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Difference between revisions of "2013 AMC 8 Problems/Problem 25"

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==Problem==
 
==Problem==
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A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?
  
−
==Solution==
+
<asy>
 +
pair A,B;
 +
size(8cm);
 +
A=(0,0);
 +
B=(480,0);
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draw((0,0)--(480,0),linetype("3 4"));
 +
filldraw(circle((8,0),8),black);
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draw((0,0)..(100,-100)..(200,0));
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draw((200,0)..(260,60)..(320,0));
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draw((320,0)..(400,-80)..(480,0));
 +
draw((100,0)--(150,-50sqrt(3)),Arrow(size=4));
 +
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));
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draw((400,0)--(440,-40sqrt(3)),Arrow(size=4));
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label("$A$", A, SW);
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label("$B$", B, SE);
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label("$R_1$", (100,-40), W);
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label("$R_2$", (260,40), SW);
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label("$R_3$", (400,-40), W);</asy>
 +
 
 +
<math>\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi</math>
 +
 
 +
==Solution 1 ==
 +
 
 +
The total length of all of the arcs is <math>100\pi +80\pi +60\pi=240\pi</math>. Since we want the path from the center, the actual distance will be subtracted by <math>2\pi</math> because it's already half the circumference through semicircle A, which needs to go half the circumference extra through semicircle B, and it's already half the circumference through semicircle C, and the circumference is <math>4\pi</math> Therefore, the answer is <math>240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>.
 +
 
 +
~[[User:PowerQualimit|PowerQualimit]]
  
 
==See Also==
 
==See Also==
−
{{AMC8 box|year=2013|before=First Problem|num-a=2}}
+
{{AMC8 box|year=2013|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:01, 30 August 2024

Problem

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

[asy] pair A,B; size(8cm); A=(0,0); B=(480,0); draw((0,0)--(480,0),linetype("3 4")); filldraw(circle((8,0),8),black); draw((0,0)..(100,-100)..(200,0)); draw((200,0)..(260,60)..(320,0)); draw((320,0)..(400,-80)..(480,0)); draw((100,0)--(150,-50sqrt(3)),Arrow(size=4)); draw((260,0)--(290,30sqrt(3)),Arrow(size=4)); draw((400,0)--(440,-40sqrt(3)),Arrow(size=4)); label("$A$", A, SW); label("$B$", B, SE); label("$R_1$", (100,-40), W); label("$R_2$", (260,40), SW); label("$R_3$", (400,-40), W);[/asy]

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

Solution 1

The total length of all of the arcs is $100\pi +80\pi +60\pi=240\pi$. Since we want the path from the center, the actual distance will be subtracted by $2\pi$ because it's already half the circumference through semicircle A, which needs to go half the circumference extra through semicircle B, and it's already half the circumference through semicircle C, and the circumference is $4\pi$ Therefore, the answer is $240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$.

~PowerQualimit

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 AJHSME/AMC 8 Problems and Solutions

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