15. (15 points) As shown in the figure, $\triangle A B C$ is an equilateral triangle with a side length of $108 \mathrm{~cm}$, insect A and insect B start from point $A$ and point $C$ respectively at the same time, and crawl along the sides of $\triangle A B C$. B crawls counterclockwise, and the speed ratio is 4:5. After meeting, A rests for 10 seconds at the meeting point, then continues to crawl at the original speed in the original direction; B does not rest, increases its speed by $20 \%$, and continues to crawl in the original direction, and they meet for the second time exactly at the midpoint of $B C$. Find the crawling speeds of insect A and B at the beginning.
【Solution】Solution: The distance traveled by A $=108 \times 2 \div(4+5) \times 4=96$ (cm), the distance traveled by B $=108 \times 2-96=120$ (cm).

The second meeting occurs at the midpoint of $B C$, so the distance traveled by A from the first meeting to the second meeting is $120-108 \div 2=66$ (cm), and the distance traveled by B is $96+108+108 \div 2=258$ (cm).

After meeting, the speed ratio of A to B $=4:(5 \times 120 \%)=2: 3$, so when A travels 66 cm, the distance B travels is $66 \div 2 \times 3=99$ (cm),

Thus, the distance B travels during A's 10-second rest is $258-99=159$ (cm),
B's initial crawling speed is $159 \div 10 \div(1+20 \%)=13.25(\mathrm{~cm} / \mathrm{s})$, and A's speed is $13.25 \div 5 \times 4=10.6$ $(\mathrm{cm} / \mathrm{s}$ )

Answer: The crawling speed of insect A is $10.6 \mathrm{~cm} / \mathrm{s}$, and the crawling speed of insect B is $13.25 \mathrm{~cm} / \mathrm{s}$.