2006 AMC 10A Problem 14

The outside diameters are $20, 19, \dots, 3$, so there are $18$ rings. A ring $1$ cm thick has inside diameter $2$ less than its outside diameter, so the holes have diameters $18, 17, \dots, 1$.

Now see how the rings stack. Walk downward from the top of the top ring: first $1$ cm of metal, then the top ring's hole of $18$ cm. The second ring hangs from the inside bottom of the first, so the top of the second ring's metal sits in the last centimeter of the first ring's hole, and the second ring's hole begins exactly where the first ring's hole ends. The same happens at every link: the holes follow one another with no gaps and no overlaps.

So the total height is
$$1 + (18 + 17 + \dots + 1) + 1 = 2 + \frac{18 \cdot 19}{2} = 2 + 171 = 173,$$
the final $1$ being the metal at the bottom of the bottom ring.

The answer is $\boxed{\textbf{(B)}\ 173}$.

Stacked-object problems reduce to "what does each object contribute to the total once overlaps are removed?" Here each ring's net contribution is its inside diameter, and only the outermost metal at the two ends is extra. Testing the rule on two rings (outer $20$ and $19$: height $1 + 18 + 17 + 1 = 37$, which you can confirm by drawing) is a quick way to trust the pattern before summing.

Sum the outside diameters, $3 + 4 + \dots + 20 = 207$, then subtract the overlap at each of the $17$ links. At each link the lower ring's top centimeter lies inside the upper ring's bottom centimeter, an overlap of $2$ cm: $207 - 34 = 173$.

• Adding the outside diameters $3 + 4 + \dots + 20 = 207$ without removing overlaps, or forgetting the $1$ cm of metal at the very top and bottom ($171$, choice (A)).
• Miscounting the number of rings as $17$ (the number of gaps) instead of $18$.