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

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== Problem 3 ==
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== Problem ==
 
Rectangle <math>ABCD</math>, pictured below, shares <math>50\%</math> of its area with square <math>EFGH</math>. Square <math>EFGH</math> shares <math>20\%</math> of its area with rectangle <math>ABCD</math>. What is <math>\frac{AB}{AD}</math>?
 
Rectangle <math>ABCD</math>, pictured below, shares <math>50\%</math> of its area with square <math>EFGH</math>. Square <math>EFGH</math> shares <math>20\%</math> of its area with rectangle <math>ABCD</math>. What is <math>\frac{AB}{AD}</math>?
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 +
<center><asy>
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unitsize(1mm);
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defaultpen(linewidth(.8pt)+fontsize(8pt));
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draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
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fill((0,20)--(0,15)--(25,15)--(25,20)--cycle,gray);
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draw((0,15)--(0,20)--(25,20)--(25,15)--cycle);
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draw((25,15)--(25,20)--(50,20)--(50,15)--cycle);
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label("$A$",(0,20),W);
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label("$B$",(50,20),E);
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label("$C$",(50,15),E);
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label("$D$",(0,15),W);
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label("$E$",(0,25),NW);
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label("$F$",(25,25),NE);
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label("$G$",(25,0),SE);
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label("$H$",(0,0),SW);
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</asy></center>
  
 
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math>
 
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math>
  
[[2010 AMC 12A Problems/Problem 3|Solution]]
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== Solution ==
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If we shift <math>A</math> to coincide with <math>E</math>, and add new horizontal lines to divide <math>EFGH</math> into five equal parts:
 +
 
 +
<center><asy>
 +
unitsize(1mm);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
 +
 
 +
draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
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fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray);
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draw((25,20)--(25,25)--(50,25)--(50,20)--cycle);
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draw((0,5)--(25,5));
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draw((0,10)--(25,10));
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draw((0,15)--(25,15));
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label("$A=E$",(0,25),W);
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label("$B$",(50,25),E);
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label("$C$",(50,20),E);
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label("$D$",(0,20),W);
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label("$F$",(25,25),NE);
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label("$G$",(25,0),SE);
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label("$H$",(0,0),SW);
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</asy></center>
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This helps us to see that <math>AD=a/5</math> and <math>AB=2a</math>, where <math>a=EF</math>.
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Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10</math>.
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== See also ==
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{{AMC12 box|year=2010|num-b=2|num-a=4|ab=A}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 18:05, 3 August 2020

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 $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$.

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 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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