2006 AMC 12A Problems/Problem 12

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?

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

Solution

The sum of the inner diameters of the rings is the series from 1 to 18. To this sum, we must add 2 for the top of the first ring and the bottom of the last ring. Thus, $\frac{18 * 19}{2} + 2 = 173 \Rightarrow B$ .

Alternatively, the sum of the consecutively increasing integers 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 $207 - 2(17) = 173 \Rightarrow B$ .

2006 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2006 AMC 12A Problems/Problem 12

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