AMC 10 · 2022 · #19

Grade 3 geometry-2d
systematic-enumerationcaseworkspatial-visualizationcellular-automatoncombinations-basic caseworksystematic-enumerationidentify-subproblemsphysical-representation ↑ Prerequisites: systematic-enumerationspatial-visualization
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Problem

Each square in a 5×55 \times 5 grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

Any filled square with two or three filled neighbors remains filled.
Any empty square with exactly three filled neighbors becomes a filled square.
All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

Suppose the 5×55 \times 5 grid has a border of empty squares surrounding a 3×33 \times 3 subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

Pick an answer.

(A)
14
(B)
18
(C)
22
(D)
26
(E)
30
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Toolkit + CCSS Solution

Understand

Restated: In a $5 \times 5$ Conway-style grid whose outer border is empty, count the initial configurations of the $3 \times 3$ inner subgrid that produce, after one transformation, a single filled square at the center.

Givens: Transformation rules: a filled square survives iff it has $2$ or $3$ filled neighbors; an empty square comes alive iff it has exactly $3$ filled neighbors; otherwise empty; Neighbors include diagonal — up to $8$ per cell; Border is always empty, so only the $9$ inner cells can ever be filled; Goal: end-state has the center cell filled and every other inner cell empty; Choices: (A) $14$, (B) $18$, (C) $22$, (D) $26$, (E) $30$

Unknowns: Number of initial $3 \times 3$ configurations leading to that one-cell end-state

Understand

Restated: In a $5 \times 5$ Conway-style grid whose outer border is empty, count the initial configurations of the $3 \times 3$ inner subgrid that produce, after one transformation, a single filled square at the center.

Givens: Transformation rules: a filled square survives iff it has $2$ or $3$ filled neighbors; an empty square comes alive iff it has exactly $3$ filled neighbors; otherwise empty; Neighbors include diagonal — up to $8$ per cell; Border is always empty, so only the $9$ inner cells can ever be filled; Goal: end-state has the center cell filled and every other inner cell empty; Choices: (A) $14$, (B) $18$, (C) $22$, (D) $26$, (E) $30$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #2 Make a Systematic List, #10 Create a Physical Representation

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 $3 \times 3$ 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.

Execute — Answer: C

#1 Draw a Diagram K.G.A.1 Step 1
  • Label the $3 \times 3$ inner grid: four corners $C_1, C_2, C_3, C_4$, four edges $E_N, E_S, E_E, E_W$, and center $M$.
  • Two facts to remember: every corner has $3$ peripheral neighbors (two edges and the center); every edge has $5$ peripheral neighbors (two corners, the two adjacent edges across the center are actually neighbors via diagonals — wait, count by adjacency: edge $E_N$ at $(1,2)$ has neighbors $(1,1), (1,3), (2,1), (2,2), (2,3)$, so $5$ neighbors).
  • Center $M$ at $(2,2)$ has all $8$ peripheral cells as neighbors.
$$\text{corner: } 3\text{ peripheral nbrs}, \text{ edge: } 5, \text{ center: } 8$$

💡 Draw the $3 \times 3$ once and count neighbors for each role — corners (sparse), edges (medium), center (everyone).

#7 Identify Subproblems 1.OA.A.2 Step 2
  • Case 1: center $M$ is initially filled.
  • For $M$ to survive it needs $2$ or $3$ filled peripherals.
  • Every other peripheral must end empty: if filled, its neighbor count (which includes $M = 1$) must not be $2$ or $3$; if empty, it must not see exactly $3$ filled neighbors.
$M$ filled $\Rightarrow$ $2 \le \#\text{filled peripherals} \le 3$

💡 Treat the center separately because it's the only cell every other cell sees.

