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 == | ||
| − | == Solution == | + | A number of linked rings, each <math>1</math> cm thick, are hanging on a peg. The top ring has an outside diameter of <math>20</math> cm. The outside diameter of each of the outer rings is <math>1</math> cm less than that of the ring above it. The bottom ring has an outside diameter of <math>3</math> 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>\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad</math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | The inside diameters of the rings are the positive integers from <math>1</math> to <math>18</math>. The total distance needed is the sum of these values plus <math>2</math> 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 = \boxed{\textbf{(B) }173}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | Alternatively, the sum of the consecutive integers 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) = \boxed{\textbf{(B) }173}</math>. | ||
| + | |||
| + | == 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 10:56, 17 December 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 
cm thick, are hanging on a peg. The top ring has an outside diameter of 
cm. The outside diameter of each of the outer rings is cm less than that of the ring above it. The bottom ring has an outside diameter of 
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]](http://latex.artofproblemsolving.com/7/5/b/75b3e762b96f49323740d90a83fdb1339789180e.png)
Solution 1
The inside diameters of the rings are the positive integers from to 
. The total distance needed is the sum of these values plus 
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. 
