Difference between revisions of "Mock AIME II 2012 Problems/Problem 7"
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==Solution 1== | ==Solution 1== | ||
'''Lemma:''' <math> x<1 </math> | '''Lemma:''' <math> x<1 </math> | ||
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'''Proof:''' We can rearrange the given equation to get <math> 3\sqrt{x}-3x=4y-2\sqrt{y}+1 </math>. Notice that the RHS is almost a perfect square. <math> 3\sqrt{x}-3x=4y-4\sqrt{y}+1+2\sqrt{y}=(2\sqrt{y}-1)^2+2\sqrt{y} </math>. Notice now that the RHS is always positive, and so we must have <math> 3\sqrt{x}-3x>0\implies \sqrt{x}>x\implies x>x^2 </math>. Since <math> x </math> is positive, we can divide by <math> x </math> to get <math> x<1 </math>. | '''Proof:''' We can rearrange the given equation to get <math> 3\sqrt{x}-3x=4y-2\sqrt{y}+1 </math>. Notice that the RHS is almost a perfect square. <math> 3\sqrt{x}-3x=4y-4\sqrt{y}+1+2\sqrt{y}=(2\sqrt{y}-1)^2+2\sqrt{y} </math>. Notice now that the RHS is always positive, and so we must have <math> 3\sqrt{x}-3x>0\implies \sqrt{x}>x\implies x>x^2 </math>. Since <math> x </math> is positive, we can divide by <math> x </math> to get <math> x<1 </math>. | ||
Latest revision as of 02:09, 5 April 2012
Problem
Given are positive real numbers that satisfy
, then the value
can be expressed as
, where
and
are relatively prime positive integers. Find
.
Solution 1
Lemma:
Proof: We can rearrange the given equation to get . Notice that the RHS is almost a perfect square.
. Notice now that the RHS is always positive, and so we must have
. Since
is positive, we can divide by
to get
.
Seeing that we can make perfect squares with the linear terms and square root terms, we try to force perfect squares. Since the coefficient of is already a perfect square, we begin with that. We have
. Now we divide both sides of the equation by
to get
. We recognize
from our original equation as
, and we substitute that back in to get
. We again recognize a template for a perfect square in the right hand fraction, so we have
. We can also right the other terms in the numerator of the right hand fraction as perfect squares in terms of
and
.
. We now see that
. We can take the square root of both sides to see the full picture.
. This equation is just the equality case for the RMS-AM inequality! Therefore, from the equality case, we must have
, from which we find that
and
. We see that these satisfy the original equality, and so their product is
.
Note: We had to prove that so that
, one of the terms in the RMS-AM inequality, is positive, otherwise the inequality is false.
Solution 2
Let and
. We have
. Moving this to one side, we have
. Completing the square, we have
The above simplifies into , and since squares are nonnegative, we have both squares equal to
, which means
and
. Thus,
and
, so
and
.