Difference between revisions of "2006 AMC 12A Problems/Problem 12"
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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> | <math>\mathrm{(A) \ } 171\qquad\mathrm{(B) \ } 173\qquad\mathrm{(C) \ } 182\qquad\mathrm{(D) \ } 188\qquad\mathrm{(E) \ } 210\qquad</math> | ||
| − | + | == Solution 1 == | |
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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>. | ||
| − | + | == 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>. | 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>. | ||
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?
![[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 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. 
