2016 AMC 12A Problems/Problem 17: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
The Original Solution 2 is bs. Someone find an alternate solution for this problem. | |||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:19, 19 January 2019
Problem 17
Let 
be a square. Let 
and 
be the centers, respectively, of equilateral triangles with bases 
and 
each exterior to the square. What is the ratio of the area of square 
to the area of square ?
Solution
The center of an equilateral triangle is its centroid, where the three medians meet.
The distance along the median from the centroid to the base is one third the length of the median.
Let the side length of the square be 
. The height of 
is 
so the distance from 
to the midpoint of 
is 
(from above) 
(side length of the square).
Since 
is the diagonal of square , ![$\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$](http://latex.artofproblemsolving.com/c/b/3/cb3262100140d3d4d9c031ed59694f5fa0ceaeba.png)
Solution 2
The Original Solution 2 is bs. Someone find an alternate solution for this problem.
See Also
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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 12 Problems and Solutions | |
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