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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>?
 +
 +
<center><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);
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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);
 +
 +
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);
 +
</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>
  
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[[2010 AMC 12A Problems/Problem 3|Solution]]
+
== Solution 1 ==
 +
 
 +
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);
 +
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));
 +
draw((0,10)--(25,10));
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draw((0,15)--(25,15));
 +
 
 +
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);
 +
label("$F$",(25,25),NE);
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label("$G$",(25,0),SE);
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label("$H$",(0,0),SW);
 +
</asy></center>
 +
 
 +
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>.
 +
 
 +
== Solution 2 ==
 +
From the problem statement, we know that
 +
<cmath>\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}</cmath>
 +
 
 +
If we let <math>a = EF</math> and <math>b  = AD</math>, we see
 +
<cmath>[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}</cmath>. Hence, <math>\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{\bold{10}}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/TLtqy-62TKo?si=-UUsxm_I0svPWCnk
 +
 
 +
~Charles3829
 +
 
 +
 
 +
==Video Solution 2 (Logic and Word Analysis)==
 +
https://youtu.be/BTDMUrT9oYo
 +
 
 +
~Education, the Study of Everything
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2010|num-b=2|num-a=4|ab=A}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 09:55, 18 September 2024

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 1

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

Solution 2

From the problem statement, we know that \[\frac{[ABCD]}{2} = \frac{[EFGH]}{5} \Rightarrow [ABCD]=\frac{2[EFGH]}{5}\]

If we let $a = EF$ and $b = AD$, we see \[[ABCD] = 2ab = \frac{2a^2}{5} \Rightarrow b = \frac{a}{5}\]. Hence, $\frac{AB}{AD} = \frac{2a}{b} = 2a(\frac{5}{a}) = \boxed{\bold{10}}$

Video Solution

https://youtu.be/TLtqy-62TKo?si=-UUsxm_I0svPWCnk

~Charles3829


Video Solution 2 (Logic and Word Analysis)

https://youtu.be/BTDMUrT9oYo

~Education, the Study of Everything

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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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