2003 AMC 12B Problem 15

Question

A regular octagon

ABCDEFGH

has an area of one square unit. What is the area of the rectangle

ABEF?

1 - \dfrac{\sqrt{2}}{2}

\dfrac{\sqrt{2}}{4}

\sqrt{2} - 1

\dfrac{1}{2}

\dfrac{1 + \sqrt{2}}{4}

Let

O

be the center of the octagon, which is the midpoint of diagonal

AE

\triangle AOB

is one of the eight triangles from the center, so it has area

\dfrac{1}{8}

Let

O

be the center of the octagon. Joining

O

to the vertices splits the octagon into

8

congruent triangles, so

\triangle AOB

has area

\dfrac{1}{8}.

Since

O

is the midpoint of

\overline{AE},

triangles

AOB

and

BOE

have equal areas, so

\triangle ABE

has area

\dfrac{1}{4}.

The rectangle

ABEF

is split by diagonal

\overline{BE}

into two congruent triangles, so

\triangle ABE

is half of it. Hence

ABEF

has area

\dfrac{1}{2}.

Thus, the correct answer is D.

Problem 15 in other years