Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 17: Line 17:
 
==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]
+
<asy>
 
unitsize(0.04cm);
 
unitsize(0.04cm);
 
import graph;
 
import graph;
Line 32: Line 32:
 
label("4",(60,-60));
 
label("4",(60,-60));
 
label("5",(87,0));
 
label("5",(87,0));
−
[/asy]
+
</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 12:10, 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 $R_1 = 100$ 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] size(8cm); 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)); [/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

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 $2\pi*2/2=2\pi$ inches, and it gains $2\pi$ 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] So, the departure from the length of the track means that the answer is $\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_25&oldid=58007"