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

m (Semi-automated contest formatting - script by azjps)
Line 1: Line 1:
== Problem 3 ==
+
SCRIPT PREVIEW BY AZJPS
 +
WILL AUTO-SUBMIT UNLESS YOU MOVE FROM WINDOW
 +
== 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>?
  
Line 20: Line 22:
 
label("$H$",(0,0),SW);
 
label("$H$",(0,0),SW);
 
</asy></center>
 
</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>
Line 30: Line 31:
 
Let <math>EF = FG = GF = HE = s</math>, let <math>AD = BC = h</math>, and let <math>AB = CD = x</math>.
 
Let <math>EF = FG = GF = HE = s</math>, let <math>AD = BC = h</math>, and let <math>AB = CD = x</math>.
  
 
+
<cmath>\begin{align*}&0.2 \cdot s^2 = hs\\
<math>0.2 \cdot s^2 = hs</math>
+
&s = 5h\\
 
+
&0.5 \cdot hx = hs\\
<math>s = 5h</math>
+
&x = 2s = 10h\\
 
+
&\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}</cmath>
<math>0.5 \cdot hx = hs</math>
 
 
 
<math>x = 2s = 10h</math>
 
 
 
<math>\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}</math>
 
  
 
=== Solution 2 ===
 
=== Solution 2 ===
Line 67: Line 63:
 
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>.
 +
 +
== See also ==
 +
{{AMC12 box|year=2010|num-b=2|num-a=4|ab=A}}
 +
 +
[[Category:Introductory Geometry Problems]]

Revision as of 21:45, 25 February 2010

SCRIPT PREVIEW BY AZJPS WILL AUTO-SUBMIT UNLESS YOU MOVE FROM WINDOW

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

Solution 1

Let $EF = FG = GF = HE = s$, let $AD = BC = h$, and let $AB = CD = x$.

\begin{align*}&0.2 \cdot s^2 = hs\\  &s = 5h\\  &0.5 \cdot hx = hs\\  &x = 2s = 10h\\  &\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}

Solution 2

The answer does not change 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