Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.
It's late, but I think I have found that the product of the two
bisectors x, y that are not the bisector of the right angle comes
out to be hypotenuse x inradius x 2 x sqrt(2). The inradius
of a right triangle with integer sides is an integer, and
therefore x and y cannot both be integers since their product
clearly is not an integer.
I used the classical formulas for the sides of a Pythagorean
triangle as 2mn, m^2n^2, and m^2+n^2 where m and n are arbitrary
positive integers such that m > n. The inradius is n(mn), and
x and y may be found using similar triangles once the inradius is
marked out on the sides.
David Shin showed below that the bisector of the right angle is never
an integer, so no more than one of the bisectors can be an integer.
Perhaps someone will give an example of a Pythagorean triangle that does have one integer bisector.

Posted by Richard
on 20050305 09:00:49 