# A simple to explain solution to this kids' geometry puzzle

A smart 10 year old asked me basically this question.

Consider a rectangle with both diagonals drawn in. Now ask if you can visit all the edges by travelling from some starting vertex and only ever going to a neighboring vertex but never visiting the same edge twice.

The answer is no, but is there an explanation would you could give to a smart 10 year old for this?

Suppose that your drawing DOES have a path that does what you say.

Any vertex is either a start or end vertex or an interior (visited along the way) vertex. If it's an interior vertex, it has an EVEN number of edges meeting it, because for each visit, there's the edge you arrived on and the edge you left on.

For a start/end vertex, it has an ODD number of edges meeting it. [But see below].

Conclusion: for a drawing with a suitable path, there are exactly two vertices with an odd number of edges meeting them. But in your example, there are 4 such vertices. Hence, no suitable path.