2009 AIME I Problems/Problem 4

Problem 4

In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ .

Solution

One of the ways to solve this problem is to make this parallelogram a straight line.

So the whole length of the line $APC$ ( $AMC$ or $ANC$ ), and $ABC$ is $1000x+2009x=3009x$

And $AP$ ( $AM$ or $AN$ ) is $17x$

So the answer is $3009x/17x = \boxed{177}$

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2009 AIME I Problems/Problem 4

Problem 4

Solution

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