Difference between revisions of "2013 AIME II Problems/Problem 15"
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<cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x</cmath> or | <cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x</cmath> or | ||
<cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x</cmath> or | <cmath> 2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x</cmath> or | ||
− | <cmath> \cos^2 B + \cos^2 C = x - \frac{1}{8} \text{------ (4)}</cmath> | + | <cmath> \cos^2 B + \cos^2 C = x - \frac{1}{8} \text{ ------ (4)}</cmath> |
Similarly adding (2) and (3) we get: | Similarly adding (2) and (3) we get: | ||
− | <cmath> \cos^2 A + \cos^2 B = x - \frac{4}{9} \text{------ (5)} </cmath> | + | <cmath> \cos^2 A + \cos^2 B = x - \frac{4}{9} \text{ ------ (5)} </cmath> |
Similarly adding (1) and (2) we get: | Similarly adding (1) and (2) we get: | ||
− | <cmath> \cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} \text{------ (6)} </cmath> | + | <cmath> \cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} \text{ ------ (6)} </cmath> |
And (4) - (5) gives: | And (4) - (5) gives: | ||
− | <cmath> \cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} \text{------ (7)} </cmath> | + | <cmath> \cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} \text{ ------ (7)} </cmath> |
Now (6) - (7) gives: | Now (6) - (7) gives: |
Revision as of 17:42, 22 June 2017
Problem 15
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
.
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 .
By the Law of Sines, we must have and
.
Now let us analyze the given:
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
.
Solution 2
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
and the rest of the solution proceeds as above.
Solution 3
Let
Adding (1) and (3) we get:
or
or
or
Similarly adding (2) and (3) we get:
Similarly adding (1) and (2) we get:
And (4) - (5) gives:
Now (6) - (7) gives:
or
and
so
is
and therefore
is
Now can be computed first and then
is easily found.
Thus and
can be plugged into (4) above to give x =
.
Hence the answer is = .
Kris17
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.