(In reply to
re: Even by Richard)
My original idea was to start with n triangles, i.e. 2n right angles.
We need 4 for the corners and 4 for each interior point.
However we can also add 2 RA by combining a one of the edges of a triangle with a point, so that for n=3, the "extra" point lies on an edge of one of the triangles but now we cannot achieve equal areas.
However I have not figured out how or if this can be generalised.
I suspect that for odd n with equal areas, because the number of right angles is not divisible by 4, they cannot be fitted into a square.

Posted by goFish
on 20060205 17:09:44 