#2 Make a Systematic List 3.OA.D.8 Step 3
  • Sub-case 1.1: $M$ has exactly $2$ filled peripherals, call them $P_1, P_2$.
  • Each $P_i$ has $1$ (from $M$) plus any filled neighbors among other peripherals; for $P_i$ to die, this total must not be $2$ or $3$, so $P_1, P_2$ must not be neighbors.
  • Also any empty peripheral neighbor of both $P_1$ and $P_2$ would see $1 + 2 = 3$ filled and be born — forbidden.
  • Enumerating non-adjacent peripheral pairs with no common empty neighbor leaves exactly two cases: the two pairs of opposite corners $\{C_1, C_3\}$ and $\{C_2, C_4\}$.
  • Other non-adjacent pairs (e.g., two corners on the same row) share the middle edge as common neighbor.
$$\text{Sub-case 1.1: } 2 \text{ configurations}$$

💡 Opposite-corner pairs are the only non-adjacent pairs with disjoint peripheral neighborhoods.

#2 Make a Systematic List 3.OA.D.8 Step 4
  • Sub-case 1.2: $M$ has $3$ filled peripherals $P_1, P_2, P_3$.
  • For each $P_i$ to die, it must have $0$ or $1$ other-peripheral filled neighbor (so $\#\text{nbrs} = 1, 2$, but we need $\ne 2$, so $0$ peripheral neighbors).
  • All three must be mutually non-adjacent.
  • But any independent set of $3$ peripheral squares produces an empty peripheral neighbor adjacent to two of them — that cell would see $1 + 2 = 3$ filled and be born.
  • So this sub-case is empty: $0$ configurations.
$$\text{Sub-case 1.2: } 0 \text{ configurations}$$

💡 Three mutually non-adjacent peripherals always leave a 'common neighbor' empty cell that gets born.

#7 Identify Subproblems 1.OA.A.2 Step 5

Case 1 total: $2$ configurations.

$$\text{Case 1 total} = 2$$

💡 Two opposite-corner pairs, nothing else.

#7 Identify Subproblems 1.OA.A.2 Step 6
  • Case 2: $M$ is initially empty.
  • For $M$ to be born it needs exactly $3$ filled peripherals $P_1, P_2, P_3$.
  • For each filled $P_i$ to die it must have $0$ or $1$ peripheral neighbor in $\{P_1, P_2, P_3\}$ (since center is empty, $\#\text{nbrs} = $ peripheral count; must $\ne 2, 3$).
  • For each empty peripheral, $\#$ filled peripheral neighbors $\ne 3$.
  • Classify the valid $\{P_1, P_2, P_3\}$ shapes by adjacency.
$M$ empty $\Rightarrow$ exactly $3$ filled peripherals, induced subgraph has max degree $\le 1$

💡 Three filled cells, no cell touches both others, no outside cell touches all three.

#2 Make a Systematic List 3.OA.A.3 Step 7
  • Shape A — three corners (mutually non-adjacent): $\binom{4}{3} = 4$ ways.
  • Each remaining empty cell sees at most $2$ filled neighbors.
  • Valid.
  • $\Rightarrow 4$ configs.
$$\text{Shape A: } 4$$

💡 Three of the four corners — pick which corner to leave out.

#2 Make a Systematic List 3.OA.A.3 Step 8
  • Shape B — two corners on the same side plus the opposite middle edge (e.g., $C_1, C_2, E_S$): mutually non-adjacent, no empty cell is adjacent to all three.
  • Choose which side has two corners: $4$ ways.
  • $\Rightarrow 4$ configs.
$$\text{Shape B: } 4$$

💡 Two corners flanking one side, with the lonely opposite edge filling in.

#2 Make a Systematic List 3.OA.A.3 Step 9
  • Shape C — one filled edge plus the diagonally opposite corner (e.g., $E_N, E_W, C_3$ where $E_N$ and $E_W$ share corner $C_1$ that is empty)?
  • Re-examine: this shape is "two edges sharing a corner plus the opposite corner." Two edges sharing a corner have adjacency degree $1$ each (they touch each other), and the lone opposite corner is non-adjacent to both.
  • The shared common-neighbor corner sees both edges plus possibly more — needs $\#$ filled neighbors $\ne 3$.
  • The shared corner sees the two filled edges ($2$), the corner across ($1$ via diagonal?
  • no, opposite corner is not adjacent to the shared corner) — wait, the empty corner at $C_1$ has $E_N$ and $E_W$ as neighbors and $M$ as neighbor; with $M$ empty and the opposite corner $C_3$ not adjacent to $C_1$, that's $2$ filled neighbors — fine.
  • Count: $4$ corners to position the empty shared corner, giving $4$ configs.
$$\text{Shape C: } 4$$

💡 Two adjacent edges + the far-diagonal corner — the V-and-dot pattern.

