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2009 AIME I Problems/Problem 4

Revision as of 19:20, 20 March 2009 by Ewcikewqikd (talk | contribs) (→‎Solution)

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 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 $3009x/17x = 177$

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