1992 AMC 12 Problem 26

Question

Semicircle

\overset{\frown}{AB}

has center

C

and radius

1.

Point

D

is on

\overset{\frown}{AB}

and

CD\perp AB.

Extend

\overline{BD}

and

\overline{AD}

to

E

and

F,

respectively, so that circular arcs

\overset{\frown}{AE}

and

\overset{\frown}{BF}

have

B

and

A

as their respective centers. Circular arc

\overset{\frown}{EF}

has center

D.

The area of the shaded “smile,”

AEFBDA,

is

(2-\sqrt2)\pi

2\pi-\pi\sqrt2-1

\left(1-\frac{\sqrt2}{2}\right)\pi

\frac{5\pi}{2}-\pi\sqrt2-1

(3-2\sqrt2)\pi

Answer: B

Concepts:sector area semicircle isosceles right triangle composite region

Difficulty rating: 2260

Small hint:

Use

AC=BC=CD=1

to identify two isosceles right triangles and determine the relevant radii

Big hint:

Express the smile as one

90^\circ

sector plus two

45^\circ

sectors, then subtract triangle

ABD

and semicircle

ADB

Solution:

Triangles

ACD

and

BCD

are isosceles right triangles, so

AD=BD=\sqrt2,

\angle ADB=90^\circ,

and

DE=DF=2-\sqrt2.

The smile is the

90^\circ

sector

EDF,

plus the congruent

45^\circ

sectors

ABE

and

BAF,

minus triangle

ABD

and the original semicircle. Its area is

\begin{aligned} &\frac14\pi(2-\sqrt2)^2 +2\left(\frac18\pi\cdot2^2\right)\\ &\qquad-\frac12(2)(1)-\frac12\pi\\ &=2\pi-\pi\sqrt2-1. \end{aligned}

Thus the correct answer is B.