Difference between revisions of "2013 AMC 8 Problems/Problem 25"
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<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> | <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 | + | ==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>. | 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>. | ||
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 
inches, 
inches, and 
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]](http://latex.artofproblemsolving.com/3/a/c/3ac0f4254d038847d9ee52d724bc0df18b1b1df4.png)
Solution 1
The total length of all of the arcs is 
. Since we want the path from the center, the actual distance will be subtracted by 
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 
Therefore, the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
