A Diagonal of Two Squares

Two squares, with side lengths A and B, are placed together such that the right side of square A touches the left side of square B and their bases are collinear. A diagonal is drawn from the bottom left corner of square A to the top right corner of square B.
  1. Find the area below the diagonal in square A, in terms of A and B.
  2. Find the smallest integers A < B, such that the area from part 1 is an integer. Repeat for A > B. Is there a restriction on the relationship between A and B to achieve this?
  3. Find A and B (A not equal to B) such that the diagonal is an integer regardless of which square is on the left. [I don't know if this can be solved with A and B both integers.]

Source: Original. Similar to a problem at the U. of Miss' Geometry Gambit Contest.

Solutions were received from Denis Borris, Radu Ionescu, and Joseph DeVincentis. Joseph's solutions follow:
1. The diagonal is actually the hypotenuse of a right triangle with
legs A+B and B.  The part that is in square A is a similar right triangle,
with legs A and AB/(A+B), so the area we want is (A^2)B/(2(A+B)).

2. A=3, B=6 and A=6, B=3.
The restriction is the A and B must be such that A+B's prime factors
are all common factors of A and B.

For this area to be an integer, we need (A^2)B to be divisible
by 2(A+B).  The only prime factors which B is going to have in common
with A+B are ones that are also factors of A.  So we need to choose
A and B such that A+B's only prime factors are common factors of
A and B, and since A+B is bigger than either A or B, that means
A+B must have at least one repeated prime factor.  So 4, 8, 9, 12,
16, 18, 20, 24, 25, 27, 28, 32, 36, ... are candidates for A+B.

For each candidate, we look for partitions into two integers which
both contain all A+B's prime factors.  Since we want A < B or A > B,
we ignore the case of A=B as with 4=2+2.  So 8=2+6 is the first
real possibility.  If A=2, B=6, we have area 24/16, and if
A=6, B=2, we have area 72/16, neither an integer.  Next is 9=3+6.
A=3, B=6 gives area 54/18 = 3, and A=6, B=3 gives area 108/18 = 6.
These are the answers we want for this part.

3. [Upon comment from one reader, Joseph asked me to remove his answer
due to errors.  This is still in need of a solution. - KD]