12. Find all pairs of positive numbers $a$ and $b$, for which the numbers $\sqrt{a b}, \frac{a+b}{2}$, and $\sqrt{\frac{a^{2}+b^{2}}{2}}$ can form an arithmetic progression.
Solution 1. We will use the characteristic property of an arithmetic progression. We have,

$$
\begin{gathered}
\frac{a+b}{2}=\frac{\sqrt{a b}+\sqrt{\frac{a^{2}+b^{2}}{2}}}{2} \Longleftrightarrow(a+b)^{2}=a b+\sqrt{2 a b\left(a^{2}+b^{2}\right)}+\frac{a^{2}+b^{2}}{2} \\
\Longleftrightarrow a^{2}+b^{2}-2 \sqrt{2 a b} \sqrt{a^{2}+b^{2}}+2 a b=0 \Longleftrightarrow\left(\sqrt{2 a b}-\sqrt{a^{2}+b^{2}}\right)^{2}=0 \\
\Longleftrightarrow 2 a b=a^{2}+b^{2} \Longleftrightarrow(a-b)^{2}=0 \Longleftrightarrow a=b
\end{gathered}
$$

Topic: "Proof of Inequalities"