2016 AMC 12A Problems/Problem 17

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 $EFGH$ , $\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$

Solution 2

The Original Solution 2 is bs. Someone find an alternate solution for this problem.

2016 AMC 12A (Problems • Answer Key • Resources)
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2016 AMC 12A Problems/Problem 17

Contents

Problem 17

Solution

Solution 2

See Also