3.5. From point $O$, 12 rays $l_{1}, l_{2}, \ldots, l_{12}$ are drawn such that each pair of adjacent rays forms an angle of $30^{\circ}$. An ant sitting on ray $l_{1}$, at some moment, started crawling along the perpendicular to ray $l_{2}$. Upon reaching $l_{2}$, it turned and crawled along the perpendicular to ray $l_{3}$, and so on. The first complete revolution around point $O$ took 1 minute. How much time will it take for the second revolution if it continues to crawl at a constant speed? What distance will it crawl to point $O$ if the first loop is $a$ m?
3.5. Suppose that the ant started its journey at point $A$ and after the first turn ended up at point $B$ (on the same ray $l_{1}$, see figure). Check that $O B / O A=(\sqrt{3} / 2)^{12}=729 / 4096$.

Prove that the length of each subsequent turn is $729/4096$ of the length of the previous one. From this, it follows that the ant will spend $729 / 4096$ min $\approx 10.7$ sec on the second turn, and for the entire path consisting of an infinite number of similar turns (with a similarity coefficient $k=729 / 4096$) 一

$\left(1+k+k^{2}+k^{3}+\ldots\right)$ min $=\frac{1}{1-k}$ min $=$

$$
=\frac{4096}{3367} \text { min } \approx 1 \text { min } 12.5 \text { sec. }
$$