In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confused about how to prove that it is really just 5. Can anyone enlighten me?
Here are 5 symmetric pentagons on a $3\times3$ grid:
A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.
@JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:
- The axis of symmetry must go through one of the 5 vertices (call it $A$)
- The other 4 vertices must be symmetric to each other about the axis of symmetry.
Now note that there are only 3 vertices to choose from for vertex $A$: The center, the edge, and the corner.
The center can have an orthogonal axis of symmetry or a diagonal one.
The edge and corner will be symmetric about the line through that vertex and the center vertex.
Putting this together, there are only four cases which leads to the 5 cases already identified:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities