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.
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.
There are some answers in http://www.mathematik.tu-bs.de/preprints/199710.ps
ReplyDeleteOr, search for "rational distances".
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.
ReplyDeleteThe 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).
Also, I have a slightly simpler (non-trigonometric) proof of the impossibility of a regular pentagon with integer coordinates.
ReplyDeleteAssume 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.