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Difference between revisions of "2013 AIME II Problems/Problem 15"

(I am posting Problem 15 (Problem LaTeXed by v_Enhance on the AoPS Forum) along with my solution.)
 
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By the Law of Sines, we must have <math>CA = \sin{B}</math> and <math>AB = \sin{C}</math>.
 
By the Law of Sines, we must have <math>CA = \sin{B}</math> and <math>AB = \sin{C}</math>.
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[geogebra]e0fd25e3d64e320f9fb1773bb77403f8b516e67e[/geogebra]
 
  
 
Now let us analyze the given:
 
Now let us analyze the given:

Revision as of 13:41, 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 $p$, $q$, $r$, and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$.

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 $BC = \sin{A}$.

By the Law of Sines, we must have $CA = \sin{B}$ and $AB = \sin{C}$.

Now let us analyze the given: $\begin{align*} \cos^2A + \cos^2B + 2\sinA\sinB\cosC &= 1-\sin^2A + 1-\sin^2B + 2\sinA\sinB\cosC \\ &= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC) \\ \intertext{Now we can use the Law of Cosines to simplify this:} &= 2-\sin^2C \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Therefore: \[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\] Similarly, \[\sin A = \sqrt{\dfrac{4}{9}}\cos A = \sqrt{\dfrac{5}{9}}.\] Note that the desired value is equivalent to $2-\sin^2B$, which is $2-\sin^2(A+C)$. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\dfrac{111-4\sqrt{35}}{72}$. Thus, the answer is $111+4+35+72 = \boxed{222}$.

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