National Math Olympiad

From $2a^2-3abt+b^2=0$ we get $t=\frac{2a^2+b^2}{3ab}$ and from $2a^2+abt-b^2=0$ 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, $t=\frac{2a^2+4a^2}{6a^2}=1$.