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

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

@@ Line 23: / Line 23: @@
 <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math>
-== Solution ==
+== Solution 1 ==
-=== 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:
-Let <math>EF = FG = GH = 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\\
- s &= 5h\\
-.5 \cdot hx &= hs\\
- x &= 2s = 10h\\
- \frac{AB}{AD} &= \frac{x}{h} = \boxed{10\ \textbf{(E)}}\end{align*}</cmath>
-=== Solution 2 ===
-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>
@@ Line 61: / Line 49: @@
 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>.
+== 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 ==

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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