Difference between revisions of "2022 AMC 10B Problems/Problem 19"
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In this case, it can be shown that there are <math>3</math> possibilities (excluding rotations and reflections). Two of them yield four cases, one yields eight, for a total of <math>16</math>. | In this case, it can be shown that there are <math>3</math> possibilities (excluding rotations and reflections). Two of them yield four cases, one yields eight, for a total of <math>16</math>. | ||
− | <math>\textbf{Subcase 2:}</math> No corners are filled | + | <math>\textbf{Subcase 2.2:}</math> 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 <math>4</math>. | 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 <math>4</math>. | ||
Our final answer is thus <math>2 + 16 + 4 = \boxed{\textbf{(C)}\ 22}</math> | Our final answer is thus <math>2 + 16 + 4 = \boxed{\textbf{(C)}\ 22}</math> | ||
~mathboy100 | ~mathboy100 |
Revision as of 11:16, 18 November 2022
Each square in a 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
grid has a border of empty squares surrounding a
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.)
Solution
There are two cases:
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
possibilities.
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.
A corner is filled
In this case, it can be shown that there are
possibilities (excluding rotations and reflections). Two of them yield four cases, one yields eight, for a total of
.
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
.
Our final answer is thus
~mathboy100