# Difference between revisions of "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 1

One of the ways to solve this problem is to make this parallelogram a straight line. So the whole length of the line is $APC$($AMC$ or $ANC$), and $ABC$ is $1000x+2009x=3009x.$

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

## Solution 3

Using vectors, note that $\overrightarrow{AM}=\frac{17}{1000}\overrightarrow{AB}$ and $\overrightarrow{AN}=\frac{17}{2009}\overrightarrow{AD}$. Note that $\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}$ for some positive x and y, but at the same time is a scalar multiple of $\overrightarrow{AB}+\overrightarrow{AD}$. So, writing the equation $\overrightarrow{AP}=\frac{x\overrightarrow{AM}+y\overrightarrow{AN}}{x+y}$ in terms of $\overrightarrow{AB}$ and $\overrightarrow{AD}$, we have $\overrightarrow{AP}=\frac{\frac{17x}{1000}\overrightarrow{AB}+\frac{17y}{2009}\overrightarrow{AD}}{x+y}$. But the coefficients of the two vectors must be equal because, as already stated, $\overrightarrow{AP}$ is a scalar multiple of $\overrightarrow{AB}+\overrightarrow{AD}$. We then see that $\frac{x}{x+y}=\frac{1000}{3009}$ and $\frac{y}{x+y}=\frac{2009}{3009}$. Finally, we have $\overrightarrow{AP}=\frac{17}{3009}(\overrightarrow{AB}+\overrightarrow{AD})$ and, simplifying, $\overrightarrow{AB}+\overrightarrow{AD}}=177\overrightarrow{AP}$ (Error compiling LaTeX. ! Extra }, or forgotten \$.) and the desired quantity is $177$.

## Solution 4

We approach the problem using mass points on triangle $ABD$ as displayed below. $[asy] pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8); draw(A--B--C--D--cycle); draw(B--D^^A--C^^M--NN); pair O=extension(A,C,B,D); pair P=extension(A,C,M,NN); dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("M",M,S); label("N",NN,NW); label("P",P,NNE); label("O",O,N); [/asy]$

But as $MN$ does not protrude from a vertex, we will have to "split the mass" at point $A$. First, we know that $AC$ bisects $BD$. Therefore, we assign equal masses to points $B$ and $D$. In this case, we assign $B$ and $D$ a mass of 17 each. Now we split the mass at $A$, so we balance segments $AB$ and $AD$ separately, and then the mass of $A$ is the sum of those masses. A mass of 983 is required to balance segment $AB$, while a mass of 1992 is required to balance segment $AD$. Therefore, $A$ has a mass of $1992+983=2965$. Also, $O$ has a mass of 34. Therefore, $\frac{AO}{AP}=\frac{2975+34}{34}=\frac{3009}{34}$, so $\frac{AC}{AP}=\frac{3009}{17}=177$.