16. (15 points) As shown in the figure, two small insects, A and B, start from the vertices $A$ and $B$ of an opaque regular pentagram fence with each side 20 cm long, and crawl along the outer side in a counterclockwise direction. Insect A crawls at 5 cm per second, and insect B crawls at 4 cm per second. Question: During the time from when A starts until it first reaches $B$, how many seconds can B see A?
【Analysis】Let the five vertices of the pentagram be marked in a counterclockwise direction as $B, B_{1}, B_{2}, B_{3}, B_{4}$, forming interval one from vertex $B$ to vertex $B_{1}$, interval two from vertex $B_{1}$ to vertex $B_{2}$, and so on to interval five. According to the problem, the condition for person B to see person A is that they must both be walking in the same interval at the same time.

Time seen in interval one: $20 \div 5=4$ (seconds);
Time seen in interval two: $20 \times 2 \div 4=10$ (seconds), $20 \times 3-10 \times 5=60-50=10$ (cm), $10 \div 5=2$ (seconds);

Situation in interval three: The time for A to reach $B_{3}$ is $(10+20+20) \div 5=10$ (seconds), the distance B moves is $10 \times 4=40$ (cm), at this point B reaches $B_{2}$, the time B can see A is 0, according to this, the problem can be solved.

【Solution】Solution:
Time seen in interval one: $20 \div 5=4$ (seconds);
Time seen in interval two: $20 \times 2 \div 4=10$ (seconds),
$20 \times 3-10 \times 5=60-50=10$ (cm),
$10 \div 5=2$ (seconds);
Time seen in interval three: 0
The total time B can see A is $2+4=6$ (seconds)
Answer: B can see A for 6 seconds.