2018 AMC 10A Problem 20

The required symmetries (four rotations and four reflections) are all the symmetries of a square. A symmetric code is determined by the colors in one fundamental wedge: the triangle bounded by a vertical line through the center and a diagonal, one-eighth of the grid. Every other cell is carried onto a wedge cell by some symmetry and must match its color.

Count the wedge cells with the center at $(0,0)$ and coordinates $-3 \le x, y \le 3$. Choose the wedge $0 \le y \le x$:

• $x = 0$: $(0,0)$, the center. $1$ cell.
• $x = 1$: $(1,0), (1,1)$. $2$ cells.
• $x = 2$: $(2,0), (2,1), (2,2)$. $3$ cells.
• $x = 3$: $(3,0), (3,1), (3,2), (3,3)$. $4$ cells.

That is $1 + 2 + 3 + 4 = 10$ independent cells, each freely black or white, so $2^{10} = 1024$ symmetric colorings.

The code must use both colors, so remove the all-white and all-black grids: $1024 - 2 = 1022$.

The answer is $\boxed{\textbf{(B)}\ 1022}$.

A pattern invariant under a symmetry group is determined by its values on one representative from each orbit of cells. The $8$ symmetries of a square fold the grid into a $45^\circ$ wedge; cells on the wedge's edges (the center, the axis $y = 0$, the diagonal $y = x$) have smaller orbits but still count as one free choice each. Counting orbits, not cells, is the whole problem; the "at least one of each color" clause is a routine subtraction at the end.

Sanity-check with orbit sizes: the center (orbit $1$), three cells on each half-axis type $(a, 0)$ (orbits of size $4$), three on the diagonals $(a, a)$ (size $4$), and three general cells $(2,1), (3,1), (3,2)$ (size $8$). Total cells $1 + 12 + 12 + 24 = 49$, confirming $10$ orbits.

• Using a quarter (the 4 by 4 corner block, 16 cells) or half of the grid instead of the eighth wedge, giving 2^16 - 2 = 65534 or 2^13 - 2.
• Forgetting to subtract the two monochrome grids, giving $1024$, or subtracting only one of them.