Difference between revisions of "2009 AIME I Problems/Problem 4"

(Solution)
(Solution)
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One of the ways to solve this problem is to make this parallelogram a straight line.
 
One of the ways to solve this problem is to make this parallelogram a straight line.
  
So the whole length of the line<math>APC(AMC or ANC), and ABC</math> is <math>1000x+2009x=3009x</math>
+
So the whole length of the line <math>APC</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x</math>
  
And <math>AP(AM or AN)</math> is <math>17x</math>
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And <math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x</math>
  
So the answer is <math>3009x/17x = 177</math>
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So the answer is <math>3009x/17x = \boxed{177}</math>
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== See also ==
 +
{{AIME box|year=2009|n=I|num-b=3|num-a=5}}

Revision as of 20:10, 20 March 2009

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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