2013 AMC 8 Problems/Problem 25

Problem 25

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 $2\pi$ because it's already half the circumference through semicircle A, needs to go half the circumference extra through semicircle B, and it's already half the circumference through semicircle C, 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)
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