Belorusija 2012

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 $$b+c=12a.\text{\hspace{1em}(1)}$$ 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, $$\frac{S(b-a)}{6a}=\frac{(a+c)}{10}.\text{\hspace{1em}(2)}$$ 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 $b(a+b)=12a^2\iff b^2+ab-12a^2=0\iff (b+4a)(b-3a)=0$. 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.