1999 AMC 12 Problem 23

Question

The equiangular convex hexagon

ABCDEF

has

AB = 1,

BC = 4,

CD = 2,

and

DE = 4.

The area of the hexagon is

\dfrac{15}{2}\sqrt{3}

9\sqrt{3}

16

\dfrac{39}{4}\sqrt{3}

\dfrac{43}{4}\sqrt{3}

Extend alternate sides to enclose the hexagon in a large equilateral triangle

Subtract three small equilateral corner triangles from the large one

Each interior angle is

120^\circ,

so extending sides

FA

and

BC,

BC

and

DE,

and

DE

and

FA

cuts off three equilateral corner triangles and forms a large equilateral triangle.

Let

EF=e

and

FA=f.

Resolving the six sides in directions separated by

60^\circ

gives

f-e=4

and

e+f=6,

so

e=1

and

f=5.

The corner triangles built on

AB, CD,

and

EF

are therefore equilateral. The large triangle has side

1+4+2=7,

while the removed triangles have sides

1,2,

and

1.

The area is

\begin{aligned} &\frac{\sqrt3}{4}\left(7^2 - 1^2 - 2^2 - 1^2\right) \\ &\quad = \frac{43\sqrt3}{4}. \end{aligned}

Thus, the correct answer is E.

Problem 23 in other years