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Sastry's Evil Bisectors (Posted on 2005-03-04) Difficulty: 5 of 5
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:

x=ab[1 - c/(a+b)]

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

x=2ab/(a+b).

Note that x cannot be an integer.

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

y=bc[1 - a/(b+c)]

and

z=ca[1 - b/(c+a)].

It suffices to show that at most one of y and z can be integral.  The fact that c=a+b should be used.


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