Difference between revisions of "2007 AMC 8 Problems/Problem 12"

 
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the area of the extensions to the area of the original hexagon?
 
the area of the extensions to the area of the original hexagon?
  
<center>[[Image:AMC8_2007_12.png]]</center>
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<asy>
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defaultpen(linewidth(0.7));
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draw(polygon(3));
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pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30);
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draw(D--E--F--cycle);
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</asy>
  
 
<math>\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5  \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1</math>
 
<math>\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5  \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1</math>
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==Solution==
 
==Solution==
 
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math>
 
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is <math>\boxed{\textbf{(A) }1:1}</math>
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==Solution 2==
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Split the hexagon into six small equilateral triangles. You will see that the six outer triangles can be folded to the hexagon, so the answer is <math>\boxed{\textbf{(A) }1:1}.</math>
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==Video Solution by OmegaLearn==
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https://youtu.be/abSgjn4Qs34?t=349
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~ pi_is_3.14
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=omFpSGMWhFc
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==Video Solution by AliceWang==
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https://youtu.be/tQXbFzxItPI
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==Video Solution by WhyMath==
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https://youtu.be/a7yMd17hP3o
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=11|num-a=13}}
 
{{AMC8 box|year=2007|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 13:14, 29 October 2024

Problem

A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

[asy] defaultpen(linewidth(0.7)); draw(polygon(3)); pair D=origin+1*dir(270), E=origin+1*dir(150), F=1*dir(30); draw(D--E--F--cycle); [/asy]

$\mathrm{(A)}\ 1:1 \qquad \mathrm{(B)}\ 6:5  \qquad \mathrm{(C)}\ 3:2 \qquad \mathrm{(D)}\ 2:1 \qquad \mathrm{(E)}\ 3:1$

Solution

The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is $\boxed{\textbf{(A) }1:1}$

Solution 2

Split the hexagon into six small equilateral triangles. You will see that the six outer triangles can be folded to the hexagon, so the answer is $\boxed{\textbf{(A) }1:1}.$

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=349

~ pi_is_3.14

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=omFpSGMWhFc

Video Solution by AliceWang

https://youtu.be/tQXbFzxItPI

Video Solution by WhyMath

https://youtu.be/a7yMd17hP3o

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 AJHSME/AMC 8 Problems and Solutions

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