4. 199 A car departs from point $O$ along a straight highway, maintaining a constant speed $v$. At the same time the car starts, a person on a bicycle departs from a point that is a distance $a$ from $O$ and a distance $b$ from the highway, aiming to deliver a letter to the car's driver. What is the minimum speed the cyclist must travel at to achieve his goal?
[Solution] Let $b>0$. If $b=0$, it means the cyclist is on the road, and the solution to the problem is obvious.

Let $M$ be the point where the cyclist is located, and $S$ be the point where the cyclist meets the car. Let $\angle M O S=\alpha, t$ be the time the cyclist spends from departure to meeting, and $x$ represent the speed of the bicycle.
In $\triangle M O S$ (see the right figure), $O S=v t, M S=x t, O M=a, \angle M O S=\alpha$. Applying the cosine rule, we get
$$\begin{aligned}
& x^{2} t^{2}=a^{2}+v^{2} t^{2}-2 a v t \cos \alpha \\
x^{2}= & \frac{a^{2}}{t^{2}}-2 a v \cos \alpha \cdot \frac{1}{t}+v^{2}
\end{aligned}$$

Let $\frac{1}{t}=S$, then
$$\begin{aligned}
x^{2} & =a^{2} S^{2}-2 a v \cos \alpha \cdot S+v^{2} \\
& =(a S-v \cos \alpha)^{2}+v^{2}-v^{2} \cos ^{2} \alpha, \\
& =(a S-v \cos \alpha)^{2}+v^{2} \sin ^{2} \alpha .
\end{aligned}$$

To find the positive value of $S$ that minimizes the positive value of $x$, i.e., $x^{2}$, we discuss two different cases:

Case 1: $\cos \alpha>0$, i.e., $\alpha$ is an acute angle. From (1), we know that when $a S-v \cos \alpha=0$, i.e.,
$$S=\frac{v \cos \alpha}{a}$$

$x$ takes its minimum value $x_{\text {min }}$, and
$$x_{\min }^{2}=v^{2} \sin ^{2} \alpha, x_{\min }=v \sin \alpha=\frac{v b}{a} .$$

Case 2: $\cos \alpha \leqslant 0$, i.e., $\alpha$ is a right or obtuse angle. In this case, there is no minimum speed for the cyclist to catch up with the car, because the smaller $S$ is, i.e., the larger $t$ is, the smaller $a S-v \cos \alpha$ and thus $x^{2}$ will be. When $t$ is infinitely large, $S$ tends to zero, and from (1), $x$ will tend to $v$.

Next, we explain the conclusion of this problem using a graph. When $\alpha<90^{\circ}$ (Figure $a$), the minimum speed of the bicycle is $v \sin \alpha$. The time the cyclist spends to catch up with the car is $t=\frac{1}{S}=\frac{a}{v \cos \alpha}$. Therefore, the distance the cyclist travels to catch up with the car $M S$ is
$$v \sin \alpha \cdot \frac{a}{v \cos \alpha}=a \operatorname{tg} \alpha.$$
This means $\angle O M S=90^{\circ}$, i.e., the cyclist must chase the car along a line perpendicular to the line $O M$.
When $\alpha \geqslant 90^{\circ}$ (Figure b), the cyclist must travel a longer distance than the car. Therefore, only when the cyclist's speed exceeds the car's speed can he catch up with the car. The larger $\angle O M S (\angle O M S=\beta)$ is, the smaller the difference in speed between the bicycle and the car will be. If the cyclist's route (a straight line) is sufficiently close to $180^{\circ}-\alpha$, then this speed difference will be sufficiently small.