AMC 10 · 2020 · #17
Grade 4 geometry-2dPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Label people 0–9; every pair is a neighbor edge or a diameter, so split cases by the number of diameters k the matching uses.
Sort the configurations by how many cross-diameters they use.
4.OA.C.5Identify SubproblemsCase k = 5: all five diameters used, everyone paired across the circle — exactly 1 way.
All diameters used — only one way.
4.OA.C.5Make A Systematic ListCase k = 4: the two leftover people are opposite ends of the unused diameter, not neighbors, so they cannot pair — impossible, 0 ways.
Dropping one diameter strands two opposite people with no legal edge between them.
4.G.A.1Draw A DiagramCase k = 3: works only when the two skipped diameters are adjacent on the diameter-cycle of 5, giving 5 ways.
The two skipped diameters must be next to each other so the leftover people land in neighbor pairs.
4.OA.C.5Draw A DiagramCase k = 2: leftover people form two arcs of 3, and 3 in a row has no perfect neighbor matching — impossible, 0 ways.
Three people in a row can't pair off using only neighbor edges.
4.G.A.1Draw A DiagramCase k = 1: pick 1 of 5 diameters, and each leftover arc of 4 pairs uniquely as neighbors — 5 ways.
Pick the single diameter, then the two arcs of 4 each pair up uniquely as neighbors.
4.OA.C.5Make A Systematic ListCase k = 0: all pairs are neighbor edges, and a 10-cycle has exactly 2 ways to match perfectly.
Two ways to pair 10 chairs around a round table — even-odd or odd-even.
4.OA.C.5Draw A DiagramAdd the surviving cases: 1 + 5 + 5 + 2 = 13, which is choice (C).
Disjoint cases — just add.
1.OA.A.2Identify SubproblemsThis 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 (C).