-\sin^2B + 2\sinA\sinB\cosC \\ &= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC) \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Now we can use the Law of Cosines to simplify this:
![\[= 2-\sin^2C\]](http://latex.artofproblemsolving.com/7/c/f/7cf7c3c03981a40597c15c7699fae31e85fe5c47.png)
Therefore: ![\[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\]](http://latex.artofproblemsolving.com/5/0/8/508923b7e07aca5705686ae09d2bccd5dd906de6.png)
Similarly, ![\[\sin A = \sqrt{\dfrac{4}{9}}\cos A = \sqrt{\dfrac{5}{9}}.\]](http://latex.artofproblemsolving.com/0/8/1/081db8166e96eb5713665592917cb1f24c2e1862.png)
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 
.
