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

Revision as of 21:50, 25 February 2010

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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*}&\begin{align*}&\begin{align*}&\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*}\end{align*}\end{align*}\end{align*} (Error compiling LaTeX. Unknown error_msg)

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

@@ Line 1: / Line 1: @@
+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>?
@@ Line 29: / Line 31: @@
 Let <math>EF = FG = GF = HE = s</math>, let <math>AD = BC = h</math>, and let <math>AB = CD = x</math>.
-<cmath>\begin{align*}&\begin{align*}&0.2 \cdot s^2 = hs\\
+<cmath>\begin{align*}&\begin{align*}&\begin{align*}&\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*}\end{align*}</cmath>
+  &\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}\end{align*}\end{align*}\end{align*}</cmath>
 === Solution 2 ===

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