2006 AMC 12A Problems/Problem 12

Problem

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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?

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

Solution

The sum of the consecutively increasing integerss from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must also be subtracted, so we get $207 - 2(17) = 173 \Rightarrow B$ .

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Preceded by Problem 11	AMC 12A 2006	Followed by Problem 13

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2006 AMC 12A Problems/Problem 12

Problem

Solution

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