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

(Solution)
(Solution)
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[[2009 AIME I Problems/Problem 4|Solution]]
 
[[2009 AIME I Problems/Problem 4|Solution]]
  
==Solution==
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==Solution 1==
  
 
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 answer is <math>3009x/17x = \boxed{177}</math>
 
So the answer is <math>3009x/17x = \boxed{177}</math>
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==Solution 2==
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Draw a diagram with all the given points and lines involved. Construct parallel lines <math>\overline{DF_2F_1}</math> and <math>\overline{BB_1B_2}</math> to <math>\overline{MN}</math>, where for the lines the endpoints are on <math>\overline{AM}</math> and <math>\overline{AN}</math>, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral <math>BB_2DF_1</math> <math>\overline{E_1E_2E_3}</math> where the points are in order from top to bottom. Clearly, by similar triangles, <math>BB_2 = \frac {1000}{17}MN</math> and <math>DF_1 = \frac {2009}{17}MN</math>. It is not difficult to see that <math>E_2</math> is the center of quadrilateral <math>ABCD</math> and thus the midpoint of <math>\overline{AC}</math> as well as the midpoint of <math>\overline{B_1}{F_2}</math> (all of this is easily proven with symmetry). From more triangle similarity, <math>E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP</math>
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<math>= \boxed{177}AP</math>.
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A diagram would be appreciated (and whoever makes it can take out the descriptive stuff).
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2009|n=I|num-b=3|num-a=5}}

Revision as of 14:35, 23 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 1

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 $3009x/17x = \boxed{177}$

Solution 2

Draw a diagram with all the given points and lines involved. Construct parallel lines $\overline{DF_2F_1}$ and $\overline{BB_1B_2}$ to $\overline{MN}$, where for the lines the endpoints are on $\overline{AM}$ and $\overline{AN}$, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral $BB_2DF_1$ $\overline{E_1E_2E_3}$ where the points are in order from top to bottom. Clearly, by similar triangles, $BB_2 = \frac {1000}{17}MN$ and $DF_1 = \frac {2009}{17}MN$. It is not difficult to see that $E_2$ is the center of quadrilateral $ABCD$ and thus the midpoint of $\overline{AC}$ as well as the midpoint of $\overline{B_1}{F_2}$ (all of this is easily proven with symmetry). From more triangle similarity, $E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP$ $= \boxed{177}AP$.

A diagram would be appreciated (and whoever makes it can take out the descriptive stuff).

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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