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Difference between revisions of "2014 AMC 10A Problems/Problem 23"

m (Some readability issues.)
m (Changed multiplication symbols. I blame OCD and boredom, even though no one will ever read this ever.)
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==Solution==
 
==Solution==
 
(not too sure about the correctness of my method, but it got the right answer.)
 
(not too sure about the correctness of my method, but it got the right answer.)
−
No clue how to draw pictures on here, so I'll give instructions.  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.  This triangle is the double-layered portion of the folded paper.  WLOG, assume the width of the paper is <math>1</math> and the length is <math>\sqrt{3}</math>.  The triangle we want to find has side lengths <math>\dfrac{2\sqrt{3}}{3}</math>, <math>\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}</math>, and <math>\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}</math>.  It is an equilateral triangle with height <math>\dfrac{\sqrt{3}}{3}*\sqrt{3}=1</math>, and area <math>\dfrac{\dfrac{2\sqrt{3}}{3}*1}{2}=\dfrac{\sqrt{3}}{3}</math>.  The area of the paper is <math>1*\sqrt{3}=\sqrt{3}</math>, and the folded paper has area <math>\sqrt{3}-\dfrac{\sqrt{3}}{3}=\dfrac{2\sqrt{3}}{3}</math>.  The ratio of the area of the folded paper to that of the original paper is thus <math>\boxed{\textbf{(C)} \: 2:3}</math>
+
No clue how to draw pictures on here, so I'll give instructions.  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.  This triangle is the double-layered portion of the folded paper.  WLOG, assume the width of the paper is <math>1</math> and the length is <math>\sqrt{3}</math>.  The triangle we want to find has side lengths <math>\dfrac{2\sqrt{3}}{3}</math>, <math>\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}</math>, and <math>\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}</math>.  It is an equilateral triangle with height <math>\dfrac{\sqrt{3}}{3}\cdot\sqrt{3}=1</math>, and area <math>\dfrac{\dfrac{2\sqrt{3}}{3}\cdot1}{2}=\dfrac{\sqrt{3}}{3}</math>.  The area of the paper is <math>1\cdot\sqrt{3}=\sqrt{3}</math>, and the folded paper has area <math>\sqrt{3}-\dfrac{\sqrt{3}}{3}=\dfrac{2\sqrt{3}}{3}</math>.  The ratio of the area of the folded paper to that of the original paper is thus <math>\boxed{\textbf{(C)} \: 2:3}</math>
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:22, 8 February 2014

Problem

A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. 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 $B$. What is the ratio $B:A$?

[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]

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

Solution

(not too sure about the correctness of my method, but it got the right answer.) No clue how to draw pictures on here, so I'll give instructions. 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. This triangle 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 $\boxed{\textbf{(C)} \: 2:3}$

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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