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

m (→‎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.
  
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So the whole length of the line<math>(AP)</math> is <math>1000+2009=3009units</math>
+
So the whole length of the line<math>(APC)</math> is <math>1000+2009=3009units</math>
  
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And <math>AC</math> will be <math>17 units</math>
+
And <math>AP</math> will be <math>17 units</math>
  
 
So the answer is <math>3009/17 = 177</math>
 
So the answer is <math>3009/17 = 177</math>

Revision as of 19:17, 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$(APC)$ is $1000+2009=3009units$

And $AP$ will be $17 units$

So the answer is $3009/17 = 177$

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