Difference between revisions of "2010 AMC 12A Problems/Problem 3"
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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}=\boxed{(E) 10}</math>. |
== See also == | == See also == | ||
Revision as of 17:04, 3 August 2020
Problem
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
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 
.
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. 
