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? | |
| − | + | [[Image:2006_AMC10A-14.png]] | |
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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> | ||
== 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 == | |
| − | == See | + | {{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}} |
| − | {{ | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 18:09, 5 February 2008
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?
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 
.
Alternatively, the sum of the consecutive integers from 3 to 20 is 
. However, the 17 intersections between the rings must be subtracted, and we also get 
.
See Also
| 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 | ||
