Difference between revisions of "2013 AIME II Problems/Problem 15"
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Let us use the identity <math>\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1</math> . | Let us use the identity <math>\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1</math> . | ||
− | Add <cmath> | + | Add <cmath> |
− | \end{align*}</cmath>, so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is | + | |
+ | 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 | ||
+ | \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} and \cos A=\sqrt{\dfrac{14}{9}-1}=\sqrt{\dfrac{5}{9}}</math> and the rest of the solution proceeds as above. |
Revision as of 21:15, 5 April 2013
Let be angles of an acute triangle with
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
.
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:
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
.
Alternate Solution
Let us use the identity .
Add to both sides of the first given equation.
Thus, as
we have
, so
is
and therefore
is
.
Similarily, we have and the rest of the solution proceeds as above.