AMC 10 · 2022 · #19
Grade 3 geometry-2dProblem
Each square in a 5×5 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 5×5 grid has a border of empty squares surrounding a 3×3 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.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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.
💡 Draw the $3 \times 3$ once and count neighbors for each role — corners (sparse), edges (medium), center (everyone).
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.
💡 Treat the center separately because it's the only cell every other cell sees.
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.
💡 Opposite-corner pairs are the only non-adjacent pairs with disjoint peripheral neighborhoods.
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.
💡 Three mutually non-adjacent peripherals always leave a 'common neighbor' empty cell that gets born.
1.OA.A.2 Step 5 Case 1 total: $2$ configurations.
💡 Two opposite-corner pairs, nothing else.
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.
💡 Three filled cells, no cell touches both others, no outside cell touches all three.
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.
💡 Three of the four corners — pick which corner to leave out.
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.
💡 Two corners flanking one side, with the lonely opposite edge filling in.
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.
💡 Two adjacent edges + the far-diagonal corner — the V-and-dot pattern.
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.
💡 L-shape (corner + one of its two adjacent edges) anchored at one corner of the grid, plus the diagonally opposite corner.
2.OA.A.1 Step 11 - Case 2 total: $4 + 4 + 4 + 8 = 20$ configurations.
- Combined with Case 1: total $= 2 + 20 = 22$.
- Answer $\textbf{(C)}\ 22$.
💡 Add the two disjoint cases — done.
K.G.A.1 Label the $3 \times 3$ inner grid: four corners $C_1, C_2, C_3, C_4$, four edges 1.OA.A.2 Case 1: center $M$ is initially filled. For $M$ to survive it needs $2$ or $3$ f 3.OA.D.8 Sub-case 1.1: $M$ has exactly $2$ filled peripherals, call them $P_1, P_2$. Each 3.OA.D.8 Sub-case 1.2: $M$ has $3$ filled peripherals $P_1, P_2, P_3$. For each $P_i$ to 1.OA.A.2 Case 1 total: $2$ configurations. 1.OA.A.2 Case 2: $M$ is initially empty. For $M$ to be born it needs exactly $3$ filled p 3.OA.A.3 Shape A — three corners (mutually non-adjacent): $\binom{4}{3} = 4$ ways. Each r 3.OA.A.3 Shape B — two corners on the same side plus the opposite middle edge (e.g., $C_1 3.OA.A.3 Shape C — one filled edge plus the diagonally opposite corner (e.g., $E_N, E_W, 3.OA.A.3 Shape D — an L (one corner + an adjacent edge) plus the diagonally opposite corn 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.1Describe 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.2Solve 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.8Solve 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.3Solve multiplication and division word problems within 100 (Counting symmetric copies of each shape by multiplying base configurations by their rotational symmetry.)2.OA.A.1Solve 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 $\textbf{(C)}$.
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