Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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+ | {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2006 AMC 10A #14]]}} | ||
+ | |||
== Problem == | == Problem == | ||
− | 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? | ||
+ | |||
+ | <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> | ||
<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> | ||
− | == Solution == | + | |
+ | == Solution 1 == | ||
+ | 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>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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 Also == | ||
− | + | {{AMC12 box|year=2006|ab=A|num-b=11|num-a=13}} | |
+ | {{AMC10 box|year=2006|ab=A|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 22:25, 21 January 2021
- The following problem is from both the 2006 AMC 12A #12 and 2006 AMC 10A #14, so both problems redirect to this page.
Contents
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?
Solution 1
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 .
Solution 2
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 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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.