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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==
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{{AMC12 box|year=2016|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Revision as of 11:44, 5 February 2016

Problem 17

Let $ABCD$ be a square. Let $E, F, G$ and $H$ be the centers, respectively, of equilateral triangles with bases $\overline{AB}, \overline{BC}, \overline{CD},$ and $\overline{DA},$ each exterior to the square. What is the ratio of the area of square $EFGH$ to the area of square $ABCD$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}$

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 $1$. The height of $\triangle E$ is $\frac{\sqrt{3}}{2},$ so the distance from $E$ to the midpoint of $\overline{AB}$ is $\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}$

$EG = 2 \cdot \frac{\sqrt{3}}{6}$ (from above) $+ 1$ (side length of the square).

Since $EG$ is the diagonal of square $EFGH$, $\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$


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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