Difference between revisions of "2016 AMC 12A Problems/Problem 17"
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Since <math>EG</math> is the diagonal of square <math>EFGH</math>, <math>\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}</math> | Since <math>EG</math> is the diagonal of square <math>EFGH</math>, <math>\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:13, 20 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 ,
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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