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Mock AIME II 2012 Problems/Problem 7

Revision as of 02:09, 5 April 2012 by Testingtesting (talk | contribs) (Created page with "==Problem== Given <math> x, y </math> are positive real numbers that satisfy <math> 3x+4y+1=3\sqrt{x}+2\sqrt{y} </math>, then the value <math> xy </math> can be expressed as <mat...")
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Problem

Given $x, y$ are positive real numbers that satisfy $3x+4y+1=3\sqrt{x}+2\sqrt{y}$, then the value $xy$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Lemma: $x<1$ Proof: We can rearrange the given equation to get $3\sqrt{x}-3x=4y-2\sqrt{y}+1$. Notice that the RHS is almost a perfect square. $3\sqrt{x}-3x=4y-4\sqrt{y}+1+2\sqrt{y}=(2\sqrt{y}-1)^2+2\sqrt{y}$. Notice now that the RHS is always positive, and so we must have $3\sqrt{x}-3x>0\implies \sqrt{x}>x\implies x>x^2$. Since $x$ is positive, we can divide by $x$ to get $x<1$.

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 $y$ is already a perfect square, we begin with that. We have $4y+1=-3x+3\sqrt{x}+2\sqrt{y}\implies 4y+4\sqrt{y}+1=3\sqrt{x}-3x+6\sqrt{y}$. Now we divide both sides of the equation by $9$ to get $\dfrac{(2\sqrt{y}+1)^2}{9}=\dfrac{\sqrt{x}-x+2\sqrt{y}}{3}$. We recognize $2\sqrt{y}$ from our original equation as $3x+4y+1-3\sqrt{x}$, and we substitute that back in to get $\dfrac{(2\sqrt{y}+1)^2}{9}=\dfrac{2x+4y-2\sqrt{x}+1}{3}$. We again recognize a template for a perfect square in the right hand fraction, so we have $\dfrac{(2\sqrt{y}+1)^2}{9}=\dfrac{4y+x+x-2\sqrt{x}+1}{3}=\dfrac{4y+x+(1-\sqrt{x})^2}{3}$. We can also right the other terms in the numerator of the right hand fraction as perfect squares in terms of $\sqrt{y}$ and $\sqrt{x}$. $\dfrac{(2\sqrt{y}+1)^2}{9}=\dfrac{(2\sqrt{y})^2+(\sqrt{x})^2+(1-\sqrt{x})^2}{3}$. We now see that $2\sqrt{y}+\sqrt{x}+(1-\sqrt{x})=2\sqrt{y}+1$. We can take the square root of both sides to see the full picture. $\dfrac{(2\sqrt{y})+(\sqrt{x})+(1-\sqrt{x})}{3}=\sqrt{\dfrac{(2\sqrt{y})^2+(\sqrt{x})^2+(1-\sqrt{x})^2}{3}}$. This equation is just the equality case for the RMS-AM inequality! Therefore, from the equality case, we must have $2\sqrt{y}=\sqrt{x}=1-\sqrt{x}$, from which we find that $x=\dfrac{1}{4}$ and $y=\dfrac{1}{16}$. We see that these satisfy the original equality, and so their product is $\dfrac{1}{64}\implies 1+64=\boxed{065}$.

Note: We had to prove that $x<1$ so that $1-\sqrt{x}$, one of the terms in the RMS-AM inequality, is positive, otherwise the inequality is false.

Solution 2

Let $\sqrt{x}=a$ and $\sqrt{y}=b$. We have $3a^2+4b^2+1=3a+2b$. Moving this to one side, we have $3a^2-3a+4b^2-4b+1=0$. Completing the square, we have

\[3\left(a-\dfrac{1}{2}\right)^2-\dfrac{3}{4}+\left(2b-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1=0\]

The above simplifies into $3\left(a-\dfrac{1}{2}\right)^2+\left(2b-\dfrac{1}{2}\right)^2=0$, and since squares are nonnegative, we have both squares equal to $0$, which means $a=\dfrac{1}{2}$ and $b=\dfrac{1}{4}$. Thus, $x=\dfrac{1}{4}$ and $y=\dfrac{1}{16}$, so $xy=\dfrac{1}{64}$ and $64+1=\boxed{065}$.

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