$20 \cdot 38$ As shown in the figure, the radius of the circle is equal to the height of the equilateral triangle $ABC$, this circle rolls along the base of the triangle while remaining tangent, with the moving point of tangency denoted as $T$, the circle intersects $AC, BC$ at $M, N$, let $n$ be the degree measure of $\angle MTN$, then for all possible positions of the circle, $n$
(A) varies from $30^{\circ}$ to $90^{\circ}$.
(B) varies from $30^{\circ}$ to $60^{\circ}$.
(C) varies from $60^{\circ}$ to $90^{\circ}$.
(D) remains $30^{\circ}$ unchanged.
(E) remains $60^{\circ}$ unchanged.
(15th American High School Mathematics Examination, 1964)
[Solution] The figure shows one position of the circle: the center is at vertex $C$ of the equilateral $\triangle A B C$, at this time
the measure of $\angle M T N$ is $60^{\circ}$.
The right figure shows another possible position of the circle: the center is on a line parallel to the base $A B$ of the equilateral $\triangle A B C$.

Extend $N C$ to intersect $\odot O$ at $D$, and connect $C M, D M$. Then the diameter through $C$ bisects the vertex angle $\angle C$ of the isosceles $\triangle C D M$.
$$
\text { Also, } \begin{aligned}
\angle C & =180^{\circ}-\angle M C N \\
& =180^{\circ}-60^{\circ}=120^{\circ},
\end{aligned}
$$

we know $\angle C D M=30^{\circ}$. Thus, the measure of $\overparen{M N}$ is $60^{\circ}$. Therefore, the answer is $(E)$.