Difference between revisions of "2010 AMC 12A Problems/Problem 3"
(→Problem 3) |
(→Solution) |
||
(20 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
− | == 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> | ||
+ | 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></center> | ||
<math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | <math>\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | ||
− | [[2010 | + | == 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: | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | == 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 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}} |
Latest revision as of 17:12, 1 April 2024
Contents
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 1
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
.
Solution 2
From the problem statement, we know that
If we let and
, we see
. 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.