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

Ewcikewqikd (talk | contribs) (→Solution) |
God of Math (talk | contribs) m (→Solution) |
||

Line 6: | Line 6: | ||

==Solution== | ==Solution== | ||

− | One of the | + | 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 , point is on so that and point is on so that . Let be the point of intersection of and . Find .

## Solution

One of the ways to solve this problem is to make this parallelogram a straight line.

So the whole length of the line is

And will be

So the answer is