Difference between revisions of "2013 AMC 8 Problems/Problem 25"
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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 shorter. Therefore, the only answer choice less than <math>240\pi</math> is <math>\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 shorter. Therefore, the only answer choice less than <math>240\pi</math> is <math>\boxed{\textbf{(A)}\ 238\pi}</math>. | ||
| − | This solution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump. | + | This solution may be invalid on other problems because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump. In this problem it works though. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=24|after=Last Problem}} | {{AMC8 box|year=2013|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:47, 14 November 2016
Contents
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 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.
![[asy] unitsize(0.04cm); import graph; draw(circle(96*dir(0),4),linewidth(1.3)); draw(circle(96*dir(-45),4),linetype("4 4")); draw(circle(96*dir(-90),4),linetype("4 4")); draw(circle(96*dir(-135),4),linetype("4 4")); draw(circle(96*dir(180),4),linetype("4 4")); draw((-100,0)..(0,-100)..(100,0)); draw((-96,0)..(0,-96)..(96,0),dotted); label("1",(-87,0)); label("2",(-60,-60)); label("3",(0,-87)); label("4",(60,-60)); label("5",(87,0)); [/asy]](http://latex.artofproblemsolving.com/c/9/7/c979ac6edc288bee6a165063d8720c6100a11a0f.png)
So, the departure from the length of the track means that the answer is 
.
Solution 2
The total length of all of the arcs is 
. Since we want the path from the center, the actual distance will be shorter. Therefore, the only answer choice less than 
is 
.
This solution may be invalid on other problems because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump. In this problem it works though.
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. 
