Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
Line 6: Line 6:
 
===Solution 1===
 
===Solution 1===
  
−
One of the ways to solve this problem is to make this parallelogram a straight lin
+
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</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x</math>
+
So the whole length of the line is <math>APC</math>(<math>AMC</math> or <math>ANC</math>), and <math>ABC</math> is <math>1000x+2009x=3009x.</math>
  
−
And <math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x</math>
+
<math>AP</math>(<math>AM</math> or <math>AN</math>) is <math>17x.</math>
  
 
So the answer is <math>3009x/17x = \boxed{177}</math>
 
So the answer is <math>3009x/17x = \boxed{177}</math>
 +
 
===Solution 2===
 
===Solution 2===
 
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>
 
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>

Revision as of 23:11, 1 January 2013

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 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. Unknown error_msg) and the desired quantity is $177$.

See also

2009 AIME I (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&oldid=50432"