2021 AMC 12A Fall Problem 14

Question

In the figure, equilateral hexagon

ABCDEF

has three nonadjacent acute interior angles that each measure

30^\circ.

The enclosed area of the hexagon is

6\sqrt{3}.

What is the perimeter of the hexagon?

4

4\sqrt{3}

12

18

12\sqrt{3}

Let the side be

s.

The three

30^\circ

tips are isosceles triangles with legs

s;

they sit on the sides of the inner triangle formed by the three reflex vertices

Each tip has area

\tfrac12 s^2 \sin 30^\circ,

and the inner equilateral triangle has side

2s\sin 15^\circ

Let the common side length be

s.

The three acute vertices are the tips of isosceles triangles with two sides

s

and apex

30^\circ;

each has area

\tfrac12 s^2 \sin 30^\circ = \tfrac{s^2}{4}.

The three reflex vertices form an inner equilateral triangle with side

2s\sin 15^\circ,

whose area is

\sqrt3\,s^2\sin^2 15^\circ.

Using

\sin^2 15^\circ = \tfrac{2 - \sqrt3}{4},

the total area is

\frac{3s^2}{4} + \sqrt3\,s^2\cdot\frac{2 - \sqrt3}{4} = \frac{s^2\sqrt3}{2}.

Setting

\tfrac{s^2\sqrt3}{2} = 6\sqrt3

gives

s^2 = 12,

so

s = 2\sqrt3

and the perimeter is

6s = 12\sqrt3.

Thus, the correct answer is E.

Problem 14 in other years