$10 \cdot 20$ Try to find two different natural numbers, whose arithmetic mean $A$ and geometric mean $G$ are both two-digit numbers, where $A$ and $G$ are such that one can be obtained by swapping the units and tens digits of the other.
(China Zhejiang Province National Junior High School League Selection Exam, 1992)
[Solution] Let the two natural numbers be $x_{1}, x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=2 A, \\
x_{1} x_{2}=G^{2} .
\end{array}\right.
$$

Thus, $x_{1}$ and $x_{2}$ are the roots of the equation
$$
x^{2}-2 A x+G^{2}=0
$$

which gives
$$
x=\frac{2 A \pm \sqrt{4 A^{2}-4 G^{2}}}{2}=A \pm \sqrt{A^{2}-G^{2}} .
$$

Therefore, $\sqrt{A^{2}-G^{2}}$ must be a natural number.
Let $A=10 a+b$, then $G=10 b+a, a, b \in\{1,2, \cdots, 9\}$.
$$
\begin{aligned}
A^{2}-G^{2} & =(10 a+b)^{2}-(10 b+a)^{2} \\
& =9 \cdot 11 \cdot(a+b)(a-b) .
\end{aligned}
$$

For $\sqrt{A^{2}-G^{2}}$ to be a natural number, $A^{2}-G^{2}$ must be a perfect square.
Thus, it must be true that
$$
11 \mid a+b \text { or } 11 \mid a-b \text {. }
$$

Since $1 \leqslant a-b \leqslant 8$, it follows that $11 \nmid a-b$. Therefore, $11 \mid a+b$.
Also,
$$
\begin{array}{c}
3 \leqslant a+b \leqslant 9+8=17 . \\
a+b=11 .
\end{array}
$$

Thus,
$A^{2}-G^{2}=9 \cdot 11^{2} \cdot(a-b)$ is a perfect square.
Hence, $a-b$ must be a perfect square.
By
$$
a-b=(a+b)-2 b=11-2 b,
$$

$a-b$ is an odd perfect square.
Furthermore, $1 \leqslant a-b \leqslant 9<3^{2}$, so
$$
a-b=1^{2}=1 \text {. }
$$

Solving (1) and (2) gives
$$
a=6, \quad b=5 .
$$

Thus,
$$
A=65, \quad G=56 \text {. }
$$
$x_{1}$ and $x_{2}$ are 98 and 32.