AMC 10 · 2022 · #18

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
A five by five grid has an empty outer border. Count the starting configurations of the inner three by three block that produce, after one transformation, a single filled square at the centre.

Pick an answer.

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

Count the neighbours

Each position has a different neighbour count.

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

When the centre survives

Write the condition when the centre started filled.

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

Count the first sub-case

The first sub-case gives two.

Sub-case 1.1: 2 configurations
4STEP 4

Count the second sub-case

The second sub-case gives none.

Sub-case 1.2: 0 configurations
5STEP 5

Total for the first case

Total up the first case.

Case 1 total = 2
6STEP 6

When the centre started empty

Now exactly three must be filled.

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

Count the first shape

Count the placements of the first shape.

Shape A: 4
8STEP 8

Count the second shape

Count the second shape too.

Shape B: 4
9STEP 9

Count the third shape

Count the third shape as well.

Shape C: 4
10STEP 10

Count the fourth shape

The fourth shape has more placements.

Shape D: 8
11STEP 11

Add them all

Adding them gives 22.

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).