See figure: http://zak08.livejournal.com/28922.html

well, obviously if the two interior triangles each are halfsize similar copies of the original, then they both will fit inside it, inscribed in one corner each, disjointly except for 1 point, and each have half the perimeter. So that is one solution. It might be feasible to describe the full set of solutions, if we knew exactly what the rules were. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)