KA-0173Counting up to symmetry
4 points, difficulty 2 of 3Level 11–12about 135sEach of the four edges of a square is painted either black or white. Two paintings count as the same if one can be rotated onto the other. How many genuinely different paintings are there?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Some paintings survive a turn unchanged.
2The strategy
Group the paintings by how many edges are black: 0, 1, 2, 3 or 4.
3The full solution
Zero black: 1 way. One black: 1. Two black: 2 (adjacent or opposite). Three black: 1. Four black: 1. Total 1+1+2+1+1 = 6.
Solution
The reliable way
Divide-by-four fails here, because some paintings are unchanged by a rotation, so classify by hand instead. Group by the number of black edges. All white: 1 painting. Exactly one black edge: any single edge can be rotated to any other, so 1. Exactly two black: the two black edges are either adjacent or opposite, and rotation cannot turn one into the other, so 2. Exactly three black: 1, by the same argument as one. All black: 1. The total is 1 + 1 + 2 + 1 + 1 = 6. The transferable idea: when symmetry is involved, case-splitting on an invariant feature - here the number of black edges - is safer than dividing.
The elegant way
The count of black edges survives any rotation, so classify by it and only the two-black case splits, into adjacent and opposite.
Why this is on the test: Dividing by the number of symmetries is the standard trap, and it is wrong exactly when some object is fixed by a symmetry.