2013 AIME II Problems/Problem 15
\[ \begin{align*}
\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}
\end{align*} \]
There are positive integers ,
,
, and
for which
where
and
are relatively prime and
is not divisible by the square of any prime. Find
.
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 .
By the Law of Sines, we must have and
.
[geogebra]e0fd25e3d64e320f9fb1773bb77403f8b516e67e[/geogebra]
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: Similarly,
Note that the desired value is equivalent to
, which is
. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of
. Thus, the answer is
.