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 | + | {{image}} |
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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</math><math>\mathrm{(E) \ } 210</math> | ||
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== Solution == | == Solution == | ||
| − | + | The sum of the inner [[diameter]]s of the rings is the series from 1 to 18. To this sum, we must add 2 for the top of the first ring and the bottom of the last ring. Thus, <math>\frac{18 * 19}{2} + 2 = 173 \Rightarrow B</math>. | |
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| − | + | Alternatively, the sum of the consecutively increasing [[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 also be subtracted, so we get <math> 207 - 2(17) = 173 \Rightarrow B </math>. | |
| − | *[[2006 AMC | + | == See also == |
| + | * [[2006 AMC 12A Problems]] | ||
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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]] | ||
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Revision as of 08:02, 15 February 2007
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?
Solution
The sum of the inner diameters of the rings is the series from 1 to 18. To this sum, we must add 2 for the top of the first ring and the bottom of the last ring. Thus, 
.
Alternatively, the sum of the consecutively increasing integers from 3 to 20 is 
. However, the 17 intersections between the rings must also be subtracted, so we get 
.
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 | |
