12. (12 points) Person A and Person B start from points A and B respectively at the same time, moving towards each other at a constant speed. When A and B meet at point C, Person C starts from B, moving at a constant speed towards A. When A and C meet at point D, A immediately turns around and reduces their speed to 80% of the original speed. When A and C arrive at A at the same time, B is still 720 meters away from A. If the distance between C and D is 900 meters, then the distance between A and B is $\qquad$ meters.
【Solution】Solution: If Party A's speed decreases to $80 \%$ of the original, and arrives at location $A$ simultaneously with Party C, then the speed ratio of Party A to Party C is: $5: 4$. Therefore, if Party A walks 900 meters from $C$ to $D$, then Party C walks from $B$ to $D$:
$$
900 \times \frac{4}{5}=720 \text{ meters }
$$

Party A walks two (full distance - 720 meters) back to $A$, while Party B walks one (full distance - 720 meters) in the same time. Party A walks the second (full distance - 720 meters) at $80 \%$ of the original speed, which is equivalent to walking $\frac{9}{4}$ (full distance - 720 meters) at the original speed. Therefore, the speed ratio of Party A to Party B is: $9: 4$. When Party A and Party B meet, Party B has walked from $B$ to $C$:
$$
900+720=1620 \text{ meters }
$$

Thus, the full distance is:
$$
1620 \times \frac{4+9}{4}=5265 \text{ meters }
$$

Answer: The distance between $A$ and $B$ is 5265 meters.
The answer is: 5265.