457. A motorcyclist and a cyclist set off towards each other from points A and B simultaneously and met 4 km from B. At the moment when the motorcyclist arrived at B, the cyclist was 15 km from A. Find the distance AB.
$\triangle$ Let the distance AB be denoted as $x$ km.

By the time they meet, the motorcyclist and the cyclist have traveled $(x-4)$ km and $x$ km, respectively, and by the time the motorcyclist arrives at $\mathrm{B}$, the distances are $x$ km and $(x-4)$ km, respectively.

Thus, the ratio of the speeds of the motorcyclist and the cyclist is, on one hand, $\frac{x-4}{x}$ (since the ratio of distances traveled by two bodies moving at a constant speed over the same time is equal to the ratio of their speeds), and on the other hand, $-\frac{x}{x-15}$. Therefore,

$$
\frac{x-4}{x}=\frac{x}{x-15}
$$

Solving this equation, we get: $x_{1}=20, x_{2}=3$. Only $x=20$ is valid.

Answer: 20 km.