Difference between revisions of "2007 AMC 8 Problems/Problem 12"
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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== | ||
| + | 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> | ||
==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}} | ||
Revision as of 01:58, 2 September 2021
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 
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. 
