2022 AMC 12B Problem 18

Question

Each square in a

5 \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 \times 5

grid has a border of empty squares surrounding a

3 \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.)

14

18

22

26

30

Answer: C

Concepts:process simulation casework

Difficulty rating: 2000

Small hint:

Only the inner

3 \times 3

can be filled; the center becoming filled needs exactly

3

filled neighbors (if it was empty) or

2

-

3

(if it was filled)

Big hint:

The binding constraint is that no border square gains exactly three filled neighbors; check that every valid configuration has exactly three filled cells

Solution:

Only the inner

3 \times 3

squares can start filled. For the center to be filled afterward, if it began empty it needs exactly

3

filled neighbors, and if it began filled it needs

2

or

3.

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the

3 \times 3.

If the center starts empty, exactly

3

of its

8

neighbors must be filled. Checking these triples up to square symmetry leaves four types. With ring coordinates centered at

(0,0),

representatives are

\begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered}

Their symmetry-orbit sizes are

8,4,4,4,

giving

20

configurations.

If the center starts filled, the same neighbor check leaves only the two configurations in which the other filled cells are opposite corner neighbors. Hence the total is

20+2=22.

Thus, the correct answer is C.