Art of Problem Solving
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

2009 AIME I Problems/Problem 4: Difference between revisions

← Older editNewer edit →
Line 6: Line 6:
==Solution==
==Solution==


One of the way 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>(AP)</math> is <math>1000+2009=3009units</math>
So the whole length of the line<math>(AP)</math> is <math>1000+2009=3009units</math>

Revision as of 18:58, 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$(AP)$ is $1000+2009=3009units$

And $AC$ will be $17 units$

So the answer is $3009/17 = 177$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_4&oldid=30839"