3.50 Two cyclists started simultaneously from the same place in the same direction. A third cyclist started from the same place 10 minutes after them. First, he overtook the first cyclist, and then continued his journey for another 20 minutes before catching up with the second cyclist. From the start to the end of the journey, each cyclist traveled at a constant speed: $a$ km/h - the first cyclist, $b$ km/h - the second. Find the speed of the third cyclist.
3.50 The graphs of the cyclists' movements are shown in Fig. 3.13.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-114.jpg?height=594&width=1004&top_left_y=184&top_left_x=150)

Fig. 3.13

Let's express the distances they have traveled: for the first $A B=\left(\frac{1}{6}+x\right) a$; for the third $A B=x v$, where $v-$ is the desired speed; $C D=\left(x+\frac{1}{3}\right) v$; for the second $C D=\left(\frac{1}{6}+x+\frac{1}{3}\right) b$. We have a system of equations with respect to $x$ and $v$.

$\left\{\begin{array}{l}\left(\frac{1}{6}+x\right) a=x v \\ \left(\frac{1}{2}+x\right) b=\left(x+\frac{1}{3}\right) v .\end{array}\right.$

By eliminating $x$, we find $v=\frac{1}{4}\left(a+3 b+\sqrt{a^{2}-10 a b+b^{2}}\right)$.

Answer: $\quad \frac{1}{4}\left(a+3 b+\sqrt{a^{2}-10 a b+b^{2}}\right) \mathrm{km} / \mathrm{h}$.