AMC 10 · 2022 · #19

Grade 3 geometry-2d
systematic-enumerationcaseworkspatial-visualizationcellular-automatoncombinations-basic caseworksystematic-enumerationidentify-subproblemsphysical-representation ↑ Prerequisites: systematic-enumerationspatial-visualization
📏 Long solution 💡 4 insights 📊 Diagram
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Problem
In a 5 × 5 Conway-style grid whose outer border is empty, count the initial configurations of the 3 × 3 inner subgrid that produce, after one transformation, a single filled square at the center.

Pick an answer.

(A)
14
(B)
18
(C)
22
(D)
26
(E)
30

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

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.

1STEP 1

Label the inner 3 × 3: four corners, four edges, center M. Neighbor counts — corner: 3, edge: 5, center: 8.

corner: 3 peripheral nbrs, edge: 5, center: 8
2STEP 2

Case 1 — center M starts filled: it survives only with 2 or 3 filled peripherals, and every other peripheral must end empty.

M filled → 2 ≤ #filled peripherals ≤ 3
3STEP 3

Sub-case 1.1 (M has 2 filled peripherals): only the two opposite-corner pairs avoid a reborn shared empty neighbor → 2 configs.

Sub-case 1.1: 2 configurations
4STEP 4

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.

Sub-case 1.2: 0 configurations
5STEP 5

Case 1 total: only the two opposite-corner pairs work → 2 configurations.

Case 1 total = 2
6STEP 6

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.

M empty → exactly 3 filled peripherals, induced subgraph has max degree ≤ 1
7STEP 7

Shape A — three corners (mutually non-adjacent): choose which corner to omit → 4 configs.

Shape A: 4
8STEP 8

Shape B — two corners on one side plus the opposite middle edge: pick which side holds the two corners → 4 configs.

Shape B: 4
9STEP 9

Shape C — two edges meeting at a corner plus the far diagonal corner: place the shared corner in 4 spots → 4 configs.

Shape C: 4
10STEP 10

Shape D — an L (corner + one adjacent edge) plus the diagonally opposite corner: 4 corners × 2 edge choices → 8 configs.

Shape D: 8
11STEP 11

Case 2: 4 + 4 + 4 + 8 = 20; with Case 1's 2, the total = 22, choice (C).

Total = 2 + 20 = 22
Answer
22
Sanity: each of the four shapes is invariant under rotation, so 4 or 8 counts (depending on whether the shape has additional reflection symmetry) make sense; Shape D has only rotational symmetry (no reflection symmetry, so multiply by 2), giving 8. The Case 1 count of 2 matches the obvious fact that only diagonal pairs of corners are 'spread out' enough to keep the center alive without spawning anything. The final 22 matches choice (C). Nearby answers (18, 26) trap students who miscount Shape D (± 4) or omit Shape C.
💡Key takeaway

Split 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).