Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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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 | + | 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? |
<math> \mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188</math><math>\mathrm{(E) \ } 210</math> | <math> \mathrm{(A) \ } 171\qquad \mathrm{(B) \ } 173\qquad \mathrm{(C) \ } 182\qquad \mathrm{(D) \ } 188</math><math>\mathrm{(E) \ } 210</math> | ||
== Solution == | == Solution == | ||
| + | The sum of the [[consecutive]]ly increasing [[integers]]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>. | ||
== See also == | == See also == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
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| − | + | {{AMC box|year=2006|n=12A|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 18:37, 31 January 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 consecutively increasing integerss from 3 to 20 is $\frac{1}{2}\(18\)\(3+20\) = 207$ (Error compiling LaTeX. Unknown error_msg). However, the 17 intersections between the rings must also be subtracted, so we get 
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See also
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| Preceded by Problem 11 |
AMC 12A 2006 |
Followed by Problem 13 |
