2006 AMC 12A Problems/Problem 12

The following problem is from both the 2006 AMC 12A #12 and 2006 AMC 10A #14, so both problems redirect to this page.

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$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]$

$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$

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 $\frac{18 \cdot 19}{2} + 2 = \boxed{\textbf{(B) }173}$ .

Solution 2

Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{\textbf{(B) }173}$ .

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.

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2006 AMC 12A Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also