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

Revision as of 15:49, 6 April 2013

Problem 15

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 $p+q+r+s$ .

Solutions

Solution 1

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 $111+4+35+72 = \boxed{222}$ .

Solution 2

Let us use the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1$ .

Add $\begin{align*} \cos^2 C+2(\cos A\cos B-\sin A \sin B)\cos C \end{align*}$ to both sides of the first given equation.

Thus, as $\begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*}$ we have $\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1, \end{align*}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$ .

Similarily, we have $\sin A =\dfrac{2}{3}$ and $\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}$ and the rest of the solution proceeds as above.

@@ Line 38: / Line 38: @@
 Thus, as <cmath>\begin{align*} \cos A\cos B-\sin A\sin B=\cos (A+B)=-\cos C ,\end{align*}</cmath>
-we have <cmath>\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1
+we have <cmath>\begin{align*} \dfrac{15}{8}-2\cos^2 C +\cos^2 C=1,
-\end{align*}</cmath>, so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is <math>\sqrt{\dfrac{1}{8}}</math>.
+\end{align*}</cmath> so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is <math>\sqrt{\dfrac{1}{8}}</math>.
 Similarily, we have <math>\sin A =\dfrac{2}{3}</math> and <math>\cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}</math> and the rest of the solution proceeds as above.

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