13.50. Three runners $A, B$ and $C$ are running on parallel tracks at constant speeds. At the initial moment, the area of triangle $A B C$ is 2, and after 5 seconds, it is 3. What could it be after another 5 seconds?
13.50. Let at the initial moment, i.e., at $t=0, \overrightarrow{A B}=\boldsymbol{v}$ and $\overrightarrow{A C}=\boldsymbol{w}$. Then at moment $t$ we get $\overrightarrow{A B}=\boldsymbol{v}+t(\boldsymbol{b}-\boldsymbol{a})$ and $\overrightarrow{A C}=\boldsymbol{w}+t(\boldsymbol{c}-\boldsymbol{a})$, where $\boldsymbol{a}, \boldsymbol{b}$ and $\boldsymbol{c}$ are the velocity vectors of runners $A, B$ and $C$. Since vectors $\boldsymbol{a}, \boldsymbol{b}$ and $\boldsymbol{c}$ are parallel, then $(\boldsymbol{b}-\boldsymbol{a}) \vee(\boldsymbol{c}-\boldsymbol{a})=0$, and therefore, $|S(A, B, C)|=|\overrightarrow{A B} \vee \overrightarrow{A C}| / 2=|x+y t|$, where $x$ and $y$ are some constant numbers. Solving the system $|x|=2,|x+5 y|=3$, we get two solutions, giving for the dependence of the area of triangle $A B C$ on time $t$ the expressions $|2+(t / 5)|$ and $|2-t|$. Therefore, at $t=10$ the area can take the values 4 and 8.