AoPSWiki

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

Personal tools

Search

Toolbox

2007 Cyprus MO/Lyceum/Problem 14

From AoPSWiki

Problem

In square $ABCD$ the segment $KB$ equals a side of the square. The ratio of areas $\frac{S_1}{S_2}$ is

$\mathrm{(A) \ } \frac{1}{3}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{1}{\sqrt{2}}\qquad \mathrm{(D) \ } ...$

Solution

$\triangle KBC$ and $\triangle KCD$ have the same heights ( $\perp BD$ ), so the ratio of their areas is simply the ratio of $KD:KB$ .

Let the length of $KB$ be $x$ . Then $KD=x\sqrt{2}-x$ , and the ratio of $S_1:S_2$ is $x\sqrt{2}-x:x$ , or $\sqrt{2}-1:1\Longrightarrow\mathrm{ D}$ .

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_Cyprus_MO/Lyceum/Problem_14"

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!

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