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

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/boxed{Congratulations! You've made it to the last problem. Now you are a great mathmatician! This is just the beginning of competition math. Other math competitions you may want to try
+
Congratulations! You've made it to the last problem. Now you are a great mathmatician! This is just the beginning of competition math. Other math competitions you may want to try
 
-AMC 10/12
 
-AMC 10/12
 
-Mathcounts
 
-Mathcounts
 
-Putnam
 
-Putnam
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-Penn State Math Competition}
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-Penn State Math Competition
  
 
-JasonDaGoat
 
-JasonDaGoat

Latest revision as of 15:17, 24 August 2025

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

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

Video Solution: https://www.youtube.com/watch?v=zZGuBFyiQrk by WhyMath

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Congratulations! You've made it to the last problem. Now you are a great mathmatician! This is just the beginning of competition math. Other math competitions you may want to try -AMC 10/12 -Mathcounts -Putnam -Penn State Math Competition

-JasonDaGoat

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