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

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(Solution 2)
Line 78: Line 78:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\
 
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 &= 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 &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\
 
A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}</cmath>
 
A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}</cmath>
 +
 +
==Alternate Solution==
 +
Credit to the Math Jam for the solution.
 +
(Diagram needed)
 +
 +
Let <math>\triangle ABX</math> be an equilateral triangle "above" segment <math>AB</math>. Now imagine having a "moving" circle of radius 1 pegged through point <math>B</math> and sliding it around as much as possible. The center of the circle must trace an arc of a circle <math>\omega_1</math> with radius 1 centered at <math>B</math>. Note that <math>\omega_1</math> passes through <math>X</math>. In addition,
 +
the point on the "moving" circle farthest from <math>B</math> must always be 2 units away from <math>B</math>, so it traces out an arc of a circle <math>\omega_2</math> with radius 2 centered at <math>B</math>. Now as we move around the circle the center will move around arc <math>AX</math> of <math>\omega_1</math>. Let <math>W</math> be the intersection of <math>AB</math> with <math>\omega_2</math>  such that <math>WA<WB</math>. Let <math>M</math> be the midpoint of <math>AB</math>, and <math>Z</math> be the intersection of <math>BX</math> with <math>\omega_2</math> such that <math>ZX<ZB</math>.
 +
Now the point on the "moving" circle farthest from <math>B</math> traces out arc <math>WZ</math> on <math>\omega_2</math>. Finally, one of the extreme positions of the moving circle is when the circle is centered at <math>X</math>. Let the circle centered at <math>X</math> with radius <math>1</math> be called <math>\omega_3</math>. Let <math>Y</math> be the intersection of <math>MX</math> and <math>\omega_3</math> such that <math>YX<YM</math>. Once the center of the moving circle has moved from <math>A</math> to <math>X</math>, the moving circle will have finished one-fourth of its total motion. Thus, <math>S</math> is made up of 4 quadrants (consult the diagram in solution 1). Focus on the upper left quadrant of <math>S</math>. The upper left quadrant is bounded by arc <math>WZ</math> of <math>\omega_2</math>, arc <math>ZY</math> of <math>\omega_3</math>, segments <math>YM</math> and <math>WM</math>. The area of this region is equal to
 +
the area of sector <math>WBZ</math> of <math>\omega_2</math> plus the area of sector <math>ZXY</math> of <math>\omega_3</math> minus the area of <math>\triangle BMX</math>. <math>\angle WBZ=60, \angle ZXY=\angle BXM=30</math>, and <math>\triangle BXM</math> is a 30-60-90 triangle. Routine calculations yield answer choice <math>\textbf {(C)}</math>.
  
 
==See also==  
 
==See also==  

Revision as of 22:34, 6 December 2015

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*}

Alternate Solution

Credit to the Math Jam for the solution. (Diagram needed)

Let $\triangle ABX$ be an equilateral triangle "above" segment $AB$. Now imagine having a "moving" circle of radius 1 pegged through point $B$ and sliding it around as much as possible. The center of the circle must trace an arc of a circle $\omega_1$ with radius 1 centered at $B$. Note that $\omega_1$ passes through $X$. In addition, the point on the "moving" circle farthest from $B$ must always be 2 units away from $B$, so it traces out an arc of a circle $\omega_2$ with radius 2 centered at $B$. Now as we move around the circle the center will move around arc $AX$ of $\omega_1$. Let $W$ be the intersection of $AB$ with $\omega_2$ such that $WA<WB$. Let $M$ be the midpoint of $AB$, and $Z$ be the intersection of $BX$ with $\omega_2$ such that $ZX<ZB$. Now the point on the "moving" circle farthest from $B$ traces out arc $WZ$ on $\omega_2$. Finally, one of the extreme positions of the moving circle is when the circle is centered at $X$. Let the circle centered at $X$ with radius $1$ be called $\omega_3$. Let $Y$ be the intersection of $MX$ and $\omega_3$ such that $YX<YM$. Once the center of the moving circle has moved from $A$ to $X$, the moving circle will have finished one-fourth of its total motion. Thus, $S$ is made up of 4 quadrants (consult the diagram in solution 1). Focus on the upper left quadrant of $S$. The upper left quadrant is bounded by arc $WZ$ of $\omega_2$, arc $ZY$ of $\omega_3$, segments $YM$ and $WM$. The area of this region is equal to the area of sector $WBZ$ of $\omega_2$ plus the area of sector $ZXY$ of $\omega_3$ minus the area of $\triangle BMX$. $\angle WBZ=60, \angle ZXY=\angle BXM=30$, and $\triangle BXM$ is a 30-60-90 triangle. Routine calculations yield answer choice $\textbf {(C)}$.

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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