6. The following is a schematic diagram of a computer shooting game. The lengths of line segments $\mathrm{CD}$, $\mathrm{EF}$, and $\mathrm{GH}$ are all 20 centimeters. $0$, $P$, and $Q$ are their midpoints and are located on the same line $A B$. $A 0=45$ centimeters, $\mathrm{OP}=\mathrm{PQ}=20$ centimeters. It is known that the speed of the small ring on $\mathrm{CD}$ is 5 centimeters per second, the speed of the small ring on $\mathrm{EF}$ is 9 centimeters per second, and the speed of the small ring on $\mathrm{GH}$ is 27 centimeters per second. At zero time, a small ring starts from the left endpoint of each of $\mathrm{CD}$, $\mathrm{EF}$, and $\mathrm{GH}$ and moves back and forth uniformly along the line segments. Now, a bullet is fired from point $\mathrm{A}$ toward $\mathrm{B}$ at a uniform speed. To pass through the three rings, what is the maximum speed of the bullet in centimeters per second?
6. Solution: The time for the small ring to pass through point 0 is $4 k+2(k=0,1,2, \cdots)$;
The time for the small ring to pass through point P is $\frac{20}{9} m+\frac{10}{9} \quad(m=0,1,2, \cdots)$;
The time for the small ring to pass through point Q is $\frac{20}{27} n+\frac{10}{27}(n=0,1,2, \cdots)$;
From the fact that the speed of the small ring on GH is exactly 3 times the speed of the small ring on EF, when the small ring on EF is at point P, the small ring on GH must simultaneously be at point Q. The bullet, after passing through the small ring at point P, reaches point Q. If it can penetrate the small ring on GH, it can only be when the small ring on GH has descended 1 time, or 2 times, or 3 times, $\cdots$ and then passes through point Q, meaning the time difference between the bullet reaching point P and reaching point Q must be $\frac{20}{27} x_{n}(n=1,2,3, \cdots)$, an integer multiple of $\frac{20}{27}$.

Since $OP=PQ$, and the bullet travels at a constant speed, the time for the bullet to travel from 0 to P should also be an integer multiple of $\frac{20}{27}$. When $k=0$, $\frac{20}{9} m+\frac{10}{9}-2=\frac{20}{9} m-\frac{8}{9}$, regardless of the value of $m$, it is not an integer multiple of $\frac{20}{27}$. Only when $k=5x+2$ (where $x=0,1,2, \cdots)$, the value of $\frac{20}{9} m+\frac{10}{9}-(4 k+2)$ satisfies being an integer multiple of $\frac{20}{27}$. Since the problem requires the maximum value, $k$ should be the smallest, taking $x=0$, at this time $k=2$.

When $k=2$, the time for the small ring to reach point 0 is $4 k+2=10$ (seconds), and the bullet from A to 0 should also be 10 seconds, with a speed of 4.5 cm/s. The time for the bullet to travel from A to P is $\frac{65}{4.5}$ seconds, i.e., $\frac{20}{9} m+\frac{10}{9}=\frac{65}{4.5}, m=6$; the time for the bullet to travel from A to Q is $\frac{85}{4.5}$ seconds, i.e., $\frac{20}{27} n+\frac{10}{27}=\frac{85}{4.5}, n=25$.

It can be seen that when the bullet's speed is 4.5 cm/s, it can penetrate the three rings, and this is the maximum speed for penetrating the three rings.