It is easy to divide a square into four congruent isosceles triangles, just draw both diagonals. But can you divide a square into four incongruent isosceles triangles?
I found a few nonsolutions with more than 4 triangles (but not exactly 4, good job Charlie.)
This one is my favorite as is has six nonisoceles triangles and one degree of freedom [no proof]:
On square ABCD construct a circle centered at A passing through B and D. Pick any point on this circle and call it E. EAB and EDA are isosceles.
The line tangent to the circle at E intersects BC at F and DC at G. BFE and DGE are isoceles. (alternately bisect angles EAD and EAB.)
Call the midpoint of FG by H. GHC and FHC are isoceles.

Posted by Jer
on 20070326 11:28:18 