by David Radcliffe
Here is the problem. Given a regular octagon in the plane, suppose that we join the vertices using line segments to form a continuous loop that visits each vertex exactly once. Is it necessarily the case that two or more of the line segments are parallel?
I claim that every such circuit contains a pair of parallel line segments. Suppose to the contrary that there exists a circuit which has no parallel segments. In order to analyze the problem mathematically, we number the vertices consecutively from 0 to 7. We also label each edge with the sum of its vertices; for example, the edge from vertex 3 to vertex 7 is labeled with 10.
Now we observe that two line segments are parallel if and only if their labels or the same, or they differ by 8. For example, the line segment between vertex 0 and vertex 4 is parallel to the line segment between vertex 1 and vertex 3, and it is also parallel to the line segment between vertex 5 and vertex 7.
Since there are 8 possible sums, there are also 8 possible directions for the line segments, and a circuit with no parallel segments must have exactly one segment in each direction. Therefore, the sum of the edge labels is 1+2+3+4+5+6+7+8+8k = 36+8k for some integer k.
But this sum must also equal 2*(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7) = 56, because the sum of the edge labels includes every vertex twice. Therefore, 36 + 8k = 56, or k = 2.5.
This is a contradiction, since k is required to be an integer. Therefore, every circuit joining the vertices of a regular octagon must contain a pair of parallel edges.
Exercise: Show that a similar argument works for any regular polygon having an even number of sides. What happens if the polygon has an odd number of sides?