36.41 A person walks from vertex $A$ to the center $O$ of square $ABCD$ (as shown by the solid line in the figure, same below), then from $O$ to the center $O_{1}$ of square $OEDF$, then from $O_{1}$ to the center $O_{2}$ of square $O_{1}GFH$, and then from $O_{2}$ to the center $O_{3}$ of square $O_{2}LHJ$, covering a total distance of $15 \sqrt{2}$ meters. The perimeter of square $ABCD$ is
(A) 48 meters.
(B) 64 meters.
(C) 72 meters.
(D) 80 meters.
(Chinese Junior High School Mathematics Competition, Nanjing, Jiangsu Province, 1991)
[Solution] Let the side length of the square be $a$, then its diagonal length is $l=\sqrt{2} a$. It is easy to see that
$$
\begin{array}{l}
A O=\frac{l}{2}, \quad O O_{1}=\frac{l}{4}, \quad O_{1} O_{2}=\frac{l}{8}, \quad O_{2} O_{3}=\frac{l}{16} \text {. } \\
\therefore \quad \frac{l}{2}+\frac{l}{4}+\frac{l}{8}+\frac{l}{16}=15 \sqrt{2} \text {. } \\
\text { i.e., } \frac{15}{16} l=\frac{15}{16} \sqrt{2} a=15 \sqrt{2} \text {, } \\
\therefore \quad a=16 \text {. } \\
\end{array}
$$

Then the perimeter of the square is
$$
4 a=4 \cdot 16=64 .
$$

Therefore, the answer is $(B)$.