2006 AMC 12A Problem 12

Question

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 other 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?

171

173

182

188

210

The top ring spans

20

cm; each lower ring adds its outside diameter minus

2

for the overlap

Total

= 20 + (17 + 16 + \cdots + 1)

The top ring spans

20

cm. Each ring below overlaps the ring above by

2

cm (twice the

1

-cm thickness), so it adds its outside diameter minus

2.

The lower rings have outside diameters

19, 18, \ldots, 3,

contributing

17, 16, \ldots, 1.

Thus the total distance is

\begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173 \text{ cm}. \end{gathered}

Thus, the correct answer is B.

Problem 12 in other years