Difference between revisions of "2010 AMC 12A Problems/Problem 3"
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== Solution == | == Solution == | ||
+ | |||
+ | === Solution 1 === | ||
+ | |||
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>. | ||
− | <math>. | + | <math>0.2 \cdot s^2 = hs</math> |
<math>s = 5h</math> | <math>s = 5h</math> | ||
− | <math>. | + | <math>0.5 \cdot hx = hs</math> |
<math>x = 2s = 10h</math> | <math>x = 2s = 10h</math> | ||
+ | <math>\frac{AB}{AD} = \frac{x}{h} = \boxed{10\ \textbf{(E)}}</math> | ||
− | <math>\ | + | === 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> | ||
+ | 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></center> | ||
+ | |||
+ | 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>. |
Revision as of 10:34, 16 February 2010
Problem 3
Rectangle , pictured below, shares
of its area with square
. Square
shares
of its area with rectangle
. What is
?
![[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]](http://latex.artofproblemsolving.com/5/4/3/54371c10b04c0bcf5a86491a02c9b7ecf851cb2d.png)
Solution
Solution 1
Let , let
, and let
.
Solution 2
The answer does not change if we shift to coincide with
, and add new horizontal lines to divide
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]](http://latex.artofproblemsolving.com/8/e/9/8e9999515009eee42667e0689b77caa28440a9c8.png)
This helps us to see that and
, where
.
Hence
.