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Berkeley Math Circle: Monthly Contest 1

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

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.)

Solution

Solution:

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.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.