Difference between revisions of "2014 AMC 10A Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
I have 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. 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 <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> | I have 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. 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 <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> | ||
+ | |||
+ | 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> | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:59, 6 February 2017
Contents
Problem
A rectangular piece of paper whose length is times the width has area . 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 . What is the ratio ?
Solution 1
I have 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. 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 and the length is . The triangle we want to find has side lengths , , and . It is an equilateral triangle with height , and area . The area of the paper is , and the folded paper has area . The ratio of the area of the folded paper to that of the original paper is thus
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>
Solution 2
Our original paper can be divided like this: After the fold across the dotted line, our paper becomes:
Since our original sheet of paper has six congruent triangles and and our new one has four, the ratio of the area is equal to
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.