2013 AMC 8 Problems/Problem 25

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$", (400,-40), W);[/asy]$

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

Video Solution for Problems 21-25

https://youtu.be/-mi3qziCuec

Video Solution

https://youtu.be/zZGuBFyiQrk ~savannahsolver

Solution 1

Since the diameter of the ball is 4 inches, $\text{radius}=2$ .

If we think about the ball rolling or draw a path for the ball (see figure below), we see that in semicircle A and semicircle C it loses $2\pi$ inches each, because $\dfrac{1}{2} 2\pi (x-2) - \dfrac{1}{2} 2\pi (x)= -2 \pi$

By similar reasoning, it gains $2\pi$ inches on semicircle 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 $\dfrac{1}{2}2\pi (100+60+80) +(-2+2-2)\cdot\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$ .

Solution 2

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 an arc of $2\pi$ because it's already half the circumference through the track, and the circumference is $4\pi$ Therefore, the answer is $240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$ .

~PowerQualimit

Solution 3

Similar to Solution 1, we notice that the center of the ball follows a different semi-circle to the actual track. For the first section, the radius of the semi-circle that the ball's center follows is, $r = 100 - 2 = 98$ , and the arc is $r \pi = 98\pi$ . For the second section, the radius of the semi-circle that the ball's center follows is $r = 60 + 2 = 62$ , and the arc is $62\pi$ . For the third section, the radius of the semi-circle that the ball's center follows is $r = 80 - 2 = 78$ , and the arc is $78\pi$ .

Hence, the total length is $98\pi + 62\pi + 78\pi = 238\pi \Longrightarrow \boxed{\textbf{(A)}\ 238\pi}$

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

Page

Toolbox

Search