2009 AIME I Problems/Problem 4

Revision as of 17:30, 20 March 2009 by Ewcikewqikd (talk | contribs) (New page: == Problem 4 == In parallelogram <math>ABCD</math>, point <math>M</math> is on <math>\overline{AB}</math> so that <math>\frac {AM}{AB} = \frac {17}{1000}</math> and point <math>N</math> is...)
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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

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 $3009/17 = 177$

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