2006 AMC 12A Problems/Problem 12

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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$ (Error compiling LaTeX. Unknown error_msg). However, the 17 intersections between the rings must also be subtracted, so we get $207 – 2(17) = 173 \Rightarrow B$.

See also


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