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? | ||
| − | < | + | <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> | ||
<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> | ||
| Line 10: | Line 15: | ||
==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> | ||
| + | |||
| + | ==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 <math>\boxed{\textbf{(A) }1:1}.</math> | ||
| + | |||
| + | ==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== | ==Video Solution by WhyMath== | ||
| − | https://youtu.be/ | + | 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:14, 29 October 2024
Contents
Problem
A unit hexagram is composed of a regular hexagon of side length 
and its 
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]](http://latex.artofproblemsolving.com/b/7/5/b756761ed7ffb05e7be7328280ba353674aecaf6.png)
Solution
The six equilateral triangular extensions fit perfectly into the hexagon meaning the answer is 
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 
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
Video Solution by WhyMath
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
