2006 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2006 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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Concepts:arithmetic sequencesummation

Difficulty rating: 1370

12.

A number of linked rings, each 11 cm thick, are hanging on a peg. The top ring has an outside diameter of 2020 cm. The outside diameter of each of the other rings is 11 cm less than that of the ring above it. The bottom ring has an outside diameter of 33 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

171171

173173

182182

188188

210210

Solution:

The top ring spans 2020 cm. Each ring below overlaps the ring above by 22 cm (twice the 11-cm thickness), so it adds its outside diameter minus 2.2.

The lower rings have outside diameters 19,18,,3,19, 18, \ldots, 3, contributing 17,16,,1.17, 16, \ldots, 1. Thus the total distance is 20+(17+16++1)=20+17182=20+153=173 cm. 20 + (17 + 16 + \cdots + 1) = 20 + \frac{17 \cdot 18}{2} = 20 + 153 = 173 \text{ cm}.

Thus, the correct answer is B.

Problem 12 in Other Years