AoPSWiki

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

Personal tools

Search

Toolbox

2009 AIME II Problems/Problem 3

From AoPSWiki

Problem

In rectangle $ABCD$ , $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ .

Solution

pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5);draw (A--B--C--D--cycle);pair E=(7,10);draw (B--E);draw (A--C);pair F=(6...

From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$ , and $ABE$ are also right triangles. By $AA$ , $\triangle FBA \sim \triangle BCA$ , and $\triangle FBA \sim \triangle ABE$ , so $\triangle ABE \sim \triangle BCA$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ . $AE=\frac{AD}{2}$ and $BC=AD$ , so $\frac {AD}{2AB}= \frac {AB}{AD}$ , or $(AD)^2=2(AB)^2$ , so $AD=AB \sqrt{2}$ , or $100 \sqrt{2}$ , so the answer is $\boxed{141}$ .

See Also

2009 AIME II (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_II_Problems/Problem_3"

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.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us