Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Slovenia

For a real number $t$ and positive real numbers $a$ and $b$ we have
$2a^2 - 3abt + b^2 = 2a^2 + abt - b^2 = 0.$
Find $t$ .

Solution

From $2a^2 - 3ab t + b^2 = 0$ we get $t = \frac{2a^2 + b^2}{3ab}$ and from $2a^2 + ab t - b^2 = 0$ we get $t = \frac{b^2 - 2a^2}{ab}$ . So, $\frac{2a^2 + b^2}{3ab} = \frac{b^2 - 2a^2}{ab}$ . Eliminating the fractions we obtain $8a^2 = 2b^2$ or $4a^2 = b^2$ . Since $a$ and $b$ are positive, we conclude that $b = 2a$ . Thus, $t = \frac{2a^2 + 4a^2}{6a^2} = 1$ .

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