$12 \cdot 31$ As shown in the figure, the length of the diameter $AB$ of a circle is a two-digit integer in decimal notation. By reversing the two digits, we get the length of the chord $CD$ which is perpendicular to the diameter $AB$. The distance from the intersection point $H$ to the center $O$ of the circle is a positive rational number. Find the length of $AB$.
(1st American Invitational Mathematics Examination, 1983)
[Solution] Let $AB = 10x + y$, where $x, y$ are numbers from $0, 1, 2, 3, \cdots, 9$, and $x \neq 0$. Then
$$
CD = 10y + x.
$$

By the Pythagorean theorem,
$$
\begin{aligned}
OH^2 & = OC^2 - CH^2 \\
& = \left(\frac{10x + y}{2}\right)^2 - \left(\frac{10y + x}{2}\right)^2 \\
& = \frac{9}{4} \cdot 11(x^2 - y^2).
\end{aligned}
$$

Since $OH$ is a rational number, and $\frac{9}{4} = \left(\frac{3}{2}\right)^2$, then $11(x^2 - y^2)$ must be a perfect square.
Let
$$
11(x^2 - y^2) = m^2.
$$

Since 11 is a prime number, then $m$ must be a multiple of 11.
Let $m = 11k$, then
$$
x^2 - y^2 = (x + y)(x - y) = 11k.
$$

Clearly,
$$
\begin{array}{l}
x + y \geq x - y, \\
0 < x + y \leq 18,
\end{array}
$$
$$
0 < x - y \leq 8.
$$

Therefore, the only possibility is
$$
\left\{\begin{array}{l}
x + y = 11 \\
x - y = 1
\end{array}\right.
$$

Solving these equations gives $x = 6, y = 5$, so $AB = 65$.