$36 \cdot 46$ As shown, the vertex $B$ of $\triangle ABC$ is at the center of the unit circle, and $A, C$ are on the circumference, $\angle ABC = 2\alpha \left(0 < \alpha < \frac{\pi}{3}\right)$. Now, $\triangle ABC$ is rotated counterclockwise within the circle according to the following method: the first time, with $A$ as the center, $B$ is moved to the circumference; the second time, with $B$ as the center, $C$ is moved to the circumference; the third time, with $C$ as the center, $A$ is moved to the circumference. This rotation continues until the 100th time. Then, the total length of the path traveled by point $A$ is
(A) $22\pi(1 + \sin \alpha) - 66\alpha$.
(B) $\frac{67}{3}\pi$.
(C) $22\pi + \frac{68}{3}\pi \sin \alpha - 66\alpha$.
(D) $33\pi - 66\alpha$.
[Solution] In these 100 movements, point $A$ does not move in the 1st, 4th, 7th, 11th, ..., 100th movements, while in the 2nd, 5th, 8th, ..., 98th movements, the distance traveled by point $A$ is the same; in the 3rd, 6th, 9th, ..., 99th movements, the distance traveled by point $A$ is the same; therefore, we only need to calculate the distance traveled by point $A$ in the 2nd and 3rd movements.
In the figure below, $A_{2}$ represents the position of point $A$ after the 2nd movement, and the other subscripts are similar.
Since $\angle A B_{1} C_{2}=\frac{2 \pi}{3}$, we have $\angle A B_{1} A_{2}=\frac{2 \pi}{3}-2 \alpha$, so the distance traveled by point $A$ in the 2nd movement is
$$
\overparen{A A_{2}}=1 \cdot\left(\frac{2 \pi}{3}-2 \alpha\right)=\frac{2 \pi}{3}-2 \alpha,
$$

Since $B_{3}$ must be on the perpendicular bisector of $C_{2} A_{3}$, and $C_{2} B_{3}=1$, $B_{3}$ is the point $B$, so the rotation angle from $\triangle C_{3} A_{2} B_{2}$ to $\triangle C_{2} A_{3} B_{3}$ is $\frac{\pi}{3}$,
thus
$$
\begin{array}{c}
\angle A_{3} C_{2} A_{2}=\frac{\pi}{3} . \\
A_{3} C_{2}=A C=2 \sin \alpha,
\end{array}
$$

Therefore, the distance traveled by point $A$ in the 3rd movement is
$$
\widetilde{A_{2} A_{3}}=2 \sin \alpha \cdot \frac{\pi}{3} \text {. }
$$

In summary, the distance traveled by point $A$ in 100 movements is
$$
33\left(\overparen{A A_{2}}+\overparen{A_{2} A_{3}}\right)=22 \pi(1+\sin \alpha)-66 \alpha \text {. }
$$

Therefore, the answer is $(A)$.