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Difference between revisions of "2013 AMC 8 Problems/Problem 25"
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==Solution 2 (answer choices)==
 
==Solution 2 (answer choices)==
  
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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>.
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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 subtracted by an arc of <math>2\pi</math> because we are taking half the circumference, and the circumference is <math>4\pi</math> Therefore, the answer is <math>240\pi-2\pi</math>=\boxed{\textbf{(A)}\ 238\pi}$.
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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.
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~[[User:PowerQualimit|PowerQualimit]]
  
 
==Solution 3==
 
==Solution 3==

Revision as of 22:00, 23 September 2023

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 (answer choices)

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 we are taking half the circumference, 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 $98\pi + 62\pi + 78\pi = 238\pi \Longrightarrow \boxed{\textbf{(A)}\ 238\pi}$

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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All AJHSME/AMC 8 Problems and Solutions

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