Difference between revisions of "2014 AMC 10A Problems/Problem 23"
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<math> \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 </math> | <math> \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 </math> | ||
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| + | ==Solution== | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2014|ab=A|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Revision as of 22:20, 6 February 2014
Problem
A rectangular piece of paper whose length is 
times the width has area 
. The paper is divided into three sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area 
. What is the ratio 
?
![[asy] import graph; size(6cm); real L = 0.05; pair A = (0,0); pair B = (sqrt(3),0); pair C = (sqrt(3),1); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2= (2*sqrt(3)/3,0); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); dot(X1); dot(Y1); draw(A--B--C--D--cycle, linewidth(2)); draw(X1--Y1,dashed); draw(X2--(2*sqrt(3)/3,L)); draw(Y2--(sqrt(3)/3,1-L)); [/asy]](http://latex.artofproblemsolving.com/3/8/3/383b46c75e8be868196ec01badfad01411d16223.png)
Solution
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
