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2016 AMC 12A Problems/Problem 17

Revision as of 15:54, 27 November 2023 by Aleyang (talk | contribs) (→‎Solution(Coordinates))

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 (Coordinates)

Note that we can also use coordinates to solve this problem. WLOG, set the side length of square $ABCD$ equal to $6$.

This makes the coordinates of the square $EFGH$ equal to $(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),$ and $(3, -\sqrt{3})$.

Using the first two points, this gives $EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12 \sqrt{3}$.

Thus, $[EFGH]=24+12\sqrt{3}$.

Because the side length of $ABCD$ is $6$, $[ABCD] = 36$.

Therefore, $\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$

- Pleaseletmewin

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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