Four cars $A, B, C$, and $D$ travel at constant speeds on the same road (not necessarily in the same direction). Car $A$ passed $B$ and $C$ at 8am and 9am, respectively, and met $D$ at 10am. Car $D$ met $B$ and $C$ at $12 \mathrm{pm}$ and $2 \mathrm{pm}$, respectively. Determine at what time $B$ passed $C$. (The times given are within a single day.)
Draw lines $A, B, C, D$ to graph the movement of the four cars, with time on the $x$-axis and distance on the $y$-axis, and let $(X Y)$ be the point where lines $X$ and $Y$ meet. Then $(A C)$ is the midpoint of the line from $(A B)$ to $(A D)$, and $(D B)$ is the midpoint of the line from $(D A)$ to $(D C)$. Thus, $(B C)$ is the intersection of two medians of the triangle with vertices at $(A B),(A D)$, and $(D C)$, so it is the centroid. But this means the distance from $(A C)$ to $(B C)$ is half the distance from $(B C)$ to $(C D)$. Since $A$ meets $C$ at 9 am and $C$ meets $D$ at $2 \mathrm{pm}$, the time between these meetings is 5 hours. The meeting of $C$ and $B$ occurs $1 / 3$ of the way from the first meeting to the second, i.e. $5 / 3$ hours after $A$ meets $C$, or at 10:40am.