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

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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>\overline{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>?
  
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<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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<math>\textbf {(A) } 2\pi + \sqrt3 \qquad \textbf {(B) } \frac {8\pi}{3} \qquad \textbf {(C) } 3\pi - \frac {\sqrt3}{2} \qquad \textbf {(D) } \frac {10\pi}{3} - \sqrt3 \qquad \textbf {(E) }4\pi - 2\sqrt3</math>
  
 
==Solution==
 
==Solution==

Latest revision as of 21:25, 22 November 2021

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 $S$?

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

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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