Difference between revisions of "2010 AMC 12A Problems/Problem 3"
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− | == Problem | + | == 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>? | ||
<center><asy> | <center><asy> | ||
unitsize(1mm); | unitsize(1mm); | ||
− | defaultpen(linewidth(.8pt)); | + | defaultpen(linewidth(.8pt)+fontsize(8pt)); |
draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); | draw((0,0)--(0,25)--(25,25)--(25,0)--cycle); | ||
Line 10: | Line 10: | ||
draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); | draw((0,15)--(0,20)--(25,20)--(25,15)--cycle); | ||
draw((25,15)--(25,20)--(50,20)--(50,15)--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></center> | </asy></center> | ||
Line 15: | Line 24: | ||
== Solution == | == Solution == | ||
− | |||
+ | 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> | ||
− | <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>. | ||
− | + | ==Video Solution 1 (Logic and Word Analysis)== | |
+ | https://youtu.be/BTDMUrT9oYo | ||
− | + | ~Education, the Study of Everything | |
+ | == See also == | ||
+ | {{AMC12 box|year=2010|num-b=2|num-a=4|ab=A}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
Revision as of 20:48, 27 October 2022
Problem
Rectangle , pictured below, shares of its area with square . Square shares of its area with rectangle . What is ?
Solution
If we shift to coincide with , and add new horizontal lines to divide into five equal parts:
This helps us to see that and , where . Hence .
Video Solution 1 (Logic and Word Analysis)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.