Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AMC 12A Problems/Problem 12"
(→‎Solution 2)
 
(33 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 +
{{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 outisde 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\qquad\mathrm{(E) \ } 210\qquad</math>
+
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?
−
== Solution ==
+
 
 +
<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 ==
 
== See Also ==
−
*[[2006 AMC 10A Problems]]
+
{{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.

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 $207 - 2(17) = \boxed{\textbf{(B) }173}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_12&oldid=167919"