\section*{Problem 3 - 151013}

Determine all possibilities for the ratio \(a: b\) of two positive real numbers \(a\) and \(b\) with \(a < b\) such that the following condition is satisfied!

The geometric mean \(\sqrt{a b}\) of these numbers is \(60\%\) of their arithmetic mean.
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Assuming for two numbers \(a\) and \(b\) with \(0 < a < b\), we have \(m = \frac{b}{a} > 1\). Therefore, 
\[
b(m+1)=\frac{10}{3} b \sqrt{m} \quad ; \quad m+1=\frac{10}{3} \sqrt{m}
\]

By squaring and subtracting \(\frac{100}{9} m\) from it, we get

\[
m^{2}-\frac{82}{9} m+1=0
\]

This quadratic equation has the solutions \(m_{1}=9\) and \(m_{2}=\frac{1}{9}\), of which only the second one satisfies the property \(m > 0\), \(\sqrt{a b}=\sqrt{9 a^{2}}=3 a\), and \(3 a\) is exactly \(60\%\) of \(5 a\).

Thus, \(a\) and \(b\) satisfy the given conditions if and only if \(a: b=1: 9\).