1995 AIME Problems/Problem 7
Contents
Problem
Given that and
where and
are positive integers with
and
relatively prime, find
Solution
From the givens,
, and adding
to both sides gives
. Completing the square on the left in the variable
gives
. Since
, we have
. Subtracting twice this from our original equation gives
, so the answer is
.
Solution 2
Let . Multiplying
with the given equation,
, and
. Simplifying and rearranging the given equation,
. Notice that
, and substituting,
. Rearranging and squaring,
, so
, and
, but clearly,
. Therefore,
, and the answer is
.
SOLUTION 3 (-synergy)
We have $1+\sinx\cosx+\sinx+\cosx = \frac{5}{4}$ (Error compiling LaTeX. Unknown error_msg)
We want to find $1+\sinx\cosx-\sinx-\cosx$ (Error compiling LaTeX. Unknown error_msg)
If we find \sinx+\cosx, we will be done with the problem.
Let $y = \sinx+\cosx$ (Error compiling LaTeX. Unknown error_msg)
Squaring, we have $y^2 = \sin^2 x + \cos^2 x + 2\sinx\cosx = 1 + 2\sinx\cosx$ (Error compiling LaTeX. Unknown error_msg)
From this we have $\sinx\cosx = y$ (Error compiling LaTeX. Unknown error_msg) and $\sinx + \cos x = \frac{y^2-1}{2}$ (Error compiling LaTeX. Unknown error_msg)
Substituting this into the first equation we have ,
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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