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

m (Solution 1)
m (Solution 1)
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Now let us analyze the given:
 
Now let us analyze the given:
  
<cmath>\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\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\
&= 2-(\sin^2A + \sin^2B - 2\sinA\sinB\cosC)
+
&= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C)
\end{align}</cmath>
+
\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 20:30, 13 March 2015

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\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\ &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) \end{align*}

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.

Solution 3

Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{     ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{     ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{                        ------ (3)}\\ \end{align*}

Adding (1) and (3) we get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C +  \sin C \cos B)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\] or \[\cos^2 B + \cos^2 C =  x - \frac{1}{8}     --- (4)\]

Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B =  x - \frac{4}{9}     --- (5)\] Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C =  \frac{14}{9}  - \frac{1}{8}     --- (6)\]

And (4) - (5) gives: \[\cos^2 C - \cos^2 A =  \frac{4}{9}  - \frac{1}{8}    --- (7)\]

Now (6) - (7) gives: $\cos^2 A  =  \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$

Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found.

Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$.

Hence the answer is = $111+4+35+72 = \boxed{222}$.

Kris17

See Also

2013 AIME II (ProblemsAnswer KeyResources)
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Problem 14
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