4. As shown in the figure, the vertex $B$ of $\triangle A B C$ is at the center of the unit circle, and $\angle A B C=2 \alpha\left(0<\alpha<\frac{\pi}{3}\right)$. Now, $\triangle A B C$ is rotated counterclockwise within the circle using the following method: First, rotate around $A$ so that $B$ lands on the circumference; second, rotate around $B$ so that $C$ lands on the circumference; third, rotate around $C$ so that $A$ lands on the circumference. This rotation continues until the 100th time. Then, the total length of the path traveled by point 1 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) $33 \pi-66 \alpha$.
Solve as shown in the figure,
First time $A \rightarrow A_{1}$, the distance traveled is 0,
Second time $A_{1} \rightarrow A_{2}$, the distance traveled is $A_{1} A_{2}$;
Third time $A_{2} \rightarrow A_{3}$, the distance traveled is $A_{2} A_{3}$. After three rounds, it returns to a similar initial position of $\triangle A B C$, and so on, for a total of 33 rounds plus one more.
And $\overbrace{1} A_{2}=A B \times\left(\frac{2 \pi}{3}-2 \alpha\right)$
$$
\begin{aligned}
= & 1 \times\left(\frac{2 \pi}{3}-2 \alpha\right), \\
A_{2} A_{3} & =A C \times \frac{\pi}{3}=2 \sin \alpha \times \frac{\pi}{3} \\
& =\frac{2 \pi}{3} \cdot \sin \alpha,
\end{aligned}
$$

Therefore, the total distance traveled by point $A$ is $33\left(\frac{2 \pi}{3}\right.$ $\left.-2 \alpha)+\frac{2 \pi}{3} \sin \alpha\right)=22 \pi(1+\sin \alpha)$
$-66 \alpha$. Hence, the correct choice is $(A)$.