Pentagons cutting Is it possible to cut a pentagon into two equal pentagons?
Pentagons are not neccesarily convex as otherwise it would be trivially impossible.
I was given this problem at a contest but cannot figure the solution out, can somebody help?
Edit: Buy "cut" I mean "describe as a union of two pentagons which intersect only on the boudaries".
 A: A few observations that constrain any possible answer:


*

*Your cut must start on a vertex and end on a side, and must have one internal vertex in between.  Otherwise, you won't get two pentagons with the cut.

*Because of the internal vertex, the two pentagons must be convex.

*Because the two pentagons after the cut are convex, the original pentagon must be convex  I'm suspecting only one reflex angle, but I'm not sure.  If my suspicion is correct, then the cut must start at this reflex angle, and one of the resulting angles must be reflex.

*The sum of the interior angles will be $540$ degrees.

*Two of the angles will be supplementary (from the intersection on the edge).

*The sum of two of the angles will be $360$ degrees (from the internal angle created by the cut).

*The sum of the two angles created at the vertex of the original pentagon will be equal to the angle at that vertex.

A: I tried playing around with this on a image editor and I realized that it is not possible. Two of the interior angles will be explementary and two will be supplementary, due to the cut starting on an edge and the requirement of an inner vertex, as John has said. These four angles add up to $540^\circ$, which is the sum of all five interior angles of a pentagon. Therefore the last angle will be $0^\circ$, which is impossible. Therefore the cut is impossible.
A: Hint: try from brief sketch or prove it is not possible

