\section*{Problem 1 - 111031}

Determine all ordered pairs \((a ; b)\) of real numbers \(a, b\) with \(a \neq 0, b \neq 0\), for which the following holds:

(1) The sum of the two numbers is 6.

(2) The sum of the reciprocals of both numbers is also 6.
Assume there is a pair of numbers \((a ; b)\) that satisfies conditions (1), (2). Then:

\[
\text { (3) } \quad a+b=6 \quad \text { and } \quad \text { (4) } \quad \frac{1}{a}+\frac{1}{b}=6
\]

From (3) it follows that \(b=6-a\) and \(a \neq 6\). From this and (4): \(\frac{1}{a}+\frac{1}{6-a}=6\).

After multiplying by \(a(6-a)\), subtracting \(6 a(6-a)\), and dividing by 6, we get \(a^{2}-6 a+1=0\). This equation has the solutions

\[
a_{1,2}=3 \pm \sqrt{9-1}=3 \pm 2 \sqrt{2}
\]

The corresponding values from (3) are \(b_{1,2}=3 \pm 2 \sqrt{2}\). Therefore, the pairs \((3+2 \sqrt{2} ; 3-2 \sqrt{2})\) and \((3-2 \sqrt{2} ; 3+2 \sqrt{2})\) can be solutions at most.

Indeed, the equations

\[
3+2 \sqrt{2}+3-2 \sqrt{2}=6 \quad \text { and } \quad \frac{1}{3+2 \sqrt{2}}+\frac{1}{3-2 \sqrt{2}}=\frac{3-2 \sqrt{2}+3+2 \sqrt{2}}{9-8}=6
\]

as well as those equations that arise by swapping \((+2 \sqrt{2})\) with \((-2 \sqrt{2})\) hold true.

\section*{Adapted from \([5]\)}