Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Find the answer

Suppose $x$ and $y$ are positive real numbers such that $x+\frac{1}{y}=y+\frac{2}{x}=3$ . Compute the maximum possible value of $xy$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Rewrite the equations as $xy+1=3y$ and $xy+2=3x$ . Let $xy=C$ , so $x=\frac{C+2}{3}$ and $y=\frac{C+1}{3}$ . Then $\left(\frac{C+2}{3}\right)\left(\frac{C+1}{3}\right)=C \Longrightarrow C^{2}-6C+2=0$ . The larger of its two roots is $3+\sqrt{7}$ .

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