Competition · AMC preparation · step 4 of 4
AMC 10 · 2022B · #19
Grade 3 geometry-2d
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Subproblems): split on whether the center starts filled or empty — the two cases need very different reasoning. If center starts filled, it must keep 2 or 3 filled peripheral neighbors and no peripheral can survive or be born. If center starts empty, it must gain exactly 3 filled neighbors and again all peripherals must end empty. Tool #1 (Diagram): use a labeled 3 × 3 picture with corners C and edges E to track neighbor counts. Tool #2 (Systematic List): once each sub-case has its constraint, walk through the small finite set of peripheral patterns. Tool #10 (Physical): coins on graph paper let you check each pattern by counting neighbors with your finger.
Label the grid cells
Label the inner 3 × 3: four corners, four edges, center M. Neighbor counts — corner: 3, edge: 5, center: 8.
Draw the 3 × 3 once and count neighbors for each role — corners (sparse), edges (medium), center (everyone).
K.G.A.1Draw A DiagramSplit off the filled-center case
Case 1 — center M starts filled: it survives only with 2 or 3 filled peripherals, and every other peripheral must end empty.
Treat the center separately because it's the only cell every other cell sees.
Handling the centre cell separately splits the count into cases that never overlap.
▸ Why?
Every arrangement either fills the centre or does not, so the two cases cover everything once.
▸ Why?
Each arrangement lands in exactly one case, so nothing is missed and nothing counted twice.
Count two filled neighbors
Sub-case 1.1 (M has 2 filled peripherals): only the two opposite-corner pairs avoid a reborn shared empty neighbor → 2 configs.
Opposite-corner pairs are the only non-adjacent pairs with disjoint peripheral neighborhoods.
3.OA.D.8Make A Systematic ListCount three filled neighbors
Sub-case 1.2 (M has 3 filled peripherals): any non-adjacent triple leaves an empty cell touching two of them, which is reborn → 0 configs.
Three mutually non-adjacent peripherals always leave a 'common neighbor' empty cell that gets born.
3.OA.D.8Make A Systematic ListTotal the filled-center case
Case 1 total: only the two opposite-corner pairs work → 2 configurations.
Two opposite-corner pairs, nothing else.
1.OA.A.2Identify SubproblemsSplit off the empty-center case
Case 2 — center M starts empty: it needs exactly 3 filled peripherals, their induced graph max degree ≤ 1, and no empty cell touching all 3.
Three filled cells, no cell touches both others, no outside cell touches all three.
1.OA.A.2Identify SubproblemsCount shape A
Shape A — three corners (mutually non-adjacent): choose which corner to omit → 4 configs.
Three of the four corners — pick which corner to leave out.
3.OA.A.3Make A Systematic ListCount shape B
Shape B — two corners on one side plus the opposite middle edge: pick which side holds the two corners → 4 configs.
Two corners flanking one side, with the lonely opposite edge filling in.
3.OA.A.3Make A Systematic ListCount shape C
Shape C — two edges meeting at a corner plus the far diagonal corner: place the shared corner in 4 spots → 4 configs.
Two adjacent edges + the far-diagonal corner — the V-and-dot pattern.
3.OA.A.3Make A Systematic ListCount shape D
Shape D — an L (corner + one adjacent edge) plus the diagonally opposite corner: 4 corners × 2 edge choices → 8 configs.
L-shape (corner + one of its two adjacent edges) anchored at one corner of the grid, plus the diagonally opposite corner.
3.OA.A.3Make A Systematic ListAdd all the cases
Case 2: 4 + 4 + 4 + 8 = 20; with Case 1's 2, the total = 22, choice (C).
Add the two disjoint cases — done.
2.OA.A.1Identify SubproblemsSplit by what the center starts as. Center filled: only two opposite-corner pairs survive (2 configs). Center empty: exactly 3 filled peripherals in one of four geometric shapes (4 + 4 + 4 + 8 = 20 configs). Total = 22, choice (C).
- Label the grid cells
- Split off the filled-center case
- Count two filled neighbors
- Count three filled neighbors
- Total the filled-center case
- Split off the empty-center case
- Count shape A
- Count shape B
- Count shape C
- Count shape D
- Add all the cases
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