AoPSWiki
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

2009 AIME I Problems/Problem 4

From AoPSWiki

Contents

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 APC(AMC or ANC), and ABC is 1000x+2009x=3009x

And 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 {3... = \boxed{177}AP.

See also

2009 AIME I (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us