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

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== Solution ==
 
== Solution ==
  
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=== Solution 1 ===
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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:
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Let <math>EF = FG = GF = HE = s</math>, let <math>AD = BC = h</math>, and let <math>AB = CD = x</math>.
 
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<cmath>\begin{align*}&\begin{align*}&0.2 \cdot s^2 = hs\\
 
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&s = 5h\\
 
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&0.5 \cdot hx = hs\\
 
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&x = 2s = 10h\\
 
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&\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}\end{align*}</cmath>
 
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=== Solution 2 ===
 
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The answer does not change 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>
 
<center><asy>
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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>.
 
This helps us to see that <math>AD=a/5</math> and <math>AB=2a</math>, where <math>a=EF</math>.
 
Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10</math>.
 
Hence <math>\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10</math>.
 +
 +
==Video Solution 1 (Logic and Word Analysis)==
 +
https://youtu.be/BTDMUrT9oYo
 +
 +
~Education, the Study of Everything
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 20:48, 27 October 2022

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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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