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== See also ==
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{{AHSME box|year=1993|num-b=1|num-a=2}}   
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[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
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Revision as of 21:33, 27 September 2014

Problem

[asy] fill((1,0)--arc((1,0),2,180,225)--cycle,grey); fill((-1,0)--arc((-1,0),2,315,360)--cycle,grey); fill((0,-1)--arc((0,-1),2-sqrt(2),225,315)--cycle,grey); fill((0,0)--arc((0,0),1,180,360)--cycle,white); draw((1,0)--arc((1,0),2,180,225)--(1,0),black+linewidth(1)); draw((-1,0)--arc((-1,0),2,315,360)--(-1,0),black+linewidth(1)); draw((0,0)--arc((0,0),1,180,360)--(0,0),black+linewidth(1)); draw(arc((0,-1),2-sqrt(2),225,315),black+linewidth(1)); draw((0,0)--(0,-1),black+linewidth(1)); MP("C",(0,0),N);MP("A",(-1,0),N);MP("B",(1,0),N); MP("D",(0,-.8),NW);MP("E",(1-sqrt(2),-sqrt(2)),SW);MP("F",(-1+sqrt(2),-sqrt(2)),SE); [/asy]

Semicircle $\widehat{AB}$ has center $C$ and radius $1$. Point $D$ is on $\widehat{AB}$ and $\overline{CD}\perp\overline{AB}$. Extend $\overline{BD}$ and $\overline{AD}$ to $E$ and $F$, respectively, so that circular arcs $\widehat{AE}$ and $\widehat{BF}$ have $B$ and $A$ as their respective centers. Circular arc $\widehat{EF}$ has center $D$. The area of the shaded "smile" $AEFBDA$, is

$\text{(A) } (2-\sqrt{2})\pi\quad \text{(B) } 2\pi-\pi \sqrt{2}-1\quad \text{(C) } (1-\frac{\sqrt{2}}{2})\pi\quad\\ \text{(D) } \frac{5\pi}{2}-\pi\sqrt{2}-1\quad \text{(E) } (3-2\sqrt{2})\pi$

Solution

$\fbox{B}$

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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