# Difference between revisions of "2010 AMC 12A Problems/Problem 3"

## Problem

Rectangle $ABCD$, pictured below, shares $50\%$ of its area with square $EFGH$. Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$. What is $\frac{AB}{AD}$?

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray); draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); draw((25,15)--(25,20)--(50,20)--(50,15)--cycle); label("A",(0,20),W); label("B",(50,20),E); label("C",(50,15),E); label("D",(0,15),W); label("E",(0,25),NW); label("F",(25,25),NE); label("G",(25,0),SE); label("H",(0,0),SW); [/asy]$

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

## Solution

If we shift $A$ to coincide with $E$, and add new horizontal lines to divide $EFGH$ into five equal parts:

$[asy] unitsize(1mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray); draw((25,20)--(25,25)--(50,25)--(50,20)--cycle); draw((0,5)--(25,5)); draw((0,10)--(25,10)); draw((0,15)--(25,15)); label("A=E",(0,25),W); label("B",(50,25),E); label("C",(50,20),E); label("D",(0,20),W); label("F",(25,25),NE); label("G",(25,0),SE); label("H",(0,0),SW); [/asy]$

This helps us to see that $AD=a/5$ and $AB=2a$, where $a=EF$. Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$.

## Video Solution 1 (Logic and Word Analysis)

~Education, the Study of Everything

 2010 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 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 All AMC 12 Problems and Solutions

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