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Difference between revisions of "2004 AMC 12A Problems/Problem 24"

(Solution with Asymptote)
 
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== Problem 24 ==
 
== Problem 24 ==
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A plane contains points <math>A</math> and <math>B</math> with <math>AB = 1</math>. Let <math>S</math> be the union of all disks of radius <math>1</math> in the plane that cover <math>\overline{AB}</math>. What is the area of <math>S</math>?
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A [[plane]] contains points <math>A</math> and <math>B</math> with <math>AB = 1</math>. Let <math>S</math> be the [[union]] of all disks of radius <math>1</math> in the plane that cover <math>\overline{AB}</math>. What is the area of <math>S</math>?
  
 
<math>\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3</math>
 
<math>\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3</math>
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==Solution==
 
==Solution==
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<asy>
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<center><asy>
 
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
 
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
  
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label("\(C\)",C,W);
 
label("\(C\)",C,W);
 
label("\(D\)",D,E);
 
label("\(D\)",D,E);
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</asy>
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</asy></center>
  
 
As the red circles move about segment <math>AB</math>, they cover the area we are looking for.
 
As the red circles move about segment <math>AB</math>, they cover the area we are looking for.
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This egg-like shape is <math>S</math>.
 
This egg-like shape is <math>S</math>.
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<asy>
+
<center><asy>
 
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
 
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
  
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label("\(1\)",B/2+3D/2,E);
 
label("\(1\)",B/2+3D/2,E);
 
label("\(1\)",B/2+3C/2,E);
 
label("\(1\)",B/2+3C/2,E);
−
</asy>
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</asy></center>
  
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Now, the hard part is over. Finding the area of this shape is not challenging.
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The area of the region can be found by dividing it into several sectors, namely
  
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<math>A = 2(Blue Sector) + 2(Red Sector) - 2(Equilateral Triangle)</math>
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<cmath>\begin{align*}
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A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\
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A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\
 +
A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\
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A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}</cmath>
  
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<math>A = 2(\frac{120^\circ}{360^\circ}\pi (2)^2) + 2(\frac{60^\circ}{360^\circ}\pi (1)^2) - 2(\frac{(1)^2\sqrt{3}}{4})</math>
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==See also==
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{{AMC12 box|year=2004|ab=A|num-b=23|num-a=25}}
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<math>A = \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2}</math>
 
−
 
 
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<math>A = 3\pi - \frac{\sqrt{3}}{2} \Rightarrow \text {(C)}</math>
 
  
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==See Also==
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[[Category:Intermediate Geometry Problems]]
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{{AMC12 box|year=2004|ab=A|num-b=23|num-a=25}}
 

Revision as of 15:34, 15 August 2008

Problem 24

A plane contains points $A$ and $B$ with $AB = 1$. Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$?

$\text {(A)} 2\pi + \sqrt3 \qquad \text {(B)} \frac {8\pi}{3} \qquad \text {(C)} 3\pi - \frac {\sqrt3}{2} \qquad \text {(D)} \frac {10\pi}{3} - \sqrt3 \qquad \text {(E)}4\pi - 2\sqrt3$


Solution

[asy] pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(circle(C,1),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("\(1\)",(0,0),N); label("\(1\)",A/2+D/2,W); label("\(1\)",A/2+C/2,W); label("\(1\)",B/2+D/2,E); label("\(1\)",B/2+C/2,E); label("\(1\)",A/2+3D/2,W); label("\(1\)",A/2+3C/2,W); label("\(1\)",B/2+3D/2,E); label("\(1\)",B/2+3C/2,E); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,E); [/asy]

As the red circles move about segment $AB$, they cover the area we are looking for. On the left side, the circle must move around pivoted on $B$. On the right side, the circle must move pivoted on $A$ However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.

This egg-like shape is $S$.

[asy] pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(arc(C,1,60,120),red); draw(arc(D,1,-120,-60),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,E); label("\(1\)",(0,0),N); label("\(1\)",A/2+D/2,W); label("\(1\)",A/2+C/2,W); label("\(1\)",B/2+D/2,E); label("\(1\)",B/2+C/2,E); label("\(1\)",A/2+3D/2,W); label("\(1\)",A/2+3C/2,W); label("\(1\)",B/2+3D/2,E); label("\(1\)",B/2+3C/2,E); [/asy]

The area of the region can be found by dividing it into several sectors, namely

\begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}

See also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AMC 12 Problems and Solutions
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