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)
(→‎Solution)
Line 8: Line 8:
 
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.
  
−
So the whole length of the line<math>APC, or AMC</math> is <math>1000x+2009x=3009x</math>
+
So the whole length of the line<math>APC(AMC or ANC), and ABC</math> is <math>1000x+2009x=3009x</math>
  
−
And <math>AP</math> will be <math>17x</math>
+
And <math>AP(AM or AN)</math> is <math>17x</math>
  
 
So the answer is <math>3009/17 = 177</math>
 
So the answer is <math>3009/17 = 177</math>

Revision as of 20:20, 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$APC(AMC or ANC), and ABC$ is $1000x+2009x=3009x$

And $AP(AM or AN)$ is $17x$

So the answer is $3009/17 = 177$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&oldid=30843"