20.97 $ABCD$ is an isosceles trapezoid, with base $AB=92, CD=19$, assuming $AD=$ $BC=x$, a circle with its center on $AB$ is tangent to $AD, BC$. If $m$ is the minimum value of $x$, then $m^2$ equals
(A) 1369.
(B) 1679.
(C) 1748.
(D) 2109.
(E) 8825.
(43rd American High School Mathematics Examination, 1992)
[Solution] Move $CD$. It is evident that when $CD$ passes through the points of tangency of the circle with the legs $AD$ and $BC$, $AD$ and $BC$ are minimized.

As shown in the figure, let the center of the circle be $O$, connect $OC$, and draw $CE \perp AB$.
$$
\begin{aligned}
\because \quad B E & =\frac{92-19}{2}=\frac{73}{2}, \\
O B & =\frac{92}{2}=46 .
\end{aligned}
$$

In $\mathrm{Rt} \triangle OCB$, $\mathrm{CB}^{2}=x^{2}=m^{2}$
$$
=O B \cdot E B=1679 \text{.}
$$

Therefore, the answer is $(B)$.