AMC 10 · 2020 · #17
Grade 4 geometry-2dProblem
There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there for the people to split up into pairs so that the members of each pair know each other?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Ten people stand equally spaced around a circle. Each knows exactly $3$ others: their $2$ neighbors plus the one directly across the circle. Count the ways to split the $10$ into $5$ pairs so that every pair is made of people who know each other.
Givens: $10$ people at equally spaced positions around a circle (label them $0, 1, \dots, 9$); Person $i$ knows person $i - 1$ and person $i + 1$ (neighbors, indices mod $10$) and person $i + 5$ (directly across, a 'diameter'); There are exactly $5$ diameters: $(0,5), (1,6), (2,7), (3,8), (4,9)$; Choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: Number of valid $5$-pair groupings of the $10$ people
Understand
Restated: Ten people stand equally spaced around a circle. Each knows exactly $3$ others: their $2$ neighbors plus the one directly across the circle. Count the ways to split the $10$ into $5$ pairs so that every pair is made of people who know each other.
Givens: $10$ people at equally spaced positions around a circle (label them $0, 1, \dots, 9$); Person $i$ knows person $i - 1$ and person $i + 1$ (neighbors, indices mod $10$) and person $i + 5$ (directly across, a 'diameter'); There are exactly $5$ diameters: $(0,5), (1,6), (2,7), (3,8), (4,9)$; Choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Tool #1 (Diagram): draw the $10$-cycle with the $5$ diameters across it — the picture instantly shows the two pair types (neighbor edge or diameter). Tool #7 (Subproblems): split into cases by how many diameters $k$ the pairing uses ($k = 0, 1, 2, 3, 4, 5$); inside each case the remaining people must pair as neighbors. Tool #2 (Systematic List): inside each case, list the configurations in order so nothing is missed and nothing is double-counted.
Execute — Answer: C
4.OA.C.5 Step 1 - Label the people $0, 1, \dots, 9$ around the circle.
- Each pair must be a neighbor edge or a diameter.
- Split the count by $k$, the number of diameter pairs used.
- The non-diameter people must pair up among themselves using only neighbor edges.
💡 Sort the configurations by how many cross-diameters they use.
4.OA.C.5 Step 2 - Case $k = 5$ (all five diameters).
- Every person is paired across the circle.
- Exactly $1$ configuration.
💡 All diameters used — only one way.
4.G.A.1 Step 3 - Case $k = 4$.
- Use four diameters; the two leftover people are the endpoints of the unused diameter — they are directly across each other, not neighbors.
- They cannot pair (the diameter is excluded by assumption and they aren't adjacent).
- Impossible.
- $0$ ways.
💡 Dropping one diameter strands two opposite people with no legal edge between them.
4.OA.C.5 Step 4 - Case $k = 3$.
- Use three diameters; the four leftover people fill two opposite arcs and must pair as neighbors.
- This works iff the two omitted diameters are 'adjacent' (consecutive among the five diameters), so each side has two neighboring people.
- The number of adjacent pairs of omitted diameters around the diameter-cycle of length $5$ is $5$.
💡 The two skipped diameters must be next to each other so the leftover people land in neighbor pairs.
4.G.A.1 Step 5 - Case $k = 2$.
- Two diameters used, six leftover people in two opposite arcs of $3$ each.
- An arc of three consecutive people on the cycle has no perfect neighbor matching (any matching leaves one person out).
- Impossible.
- $0$ ways.
💡 Three people in a row can't pair off using only neighbor edges.
4.OA.C.5 Step 6 - Case $k = 1$.
- Choose $1$ of $5$ diameters ($5$ choices).
- The eight leftover people sit on two opposite arcs of $4$ consecutive people each.
- Each arc of $4$ has exactly $1$ neighbor-pair perfect matching.
- Total $5 \cdot 1 = 5$.
💡 Pick the single diameter, then the two arcs of $4$ each pair up uniquely as neighbors.
4.OA.C.5 Step 7 - Case $k = 0$ (no diameters).
- All five pairs are neighbor edges on the $10$-cycle.
- A $10$-cycle has exactly $2$ perfect matchings: $(0,1)(2,3)(4,5)(6,7)(8,9)$ and $(1,2)(3,4)(5,6)(7,8)(9,0)$.
💡 Two ways to pair $10$ chairs around a round table — even-odd or odd-even.
1.OA.A.2 Step 8 - Add the contributing cases: $1 + 5 + 5 + 2 = 13$.
- That matches choice (C).
💡 Disjoint cases — just add.
4.OA.C.5 Label the people $0, 1, \dots, 9$ around the circle. Each pair must be a neighbo 4.OA.C.5 Case $k = 5$ (all five diameters). Every person is paired across the circle. Exa 4.G.A.1 Case $k = 4$. Use four diameters; the two leftover people are the endpoints of t 4.OA.C.5 Case $k = 3$. Use three diameters; the four leftover people fill two opposite ar 4.G.A.1 Case $k = 2$. Two diameters used, six leftover people in two opposite arcs of $3 4.OA.C.5 Case $k = 1$. Choose $1$ of $5$ diameters ($5$ choices). The eight leftover peop 4.OA.C.5 Case $k = 0$ (no diameters). All five pairs are neighbor edges on the $10$-cycle 1.OA.A.2 Add the contributing cases: $1 + 5 + 5 + 2 = 13$. That matches choice (C). Review
Reasonableness: The answer $13$ sits in the middle of the offered range $11$ to $15$, exactly where casework with two large contributing buckets (the $k = 1$ and $k = 3$ cases give $5$ each) plus the small $k = 0$ and $k = 5$ buckets ($2 + 1 = 3$) lands. The two impossible cases ($k = 2$ and $k = 4$) make sense from the picture: the leftover arcs have odd length or strand opposite endpoints, so no neighbor matching exists.
Alternative: Tool #2 (Systematic List) directly: list every perfect matching of the graph with vertices $0$-$9$ and edges $\{i, i+1\}$ and $\{i, i+5\}$. Order the matchings by which edge contains vertex $0$ (three choices: $\{0,1\}, \{0,9\}, \{0,5\}$) and recursively pair the remaining vertices. Counting yields the same $13$.
CCSS standards used (min grade 4)
1.OA.A.2Solve word problems involving three whole numbers whose sum is within 20 (Adding the contributions $1 + 5 + 5 + 2 = 13$ across cases.)4.OA.C.5Generate a number or shape pattern following a given rule (Counting configurations in each case by following the rule 'every pair is a neighbor edge or a diameter'.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Drawing the $10$ points around the circle with the $5$ diameters to see which leftover arcs can be perfectly matched by neighbors.)
⭐ This AMC 10 problem only needs Grade 4 case-by-case counting you already know — draw the $10$ people around the circle and split by how many cross-diameter pairs the matching uses ($k = 0, 1, 2, 3, 4, 5$). The cases give $2 + 5 + 0 + 5 + 0 + 1 = 13$ valid pairings. The answer is $\textbf{(C)}$.
⭐ This AMC 10 problem only needs Grade 4 case-by-case counting you already know — draw the $10$ people around the circle and split by how many cross-diameter pairs the matching uses ($k = 0, 1, 2, 3, 4, 5$). The cases give $2 + 5 + 0 + 5 + 0 + 1 = 13$ valid pairings. The answer is $\textbf{(C)}$.