Difference between revisions of "2013 AIME II Problems/Problem 15"
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Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>. | Therefore: <cmath>\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.</cmath> Similarly, <cmath>\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.</cmath> Note that the desired value is equivalent to <math>2-\sin^2B</math>, which is <math>2-\sin^2(A+C)</math>. All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of <math>\dfrac{111-4\sqrt{35}}{72}</math>. Thus, the answer is <math>111+4+35+72 = \boxed{222}</math>. | ||
+ | |||
+ | |||
+ | ==Alternate Solution== | ||
+ | Let us use the identity <math>\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1</math> . | ||
+ | |||
+ | Add <cmath> | ||
+ | \end{align*}</cmath>, so <math>\cos C</math> is <math>\sqrt{\dfrac{7}{8}}</math> and therefore <math> \sin C</math> is |
Revision as of 21:03, 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