AMC 10 · 2006 · #14
Grade 4 geometry-2dA 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?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Rings hang linked on a peg, one below the next. Every ring is $1$ cm thick. The outside diameters shrink by $1$ cm each step: the top ring is $20$ cm across the outside and the bottom ring is $3$ cm across the outside, so the outside diameters run $20, 19, 18, \dots, 3$. We measure the straight-line distance from the very top of the top ring down to the very bottom of the bottom ring.
Givens: Each ring is $1$ cm thick (the metal itself); Top ring outside diameter is $20$ cm; Each ring below has outside diameter $1$ cm smaller than the one above; Bottom ring outside diameter is $3$ cm; Answer choices: (A) 171, (B) 173, (C) 182, (D) 188, (E) 210
Unknowns: The total vertical distance from the top of the top ring to the bottom of the bottom ring
Understand
Restated: Rings hang linked on a peg, one below the next. Every ring is $1$ cm thick. The outside diameters shrink by $1$ cm each step: the top ring is $20$ cm across the outside and the bottom ring is $3$ cm across the outside, so the outside diameters run $20, 19, 18, \dots, 3$. We measure the straight-line distance from the very top of the top ring down to the very bottom of the bottom ring.
Givens: Each ring is $1$ cm thick (the metal itself); Top ring outside diameter is $20$ cm; Each ring below has outside diameter $1$ cm smaller than the one above; Bottom ring outside diameter is $3$ cm; Answer choices: (A) 171, (B) 173, (C) 182, (D) 188, (E) 210
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #5 Look for a Pattern
Adding up the outside diameters ($20 + 19 + \dots + 3$) overcounts, because linked rings overlap where one hangs inside the other. Tool #7 (Identify Subproblems) says: cut the total height into pieces that do not overlap. Tool #1 (Draw a Diagram) shows what the pieces are — each ring's top bar tucks inside the hole of the ring above it, so what truly stacks up are the inside diameters, bracketed by just the top bar of the first ring and the bottom bar of the last. Tool #5 (Look for a Pattern) then adds the inside diameters fast: they are exactly the whole numbers $1$ through $18$, which pair up neatly. The whole trick is measuring inside holes, not outside edges.
Execute — Answer: B
4.OA.C.5 Step 1 Switch from outside to inside diameters
- The metal is $1$ cm thick, so it eats $1$ cm off each side of the hole.
- That makes every ring's inside diameter equal to its outside diameter minus $2$.
- The top ring: $20 - 2 = 18$.
- The bottom ring: $3 - 2 = 1$.
- Since the outside diameters step down through $20, 19, \dots, 3$, the inside diameters step down through $18, 17, \dots, 1$ — the whole numbers from $1$ to $18$.
💡 Each ring's hole is what the next ring drops into, so the hole size is what really matters, not the outer edge.
4.MD.A.2 Step 2 See how linked rings overlap
- Picture two neighboring rings.
- The lower ring hangs through the upper ring, and its top bar rests right against the bottom bar of the ring above.
- So the lower ring's top bar sits *inside* the hole of the upper ring — that overlap is already counted once you count the upper ring's inside hole.
- Going all the way down, only the inside holes stack up.
- The only metal not swallowed by a hole above is the top bar of the very top ring ($1$ cm) and the bottom bar of the very bottom ring ($1$ cm).
- So the total height is the top bar, plus every inside diameter stacked in a column, plus the bottom bar.
💡 Where two rings touch, one ring's bar hides inside the other's hole, so that length only gets counted a single time.
4.NBT.B.5 Step 3 Add the inside diameters by pairing
- Add $1 + 2 + 3 + \cdots + 18$ by pairing the ends: $1 + 18 = 19$, $2 + 17 = 19$, $3 + 16 = 19$, and so on.
- The $18$ numbers make $9$ pairs, each summing to $19$.
- So the total is $9 \times 19 = 171$.
💡 Pairing the smallest with the largest makes every pair the same size, turning a long sum into one quick multiplication.
4.NBT.B.4 Step 4 Add the two end bars
- Now add the leftover metal: the top bar of the top ring and the bottom bar of the bottom ring, each $1$ cm.
- That gives $171 + 1 + 1 = 173$ cm from the top of the top ring to the bottom of the bottom ring, which is choice $\textbf{(B)}$.
💡 The holes stack in the middle; you only add one extra centimeter of solid metal at the very top and one at the very bottom.
4.OA.C.5 The metal is $1$ cm thick, so it eats $1$ cm off each side of the hole. That mak 4.MD.A.2 Picture two neighboring rings. The lower ring hangs through the upper ring, and 4.NBT.B.5 Add $1 + 2 + 3 + \cdots + 18$ by pairing the ends: $1 + 18 = 19$, $2 + 17 = 19$, 4.NBT.B.4 Now add the leftover metal: the top bar of the top ring and the bottom bar of th Review
Reasonableness: Sanity-check the two traps. Adding the inside diameters alone gives $171$, which is choice (A) — that forgets the top and bottom bars, so it must be a bit small. Adding the outside diameters $3 + 4 + \cdots + 20 = 207$ is far too big, because it counts every overlap twice; note $207$ is not even a choice, warning you overlaps must be removed. Our answer $173$ sits just above $171$ by the $2$ cm of end bars, which is exactly the small correction the picture demands. Also check the count: outside diameters $3$ through $20$ is $18$ rings, matching $18$ inside diameters, so nothing was dropped.
Alternative: Start from the outside diameters instead. Their sum is $3 + 4 + \cdots + 20 = 207$. Each of the $17$ links between neighboring rings hides an overlap of $2$ cm (the two touching bars, $1$ cm each), so subtract $17 \times 2 = 34$: $207 - 34 = 173$. Same answer, reached by removing the double-counted overlaps rather than by stacking inside holes.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Turning the stepping-down outside diameters $20, 19, \dots, 3$ into the inside diameters $18, 17, \dots, 1$, the whole numbers $1$ to $18$.)4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Breaking the top-to-bottom distance into non-overlapping lengths: one top bar, the stacked inside diameters, and one bottom bar.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Summing $1 + 2 + \cdots + 18$ as $9$ equal pairs of $19$, computed by $9 \times 19 = 171$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Subtracting $2$ for each inside diameter and adding the two end bars, $171 + 1 + 1 = 173$.)
⭐ Linked rings overlap, so measure the inside holes that truly stack — here the holes are $1$ through $18$ (sum $171$), plus one centimeter of metal at the very top and very bottom, giving $173$.
⭐ Linked rings overlap, so measure the inside holes that truly stack — here the holes are $1$ through $18$ (sum $171$), plus one centimeter of metal at the very top and very bottom, giving $173$.
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