2022 AMC 10B Problems/Problem 19

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.

$[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible); } ds((1,1)); ds((2,1)); ds((3,1)); ds((1,3)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((10,2)); ds((11,1)); ds((11,0)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]$ 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.) $[asy] import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,gray+opacity(0.5),invisible); } for (int i = 1; i < 4; ++ i) { for (int j = 1; j < 4; ++j) { label("?",(i + 0.5, j + 0.5)); } } for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("Initial", (2.5,-1)); draw((6,2.5)--(8,2.5),Arrow); ds((11,2)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } label("Transformed", (11.5,-1)); [/asy]$ $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$

Solution

There are two cases:

$\textbf{Case 1:}$ The center is filled in the initial configuration In this case, one or two more squares must be filled in the initial configuration. It is easy to show that the only configuration that results in additional squares for the next turn is the configuration with all squares along the diagonal. This case yields $2$ possibilities.

$\textbf{Case 2:}$ The center is not filled In this case, exactly three squares outside the center must be filled. Additionally, these three squares must all be adjacent to only one square, the center square.

$\textbf{Subcase 2.1:}$ A corner is filled In this case, it can be shown that there are $3$ possibilities (excluding rotations and reflections). Two of them yield four cases, one yields eight, for a total of $16$ .

$\textbf{Subcase 2.2:}$ No corners are filled In this case, there is only one way to do it (excluding rotations and reflections). There are a total of four ways to rotate this permutation, so this case has a count of $4$ . Our final answer is thus $2 + 16 + 4 = \boxed{\textbf{(C)}\ 22}$

~mathboy100

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2022 AMC 10B Problems/Problem 19

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