B5. Charlie chooses one cell from a blank $n \times n$ square grid and shades it. The resulting grid has no lines of symmetry.
In terms of $n$, how many different cells could be shaded?
SolUTION
If the cell shaded is reflected in a line of symmetry of the blank grid, it must be reflected to a different cell or else that line is a line of symmetry of the resulting grid. In addition, shading a single cell cannot introduce a new line of symmetry because reflecting in any other line will cause at least one cell to reflect to a position outside of the grid. Therefore it must be that the cell shaded is such that no line of symmetry of the blank grid passes through it.

We will first count the number of cells which do have a line of symmetry passing through them. This count will differ depending on the parity of $n$, that is whether $n$ is even or odd. We can then subtract this from the total number of cells, which is $n^{2}$ in both cases.
For $n$ even, the only lines of symmetry of the blank $n \times n$ grid that pass through cells of the grid are the two main diagonals as pictured. Each of these lines pass through $n$ cells, and no cell lies on both lines. Hence there are $2 n$ such cells when $n$ is even.

For $n$ odd, there are four lines of symmetry of the blank $n \times n$ grid. Two are the main diagonals and the other two are horizontal and vertical lines which pass through the centre of the grid as pictured. These four lines each pass through $n$ cells but the centre cell lies on all of them. Hence there are $n+n+n+n-3=4 n-3$ such cells when $n$ is odd.

If $n$ is even, then the number of cells which can be shaded in order to avoid a line of symmetry is $n^{2}-2 n$.

If $n$ is odd, then the number of cells which can be shaded in order to avoid a line of symmetry is $n^{2}-(4 n-3)=n^{2}-4 n+3$.