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

(→‎Solution)
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\[ \begin{align*}
+
Let <math>A,B,C</math> be angles of an acute triangle with
 +
<cmath> \begin{align*}
 
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
 
\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}
 
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}
−
\end{align*} \]
+
\end{align*} </cmath>
 
There are positive integers <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> for which <cmath> \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, </cmath> where <math>p+q</math> and <math>s</math> are relatively prime and <math>r</math> is not divisible by the square of any prime.  Find <math>p+q+r+s</math>.
 
There are positive integers <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> for which <cmath> \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, </cmath> where <math>p+q</math> and <math>s</math> are relatively prime and <math>r</math> is not divisible by the square of any prime.  Find <math>p+q+r+s</math>.
  
Line 12: Line 13:
 
Now let us analyze the given:
 
Now let us analyze the given:
  
−
<math>\begin{align*}
+
<cmath>\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*}</cmath>
  
 
Now we can use the Law of Cosines to simplify this:
 
Now we can use the Law of Cosines to simplify this:

Revision as of 13:43, 4 April 2013

Let $A,B,C$ be angles of an acute triangle with \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)
\end{align*} (Error compiling LaTeX. Unknown error_msg)

Now we can use the Law of Cosines to simplify this:

\[= 2-\sin^2C\]


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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