2009 AIME I Problems/Problem 4

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, $E_1E_3 = \frac12\cdot\frac {3009}{17}MN\implies AE_2 = \frac12\cdot\frac {3009}{17}AP\implies AC = 2\cdot\frac12\cdot\frac {3009}{17}AP$ $= \boxed{177}AP$ .

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2009 AIME I Problems/Problem 4

Contents

Problem 4

Solution

Solution 1

Solution 2

See also