Problem 1850

Answer: 13.30.
Let the distance between $A$ and $B$ be equal to $S$ (km), the speeds of Pedestrian, Cyclist and Motorcyclist be equal to $a$, $b$ and $c$ (km/h), respectively. By condition (the first meeting is the meeting of Cyclist and Motorcyclist), it follows that $a < b < c$. Let Pedestrian arrive at point $D$ at that moment when Cyclist and Motorcyclist meet at point $C$. By condition, $AD = \frac{S}{6}$. Then $\frac{AD}{a} = t_1 = \frac{S}{6a}$. Since $AC + (AB + BC) = 2S$ we have $t_1 = \frac{2S}{b+c}$. So, $\frac{S}{6a} = \frac{2S}{b+c}$, hence
$$b+c=12a. \tag{1}$$
From $t_1 b = \frac{Sb}{6a}$ it follows that $AC = \frac{Sb}{6a}$, so $CD = AC - AD = \frac{Sb}{6a} - \frac{S}{6} = \frac{S(b-a)}{6a}$. Therefore,
$$\frac{S(b-a)}{6a} = \frac{(a+c)}{10}. \tag{2}$$
By condition, to arrive at $C$ (the point of meeting with Cyclist) Pedestrian need the time $t_2 = \frac{AC}{a} = \frac{Sb}{6a^2}$. On the other hand, $t_2 = \frac{2S}{a+b}$, so $b(a+b) = 12a^2 \Leftrightarrow b^2 + ab - 12a^2 = 0 \Leftrightarrow (b+4a)(b-3a) = 0$. Since $b+4a > 0$ ($a$ and $b$ are positive), we have $b-3a = 0$, i.e. $b = 3a$. Now from (1) it follows $c = 9a$, and (2) gives $S = 3a$. Therefore, $AC = \frac{Sb}{6a} = \frac{3a \cdot 3a}{6a} = 1.5a$. It means that Pedestrian and Cyclist meet at point $C$ in $t_2$ hours after the start, $t_2 = \frac{1.5a}{a} = 1.5$. Moreover, for Motorcyclist we see $1.5 \cdot c = 13.5 \cdot a = 4 \cdot 3a + 1.5 \cdot a = 4S + 1.5a = AB + BA + AB + BA + AC$, which means that after 1.5 hours after the start he arrive at point $C$. Therefore, Pedestrian, Cyclist and Motorcyclist meet at point $C$ at 13.30.