If the altitude h_a divides the side a into the segments a_b and a_c, by Pythagoras we have:
a_b = sqrt(b^2 - h_a^2) and
a_c = sqrt(c^2 - h_a^2)
But (a_b + a_c) or (a_b - a_c), depending on whether angle C is acute or obtuse, is equal to an integer, namely a - the radicands being two different integers.
Thus, not only the sides and altitudes, but also the segments of the sides determined by the altitudes are integers.
The sides of a rectangular triangle, if they are integers, are determined by the formula
(a^2 + b^2)^2 = (a^2 - b^2)^2 + (2ab)^2, a and b being integers.
For a=2, b=1, the hypotenuse is 5 and the sides, 4 and 3, or any integral multiple of these values. Other groups are 13, 12, and 5; 17, 15, and 8; 25, 24, and 7; and so on.
Assume that all rectangular triangles in our problem are members of the family 5, 4, 3. The, because h_a must split our triangle into two different rectangular triangles, h_a = 4jn = 3kn and the smallest values for j and k are 3 and 4 respectively.
Thus h_a = 12n, b = 5jn = 15n, c = 5kn = 20n, a_b = 3jn = 9n, a_c = 4kn = 16n, and a = 25n or a = 7n, depending on whether h_a falls inside or outside the original triangle.
Take the smaller value a = 7n. Twice the area of our triangle is a*h = 84n^2; the altitude h_b = 84n^2/15n = 28n/5, and h_c = 84n^2/20n = 21n/5.
Thus, if we take n = 5, all sides and altitudes decome integers: a = 35, b =75, c = 100, h_a = 60, h_b = 28, h_c = 21, and the area of the triangle is 1050 square units.
The above calculation with a = 25n, or with other groups, leads to larger areas for the triangle.
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