4. As shown in the figure, the vertex $B$ of $\triangle A B C$ is at the center of the unit circle, and $A, C$ are on the circumference, $\angle A B C=2 a\left(0<a<\frac{\pi}{3}\right)$. Now, $\triangle A B C$ is rotated counterclockwise within the circle, with the specific method as follows: 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, and so on until the 100th time. Then, the total length of the path traveled by point $A$ is
(A) $22 \pi(1+\sin a)-66 a$
(B) $22 \pi+\frac{68}{3} \pi \sin a-66 a$
(C) $\frac{67}{3} \pi$
( D) $33 \pi-66 a$
4.【Analysis and Solution】When $\triangle A B C$ rotates counterclockwise in the circle, successively centered at $A, B, C$ (i.e., at $A, B_{1}, C_{2}$ in the figure), until point $B$ returns to the center of the circle, the path traced by point $A$ is the circular arcs $\overparen{A A}_{1}$ and $\overparen{A_{1} A_{2}}$, with radii of $1$ and $AC$ respectively, and the side length of $AC$ is $2 \sin a$.

Since $\triangle A B_{1} B$ and $\triangle B_{1} C_{2} B$ are equilateral triangles, and $\angle A_{1} B_{1} C_{2}=2 a$, the central angle corresponding to $\overparen{A A_{1}}$ is $\left(\frac{2}{3} \pi-2 a\right)$. When $\triangle A_{1} B_{1} C_{2}$ rotates for the third time centered at $C_{2}$, its side $C_{2} B_{1}$ rotates $\frac{\pi}{3}$ to reach $C_{2} B$, so the central angle corresponding to $\overparen{A_{1} A_{2}}$ is $\frac{\pi}{3}$. Thus, the distance traveled by point $A$ after each three rotations of $\triangle A B C$ is:
$$
\left(\frac{2}{3} \pi-2 a\right) \times 1+\frac{\pi}{3} \times 2 \sin a=\frac{2}{3} \pi(1+\sin a)-2 a \text {. }
$$

Therefore, the distance traveled by point $A$ after 100 rotations of $\triangle A B C$ is:
$$
s=\frac{100-1}{3}\left[\frac{2}{3} \pi(1+\sin a)-2 a\right]=22 \pi(1+\sin a)-66 a \text {. }
$$