Right Triangles in a Square

Draw two lines out from a corner of a square to the opposite sides. Connect the ends of these two lines with a third line, perpendicular to one of the two lines, to form a right triangle. (You may have to adjust the location of the first lines slightly.) Surrounding this central right triangle are three more right triangles.
  1. Find such a figure, such that the side lengths of all four triangles are all integers.
  2. I expect the previous answer, if it exists, will use some large integers. For a smaller problem, find such a figure, such that just the side lengths of the square and the internal triangle are integers.
  3. What is the side length of the smallest square which can contain a 3-4-5 right triangle? A 5-12-13? Is it possible to generalize this?
  4. Label the square ABCD, with a side length of s. One acute angle of the central right triangle is at point A. What is the maximum distance the other acute angle can be away from point C?

Source: Original.

Solutions were received from Bob Odineal and Denis Borris. Update 7/26/00: Henry Bottomley added to parts 2 and 4.
  1. Denis says it's impossible. He references Kevin Brown's site on Concordant Forms as a mathematical basis. The equations involved in the square above can be massaged into the forms expressed in that explanation.
  2. Also impossible by same proof as #1. Henry Bottomlet adds:
    To get some labels: 
    Label the square ABCD, with a side length of s, where two lines leave A, there
    is a right angle of the internal triangle at a point on BC labeled E and an
    acute angle of the internal triangle on CD labeled F. Note that triangles ABE
    and ECF are similar. 
    2.  If there is not integer solution to 1 then there is no integer solution
    to 2 
    since an integer solution to 2 would be a rational solution to 1 [call length
    AE b, EF c and FA d : so EC=sc/b, BE=s(b-c)/b, CF=sc(b-c)/b^2, FD=s(b^2-bc+c^2)/b^2]
    and would thus lead to an integer solution to 1 (possibly bigger).
  3. General formula: b^2 / sqrt[b^2 + (b - a)^2], SO:
    for 3- 4- 5 case: 16 / sqrt(17)
    for 5-12-13 case: 144 / sqrt(193)
  4. Henry Bottomley: s/4, since if BE=x then EC=s-x and CF=x(s-x)/s=x-x^2/s which has a maximum value of s/4 when x=s/2

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