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Difference between revisions of "2006 AMC 10B Problems/Problem 19"

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m (<asy>s by dragon96)
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A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?
 
A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <math>AB</math> and <math>CB</math> are extended past <math>B</math> to meet the circle at <math>D</math> and <math>E</math>, respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting <math>D</math> and <math>E</math>?
  
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[[Image:2006amc10b19.gif]]
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<!-- [[Image:2006amc10b19.gif]] -->
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<asy>defaultpen(linewidth(0.8));
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pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;
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fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray);
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clip(B--Arc(O, 2, 30, 60)--cycle);
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draw(Circle(origin, 2));
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draw((-2,0)--(2,0)^^(0,-2)--(0,2));
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draw(A--D^^C--E);
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label("$A$", A, dir(point--A));
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label("$C$", C, dir(point--C));
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label("$O$", O, dir(point--O));
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label("$D$", D, dir(point--D));
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label("$E$", E, dir(point--E));
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label("$B$", B, SW);</asy>
 
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})
 
<math> \mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})
 
\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math>
 
\qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math>
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The shaded area is equivalent to the area of sector <math>DOE</math>, minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>.  
 
The shaded area is equivalent to the area of sector <math>DOE</math>, minus the area of triangle <math>DOE</math> plus the area of triangle <math>DBE</math>.  
  
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Using the Pythagorean Theorem:
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Using the [[Pythagorean Theorem]], <math>(DA)^2=(CE)^2=2^2-1^2=3</math> so <math>DA=CE=\sqrt{3}</math>.
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<math>(DA)^2=(CE)^2=2^2-1^2=3</math>
 
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<math>DA=CE=\sqrt{3}</math>
 
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Clearly, <math>DOA</math> and <math>EOC</math> are <math>30-60-90</math> triangles with <math>\angle EOC = \angle DOA = 60^\circ </math>.
 
  
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Since <math>OABC</math> is a square, <math> \angle COA = 90^\circ </math>.
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Clearly, <math>DOA</math> and <math>EOC</math> are <math>30-60-90</math> triangles with <math>\angle EOC = \angle DOA = 60^\circ </math>. Since <math>OABC</math> is a square, <math> \angle COA = 90^\circ </math>.
  
 
<math>\angle DOE</math> can be found by doing some subtraction of angles.  
 
<math>\angle DOE</math> can be found by doing some subtraction of angles.  
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The area of triangle <math>DOE</math> is <math> \frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1 </math>.
 
The area of triangle <math>DOE</math> is <math> \frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1 </math>.
  
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Since <math>AB=CB=1</math> , <math>DB=ED=(\sqrt{3}-1)</math>.  
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Since <math>AB=CB=1</math> , <math>DB=ED=(\sqrt{3}-1)</math>. So, the area of triangle <math>DBE</math> is <math>\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}</math>. Therefore, the shaded area is <math> (\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Longrightarrow \boxed{\mathrm{(A)}} </math>
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So, the area of triangle <math>DBE</math> is <math>\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}</math>.  
 
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Therefore, the shaded area is <math> (\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Rightarrow A </math>
 
  
 
== See Also ==
 
== See Also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}}
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*[[2006 AMC 10B Problems/Problem 18|Previous Problem]]
 
−
 
 
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*[[2006 AMC 10B Problems/Problem 20|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Problems]]
 +
[[Category:Circle Problems]]

Revision as of 23:28, 20 August 2011

Problem

A circle of radius $2$ is centered at $O$. Square $OABC$ has side length $1$. Sides $AB$ and $CB$ are extended past $B$ to meet the circle at $D$ and $E$, respectively. What is the area of the shaded region in the figure, which is bounded by $BD$, $BE$, and the minor arc connecting $D$ and $E$?

[asy]defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2)); draw(A--D^^C--E); label("$A$", A, dir(point--A)); label("$C$", C, dir(point--C)); label("$O$", O, dir(point--O)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$B$", B, SW);[/asy] $\mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3}) \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3}$

Solution

The shaded area is equivalent to the area of sector $DOE$, minus the area of triangle $DOE$ plus the area of triangle $DBE$.

Using the Pythagorean Theorem, $(DA)^2=(CE)^2=2^2-1^2=3$ so $DA=CE=\sqrt{3}$.

Clearly, $DOA$ and $EOC$ are $30-60-90$ triangles with $\angle EOC = \angle DOA = 60^\circ$. Since $OABC$ is a square, $\angle COA = 90^\circ$.

$\angle DOE$ can be found by doing some subtraction of angles.

$\angle COA - \angle DOA = \angle EOA$

$90^\circ - 60^\circ = \angle EOA = 30^\circ$

$\angle DOA - \angle EOA = \angle DOE$

$60^\circ - 30^\circ = \angle DOE = 30^\circ$

So, the area of sector $DOE$ is $\frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3}$.

The area of triangle $DOE$ is $\frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1$.

Since $AB=CB=1$ , $DB=ED=(\sqrt{3}-1)$. So, the area of triangle $DBE$ is $\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}$. Therefore, the shaded area is $(\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Longrightarrow \boxed{\mathrm{(A)}}$

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AMC 10 Problems and Solutions
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