2015/16 AHSMC Part 1 Problem 15

Problem 15 of 16

Points $E$ and $F$ are on the sides $BC$ and $CD$ respectively of the parallelogram $ABCD$ such that $\dfrac{EB}{EC} = \dfrac{2}{3}$ and $\dfrac{FC}{FD} = \dfrac{1}{4}$ . Let $M$ be the intersection of $AE$ and $BF$ . The value of $\dfrac{AM}{ME}$ is equal to

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Solution

The parallel line to $FB$ through $C$ intercepts $AB$ at $P$ and $AM$ intercepts $CP$ at $Q$ . Then $\frac{MQ}{ME} = \frac{ME + EQ}{ME} = 1 + \frac{EQ}{ME} = 1 + \frac{EC}{EB} = 1 + \frac{3}{2} = \frac{5}{2}$ and $\frac{AM}{MQ} = \frac{AB}{BP} = \frac{DC}{FC} = \frac{DF + FC}{FC} = 1 + \frac{DF}{FC} = 4 + 1 = 5.$ Hence $\frac{AM}{ME} = \frac{AM}{MQ} \cdot \frac{MQ}{ME} = 5 \cdot \frac{5}{2} = 12.5.$ The answer is (D).

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