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

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== Problem ==
 
== 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?  
 
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?  
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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));
 +
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));
 +
D(CR((0,-49.5),1.5));
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D(CR((0,-49.5),1.0));
  
[[Image:2006_AMC10A-14.png]]
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D((12,-10)--(12,10)); MP('20',(12,0),E);
 
+
D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);</asy>
 
<math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math>
 
<math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math>
  

Revision as of 14:18, 20 August 2011

The following problem is from both the 2004 AMC 12A #12 and 2004 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] $\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad$

Solution

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 $\frac{18 \cdot 19}{2} + 2 = 173 \Rightarrow \mathrm{(B)}$.

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 $207 - 2(17) = 173$.

See Also

2006 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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