2006 AIME II Problem 1

Question

In convex hexagon

ABCDEF,

all six sides are congruent,

\angle A

and

\angle D

are right angles, and

\angle B,

\angle C,

\angle E,

and

\angle F

are congruent. The area of the hexagonal region is

2116(\sqrt{2} + 1).

Find

AB.

Answer

Each of angles

B,

C,

E,

F

measures

135^\circ.

Draw diagonals

BF

and

CE.

These diagonals cut the hexagon into two right isosceles triangles and a rectangle with side lengths

AB

and

AB\sqrt{2}.

The angles of a hexagon sum to

720^\circ,

so each of the four congruent angles measures

\frac{720 - 2 \cdot 90}{4} = 135

degrees. Let

AB = x.

The diagonals

BF

and

CE

cut off the right isosceles triangles

FAB

and

CDE,

each with legs

x

and hypotenuse

x\sqrt{2},

and the

135^\circ

angles guarantee that the remaining piece

BCEF

is a rectangle with sides

x\sqrt{2}

and

x.

Hence the area is

\begin{aligned} &2 \cdot \frac{1}{2}x^2 + x \cdot x\sqrt{2} \\ &= x^2(1 + \sqrt{2}) \\ &= 2116(\sqrt{2} + 1), \end{aligned}

so

x^2 = 2116

and

AB = x = 46.

Problem 1 in other years