2009 AIME I Problems/Problem 4

Revision as of 20:10, 20 March 2009 by Kubluck (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 $3009x/17x = \boxed{177}$

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions
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