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 Sastry's Evil Bisectors (Posted on 2005-03-04)
Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.

 No Solution Yet Submitted by owl Rating: 4.2000 (5 votes)

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 Product | Comment 7 of 10 |
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^2-n^2, and m^2+n^2 where m and n are arbitrary positive integers such that m > n.  The inradius is n(m-n), 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 2005-03-05 09:00:49

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