Difference between revisions of "1992 AHSME Problems/Problem 26"

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== Solution ==
 
== Solution ==
<math>\fbox{B}</math>
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<math>\fbox{B}</math> The area of the entire outer shape is the area of sector <math>ABE</math>, plus the area of sector <math>ABF</math>, minus the area of triangle <math>ABD</math> (since it is part of both sectors), plus the area of sector <math>DEF</math>. We know <math>AC = CD = 1</math>, so the sector angles for <math>ABE</math> and <math>ABF</math> are <math>45</math> degrees, and the radius of both of them is <math>2</math>. The radius of <math>DEF</math> is <math>DE = BE - BD = 2 - BD</math>, and <math>BD</math> can be found using Pythagoras in triangle <math>BCD</math>, giving <math>BD = \sqrt{2}</math> and <math>DE = 2 - \sqrt{2}</math>, so after doing all the calculations, the area of the entire outer shape is <math>\pi(\frac{5}{2} - \sqrt{2}) - 1</math>. To get the area of the smile, we need to subtract the area of semicircle <math>ABD</math>, which is <math>\frac{1}{2} \pi 1^2 = \frac{\pi}{2}</math>, so the answer is <math>\pi(\frac{5}{2} - \frac{1}{2} - \sqrt{2}) - 1</math> = <math>2\pi - \pi \sqrt{2} - 1</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:29, 20 February 2018

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

$\fbox{B}$ The area of the entire outer shape is the area of sector $ABE$, plus the area of sector $ABF$, minus the area of triangle $ABD$ (since it is part of both sectors), plus the area of sector $DEF$. We know $AC = CD = 1$, so the sector angles for $ABE$ and $ABF$ are $45$ degrees, and the radius of both of them is $2$. The radius of $DEF$ is $DE = BE - BD = 2 - BD$, and $BD$ can be found using Pythagoras in triangle $BCD$, giving $BD = \sqrt{2}$ and $DE = 2 - \sqrt{2}$, so after doing all the calculations, the area of the entire outer shape is $\pi(\frac{5}{2} - \sqrt{2}) - 1$. To get the area of the smile, we need to subtract the area of semicircle $ABD$, which is $\frac{1}{2} \pi 1^2 = \frac{\pi}{2}$, so the answer is $\pi(\frac{5}{2} - \frac{1}{2} - \sqrt{2}) - 1$ = $2\pi - \pi \sqrt{2} - 1$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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