We can divide a square into 3 triangles by joining two adjacent
vertices to the midpoint of the opposite side. This gives us two
congruent triangles while the larger is twice the area of each of the
other two.
An initial thought was that all triangles would have to be
congruent, which would mean that the large triangle would have to be
bisected yielding 4 identical triangles. However I note that this
large triangle can be divided into two equal areas by joining one of
vertices (which it shares with the square) to the midpoint of the
opposing side.
I am sure that this is not going to assist in the solution, but I felt that the observation might be of interest.

Posted by brianjn
on 20060204 17:57:27 