$36 \cdot 40$ An equilateral triangle $A B P$ with side length 2 inches (as shown in the figure) is placed inside a square $A X Y Z$ with side length 4 inches, such that point $B$ is on side $A X$. The triangle is first rotated clockwise around point $B$, and then rotated clockwise around $P$, continuing in this manner along the edges of the square.
until $P$ returns to its original position. The length of the path traveled by vertex $P$ in inches is
(A) $20 \pi / 3$.
(B) $32 \pi / 3$.
(C) $12 \pi$.
(D) $40 \pi / 3$.
(E) $15 \pi$.
(23rd American High School Mathematics Examination, 1972)
[Solution] It is clear that the triangle has rotated $24=8 \cdot 3$ times before point $P$ returns to its original position. Among these, 8 rotations are around vertex $P$, during which $P$ does not move.

In the other 16 rotations, 8 are around the midpoint of a side of the square, sweeping through $120^{\circ}$ or $\frac{2}{3} \pi$ radians with the distance to point $P$ as the radius.
$P$ point moves $\quad P$ point does not move $\quad P$ point moves
The remaining 8 rotations are around a corner of the square, sweeping through $30^{\circ}$ or $\frac{1}{6} \pi$ radians with the distance to point $P$ as the radius.
Thus, the total angle swept with the distance to point $P$ as the radius is
$$
8\left(\frac{2}{3} \pi+\frac{1}{6} \pi\right)=\frac{20}{3} \pi \text { radians. }
$$

Therefore, the length of the path traveled by point $P$ is $\frac{40}{3} \pi$ inches.
Hence, the answer is $(D)$.