## Wednesday, January 31, 2007

### A pentagon problem

Prove or disprove:
There exists no pentagon in the plane all of whose lengths (sides and diagonals) are rational.
Passed on to me by a friend. And no, I don't know the answer.

One fact that is known: no regular pentagon in the plane can have integer coordinates.

1. There are some answers in http://www.mathematik.tu-bs.de/preprints/199710.ps

Or, search for "rational distances".

2. I can answer this because I once solved a somewhat related puzzle: find the smallest hexagon with integer sides/diagonals which can be inscribed in a circle.

The answer to that puzzle is the inscribed hexagon of side lengths 3-5-3-5-3-5 (yielding diagonals of length 7 and 8). Picking 5 of the hexagon's vertices yields the desired pentagon.

I discovered the hexagon by computer program (by enumerating triangles per circumradius and applying Ptolemy's Theorem).

3. Also, I have a slightly simpler (non-trigonometric) proof of the impossibility of a regular pentagon with integer coordinates.

Assume the contrary, and take the smallest such pentagon with one vertex at the origin. The clockwise-adjacent vertex is without loss of generality either (odd,odd), (odd,even), or (even,even).

The (odd,odd) case corresponds to a side length whose square is 2 mod 4. The next clockwise-vertex must then necessarily be (even,even). Following parities around the pentagon leads to a contradiction.

A similar argument rules out the (odd,even) case. The (even,even) case yields all even coordinates, which leads to a violation of the pentagon's minimality.

This proof works for any regular n-gon with n odd.