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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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 Thoughts | Comment 6 of 10 |

One may prove by the law of cosines that the length of the angle bisector from C in triangle ABC with side lengths a, b, c, is given by:

x²=ab[1 - c²/(a+b)²]

If we let C be the right angle, and let x be the length of the angle bisector from C, we find then that

x²=2a²b²/(a+b)².

Note that x cannot be an integer.

Let y and z be the lengths of the angle bisectors from A and B, respectively.

y²=bc[1 - a²/(b+c)²]

and

z²=ca[1 - b²/(c+a)²].

It suffices to show that at most one of y and z can be integral.  The fact that c²=a²+b² should be used.

 Posted by David Shin on 2005-03-05 06:53:25

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