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

Ewcikewqikd (talk | contribs) (→Solution) |
Ewcikewqikd (talk | contribs) (→Solution) |
||

Line 12: | Line 12: | ||

And <math>AP(AM or AN)</math> is <math>17x</math> | And <math>AP(AM or AN)</math> is <math>17x</math> | ||

− | So the answer is <math> | + | So the answer is <math>3009x/17x = 177</math> |

## Revision as of 19:20, 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 is

So the answer is