2013 AIME II Problems/Problem 15

Revision as of 13:41, 4 April 2013 by ProbaBillity (talk | contribs) (Solution)

\[ cos2A+cos2B+2sinAsinBcosC=158 andcos2B+cos2C+2sinBsinCcosA=149 \] 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: $cos2A+cos2B+2\sinA\sinB\cosC=1sin2A+1sin2B+2\sinA\sinB\cosC=2(sin2A+sin2B2\sinA\sinB\cosC)\intertextNowwecanusetheLawofCosinestosimplifythis:=2sin2C$ (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 $111+4+35+72 = \boxed{222}$.