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Difference between revisions of "2014 AMC 10A Problems/Problem 23"
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==Problem==
 
==Problem==
  
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A rectangular piece of paper whose length is <math>\sqrt3</math> times the width has area <math>A</math>. The paper is divided into three equal 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 <math>B</math>. What is the ratio <math>B:A</math>?
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A rectangular piece of paper whose length is <math>\sqrt3</math> times the width has area <math>A</math>. The paper is divided into three equal 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 <math>B</math>. What is the ratio <math>\frac{B}{A}</math>?
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
  
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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>
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<math> \textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{5}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{3}{4}\qquad\textbf{(E)}\ \frac{4}{5} </math>
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[[Category: Introductory Geometry Problems]]
 
  
 
==Solution 1==
 
==Solution 1==
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Since our original sheet of paper has six congruent <math>30-60-90</math> triangles and and our new one has four, the ratio of the area <math>B:A</math> is equal to <math>4:6\implies \boxed{\textbf{(C)} \: 2:3}</math>
 
Since our original sheet of paper has six congruent <math>30-60-90</math> triangles and and our new one has four, the ratio of the area <math>B:A</math> is equal to <math>4:6\implies \boxed{\textbf{(C)} \: 2:3}</math>
  
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==Solution 3==
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== Video Solution by Pi Academy ==
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Note that the bottom right corner is folded onto the top left corner. It follows that the overlap is an equilateral triangle with area one-third of the original rectangle, and the answer is <math>1-\frac13:1\implies \boxed{\textbf{(C)} \: 2:3}.</math>
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https://youtu.be/ql_90z1m8Qs?si=D8nF9MULxipqoxmc ~ Pi Academy
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== Video Solution by Richard Rusczyk ==
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 +
https://artofproblemsolving.com/videos/amc/2014amc10a/377
  
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~dolphin7
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~ ripkobe_745
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https://youtu.be/jYRA3thX4pI
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 +
~suprboygamer_jimpop(direct youtube link)
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2014|ab=A|num-b=22|num-a=24}}
 
{{AMC10 box|year=2014|ab=A|num-b=22|num-a=24}}
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[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:44, 7 October 2024

Problem

A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into three equal 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 $B$. What is the ratio $\frac{B}{A}$?

[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]

$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{5}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{3}{4}\qquad\textbf{(E)}\ \frac{4}{5}$

Solution 1

Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the paper is $1$ and the length is $\sqrt{3}$. The triangle we want to find has side lengths $\dfrac{2\sqrt{3}}{3}$, $\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}$, and $\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}$. It is an equilateral triangle with height $\dfrac{\sqrt{3}}{3}\cdot\sqrt{3}=1$, and area $\dfrac{\dfrac{2\sqrt{3}}{3}\cdot1}{2}=\dfrac{\sqrt{3}}{3}$. The area of the paper is $1\cdot\sqrt{3}=\sqrt{3}$, and the folded paper has area $\sqrt{3}-\dfrac{\sqrt{3}}{3}=\dfrac{2\sqrt{3}}{3}$. The ratio of the area of the folded paper to that of the original paper is thus $\boxed{\textbf{(C)} \: 2:3}$

[asy]import graph; unitsize(3cm); 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(B--D,dashed); draw(X1--Y1,dashed); draw(Y2--X1--D, dotted); draw(X2--Y1--B, dotted);[/asy]

Solution 2

Our original paper can be divided like this: [asy] import graph; unitsize(3cm); 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(Y2--X1--D, dotted); draw(X2--Y1--B, dotted);[/asy] After the fold across the dashed line, our paper becomes:

[asy] import graph; unitsize(3cm); real L = 0.05; pair A = (0,0); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2 = (sqrt(3)/6,0.5); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); pair Z1 = (sqrt(3)/2,1.5); dot(X1); dot(Y1); draw(X1--A--D--Z1--Y1, linewidth(2)); draw(X1--D--Y1); draw(X1--Y1, dashed); draw(Y2--X1,dotted); draw(X2--((sqrt(3)/6 + L/sqrt(3)),(0.5+L/2))); draw(Y2--(sqrt(3)/3,1-L));[/asy] Since our original sheet of paper has six congruent $30-60-90$ triangles and and our new one has four, the ratio of the area $B:A$ is equal to $4:6\implies \boxed{\textbf{(C)} \: 2:3}$

Video Solution by Pi Academy

https://youtu.be/ql_90z1m8Qs?si=D8nF9MULxipqoxmc ~ Pi Academy

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc10a/377

~ ripkobe_745

https://youtu.be/jYRA3thX4pI

~suprboygamer_jimpop(direct youtube link)

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

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