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.

1992 AMC 12 Problem 26

Answer: B

Difficulty rating: 2260

Small Hint:

Big Hint:

Solution:

Problem 26 in Other Years