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 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 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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<math> \mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188\qquad \mathrm{(E) \ }  210</math>
  
<math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math>
 
 
== Solution ==
 
== Solution ==
 
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>.
 
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>.
 
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>.
  
== See Also ==
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{{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}}
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== See also ==
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{{AMC12 box|year=2006|ab=A|num-b=11|num-a=13}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 18:41, 5 February 2008

Problem


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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?

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

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
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