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 2(No Advanced Computation): 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 b)
(→‎Solution 2)
(25 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 
==Problem==
 
==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 <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> 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?
 
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 <math>R_1 = 100</math> inches, <math>R_2 = 60</math> inches, and <math>R_3 = 80</math> 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>
 
<asy>
 
pair A,B;
 
pair A,B;
Line 21: Line 22:
  
 
<math>\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi</math>
 
<math>\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi</math>
 +
 +
==Video Solution for Problems 21-25==
 +
https://youtu.be/-mi3qziCuec
 +
 +
==Video Solution==
 +
https://youtu.be/zZGuBFyiQrk ~savannahsolver
  
 
==Solution 1==
 
==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 <math>2\pi*2/2=2\pi</math> inches, and it gains <math>2\pi</math> inches on B.  
+
Since the diameter of the ball is 4 inches, <math>\text{radius}=2</math>.  
 +
 
 +
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 <math>2\pi</math> inches each, because <math>\dfrac{1}{2} 2\pi (x-2) - \dfrac{1}{2} 2\pi (x)= -2 \pi</math>
 +
 
 +
By similar reasoning, it gains <math>2\pi</math> inches on semicircle B.  
 
<asy>
 
<asy>
 
unitsize(0.04cm);
 
unitsize(0.04cm);
Line 40: Line 51:
 
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>\dfrac{1}{2}2\pi (100+60+80) +(-2+2-2)\cdot\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>.
 +
 
 +
==Solution 2 ==
 +
 
 +
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 <math>2\pi</math> because it's already half the circumference through semicircle A, which needs to go half the circumference extra through semicircle B, and it's already half the circumference through semicircle C, and the circumference is <math>4\pi</math> Therefore, the answer is <math>240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>.
 +
 
 +
~[[User:PowerQualimit|PowerQualimit]]
  
−
==Solution 2==
+
==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, <math>r = 100 - 2 = 98</math>, and the arc is <math>r \pi = 98\pi</math>.
 +
For the second section, the radius of the semi-circle that the ball's center follows is <math>r = 60 + 2 = 62</math>, and the arc is <math>62\pi</math>.
 +
For the third section, the radius of the semi-circle that the ball's center follows is <math>r = 80 - 2 = 78</math>, and the arc is <math>78\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>.
+
Hence, the total length is <math>98\pi + 62\pi + 78\pi = 238\pi \Longrightarrow \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.
 
  
 
==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 18:28, 21 December 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

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, which 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 $98\pi + 62\pi + 78\pi = 238\pi \Longrightarrow \boxed{\textbf{(A)}\ 238\pi}$

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=208067"