#2 Make a Systematic List 3.OA.A.3 Step 10
  • Shape D — an L (one corner + an adjacent edge) plus the diagonally opposite corner.
  • E.g., $C_1, E_N, C_3$: $C_1$ and $E_N$ are adjacent (degree $1$ each), $C_3$ is non-adjacent to both.
  • Check empty-cell rule: the middle edge $E_S$ has $C_3$ as a neighbor and $M$ empty — only $1$ filled neighbor.
  • The empty corner adjacent to $C_3$ has $1$ filled neighbor.
  • No empty cell sees $3$ filled.
  • $4$ corners $\times 2$ choices of which adjacent edge $= 8$ configs.
$$\text{Shape D: } 8$$

💡 L-shape (corner + one of its two adjacent edges) anchored at one corner of the grid, plus the diagonally opposite corner.

#7 Identify Subproblems 2.OA.A.1 Step 11
  • Case 2 total: $4 + 4 + 4 + 8 = 20$ configurations.
  • Combined with Case 1: total $= 2 + 20 = 22$.
  • Answer $\textbf{(C)}\ 22$.
$$\text{Total} = 2 + 20 = 22$$

💡 Add the two disjoint cases — done.

[1] #1 K.G.A.1 Label the $3 \times 3$ inner grid: four corners $C_1, C_2, C_3, C_4$, four edges
[2] #7 1.OA.A.2 Case 1: center $M$ is initially filled. For $M$ to survive it needs $2$ or $3$ f
[3] #2 3.OA.D.8 Sub-case 1.1: $M$ has exactly $2$ filled peripherals, call them $P_1, P_2$. Each
[4] #2 3.OA.D.8 Sub-case 1.2: $M$ has $3$ filled peripherals $P_1, P_2, P_3$. For each $P_i$ to
[5] #7 1.OA.A.2 Case 1 total: $2$ configurations.
[6] #7 1.OA.A.2 Case 2: $M$ is initially empty. For $M$ to be born it needs exactly $3$ filled p
[7] #2 3.OA.A.3 Shape A — three corners (mutually non-adjacent): $\binom{4}{3} = 4$ ways. Each r
[8] #2 3.OA.A.3 Shape B — two corners on the same side plus the opposite middle edge (e.g., $C_1
[9] #2 3.OA.A.3 Shape C — one filled edge plus the diagonally opposite corner (e.g., $E_N, E_W,
[10] #2 3.OA.A.3 Shape D — an L (one corner + an adjacent edge) plus the diagonally opposite corn
[11] #7 2.OA.A.1 Case 2 total: $4 + 4 + 4 + 8 = 20$ configurations. Combined with Case 1: total $

Review

Reasonableness: 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 ($\pm 4$) or omit Shape C.

Alternative: Tool #10 (Physical Representation): grab a $3 \times 3$ grid of coins. Place $3$ heads in any pattern and check the rules by hand. With only $\binom{9}{3} + \binom{9}{2} \cdot \dots$ initial configurations involving the center (a few dozen), brute force is feasible if you're systematic. The toolkit/casework approach is cleaner but the physical check is the trustworthy backup.

CCSS standards used (min grade 3)

  • K.G.A.1 Describe positions of objects using above, below, beside, in front of (Labeling the $3 \times 3$ inner grid as corners, edges, center and tracking which cells are neighbors.)
  • 1.OA.A.2 Solve word problems involving three whole numbers whose sum is within 20 (Adding up filled-neighbor counts ($1 + 2 = 3$, etc.) to check the birth and survival rules.)
  • 3.OA.D.8 Solve two-step word problems using four operations within 100 (Combining sub-case constraints (non-adjacency, no common neighbor) to filter valid patterns.)
  • 3.OA.A.3 Solve multiplication and division word problems within 100 (Counting symmetric copies of each shape by multiplying base configurations by their rotational symmetry.)
  • 2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Summing the sub-case totals $2 + 4 + 4 + 4 + 8 = 22$.)

⭐ 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 $\textbf{(C)}$.

⭐ 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 $\textbf{(C)}$.