# 2009 AIME I Problems/Problem 4

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## 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 way to solve this problem is to make this parallelogram a straight line. So the whole length of the line$(AP)$ is $1000+2009=3009units$ And $AC$ will be $17 units$ So the answer is $3009/17 = 177$