2016 AMC 12A Problem 17

Question

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?

1

\dfrac{2+\sqrt{3}}{3}

\sqrt{2}

\dfrac{\sqrt{2}+\sqrt{3}}{2}

\sqrt{3}

The center of an equilateral triangle lies

\frac13

of its height from the base

Compare diagonals of the two squares; the area ratio is the square of the length ratio

Let square

ABCD

have side length

6.

Each equilateral triangle has height

3\sqrt3,

and its center lies

\frac13

of that height, namely

\sqrt3,

from the square’s side.

Square

ABCD

has diagonal

6\sqrt2.

Square

EFGH

has diagonal equal to the side of

ABCD

plus twice

\sqrt3,

namely

6+2\sqrt3.

The area ratio is the square of the ratio of diagonals:

\begin{gathered} \left(\dfrac{6+2\sqrt3}{6\sqrt2}\right)^2\\ =\left(\dfrac{3+\sqrt3}{3\sqrt2}\right)^2\\ =\dfrac{12+6\sqrt3}{18}\\ =\dfrac{2+\sqrt3}{3}. \end{gathered}

Thus, the correct answer is B.

Problem 17 in other years