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, $\angle A B C=2 \alpha(0<\alpha<\frac{\pi}{3})$. 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 rotation. Then, the total length of the path traveled by point $A$ is
A. $22 \pi(1+\sin \alpha)-66 \alpha$
B. $22 \pi+\frac{68}{3} \pi \sin \alpha-66 \alpha$
C. $\frac{67}{3} \pi$
D. $33 \pi-66 \alpha$
Answer
As shown in the figure, during the first rotation (around point $A$), the black $\triangle A B C$
$\rightarrow$ blue $\triangle A B C$, point $B$ rotates from the center of the circle to the circumference, $B B=1$, the rotation angle is $60^{\circ}$.
During the second rotation (around point $B$), blue $\triangle A B C \rightarrow$ yellow
$\triangle A B C$, the rotation path passes through the center $B$, the blue $\triangle A B C$ and the yellow $\triangle A B C$ are symmetrical about the line $B B$, the rotation angle $\beta=120^{\circ}-2 \alpha$.
By symmetry, during the third rotation (around point $C$), yellow $\triangle A B C \rightarrow$ green $\triangle A B C$ is equivalent to the reset of the first rotation, the rotation angle is $60^{\circ}$.
Since $A C=2 \sin \alpha$, the total length of the path that point $A$ travels in three rotations is $\frac{2 \pi}{3}-2 \alpha+\frac{2 \pi}{3} \sin \alpha$. Rotating 100 times is equivalent to 33 cycles (the last rotation is around point $A$), so the total length of the path is $22 \pi(1+\sin \alpha)-66 \alpha$, hence the answer is $A$.