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Difference between revisions of "2006 AMC 12A Problems/Problem 12"

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{{duplicate|[[2006 AMC 12A Problems|2004 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2004 AMC 10A #14]]}}
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2006 AMC 10A #14]]}}
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== Problem ==
 
== Problem ==
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A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outisde [[diameter]] of 20 cm. The outside diameter of each of the outer rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the [[distance]], in cm, from the top of the top ring to the bottom of the bottom ring?
 
  
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[[Image:2006_AMC10A-14.png]]
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A number of linked rings, each <math>1</math> cm thick, are hanging on a peg. The top ring has an outside diameter of <math>20</math> cm. The outside diameter of each of the outer rings is <math>1</math> cm less than that of the ring above it. The bottom ring has an outside diameter of <math>3</math> cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
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<asy>
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size(7cm); pathpen = linewidth(0.7);
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D(CR((0,0),10));
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D(CR((0,0),9.5));
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D(CR((0,-18.5),9.5));
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D(CR((0,-18.5),9));
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MP("$\vdots$",(0,-31),(0,0));
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D(CR((0,-39),3));
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D(CR((0,-39),2.5));
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D(CR((0,-43.5),2.5));
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D(CR((0,-43.5),2));
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D(CR((0,-47),2));
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D(CR((0,-47),1.5));
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D(CR((0,-49.5),1.5));
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D(CR((0,-49.5),1.0));
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D((12,-10)--(12,10)); MP('20',(12,0),E);
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D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);</asy>
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<math>\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad</math>
  
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<math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math>
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== Solution 1 ==
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== Solution ==
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The inside diameters of the rings are the positive integers from <math>1</math> to <math>18</math>.  The total distance needed is the sum of these values plus <math>2</math> for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an [[arithmetic series]], the answer is <math>\frac{18 \cdot 19}{2} + 2 = \boxed{\textbf{(B) }173}</math>.
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The inside diameters of the rings are the [[positive integer]]s from 1 to 18.  The total distance needed is the sum of these values plus 2 for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an [[arithmetic series]], the answer is <math>\frac{18 \cdot 19}{2} + 2 = 173 \Rightarrow \mathrm{(B)}</math>.
 
  
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Alternatively, the sum of the consecutive [[integer]]s from 3 to 20 is <math> \frac{1}{2}(18)(3+20) = 207 </math>.  However, the 17 [[intersection]]s between the rings must be subtracted, and we also get <math> 207 - 2(17) = 173</math>.
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== Solution 2 ==
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Alternatively, the sum of the consecutive integers from 3 to 20 is <math> \frac{1}{2}(18)(3+20) = 207 </math>.  However, the 17 [[intersection]]s between the rings must be subtracted, and we also get <math> 207 - 2(17) = \boxed{\textbf{(B) }173}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2006|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2006|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Latest revision as of 10:56, 17 December 2021

The following problem is from both the 2006 AMC 12A #12 and 2006 AMC 10A #14, so both problems redirect to this page.

Problem

A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]

$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$

Solution 1

The inside diameters of the rings are the positive integers from $1$ to $18$. The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is $\frac{18 \cdot 19}{2} + 2 = \boxed{\textbf{(B) }173}$.

Solution 2

Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$. However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{\textbf{(B) }173}$.

See Also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 AMC 12 Problems and Solutions
2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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