2006 AMC 10A Problem 14

P11-15~4 minPrint

Sequences & Series Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

Key insight

Each ring hangs on the one above, so the holes line up end to end. Add the holes, 18 down to 1, plus 1 cm at each end.

Solution

Try two rings first to see how they stack, then add them all up.

1. The rings. The outside diameters are $20, 19, 18, \ldots, 3$ . That is $20 - 3 + 1 = 18$ rings.
2. The holes. Each ring is $1$ cm thick, so its hole is $2$ cm less than its outside diameter ( $1$ cm of metal on each side). The holes are $18, 17, \ldots, 1$ cm.
3. Two rings. Go down from the top of the top ring: $1$ cm of metal, then its $18$ cm hole. The second ring hangs on the bottom of that hole, so the top of the second ring's metal sits in the last centimeter of the first hole. That means the second ring's hole starts exactly where the first hole ends. Then comes the $17$ cm hole of the second ring, then its $1$ cm of bottom metal. Two rings measure $1 + 18 + 17 + 1 = 37$ cm.
4. All the rings. The same thing happens at every link: the holes follow one after another with no gap and no overlap. Only the top metal of the first ring and the bottom metal of the last ring are extra:
$1 + (18 + 17 + \cdots + 1) + 1.$
5. Add. $1 + 2 + \cdots + 18 = \dfrac{18 \times 19}{2} = 171$ . The total is $171 + 2 = 173$ cm.

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

Why this works

For things stacked or linked together, ask what each piece adds to the total once the overlaps are taken out. Here each ring adds exactly its hole, and only the two outer bits of metal are extra. Checking the rule on two rings before summing all eighteen is a quick way to trust it.

Alternative approach

Add all the outside diameters: $3 + 4 + \cdots + 20 = 207$ . This counts the places where two rings overlap twice, so take those away. At each link, the top of the lower ring sits $1$ cm above the inside bottom of the upper ring, which is $2$ cm above the upper ring's outside bottom. So neighboring rings overlap by $2$ cm. There are $17$ links, so the height is $207 - 17 \times 2 = 173$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(B)}\ 173}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)