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

(Solution)
(Solution 2)
(13 intermediate revisions by 8 users not shown)
Line 2: Line 2:
 
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;
 
size(8cm);
 
size(8cm);
 +
A=(0,0);
 +
B=(480,0);
 
draw((0,0)--(480,0),linetype("3 4"));
 
draw((0,0)--(480,0),linetype("3 4"));
 
filldraw(circle((8,0),8),black);
 
filldraw(circle((8,0),8),black);
Line 11: Line 14:
 
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));
 
draw((260,0)--(290,30sqrt(3)),Arrow(size=4));
 
draw((400,0)--(440,-40sqrt(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>
 +
 +
<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>
 +
 +
==Solution 1==
 +
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>
 +
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>
 
</asy>
 +
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 (answer choices)==
 +
 +
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>.
 +
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.
  
<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>
+
==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>.
  
==Solution==
+
Hence, the total length is <math>98\pi + 62\pi + 78\pi = 238\pi \Longrightarrow \boxed{\textbf{(A)}\ 238\pi}</math>
The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball, 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. 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>.
 
  
 
==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 13:39, 12 January 2021

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]

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

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 shorter. Therefore, the only answer choice less than $240\pi$ is $\boxed{\textbf{(A)}\ 238\pi}$. 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.

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 (ProblemsAnswer KeyResources)
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