Difference between revisions of "2013 AMC 8 Problems/Problem 25"
Lord.of.AMC (talk | contribs) (→Solution) |
Lord.of.AMC (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses <math>2\pi*2/2=2\pi</math> inches, and it gains <math>2\pi</math> inches on B. | The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses <math>2\pi*2/2=2\pi</math> inches, and it gains <math>2\pi</math> inches on B. | ||
− | + | <asy> | |
unitsize(0.04cm); | unitsize(0.04cm); | ||
import graph; | import graph; | ||
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label("4",(60,-60)); | label("4",(60,-60)); | ||
label("5",(87,0)); | label("5",(87,0)); | ||
− | + | </asy> | |
So, the departure from the length of the track means that the answer is <math>\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>. | So, the departure from the length of the track means that the answer is <math>\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>. | ||
Revision as of 13:10, 27 November 2013
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?
Solution
The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B. So, the departure from the length of the track means that 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.